Earth's baseline black-body model – "a damn hard problem"

The Earth only has an absorbing area equal to a two dimensional disk, rather than the surface of a sphere.

By Robert G. Brown, Duke University (elevated from a WUWT comment)

I spent what little of last night that I semi-slept in a learning-dream state chewing over Caballero’s book and radiative transfer, and came to two insights. First, the baseline black-body model (that leads to T_b = 255K) is physically terrible, as a baseline. It treats the planet in question as a nonrotating superconductor of heat with no heat capacity. The reason it is terrible is that it is absolutely incorrect to ascribe 33K as even an estimate for the “greenhouse warming” relative to this baseline, as it is a completely nonphysical baseline; the 33K relative to it is both meaningless and mixes both heating and cooling effects that have absolutely nothing to do with the greenhouse effect. More on that later.

I also understand the greenhouse effect itself much better. I may write this up in my own words, since I don’t like some of Caballero’s notation and think that the presentation can be simplified and made more illustrative. I’m also thinking of using it to make a “build-a-model” kit, sort of like the “build-a-bear” stores in the malls.

Start with a nonrotating superconducting sphere, zero albedo, unit emissivity, perfect blackbody radiation from each point on the sphere. What’s the mean temperature?

Now make the non-rotating sphere perfectly non-conducting, so that every part of the surface has to be in radiative balance. What’s the average temperature now? This is a better model for the moon than the former, surely, although still not good enough. Let’s improve it.

Now make the surface have some thermalized heat capacity — make it heat superconducting, but only in the vertical direction and presume a mass shell of some thickness that has some reasonable specific heat. This changes nothing from the previous result, until we make the sphere rotate. Oooo, yet another average (surface) temperature, this time the spherical average of a distribution that depends on latitude, with the highest temperatures dayside near the equator sometime after “noon” (lagged because now it takes time to raise the temperature of each block as the insolation exceeds blackbody loss, and time for it to cool as the blackbody loss exceeds radiation, and the surface is never at a constant temperature anywhere but at the poles (no axial tilt, of course). This is probably a very decent model for the moon, once one adds back in an albedo (effectively scaling down the fraction of the incoming power that has to be thermally balanced).

One can for each of these changes actually compute the exact parametric temperature distribution as a function of spherical angle and radius, and (by integrating) compute the change in e.g. the average temperature from the superconducting perfect black body assumption. Going from superconducting planet to local detailed balance but otherwise perfectly insulating planet (nonrotating) simply drops the nightside temperature for exactly 1/2 the sphere to your choice of 3K or (easier to idealize) 0K after a very long time. This is bounded from below, independent of solar irradiance or albedo (or for that matter, emissivity). The dayside temperature, on the other hand, has a polar distribution with a pole facing the sun, and varies nonlinearly with irradiance, albedo, and (if you choose to vary it) emissivity.

That pesky T^4 makes everything complicated! I hesitate to even try to assign the sign of the change in average temperature going from the first model to the second! Every time I think that I have a good heuristic argument for saying that it should be lower, a little voice tells me — T^4 — better do the damn integral because the temperature at the separator has to go smoothly to zero from the dayside and there’s a lot of low-irradiance (and hence low temperature) area out there where the sun is at five o’clock, even for zero albedo and unit emissivity! The only easy part is to obtain the spherical average we can just take the dayside average and divide by two…

I’m not even happy with the sign for the rotating sphere, as this depends on the interplay between the time required to heat the thermal ballast given the difference between insolation and outgoing radiation and the rate of rotation. Rotate at infinite speed and you are back at the superconducting sphere. Rotate at zero speed and you’re at the static nonconducting sphere. Rotate in between and — damn — now by varying only the magnitude of the thermal ballast (which determines the thermalization time) you can arrange for even a rapidly rotating sphere to behave like the static nonconducting sphere and a slowly rotating sphere to behave like a superconducting sphere (zero heat capacity and very large heat capacity, respectively). Worse, you’ve changed the geometry of the axial poles (presumed to lie untilted w.r.t. the ecliptic still). Where before the entire day-night terminator was smoothly approaching T = 0 from the day side, now this is true only at the poles! The integral of the polar area (for a given polar angle d\theta) is much smaller than the integral of the equatorial angle, and on top of that one now has a smeared out set of steady state temperatures that are all functions of azimuthal angle \phi and polar angle \theta, one that changes nonlinearly as you crank any of: Insolation, albedo, emissivity, \omega (angular velocity of rotation) and heat capacity of the surface.

And we haven’t even got an atmosphere yet. Or water. But at least up to this point, one can solve for the temperature distribution T(\theta,\phi,\alpha,S,\epsilon,c) exactly, I think.

Furthermore, one can actually model something like water pretty well in this way. In fact, if we imagine covering the planet not with air but with a layer of water with a blackbody on the bottom and a thin layer of perfectly transparent saran wrap on top to prevent pesky old evaporation, the water becomes a contribution to the thermal ballast. It takes a lot longer to raise or lower the temperature of a layer of water a meter deep (given an imbalance between incoming radiation) than it does to raise or lower the temperature of maybe the top centimeter or two of rock or dirt or sand. A lot longer.

Once one has a good feel for this, one could decorate the model with oceans and land bodies (but still prohibit lateral energy transfer and assume immediate vertical equilibration). One could let the water have the right albedo and freeze when it hits the right temperature. Then things get tough.

You have to add an atmosphere. Damn. You also have to let the ocean itself convect, and have density, and variable depth. And all of this on a rotating sphere where things (air masses) moving up deflect antispinward (relative to the surface), things moving down deflect spinward, things moving north deflect spinward (they’re going to fast) in the northern hemisphere, things moving south deflect antispinward, as a function of angle and speed and rotational velocity. Friggin’ coriolis force, deflects naval artillery and so on. And now we’re going to differentially heat the damn thing so that turbulence occurs everywhere on all available length scales, where we don’t even have some simple symmetry to the differential heating any more because we might as well have let a five year old throw paint at the sphere to mark out where the land masses are versus the oceans, and or better yet given him some Tonka trucks and let him play in the spherical sandbox until he had a nice irregular surface and then filled the surface with water until it was 70% submerged or something.

Ow, my aching head. And note well — we still haven’t turned on a Greenhouse Effect! And I now have nothing like a heuristic for radiant emission cooling even in the ideal case, because it is quite literally distilled, fractionated by temperature and height even without CO_2 per se present at all. Clouds. Air with a nontrivial short wavelength scattering cross-section. Energy transfer galore.

And then, before we mess with CO_2, we have to take quantum mechanics and the incident spectrum into account, and start to look at the hitherto ignored details of the ground, air, and water. The air needs a lapse rate, which will vary with humidity and albedo and ground temperature and… The molecules in the air recoil when the scatter incoming photons, and if a collision with another air molecule occurs in the right time interval they will mutually absorb some or all of the energy instead of elastically scattering it, heating the air. It can also absorb one wavelength and emit a cascade of photons at a different wavelength (depending on its spectrum).

Finally, one has to add in the GHGs, notably CO_2 (water is already there). They have the effect increasing the outgoing radiance from the (higher temperature) surface in some bands, and transferring some of it to CO_2 where it is trapped until it diffuses to the top of the CO_2 column, where it is emitted at a cooler temperature. The total power going out is thus split up, with that pesky blackbody spectrum modulated so that different frequencies have different effective temperatures, in a way that is locally modulated by — nearly everything. The lapse rate. Moisture content. Clouds. Bulk transport of heat up or down via convection. Bulk transport of heat up or down via caged radiation in parts of the spectrum. And don’t forget sideways! Everything is now circulating, wind and surface evaporation are coupled, the equilibration time for the ocean has stretched from “commensurate with the rotational period” for shallow seas to a thousand years or more so that the ocean is never at equilibrium, it is always tugging surface temperatures one way or the other with substantial thermal ballast, heat deposited not today but over the last week, month, year, decade, century, millennium.

Yessir, a damn hard problem. Anybody who calls this settled science is out of their ever-loving mind. Note well that I still haven’t included solar magnetism or any serious modulation of solar irradiance, or even the axial tilt of the earth, which once again completely changes everything, because now the timescales at the poles become annual, and the north pole and south pole are not at all alike! Consider the enormous difference in their thermal ballast and oceanic heat transport and atmospheric heat transport!

A hard problem. But perhaps I’ll try to tackle it, if I have time, at least through the first few steps outlined above. At the very least I’d like to have a better idea of the direction of some of the first few build-a-bear steps on the average temperature (while the term “average temperature” has some meaning, that is before making the system chaotic).

rgb

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JeffC
January 12, 2012 8:22 am

seems to me that the 33 degree greenhouse effect is wildly overstated …

tallbloke
January 12, 2012 8:22 am

Great article Dr Brown!
I guess it depends what you are setting out to prove as to whether all the effort is worthwhile. perhaps Hans Jelbring made a smart move with his model atmosphere and isometrically heated planet surface in his 2003 paper:
http://tallbloke.wordpress.com/2012/01/01/hans-jelbring-the-greenhouse-effect-as-a-function-of-atmospheric-mass/
At least if he is right you won’t have to worry about including radiative effects of GHG’s

Randall Harris
January 12, 2012 8:25 am

It is a very tough problem. I hope you can get some sleep tonight.

G. Karst
January 12, 2012 8:29 am

Yes, it truly is a Gorgon’s knot. We are all waiting for an Alexander, to appear, with a very sharp sword. GK

January 12, 2012 8:29 am

If the IPCC had spent $1b trying to do this before they started telling everyone what the solution was, we may have not only got somewhere by now but also had some respect for them. As it is, I get the feeling those telling us how to solve the problem don’t even know what the question is — or even where to start looking for the question.

Victor Barney
January 12, 2012 8:29 am

BERESHITH 8:22…as long as the earth remains, seedtime and harvest, and cold and heat, and winter and summer, and day and night shall not cease.” AND, it doesn’t matter if you believe it or not because it’s going to occur anyhow… because that’s exactly what’s written!

Francois
January 12, 2012 8:30 am

Are you serious? You know there are a few books which might help you understand how the system works.

January 12, 2012 8:30 am

Would it be impossile to build a physical model?

Mike M
January 12, 2012 8:30 am

But but.. the ‘settled science’ is so settled that we’re already spending over $2.5 billion per year to ‘combat climate change’. Ain’t no physics in the universe gonna slow down a gravy train with that much inertia.

Jim Cripwell
January 12, 2012 8:30 am

When I see those magic words “lapse rate”, I hope that what you are talking about makes sense.

pd
January 12, 2012 8:31 am

And then you haven’t even mentioned that the rotation of the earth around the sun isn’t a perfect circle, which influences the amount of radiation from the sun on the earth.

michael hart
January 12, 2012 8:34 am

The problem is [well, one of the many problems], that I fully expect climate modellers to say “Yes, of course we know all this. We take it fully into account. Now, run along, while we tell politicians how to run the world.”
But where can us lesser mortals examine the algorithms and computer code, not to mention the assumptions about which bits can be safely ignored?

kwik
January 12, 2012 8:35 am

You need to write a CSharp program and combine it with a ball made in WPF. WPF (Windows presentation Foundation) is perfect for this. We use it in my business every day.
If we could get some money from some rich guy, and I could take a 2 years leave from my day job, I would love to join you.
We could patch up the globe with small triangles with texture. The globe can be made in any 3-D package and exported to xaml. You can then import it into a WPF project and then write a CSharp program that starts turning the globe….and start calculating stuff…..ahh…a fantastic project…. Click “play” and it comes alive….
But for crying out loud; Dont make a “Report for Policy-Makers” when we are finished!!!!!

Simpleton
January 12, 2012 8:39 am

I’d be tempted to try the Earth as a simple disc (ie, what the sun ‘sees’), flipping over once per day, to give a global average. Then let temperature rise by 1 degree for each doubling of CO2. Then use average albedo, clouds increasing with temperature (more water vapour), hence more albedo and less temperature. Fudge something in for leads and lags if you must.
I’m interested to see if more CO2 eventually ends up with a faster cooling Earth, and therefore a route to the next ice age.
My view is that if the actual situation is too complicated to model, then move to a simpler picture and play around with it.

higley7
January 12, 2012 8:39 am

Looking back at the historical record for clues regarding the character or behavior of these many factors might helpt he effort, at least to decide whether a given addition is going in the right direction.
I like to point out that maybe kids are right to not like vegetables. After all, everybody who ate green peas during the Civil War died

Arno Arrak
January 12, 2012 8:39 am

You are on the right track. Have you looked at what Ferenc Miskolczi has done? You should because he has some very important conclusions about the greenhouse effect.

Joules Verne
January 12, 2012 8:41 am

@rgb
I’m surprised at your surprise. Anyone who’s been paying attention to this blog knows that the 255K baseline is an ideal grey sphere which has no mass and superconducts. The only difference between gray and black body is albedo. The only difference between sphere and uniformly lit plane is angle of incidence adjustment.
We can at least use some good old experimental science to get us to what an airless world made out of the same rocks as the earth at the same distance from the sun does as far as average surface temperature. At mid-latitudes on the moon that measured number is 250K.
The moon’s slower spin and thermal conductivity of rocks combine to lower its temperature some more from a gray body and the earth’s faster spin would then serve to make it closer to the black body. So I guess what I’m saying is that arguing with 255K as a baseline is probably something only cranks, pikers, and pedants find unsatisfactory for most purposes.
is of course massless and superconducting. I prefer to at least say, for laymen, that the ideal grey sphere is spinning so fast the temperature is equal at every point. An ideal gray body is pretty darn basic physics. High school level stuff int the NYS regents science course I took in the 1970’s anyway. It’s been described here on Watt’s Up With That many times too.
The problem is the body is painted with all kinds of colors other than levels of grey (albedo). And even the grey changes in a very poorly characterized manner. Then there’s like an orchestra playing with different frequencies of light instead of sound and a fairly large assortment of different arrangements of matter that interacts with it to figure out just the radiative part. Then, as long as this remains a water planet, the three phases of water and other unique properties like high latent heat capacity and solid form lighter than liquid to deal with. Then there’s convection up the wazoo which doesn’t happen on grey bodies and has a large effect on temperature stratification. The grey body is an anchor and of course it must be understood what a grey body is and everything that can act to change things.

January 12, 2012 8:42 am

Thank you Dr. Brown. For us non physicists a lucid description. Like all good scientific investigations it raises way more questions then answers. It also illustrates better then I have ever been able to do how dependent any of these numeric models, even the ones we have confidence in, are to voracity of the underlying empirical measurements.

Stephen Wilde
January 12, 2012 8:42 am

Excellent article.
One question:
How do we know that the Earth is any warmer than it would be without greenhouse gases if the standard assumptions are so obviously inappropriate and/or incomplete ?
I don’t believe there is any model anywhere that is based on empirical data rather than flawed guesswork.

Dolphinhead
January 12, 2012 8:45 am

If we absorb radiation as a disk and radiate away as a sphere the cards are already stacked in favour of cooling. Without the oceans we would be icicles.

Larry Kirk
January 12, 2012 8:48 am

Deliciously put! With that many variables (ie. more than about two), nobody could ever HOPE to model the system, or the effect on it of changing a single, relatively minor factor. Anybody who thinks that they can has got to be kidding themselves. (Nobody but an economist, that is. But they do not have a particularly good track record, knighted or otherwise)

TerryS
January 12, 2012 8:48 am

Re: Francois

You know there are a few books which might help you understand how the system works.

List a few for me please. I would like to read them.

Joe Soap
January 12, 2012 8:52 am

Just out of curiosity what is the geothermal input?
Or heat from magnetic flux eddy-currents?
Or the energy input from lunar gravity moving the oceans?
Maybe these are very small but it would be good to see some estimation of them.

crosspatch
January 12, 2012 8:59 am

One question I have had is the impact of changes in solar UV on the troposphere by indirect means via changes in stratospheric heating. For example, if there is an increase in UV, the stratosphere experiences more heating. If the stratosphere is warmer than this means the top of the troposphere is warmer. If the tropopause is warmer, then if I due the adiabatic lapse to the surface, the surface will be warmer. If the temperature at the tropopause cools, then figuring the temperature down the column to the ground also cools. Unless — the affect from stratospheric changes is a change in altitude of the tropopause. If the stratosphere warms, the tropopause happens at a lower altitude — it finds the temperature inversion “sooner” as stuff is convecting upwards. In that case, since the troposphere is now “thinner” and the stratosphere is “thicker” the change is compensated for and the temperature at the surface is unchanged.
We are currently experiencing less UV than usual with a cooler Sun. That would be reflected in a cooler stratosphere. That might be reflected in a rise of the tropopause (or might not if the troposphere also cools, we are talking about temperature deltas here, not absolute temperatures). The whole thing is like squeezing a water balloon.
Nice article. I think the models currently assume an infinitely thick atmosphere with no convection, radiative heat transfer through it with no evaporation, condensation, clouds, etc.

January 12, 2012 9:00 am

seems to me that the 33 degree greenhouse effect is wildly overstated …
Well, or understated. I thought I had a relatively simple argument that would have suggested that the true baseline should be more than 33 degrees; temperature differentiation favors faster cooling, so both a static perfectly insulating sphere and a rotating sphere with heat capacity and with poles would respectively cool relative to the superconducting sphere and warm relative to the non-rotating sphere. But then I head that voice — dooo theee integrallll. Assume make an ass outa u and me. The problem really is — does the rotating sphere with heat capacity warm relative to the superconducting sphere? Intuitively, I’d say no, it still cools. But at this point I want to do the integrals. Which means first I have to derive them, which I can’t do right now because I’m about to be ass-deep in alligators teaching (really, I already am).
So it might take me weeks or even months to do so, although I don’t think it is that hard. I can probably use e.g. octave/matlab to do them numerically, although the rotating sphere with heat capacity technically requires the solution of a set of time dependent ODEs as a point at a given latitude rotates, through enough rotations to approach a steady state. Basically you have something like dQ/dt for a surface element equals dP_in/dt – dP_out/dt, = CdT/dt, where dP_in/dt is \vec{S}\cdot \hat{n} dA for incoming Poynting vector from sun, dP_out/dt is blackbody power out of dA, and C is the heat capacity of dA. \hat{n}(t) is an outward directed normal (as a function of time as the sphere rotates). The solar flux is modulated by a periodic square wave so it is zero as the point goes darkside.
One should be able to start this from any temperature distribution and spin forward to equilibrium as a function of theta, and only have to do this for the upper half sphere as it is symmetric. Then one has T(\theta, \phi, etc) and one can plot, integrate to find averages, and so forth, for different values of this and that. I think it would be very educational to do this and would take the guesswork out of the question “what does water do” or “what does an atmosphere do” to the sphere (relative to superconducting or insulating static sphere). With a bit more work, one could probably add in depth and do the vertical heat equation as well (only conduction for some conductance) — I think octave would still solve it in less than eternity, although it might well be hours per sphere. A small cluster and you could do a whole range of spheres in a day of compute time and generate pretty pictures without resorting to C coding.
So I don’t pretend to be able to guess the answer. Time to do the work instead.
rgb

Allanj
January 12, 2012 9:00 am

Dr Brown beautifully demonstrates the complexity of the problem. I don’t think we know how much of that complexity is embodied in the IPPC models but I suspect not enough. I also know from experience that you can reach a point of complexity in models that you cannot really understand what your own models are doing. Then you start running sensitivity analyses until they become so complex you don’t understand them.
It is a wonderful world of uncertainty. It is fun to play with. Just don’t bet the farm on the results.

A physicist
January 12, 2012 9:00 am

Professor Brown, please let me express my appreciation for this outstanding WUWT guest post!
Regarding the physics of radiation transport, please allow me to recommend the web page that the American Institute of Physics maintains on this topic, titled Basic Radiation Calculations.
One merit of the AIP’s page is that it covers the history in parallel with the physics: definitely it’s taken a long time and a huge effort to work through these details.
It made me LOL to read, halfway down the Basic Radiation Calculations web page, this passage:

“Dear reader: You have made your way into one of the most difficult corners of this experimental site, and it would be very useful to know why. Would you take just three minutes to answer a few questions? Please click here”

Professor Brown, you have earned everyone’s appreciation, and please accept my sincere thanks, for working to bring a comparable level of understanding to the skeptical community here on WUWT. Your post shows rational skepticism at its finest and best!

Joules Verne
January 12, 2012 9:04 am

@rgb (con’t)
So I do agree the ideal grey body baseline is horrible. I prefer to start with the moon as the baseline. That gives us an actual sphere made out of the same rocks as the earth at the same distance from the sun. The earth spins faster than the moon which makes it several degrees warmer which is fairly straightfoward.
So here’s what I say. The next biggest difference between earth and moon is that earth basically presents as a ball of brine instead of a ball of rocks. There’s a lot of radiative heating/cooling implications in just that alone. Water is very much different from rocks in physical properties.

January 12, 2012 9:06 am

Bravo Dr. Brown.
The 33 K figure is based on treating the earth as a spherical cross section, which is obviously completely wrong. The AGW’ers are off by over 100 K to start with and they are trying to make predictions accurate to tenths of degrees : )

January 12, 2012 9:09 am

When I see those magic words “lapse rate”, I hope that what you are talking about makes sense.
It does. The lapse rate is just the rate that the temperature of the atmosphere drops off with height. As the Earth’s surface cools via radiation, the vertically stratified volume over a given piece of surface (including the surface) radiates at different temperatures if/when the frequencies associated with those temperatures become unblocked by “high albedo” GH gases that have a large absorption cross section there. CO_2, for example, tends to radiate from near the top of the troposphere where it is much colder. Colder means that it radiates less power (in this part of the spectrum) than the warm surface underneath would have due to BB radiation alone. Radiating away less heat creates a differential warming (less cooling). In Caballero’s nice online book, you can look at figure 5.15 and see- the measurements of the IR spectrum from the Sahara Desert that perfectly illustrate the point, and the book itself defines and derives e.g. the adiabatic dry air lapse rate.
It’s complicated, in other words — very complicated when you add clouds, varying albedo, convection, wet air, global circulation — but not senseless or unreasonable.
rgb

Joel Shore
January 12, 2012 9:14 am

Robert,
(1) Arthur Smith has already tackled the issue of the temperature variation on a planet with some rotation rate and some uniform heat capacity: http://arxiv.org/abs/0802.4324
(2) While the issue of average temperature is complicated, it is not quite as bad as you seem to think. First of all, we have Holder’s Inequality, which tells us that the 4th root of the average of T^4 is greater or equal to the average of T for any distribution of temperature T. (The equality occurs in the case of a uniform temperature distribution, and it turns out that the current Earth is close enough to having a uniform temperature that the difference between these two ways of computing average temperature is small.)
Second of all, what is probably the best way to characterize the radiative greenhouse effect is by considering the average emission (in W/m^2) of the planet’s surface (given its actual temperature distribution) vs the average intensity (in W/m^2) that is absorbed by the planet and its atmosphere. So, for example, for Earth we know that the average intensity absorbed is ~240 W/m^2 and the average emission is ~390 W/m^2 or so. We also know that the average surface temperature of the Earth is ~288 K. In the absence of a radiative greenhouse effect, an Earth (with everything otherwise the same, including the albedo) would have to an average temperature such that it is emitting 240 W/m^2. The highest average temperature that a blackbody can have and emit 240 W/m^2 is 255 K (by Holder’s Inequality). To the extent that the temperature distribution is non-uniform, the average temperature could be lower…even considerably lower for very non-uniform distributions. (To the extent that the Earth is not a perfect blackbody emitter in the infrared, the average temperature could be a little higher…but in fact the emissivity of the Earth is very close to 1 over the relevant wavelengths.)

Bryan
January 12, 2012 9:15 am

Great article.
I like your ‘well, lets try to figure it out from first principles approach’.
Most others slide past the awkward bits leaving the reader feeling inadequate.
A sign of a good teacher!
I thought you might be interested in a new peer reviewed paper by Gerhard Kramm and Ralph Dlugi.
Its a review of the current understanding of the atmosphere effect.
Of particular interest
A detailed application of Keplers Laws to Earth/Sun situation.
Equation 2.17 which includes a ground flux contribution as well as the radiative contributions.
The energy reservoir diagram Fig 11
The irradiance overlap area Fig 5 which contradicts some IPCC presentations.
http://www.scirp.org/journal/PaperInformation.aspx?paperID=9233

January 12, 2012 9:15 am

So here’s what I say. The next biggest difference between earth and moon is that earth basically presents as a ball of brine instead of a ball of rocks. There’s a lot of radiative heating/cooling implications in just that alone. Water is very much different from rocks in physical properties.
I think you’re right, although it isn’t just water — water alone is just a matter of changing the heat capacity per unit area of the surface in the model series I suggest. It has to be water with circulation, I think. And as I said, until I formulate and do the integrals I won’t try to guess the answer. Too difficult, and there is Kirchoff lurking behind Arrhennius and Stefan-Boltzmann. Minimizing temperature differences warms, but only up to where there is none and you’re back to the superconducting limit, I think. But I want to do the integrals and see.
rgb

Mydogsgotnonose
January 12, 2012 9:15 am

Read up Misckolczi’s paper: he’s done a lot of the maths.

Larry Kirk
January 12, 2012 9:20 am

What one could in theory do, would be to simply change one of the variables of this immensely complex system and then sit back for an appropriate length of time and observe the effect that this change had had.
Not in the case of this planet however, as we are already dealing with a dynamic, ever-changing system, of which we do not know the detailed history, nor the full nature or history of the numerous factors that acted upon it to date.
So, looking forward: as we do not know how this constantly changing system would have changed anyway, without our intervention, then we can never deduce the effect that our interference with one single variable has had.
And that is the flaw in global warming modelling. We do not understand the pre-existing system, let alone the effects of our intervention. And we never could, because it is too complex.
Economics seems equally flawed, and equally subordinate to political expediency and the strange human desire to always have been ‘right’, regardless of the facts.

Joules Verne
January 12, 2012 9:21 am

@rgb
I don’t think there’s much argument that building a climate model from the bottom up is hard. Climate boffins do it from the top down. This drastically simplifies things as it’s all radiation at that point and these days you can measure it with satellites. Believe it not though we still can’t get satisfactory agreement between different methods of global average albedo better than a few percent accuracy. About the the only thing all the methods agree on is it’s not constant from year to year. The more sophisticated models take obvious seasonal variation into account but from year-to-year it holds constant. So it becomes a prime candidate for a fudge factor which can be tweaked to make your model more agreeable with history. A variation in global average albedo of just 1% is equivalent to the net of all anthropogenic forcing estimates. Experiments measuring albedo, those few that have been running for long, find albedo variation of that amount from year to year.

Joel Shore
January 12, 2012 9:23 am

Stephen Wilde says:

How do we know that the Earth is any warmer than it would be without greenhouse gases if the standard assumptions are so obviously inappropriate and/or incomplete ?

Because we know that the Earth + atmosphere are absorbing ~240 W/m^2 from the sun and that the Earth’s surface is emitting ~390 W/m^2 or so. The only way that this can happen without rapid cooling is if the Earth’s atmosphere absorbs some of the radiation emitted by the surface, which means there is a radiative greenhouse effect. (People like you have desperately tried to explain it other ways, but alas none of those ways use correct physics principles.)
This notion that the atmosphere is absorbing the surface radiation is in fact seen to be empirically-correct by looking at the emissions from the Earth as seen from satellites in space. They see an Earth emitting only ~240 W/m^2 and the spectrum shows that the reason it is emitting less than the 390 W/m^2 that the surface is emitting is because certain wavelengths are getting strongly absorbed. These wavelengths correspond precisely to those wavelengths at which the various greenhouse gases (and clouds) absorb radiation.

Viv Evans
January 12, 2012 9:23 am

Yep – a damn hard problem, and that’s without sticking stuff into the model which mess up things like volcanoes or plants (forget SUV-driving humans for the moment, they mess up everything anyway!!!) and all the small critters in the oceans doing their things as well …
I thoroughly enjoyed reading this post!

January 12, 2012 9:32 am

Start from the definitive facts, of my Venus/Earth temperatures comparison, and don’t lose sight of them, ever. At least half of what is written in this article as unquestioned assumption, I consider rank speculation. The Venus/Earth comparison demolishes all climate models with the simple facts. Someone earlier likened the problem to the Gordian (not “Gorgons”) Knot, and looks for an Alexander to cut it. Well, you cut it by not jumping on the radiative transfer theory as holy writ, and by not ignoring the Venus/Earth facts. I submitted the following comment to tallbloke’s web site just this morning:
In the context of public debate, the “greenhouse effect” is not about radiative equilibria with and without an atmosphere, or even with or without “greenhouse gases” in the atmosphere. It is about whether atmospheric temperature at the surface (or at any given pressure level) increases with an increase in atmospheric carbon dioxide. I am astonished that even skeptics cannot focus upon this obvious fact in the real world; everyone can’t seem to stop themselves from launching into radiative transfer theory arguments. And in that context, of the public and political debate — that irreducible, unarguable reality — the FACT (not theory) is, my utterly simple and transparent comparison of the temperatures in the atmospheres of Venus and Earth demonstrates there is no such greenhouse effect, whatsoever. All the supposedly learned theorizing by one and all is precisely worthless, because everyone uses it to ignore the simple, definitive fact that disproves the tyrannously-promulgated carbon dioxide greenhouse effect, and reveals the radiative transfer theory as unconnected [or better, disconnected] from the real thermodynamics of the atmosphere. (And that last should be obvious, since the radiative theory ASSUMES a fixed temperature distribution, with every incremental layer of the atmosphere at thermal equilibrium. So the radiation levels in the atmosphere are, by that assumption, the EFFECT, not the CAUSE, of the thermodynamics — the radiative EFFECT of the gravitationally-imposed tropospheric lapse rate.)
Where I stand at the moment (and don’t even bother trying to change my mind, I am still learning, and I will develop my understanding on my own, unless and until I see “experts” recognize what I have done, and do better than I have already done): You can’t consider the Earth’s surface a blackbody, or each differential layer of the atmosphere as contributing as a “gray body” (blackbody times emissivity) without the assumption of detailed thermal equilibrium (radiative transfer theorists, are you listening?). More specifically, you can’t do it in the presence of convection and conduction of heat energy between the layers, and expect to get the thermodynamics right. That’s why the blackbody is traditionally described in terms of an enclosed cavity, held at constant temperature, with just a small hole to allow transfer of radiation (no convection, no conduction) into and out of it (Christopher Monckton followers, are you listening?). And even with such detailed thermal equilibrium as the radiative transfer theory imagines, you can only recover the observed radiation levels, and pretend you know the thermodynamics from them; you can’t predict their thermodynamic effect, because the cause-and-effect works the other way — the thermodynamics (in the presence of incident solar infrared irradiation) gives the temperatures and those determine the measured radiation levels. And the presence of CO2 and other IR-active gases only funnels the radiation portion (but not the convection and conduction portions) of atmospheric heat transfer through them, it doesn’t necessarily heat the atmosphere (in particular, in passing from the surface upward, as my Venus/Earth analysis demonstrates). The Gordian Knot does not exist.

tallbloke
Reply to  Harry Dale Huffman
January 12, 2012 10:08 am

Harry Dale Huffman,
Dont worry, your comment was noted. Just been too busy at work to reply.
Call back soon.

January 12, 2012 9:32 am

I’m surprised at your surprise. Anyone who’s been paying attention to this blog knows that the 255K baseline is an ideal grey sphere which has no mass and superconducts. The only difference between gray and black body is albedo. The only difference between sphere and uniformly lit plane is angle of incidence adjustment.
Well there you have it. I haven’t been “Paying attention to this blog” or reviewing climate physics in detail until, well, yesterday. Having a topical textbook really helps. All the meandering above was my attempt to mentally organize some of what I’ve learned so far from it — I didn’t really intend for it to be a toplevel post but rather the last post I was going to make in a thread.
Now I may have to make this one the last. I”ve got a ton of work to do (of the teaching variety) before I can come back to WUWT, although it is addictive.
I’ll make one last comment. Do not assume (and this goes for everybody) that all of the physics in the standard models is ill conceived or overtly wrong. The real differences between well-educated skeptics (where I am not one, not yet, not in the right physics in the right context) and dilettantes is that the former tend to think that the GCMs are “mostly right” in their physics, but are getting some pretty subtle stuff wrong, notably climate sensitivity. None of them doubt the greenhouse effect, most of the good ones include the ocean and so on. Perhaps not at the right level of detail — not all of the work done is of high quality — but it is probably mistaken in small details with large effects, with certain notable possible exceptions.
So I’m pretty “skeptical” that current climate science gets the kind of stuff I’m considering above in the first few steps at all wrong — I just want to work through it on my own as “homework”, to learn it properly. Far better to derive things yourself than to trust or rely on a textbook, better still to do both, derive it yourself and learn to understand it and then check your work against texts and resolve the differences, learn from your errors (and look for possible errors in the accepted literature — they are not unknown:-).
Believing that you know the answer before you work it through is OK, right up to the point where it becomes self-fulfilling prophecy and you make choices that make your beliefs work out. Confirmation bias (and its close cousin, cherrypicking) are often just as seductive to skeptics as they are to non-skeptics — they are just polarized the opposite, contrary way. The best thing to do is just plain work through it. When I have time, now, I will.
rgb

Mac the Knife
January 12, 2012 9:32 am

Excellent summary!
You ‘painted’ a very complex thought experiment with sufficient fidelity to reality and attention to detail that I could visualize contributions from each added element as you constructed the global warming picture. I have a much clearer mental picture of the problem and a better grasp of the daunting complexities associated with any ‘modeling’ attempts, as a result. I’m going to read this through several times more, to reinforce my understanding and appreciation of your ‘modern physics masterpiece’. As I’m malingering at home with a minor bout of ‘flu’ today, this will provide appropriate distraction, education, and entertainment.
Thank You!!!
MtK

January 12, 2012 9:34 am

At Earth orbital distance from the Sun the total incoming energy provided by the Sun is 1366 / 4 = 341.5 W m-2 .
(solar constant divided by 4 to take day/night and spherical shape into account).
Without an atmosphere, reflection and absorption of the solar irradiance would only take place at the Earth surface, composed by ice, rocks and sand. With an assumed albedo of 0.4 (reflecting power of a surface) the earth surface would absorb 341.5*(1-0.4) = 204.9 W m-2 and re-emit this energy to the outer space.
According to the Stefan- Boltzmann law, the mean temperature at the earth surface would establish itself at:
T= (E / ε σ)^¼ = (204.9/(1 x 5.6710-8))^ ¼ = 245.2 K (-28°C)
Obviously, large differences would be observed at different latitudes and wide changes would take place during every day-night cycle, including some ice melting and water freezing (but with water evaporating an atmosphere would exist… no water is left on the surface of planet Mars).
Also, the assumed albedo is quite hypothetical since we have no clue about what would be the Earth’s surface if no atmosphere were present.
Actually, the mean Earth surface temperature is approximately 15 °C, or 43 K higher than the assumed naked planet. If you assume 33 K it is also within the ballpark.
This is the warming contribution of the atmosphere, mostly by absorbing some of the INcoming sunlight (UV/visible) AND some of the OUTgoing long wave radiation from the Earth surface (by so-called greenhouse gases such as water and carbon dioxide).
Actually the mean surface temperature doesn’t make any physical sense in this hypothetical calculation. The surface temperature would swing between extremes over the night/day cycle (incoming radiation between 0 and 1366 W m-2), with the heat capacity of the material at the surface contributing to a dampening effect.
For a simple 2 layers atmospheric model look at: http://climate.mr-int.ch/TwoLayersClimateModel.html

DeWitt Payne
January 12, 2012 9:38 am

Robert,
Hölder’s Inequality requires that the average temperature on a non-isothermal sphere at radiative equilibrium, assuming the same emissivity and incident power level, must be less than the average temperature of an isothermal sphere. That’s something that G&T do correctly ( http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf Section 3.7.4 ).

January 12, 2012 9:39 am

ask scaffeta, he models all that with one function.
ha.
sounds like you are on your way to building a GCM

Thermodynamic professor, power plant manager
January 12, 2012 9:41 am

I’m sure that thermodynamic atmosphere effect explains everything much better, it’s the only method that is based on physical facts. If you try to model earths temperature with this very clear Anthonys description you can argue everything until you are dead or used 1 bn$;) Backradiation is bullshit from cooler gas to warmer surface, physically impossible! Gases can’t heat up with infrared radiation without incredible W/m2, only surfaces.

David L.
January 12, 2012 9:44 am

All this reminds me once again why we talk about “average temperature”? The number, as has been mentioned on this site many times, is meaningless. Let’s talk about my house. I have a wood stove in the living room and it pretty much heats my entire house. I have a grate in the ceiling above the wood stove and being a stone farmhouse there is convection throughout the main part of the house. The kitchen is on the opposite side end of the house from the woodstove. Our TV room is perpendicular to this arrangement and the bedroom is above the TV room on the second floor. Now I like it cool and my wife likes it warm so we are constantly changing the draft on the stove or kicking on the oil burner for extra heat, especially in the bedroom which has it’s own heating zone since it gets very little convection from the woodstove.
Now my wife and I never talk about the average temperature of our house. Typically when it’s 20F outside, the equilibrium temperature distribution in my house is: The stove is 400F, the living room is 80F, the kitchen is 64F, the TV room is 70F, the bedroom is 62F, and the second floor is 67F. Oh, by the way, inside the fridge is 30F and the freezer is 10F. The basement is 56F. The attic is 32F. Some of the house is insulated by stone, some by modern R33 fiberglass…some by 1940’s R8 insulation.
So when I’m cold in the bedroom I kick on the furnace. When I’m too warm in the living room I close the woodstove damper. When the milk goes bad in the fridge I turn down the thermostat in the fridge. My point is that no one would ever think of this system as an average temperature. It has no meaning. There are different temperatures and gradients all over that house for real physical reasons…a good physicist may actually even attempt to model it and get it close…maybe with a degree or two?
Why does the average not matter? Why should it? If I did somehow calculate an everage home temperature what would it mean? I can’t act on that information. If the average temperature of the house increased because the fridge broke and warmed up, then that’s not helping my wife who says she’s freezing in the living room. I can’t say “hey, the average home temperature is actually going up so you can’t be freezing in there”. What’s important is knowing the actual temperatures at the various locations and how they change or their distribution changes over time.
One last point: I couldn’t ascertain an average temperature of my home to better than a few degrees. I certainly couldn’t tell if that average changed by a few degrees or even if it mattered. And they think they can 1) get a handle on the average temperature of the entire planet to +/- 0.1 and 2) that it provides any useful information?

AC
January 12, 2012 9:44 am

This may be simplistic, but would it be fair to say that the 33K of temp attributed to CO2 is really 33K of temp attributed to the non simple BlackBody?
What I mean is if we assume the black body gets us to 255K, then all the complexity (including, but not limited to rotation, circulation, CO2 ppmv, atmosphere and water ingeneral, etc) makes up that 33K? Or did I miss something.
I’m wondering if each comonnent mentioned in the article can be thought about AFTER a baseline is used. I know that PV=nRT is for ‘ideal gasses’ and non of them exist, but it is a decent enough guide to teach in intro Chem classes. is the T_b=255K derived in a similar fashion?

jim hogg
January 12, 2012 9:44 am

A roadmap to a possible approximate answer from someone who knows where to start: our messy and incredibly complex reality. A pleasure to read a first rate and honest mind at work.

Great Greyhounds
January 12, 2012 9:46 am

Ahhh! Those fine words… “Do the Integral!”
Thank You for the explanation of how a model should be developed, but the big question for me is: How do we hope to test the output of the Model?
I worked for a number of years doing E-M simulations for Antennas and Radomes, and at least we had the ability to build real models to test our simulations against!

January 12, 2012 9:52 am

RE: R. Brown, 9:00am

But then I head that voice — dooo theee integrallll. Assume make an ass outa u and me. The problem really is — does the rotating sphere with heat capacity warm relative to the superconducting sphere? Intuitively, I’d say no, it still cools. But at this point I want to do the integrals.

Another big complication is the heat capacity effects of water turning into ice. First, there is the simple heat of fusion where heat is released while maintaining a temperature of 0 degC while water changes from liquid to solid. But then you form a crust of ice whose surface can be much colder than the water under the ice. I cannot imagine the differential equations describing these effects filled with discontinuities, much less attempt to integrate them.
I very much appreciate you contributions here over the past week. I envy your students at Duke.

Carrick
January 12, 2012 9:58 am
Kevin Kilty
January 12, 2012 9:59 am

Dr. Brown,
Some time ago I read a paper by Gerlich and Tscheuschner, who pointed out a number of inconsistencies in the usual presentation of the Greenhouse effect. Some of these are trivial (such as the atmospheric heating is not the way a greenhouse works) but others are quite substantive. The article inspired me to think about putting together a more realistic model of a rotating earth that could store heat in the crust and ocean, but it is a problem that presents huge difficulty and may not be worth the effort anyway–Gerlich and Tscheuschner suggest it is intractable.
Let’s start with a point we can probably all agree with, which is, the model of a uniformly radiating earth (i.e. uniform temperature) illuminated on one side and reflecting some fraction of the radiation is an unphysical model for the problem at hand. Without atmosphere it is 255K and with it is 288K–who cares?
What is of real importance is the actual temperature distribution of the earth, because that is the radiator we have to work with. At one time Victor Starr pointed out that the polar regions are the radiator for Earth and they are efficient in this task because they are so large–i.e. a large, cool radiator can work as well as a small hot one. But surely it is important that some radiation in the tropics and subtropics switches back and forth from surface to cloud tops, a huge change in radiator temperature, or that high plateaus, with little water vapor above radiate very efficiently, and so forth.
We waste a lot of breath huffing and puffing over the mean temperature of the planet and whether or not it is increasing or decreasing. I suggest that this arithmetical mean temperature is beside the point because it is not connected with the radiator itself. Because of the disconnection between the physical radiator and the arithmetic mean, there are many mean temperatures that are consistent with an Earth in equilibrium, each one a function of the internal dynamics of the oceans, atmosphere, and crust. Even when the forcing is not changing we may see mean temperature vary.
So, why should we even care about your build-a-bear effort, when it will likely turn out to contain options that make the problem tractable rather than features that are essential? What are the essential features, bare minimum, of a realistic model of how the Earth gains and rejects radiative heat? Maybe we could call it the build-a-bare model.

JPeden
January 12, 2012 10:05 am

Francois says:
January 12, 2012 at 8:30 am
Are you serious? You know there are a few books which might help you understand how the system works.
Right, in fact it’s so simple that the CO2 = CAGW Climate Scientists haven’t got even one relevant empirical prediction right yet, and are hard pressed to even be able to “explain” any more than about the first 20 of the last 32 years of the most recent past, another one of their records which is continuing to get even worse as we speak.
Right

Kevin Kilty
January 12, 2012 10:06 am

steven mosher says:
January 12, 2012 at 9:39 am

sounds like you are on your way to building a GCM

Except the GCM will parameterize the interesting smaller-scale physics in the problem, and we won’t know if this has a small or large impact on the solution.

Gary Pearse
January 12, 2012 10:19 am

“Basically you have something like dQ/dt for a surface element equals dP_in/dt – dP_out/dt, = CdT/dt, where dP_in/dt is \vec{S}\cdot \hat{n} dA for incoming Poynting vector from sun, dP_out/dt is blackbody power out of dA, and C is the heat capacity of dA. \hat{n}(t) is an outward directed normal (as a function of time as the sphere rotates). The solar flux is modulated by a periodic square wave so it is zero as the point goes darkside.”
Francois says:
January 12, 2012 at 8:30 am
Are you serious? You know there are a few books which might help you understand how the system works.
Dr. Brown is serious, Francois, it is this hard. So pitch in and give him a hand.

jorgekafkazar
January 12, 2012 10:26 am

Ocean albedo is in itself a huge problem. Not only does the reflectance vary [zero to 1.00] with zenith angle, it also is a function of wind velocity and direction, tides, sea foam, temperature, humidity, viscosity, density, and, so help me, plankton content. Assuming “average” values for any or all of these variables will NOT produce a dependable model, similar to the situation with clouds.

Theo Goodwin
January 12, 2012 10:35 am

Wonderful article, Dr. Brown. Your presentation is lucid and clear. I wish that this kind of work had been available twenty years ago. I thank you for doing it now.

Dan Kirk-Davidoff
January 12, 2012 10:37 am

Michael Hart:
Many climate modeling groups make their code public. NCAR’s is here: http://www.cesm.ucar.edu/models/cesm1.0/
GFDL’s is here: http://data1.gfdl.noaa.gov/~arl/pubrel/m/am2/doc/
NCAR’s is not too hard to download, set up and run, if you have the time and inclination, and a good sized computer.

Neo
January 12, 2012 10:37 am

The Earth radiates it’s own heat generated at the core.
The function of how it is emitted at the crust is quite complicated because it depends both the local surface material and geological faults.
It might be interesting for map AGW hot spots with known high geothermal emissions.

January 12, 2012 10:40 am

Dr. Brown, It is kinda complicated. I have a question that sounds off topic but is on topic to a degree. When I started thinking about how to build a better model, I was struck by the stability of the tropopause minimum temperature. Yes, stability is a bit relative in this case, but it tends to bounce back nicely. The coldest temperature on Earth and in its lower atmosphere is around 184K degrees -89C, with the tropopause making excursions to -95 C from time to time, but bouncing back rather quickly to below -90C. Oddly, Venus, the granddaddy of greenhouse effect, has a black body temperature of about 184K which is around 65Wm-2.
This seems to be a non-thermal limit to the atmospheric effect (the better term since a lot of crap is going on). Since Venus and Earth don’t have a heck of a lot in common other than mass, this is circumstantial evidence that the gravitational constant may be the prime suspect. Crazy talk right?
So once you iron out all the issues you are looking into now, be aware that there are possibly a few more shoes to fall. If the approximate 65Wm-2 is a gravitationally set limit, not numerical happenstance, it adds some neat thermomagnetic and thermoelectric atmospheric chemistry to the problem which may involve ozone and CO2 and/or CH4. Which is totally bassackwards to the Venus atmosphere building mass with runaway greenhouse effect. I believe someone theorized that Earth and Mars started out like Venus, but picked a different path. He may have been smarter than most of his buddies thought.

Tim Folkerts
January 12, 2012 10:49 am

It is great to have another voice of a physicist here.
The problem is indeed complicated, but not insurmountable. As Dr. Brown noted (“I just want to work through it on my own as “homework”, to learn it properly”), the sort of things he is doing are the “homework” to even start to discussing the topics. Anyone who wants to intelligently discuss the climate and the greenhouse effect really needs to understand the science at the level he is addressing.
One thing that often bothers me is the assumption by many people that scientists in the field have not thought of these things before. Nothing that Dr. Brown said is “new” or “unknown”. These would be the sorts of problems that grad students in the field would work through in their coursework. (The source that Dr Brown references is, after all, a textbook with sample problems.) So this is “basic” science for those specializing in climate modeling.
Finally, even though the topic is difficult and certainly not “settled” as a whole, many individual parts of he problem are “settled”.
1] the starting scenario (“Start with a nonrotating superconducting sphere, zero albedo, unit emissivity, perfect blackbody radiation from each point on the sphere. What’s the mean temperature?”) would give a planet with a uniform temperature of ~ 278.5 in earth’s orbit around our sun.
2] raising the albedo would lower the temperature of the planet (to ~ 255 K for the earth’s current ~ 0.3 albedo)
3] lowering the emissivity would raise the temperature of the planet (but experimentally the emissivity is very close to 1, so this is not a major concern).
4] a nonuniform temperature would lower the average temperature of the planet. (so any conduction or convection or fast rotation would help make the temperature more uniform and hence warmer on average).
5] GHGs high up in the atmosphere where it is colder would raise the surface temperature.
These are all “settled” in terms of the general affect on global temperatures. Of course, the details and exact extent are not “settled” (or we could perfectly predict weather and climate).
One big challenge is that some things have two or more effects. For example, more H2O in the atmosphere adds GHG and would warm the atmopshere [5], but it would also lead to more clouds which raises the albedo and cools the planet [2].

January 12, 2012 10:50 am

Harry Dale Huffman said @ January 12, 2012 at 9:32 am

And in that context, of the public and political debate — that irreducible, unarguable reality — the FACT (not theory) is, my utterly simple and transparent comparison of the temperatures in the atmospheres of Venus and Earth demonstrates there is no such greenhouse effect, whatsoever. [Emphasis mine].

It is a fact that p = it is true that p, or it is the case that p.
A theory is a set of propositions providing an explanation of a subject matter. Even a single proposition can be a theory. Such as it is true that p, or it is the case that p.
Thus you contradict yourself rendering whatever else you are saying incoherent.

January 12, 2012 10:52 am

“It can scarcely be denied that the supreme goal of all theory is to make the irreducible basic elements as simple and as few as possible without having to surrender the adequate representation of a single datum of experience. ” – A. Einstein
‘Chaos Theory’ teaches that even the simplest of non-linear systems can be ultimately unpredictable, and the earth is a definitively non-linear system.

January 12, 2012 10:53 am

Robert. What a wonderful world !
What you have shown is that the Earth, or any other planet for that matter, can never reach thermal equilibrium since energy is constantly flowing through the system. A single black body temperature is therefore an idealized myth. For the same reason it is also a fairly gross simplification to measure or to predict just one single globally averaged temperature (GAT). There are plenty of non anthropogenic phenomena which continuously change this average temperature. For example plate tectonics including the raising of the Tibetan plateau by 5 km which not only reduced the surface temperatures but also increased albedo. Then we know that albedo is anyway constantly changing due to clouds, let alone volcanic ash and deforestation.
The greenhouse effect itself is also changing on an hourly basis as water vapor enters the atmosphere and then rains out again. Into this mixture we then introduce over the last 260 years about a 30% increase in one greenhouse gas – CO2. We go back and measure GAT over the last 160 years of weather station measurements and assume that the observed rise of about 0.6C is just due to CO2. Well maybe they are right and it is all due to CO2. However, this can never be proved until measurements or experiments are designed which can eliminate all other effects except CO2. There are only a few places on Earth where such an experiment may even be possible. Firstly it would have to be somewhere with zero or stable water vapor at least for a fixed period each year far away from civilization. Ideally we would need a reliable weather station with a long historic time span. Does such a place exist ? I wonder if there is perhaps just one long term weather station somewhere in the Sahara for example ?

January 12, 2012 10:54 am

Dr Brown,
You’ll go insane trying to develop a deterministic model because the boundary conditions are effectively indeterminate. Stochastic. As you have observed; non-linearities are rife in the climate system; beyond T^4 effects. Especially convective heat transfer and the phases change of water, in all their phases make a big difference.
Then there are topographical effects; mountain-sides, canyons, hills … which change the angle of incidence of light dramatically throughout the day; and hence when and how much heat is stored and then released. IIRC, even water has different reflectivity at low angles of incidence.. When smooth. Wave action makes things a great deal more complicated. It’s not valid to assume that the average will be the same..
Vegetation is also highly variable in terms of albedo and the energy which it stores and how/when it releases that energy. Seasonal variations are substantial. And if you’ve ever looked out of the window of an aircraft flying at altitude, you may have noticed that the green forests appear to be almost black from directly above yet green from the sides at a distance. Plants orientate themselves (leaves turn) in order to optimise the balance between photosynthetic takeup and loss of water. Vegetation changes seasonally with species becoming dominant during growing their peak; and for those who stimulate the climate neurons; their seasonal stasic, decline and death are also important because that has a strong effect on albedo.
There are very many factors which are known to have an effect on weather and hence climate. Many of those known factors can be described in isolation, under controlled conditions (i.e. with reality adjusted to impose assumptions). But because the climate system is non-linear, the principles of super-position cannot be applied to combine two or more non-linear factors. And that’s without even considering the coupling effects between the variables; which also tend to be non-linear in magnitude.
Keep in mind that climate is entirely synthetic; an arbitrary statistical artifact, based on poor statistics. Only weather is real.
Although one can build simplified models of the climate system, those work on bulk averages (and fantastic assumptions). But those bulk averages imply that the condition of the climate system withing each cell is uniform. Which might be a reasonable approximation if it’s say 1000 cubic metres, but in the case of the models, it’s more like 1000 cubic kilometres for the smallest cell size. (Order of magnitude) And it’s extremely unlikely that conditions will be uniform throughout.
It is the internal differences in conditions that act as a perturbation (“disturbance”) to provide interesting weather. The non-linear response to the perturbation is impossible to determine a priori without knowing the boundary conditions; the state of the system.

Fred H. Haynie
January 12, 2012 10:55 am

To simplify the energy flow problem pick the South Pole during the six months of night where it has very little atmospheric water vapor to cause a “greenhouse effect” and see if the rise in CO2 in the last thirty years has had any significant effect on the rate of energy radiated to space. Energy is being delivered via wind and lost via radiation to space. There is no direct input from the sun to complicate the balance. Also, nights and days are six months long. The air is thinner at that altitude and the TOA will be closer to the surface.

Bill Illis
January 12, 2012 10:58 am

How about if one makes it a non-rotating planet first with some set heat capacity (and I’m not sure there is really a limit to how much heat a surface or gas can absorb).
How hot does it get at the equator at the spot that is directly facing the Sun.

hswiseman
January 12, 2012 10:59 am

Lubos “crickets,” Lubos “crickets,” Bueller…

January 12, 2012 11:00 am

Robert Brown said @ January 12, 2012 at 9:32 am

So I’m pretty “skeptical” that current climate science gets the kind of stuff I’m considering above in the first few steps at all wrong — I just want to work through it on my own as “homework”, to learn it properly. Far better to derive things yourself than to trust or rely on a textbook, better still to do both, derive it yourself and learn to understand it and then check your work against texts and resolve the differences, learn from your errors (and look for possible errors in the accepted literature — they are not unknown:-).

And as Joseph Joubert once said: to teach is to learn twice (except he said it in French).

Peter Miller
January 12, 2012 11:02 am

Mr Watts
It is all very simple, trust us, we know!.
Let us explain: rising carbon dioxide levels cause lots and lots of heat and that changes climate and that means our children are all going to die horribly unless we do something about it now. Most important we need more grants. Then we need to create lots of nice new taxes and send the money to the nice people who lead the countries in the Third World.
You are making this simple situation much too complex and confusing, we would never dream of considering the things you discuss in your article. KISS has always been our motto, we don’t want to confuse those nice politicians, after all they are very simple people who don’t want to try and understand complex issues.
So, just remember this simple motto: carbon dioxide very bad, climate scientists very good – climate scientists understand, you don’t, and they can always be trusted to tell the truth.
Yours affectionately
The Team
/sarc

Wenson
January 12, 2012 11:06 am

As I mentioned in the another thread, Earth is not a true black body. Using “average T” would “increases” the earth temperature. And I don’t believe CO2 has any meaningful effect on earth temperature.
I’m a layman. Here is more question about CO2.
In anyway can I think a photo is much smaller than a Co2 molecule?
If so
How many photos will miss the target of CO2 molecules? Since CO2 only 0.3% , 0.4% in the air plus there a lots of empty space ( to the photo and air molecules) in the air.
What is the percentage of the photo will miss the target CO2? (at each “layer “of the air)
(In layman’s thinking, Only very very small portion of the photos will hit the CO2 molecules.)
the following is a layman’s another thinking. (based on NASA uses photo to drive a spaceship)
What happens After a photo hit CO2 molecule. (Assume a moving photo has momentum,layman’s photo just a glass ball now)
will the CO2 molecule change speed?
If the CO2 molecule really had speed change, then its energy changed. Assume CO2 molecule randomly moving.
Case 1. CO2 molecule moving faster,
Case 2. CO2 molecule moving slower.
Case 3. CO2 molecule changes moving direction only.
In which case the CO2 molecule will “reflect” a photo out? What is the “reflected” photo’s energy compare to the incident photo?
(Layman’s thinking: maybe not all the “hits” generate a “reflection “. If “reflection” occurs, the energy of the photo would be less than the incident photo’s energy. Because part of the energy transferred to molecule’s kinetic energy. How much less?)
And a molecule with increased speed will dissipate energy more quickly. ( how to relate the speed change to the temperature change? )
Layman’s conclusion: CO2 cannot have big “green house” effect.
I like to learn what is wrong in my thoughts.
Wenson

wayne
January 12, 2012 11:07 am

Robert Brown:
Hi Robert. Just read you post and later comment (and kwik’s).
http://wattsupwiththat.com/2012/01/12/earths-baseline-black-body-model-a-damn-hard-problem/#comment-861864
Guess you know I was the one that jumped into the Unified Theory of Climate thread and coded up that simple integration in c. Kwik’s thoughts are very close to mine. I have already ported that to CSharp and have had thoughts along the same lines: how far could you carry the next steps to an integrated surfaces so all of the pertinent parameters which could be altered to give us all of the variant questions you so well laid out in words.
It would be quite easy to program such a model, staying strictly in CSharp for portability. The integrations to me would pose no problem. I already have in CSharp a set of very fast adaptive Simpson 3/8 interpolators, Runge-Kutta of various orders and my favorite Yashida 6th order sympletic integrator for minimal energy error terms, fast and very precise. After all, all integrators will have round-off error, it’s just where you want them minimize, in the spatial terms or the energy terms.
But the jump to multiple vertical layers is more challenging memory wise and performance. Performance is why some flavor of c would be preferred to MatLab or other script language.
Yes, I have been pondering on these very same questions, could you simulate this whole question that Dr. Nikolov brought up and carry it further towards reality.
Good post Robert.

stumpy
January 12, 2012 11:08 am

Its a key issue, as the greenhouse gasses are assigned their importance on the 33k warming they may or may not actually make – no one never actually took the time to do it right though!

Phil.
January 12, 2012 11:11 am

Stephen Wilde says:
January 12, 2012 at 8:42 am
Excellent article.
One question:
How do we know that the Earth is any warmer than it would be without greenhouse gases if the standard assumptions are so obviously inappropriate and/or incomplete ?

It’s quite straightforward. The emission spectrum of the Earth from space is grey body with numerous missing bands which can be unambiguously assigned to the GHGs (CO2, H2O, O3, CH4 & N2O), the spectrum of the Earth absent those GHGs would be grey body but would have to have the same area under the curve which requires a lower temperature. Therefore there is a GHE due to the presence of those gases.

George E. Smith;
January 12, 2012 11:11 am

“””””Thermodynamic professor, power plant manager says:
January 12, 2012 at 9:41 am
I’m sure that thermodynamic atmosphere effect explains everything much better, it’s the only method that is based on physical facts. If you try to model earths temperature with this very clear Anthonys description you can argue everything until you are dead or used 1 bn$;) Backradiation is bullshit from cooler gas to warmer surface, physically impossible! Gases can’t heat up with infrared radiation without incredible W/m2, only surfaces. “””””
So Thermodynamic Professor, whether or not one believes that the ordinary atmospheric mono or homo-diatomic gases: Argon, Nitrogen, Oxygen (maybe Hydrogen) etc, do, or do not absorb and emit Electromagnetic Radiation under ordinary atmospheric conditions (circa STP), it is an experimental fact that the earth atmosphere DOES emit EM radiation, and there is no theoretical physical basis for believing that such atmospheric emission is not locally isotropic. That is, there is no preferred direction of emission; all directions are equally likely. And some declare on their mother’s grave, that it is solely the triatomic (or more) or hetero-diatomic molecules generally grouped as GHGs that emit this isotropic atmospheric EM radiation.
So whether you or I like it or not, it is an experimental fact that at all levels in the atmosphere, EM radiation IS emitted downwards; and those emitting GHGs can have no a priori knowledge or information as to the Temperature of the earth surface, or whether that Temperature is lower or higher than the local Temperature of the emitting atmospheric gas, at the time such radiation leaves its originating gas molecule.
So now what is it; in your view; as a Thermodynamic Professor, that instructs that radiation that it may not land on the surface, because that surface happens to be at a higher Temperature than was the emitting gas molecule; about which the near surface arriving EM radiation, carries absolutely no information, by which to inform the surface that it originated from a colder gas molecule.
Or is it perhaps that such matters are managed, as in your role as a power plant manager, rather than the laws of Physics, which would be relevent to a Thermodynamics Professor.
How it it that EM radiation emitted from the earth at averaged 288 K Temperature, is safe to land on the moon at average Temperature 240 K but the same solid angle beam may NOT land on the sun, at its surface Temperature of about 6,000 K ??
Some of us non thermo-dynamic professors, and non power plant managers, are anxious to learn the answer to that from you ?

Septic Matthew
January 12, 2012 11:23 am

Francois: Are you serious? You know there are a few books which might help you understand how the system works.
First off, how the system works is not known completely, though a great deal is known.
Second, the main questions are the rates of energy transport through the many parts of the system, and most of those rates are not known exactly. Without knowing those rates, even the notional equilibrium for a constant input can’t be computed accurately. The author has shown that there are substantial reasons for believing that the widely accepted figure is significantly inaccurate.
Robert G. Brown, that’s a good summary. I think it’s something everyone sort of “knows”, but has decided to ignore. It is worth repeating from time to time, until everyone asks “Where has it been shown that the errors in the simplified approximations are negligible.”

wayne
January 12, 2012 11:24 am

Robert, what I was trying to say is if you should ever need any help, let me know. I have written six different solar system simulations able to hold accuracy up to about one thousand years, not a NASA Horizon system clone, but close, and I already know many of the problems such a program would entail when you carry it into the non-linearities.

Lars P.
January 12, 2012 11:27 am

Thank you for the article! Great, we need more of these!
I wonder does the 33°K difference play any role in the assumptions of greenhouse warming for the models?
Hm, what would be the temperature on a rotating sphere covered with water – as stated in the article, before considering the atmosphere and the land, simply 4 km deep ocean everywhere.

Joseph Dunn
January 12, 2012 11:28 am

Robert Brown says:
January 12, 2012 at 9:00 am
…I can probably use e.g. octave/matlab to do them numerically, although the rotating sphere with heat capacity technically requires the solution of a set of time dependent ODEs as a point at a given latitude rotates, through enough rotations to approach a steady state….

You might want to check out the following paper, which works through a model similar to the one you describe.
Vasavada, A. R., D. A. Paige, and S. E. Wood, Near-surface temperatures on Mercury and the Moon and the stability of polar ice deposits, Icarus, 141, 179-193, 1999
Available online at:
http://www.lpl.arizona.edu/~shane/PTYS_395_MERCURY/reading/vasavada_etal_icarus_1999.pdf

January 12, 2012 11:31 am

Bernd Felsche said @ January 12, 2012 at 10:54 am

Keep in mind that climate is entirely synthetic; an arbitrary statistical artifact, based on poor statistics. Only weather is real.

So the trees I planted 30 years ago that reduce windspeed and evaporation from the soil, thus warming it, haven’t affected the climate on my property? It seems odd to me that I grow more grass as a result of something “entirely synthetic; an arbitrary statistical artifact, based on poor statistics”. Perhaps you meant global climate. I always have trouble imagining something to be both local and global.

jorgekafkazar
January 12, 2012 11:33 am

Joe Soap says: “Just out of curiosity what is the geothermal input?…”
I’ve done the estimate and it’s negligible. I’ve also steam heated a cylinder of dirt 12′ deep by 3’Φ and still found the spot quite hot two weeks later. Earth is a great insulator.
Francois says: “…there are a few books which might help you understand how the system works.”
Then you must have read these books, Francois, and understood them. Please tell us in your own words what they say. Maybe Anthony will want to post what you have to say.
crosspatch says: “One question I have had is the impact of changes in solar UV…
You may be onto something. I suspect the impact is significant, despite Leif having said the density of the ionosphere (rather than the stratosphere, as you suggest) is too low to have an effect. But you can’t shoot a photon through the ionosphere without hitting something…
Joel Shore says: “…Holder’s Inequality, which tells us that the 4th root of the average of T^4 is greater or equal to the average of T for any distribution of temperature T. (…it turns out that the current Earth is close enough to having a uniform temperature that the difference between these two ways of computing average temperature is small.)”
True, but when doing the energy balance calculations, any error in Tavg must later be taken to the 4th power, which makes it large once more. Remember, we’re looking at a very few W/m² of putative imbalance in all this global warming nonsense.
David L. says: “…“average temperature”? The number, as has been mentioned on this site many times, is meaningless.”
Yes, as in Joel Shore’s hand waving, above. It’s the energy flows that have meaning. It’s all too easy to brush the hair-splitting nature of the GHG problem under the rug by talking temperature or, better yet, anomaly.

NoAstronomer
January 12, 2012 11:34 am

tl;dr version:
A black body model won’t, can’t, tell you *anything* about the temperature at the Earth’s surface.
Mike.

January 12, 2012 11:36 am

Geckko says:
January 12, 2012 at 8:30 am
Would it be impossile to build a physical model?

There was a book that described this. I believe it was “The Hitchhiker’s Guide To The Galaxy”. If I recall correctly the scale was 100cm to the meter.

George E. Smith;
January 12, 2012 11:36 am

“”””” Bill Illis says:
January 12, 2012 at 10:58 am
How about if one makes it a non-rotating planet first with some set heat capacity (and I’m not sure there is really a limit to how much heat a surface or gas can absorb).
How hot does it get at the equator at the spot that is directly facing the Sun “””””
Bill, the highest officially recorded in shadow atmospheric Temperature, was about 136 deg F (57.8 deg C) somewhere in North Africa. US troops in Iraq apparently often went on patrols in air Temperatures of 130 deg F. I take +60 deg C (140 deg F) to be a common dry desert surface Temperature max. and it has often been reported of black top surfaces reaching +90 deg C
At 333 K BB Temperature rather than 288 K to BB total radiant emittance is 1.787 times higher than Trenberth’s 390 W/m^2.
More importantly, the emission spectrum peak wavelength shifts to a shorter wavelength, further removed from the CO2 15 micron band, reducing the CO2 effect and the peak spectral emittance at that shorter wavelength is 2.067 times what it is at 28 K since that goes as T^5, not T^4.
George
PS when that highest record Temperature occurred, it was of course Winter midnight in Antarctica and at Vostok Station; and the lowest Temperature that has been recorded there was below -128 deg F -88.9 deg C. Anecdotal reports of lower Temperature further up the hill are of course not official numbers. I take -90 deg C (-130 deg F) to be the low limit, and note that whole range could occur simultaneously, and therefore there would be an infinity of places on the earth that can have ANY Temperature within those extreme limits, and all at the same time. So much for hundredths of a degree changes being relevent.

Lady Life Grows
January 12, 2012 11:37 am

All of the above ignore one more highly relevant factor. Climate scientists, alarmists and skeptics alike, usually assume that all the temperature at the surface is due to incoming solar irradiance (except for very trivial lunar and stellar, etc. inputs).
The majority of the heat at the Earth’s surface is due to its radioactive core, especially U-238. Since the half-life of uranium is about 4.5 billion years, that is a factor in changes since the Hadean, or even Cambrian ages, but is effectively a constant for the duration of human existence on the planet.
It is a constant that should be included in estimates and calculations.

Septic Matthew
January 12, 2012 11:41 am

Tim Folkerts: These are all “settled” in terms of the general affect on global temperatures. Of course, the details and exact extent are not “settled” (or we could perfectly predict weather and climate).
Dr. Brown’s point is that the details and “exact extent” are not known accurately. Thus you affirm his main point, and affirm my point above that everyone already knows it. It bears repeating: even something relatively simple like the mean temperature of the earth without atmospheric GHGs can not be computed accurately from the known science.

wayne
January 12, 2012 11:43 am

Why not a game. Gore’s got a game, let’s make a game. As Robert was saying, a tinker-toy put together GCM astronomic body climate simulator for any planet/moon system. Why not. Let’s out do him. A physics law tutorial of kids and parents alike. It seems to have some merit.
If you could simulates all known bodies as Ned laid out in his paper, with the same parameters and functions, that actually performs very close to what is seen and measured on all bodies, I would trust that hugely over the current GCMs being used to solely point at CO2. I think most citizens of this world would too.

Ian W
January 12, 2012 11:45 am

It would have been nice to see a mention of enthalpy. Atmospheric temperature is meaningless as a way of quantifying heat content without knowing the enthalpy of the atmosphere at that point and that is largely dependent on the water content of the atmosphere and the state the water is in ice, liquid or vapor. It then goes without saying that ‘average’ atmospheric temperatures are completely meaningless as an input into energy content.
The main problem is that with a chaotic system of chaotic sub-systems any ‘simplification’ or assumptions of linear behavior lead to results that are rapidly divergent from the ‘real world’. Even mesoscale modeling of the atmosphere 10 km around a point with best feasible start parameters from normal observations plus LIDAR and RADAR etc., is only feasible out to around 30 minutes with any accuracy. Yet the standard mathematician’s approach is to simplify, as demonstrated by many suggestions on this thread. The assumption that those mathematical simplifications have little effect or only a known effect that can be calculated out afterward, is almost certainly false and impossible to validate.

Editor
January 12, 2012 11:46 am

Dr. Brown, as someone else said, I envy your students.
All that you have said is true. However, let me suggest that there are some things we can conclude despite the difficulties.
First, we have a pretty good handle on how much energy hits the earth system after albedo. It’s on the order of 240 W/m2. I often model the simplified, theoretically perfect situation as an airless spherical blackbody in space heated from the inside by unspecified nuclear reactions, with 240W/m2 emitted everywhere on the surface of the sphere. This has an equivalent blackbody temperature on the order of -18°C. Using this model avoids some of the conceptual difficulties with a rotating planet heated from the outside. I will call the Stefan-Boltzmann temperature of such a sphere the “theoretical S-B temperature” of the planet given the 240 W/m2 radiation.
Now you have done an excellent job of discussing all of the ways that an actual rotating planet with axial tilt will diverge from the theoretical number of -18°C. The important point, as you pointed out, is that in all cases, assuming total emitted radiation stays constant, any divergence from a perfectly even surface distribution of emitted radiation will lower the average temperature from the theoretical S-B temperature.

Knowing this inequality, that all temperature swings and variations only cause average cooling and never warming, will let us draw important conclusions.
Second, we have a pretty good handle on how much the earth surface is emitting. It’s on the order of 400 W/m2. It is estimated from temperatures taken on the surface as well as from satellite observations.
This plus the first fact allows us to put a minimum value on the heating which comes from the “greenhouse effect”. The earth is warmer than its theoretical S-B temperature by at least thirty degrees Celsius. As you point out, it is assuredly more than that. How much more? “Dooo theee integral”, as the wise man said … but for some purposes, it doesn’t matter how much more. That gives us the minimum amount of the greenhouse effect.
This becomes important given the recent prominence of such hypotheses as Jelbring’s and Nikolov’s. As I understand the claims of their proponents, these are said to explain that thirty degrees of warming above the theoretical S-B temperature. However, they say the thirty degrees C of warming is from some atmospheric gravitational effect which doesn’t depend on GHGs.
Now, I hold the following, based on your statements above. My claim is much more general than the specific hypotheses of Jelbring and Nikolov:
If the atmosphere is transparent (contains no GHGs), there is no way for any atmospheric gravitational effect to raise the average surface temperature of a planet above the theoretical S-B temperature.
The proof is by contradiction. Assume as above a planet heated from the inside by radiation. Give it a perfectly transparent, GHG-free atmosphere.
If any such atmospheric or gravitational or any other effect existed, and the surface temperature were raised above the theoretical S-B temperature by such an effect, the inequality noted above means that the outgoing radiation would have to increase.
But since the atmosphere is perfectly transparent, that means that the surface would be radiating more than it is absorbing. This is a perpetual motion machine, and as such a violation of conservation of energy. Q. E. D.
Professor Brown, many thanks for your contributions. I’d be interested in your comments on my proof that no possible atmospheric-based mechanism can push a planet warmer than its theoretical S-B temperature, and I know you are late for class. If you have time.
All the best,
w.

Septic Matthew
January 12, 2012 11:50 am

Harry Dale Huffman: In the context of public debate, the “greenhouse effect” is not about radiative equilibria with and without an atmosphere, or even with or without “greenhouse gases” in the atmosphere. It is about whether atmospheric temperature at the surface (or at any given pressure level) increases with an increase in atmospheric carbon dioxide. I am astonished that even skeptics cannot focus upon this obvious fact in the real world; everyone can’t seem to stop themselves from launching into radiative transfer theory arguments.
I agree, but addressing the main issue entails addressing a lot of other issues. The effect of additional CO2 can’t be known if the rest of the system is not known well enough.

Scarface
January 12, 2012 11:53 am

@Wenson: It’s worse than you thought. CO2 is only 0,04% of the air!

John West
January 12, 2012 11:55 am

What is the rational behind the statement: “where it is emitted at a cooler temperature”? I’m not convinced that the Stefan-Boltzmann Law applies to GHG’s any more than it does fluorescent or neon lights. It seems to me that a CO2 (or H2O) molecule can emit IR at a rate not proportion to it’s temperature as long as the temperature is above the “freezing” point of it’s vibration modes (degrees of freedom), such that temperature is merely a threshold variable to IR emissions from GHG’s.
Consider the following:
1) The temperature of a gas is a measure of the kinetic energy of translation not vibration. “Fundamentals of Modern Physics”
2) The specific heat of a gas increases proportionally to degrees of freedom availability. More energy is required to increase the temperature of a gas as degrees of freedom other than translation are “unfrozen” suggesting energy stored in non-translational modes do not effect temperature directly.
http://theory.phy.umist.ac.uk/~judith/stat_therm/node81.html
3) Even though equipartition of energy occurs within a mass of gas, the likelihood for a collision capable of imparting a translational motion from a vibration motion is the same as the likelihood for a collision capable of imparting a vibration motion from a translational motion for a given set of circumstances. Whether the atmosphere would be heated or cooled is dependant on the probability of a collision capable of transferring motion to/from translation to/from vibration encountering the opposite condition; or to put another way, the proportion of GHG to IR input. Consider the atmosphere if the Earth did not radiate IR; any collisions resulting in vibration motion could be emitted as IR, reducing the overall translational motion of the atmosphere, thus cooling it. Considering the other extreme where the Earth radiated so much IR that all available vibration modes were always immediately exited by IR input; then equipartion of energy would be “averaging” energy from vibration induced by IR input into translational movement through collisions thus heating the atmosphere. Obviously, the actual atmosphere is somewhere in between these two extremes. Whether GHG’s heat or cool the atmosphere depends on the amount of GHG’s (increases may result in less direct heating), the amount of IR input, and the availability of vibration modes. Only the availability of vibration modes is dependant upon temperature and then only as a threshold.
4) Absorption and emission of specific bands of IR by GHG’s corresponding and limited to vibration modes also suggests that IR absorption and emission by GHG’s are not black body emissions but instead exhibit this characteristic of “cold radiators”.
5) Earth’s atmosphere temperature profile does not correlate with GHG concentration appreciably warming the atmosphere.
http://regentsearth.com/Illustrated%20ESRT/Page%2014%20(Properties%20of%20Atm.)/ESRT10-Properties%20of%20Atmosphere.jpg
6) Anecdotal evidence: IR heaters and lamps do not heat the air in a room directly but heat the IR absorbing surfaces exposed to its output.
I conclude from the above that IR emissions from the atmosphere are not directly proportional to its temperature and therefore cannot be black/grey body emissions or Stefan-Boltzmann Law would be violated. Therefore, a temperature increase is not required for an increase in atmospheric IR emissions (atmospheric radiance is not proportional to temperature). How can an increase in GHG mass in the atmosphere cause an increase in GHE (back radiation) prior to any significant atmospheric temperature increase if a temperature decrease (emitting from higher, colder position) reduces the radiance of the atmosphere? Until I see some evidence to the contrary, I maintain that no temperature increase is required in order to emit an additional amount radiation from GHG’s both down and up.

GregO
January 12, 2012 12:07 pm

Dr Brown,
Thanks for your thought provoking post. I am a big fan of (attempting) to put into words and logical arguments tough modelling problems. An excellent, though non-related text on the topic of CFD is “Computational Fluid Dynamics” by John D. Anderson, Jr and I only mention this text because Anderson illuminates the wall-to-wall Navier-Stokes equations of CFD with actual physical and logical descriptions. For me, this really brought the topic to life. You have done the same here in this post for modelling GHG and I appreciate your work.
Carrick – thanks for the link; I’ve printed out the paper and will study it. As usual, you rock.
Francois says:
January 12, 2012 at 8:30 am
“Are you serious? You know there are a few books which might help you understand how the system works.”
Francois, I’m a mechanical engineer only interested in climate change since Climategate 1.o but since then have read over 800 pages (on last count) of scientific papers and one text book on climate change, and (14) more books on the topic – not necessarily highly technical books. I come away finding that mankind has far more impact on humanity than we on this dead orb circling a massive star while embedded in a complex solar-system. If you can suggest “a few books” that would not be beyond the understanding of a graduate-level mechanical engineer I am sincerely interested in knowing the titles. BTW, of those books I read, the textbook was the most unsatisfying.

Scarface
January 12, 2012 12:11 pm

Dear Mr. Eschenbach,
Please allow me to combine the things you said with a comment of Neo, to ask you a question:
You said:
First, we have a pretty good handle on how much energy hits the earth system after albedo. It’s on the order of 240 W/m2. This has an equivalent blackbody temperature on the order of -18°C.
Second, we have a pretty good handle on how much the earth surface is emitting. It’s on the order of 400 W/m2. This plus the first fact allows us to put a minimum value on the heating which comes from the “greenhouse effect”.
The earth is warmer than its theoretical S-B temperature by at least thirty degrees Celsius.
Neo says:
January 12, 2012 at 10:37 am
The Earth radiates it’s own heat generated at the core.
My question is:
How much of that 30 degrees could earth have produced by itself?
I’m looking forward to your response. Thanks in advance!
Kind regards,
Scarface

Tim Folkerts
January 12, 2012 12:17 pm

Lady Life Grows says: “The majority of the heat at the Earth’s surface is due to its radioactive core …”
All estimates I have seen suggest that the geothermal heat flow is less than 1 W/m^2 (compared to 100’s of W/m^2 for the sun). People are welcome to find their own estimates, but I am 99.999% sure they will find that geothermal contributions are very far from being “the majority”.

Kelvin Vaughan
January 12, 2012 12:20 pm

Dont forget the earths magnetic field!

Stephen Wilde
January 12, 2012 12:24 pm

“I maintain that no temperature increase is required in order to emit an additional amount radiation from GHG’s both down and up.”
Would it follow that acceleration of energy flow by radiation upward to space would offset deceleration of energy flow by radiation downward to the surface for a zero net effect ?

Lars P.
January 12, 2012 12:32 pm

Michel de Rougemont says:
January 12, 2012 at 9:34 am
At Earth orbital distance from the Sun the total incoming energy provided by the Sun is 1366 / 4 = 341.5 W m-2 .
(solar constant divided by 4 to take day/night and spherical shape into account).
Without an atmosphere, reflection and absorption of the solar irradiance would only take place at the Earth surface, composed by ice, rocks and sand. With an assumed albedo of 0.4 (reflecting power of a surface) the earth surface would absorb 341.5*(1-0.4) = 204.9 W m-2 and re-emit this energy to the outer space.
According to the Stefan- Boltzmann law, the mean temperature at the earth surface would establish itself at:
T= (E / ε σ)^¼ = (204.9/(1 x 5.6710-8))^ ¼ = 245.2 K (-28°C)
Michel, how does it look like with a water planet with 12 hours of 1366/2 = 683 W m-2 and 12 hours night?
Water emisivity and albedo are known. Ignore any atmosphere, clouds, evaporation but waters heat capacity, keeping also in mind that light is warming into depth.

Stephen Wilde
January 12, 2012 12:37 pm

“But since the atmosphere is perfectly transparent, that means that the surface would be radiating more than it is absorbing. This is a perpetual motion machine, and as such a violation of conservation of energy. Q. E. D.”
No atmosphere is transparent to conduction.
Conduction being a slower process than radiation an increase in the amount of conduction relative to radiation is what causes the equilibrium temperature to rise.
A planet with no atmosphere receives radiation in and sends it back out again virtually instantly.
The more mass in the atmosphere the more conduction takes place, the slower is the rate of energy loss to space and the higher the equilibrium temperature must rise.
Gravitationally induced pressure at the surface increases density of the gas at the surface by reducing volume so as to make conduction occur sooner/faster than if the gas were less dense which increases the greenhouse effect of the atmosphere.
That is the true greenhouse effect and it is nothing to do with radiative characteristics of individual gas molecules.
It is a matter of mass and gravitationally induced pressure reducing outward radiation in favour of an increase in conducted energy accumulating in the atmosphere.

David L.
January 12, 2012 12:42 pm

Temperature swings are cyclical at least daily and yearly (if not decadal, etc.). Daylight temps are higher and night temps are cooler. This daily cycle is superimposed on a yearly cycle where the temperature is coldest in the winter and warmest in the summer. Yet all this cyclic temperature behavior is boiled down to an average. All the information is in the cycle, not in the average.
It’s analogous to an AC voltage: a sine wave or “alternating current” that swings from -170V to 170V. You can measure Peak, (170V), peak to peak (340V) or RMS (120V) voltage. What don’t they measure? The average voltage. What is the average? Zero volts. Which climate scientist would stick their finger in a wall socket when told the average voltage was zero?

Russ in Houston
January 12, 2012 12:44 pm

Why are we worried about the temperature of the sphere. We live in the atmosphere near the surface of the sphere. Also, average global temperatures have no value other than for curiosities sake.

January 12, 2012 12:49 pm

Too complicated. Start like this: Earth is on average warmer than Moon. Earth day (+20°C) is much cooler than Moon day (+130°C), so the Earth night (+10°C) is much, much warmer than Moon night (-230°C). Why?
Some say it is a thick arrow of IR radiation, blazing on the Earth surface from black cold sky. At least Kiehl-Trenberth diagram says so. That every-present beam of downward IR radiation is so powerful, that our night has 10°C instead of -230°C. Must be so, since our day is, cough cough, mightily cooled by abundant “greenhouse gas” in form of clouds, ice/snow and evaporation. So that invisible stream of energy, falling at our heads in a speed of light is at every moment warming the night surface by incredible 240°C!!! Dare you to switch it off for a moment, like hiding under a roof; you would freeze instantly! And what is even more awesome, the whole stream of energy, similar to heat wave produced by nuclear explosion or what, is just a less half of recycled outgoing radiation, since the night surface of 10°C, well, also radiates something upwards. It means, however, that the “something” leaving the night surface from under our feet is more than twice powerful as the night sky downward radiation! Are you roasting already? Feel the heat?
Sorry gents, this radiation nonsense is just nonsense. At night, I am warmed by a bulk atmosphere, nitrogen and oxygen, which is keeping the daily heat.
PS. Is there any consensus what the elephant in the room (nitrogen and oxygen warmed by conduction and convection from the surface warmed by Sun) actually do? If they radiate IR (as all matter with temperature >0K should do), good luck in recognizing the IR coming from one molecule of CO2, compared to 25,000 other molecules (this is the ratio of CO2 above the “safe” 350 ppm level). If they do not radiate IR, they simply keep the daily accumulated heat, working as a thick blanket.
PPS. How is that 17W in Kiehl-Trenberth diagram, assigned to thermals, calculated? Is it a physically reasonable number, compared to 333W of outgoing radiation from the surface?

B.Klein
January 12, 2012 12:49 pm

Dr. Brown has done a good job of restating a very simple statement : The Earth is not a “theoretical Black body ” of course this was also stated by Gerlich and Tscheuschner in the abstraction of their paper (see below) in 2009.
Now lets see if Dr. Brown can come up with a similar analysis to show that the Hypotheses of the “greenhouse gas effect” is a fairy-tale – a hoax. below is a list of references for him to ponder.
List of references:
The paper “Falsification of the Atmospheric CO2 greenhouse effect within the frame of physics” by Gerhard Gerlich and Ralf D. Tscheuschner is an in-depth  examination of the subject.  Version 4 2009
Electronic version of an article published as International Journal of Modern Physics
B, Vol. 23, No. 3 (2009) 275{364 , DOI No: 10.1142/S021797920904984X, c World
Scientific Publishing Company, http://www.worldscinet.com/ijmpb.
Report of Alan Carlin of US-EPA  March, 2009 that shows that CO2 does not cause global warming.
 
Greenhouse Gas Hypothesis Violates Fundamentals of Physics” by Dipl-Ing Heinz Thieme This work has about 10 or 12 link
that support the truth that the greenhouse gas effect is a hoax.
R.W.Wood
 from the  London, Edinborough and Dublin Philosophical Magazine , 1909, vol 17, p319-320. Cambridge UL shelf mark p340.1.c.95, i
The Hidden Flaw in Greenhouse Theory
By  Alan Siddons
from:http://www.americanthinker.com/2010/02/the_hidden_flaw_in_greenhouse.html at March 01, 2010 – 09:10:34 AM CST
The below information was a foot note in the IPCC 4 edition. It is obvious that there was no evidence to prove that the ghg effect exists.
 
“In the 1860s, physicist John Tyndall recognized the Earth’s natural greenhouse effect and suggested that slight changes in the atmospheric composition could bring about climatic variations. In 1896, a seminal paper by Swedish scientist Svante Arrhenius first speculated that changes in the levels of carbon dioxide in the atmosphere could substantially alter the surface temperature through the greenhouse effect.”
 
After 1909 when R.W.Wood proved that the understanding of the greenhouse effect was in error and the ghg effect does not exist. After Niels Bohr published his work and receive a Nobel Prize in Physics in 1922. The fantasy of the greenhouse gas effect should have died in 1909 and 1922. Since then it has been shown by several physicists that the concept is a Violation of the Second Law of Thermodynamics.
 
Obviously the politicians don’t give a dam that they are lying. It fits in with what they do every hour of every day .Especially the current pretend president.
Paraphrasing Albert Einstein after the Publishing of “The Theory of Relativity” –one fact out does 1 million “scientist, 10 billion politicians and 20 billion environmental whachos-that don’t know what” The Second Law of thermodynamics” is.
ILEUniversity of Pennsylvania Law School
INSTITUTE FOR LAW AND ECONOMICS
A Joint Research Center of the Law School, the Wharton School,
and the Department of Economics in the School of Arts and Sciences
at the University of Pennsylvania
RESEARCH PAPER NO. 10-08
Global Warming Advocacy Science: a Cross Examination
Jason Scott Johnston
UNIVERSITY OF PENNSYLVANIA
May 2010
This paper can be downloaded without charge from the
Social Science Research Network Electronic Paper Collection:
http://ssrn.
Israeli Astrophysicist Nir Shaviv: ‘There is no direct evidence showing that CO2 caused 20th century warming, or as a matter of fact, any warming’ link to this paper on climate depot.
[1] [2] Slaying the Sky Dragon – Death of the Greenhouse Gas Theory [Kindle Edition]
Tim Ball (Author), Claes Johnson (Author), Martin Hertzberg (Author), Joseph A. Olson (Author), Alan Siddons (Author), Charles Anderson (Author), Hans Schreuder (Author), John O’Sullivan (Author)
 
Israeli Astrophysicist Nir Shaviv: ‘There is no direct evidence showing that CO2 caused 20th century warming, or as a matter of fact, any warming’ link to this paper on climate depot
Web- site references:
http://www.americanthinker.com                                            Ponder the Maunder
wwwclimatedepot.com
icecap.us
http://www.stratus-sphere.com
SPPI
The Great Climate Clash -archives December, 2010 , G3 The greenhouse gas effect does not exist.(  peer reviewed and revised  but not yet released).
Wood is correct: There is no Greenhouse Effect
Posted on July 19, 2011 by Dr. Ed
Repeatability of Professor Robert W. Wood’s 1909 experiment on the Theory of the Greenhouse (Summary by Ed Berry. Full report here or here. & PolyMontana.)
by Nasif S. Nahle, June 12, 2011
University Professor, Scientific Research Director at Biology Cabinet® San Nicolas de los Garza, N. L., Mexico.
many others are available.
The bottom line is that the facts show that the greenhouse gas effect is a fairy-tale and that Man-made global warming is the World larges Scam!!!The IPCC and Al Gore should be charged under the US Anti-racketeering act and when convicted – they should spend the rest of their lives in jail for the Crimes they have committed against Humanity.
The only thing more dangerous than ignorance is arrogance.”
—Albert Einstein
“Democracy is two wolves and a lamb deciding what to have for dinner. Liberty is a well-armed lamb.”   Benjamin Franklin
 [1]HTML:
 [2]HTML:

adolfogiurfa
January 12, 2012 12:51 pm

Worst of all, the Earth has no lid!

Fernando (in Brazil)
January 12, 2012 12:51 pm

So, for example, for Earth we know that the average intensity absorbed is ~240 W/m^2 and the average emission is ~390 W/m^2 or so….
If this is true.
I ask.
? How many Earths are needed to increase the sun’s surface temperature in 100K???

wayne
January 12, 2012 12:56 pm

John West says:
January 12, 2012 at 11:55 am
How can an increase in GHG mass in the atmosphere cause an increase in GHE (back radiation) prior to any significant atmospheric temperature increase if a temperature decrease (emitting from higher, colder position) reduces the radiance of the atmosphere? Until I see some evidence to the contrary, I maintain that no temperature increase is required in order to emit an additional amount radiation from GHG’s both down and up.
– – – –
John, think you hit it on the nose. One additional viewpoint I might add, the way I see it, is that in all aspects but one radiation in an atmosphere is but a fast conductor. The exception is radiation can leave the atmosphere into space of course. But within the thick of the gases, radiation performs everything conduction performs but on a larger, faster scale and depending on the concentration of other molecules (usually the same but not always) with the exact frequencies to absorb the radiation.
Radiation therefore allows an atmosphere to equalize faster that without it. I see like you seem to see that there is no magic concerning radiation at the surface-atmosphere interface. Radiation just bounces from molecule to molecule over many meters instead of directly from molecule to molecule as seen in conduction.
Mix that in with the thermalization, re-excitation and the equipartition points you made and you might see it the way I do. Phil seems to see radiation in a different manner, totally separate that the other means of transfer… like it had a magical aspect compared to conduction, I see none (but being able to leave that is).
I have a hard time to get others to view it that way, but that view is much simpler to visualize while remains proper in physics.

gbaikie
January 12, 2012 1:02 pm

Earth’s baseline black-body model – “a damn hard problem”
And Willis quote from pg 3:
“But in practice, the fact that you can buy a handheld remote thermometer, which uses the strength of IR radiation plus the S-B equation to measure the temperature of common objects around us means that S-B almost holds almost everywhere. Most things radiate at a level which is quite close to their theoretical S-B radiation of epsilon sigma T^4.”
And post I didn’t post. Which could summed up as the temperature you take isn’t the energy radiated. It involved warm concrete and evaporation rate. But anyhow, that’s on my mind.
So there no doubt about S-B equation measuring temperature. No doubt about energy must balance- incoming energy “absorbed” must roughly equal out going radiation. If they don’t balance it means the planet is warming or cooling.
What we do know without doubt is the earth balance does not change in a day or week- the whole planet does not get warmer or cooler is short period time- it’s at most like .001 K for very fast warming or cooling globally if occurs in a less than week’s time
But as obviously know the each day it warms and cools- it’s always warming or cooling- it’s the sum total which isn’t changing- it’s warming in one part of planet and cooling at same time in another.
In post I didn’t post I was attempting to argue that a warmed piece sidewalk which was 20 C warmer than same sidewalk in the shade, was radiating significant amount energy and the shaded wasn’t radiating any or small amount. the proof was that if shaded the warm sidewalk and put few grams of water on it, it evaporate the water much quicker than sidewalk always in shade.
In other words I could measure how heat the warmed sidewalk emitted. Or something at temperature of 400 watts per meter [shaded] will evaporate less than warmed sidewalk which was emitting 500 watts. Or in other words I going to propose a test if the 400 watt surface could do work as compared something something emitting 100 watts more. 2 square meters at 400 watts “should” do more work than 1 sq meter at 500 watts? But it’s flawed because doesn’t make anything clearer. Hence the non post.
Or surface is covered little pyramids thereby at least doubling the surface area, come anywhere close to radiating twice the energy? it will radiate more, that pretty I am certain, but a lot more?
Just to be clear, if surface is in sunlight, and it’s reached about hot as it going to get, it will radiate the same amount energy it is receiving from the sun. So sidewalk in sunlight will radiate the solar energy- it will radiate 1000 watts minus maybe 10 watts it is absorbing [heat conducted thru concrete and very amount to ground below it]. So incoming sunlight 1000 watts, and sidewalk in sun radiating around 1000 watts. When you shade the sun and sidewalk is reflecting less sunlight, but continues to radiate around 900 watts of energy.- and this recently shaded sidewalk can lots of work. It has power, it has watts of energy- maybe roughly 100,000 watts seconds of work it can do.
The always shaded sidewalk won’t do work. Put both in space in which they can radiate to 2 K, and both can do work.
Or pour liquid nitrogen on either and they both can do work. But my point is the shaded sidewalk with same temperature as the air, doesn’t do work. So it can be said to be at 400 watts. But 400 watts per second of energy it is not emiting- it is possible to emit it, but it’s not doing it.
So of course this point, we get: “Of course, this is the greenhouse effect, you idiot!!!
Measure the sky it’s 400 watts and is warming the sidewalk or preventing it from radiating it’s energy!!!”
But this isn’t really true. “It” might doing a little bit.
You have similar thing if in space. Yes it cools, how much power does it cool at? It’s a question of the conduction rate of the material.
If the magical perfect bodybody, it’s not question- it radiates 400 watts of energy per second. But this concrete we talking about it, not magical material. The magical material isn’t just colored black, it conducts heat perfectly- which isn’t vaguely a descriptive of concrete. Copper or diamond get somewhat close to a blackbody- as far as conductivity.
Now, I can do better than a black body as far as losing heat/doing work. Liquid Nitrogen, will cool that sucker fairly fast, not immediately, but explosively. Simple water will also cool it fairly fast- and water in vacuum of space is probably better than liquid nitrogen.
One way of saying this is the sun in terms of radiating heat is pretty good, but what it heat up isn’t as good at radiating heat. And the closer you are the the sun the better it is at radiating- it heats things faster.
So Greenhouse gases and/or conduction of air might be slowing an shaded concrete to from cooling, but even in space it doesn’t radiate at 400 watts- it’s that temperature, but temperature can’t measure the work it does, different things at same temperature can do more work per square meter- and the fictional black body does the most work.

More Soylent Green!
January 12, 2012 1:03 pm

kwik says:
January 12, 2012 at 8:35 am
You need to write a CSharp program and combine it with a ball made in WPF. WPF (Windows presentation Foundation) is perfect for this. We use it in my business every day.
If we could get some money from some rich guy, and I could take a 2 years leave from my day job, I would love to join you.
We could patch up the globe with small triangles with texture. The globe can be made in any 3-D package and exported to xaml. You can then import it into a WPF project and then write a CSharp program that starts turning the globe….and start calculating stuff…..ahh…a fantastic project…. Click “play” and it comes alive….
But for crying out loud; Dont make a “Report for Policy-Makers” when we are finished!!!!!

Let’s make a report to the policy-makers before we finish!

Brian H
January 12, 2012 1:09 pm

clivehbest says:
January 12, 2012 at 10:53 am

Does such a place exist ? I wonder if there is perhaps just one long term weather station somewhere in the Sahara for example ?

There was, but Lawrence’s Bedou reassembled it into a couple of muskets. And one ceremonial daggar.

Brian H
January 12, 2012 1:10 pm

[Yes, I know, wrong desert. But somebody did it!!]
😉

January 12, 2012 1:18 pm

George E. Smith; says:
January 12, 2012 at 11:11 am
“…that instructs that radiation that it may not land on the surface, because that surface happens to be at a higher Temperature than was the emitting gas molecule; …”
Mr. Smith, it is agreed that the EM can land on a warmer surface the question is does it get absorbed, as some say, there by increasing the W/m^2.
Sir, if I have 2 torches at 1000 C and point one at a spot what is the max temperature that spot can get? If I then point the other torch at the same spot at the same time what is the max temperature the spot can get? When I remove one of the torches what is the max temperature the spot can get?
I was taught, years ago admittedly, that a higher energy state object does not absorb lower energy. Maybe poorly said but I think you understand.

Thermodynamic professor, power plant manager
January 12, 2012 1:19 pm

George E. Smith; says:
January 12, 2012 at 11:11 am
So Thermodynamic Professor, whether or not one believes that the ordinary atmospheric mono or homo-diatomic gases: Argon, Nitrogen, Oxygen (maybe …..
Sorry but we have never managed to warm up anything with colder objects or gases, the nonsense you are talking about. This is based purely for Thermodynamic laws. It would of course be very nice if such unphysical [snip] happens, our power plant production costs would be much lower;)

January 12, 2012 1:25 pm

Let me add more to your eq.
The Earth’s core is molten and heat rises to the surface unevenly. We have fewer data points on this heat transfer than on the solar flux. Mid ocean ridges are literally boiling with heat (trapped because the pressure can raise the temperature of the water WELL above 100°C (boiling at sea level).
No one has disproved that El Nina/El Nino are not actually due to fluctuations in heat from the dense number of ridges in that area of the Pacific. Speaking of Coriolis effects, why aren’t rotation directions of the PDO and ANO etc driven by super hot water coming from the depths instead of solar heating at the surface????

Stephen Wilde
January 12, 2012 1:26 pm

“The emission spectrum of the Earth from space is grey body with numerous missing bands which can be unambiguously assigned to the GHGs (CO2, H2O, O3, CH4 & N2O), the spectrum of the Earth absent those GHGs would be grey body but would have to have the same area under the curve which requires a lower temperature. Therefore there is a GHE due to the presence of those gases.”
That could also happen if the initial warming was from surface to atmosphere via conduction could it not ?
Thus there is a GHE effect but not necessarily caused by those gases.The energy that would have been in the missing wavelengths could just be offset by an increase in other outgoing wavelengths.No one is denying that GHGs radiate 50% upwards are they ?
Suppose that the radiative effect of GHGs were net zero because they send as much energy out of the system as they retain within the system. (non GHGs retain 100% within the system).
What does it matter if GHGs utilise different spectral bands to those used by the surface ?
They could be acquiring all of their energy from the other non GHG molecules which can be heated by conduction from the surface and would pass it on to the GHGs via conduction. Once the GHGs reach the ambient temperature for other molecules at the same height (via conduction) they cannot get any warmer from longwave coming up from the surface so they would have no more ‘blocking’ ability than the non GHGs already have (100%). Meanwhile they are busily radiating upward.
They may pass 50% of their energy back down again but by radiation rather than by conduction and unlike non GHGs that 50% passed down would be matched by 50% sent out of the system altogether which is something that non GHGs cannot achieve.
I think it is conduction intervening between radiation coming into the system and radiation going out of the system that creates the greeenhouse effect and not radiative physics.

Jean Parisot
January 12, 2012 1:35 pm

How does the ovoid shape of the surface effect the energy model, the seasonal tilt angles do not have the same surface areas presented to the sun.

Robert Clemenzi
January 12, 2012 1:36 pm

John West says:
January 12, 2012 at 11:55 am

Whether GHG’s heat or cool the atmosphere depends on the amount of GHG’s (increases may result in less direct heating), the amount of IR input, and the availability of vibration modes.

That ignores the energy added by conduction. Because the amount of energy emitted is always greater than the amount absorbed, increasing greenhouse gases will always reduce the temperature of the atmosphere. (Does not apply above the troposphere.)

Robert Clemenzi
January 12, 2012 1:39 pm

Tim Folkerts says:
January 12, 2012 at 10:49 am

5] GHGs high up in the atmosphere where it is colder would raise the surface temperature.
These are all “settled” in terms of the general affect on global temperatures.

I fully agree with points 1 thru 4, but 5 is a problem. First, what do you mean by “high”?
Below the tropopause? Above it? Also which GHG are you suggesting?
In general, it is my understanding that GHGs cool the atmosphere, with the obvious exception of the stratosphere. It is also my understanding that GHGs below about 2 km warm the surface, and that those higher up have no effect on the surface temperature.

January 12, 2012 1:39 pm

Robert Brown (and others)
Thank you. It’s great to see fundamentals coming up for rethink with clarity. Two points
(1) Two others have referred to Miskolczi here, suggesting you might be interested to check him out. I second that. I still don’t understand his theory, but I do recognize that it has an extraordinarily close fit with actual measurements both from Earth and from Mars. This makes me think he is really onto something significant, if only he could be explained to duffers.
(2) What is the significance of the “W” shape temperature profile of our atmosphere? Is this prime evidence of the GHG effect of ozone formed at these heights? and by extension, proof of the GHG effect of CO2 which tends to lie low?

timg56
January 12, 2012 1:53 pm

Robert Brown shows why he is one of the commentors I most admire here.
(So much so I’m willing to ignore he’s a Dukie.)

January 12, 2012 2:01 pm

Hölder’s Inequality requires that the average temperature on a non-isothermal sphere at radiative equilibrium, assuming the same emissivity and incident power level, must be less than the average temperature of an isothermal sphere. That’s something that G&T do correctly ( http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf Section 3.7.4 ).
As always, an awesome post (although I know of the inequality and derived a special form of it last week, I am infected with G&T’s skepticism about ALL of the physics now and want to see it all for myself). I’m about halfway through G&T, getting a bit bogged down in the rhetoric but looking forward to chapter 5 and the promised (hopefully correct) physics. I’ve already concluded that while top-of-atmosphere measurements do mean something with respect to the absorption properties of CO_2 (and probably with the warming) that a) it ain’t like “a greenhouse” except in the most metaphorical of senses; and b) the physics is way more difficult than even Caballero suggests. How it all works out — I’m open minded.
rgb

January 12, 2012 2:03 pm

Thank you. It’s great to see fundamentals coming up for rethink with clarity. Two points
Neither of which I’m prepared to answer. My reading list is (still) full to the brim, and I have no good idea of why there is a stratosphere and a troposphere and an exosphere and all that. I’ve read a few things that purport to explain parts of it, but none of them have completely convinced me, in part because I suspect one needs a “wholistic” explanation, not an explanation by parts.
rgb

January 12, 2012 2:06 pm

It is also my understanding that GHGs below about 2 km warm the surface, and that those higher up have no effect on the surface temperature.
Although there may be some surprising (?) evidence that water vapor content in the stratosphere may play some role. But separating cause and effect is difficult here. Relatively warm water vapor, very high up — what’s not to like?
rgb

Robert Clemenzi
January 12, 2012 2:10 pm

Lucy Skywalker says:
January 12, 2012 at 1:39 pm

(2) What is the significance of the “W” shape temperature profile of our atmosphere?

That is caused by greenhouse gases. The tropopause is cold because the water vapor emission bands are transparent above that point. The primary greenhouse gas in the stratosphere is CO2 (380 ppm). Water vapor is 5 ppm and ozone is less than 1 ppm. CO2 in the stratosphere emits radiation toward the surface. At the tropopause, that radiation is absorbed by other CO2 molecules, those heat the local atmosphere, which then heats the water molecules, which finally emit the heat to space.
None of the stratosphere heat makes it below the tropopause because the air pressure makes the spectra bands more opaque as the pressure increases.
I have written a full description of this here.

January 12, 2012 2:11 pm

How does the ovoid shape of the surface effect the energy model, the seasonal tilt angles do not have the same surface areas presented to the sun.
An excellent question. The answer is that as far as energy influx is concerned for the simple models, one has to consider the flux of the poynting vector (an integral over the surfaces) associated with the vector-directed intensity distributions. This really can’t be done by pretending that everything is one-dimensional as is commonly done (as G&T, linked above, point out). This makes the math considerably more difficult, but not intractable numerically if you know the shape and rotation and tilt and all that. Just a major pain of direction cosines, I guess.
Things get even nastier, as one might imagine, once you add a relatively thick atmosphere and a nice thick asymmetrically distributed ocean and mountains and plains and forests and rocks and swamps and ice and clouds and circulation. I’m waiting to see if G&T have any suggestions for a decent decomposition of the problem, but I’m prepared for the answer to be “no”.
rgb

Bart
January 12, 2012 2:29 pm

Joel Shore says:
January 12, 2012 at 9:14 am
…So, for example, for Earth we know that the average intensity absorbed is ~240 W/m^2 and the average emission is ~390 W/m^2 or so… [as] in fact the emissivity of the Earth is very close to 1 over the relevant wavelengths.
In fact, we do not know this, because we have not tested under conditions reflecting the true environment.
B.Klein says:
January 12, 2012 at 12:49 pm
“After 1909 when R.W.Wood proved that the understanding of the greenhouse effect was in error and the ghg effect does not exist.”
He found that it did not exist in a laboratory setting. But, in a dynamic atmosphere which results in Doppler broadening of the spectral absorption bands, any GHG effect is likely more pronounced.
Both of these evidences, both pro- and con-, rely on laboratory experiments which do not represent the actual physical environment.

Myrrh
January 12, 2012 2:33 pm

Robert Brown says:
January 12, 2012 at 9:00 am
I think it would be very educational to do this and would take the guesswork out of the question “what does water do” or “what does an atmosphere do” to the sphere (relative to superconducting or insulating static sphere).
Why not start from where water has been included?
Earth temperature 15°C
Earth without any atmosphere: -18°C
Earth with atmosphere but without water, therefore no water cycle: 67°C
Therefore, the plus 33°C ‘warming’ from -18°C to 15°C comes via a 52°C cooling through the Water Cycle.
And there’s also that forty years ago NASA had to junk Stefan-Boltzmann to get real moon temps estimated for the landings, somewhere on-line, as they needed three dimensions not the flat earth of SB and they had to include that thermal energy absorbed from the Sun penetrated and was released later.

Anders L.
January 12, 2012 2:33 pm

“I can’t figure this out. Considering how smart I am, it must mean that noone else can, either.”

Septic Matthew
January 12, 2012 2:38 pm

Willis Eschenbach: First, we have a pretty good handle on how much energy hits the earth system after albedo. It’s on the order of 240 W/m2.
Second, we have a pretty good handle on how much the earth surface is emitting. It’s on the order of 400 W/m2.

I have a feeling that you have left out something important, or I have forgotten something important — I note you wrote “earth system” in the first sentence and “earth surface” in the second. If this were true as written, we would have a pretty good handle on the earth losing 160 W/m^2. What do you mean by “on the order of”? that the error can’t be more than 1%? 10%? 100%? Since electronic communication can sound snarky when not intended so, let me apologize in advance if this sounds snarky.

J. Bob
January 12, 2012 2:39 pm

Francois says:
January 12, 2012 at 8:30 am
“Are you serious? You know there are a few books which might help you understand how the system works”.
I guess, I would have to say are you serious? Have you ever had to model a multidimensional fluid model ( i.e. including Naiver-Stokes )? Not to mention the multiple layers of radiation and convective heat transfer in the atmosphere and ocean.
Even simple models, can take hours on a Cray. We had to run these on the night shift, to keep costs down.

January 12, 2012 2:40 pm

The majority of the heat at the Earth’s surface is due to its radioactive core, especially U-238. Since the half-life of uranium is about 4.5 billion years, that is a factor in changes since the Hadean, or even Cambrian ages, but is effectively a constant for the duration of human existence on the planet.
The majority of the heat flux out is due to U-238? Doing what? As you say, the half-life is 4.5 billion years, meaning that it is really not terribly radioactive. It’s decay channels are also not terribly energetic — this isn’t fission, it is alpha decay, IIRC.
Also, does this make sense? Why is it cold at the poles and hot at the equator, if this is true? Why does it get hot in the day and cold at night? I’m open to there being more geothermal heat (not terribly uniformly distributed) than is currently estimated because there is a lot of unexplored territory out there in the world even today (on the ocean floor, for example). But places where there is an unmistakable surplus of geothermal energy — Iceland, Yellowstone, near active volcanoes — don’t even warm the ground in general enough to prevent winter locally anywhere but right on top of a hot spot.
I think this fails the common sense test, as well as contradicting a whole lot of measurements by hapless geologists who have no dog in the AGW fight, sorry. Although I’d be happy to be convinced otherwise if you can provide me with a quantitative basis for your estimate. Sure, the Earth’s core is very hot, because it is wrapped in a big blanket of insulation — the crust and mantle. What heat is generated in there, stays in there a long, long time, but we live on the outside of the blanket and are heated and chilled with the vagaries of, well, the weather!
rgb

Truthseeker
January 12, 2012 2:47 pm

thepompousgit says:
January 12, 2012 at 10:50 am
Since you cannot seem to even comment in complete sentences, I do not think that you calling Harry Huffman’s comment incoherent has any validity whatsoever.

January 12, 2012 2:49 pm

Any wealthy folks listening? Here’s my proposal. Fund someone like Prof Lindzen to design a research programme including physics and statistics and computing to review fundamental issues relevant to the current controversies around climate variation and its causes. Work such as that outlined by Prof Brown is exactly the kind of thing I have in mind. The aim? To bring more 1st class science and analysis into a field taken over by superficial analyses, tawdry manipulations (just look at the dross in CG1 and CG2 for example), and pandering to destructive political activists.

January 12, 2012 3:02 pm

Ian W says: several things I don’t repeat…
but reply to as “Damn skippy, you are so very right….”
However, Enthalpy, the Earth’s climate viewed as a self-organizing heat engine, and chaotic fluid dynamics of open systems are for later. First I need to take baby steps… even if they follow in the footsteps of others or if future things will confound them.
rgb

January 12, 2012 3:03 pm

And after all that figgerin., you just have weather, not climate.

Joe Postma
January 12, 2012 3:03 pm

Robert Brown:
You might be interested in reading my critiques of standard theory here:
http://principia-scientific.org/publications/Copernicus_Meets_the_Greenhouse_Effect.pdf
here:
http://principia-scientific.org/publications/The_Model_Atmosphere.pdf
and here:
http://www.tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf.
As to your post and the questions/analysis you present, I have developed the code and theory for modeling the Earth in real time just as you have described – modelling, that is, the differential equation of heat input and output as a function of time, latitude, eccentricity, local albedo, local extinction, etc. This is the question and task I left posed at the end of my 2nd paper, linked above. My background is astrophysics so I already had on hand the theory of how to do much of this…as it is described in the Astronomical Almanac, but for a different context of course.
It can be simplified for demonstrative purposes to an analytically solvable equation; otherwise, a numerical approach would not be difficult to perform, as all the equations are intrinsically smooth, aside from any perturbations one might eventually want to introduce, for example, a cloud passing over-head.
Such a model gives MUCH greater insight into the real physics of the Earth-Sun interaction. Treating the Earth as flat, being shone with a cold Sun at -18C, just can not be a philosophically valid model or paradigm. For example, if Sunshine is -18C, then why does water exist? Due to the GHE? Of course not. Direct, real-time, non-averaged into night-time sunlight, is more than energetic enough to melt ice into water, and thus dive a climate system. The instantaneous solar input to the Earth actually has a heating potential of anywhere between +30C to +121C. And this DOES lead to an average output from the Earth corresponding to -18C, or 240 W/m2, in accordance with the expected conservation of total energy.
Anyway, if you would like to work on this together, let me know and I will look up your email on your university’s website and contact you. Or you may find and contact me, at the University of Calgary.
Regards
Joe Postma
MSc Astrophysics

January 12, 2012 3:15 pm

As the ‘standard model’, used by warmistas, gives a -18 deg C temperature due to solar irradiation, how on earth did earth ever manage to unfreeze? so-called GHGs would be at very low levels in the atmosphere – being either ice or trapped/dissolved in the ice. Also they mistake conservation of energy – which is real – as conservation of power which is meaningless.

Joe Postma
January 12, 2012 3:19 pm

“Myrrh says:
January 12, 2012 at 2:33 pm
And there’s also that forty years ago NASA had to junk Stefan-Boltzmann to get real moon temps estimated for the landings, somewhere on-line, as they needed three dimensions not the flat earth of SB and they had to include that thermal energy absorbed from the Sun penetrated and was released later.”
That’s quite interesting. In the model I have been developing, it seems apparent that such a term needs to be introduced in order to capture all of the physics…you can’t ignore the massive thermal mass and extremely long time-constant of “dirt”. You also therefore question the idea that the ground surface has zero contribution from geothermal energy – if this were true, the ground would go down to zero kelvin at some point. But in fact it never actually gets all that cold beneath the ground surface, in fact it is MUCH higher than 0 Kelvin! You have to include that…and it WILL raise the temperature. The same effect from the atmosphere is known to be much less significant, having a much, MUCH shorter time-constant, and much less thermal energy stored within in.
Hopefully this is something that Robert and I could work on together…

Dave Wendt
January 12, 2012 3:39 pm

“You have to add an atmosphere. Damn. You also have to let the ocean itself convect, and have density, and variable depth. And all of this on a rotating sphere where things (air masses) moving up deflect antispinward (relative to the surface), things moving down deflect spinward, things moving north deflect spinward (they’re going to fast) in the northern hemisphere, things moving south deflect antispinward, as a function of angle and speed and rotational velocity. Friggin’ coriolis force, deflects naval artillery and so on”
The problem I have always had with all these efforts to reduce the Earth to a simplistic radiative energy model is that there are very significant energetics continually ongoing on the planet that are not captured in the calculation. Every time a molecule moves from coordinate abc to coordinate xyz there is energy involved. Most of everything in the atmosphere, the oceans, lakes rivers, even the solid earth is in constant motion. Admittedly some, maybe most, of that motion is driven by gravity and the rotation of the planet, but some of it is driven by the conversion of the radiative input of insolation from the Sun into kinetic energy. Is anyone willing to suggest we know enough about what that number is to conclude that it is insignificant, particularly given the almost infinitesimal marginal changes we are trying to estimate in the climate?
Then there is the real difference that differentiates the Earth, from simple S/B models or from any other planetary body or satellite we are aware of. The Earth is covered not only by a large atmosphere and massive oceans, it is covered by LIFE. Almost everything that lives, grows, reproduces, and moves by swimming, slithering, crawling, hopping, walking, flying or any other mode of motivation, ultimately draws the energy to do so from the energy falling on the planet from the Sun and most of it is accomplished without any accompanying radiative signature. Again I would ask is anyone willing to suggest that we can create even a ballpark estimate of the energy use of life on this planet? Given all we don’t know about that life, I would suggest that even our best efforts couldn’t come within 1000 Superdomes of an accurate estimate.

Ian W
January 12, 2012 3:52 pm

John West says:
January 12, 2012 at 11:55 am
What is the rational behind the statement: “where it is emitted at a cooler temperature”? I’m not convinced that the Stefan-Boltzmann Law applies to GHG’s any more than it does fluorescent or neon lights. It seems to me that a CO2 (or H2O) molecule can emit IR at a rate not proportion to it’s temperature as long as the temperature is above the “freezing” point of it’s vibration modes (degrees of freedom), such that temperature is merely a threshold variable to IR emissions from GHG’s.
Consider the following:
1) The temperature of a gas is a measure of the kinetic energy of translation not vibration. “Fundamentals of Modern Physics”
2) The specific heat of a gas increases proportionally to degrees of freedom availability. More energy is required to increase the temperature of a gas as degrees of freedom other than translation are “unfrozen” suggesting energy stored in non-translational modes do not effect temperature directly.
http://theory.phy.umist.ac.uk/~judith/stat_therm/node81.html
3) Even though equipartition of energy occurs within a mass of gas, the likelihood for a collision capable of imparting a translational motion from a vibration motion is the same as the likelihood for a collision capable of imparting a vibration motion from a translational motion for a given set of circumstances. Whether the atmosphere would be heated or cooled is dependant on the probability of a collision capable of transferring motion to/from translation to/from vibration encountering the opposite condition; or to put another way, the proportion of GHG to IR input. Consider the atmosphere if the Earth did not radiate IR; any collisions resulting in vibration motion could be emitted as IR, reducing the overall translational motion of the atmosphere, thus cooling it. Considering the other extreme where the Earth radiated so much IR that all available vibration modes were always immediately exited by IR input; then equipartion of energy would be “averaging” energy from vibration induced by IR input into translational movement through collisions thus heating the atmosphere. Obviously, the actual atmosphere is somewhere in between these two extremes. Whether GHG’s heat or cool the atmosphere depends on the amount of GHG’s (increases may result in less direct heating), the amount of IR input, and the availability of vibration modes. Only the availability of vibration modes is dependant upon temperature and then only as a threshold.
4) Absorption and emission of specific bands of IR by GHG’s corresponding and limited to vibration modes also suggests that IR absorption and emission by GHG’s are not black body emissions but instead exhibit this characteristic of “cold radiators”.
5) Earth’s atmosphere temperature profile does not correlate with GHG concentration appreciably warming the atmosphere.
http://regentsearth.com/Illustrated%20ESRT/Page%2014%20(Properties%20of%20Atm.)/ESRT10-Properties%20of%20Atmosphere.jpg
6) Anecdotal evidence: IR heaters and lamps do not heat the air in a room directly but heat the IR absorbing surfaces exposed to its output.
I conclude from the above that IR emissions from the atmosphere are not directly proportional to its temperature and therefore cannot be black/grey body emissions or Stefan-Boltzmann Law would be violated. Therefore, a temperature increase is not required for an increase in atmospheric IR emissions (atmospheric radiance is not proportional to temperature). How can an increase in GHG mass in the atmosphere cause an increase in GHE (back radiation) prior to any significant atmospheric temperature increase if a temperature decrease (emitting from higher, colder position) reduces the radiance of the atmosphere? Until I see some evidence to the contrary, I maintain that no temperature increase is required in order to emit an additional amount radiation from GHG’s both down and up.

Reference your points. We actually have geostationary satellites watching the IR output from the Earth and available to anyone to look at.
See:
GOES East Water vapor imagery – this shows where there are front’s and other upwelling moist convection
http://www.ssd.noaa.gov/goes/east/natl/flash-wv.html
GOES East Rainbow IR imagery – this shows where there is IR being emitted to space and seen by the satellite.
http://www.ssd.noaa.gov/goes/east/natl/flash-rb.html
Note how the dry areas in the water vapor map do not show emission of IR in the IR imagery.
Note how the upwelling moist air is emitting IR at high ‘cool’ levels – this is presumably heat being emitted as water changes state condensing then freezing. This energy output is NOT shown by Stefan Boltzmann maths nor is it linked to ‘temperature’.
NASA/NOAA state that this convective transport to the tropopause is far greater than radiation from the surface. Looking at the surface and trying to use SB would appear to be incorrect it is looking inside the hohlraum.

u.k.(us)
January 12, 2012 3:52 pm

@ Robert G. Brown, Duke University
==============
Good stuff, keep it coming.
You’ve got a great writing style, and I look forward to any future contributions.
I can almost understand some of it 🙂

Editor
January 12, 2012 4:32 pm

Septic Matthew says:
January 12, 2012 at 2:38 pm
Willis Eschenbach:

First, we have a pretty good handle on how much energy hits the earth system after albedo. It’s on the order of 240 W/m2.
Second, we have a pretty good handle on how much the earth surface is emitting. It’s on the order of 400 W/m2.

I have a feeling that you have left out something important, or I have forgotten something important — I note you wrote “earth system” in the first sentence and “earth surface” in the second. If this were true as written, we would have a pretty good handle on the earth losing 160 W/m^2.

Thanks, Matthew. Not sure what the point is here. The earth doesn’t lose 160 w/m2 because of absorption in the atmosphere.

What do you mean by “on the order of”? that the error can’t be more than 1%? 10%? 100%? Since electronic communication can sound snarky when not intended so, let me apologize in advance if this sounds snarky.

What that means is that I’m tired of giving the best estimate and having someone bust me saying something like “it’s not 240 W/m2, you idiot, it’s 239, how could you make such an egregious error, or are you just lying on principle?” or the like..
Errors in both of those figures are likely within 10% of the true value, but as far as I know, we have no hard data on the error figures. I am merely trying to give a rough idea of the size of the hill that alternative theories have to climb. The exact numbers are not important for my discussion.
w.

George E. Smith;
January 12, 2012 4:50 pm

“”””” Thermodynamic professor, power plant manager says:
January 12, 2012 at 1:19 pm
George E. Smith; says:
January 12, 2012 at 11:11 am
So Thermodynamic Professor, whether or not one believes that the ordinary atmospheric mono or homo-diatomic gases: Argon, Nitrogen, Oxygen (maybe …..
Sorry but we have never managed to warm up anything with colder objects or gases, the nonsense you are talking about. This is based purely for Thermodynamic laws. It would of course be very nice if such unphysical [snip] happens, our power plant production costs would be much lower;) “””””
Well Prof Thermo, absolutely nowhere in my post did I state either specifically, or by inference, that “we managed to warm up anything with colder objects or gases.” If that is nonsense, you are the one who said it, not me.
I simply enquired how a surface can possibly tell the Temperature of an object or gas that is emitting Electromagnetic Radiation, since such radiation carries NO Temperature information whatsoever. Nor can the source determine the Temperature of some remote surface, to decide whether it is legal for it (the object or gas) to emit EM radiation or not. Since the emission of EM thermal radiation from a body or gas is of necessity isotropic, that would imply that such a body or gas can either emit EM radiation towards ANY other surface or object in any direction, at any surface Temperature at all, or else it must omit nothing, if by your assertion, a warmer surface in some direction prohibits the emission of such radiation. Or alternatively, if such a warmer surface does not prohibit isotropic emission from some remote colder source ( could be light years distant), what is the mechanism by which it prevents such radiation from landing; and if it doesn’t land; where does it go.
Is it possible to examine the spectrum of EM radiation emitting or reflecting from some surface; say the earth, and sort out which is emission and which is “reflected” rogue radiation that tried to land from some rejected colder source (which also might be light years away. ??
So why don’t you let us all in on what academic institution you Profess Thermodynamics instruction at; so we can all enroll in your classes.

beng
January 12, 2012 5:31 pm

The 33K “difference” obviously isn’t right. If you take away the GHG water vapor, they’ll be no more clouds (or snow/ice). Also no more latent heat available to cool the ocean surface & increase convection. Water (as liquid/solid or vapor) serves a dual purpose — both warming & cooling.
I’m not sure what it is, but it isn’t just the simple difference between the surface temp & the temp at the average radiating “surface” in the troposphere/tropopause — 288K-255K.

Bill H
January 12, 2012 5:37 pm

Mike M says:
January 12, 2012 at 8:30 am
But but.. the ‘settled science’ is so settled that we’re already spending over $2.5 billion per year to ‘combat climate change’. Ain’t no physics in the universe gonna slow down a gravy train with that much inertia.
——————————————————————————————
One Super volcano would do the trick…..
🙂

richard verney
January 12, 2012 5:41 pm

Joel Shore says:
January 12, 2012 at 9:23 am
“….Because we know that the Earth + atmosphere are absorbing ~240 W/m^2 from the sun and that the Earth’s surface is emitting ~390 W/m^2 or so….”
Joel
Please will you provide details of the latitude and longitude of where on planet Earth, the Earth is absorbing 240W per sq.m and the date and time when that occurs.
Please will you provide details of the latitude and longitude of where on planet Earth, the Earth/s surface is emitting 390W per sq.m and the date and time when that occurs.

January 12, 2012 5:45 pm

Earth is 259 trillion cubic miles of mostly MOLTEN rock with a temperature above 2500F. Uranium-235 is approximately 4 PPM, or 700,000 cubic miles; Thorium app 1.2 million cubic miles. These fissionable materials are NOT limited to standard half-life decay and are in fact variable in time and location. Estimates of this force had a magnitude increase from ~6 terawatts to 66 TW with the KamLAND experiment, see “First Measurements of Earth’s Core Radioactivity”, New Scientist, July 27, 2005. Fission decay can involve up to a dozen unstable daughter atoms which also release heat, see “Potassium-40 Heats Earth’s Core”, Physics World, May 07, 2003. Massive amounts of geothermal energy are released, but hidden in plain sight. Some geothermal energy is disgused at ocean floor vents, per “Earth’s Missing Geothermal Flux”. Additional fission energy is converted to hydrocarbon molecular energy as described in “Fossil Fuel is Nuclear Waste” and “Earth’s Elemental Petrol Production”.
We live on a far more complicated and interesting planet that we are FORCED to believe. Please find and share Truth.
Joseph A Olson, PE

Brian H
January 12, 2012 5:50 pm

Stephen Wilde says:
January 12, 2012 at 1:26 pm

They could be acquiring all of their energy from the other non GHG molecules which can be heated by conduction from the surface and would pass it on to the GHGs via conduction. Once the GHGs reach the ambient temperature for other molecules at the same height (via conduction) they cannot get any warmer from longwave coming up from the surface

Individual molecules can’t be said to have any temperature; they move at a huge range of speeds, depending on the net ‘vector’ after their latest collision, from nearly dead stop to kps. It’s the gas-mass that has a characteristic number and violence of collisions which can be said to have a temperature.
So there’s no problem in any CO2 molecule, at any speed, absorbing radiation from the surface (or other CO2 molecules, or anywhere) and gaining vibrational energy. But the lag before it gets hit by another molecule (O2, N2 probably) and exchanging thermal energy with it is far shorter than the probable time to re-emit. So, as Cao Jinan points out, most of the radiation from CO2 results from energy acquisition from a recent collision, not from absorbed radiation, which has long since been thermally passed on: the odds are many orders of magnitude greater, actually. Net-net, CO2 thermalizes incoming radiation, and occasionally radiates acquired thermal radiation. Since O2 and N2 don’t do the latter, CO2 constitutes a “hole in the bucket” at higher altitudes, radiating (leaking) thermal energy to space that would otherwise hang around longer (raising local/average temperatures).
The question of where the O2-N2 mix gets most of its thermal energy is thus key. Since H2O is so much more common in the atmosphere, and picks up far wider up-welling and downwelling frequency bands, it’s the obvious suspect, aside from its potent role in using its huge latent heat phase changes to cool the surface and heat (at cloud level) the atmosphere.
As a heating tool, CO2 is very small — micro-small — potatoes.

Brian H
January 12, 2012 5:54 pm

Edit: “radiating (leaking) thermal energy to space that would otherwise hang around longer.”

January 12, 2012 5:58 pm

@Lucy Skywalker says: January 12, 2012 at 1:39 pm
“….CO2 in the stratosphere emits radiation toward the surface. ….”
Quick question – perhaps a dumb one (I have checked you link but am none the wiser) – does the CO2 in the troposphere really only emit radiation towards the surface, heating the CO2 below and thence H20 in the tropospause, so the H20 then can emit it to space? I’d expect the majority of it was being emitted (to space) via CO2 in the stratosphere?
Or does this just refer to a trapping mechanism for that portion of the radiation going downwards?

Myrrh
January 12, 2012 5:59 pm

Joe Postma says:
January 12, 2012 at 3:19 pm
“Myrrh says:
January 12, 2012 at 2:33 pm
And there’s also that forty years ago NASA had to junk Stefan-Boltzmann to get real moon temps estimated for the landings, somewhere on-line, as they needed three dimensions not the flat earth of SB and they had to include that thermal energy absorbed from the Sun penetrated and was released later.”
That’s quite interesting. In the model I have been developing, it seems apparent that such a term needs to be introduced in order to capture all of the physics…you can’t ignore the massive thermal mass and extremely long time-constant of “dirt”. You also therefore question the idea that the ground surface has zero contribution from geothermal energy – if this were true, the ground would go down to zero kelvin at some point. But in fact it never actually gets all that cold beneath the ground surface, in fact it is MUCH higher than 0 Kelvin! You have to include that…and it WILL raise the temperature. The same effect from the atmosphere is known to be much less significant, having a much, MUCH shorter time-constant, and much less thermal energy stored within in.
Hopefully this is something that Robert and I could work on together…
=====================
I really would like to see a proper basic model, desperately needed to have something to pin all the variables on.., I hope this collaboration or one like it takes off.
Also to bear in mind that water wasn’t a problem for NASA’s rejigging to get Moon temps, and the Earth’s great ocean with its high heat capacity has the ability to store the heat energy it gets direct from the Sun without ostensibly showing any great difference in temperature and regardless the atmospheric temp changes going on surrounding it – it takes longer to heat up and so correspondingly longer to cool down, higher heat capacity than dirt.
And, I’m very much in agreement with http://wattsupwiththat.com/2012/01/12/earths-baseline-black-body-model-a-damn-hard-problem/#comment-862218
Especially re his LIFE, which other planets don’t have and didn’t have to be considered in the moon landings. Adding to Dave Wendt – there are two distinct categories of energies reaching Earth, Heat and Light. Heat, the direct thermal energy of the Sun on the move, transfer, via radiation is that which directly heats land and oceans, and us as water is the great absorber of thermal energy and we’re mostly water, so to some extent all Life absorbs direct thermal energy from the Sun, but, we also manufacture our own via the energy we get from food in burning it for work and this food chain has plant life at its base which manufactures food direct from Light, sugar from visible, and in burning this for work releases heat into the atmosphere via transpiration. Photosynthesis isn’t limited to the land, some 90% of the oxygen in the atmosphere is said to be produced by photosynthesis in the ocean.
As if all that isn’t complicated enough addition, it’s the variables created by all the heat exchanges through convection in the heavy fluid gaseous atmosphere around us, which gives us our weather and great wind systems around the globe, which makes me get out the popcorn..
Good luck.. 🙂

KevinK
January 12, 2012 6:02 pm

Dr. Brown;
VERY WELL PUT, thank you.
In the engineering fields we consider these “intractable” analytic problems, meaning that we cannot ever hope to solve them analytically. So, to make forward progress we turn to alternative approaches.
As a simple example;
You have a complex Earth orbiting satellite with mechanical, optical, electrical and other components that are held together with an assortment of nuts/bolts/glue, etc. (sorry, no rubber bands allowed).
You wish to launch it into space on top of a big roaring rocket that is going to shake Holy H—l out of it before it reaches the vibration free nearly zero-G altitude where it will operate.
How do you ascertain that it will survive the launch ?
Well here is how;
You write down everything you and others have learned from launching other things into space.
You do not repeat their mistakes, and copy what worked for them (if applicable)
Then, (get this all you true believers in the value of computer models); you make a test copy (aka a Qualification Model) and you SHAKE IT EVEN HARDER THAN the rocket ever could.
Yes indeed, that’s how’s it’s done, imagine a big honking table that you bolt your satellite to and the table shakes Holy H—l out of it and then some.
If it survives (about 95% of the time) and if you make a Flight Model using exactly the same extremely well documented assembly processes it is “very likely” that your satellite will make it into space and operate as advertised. But some still fail (lately about 5%, historically less than 1%).
Nobody attempts to analytically model the strength of all those mechanical interfaces when subjected to vibration, it would take forever and nobody would believe the “heritage” of your computer models (i.e. has it worked in the past ?) anyway.
Ok, as another example, how can you determine how long your satellite will operate once in orbit ?
Satellites operate in a very rough environment where radiation (those nasty cosmic rays) strike semiconductor junctions and render them useless, atomic oxygen rips material away from optical coatings and the occasional micrometeorite slices through a critical power supply wire.
Well, you could just make lots of satellites with different designs and wait a few years to see how long they last while on orbit. But given the cost of satellites (50 million and up to lots and lots of millions of dollars) this would be wasteful. Also, since modern satellites often last 5-10 plus years (Chandra was launched in 1999 and is still functioning) this would take decades to get useful information.
So we use (drumroll please) COMPUTER MODELS, these models are fairly simple and use empirically derived data (we put a satellite up with the Space Shuttle to do nothing but expose different materials to the space environment for years, then retrieved it to find out which lasted the longest) to predict how long a satellite designed with certain materials will last. We also predict how often a micrometeorite can be reasonably expected to take out an important electrical cable.
So we come up with an analytical prediction (i.e. a computer model) to predict the lifetime of the satellite. These predictions have been aligning pretty well with the performance of actual space vehicles, but we still get a mission failure every once in a while.
Sorry this has become quite a bit of a post, but the point is, you need to carefully assess when a computer model helps and when it just fools you.
The climate models which have been programmed to explicitly behave in correspondence with a hypothesis (“The Greenhouse Effect”) are just a silly and expensive attempt to beat an intractable analytical problem into submission with more computer power. I believe they are useless.
Cheers, Kevin.

LazyTeenager
January 12, 2012 6:03 pm

So Robert reckons he can work out the whole of atmospheric physics in one night and then discovers it is complicated.
People smarter than Robert have been working on this for a century or so, so I figure they will be able to tell him it is even more complicated than what he has discovered so far.
But I get the uncomfortable feeling this argument is going to devolve into a bit of logic that looks like this:
1. We do not understand everything about how blackbody radiation determines the baseline temperature.
2. Therefore we know nothing about how blackbody radiation determines the baseline temperature.
3. Therefore we know with absolute certainty that the blackbody radiation determines the baseline temperature to be 25C.
4. Therefore the greenhouse effect does not exist.
Please oh please prove me wrong.

Bill Illis
January 12, 2012 6:03 pm

Robert’s post mentions it takes “time” to raise the temperature of the components of the Earth system.
I thought I would point out some of those lags.
– daytime temperatures peak 3.5 hours after the peak of the solar radiation.
– water evaporates from the surface (taking some heat with it) and it eventually falls back to Earth as rain or snow (after dumping off its heat in the atmosphere) anywhere from 2 hours to an average 9 days later (side question, what is the temperature of layer where rain or snow forms versus the temperature of where it evaporates from).
– land temperatures lag behind the solar cycle by an average 34 days (the warmest day of the year is July 25th and solar radiation peaks about June 21) (the accumulation rate is about 0.5 W/m2 per day from the coldest day of the year, January 25th, to the warmest day of the year, July 25th).
– lakes and rivers peak about 45 days behind the seasonal cycle.
– ocean temperatures lag behind solar radiation by about 45 to 70 days.
– it takes 1500 years to warm up the entire ocean versus the Earth Surface temperature.
– it takes about 10,000 years to completely melt out a continental ice age.
– Venus is so much hotter than it really should be.
In other words, the components of the Earth system slowly accumulate and slowly release energy received from the Sun.
The Stefan-Boltzmann equation includes a mathematical expression called “Emissivity” but this is mostly ignored and mostly not understood that it contains a “time” element in it that must be taken into account. It is entirely possible that emissivity varies over time and over energy levels and it is entirely possible that an object can go on forever having an imbalance between the energy it is absorbing and the energy it is releasing (it can certainly last up to 10,000 years).
Emissivity needs to be further defined. It has a non-Zero effect on temperatures but it is treated as if it is Zero. For example, the lag in temperatures of the Earth system’s components means that they slowly accumulate very small amounts of energy per second. This could go on forever or it may reach an equilbrium level at some point.
The Greenhouse Effect may simply be a function of Emissivity.

Bill H
January 12, 2012 6:08 pm

Brian H says:
January 12, 2012 at 5:50 pm
“Net-net, CO2 thermalizes incoming radiation, and occasionally radiates acquired thermal radiation. Since O2 and N2 don’t do the latter, CO2 constitutes a “hole in the bucket” at higher altitudes, radiating (leaking) thermal energy to space that would otherwise hang around longer (raising local/average temperatures). ”
———————————————————————————
This is the key as to why CO2 is actually a cooling device rather than a heat storing one.. as the levels increase it has the black body effect increase… much more than the intake it keeps under solar input.. its that darkside net loss that is killing the AGW theory dead.. but they refuse to look at it…

Surfer Dave
January 12, 2012 6:10 pm

@Tim_Focketts
I looked up the geothermal heat flux at the surface and found that there is a paucity of real measurements globally. The USA has been studied reasonably well and while the average on continental USA is estimated at about 250mWm-2 there are hot spots up to 15Wm-2. That was from the Southern Methodist University website, very instructive. No-one really knows what the global average is, especially given that 70% of the planet is underwater. It is assumed that the flux on continental USA is representative of the flux on the entire planet. Some measurements in Australia, South Africa and Russia seems to confirm this but even so there are hot spots with fluxen well above 1Wm-2.
Given that seems like quite a low rate, how come the temperature below about 10m is 22C (295K) and increasing with depth? Is that 22C from heat that has gone down from the surface, or is it the balance between the steady outward flux from the geothermal sources (which include radioactive decay, heat from the earth’s EM currents, gravitational friction internally and the residual heat from planetary formation) and the inward flowing heat from the insolation? Would that internal temperature at 10m deep change in the absence of an atmosphere?
Also, on the original topic, one point to remember is that most of the physics of heat conductance in materials assumes that the heat flux is perpendicular to the surface of a flat plane. This is probably an acceptable simplification because in a uniform infinite plane the generally spherical flow of heat (either convective or radiative) from a point is balanced out by all the surrounding points so that the nett flow can be assumed to be perpendicular. My point is this, on a sphere each point has it own perpendicular that all converge on the centre of the sphere. The assumption of flux parallel to a flat surface is not valid for a sphere. I suspect that in some sense a sphere acts like a lense for the heat flow, the density of the flux and thus the temperature of the material of the sphere increases towards the centre. I know the earth seems “big” so that the flat plane analogy sort of makes sense, but that’s probably not correct.

Brian H
January 12, 2012 6:10 pm

Edit: “occasionally acquired thermal radiation energy.”

Brian H
January 12, 2012 6:12 pm

ARRrr, he said like a pirate but a week early:
Edit edit “and occasionally radiates acquired thermal radiation energy.”

LazyTeenager
January 12, 2012 6:16 pm

Michael Hart says
But where can us lesser mortals examine the algorithms and computer code, not to mention the assumptions about which bits can be safely ignored?
———–
There is some GCM source code dowloadable but these codes, even the old obsolete ones are rather large, and beyond mere mortal understanding.
In fact they are probably so large they are beyond the in-depth understanding of people who use them and challenging to understand for people who wrote them.
The assumptions, math and experimental work behind each individual component of a GCM program the will be found in the scientific literature and is also likely beyond your understanding. Each such thing represents 5 years work for 1-5 technical specialists.
So, in short, you can download the code but the exercise will not be useful to you.
Just learning how to get the data and create the input files is going to take several weeks of study and work.

eyesonu
January 12, 2012 6:26 pm

@ Robert Brown
I sense that you have a class of ‘students’ here at WUWT greater than you can imagine! Some are commenters and many more are ‘lurkers’. I am of the ‘lurker’ class and will not miss a lesson and strive to learn every bit of knowledge that you offer. I will not get an ‘A’ as no written exam is required. The ‘test’ will come when I refute the bullshit of the many uninformed.
I sincerely thank you. You still get no apple!

Tim Folkerts
January 12, 2012 6:26 pm

Bill, I like what you say, but I don;t see how emissivity is going to save the day.
>Emissivity needs to be further defined.
emissivity = (Actual thermal radiation) / (ideal thermal radiation)
The definition is pretty straightforward. This definition can be applied at each wavelength or (more commonly) as a weighted average over all wavelengths. I don;t see what more definition is needed.
>but it is treated as if it is Zero …
If anything, it is treats as if it is exactly 1.00. Measurements show that it is indeed pretty close to 1 for many parts of the earth (water, many soils, many leaves … ). I understand the the emissivity is included in many calculations, but this will only be a correction of a few percent — far too little to account for major effects.
>the lag in temperatures of the Earth system’s components means
>that they slowly accumulate very small amounts of energy per second.
This is most certainly true. That is why people worry about small changes in atmospheric IR radiation having long-term effects on climate.
But it is also most certainly true that a few percent error in emissivity cannot account for the difference between outgoing IR seen from space (~ 240 W/m^2 on average, with clear effects of CO2 and H2O) and outgoing IR as seen from just above earths surface (~ 390 W/m^2 on average, with a nearly blackbody spectrum).

LazyTeenager
January 12, 2012 6:32 pm

Thermodynamic professor, power plant manager on January 12, 2012 at 9:41 am said:
Backradiation is bullshit from cooler gas to warmer surface, physically impossible!
——–
You don’t need billions of dollars to prove this statement to be false.
Just go down to your local electronics shop and use $100 to buy an IR thermometer.
Point it at the sky. Measure the temperature. You have just measured the long wave down welling infrared radiation.
Case closed.
But if that is not enough consider unglazed thermal collectors used for swimming pool heaters. The LWIR from the sky figures in the efficiency calculations for those.
Score: checkout chick 1, professor 0.
So professor needs to go back to school and learn the laws of thermodynamics properly this time

eyesonu
January 12, 2012 6:35 pm

I should also add that I am also a student of the several ‘associate’ teachers here @ WUWT University.

Tim Folkerts
January 12, 2012 6:38 pm

Surfer Dave says:
>Given that seems like quite a low rate, how come the temperature
>below about 10m is 22C (295K) and increasing with depth?
>Is that 22C from heat that has gone down from the surface,
Certainly not.
>or is it the balance between the steady outward flux from the geothermal sources
Yes, the early earth was much hotter and has been cooling for a long time. But there are literally miles of insulation, so the heat flow will be rather small.
>Would that internal temperature at 10m deep change in the absence of an atmosphere?
Certainly it would be a bit cooler. I strongly suspect the temperature gradient would stay about the same, so you could pretty much subtract ~ 33 K (give or take a bit depending on exactly how you want to model the earth with no atmosphere) from the temperature at any depth.
(Of course, over the course of billions of years, this could have a significant effect all by itself. An earth that NEVER had an atmosphere would presumably be cooler than an earth that lost its atmosphere recently. )

KevinK
January 12, 2012 6:43 pm

Dr. Brown;
My last post got a bit verbose, so I split it into two parts for the comfort of any that wish to read further.
While you have captured many of the concerns with respect to our ability to predict the temperature of the Earth with/without “GHGs” I believe you may have missed a few important ”wrinkles” in the problem. I will suggest just two for everybody’s consideration.
First, the Albedo of a surface is HIGHLY dependent on the angle of the arriving optical radiation. This is well known and has been adequately characterized by the Bi-Directional Reflectivity Function, aka the BDRF. This is well known among those folks that design Earth Imaging Satellites. This surely seems to call into question the simplifying assumption that the spherical Earth can be replaced by a flat disk, IF only it was this simple.
Secondly, I see no discussion amongst the climate science community of the speed at which heat travels through the materials comprising the atmosphere of the Earth. It is well understood among engineers that heat propagates through different materials at different speeds. For example, most computers use aluminum as a “heatsink” to remove heat from the microprocessor. Like most engineering decisions this is a tradeoff between cost and performance. Aluminum is less costly and provides a reasonable level of performance for most applications. However, when higher performance is desired without concern for cost Copper is used instead of Aluminum. Why is that some may ask? well… it’s because the thermal diffusivity of copper is higher than that of aluminum. Thermal diffusivity is (from a system perspective) an effective measure of the “speed of heat” through a material. Heat flows through copper at a higher speed, thus allowing higher performance. In some applications synthetic diamond is applied to give even higher “speeds of heat” when the cost is justified.
My hypothesis is that increases in “GHGs” are displaced by decreases in “non-GHGS” (after all there are only 1,000,000 ppmv of gases in the atmosphere,by definition). Heat travels through “non-GHGs” at the “speed of heat”. Heat (or its equivalent IR radiation) travels through “GHGs” at close to the “speed of light” (albeit there are some slight delays introduced from a few (10-20max) short side trips backs towards the surface as “backradiation”). Hence the addition of “GHGs” to the atmosphere only make the gases in the atmosphere warm up faster, or cool down faster after any change in the energy input to the system (i.e. sunrise/sunset/accumulation of clouds/dissipation of clouds). This clearly indicates that no “higher equilibrum” temperature can be caused by “GHGs”.
Cheers, Kevin.

Tim Folkerts
January 12, 2012 6:43 pm

One more thing, Surfer Dave,
Geothermal energy fluxes might well be a few W/m^2 rather than a few 0.1 W/m^2 — I am certainly not an expert on the topic. This would be interesting and it would have some bearing on energy balances. But even the highest estimates you gave sill less than 10% of the power flows attributed to GHG’s so geothermal energy [transmission] is still a perturbation, not a replacement for GHG’s and atmospheric effects.

Bill Illis
January 12, 2012 6:53 pm

Tim Folkerts says:
January 12, 2012 at 6:26 pm
But it is also most certainly true that a few percent error in emissivity cannot account for the difference between outgoing IR seen from space (~ 240 W/m^2 on average, with clear effects of CO2 and H2O) and outgoing IR as seen from just above earths surface (~ 390 W/m^2 on average, with a nearly blackbody spectrum).
———
Crunch the numbers on a per second basis since the energy coming in during the daytime is actually an average 956 joules/second/m2. How much does the surface energy level change? At night, basically zero is coming in. How much does the surface energy level change (per second).

Surfer Dave
January 12, 2012 7:11 pm

@Tim_Folkerts
Thanks Tim, useful
Given that Hansen’s recent “Imbalances” came up with an imbalance of 500mWm-2 then I would think that a geothermal heat flux of almost the same magnitude as the “imbalance” should be brought into the complete story?

Tim Folkerts
January 12, 2012 7:14 pm

Dr Brown is doing his homework. I’ve been doing some homework as well to understand this all a bit better. For all those claiming that CO2 would only have a small warming effect — or even a cooling effect — here is your homework. Explain the shape of the IR spectrum as observed from space looking down and from the surface looking up. (there are good examples here: http://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/)
* Explain the shape of the dotted lines labeled with different temperatures
* Explain the cause of the various “bites” in the spectrum.
* Estimate the order of magnitudes of the GHG effects at a given place and time (ie when a specific satellite image was taken). .
This is what you need to be able to do to start discussing the GHE intelligently. Otherwise, you are free to keep idly speculating, but don’t expect much support except from other idle speculators.

eyesonu
January 12, 2012 7:21 pm

While I’m on the topic of ‘teachers, let me suggest this as follows. Every teacher/professor of physics, thermodynamics, ‘climate science’, real science, meteoroligical, or other related disipline should direct his/her students to study the past several days of WUWT as related to this discussion/topic.
The topic is very important and the forum is a model for exchange of knowledge.

AusieDan
January 12, 2012 7:39 pm

Hi Willis
I think that you may be missing the possibility that the extra heat at the surface, is just borrowed by the work due to gravity, from the higher up regions of the atmosphere.
If you work through the data for the various planets and moons, you will see how it works very simply, regardless of the percentage of GHGs in the atmosphere of the planet or moon under study.

johndo9
January 12, 2012 8:33 pm

For Dr Brown or anyone wanting to run the integral surface temperature.
Just back on the atmosphere less world (moon average temperature) problem.
The very small amount of heat coming from radioactive decay, (I thought about 0.5 watt/sq metre while someone above suggested 1 watt/sq metre) limits the minimum temperature of the never sun exposed parts.
The lowest temperature then would be around 50 to 60 K. It may add 20 to 30 K to the average for the moon.
On earth where the lowest surface temperature may be around 180 K the contribution from the interior of 0.5 to 1 watt/sq metre is quite insignificant.

January 12, 2012 8:48 pm

I second the request of Joules Verne 1/12 09:21 for greater information on the uncertainty in the albedo. Everytime I see a value, it is 30%. Ok, is that 3E-01 or is that 3.00000E-01? How many significant digits is that “30%” good to?
But there is another more basic question I have about the albedo. For this I have to return to “square 3.” Why do we account for albedo only on the direct-from-sun energy, but everyone seems to ignore it on surface-to-sky path?
Please humor me for a moment…. I haven’t seen this concept explained.
Let’s set up a thought experiment where we normalize energy from the Sun as 1.
That energy hits some atmospheric phenomena and 30% of it (call it A) is reflected as albedo into space. That leaves (1 – A)=B passing through the phenomena to strike the ground.
All of that B that hits the ground must reradiate (over some time) or else the ground would continuously heat up. So we also have an upward amount of energy B [so far so good..]… that strikes the underside of those albedo phenomena…. And then what happens? Almost everything I read has B just passing through into space as if the phenomena didn’t exist. Really? That’s quite a trick.
Let us suppose, in a general formulation, that upward B radiation encounters the phenomena and a fraction “a” is reflected back downward (and 1-a continues into space). ( a does not necessarily equal A ). Now we have some extra energy (b*a) traveling back down on a second leg…. By superposition, downward b*a is hits the surface and must be returned upwards, only to meet with the partial reflector and we have a third downward leg b*a^2, to be repeated in an infinite series.
By this “reverberation” within a partially trapped wave guide, we have energy striking the surface in the series: b + b*a + b*a^2 + …b*a^n. Rewrite as: b*(1 + a + a^2 + … a^n) which infinite series (if a<1) can be replaced with b*(1/(1-a)), or b/(1-a).
Finally, lets remember that b = 1-A. So that means the energy hitting the ground can be calculated as (1-A)/(1-a). What if the albedo phenomena reflects upward and downwards equally well? A=a. Then the ground energy is not (1-A) but it is 1.0000. The actual value of the albedo doesn’t matter if A=a.
I did this in the general case because it is quite possible that because of changes in spectra of the direct energy and the ground reflected energy, “a” may very will not be equal to “A”. In the steady-state, “divide insolation by 4”, without day-night, world, I can be talked into a A.
But I am having an increasingly hard time accepting a=0, which seems to be the value for “a” used by default in many examples. Where have I gone off the rails?
Finally, I cannot quit without returning to the real world where insolation is really divided by 2, we have day and night, and albedo is a function of time and season. You can warm the morning in a cloudless Pacific, then bring on the afternoon clouds as Nature’s thermostat (Willis: WUWT 6/7/11) and bask in the warmth of ( a = A approximately.)

January 12, 2012 8:51 pm

formatting correction to above: “I can be talked into “a” not equal to “A” .

January 12, 2012 9:01 pm

Embellishment on the trapped waveguide….
Have I short changed the earth’s total albedo? Nope.
The escaped energy off the first leg is B*(1-a). What leaks through the second bounce is B*a*(1-a). What leaks on the third leg is B*a^2*(1-a). Which again becomes an infinite series of B*(1-a)*(1+a+a^2+..) which rewrites to B*(1-a)*(1/(1-a)) which reduces to B, the same result you get if you treat the albedo as transparent when going up.

Jim D
January 12, 2012 9:32 pm

It is very simple. Net incoming solar radiation for a spherical earth with albedo 0.3 is 240 W/m2.
Black-body temperature required to radiate 240 W/m2 is 255 K. QED. Any questions?

Joe Postma
January 12, 2012 9:34 pm

I think there must be something wrong with the way standard theory treats the geothermal contribution to energy, whether it be from past absorption from sunlight energy or true geo-energy.
1 Watt/m2 is 65 Kelvin. The ground beneath the surface is NOT 65 Kelvin. If I take a shovel and dig a hole 1 meter deep, the newly exposed ground WILL be radiating at its temperature, say 5C or 338 W/m2. So, that means the ground went from contributing 1 W/m2, suddenly to 338 W/m2, just because I dug a hole. Something is amiss here, and it isn’t with the fact that exposed ground will radiate fully according to its temperature.
Look at it this way: the geothermal temperature contribution coming from deep below will, at some depth nearer to the surface, blend with the average temperature of the soil being induced by sunlight. In other words, going down beneath the surface, we expect the soil temperature to drop until it stabilizes, and then start rising again, due to the temperature from geo-energy. This merge-point temperature is not anywhere near 60K. In fact, I think for most of the planet, the merge-point temperature is actually quite shallow in depth, relatively very near the surface, and also well above 273K.
This merge-point temperature, being supported by below from the geo-energy, or from past absorbed solar energy, or both, presents a baseline temperature upon which the temperature induced by solar radiation at the surface will vary. In the real-time differential heat-flow equation I’ve been developing, which models all the other parameters of the system, this baseline temperature can to be taken into account. (BTW, this model reproduces the daily temperature lag to solar insolation, as well as the seasonal lag.)
In effect, there is SUPPORT to the surface temperature coming from beneath – that means that the night-time will not cool as fast, and warming in the day-time from solar insolation will also start from a base plateau…not 0K. The effect of this is to raise the average temperature at the surface, given that the real-time solar insolation reaches noon-time heights of up to +121C (factoring in extinction reduces that, but the value is still anywhere between 50C and 90C).
So however they’re calculating the negligible 1 W/m2 from geo-energy, it doesn’t make sense at all if you just think of the actual temperature the ground is beneath the surface, and why it has this temperature to begin with. In a physical heat-flow equation which models the actually accumulated and existing energy, or the actual direct temperature, you simply can not say that the sub-surface has near-zero temperature and/or near-zero internal energy – because the simple fact is that it doesn’t. Their is a GREAT DEAL of energy contained in the soil and sub-surface, and it is generally much closer to 0C than to 0K.
What’s really interesting is that if I DO incorporate a term for the sub-surface temperature and energy, the response of the real-time differential heat-flow equation acts just like what you expect current theory to say about the greenhouse effect: it raises the surface temperature.
What the atmosphere comes out as, in any solution to this heat-flow equation with or without the geo-temperature included, is that it is a very simple responding parameter – the temperature of the near-surface air is simply something that responds to the direct-surface temperature, and when night falls, the whole system simply cools according to its thermal time-constant. When the Sun comes back up, the surface starts warming and THEN the near-surface atmosphere starts warming; when the Sun passes the point in the afternoon when its incident-angle insolation temperature matches that of the reached-surface temperature, the system plateau’s and starts cooling again. This is the origin of the daily temperature lag. The origin of the seasonal lag comes in from the soil having such a large thermal capacity, and therefore, much slower response to the seasonal variation in solar insolation, as the Sun slowly swings up and down over the celestial equator.
Some things to think about…

DR
January 12, 2012 9:36 pm

Illis
I always find your posts interesting and informative. These two sentences caught my eye:

– it takes 1500 years to warm up the entire ocean versus the Earth Surface temperature.
– it takes about 10,000 years to completely melt out a continental ice age.

I am interested in understanding more how this relates to CWP. It would seem to me that sea level rise is a direct function of this “lag” of 10,000 in that we should not be surprised that glaciers wax and wane, and sea levels continue to rise, but rather be very worried when glaciers advance and sea levels drop.
So where are we in the geological time frame with regard to the “10,000 years to completely melt out a continental ice age”? What happens after that? I wonder if Nir Shaviv’s hypothesis on the spiral arm/GCR explanation for interglacial periods is valid. IIRC he postulates we are near the end of the current warm period and may begin transitioning to the next ice age.
Looking at the various cooling/warming periods, each warm period looks to be a step below the previous…..RWP > MWP > CWP. Is the inertia from the Holocene losing its mojo? Sorry Mike, the hockey stick is dead.

Stephen Wilde
January 12, 2012 9:39 pm

Any atmosphere, whether composed of GHGs or not reduces the ability of the surface to radiate to space by diverting some of the surface energy to conduction into the atmosphere which warms up to match the surface temperature.
Applying the Ideal Gas Law then redistributes the energy in the atmosphere to create a temperature gradient from surface to space.
Due to the atmosphere having the highest temperature just above the surface the effect of the Ideal Gas Law feeds back to the surface by reducing the rate at which the surface can conduct and radiate energy upward with the result that the surface can then itself achieve a higher temperature in reponse to the same level of solar energy input.
Thus the equilibrium temperature at the surface rises as a result of atmospheric density and pressure.
That is the Greenhouse Effect.
The radiative abilities of the atmospheric gases are relevant to the patterns of energy movement between surface and space but do not affect the surface temperature unless they also significantlly increase atmospheric mass.

u.k.(us)
January 12, 2012 10:01 pm

Wow, the colossal waste of ingenuity, perpetrated by activists.

January 12, 2012 10:28 pm

Jim D says:
January 12, 2012 at 9:32 pm
It is very simple. Net incoming solar radiation for a spherical earth with albedo 0.3 is 240 W/m2.
Black-body temperature required to radiate 240 W/m2 is 255 K. QED. Any questions?>>>
Yes. Do you understand that this number is invalid except for the case where the temperature of the earth is 100% uniform around the entire globe?
Please allow me to illustrate.
Earth is 255K across the entire planet. P= 5.67*10^-8*255^4 = 240 w/m2.
Now let’s suppose an Earth that is 310K over exactly half the planet and 200K over the other half.
(310 + 200)/2 = 255K “average”.
BUT
For T = 310K P= 5.67*10^-8*310^4 = 524 w/m2
For T = 200K P= 6.57*10^-8*200^4 = 91 w/m2
“average” P = (524 + 91)/2 = 307.5 w/m2
Uh oh. We’ve got two scenarios, each with an “average” T of 255K, but one radiates at an “average” of 240 w/m2 and the other radiates at an “average” of 307.5 w/m2.
The larger the temperature distribution, the more pronounced this becomes. For example, if we used a T of 350K and 160K, we’d still get an “average” of 255K, but the average P would rise to 444 w/m2!
In other words, the larger the temperature distribution, the MORE w/m2 it takes to maintain the “average” temperature of 255K. Put more simply, given that the temperature distribution of the earth ranges from -80C at the south pole to +40C at the equator on any given day, that the day/night cycle imposes a 20 degree temperature swing nearly every day, and that temperate zones swing by 60 degrees or more annually, there is no way the 240 w/m2 is anywhere NEAR enough of an “average” radiance to support an “average” temperature of 255K!
If the average temperature of the earth, with the kind of variance I just pointed out, actually was 255K, it could not possibly be supported in radiative balance by a mere 240w/m2. conversely, if the absorbed radiance actually is 240 w/m2, then the “average” temperature cannot possibly be anywhere near as high as 255K!
Since we at least have some arguments to suggest that 240 w/m2 is a not bad guestimate, then we can only conclude that 255K is WAY TOO HIGH as an estimate of temperature via SB Law. If that is the case, then the oft quoted 33 degrees from GHE arrived at by subtracting 255K from 288K cannot possibly be right.

January 12, 2012 10:47 pm

Robert Brown wrote:

However, Enthalpy, the Earth’s climate viewed as a self-organizing heat engine, and chaotic fluid dynamics of open systems are for later. First I need to take baby steps… even if they follow in the footsteps of others or if future things will confound them.

Here’s a comment I wrote earlier.
The only reason that I can see why one would spend time taking those baby steps; if all you have down the road is broken glass, mirrors and wolves; is to gauge the expanse of folly. 😉
Of course, I don’t get paid to do climate research.
Enthalpy is only about the amount of heat stored. It says nothing about how the weather (thus climate) works as a heat engine. If you “measure” the enthalpy, you can tell if the heat content is rising. I you only measure temperature of one or two components at their extremes, it tells you nothing about the state of the real world. Those temperatures are only useful as Lotto numbers and for political purposes.

Bob Fernley-Jones
January 12, 2012 10:53 pm

Robert G. Brown
Robert, whilst I totally agree with the wide sentiments here that your article is very interesting, after some pause in posting this, I’d like to say that all that you have achieved so far is to give yourself and others a headache. Yes, it is indeed all very complicated, but how about tackling a small part of it at a time? For instance this silly business of some experts asserting that there is an effective radiative T for an airless Earth, when albedo distribution and thermal characteristics of regolith and rocks/geology and stuff galore such as volcanism over what age, is purely speculative for an airless Earth.
Oh OK, let’s look at the moon instead where Willis asserts its albedo is 0.11, ho hum.
What IS clear is that there is a huge variation in surface T, and that under the “Noon solar hotspot”, the rate of heat loss must be comparatively huge, (per T^4), compared with the rest of the spherical surface area of the moon, which is simplistically argued to share and shed the insolation uniformly per unit area!!!
Furthermore, might I suggest, (because I’m past my prime in the maths), perhaps you or your students might take-on the task of doing an integration of insolation and emission over the entire lunar surface, based on Willis’s albedo, and a range of plausible thermal characteristics for the lunar regolith? It would be most interesting to see!
I’ve had some interesting intercourse with Willis starting in the link below, but after several exchanges he seems to have taken a premature withdrawal:
http://wattsupwiththat.com/2012/01/08/the-moon-is-a-cold-mistress/#comment-860094
Oh, and I think he also does not understand the warming effect of even a transparent atmosphere, but one thing at a time!

ferd berple
January 12, 2012 11:22 pm

davidmhoffer says:
January 12, 2012 at 10:28 pm
Since we at least have some arguments to suggest that 240 w/m2 is a not bad guestimate, then we can only conclude that 255K is WAY TOO HIGH as an estimate of temperature via SB Law. If that is the case, then the oft quoted 33 degrees from GHE arrived at by subtracting 255K from 288K cannot possibly be right.
Which pretty much explains why the IPCC climate models went off the rails 12+ years ago.

ferd berple
January 12, 2012 11:37 pm

Bob Fernley-Jones says:
January 12, 2012 at 10:53 pm
Furthermore, might I suggest, (because I’m past my prime in the maths), perhaps you or your students might take-on the task of doing an integration of insolation and emission over the entire lunar surface
Such a task is much better suited to a computer. It could get the wrong answer much quicker than a class of students.

Joe
January 12, 2012 11:57 pm

Dr. Brown,
I have been waiting a long time for someone to start this discussion. Yours was a fantastic explanation that highlights the incredible complexity required of any model for the earth’s surface and atmospheric system. An integrated circuit designer may spend a day or more in processor time to simluate a few microseconds of a slice of circuit that covers a volume a square millimeter by a few microns deep. Simulating an IC with well constrained operating conditions is far simpler than a first principles simulation of the thermodynamics of atmospheric gasses or ocean water on the same fraction-of-a-square-millimeter volume. Forget simulating the entire earth! But, simple models can constrain the allowed degrees of freedom.
One operating principle of the system can be derived from the numerous comments above this one. All of the thermal models explained above used 240W/m2 or 341W/m2 but the real number is 1366W/m2. That is the insolation upon the square meter of earth’s atmosphere at high noon at the equator, ignoring the earth’s tilted axis. At that point on the earth at that time, the incoming energy to deal with is the full 1366. At the pole it is the microwave background radiation equivalent to 3K with zero coming from the sun. Same for the night side.
In a simplified model, the breakeven point where the 1366 drops by a factor of 4 due to lattitude to match the average insolation across the globe occurs at 75.5 degrees north and south. (arccos 0.25) Thus, with no atmosphere, the points on the earth that match the average insolation for the globe occur at 75 degrees north and south. South of that lattitude, the incoming radiation is higher than the assumed average and north of that lattitude it is lower, decreasing to zero at the pole. With the atmosphere absorbing or reflecting half of the incoming energy, as noted by NASA on their web site, the breakeven lattitude drops to 60 degrees north and south. (arccos 0.5)
The conclusions to be drawn from this extension of the standard 1366/4 model are:
1) The local average temperature at lattitudes less than 60 degrees is lower than that necessary to balance incoming solar radiation.
2) The local average temperature at lattitudes greater than 60 degress is higher than that necessary to balance incoming insolation.
NOTE: NASA publishes a map of satellite measurements that actually show this differential.
3) The primary mechanism responsible for maintaining the incredible 4-billion-year stability of our system must be the physical transfer of heat from the equator to the poles where it is radiated away. This means weather and ocean currents physically moving heat are the primary means of thermal regulation of the earth’s surface. The carbon dioxide content of the atmosphere probably affects the efficiency of that transfer but it is the physical transfer that counts, not radiative balance at each point on the earth’s surface.
4) The poles, even at -90C, are still 177K above the 3K microwave background so they are not cold, they are frying pan hot and radiate heat into space at a prodigious rate across the largest temprature differential on the globe.
5) The Arcitc is unique because it has liquid water underlying all of its ice. The thermal differential of that water to outer space is even larger at 270C. Ice is an effective insulator so it prevents the radiation of surface thermal energy into space, a stopper in the drain so to speak. The incredible conclusion from this line of thought is that if the Arctic ice melts completely, it will not result in the world warming up but just the opposite. Without the insulation blanket at one of the two primary net-positive thermal radiation points on the earth’s surface, the earth will cool rapidly when the Arctic goes ice free as I am sure it has many times in the past. This effect provides the sign inversion in the system equation necessary to induce the oscillations we see.
6) I did not even try to throw in an analysis of the impact of the rotation rate of the earth but it can be quickly appreciated that it is an essential contributor to the dynamics of our system and helps limit the temperature variation. The temperature due to the sun can never exceed the boiling point of water anywhere on the earth’s surface at any time ever or we would be doomed when all water evaporated into the atmosphere. If the atmosphere did not block 50% of the incoming radiation and the earth did not rotate, that square meter at the equator would equilibrate at a temperature above 100C.
NOTE: The boiling point of water is the difference between Venus and Earth. Venus gets twice as much energy and its water boiled. Water vapor rose to the top of the atmosphere where UV broke it apart. The hydrogen drifted into space while the free oxygen combined into something more stable like carbon dioxide. All of Venus’ carbon is still in its atmosphere. On Earth, water stayed liquid and close to the surface away from the UV. Life formed, captured the carbon, and buried in the ground. All of Earth’s carbon is in its limestone, not its atmosphere. 700GT of carbon is in our atmosphere. 100,000,000GT of carbon is sequestered in the earth. Good job Life!
Conclusion: models using thermal radiation balance are static models that simply cannot represent the earth’s system. More importantly, averages don’t count. As you pointed out, Dr. Brown, energy input varies with lattitude and the length of the day. The fact that a large portion of the earth’s surface averages a temperature below the geometric mean for its incoming energy indicates that static thermal balance is not the primary determinant of surface temperature. The excess incoming energy must be removed physically by the weather and ocean currents to the colder reaches of the earth.

ferd berple
January 12, 2012 11:58 pm

Jim D says:
January 12, 2012 at 9:32 pm
Black-body temperature required to radiate 240 W/m2 is 255 K. QED. Any questions?
Why are the hottest places on earth those land areas that are below sea level? Why are deep mine shafts hotter still. It can’t be GHG radiation at the bottom of a mine shaft.

Robert Clemenzi
January 13, 2012 12:31 am

markx says:
January 12, 2012 at 5:58 pm

Robert Clemenzi says: January 12, 2012 at 2:10 pm
“….CO2 in the stratosphere emits radiation toward the surface. ….”
Quick question – perhaps a dumb one (I have checked you link but am none the wiser) – does the CO2 in the troposphere really only emit radiation towards the surface, heating the CO2 below and thence H20 in the tropospause, so the H20 then can emit it to space? I’d expect the majority of it was being emitted (to space) via CO2 in the stratosphere?
Or does this just refer to a trapping mechanism for that portion of the radiation going downwards?

No. In the atmosphere everything radiates in all directions. However, since those gases mainly radiate in the frequencies where the atmosphere is already opaque, that radiation is better than 90% reabsorbed in 100 meters or so. Near the band edges, the distance to absorb 90% increases.
At the tropopause, the water molecule mixing ratio is about 80 ppm. In the stratosphere, it drops to about 5 ppm. About 1km below the tropopause (in the troposphere), it is about 160 ppm. (These are very approximate numbers.) As a result, water vapor radiation toward the Earth is easily absorbed and energy toward space is not absorbed. This is the reason that the tropopause is colder than the air both above and below it.
CO2 is different because the mixing ratio is the same (about 380 ppm) from the surface to the mesopause. As a result, any emitted photons only travel a few 100 meters before being reabsorbed. In the stratosphere, the top is quite warm (just below freezing) and the bottom (the tropopause) is very cold. As a result, and because net heat flow is always from hot to cold, the “net” CO2 emission is toward the surface. The fact that the “typical” tropopause has an isothermal thickness of 10 km or so also supports the conclusion that heat flows from the stratosphere toward the tropopause.
CO2 will also emit the energy at the band edges to space since the spectra line widths are pressure sensitive. This partially explains the central spike seen in spectra taken from space.

Stephen Wilde
January 13, 2012 12:59 am

Further to my post of 9.39 pm.
It follows that IR sensors pointed at the sky are not measuring downwelling IR from GHG molecules higher up.
All they are measuring is the warmth of the air molecules directly in front of the sensor and those warmer molecules (whether GHGs or not) have been warmed by operation of the Ideal Gas Law (which automatically causes warmer molecules to be found lower down in the atmosphere) and NOT by so called downwelling IR.
There is no need to propose any downwelling IR at all. The warmth is already present in the lower molecules by virtue of the Ideal Gas Law working via pressure and density.

wayne
January 13, 2012 1:15 am

Bob Fernley-Jones says:
January 12, 2012 at 10:53 pm
… , perhaps you or your students might take-on the task of doing an integration of insolation and emission over the entire lunar surface, based on Willis’s albedo, and a range of plausible thermal characteristics for the lunar regolith? It would be most interesting to see!
——
Bob, I’d rather he spend the time writing more here at WUWT. Dr. Brown has covered all of the various aspects you will find over some 200,000 comments of last few years plus many never mentioned and has clearly and graciously laid it all out in one single post. I’m elated!
Sounds like you need to address Joe Postma who sounds like he has that already programmed, or, if he can’t easily convert his to the moon, I am writing what you are asking for, as I type this. Seems best to conserve our efforts and not get spread too thin with multiple duplications of the same thing. Mine is across any body, you just set the many parameters defining the orbit, body and the type of analysis of temperatures you wish to run. Keeping it simple for now (kind of), any grid size limited by memory, and am not going to create multilayered surfaces and atmospheres yet (if ever)… ( 🙁 have no supercomputer ). Since this morning I already have the dialogs written, cell/data structures created, integration comes next, just hang on. Comment to Joe, you might not have to wait the week.
See: http://wattsupwiththat.com/2012/01/12/earths-baseline-black-body-model-a-damn-hard-problem/#comment-862486

Robert Clemenzi
January 13, 2012 1:47 am

Tim Folkerts says:
January 12, 2012 at 7:14 pm

For all those claiming that CO2 would only have a small warming effect — or even a cooling effect — here is your homework. Explain the shape of the IR spectrum as observed from space looking down and from the surface looking up.
* Explain the shape of the dotted lines labeled with different temperatures
* Explain the cause of the various “bites” in the spectrum.
* Estimate the order of magnitudes of the GHG effects at a given place and time (ie when a specific satellite image was taken).

Your suggested image at http://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/ is garbage (extremely low resolution). I suggest using http://maths.ucd.ie/met/msc/PhysMet/PhysMetLectNotes.pdf Figure 5.17 from Petty (2004).
“the dotted lines labeled with different temperatures” represent expected blackbody emissions at those temperatures.
In both images, 600 to 780 cm-1 is the CO2 emission from the stratosphere. These are almost the same temperature in both the desert and ice cap images, proving that the surface temperature has no effect. In both images, there is a sharp spike in the middle of the CO2 feature. This is from the warmer parts of the stratosphere or the lower mesosphere. I have seen additional satellite data where clouds were present, and this feature was still present, proving that it is an emission feature from above the clouds.
1050 cm-1 is due to ozone. It is much warmer than the 600 to 780 cm-1 CO2 emission indicating that it is from higher in the stratosphere. It is also still present on cloudy days.
In figure B, Ice Sheet, emissions on the 180K contour are from the ice. It is obvious that he CO2 and ozone features are produced by gases much warmer than the surface.
In figure A, Sahara desert, 40 to 550 cm-1 is from water vapor in the troposphere. (I know that image starts at 400 cm-1.) This crosses many temperature contours because the band width gets narrower with increased altitude. The warmer points are lower in the atmosphere. The source of the 150 cm-1 value (not shown) should be near the tropopause. 800 to 950 cm-1 is the atmospheric window where IR radiation is not absorbed in the atmosphere. As a result, it indicates the temperature of the surface. 1300 to 1800 cm-1 is more water vapor – colder in the center of the band, warmer at the edge.
Even these spectra have a fairly low resolution since the instruments have only a moderate line resolution. In particular, the height (temperature) of the CO2 and Ozone peaks is probably much higher than shown. In addition, the bottoms of those 2 bands appear to follow blackbody contours and I don’t understand how that is possible.
“Estimate the order of magnitudes of the GHG effects” – sorry, I don’t understand the question.

January 13, 2012 1:59 am

Dr. Brown
I love the physics of this and it is why I became a skeptic, the models for the behavior of IR absorbing and emitting gasses were simply wrong. You mentioned in passing the QM aspects, which most people ignore but are the dominant factor related to absorption and emission of energy in IR absorbing gasses.
There is a great textbook that has the pertinent equations for CO2 and other IR absorbing gasses.
“The Quantum Theory of Light” by Loudon, page 81-90.
That gives you want you need for the QM treatment of IR absorption/Emission.

richard verney
January 13, 2012 2:33 am

davidmhoffer says:
January 12, 2012 at 10:28 pm
////////////////////////////////////
David
Good to see some sanity in all of this. Averages only distort what is going on. The claimate science community uses averages because this simplifies matters but they overlook that this inevitably leads to the wrong result. It is imperative to get away from averages in all aspects of this science. There is no such thing as global warming (some areas may be warming, others are staying bvroadly static and some areas are cooling). One cannot begin to understand the climate and how it may behave in the future, what problems or benefits changes may bring until one looks at it locally.
I think that your extremes of temperatures, if anything underestimate matters. But even in equitorial areas, it can be cool, eg., the Himalayas. Further, the Earth is only ‘spherical’ over the oceans, over land it is far from smooth and has nothing like a spherical surface area (albeit it will both absorb and radiate heat over this larger non spherical area).
One needs to know where the energy is being received, in what quantity , its local albedo and its latent heat capacity. The oceans respond differently to land and they retain and give up their heat in very different manners, not forgetting the phase changes in the water cycle and the latent heat involved.
The entire Earth BB model is fundamentally flawed and until this is corrected, it is impossible to get a proper and accurate handle on the radiative budget (which in any case is not the whole story).

richard verney
January 13, 2012 3:15 am

Surfer Dave says:
January 12, 2012 at 6:10 pm
@Tim_Focketts
///////////////////
As regards the oceans, don’t forget that the oceanic crust is far thinner than the continental crust ( circa 7 to 10 km thick verses circa 36 – 40km) such that one would expect more geothermal heat to make its way through the oceanic crust AND due to the depth of the ocean (average circa 3,800 meters but extending to nearly 11,000 metres), the seabed is nearer the hot core than continenetal land again suggesting that more geothermal heat would make its way through the oceanic crust in order to heat the deep oceans. To some extent, these differences may be countered by different density of crust material (generally the oceanic crust is more dense).
I have for years been suggesting that consideration needs to be given as to whether the oceans are effectively sitting on a warm hotplate and this is contibuting to keeping the deep ocean warm.
I do not know how large the effect is but I do consider it to be an area which is overlooked especially since this pertains to over 70% of the surface area of the Earth. .

richard verney
January 13, 2012 3:26 am

Don’t forget that as regards land, we are not measuring the temperature at the surface but rather at about 1.5 metres above the surface.
Are there any studies dealing with temperature profiles between the surface and the height at which global land temperatures are taken? In other words, if we were to measure temperature (using the same equipment but scaled down stevenson scrrens) at 1 cm, 5 cm, 10cm, 15 cm etc through to 1.5 metres what results would be achieved? Likewise if we were to use an IR thermometer pointing at the ground compared to the adjacent weather station? Perhaps Mr Gates will know since he is always a good source of information and I would welcome any comments that he may have on this specific issue.
I would like to know whether there is any substantial difference and if so to consider the implications of this.

GabrielHBay
January 13, 2012 3:29 am

Phil. says:
January 12, 2012 at 11:11 am
Stephen Wilde says:
January 12, 2012 at 8:42 am
Excellent article.
One question:
How do we know that the Earth is any warmer than it would be without greenhouse gases if the standard assumptions are so obviously inappropriate and/or incomplete ?
It’s quite straightforward. The emission spectrum of the Earth from space is grey body with numerous missing bands which can be unambiguously assigned to the GHGs (CO2, H2O, O3, CH4 & N2O), the spectrum of the Earth absent those GHGs would be grey body but would have to have the same area under the curve which requires a lower temperature. Therefore there is a GHE due to the presence of those gases.
—————————————————————————————
Well, sorry, but I do not buy this “straightforward” explanation. Just as an example, how does one know that the energy of the missing bands did not “change colour”, so to speak, and exit under the cloak of one or more of the other bands? Like 10 women with full spectrum of hair colours enter a salon and after a while only blondes emerge… can one conclude that there is a serial killer favouring brunettes lurking in the salon? Not unless you do a headcount. Just noting the colour is meaningless… And even with a reduced headcount, when it comes to cause, it may actually be blondes disappearing but everyone else being turned to blonde to the confusion of the observer. I agree with other posters here that NOTHING is straightforward and I suspect that Phil is deluding himself. Just because his reply sounds scientific does not mean it is correct.

michael hart
January 13, 2012 3:47 am

Lazyteenager, Dan Kirk-Davidoff has kindly posted a couple of GCM links for me. You’re quite correct about the code likely being little practical use to me. I’ve had similar experiences with far simpler free programs from academic sources [and I only own a laptop, not a Cray].
I am more interested to look at accompanying documentation to gain insights into how they go about performing the calculations they do, and how open the authors are about assumptions made [as these are frequently not obvious, and are easy to forget].

Joel Shore
January 13, 2012 6:35 am

Stephen Wilde says:

Any atmosphere, whether composed of GHGs or not reduces the ability of the surface to radiate to space by diverting some of the surface energy to conduction into the atmosphere which warms up to match the surface temperature.

Stephen has freed himself of the limitation of actually having to come up with explanations that use correct principles of physics. Apparently, he finds correct physics to be too constrictive.
What correct physics principles would tell you is that a surface radiates according to its temperature (and its emissivity, which is a property of the surface). There is no “diversion” of surface energy to conduction. Any conduction that occurs is in addition to whatever radiative transfer occurs due to the surface’s temperature.

HankHenry
January 13, 2012 6:49 am

When speaking of “surface temperature” of the earth what is usually meant is surface air temperature. The textbooks quote it as 14° C (or 288° K). It seems there are several surface temperatures, for instance: the top of atmosphere surface, the surface air temperature, or the surface temperature integrating in the temperature of land and ocean. If the third choice is the correct one then 14° C is too high because the volume of the ocean is huge (the weight of the atmosphere amounts only 33 feet of water) and the typical temperature in the depths is something like 4° C. In some ways the top of atmosphere temperature seems the proper choice but the problem is that you are then comparing a gas temperature to the black body temperature. I believe there is a presumption in the notion of black body temperature that your speaking of a *solid* black body. I am not sure if there are good measurements of temperature at the top of the atmosphere where radiation escapes. It also seems that if you are using top of atmosphere temperature you need to take atmospheric lapse into consideration.

HankHenry
January 13, 2012 6:56 am

There is a nice infrared picture of the moon taken during an eclipse that shows that the surface temperature of the moon is not at all uniform.
http://apod.nasa.gov/apod/ap050423.html

Joe Postma
January 13, 2012 7:08 am

“HankHenry says:
January 13, 2012 at 6:49 am
When speaking of “surface temperature” of the earth what is usually meant is surface air temperature. The textbooks quote it as 14° C (or 288° K). It seems there are several surface temperatures…….. It also seems that if you are using top of atmosphere temperature you need to take atmospheric lapse into consideration.”
—————————-
One other useful temperature to look at is the integrated average emission temperature of the entire planet. This is reported to be -18C, just as the law of conservation of energy and the S-B Law predicts. As you point out, applying the lapse rates via the Ideal Gas Law in the troposphere explains the rest of the temperature distribution, including that of the near-ground “surface” temperature, of 14C. And it does this without need for back-heat temperature self-amplification.
The paradigmatic problem with standard, now defunct, theory is that it doesn’t properly define, or properly understand, the exact question you posed.
For reference I will post the links again:
This first paper describes how to incorporate lapse-rates with the average emission temperature:
http://www.tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf
This one takes apart the logical errors of standard “self-heating theory”, and presents a new cognitive-physical model which we should start using to characterize the system:
http://principia-scientific.org/publications/The_Model_Atmosphere.pdf
This one summarizes in a short, and very readable and fun, the entire paradigmatic issue:
http://principia-scientific.org/publications/Copernicus_Meets_the_Greenhouse_Effect.pdf
I have continued developing the model introduced in the second paper and now have a differential heat-flow equation which can describe the heat and energy flow for any given square meter of surface in real time (daily, seasonal, yearly, etc, variations in local conditions, inputs and outputs). I will try to publish a new paper describing the theory soon.

Bryan
Reply to  Joe Postma
January 13, 2012 8:07 am

Joe Postma says
“I have continued developing the model introduced in the second paper and now have a differential heat-flow equation which can describe the heat and energy flow for any given square meter of surface in real time (daily, seasonal, yearly, etc, variations in local conditions, inputs and outputs). I will try to publish a new paper describing the theory soon.”
Look forward to your paper.
How to determine the surface temperature with radiative and ground flux contributions is the missing link in atmospheric theory.
Kramm and Dlugi are working on a similar approach.
See equation 2.17
http://www.scirp.org/journal/PaperInformation.aspx?paperID=9233

January 13, 2012 7:13 am

Joe Postma says:
January 12, 2012 at 3:03 pm
Robert Brown:
You might be interested in reading my critiques of standard theory here:

I am, and the second one, at least (which I started with) is very well written indeed, very clear. I’m still just at the “Fictions at the Boundary Conditions” section, but there are two things that I’m already itching to do, both of them “trivial” but potentially revealing. Note well that they both only address the (somewhat silly) basic model that assumes that = ^4 for a fluctuating T(t), but let’s go with that for the moment.
* The albedo \alpha is not constant, nor is it “equal to 0.3”, any more than “\pi = 3”. dT_s/d\alpha is easy to (algebraically) compute and/or plot. If one writes \alpha properly as \alpha \pm \Delta \alpha where \Delta \alpha is your choice of the uncertainty in \alpha or the observed secular variance of \alpha (assuming that as an astrophysicist you can read latex as easily as I write it:-) then one adds (at the very least) some extremely useful error bars (or fluctuation limits) to T_s. (Sadly, I don’t think this interface supports any easy way for me to insert the rendered algebra inline other than using html for the characters, which I’m allergic to doing, so I hope the rest of you can follow as well).
When I do this (from your equation 14 as good as any of them) on the back of what would be an envelope except that I keep my whitepaper clipboard with me at all time, I get (and bear in mind that I suck at arithmetic):
\Delta T_s = – 91 \Delta \alpha (degrees K)
(this is at \alpha = 0.3, T_s = 255K — the formula is actually \Delta T_s = -1/4 T_s/(1-\alpha)). Variations on the order of 0.01 in \alpha thus produce variations on the order of a degree K in T_s, even in this rather abusive treatment. Note well that this is on the order of the supposed effect of CO_2 doubling over the last 100 years, meaning that even in this simple stupid treatment first order variation in the albedo is a confounding effect.
* The thing that really struck me about the simplified GHG model that you presented is that there’s “not a thing wrong with it” as a very crude descriptive energy flow. Sure, it’s missing details (like the way that the atmosphere doesn’t absorb as opposed to reflect any of the incident solar flux ) but bear with me. The assumption is that the sunlight reaches the surface and warms it to one temperature (that radiates), that some of the energy is transferred to the atmosphere and radiated at a different temperature, and that the whole system has to self-consistently achieve detailed balance.
For the sake of argument, let’s go with this. Why not? f is then the parameter that describes the total energy transfer from the surface to the atmosphere. Forget the fact that it in the algebra it is presumed that this transfer is to all be blackbody radiative transfer and absorption (which is on the face of it absurd and even in standard models is modulated by e.g. the real spectra and absoption/emission bands and so on). The point is that heat is transferred to the atmosphere as a monotonic function of T_s (so as T_s increases the transferred heat increases) and that it is radiated from the atmosphere at a different temperature than T_s because the mean temperature of the atmosphere, T_a, is always lower.
All this means is that any monotonic coupling whatsoever between the surface and an atmosphere that warms the atmosphere ultimately warms the surface. This makes sense! It doesn’t make a damn bit of difference whether the atmosphere has GHGs in it or not — the transfer from the surface to the atmosphere need not be radiative although there is nothing at all wrong with some of it being radiative. The only thing that matters is that the atmosphere take up some of the heat delivered to the surface by any means whatsoever and then symmetrically radiate it up or down at a lower temperature.
This is very close to what I was getting from Caballero, but hadn’t quite been able to grasp because I was distracted by the fallacy that non-GHG-containing atmospheres don’t radiate. Of course they do. All that matters is that the radiating atmosphere, containing some sustained fraction of the energy of the sun, be thermally equilibrated (approximately).
Wow, epiphany. I do believe that without reading further I could just write down a modified (and equally “dumb”) model that takes incoming radiation, transfers a parametric fraction of the received heat to the atmosphere my non-surface-radiation means (adding back in the missing absorption, since it still comes out of the energy budget for the surface as if it were “transfer”) and the rest by radiation absorption, and get a perfectly lovely equation for T_s that lets one smoothly trade off “the Greenhouse Effect” (radiative absorption f\sigma T^4) against energy flux transferred by conduction/convection. Since in the end one simply fits T_s to the data, that part cannot fail, can it?
I look forward to reading the rest, and the third paper as well. As for working together, that would be lovely but let’s see how much time I end up having after I get to work proper next week and start to have student contact. I’m uncertain as to how much surplus energy I’ll have and don’t want to promise something and then end up disappointing you. But if you send me email at rgb at phy dot duke dot edu, I’ll save your address and if I have time to cook something up I’ll certainly communicate it to you.
I was in brief communication with Steve McIntyre yesterday (who like my “build-a-bear” climate model idea:-) and the idea surfaced that perhaps it would be a good idea to build an open climate model. Here’s what I would propose. We (collectively, not just you and/or I) write a modular open source software package to facilitate the general understanding of climate, with everything open and subject to modification and criticism. Anybody can download it. Anybody can make local modifications. If it is Gnu Public Licensed (viral) software (perhaps with a small codicil requiring that any modifications made that lead to a publication derived from it be published, made available according to the usual GPL rules) it both levels the playing field and establishes a common basis for climate modelling.
I’m far more inclined to do something like this (where I don’t have to do all of the work:-) than to try to build the whole thing from scratch myself. Since I am (I humbly postulate) a WGE on beowulfery and parallel computing, we could even make the (relevant parts of the simulation/statistical parts) of it parallel out of the box, ready to run on a compute cluster in a reasonably scalable way.
The fact that you are an astrophysicist makes me smile. I’ve thought for some time that physical climate modelling is something that should be done by astrophysicists working with condensed matter theorists working with computational fluid dynamicists working with statistical mechanics, perhaps with a few quantum mechanics and complex systems humans thrown in for good measure. Any single physicist knows something about all of this, but can’t possibly be an expert in it all. The traditional “I work over here all by myself with a one or two collaborators and a grant” model just doesn’t look big enough to be able to crack this particular nut.
Crowd-sourced physical climatology. I like it.
rgb

January 13, 2012 7:13 am

LazyTeenager says:
January 12, 2012 at 6:32 pm
Sir, since path and gradient are the basic requirements for any heat transfer, with gradient sloping down from higher T to lower T. Can you explain how these are satisfied by your understanding of LWIR from above.

Pierre R Latour
January 13, 2012 7:18 am

GHG Theory 33C Effect Whatchamacallit
Robert Brown is catching on.
GHG Theory was invented to explain a so-called 33C atmospheric greenhouse gas global warming effect. In 1981 James Hanson stated the average thermal T at Earth’s surface is 15C (ok) and Earth radiates to space at -18C (ok). Then he declared the difference 15 – (-18) = 33C (arithmetic ok) is the famous greenhouse gas effect. This is not ok because there is no physics to connect these two dissimilar numbers. The 33C are whatchamacallits. This greenhouse gas effect does not exist.
Here is the science for what is happening. Thermal T is a point property of matter, a scalar measure of its kinetic energy of atomic and molecular motion. It is measured by thermometers. It decreases with altitude. The rate of thermal energy transfer by conduction or convection between hot Th and cold Tc is proportional to (Th – Tc).
Radiation t is a point property of massless radiation, EMR, a directional vector measure of its energy transmission rate per area or intensity, w/m2, according to the Stefan-Boltzmann law. It is measured by pyrometers and spectrometers. Solar radiation t increases with altitude. Black bodies are defined to be those that absorb and radiate with the same intensity and corresponding t. Real, colorful bodies reflect, scatter, absorb, convert and emit radiant energy according to the nature of the incident radiation direction, spectrum and body matter reflectivity, absorptivity, emissivity and view factors. The rate of EMR energy transfer from a hot body, th, is Q, w = 5.67A*e*(th + 273)**4. But it may not be absorbed by all bodies that intercept it, as GHG theory assumes.
Above Earth’s stratosphere, thin air T is rather cold, about -80C. Yet solar radiation t is rather hot, about 120C. So spacesuits have thermal insulation and radiant reflection. The difference 200C is meaningless. On a cold, clear winter day on snowcapped mountains, dry air T = -10C and radiation t = 50C.
Much of GHG theory fails to make clear distinctions between these two different kinds of temperature, T and t. One temperature, t, is analogous to velocity, 34 km/hour north; the other, T, is analogous to density, 1 kg/liter. So 34 km/hour – 1 kg/liter is indeed 33 whatchamacallits by arithmetic, but nobody will ever know what a whatchamacallit is because velocity and density are not connected by nature.
To clarify this enormous intellectual flaw, take boiling point of water is 100C (true) and freezing point is 32F (true), subtract 100 – 32 = 68 (correct arithmetic) and declare atmospheric pressure is 68 psia. The declaration is false because a) the difference between C and F has no meaning, b) there is no physics to connect 68 to pressure, psia, and c) atmospheric pressure is actually 14.7 psia. That 33C greenhouse gas effect that has everybody so upset and is researched ad nausea to death is not an effect, merely an easily explained pair of facts.
Therefore, it is quite true the 33C greenhouse gas effect defined by James Hanson in 1981 as thermal T = 15C at surface minus radiant t = -18C to space is whatchamacallit nonsense. Everybody knows you can’t compare apples to eggs; except perhaps Greenhouse Gas theorists. Since this is irrefutable logic, no experiment is called for. Logic trumps nonsense; that is why humans invented it around 400bc. No one needs to prove or disprove the existence of whatchamacallits. They are not even imaginary. There is no greenhouse in the sky.
Planetary atmospheres reflect, scatter, transmit, absorb, emit and diminish stellar radiation intensity at the surface according to Beer-Lambert Law, 121C incident to Earth’s stratosphere to 15C at surface. Thermal T of atmospheres increase as gravity compresses gas and converts potential energy to kinetic energy closer to the surface, -80C in stratosphere to 14.5C at surface. Therefore atmospheres cause the surface to be colder than it would be if atmosphere were thinner or non-existent. The more O2 is exchanged for higher heat capacity CO2, the colder the surface radiation intensity temperature. Atmospheres are refrigerators, not blankets.
GHG theory postulates back-radiation from cold atmospheric CO2 is absorbed by the surface, heating it more. This violates Second Law of thermodynamics (energy can only be transferred from hot to cold bodies), leading to creation of energy, a violation of the First Law of thermodynamics (energy conservation), and the impossible perpetual motion machine AGW promoters need to cause eternal global warming.
CO2 does not trap radiation; like all molecules, it absorbs some incident radiation according to its absorption spectrum and promptly emits it according to its emission spectrum. CO2 is not a pollutant; it is inert green plant food. CO2 should not be curtailed, starving Earth’s flora. Minor solar driven global warming from 1974 to 1998 has stabilized through 2011. CO2 has nothing to do with global warming; it actually cools Earth. Arctic ice does not melt because of global warming, increasing T; it melts when average T > 0, at rate proportional to T, no matter whether T is increasing or decreasing.
This essay has seven scientific facts (33C whatchamacallit, no blanket, no back-radiation, CO2 no trap, CO2 inert food, no AGW, ice melts), each of which refute GHG and AGW. It has not been peer reviewed because it is well known to professional physicists and engineers; it does not merit a research paper, or research, or experiments. Logic just needs clear definitions and common sense, not government spending and regulation.

January 13, 2012 7:20 am

Ah. Your figure 3 from your third paper here is exactly what I want to compute, but correctly and parametrically, with what is still a nearly analytical and dimensionally reduced model. I’m not quite ready to try to solve a system of time dependent PDEs over the entire volume, but eventually that is where one has to go. For starters, actually making your figure 3 quantitative in a two color map polar view for a series of model parameters would be simply peachy, don’t you think? And quite doable. Pretty easy, even.
rgb

January 13, 2012 7:45 am

There is a great textbook that has the pertinent equations for CO2 and other IR absorbing gasses.
“The Quantum Theory of Light” by Loudon, page 81-90.
That gives you want you need for the QM treatment of IR absorption/Emission.

Funny that you should mention it, but around twenty years ago I worked in Quantum Optics (before being distracted into magnetism and critical phenomena, in part because the Optical Bloch equations and the equations of magnetism are seductively similar in certain contexts and I was postulating a possible dynamical second order phase transition for a very nonlinear quantum system). Loudon was one of my “bibles” of the time. My colleague and lifetime collaborator, Mike Ciftan, was one of the early co-discoverers of the laser — he observed the nonlinear gain in ruby before the laser work was published, but didn’t quite know what he had.
I’ll dig out my old copy and take a look (if I can still find it in my office, which is a mess, and if I didn’t loan it to a student over the years in between, sigh). It was right behind Allen and Eberly, and in front of Knight and a few others. One of my favorite Loudonisms from that book is his observation that spontaneous decay is understandable in a coupled oscillator model — in a closed system it isn’t really irreversible, but the lifetime for return diverges with system scale quite rapidly. These things all lead to a Generalized Master Equation (GME) approach to quantum optics as outlined IIRC by Agarwal, which brings it all back in context.
The GME is the “correct” formal description for the dynamics of an open subsystem of a larger Universe. This is true even if you make classical or semiclassical approximations (a la Jaynes) to bring the problem within the range of computability. As you make these approximations, depending on the level of detail you retain you end up with e.g. a Langevin equation, which I would argue is formally the right way to deal with the Earth in climate models — a set of coupled ODEs (or better, PDEs) describing the gross time evolution with a stochastic noise term, best studied with a dynamical Monte Carlo (of exactly the same sort I used in quantum optics, but the approach is quite general — set up a Markov process running in parallel with DE solution and with a correspondence between real time and “Monte Carlo Time”.
To make it concrete, start with a purely model rotating earth in sunlight, with heat capacity, and then add a noise term that causes (nucleates) albedo fluctuations. What does this do to the resulting temperature distribution? Break the surface up into equal area cells at some granularity and admit some sort of heat transport laterally (forbidden in the original model). What does this do? Give the heat transport some “inertia” (or hell, go ahead and make it crudely hydrodynamic with coriolis forces, a bullet that sooner or later you have to bite). What does this do? Adding the noise terms helps you identify instabilities, the additional dynamics, and maybe even the self-sustaining oscillations.
This is the kind of thing that would be a signature of being on the right track. A general model that spontaneously broke symmetry and created long-lived oscillations like the ENSO, PDO, AO, NAO etc would suggest that it is starting to have a lot of the right physics in it.
rgb

wayne
January 13, 2012 7:49 am

Robert Brown:
“Now make the surface have some thermalized heat capacity — make it heat superconducting, but only in the vertical direction and presume a mass shell of some thickness that has some reasonable specific heat. ”
Shouldn’t the word in italics be horizontal in that case? You might want to fix it in the top-post. Caught while coding those, but that is a curious combination, might just program it anyway, no cold poles.

January 13, 2012 7:50 am

There is a nice infrared picture of the moon taken during an eclipse that shows that the surface temperature of the moon is not at all uniform.
Interestingly, there is surprisingly little right-left asymmetry, suggesting that over this sort of timeframe either little cooling has occurred or else it was overall so fast that it has already come into cool side equilibrium. I’d guess the former. R. Gates? Weren’t you suggesting that overall the temperature drop was fast enough to cause a substantial asymmetry (as points on one or the other side of this picture should have been “in the dark” for hours longer than points on the other)?
rgb

January 13, 2012 7:52 am

Must… stop… reading… Work… to… do… Arrrrgggh.
rgb
(Maybe I’ll come back later to catch up, sorry, gotta have a life and all that for a bit.)

RACookPE1978
Editor
January 13, 2012 8:02 am

It appears essential that the atmosphereless, rotating, spherical, partially-gray “blackbody” problem be solved first. The “blackbody” idealized simulation that begins most CAGW scenarios is not worthless, but it does NOT even begin to describe ANY so-called “average” real world, much less the actual earth. Further, any CAGW model scenario that begins with an “average” radiation over an “average” flat-earth earth albedo over an “average” (non-rotating) day is worse than worthless.
This rotating “graybody” has, to a rough degree, been started by several writers that apply the moon’s approximate (average) albedo and (average) rotation to its theoretical blackbody characteristics, then try to factor the dwell delay of soil mass to “slow” the moon’s cooling each night. But, these early approximations need “calibrating” against actual lunar temperature sensor depth, actual lunar dust heat transfer coefficients, better heat capacity assumptions for the dust/rock combination at each lunar probe site, and accurate “calibration” of actual lunar temperatures against each site’s latitude. Also, the albedo changes on each face from the flat/darker lunar “Mares” to the crater-pocked rougher and lighter highlands need to become a part of this realworld “graybody” lunar calculator. (The varying “albedo-by-location-and-geography” factors for the moon will translate directly to the final earth’s albedo and physical-factors-by-location for the earth’s oceans and land masses.)
A program that can accurately back-calculate the moon’s temperature at all latitudes as it slowly rotates through the sun (simply and directly receiving radiation energy, moving it through conduction, and then re-radiating that energy WITHOUT the interferes of any convection, season changes, greenhouse gasses, surface gasses, winds and surface liquid phase changes!) is then a real-world start as we try to move from an idealized Einsteinian “thought experiment” of infinitely accelerating weightless elevators carrying twin brothers moving ever-closer to the speed of light to an design solution for temperatures within atmospheres. Adding surface soil heat capacities and depths, lunar soil emissivities, and actual rotation to the lunar calculator is the programming beginning to doing that for the earth’s calculator. Adding the complexities of varying solar radiation with day-of-year and latitude of the grid, and changing grid size with latitude for a lunar graybody program writes the same equations that need to be used for the earth’s more complex problem. After all,should not one be able to solve a “simple” problem without atmosphere and with no water before one begins “solving” a radiation problem with changing greenhouse gasses?
Then, after one’s computer program can correctly plot the moon’s actual temperatures in space across all latitudes across each lunar “day” exposure to the sun and the cold blackness of space, then one can begin to approximate the earth’s more complex geography of:
– vastly different albedoes (that vary somewhat with temperature and vary greatly with time-of-year!) on land from ground cover and geography (ice to tundra to prairie/steppe grasses to mixed forest/crops, to forests, to rocks, to deserts, to jungles, swamps and wetlands;
– vastly differing chemical reactions to heat (oceans and bare land and ice-covered land);
– vastly different absorption, reflection, re-radiation and movement of the surface heat energy and sub-surface heat flux
Now, one gets grossly simplified half-disk “averages” for radiation received, radiation emitted, and even worse whole-earth assumptions for average albedo. Polar circumstances are ignored – but the models are used to prove future arctic temperatures will rise catastrophically.

Tim Folkerts
January 13, 2012 8:09 am

HankHenry says: “There is a nice infrared picture of the moon …”
That is a cool image — thanks. I would warn that the conclusion that “the surface temperature of the moon is not at all uniform” is not6 necessarily true. False-color images like this often have the contrast enhanced to hightlight particular information. The difference in contrast in such a picture could correspond to ar range of only 1 C, or it could correspond to a range of 100 C. Without knowing the scale, it is difficult to conclude how uniform or non-uniform the temperature is.

Joe Postma
January 13, 2012 8:12 am

Hi Robert,
My strength is in analyzing logic and developing conceptual models that conform to reality. Such is why I was able to disassemble the standard “flat-earth” model of the GHE, and expose that it TRULY IS “flat earth” science! It’s not just in the toy models…it’s in the entire paradigm. The Earth ain’t flat. The Sun ain’t cold. Anyway…
Agreed on your point about albedo. An entirely miniscule change in albedo more than overshadows the supposed effect CO2 is pretended to have. When you consider that the average actual-surface albedo of the Earth is around 0.12, and that it is clouds which raise the average albedo to 0.3, then that means that very small changes in cloud cover dominate the average albedo, and therefore the average temperature of the Earth. This is one reason why the cosmic-ray connection in so important, and why solar activity is so important. It isn’t merely as simple as the variation in solar insolation, which is known to be finite but small…it is also the variation in solar magnetic activity which has a sort of “back-door” approach to affecting the climate. And of course this is aside from the fact that standard alarmist climate science can’t differentiate between the importance and meaning of correlation vs. causation.
What I will do then is try to finish the paper I am working on which describes the theory and math of this new model. I have taken the conceptual model from my paper – the model with the pretty colors – and written out the equations that would describe the insolation and the output for any given latitude and in real-time, for a spherical and rotating Earth. These equations then simply go into a differential heat-flow equation which I will also describe. I should be able to get enough of the idea down that others could then take it up and develop it further – I see the big picture but I get bogged down in details, but if the theory is good, then others could take it up from there.
If you just model the heat input and output, i.e. the energy, in a general heat-flow equation (identical to an RC-circuit equation with a time-dependent voltage), you don’t need to worry about all the intricate details of spectral absorption & etc etc etc. Just like you don’t need to model the intricate details of the workings of the capacitor in order to get a good model of the voltage in an RC-circuit. You just model the general conditions of the heat flow itself, given the known and measured parameters which affect & effect such. For WEATHER modelling, yes you might need more than that. But my intent is to present a new, general, fairly simple model which captures the boundary conditions of the real system. Which is something that the existing paradigm does not, because of the flat-Earth, cold-Sun problem it has.
I mean think about that – with Sunlight at only -18C, that means that the ENTIRE mass of the oceans is turned from ice into liquid by the greenhouse effect in the atmosphere. The tiny thin little atmosphere with an average temperature of -18C has a greenhouse effect which melts the entire mass of the ocean water to a higher temperature. As said: fundamental logical problems in the paradigm.
Great idea about an open-source project for a climate model. That would automatically be better science than the secret sauces of existing models.
(in software lingo, “sauce” = “source code” 😉 )

Joe Postma
January 13, 2012 8:27 am

“Bryan says:
January 13, 2012 at 8:07 am
Look forward to your paper.
How to determine the surface temperature with radiative and ground flux contributions is the missing link in atmospheric theory.
Kramm and Dlugi are working on a similar approach.
See equation 2.17
http://www.scirp.org/journal/PaperInformation.aspx?paperID=9233
——————————–
Gave a skim of that paper and it looks as though they are doing essentially what I’ve been thinking about, but in much greater and advanced detail. I’ll still write my own paper because it will still capture the essential features and logic of the new real-world paradigm, and in language and much simpler math that more people will be able to follow. Thanks for the link.

eyesonu
January 13, 2012 8:36 am

richard verney says:
January 13, 2012 at 3:15 am
A partial quote from your comment above of which should be read in its entirity:
“I have for years been suggesting that consideration needs to be given as to whether the oceans are effectively sitting on a warm hotplate and this is contibuting to keeping the deep ocean warm”.
==========
I have given thought to the same concept often. Adding to this potential ‘hot plate’ could also include an enormous amount of heat generated by the friction of the movement of the tetonic plates.
I don’t recall the temperature at the bottom of the Macondo (Deepwater Horizon) well, but it was considerable. I think it may have been on the order of 450 F at about 35,000 ft below mean sea level. An undersea ‘hot plate’ is an important factor in the sea’s energy budget it would seem. Heat from the friction of the plates movement could be tremendous.
If you can’t tax it, best not to discuss it?

January 13, 2012 8:52 am

Pierre R Latour says:
January 13, 2012 at 7:18 am
“But it may not be absorbed by all bodies that intercept it, as GHG theory assumes.”
Concur. I have mentioned this several times here at WUWT.
Another item that really baffles me is the “bait and switch tactic” of going from radiative heat transfer equations to get 240 W/m^2 to energy balance. Why not do the entire problem using what they started with radiative heat transfer equations?

Tim Folkerts
January 13, 2012 8:56 am

Dr Brown says:
>I was distracted by the fallacy that non-GHG-containing atmospheres don’t radiate.
>Of course they do.
This is only a “partial fallacy” (if such a thing is not too much of an oxymoron). To me this is akin to saying “I was distracted by the fallacy that relativity doesn’t apply to bullets. Of course it does”. The principles apply, but the next question is about orders of magnitude. No one uses relativity to calculate the trajectory of a bullet, because any relativistic corrections are orders of magnitude smaller than the newtonian predictions.
From everything I have seen, N2 radiates orders of magnitude less IR energy that CO2 or H2O. Even given that there is much more N2 in the atmosphere, the numbers still suggest that GHGs radiate much more. And simple satellite data shows that when you look down, you see very nearly 1) a blackbody radiation curve at the temperature of the ground and 2) “bites” taken out of this curve by cooler GHGs high in the atmosphere. There is no “signature” of the cool N2, so there is no noticeable radiation from the N2.
Beyond this, there is “simple quantum mechanics” (again a bit of an oxymoron) that predicts symmetric diatomic molecules will not have any vibration modes or rotation modes at the energy levels involved that could absorb or emit IR radiation. I believe that Rodrigo Caballero’s Lecture Notes on Physical Meteorology ( http://maths.ucd.ie/met/msc/PhysMet/PhysMetLectNotes.pdf ) discusses all this, but I can’t seem to open it right now. (Maybe all the traffic from WUWT has overloaded the server).
So, yes it is absolutely correct that N2 radiates. But my understanding of theory and experiment suggests that N2 is orders of magnitude less effective at this than CO2. So if you had a container with a mixture of hot N2 & CO2, the N2 could lose some energy via radiation, but it would lose more by transferring energy to CO2 via collisions, and then having the CO2 radiate the energy to space.

Stephen Wilde
January 13, 2012 9:04 am

Joel Shore said:
“What correct physics principles would tell you is that a surface radiates according to its temperature (and its emissivity, which is a property of the surface). There is no “diversion” of surface energy to conduction. Any conduction that occurs is in addition to whatever radiative transfer occurs due to the surface’s temperature.”
That begs the question as to how the surface temperature gets set in the first place.
Start with an atmosphere free planet. Solar energy comes in and goes out pretty much instantly.
Add an atmosphere and the only way for the atmosphere to acquire the same temperature as the surface is to take energy from the surface via conduction. Energy must therefore be ‘diverted’ to conduction from upward radiation that would otherwise have occurred.
In due course the atmosphere equalises with the surface (if all other things were to be equal).
However the Ideal Gas Law then kicks in and due to gravity and pressure all the warmest molecules in the atmosphere are to be found directly above the surface with a declining upward temperature gradient.
Now those molecules at the surface are the warmest atmospheric molecules of all. Their wamth exceeds that of the average for the whole atmospheric column.
As a result they inhibit the upward radiation and conduction from the surface more than would have been achieved if the air molecules above the surface had simply been at the cooler average temperature for the whole atmospheric column. Obviously the warmer the molecules just above the surface become the warmer the surface needs to get in order to push enough energy past those warmer molecules so as to achieve the necessary radiative output to space to match incoming solar energy.
In disproportionately inhibiting upward radiation and conduction those warmest atmospheric molecules raise the equilibrium temperature of the surface beyond that which would have been expected from the S-B equations.

eyesonu
January 13, 2012 9:06 am

@ eyesonu
January 13, 2012 at 8:36 am
The Macondo well depth should read approx 24,000 ft below mean sea level rather than 35,000 ft. I still haven’t found the bottom hole oil temp.

January 13, 2012 9:19 am

Joe Postma says:
January 13, 2012 at 8:12 am
“I mean think about that – with Sunlight at only -18C, that means that the ENTIRE mass of the oceans is turned from ice into liquid by the greenhouse effect in the atmosphere. The tiny thin little atmosphere with an average temperature of -18C has a greenhouse effect which melts the entire mass of the ocean water to a higher temperature. As said: fundamental logical problems in the paradigm.”
Mr. Postma add this to your thoughts. The corona of the sun is 1-3 million K, but nobody would suggest that the corona heats the sun. If that were true we should be using that temperature (1-3E6K) to figure our surface temperature not the surface of the sun’s 5700K.

HankHenry
January 13, 2012 9:30 am

Joe Postma,
Thank you for the links. It will take time for me to digest them. What I always want to account for in these discussions is radiation energy that penetrates oceans. I understand that it reradiates from the surface but what’s the real temperature that the ocean radiates at if the total ocean is only 5 or 6 or 7 degrees C.
I also believe that it is understood that heat radiated away at the poles causes chilled water at the surface to convect downward. Hence I want to say that the surface temperature at the poles does not indicate how much is truly radiating away because there is gradual “deep sixing” of water at a lower temperature. The ocean is not a black body.
Once I digest what you have to say about lapse rate this may get clarified for me.

RACookPE1978
Editor
January 13, 2012 9:30 am

wayne says:
January 13, 2012 at 7:49 am
(replying to)
Robert Brown:
“Now make the surface have some thermalized heat capacity — make it heat superconducting, but only in the vertical direction and presume a mass shell of some thickness that has some reasonable specific heat. ”
Shouldn’t the word in italics be horizontal in that case? You might want to fix it in the top-post. Caught while coding those, but that is a curious combination, might just program it anyway, no cold poles.

1. “Now make the surface have some thermalized heat capacity — make it heat superconducting, but only in the vertical direction and presume a mass shell of some thickness that has some reasonable specific heat. ”
That term must be “conducting”, not superconducting”.
2. Should the heat “flow” be horizontal? Well, no. To see why, please change your “design” from a laboratory-size 1x1x1 meter cube into something reasonable for a whole radiating body of real-world size.
To explain: Take a 10 km x 10 km square, at any latitude, and compare two scenarios for a 10 km x 10 km square: the moon and the earth.
On earth, the temperature about 30 feet (10 meters) below any land surface stays a constant 60-65 degrees F, regardless of season of year (long term surface temperature changes) or time of day (short term radiation heat flow changes). Only as you go very deep (more than 600-1000 feet) do you begin seeing temperature increases from our volcanic interior. Therefore, the heat capacity dwell times (change in stored heat energy with time due to changes in radiation received, radiation emitted, wind, humidity, air temperature, albedo, ice-melting, etc) only happens in the top 10 meters of soil.
Obviously, a difference in temperature is required for heat transfer of energy, and, as long as your 10×10 km square does not cross from ocean to land, the difference between any two adjacent squares of land is very, very low compared to all other differences: especially those difference between the 10km x 10 km surface “up” to the air above that surface. But notice that you can limit the depth of your analysis “cube” to the depth from a greatly changing surface (the top 6 inches to 1 ft) down to a “constant” temperature lower end. All this means that your 10 km x 10 km square need only be 10 meters deep, and you can “program”‘ that lower surface to be a constant temperature in your differential equation parameter setups. Also, since your “heat exchange” through each side is a function of area, each side of your “analysis cube” is 10,000 meters x 10 meters. The “upper surface” and “lower surface” of your cube is much larger (10,000 x 10,000 meters). transmit many thousand times more energy than do the sides of the “cube”.
Water-covered and ice-covered 10×10 km squares are different: they can move heat up from the “bottom” surface by conduction, convection from that water underneath; and then move that energy even further by currents under the “cube” moving the heated (or cooled) water. Further, ice-covered 10×10 km squares put another insulating boundary between the top of the water and the air, which will then prevent evaporation of the water. The water (or ice) will reflect and absorb very different amounts of energy depending on the angle of the received radiation. At very low angles of direct solar radiation, such as the arctic ocean’s ice-covered areas up north as the ice melts, both ice and water reflect equally well. At lower altitudes, ice reflects much more energy than smooth water. Both ice and water emit nearly identical radiation thermally. Indirect radiation, coming in at higher angles than direct radiation in the arctic, will be more strongly absorbed than direct radiation by water-covered 10x10m km surfaces. makes things difficult, doesn’t it? 8<)
On the moon, things are much simpler: There would only be three kinds of surface conditions to program:
– solid, flat, smooth, dark rock with few cracks (the plains),
– largely solid gray rocky surfaces with some cracks (the highlands and mountains) and highly irregular surfaces like canyons, hills and mountains.
-highly cracked and dusty very small solids.
Each may be made of similar material, but the heat transfer and heat capacity and albedo of each of the three surfaces is very different. A dust-filled surface made up of small grains will be almost like an insulator, but a solid "lunar "sea" is going to transmit heat very well. The rocky mountains and highlands are going to be somewhat in the middle.
The moon is not expected to have a molten core, so we would not expect to see any heat increase with depth as you look at your 10 km x 10 km x 10 m "cube". The bottom of the craters at the lunar poles where the sun never shines should come out very, very cold – They can radiate "out" every second, but the only heat that can come into such a spot can come from what (very) little bit gets transmitted from the rocks up at the top lip of each crater where the sun does shine. The rest of the moon's "steady state" temperature will need to be calculated based on radiation losses from the upper surface and thermal dwell times – I don't think it is known right now

January 13, 2012 9:33 am

:

First, from east to west, the geothermal gradient changes from values between 0.025 and 0.03 K/m (0.014 and 0.016F/ft) off the AlabamaMississippi shore to lower values of 0.0150.025 K/m (0.0080.014F/ft) off eastern Louisiana and to higher values of 0.030.06 K/m (0.0160.033F/ft) off western Louisiana through Texas. Second, thermal gradients tend to be lower toward the outer continental shelf (less than 0.02 K/m [0.0112F/ft]). Abstract: Nagihara et al 2008

Forget Macando, it wasn’t a hot well. There was a very interesting paper at an AAPG convention in 2010 about the “Will K” well, as a deep-high temperature test. (cannot find my notes at the moment.) It was unsuccessful and caused the company to revise their thermal cutoff from 350 deg F to 325 deg F. Above that temperature, permiability was nil.

Tim Folkerts
January 13, 2012 9:39 am

Joe Postma makes me smile (but shake my head) when he says:
>My strength is in analyzing logic and developing conceptual models
>that conform to reality.
but then also says:
>1 Watt/m2 is 65 Kelvin. The ground beneath the surface is NOT 65 Kelvin.
>If I take a shovel and dig a hole 1 meter deep, the newly exposed ground
>WILL be radiating at its temperature, say 5C or 338 W/m2. So, that means
>the ground went from contributing 1 W/m2, suddenly to 338 W/m2, just
>because I dug a hole.
It is clearly time to go back and rework that model and that logic. I would suggest figuring out this simple situation before tackling the greenhouse effect.

Joe Postma
January 13, 2012 9:40 am

Anything with a temperature radiates…in the case of non-spectral gases like N2 or O2, the radiation will arise from inter-molecular collisions. Perhaps we haven’t explored the spectrum at far enough wavelengths to see this emission; perhaps this emission is what helps constitute the entire profile of the “black-body” output curve of the Earth as seen from space in any case.
However, there is another important point to consider, which Alan Siddon’s has discussed elsewhere: if the spectraly-neutral gases like O2 and N2 don’t radiate, that means that they collect heat-energy from the solar-heated surface by conduction, and then hang on to that heat: they can’t shed it, they can’t radiate it away spectraly, they just hold on to it. A “GHG”, on the other hand, once having absorbed heat energy from outside into its internal vibration, can then shed that energy by radiating it away. Not being able to lose and radiate the energy away, vs. being able to, should be the difference between a heat-trapping gas and a heat-shedding gas.
In fact, given my perspective from astrophysics, this is exactly the theory that we use to explain how interstellar gas-clouds are able to overcome the thermal response from gravitational collapse (potential energy converts to kinetic = temperature), and continue to collapse to form stars. The spectraly-emitting molecules in the gas (like CO2, but typically CO and others) provide a “vector” through which the thermal energy build-up of the collapsing cloud can escape the cloud. They absorb thermal energy via collision into internal degrees of freedom, then radiate that energy away, out of the cloud. This effectively “damps” the thermal response and then causes cooling. This allows the cloud to collapse into a star.
The thing is, the molecules in these gas cloud don’t just radiate outwards, they also radiate inwards, i.e., back-radiate. If I was thinking as a greenhouse effect alarmist, I would then have to think that the back-radiation from the molecules causes FURTHER temperature increase internally to the gas cloud, because half of the radiation will be directed inward. This would have the opposite effect of helping the cloud to cool to assist its collapse!
So we have two scenarios in which the exact same physical processes are going on, but in one it is theorized to cause warming (alarmist GHE) and in the other it is theorized to cause cooling (modern astrophysical theory). It’s one thing to question colleagues on conformism in climate science; it would be another thing entirely if I asked them to completely re-write the standard theory of star formation: sorry, but your molecules now cause heating, not cooling, you need to find another way to explain how stars form out of gravitationally-collapsing gas clouds.
[Some of the generation 1 stars which formed right after the big bang would have been extremely massive, and these can collapse in any case; thus having seeded the interstellar/galactic medium with heavier elements and molecules, the stage was set for later low-mass star formation like our Sun via the assistance from radiating molecules like CO2 (etc).]

RACookPE1978
Editor
January 13, 2012 9:52 am

Joe Postma says:
January 13, 2012 at 8:12 am
Agreed on your point about albedo. An entirely miniscule change in albedo more than overshadows the supposed effect CO2 is pretended to have. When you consider that the average actual-surface albedo of the Earth is around 0.12, and that it is clouds which raise the average albedo to 0.3, then that means that very small changes in cloud cover dominate the average albedo, and therefore the average temperature of the Earth. This is one reason why the cosmic-ray connection in so important, and why solar activity is so important. It isn’t merely as simple as the variation in solar insolation, which is known to be finite but small…it is also the variation in solar magnetic activity which has a sort of “back-door” approach to affecting the climate. “

In trying to capture (or analyze) the effects of a changing cosmic radiation levels, and solar TSI and UV levels, and global cloud cover averages (global albedoes); look also at the earth’s magnetic pole positions and earth’s magnetic field strengths.
The recent changes in global temperature between 1600 to 2010 correspond to the position and movement of the south magnetic pole’s latitude as it moves away from the Antarctic coast out into the south Atlantic waters. Notably, the only area of the Antarctic heating up is the small strip of land sticking out into those waters. Up north, that overly simplified relationship doesn’t seem to be the case since today’s north magnetic pole is very close to where it was in the 1600’s.

HankHenry
January 13, 2012 10:01 am

On the subject of variability of albedo:
http://www.sciencedaily.com/releases/2004/05/040527233052.htm
It seems scientifically established that earth’s albedo varies.

January 13, 2012 10:08 am

Anything with a temperature radiates…in the case of non-spectral gases like N2 or O2, the radiation will arise from inter-molecular collisions. Perhaps we haven’t explored the spectrum at far enough wavelengths to see this emission; perhaps this emission is what helps constitute the entire profile of the “black-body” output curve of the Earth as seen from space in any case.
To clarify this a bit, if one looks at the actual spectrum associated with the cold top of atmosphere, one doesn’t see “CO_2 lines”, one sees pretty much a BB curve, but at a colder temperature in the general IR window. There is clearly a reduction of ground level IR at T_s and its replacement by atmospheric IR at T_a. That’s the physical reality of actual measurements. This is why I have very carefully avoided giving any impression that I “deny” that the “greenhouse effect” or “atmospheric warming effect” in general terms exists.
However, the data tells us something else as well. It is not CO_2 emissions lines. It is well-thermalized emissions from a colder radiator. It clearly does not come from CO_2, which is 0.03% of the atmosphere, recall. It comes from all of the air, well mixed.
That’s why Joe’s treatment kicked off the thought that all that is important is energy transfer, not how it happens.
rgb

Septic Matthew
January 13, 2012 10:16 am

Willis Eschenbach says:
January 12, 2012 at 4:32 pm
Thanks. I was almost sure that I was forgetting something. It’s weird that in thinking through complexities, I sometimes forget something obvious that I already know.
RGB, I thank you and others for the comments on this thread.
The spatio-temporal averages simply can’t be depended on to produce accurate approximations to needed quantities. (a) every place on earth has a different temperature from the global mean temperature almost all the time; (b) every place on earth has a different solar irradiance (measured at earth surface and at a higher level subtended by the small earth “place”) almost all of the time; (c) every place on earth has a different radiation, evaporation, advection and convection of energy from the means of these things, almost all the time, and this is true at every altitude. Illustrative calculations show that mean temperature, mean insolation, and mean radiation into space don’t relate meaningfully to each other. Given this, predicting the effect of adding CO2 is impossible at this time; CO2 will have different effects at different times of day, over different parts of earth, at different seasons, and at different altitudes, and a mean effect can’t be computed from some sort of mean (or total) change. As someone else wrote, it is necessary to know all of the rates of energy flow, not their thermodynamic equilibria, in order to estimate/predict the effect of adding CO2 to the present atmosphere. At least that’s what it looks like to me, and I think RGB and others have made the case well.

eyesonu
January 13, 2012 10:31 am

Stephen Rasey says:
January 13, 2012 at 9:33 am
:
First, from east to west, the geothermal gradient changes from values between 0.025 and 0.03 K/m (0.014 and 0.016F/ft) off the AlabamaMississippi shore to lower values of 0.0150.025 K/m (0.0080.014F/ft) off eastern Louisiana and to higher values of 0.030.06 K/m (0.0160.033F/ft) off western Louisiana through Texas. Second, thermal gradients tend to be lower toward the outer continental shelf (less than 0.02 K/m [0.0112F/ft]). Abstract: Nagihara et al 2008 …………….
===============
Thank you for your response. If the temp gradient were to become less as the outer continental shelf ia approached, would that imply that possibly the heat from the subsurface of the sea floor is being removed by the ocean waters? If so, then the deeper into the ocean that you go (closer to the earth core heat source) could show a lesser gradient that would prove the above suggestion that heat enters the ocean from a ‘hot plate’ effect as considered by richard verney (January 13, 2012 at 3:15 am).
I do not consider this to be completely off topic to the original article/post as the arguments concerning the atmospheric energy budget are all inclusive yet some seem to only focus on incoming solar radiation for a base line and in fact there may be another greater influence here on earth.
This is a most interesting discussion that has now spanned several threads. Keep it coming!

Bill Illis
January 13, 2012 10:50 am

Just noting there are large differences in the looking up (back-radiation) and looking down (emission) spectra when clouds are present.
Low level clouds create a perfect blackbody spectrum for the back-radiation and higher clouds have a little more atmospheric window in them but far, far less than the clear-sky back-radiation spectrum.
When viewing the spectrum looking down from space and there is low cloud, the outgoing radiation is emitting from 10km high at 220K in the CO2 spectrum and otherwise it is emitting from 3 kms high at the cloudtops at 265K.
When clouds are not present, the spectrum is the same 10 kms high at 220K for the CO2 spectrum and now we have atmopsheric windows emitting at 288K or the surface along with water vapour and methane bands emitting at around 250K or 5 kms high.
The spectra are governed by the temperature of the layer where emission starts to become possible directly to space in that particular spectrum.
Clouds make a big difference on the looking up or looking down spectrum and clouds are present up to 65% of the time. Furthermore, nobody seems to know that Modtran has cloud cover options in it (clear-sky is selected as the default) since these options are NEVER charted on the internet.

Bart
January 13, 2012 10:51 am

Tim Folkerts says:
January 13, 2012 at 8:56 am
“And simple satellite data shows that when you look down, you see very nearly 1) a blackbody radiation curve at the temperature of the ground and 2) “bites” taken out of this curve by cooler GHGs high in the atmosphere. There is no “signature” of the cool N2, so there is no noticeable radiation from the N2.”
This is apparent. The question I have is, where does the “bite” get taken out?
What data do we have showing emission curves versus altitude? Is it even possible to determine this?
Is the emission curve from space uniform over the Earth? Or, is it an average over the entire Earth? If it is non-uniform, over what regions is the bite most pronounced?
What accounts for the fact that the two regions before and after the “bite” do not conform to the same blackbody curve (one side is representative of a ~300K isocline, and one ~275K).
And, on a side note, what is the “signature” of radiation from non-GHGs due to collisions? Is it necessarily the same as for spontaneous emission?

January 13, 2012 10:54 am

Earth Shattering Aha! Moment To Follow
Before reading the balance of my comment, I’d ask that everyone first read (or re-read as the case may be) Joe’s comment upthread:
Joe says:
January 12, 2012 at 11:57 pm
When you have read that and come back, I will explain how Joe’s comment fits perfectly with Nikolov and Zellar, and how the combination of the two answers the vast bulk of the criticisms of N&Z. Now only shall I show how Joe’s explanation of the manner in which the poles act as a thermostat supports N&Z, I shall provide links to observational data that support both their assertions. Please read fast because I’m really excited about this and can hardly wait!
tick tick tick
Welcome back!
I’ll begin by paraphrasing what I consider to be one of two seminal points made in Joe’s comment. The Earth is a net absorber of energy in the tropics, and a net emitter of energy in the high latitude and arctic zones. Joe’s contention that this is the case is supported in spades by observational data. Possibly the easiest depiction of this data that confirms exactly what Joe has said is this graphical representation of Earth’s net radiation as measured by ERBE:
http://eos.atmos.washington.edu/cgi-bin/erbe/disp.pl?net.ann.
As can be seen from this graphic, the tropics are a net absorber of energy, and the poles a net emitter of energy, just as Joe said. In fact, Joe also calculated that the “break even” point between absorption and emission would be 60 degrees latitude, and in fact, ERBE data shows that the approximate breakeven point is about 60 degrees (north and south).
If you’ve followed this far, it shouldn’t be a very big leap from understanding the physics as Joe has explained it to understanding that this is the mechanism to support the existence that provides a thermostatic regulation of the Earth’s temperature, and further, that this thermostatic regulation fits precisely with N&Z, and furthermore explains precisely why the amount of greenhouse gas in the atmosphere is, in fact, immaterial to the equilibrium temperature of Earth. Further, I’ll provide observational data to support that as well, and then tie it all back to N&Z.
From Joe’s comment upthread:
“The primary mechanism responsible for maintaining the incredible 4-billion-year stability of our system must be the physical transfer of heat from the equator to the poles where it is radiated away.”
Precisely. The laws of thermodynamics require this statement to be true. The amount of net loss of energy in high latitudes must balance, to the last photon, the net absorption in low latitudes. Not only does this satisfy the laws of thermodynamics, it also explains why the concentration of GHG’s in the atmosphere is, in fact, immaterial.
Energy is moved from the tropics to the poles to satisfy the energy balance overall by multiple mechanisms. These include oceanic currents, atmospheric currents, and yes, “back radiation” from GHG’s. However, for net absorption to equal net emission, the exact transport mechanism actually doesn’t matter. The latitudes beyond 60 North and South are required by the laws of physics to warm up to a temperature that results in their net emission to space to exactly match the net absorption of the lower latitudes. The exact mechanism is immaterial because the amount that the high latitudes must increase in temperature such that their net loss to space balances the net absorption by the low latitudes is the same regardless of mechanism.
In other words, if the transport mechanism that moves energy from hot (net absorption areas) to cold (net loss areas) was 100% due to GHG’s, thermal equilibrium temperature of the emission zones would be exactly the same as if the transport was 100% atmospheric currents. The mix of transport mechanisms makes no difference to the equilibrium temperature that the high latitudes must reach to satisfy the laws of thermodynamics. Having additional transport capacity (in the form of increased amounts of GHG’s for example) doesn’t change the equilibrium temperature of the high latitudes, it only changes how fast that equilibrium temperature is achieved. The equilibrium temperature itself however, is regulated by the amount of net absorption in the low latitudes that must be balanced by net loss in the high latitudes.
The presence of increased amounts of GHG’s would certainly increase the temperature of the low latitudes by capturing some amount of outbound radiance that otherwise would have escaped to space. But this only serves to alter the path of energy loss to space rather than the amount of energy loss. Since the amount of absorption is NOT governed by GHG’s, neither can the amount of emission be governed by GHG’s. The net increase in temperature achieved by GHG back radiation in the net absorption latitudes does increase the temperature of those areas. That in turn creates a greater temperature differential between low latitudes and high latitudes. The greater that temperature differential, the faster transport mechanisms such as air currents must work. In other words, for every joule of energy that the GHG’s return to earth via back radiation in the area of net absorption, there must be exactly, and precisely, one extra joule of energy moved via any of the other mechanisms (air currents, oceanic currents, etc) to the areas of net loss. The ONLY way for the laws of thermodynamics to be satisfied is for that energy balance to exist.
Increased GHG’s do not change the energy balance because they have nothing to do with it in the first place. All increased GHG’s do is intercept energy that would otherwise have escaped to space directly, and caused a shift in temperature distribution such that the intercepted energy escapes from a higher latitude than it otherwise would have. We can see that this is the case simply by turning to NASA/GISS broken down by latitude. Two years ago I wrote an article on this exact matter. Here is the graph from that article showing global “average” temperature variation versus arctic temperature variation:
http://knowledgedrift.files.wordpress.com/2010/01/global-versus-equatorial-versus-arctic1.png
Note that the rise and fall in temperatures in the arctic zones not only exhibits far greater variability than the rise and fall in global temperatures, but also, that the rise and fall in the arctic zones (this is important) FOLLOWS the rise and fall in global temperatures. In other words, any change that causes global temperatures to increase is only temporary. Since the low latitudes receive the vast bulk of the energy absorbed from the Sun in the first place, any warming, from anything including GHG’s, must result in an increase of temperature differential between the net absorption areas and the net loss areas of earth. That increased differential increased the rate at which energy flows from hot to cold. The increased flow of energy to cold areas lags the initial increase in temperature that began the process, but ends when thermal equilibrium is re-established due to cold areas warming enough to compensate for the increased absorption in warm areas.
In other words, the change induced by changes in (for example) GHG concentrations do NOT change the over all energy balance of the Earth except on a temporary basis. What they DO change is the distribution of the net loss to space, forcing more of it to occur at the high latitudes and less at the low latitudes. Since energy flow in w/m2 varies with temperature in degrees Kelvin raised to the power of 4, this induces a mythical warming trend into the temperature data itself. If the planet is “warming” then there must be an energy imbalance. But since, as I’ve just shown, the energy balance is governed only by the total amount of energy absorbed in the first place, GHG’s only change the place on earth where that energy is lost. Due to the relationship of P to T^4, by moving some of that energy loss to the high (read cold)latitudes, the temperature change as a function of “average” temperature goes up, but the energy balance over the long term doesn’t change at all (except on a temporary basis).
By averaging T instead of T^4, we’ve fooled ourselves into seeing a temperature trend that seems indicative of an energy imbalance. We’re seeing nothing of the sort. What we are seeing is a redistribution of energy about the planet which results in the same exact energy balance which is ultimately governed ONLY by the amount of energy absorbed in the first place!
Averaging T introduces a positive trend that is entirely due to a simple math error. The most colossal simple math error in the history of human kind.
Now back to N&Z. Their equations rely only upon the amount of insolation and the surface pressure to predict “average” temperatures of celestial bodies. The main criticism of N&Z has been that the back radiation of GHG’s is real, and can be measured. By ignoring that, the accusation leveled at N&Z has been that the laws of thermodynamics have been breached.
Nothing of the sort has happened. The laws of thermodynamics require that net absorption and net loss balance, and balance exactly. Any change in the transport mechanisms of energy from cold areas to warm areas (and GHG’s would be one of those mechanisms) is exactly that. A change in the transport mechanism. The net energy balance does not, and according to the laws of thermodynamics, CANNOT change. The ONLY thing that can change is the distribution of that energy loss. As the distribution of the energy loss becomes more pronounced at high latitudes and less pronounced at low latitudes, the practice of averaging T suggests that an energy imbalance is in place and net warming of the earth happening. In fact, all that has happened is that the warm areas of the earth are losing less energy to space directly, and transferring that exact same amount of energy to be released by cold areas instead.
Averaging T produces the illusion of a global warming trend because over the long term, the average of P does not, and according to the laws of thermodynamics, cannot change. GHG’s are immaterial to the average of P. Which is why N&Z are bang on the money, and why all the criticisms of their equations from the point of view of the laws of thermodynamics are completely hollow. The observed data supports exactly Joe’s premise, my premise, and that of N&Z. It is the notion of GHG’s causing a permanent change in the Earth’s energy balance that are the violation of the laws of thermodynamics, and this violation has been completely hidden from view by the shear insanity of trying to measure an average trend of T when we’ve known for over a century that energy balance can ONLY be computed by the average of T^4.
A huge thank you to Robert Brown for laying the foundation for the best rational discussion of the actual physics I have seen to date. Thank you to Joe for providing the clear explanation of the manner in which the high latitudes participate as the ultimate regulators of the earth’s energy balance and for reminding me of the very concept I had proposed two years ago and then forgotten until now. Thank you to N&Z for publishing the formula which I believe will ultimately prevail for many reasons but most importantly, that when you consider their formula within the big picture, it is N&Z that preserve the laws of thermodynamics, and it is the warming attributed to GHG’s that violates them.
And a special thank you to Joel Shore. You and I have carried on a debate tangential to my Aha! moment in multiple threads, and in private correspondence. It got heated a couple of times, but without the challenges you presented to my thought process, I would not have sharpened my understanding of the issues sufficiently to have made the connection between all the pieces presented by Robert Brown, Joe, N&Z, and tied them together.
dmh

January 13, 2012 10:56 am

HankHenry says:
January 13, 2012 at 10:01 am
On the subject of variability of albedo:
http://www.sciencedaily.com/releases/2004/05/040527233052.htm
It seems scientifically established that earth’s albedo varies.

And from my back-of-the-envelope computation, inverted, that suggests that the average albedo has varied by roughly 0.02 upward. That’s enough to completely cancel even the hypothesized AGW due to CO_2. That’s why the CAGW enthusiasts seem to hate the GCR hypothesis. It’s so strange — you’d think that they’d be happy, the sky isn’t falling and we’ll all live instead of drown and burn in Hansen’s Boiling Seas. It’s enough to eliminate the problem in the worst case scenario, because the increase in albedo can only be due to clouds, and that means that climate feedback due to increased moisture has the wrong sign in their models. Oops.
rgb

Joel Shore
January 13, 2012 11:00 am

Stephen Wilde says:

That begs the question as to how the surface temperature gets set in the first place.

It isn’t even relevant how the surface temperature gets set: The fact that the surface temperature is such that the surface is emitting ~390 W/m^2 tells us that there must be elements in the atmosphere that absorb some of this radiation.

However the Ideal Gas Law then kicks in and due to gravity and pressure all the warmest molecules in the atmosphere are to be found directly above the surface with a declining upward temperature gradient.

The ideal gas law does not set the temperature…and it does not determine the lapse rate. If it did, the lapse rate would be at the adiabatic lapse rate in the stratosphere too. The reason that the lapse rate in the troposphere is around the adiabatic lapse rate is a combination of two factors: Radiative (and conduction from the surface) effects that alone would cause an even steeper lapse rate coupled with the fact that lapse rates steeper than the adiabatic lapse rate are instable to convection. Hence, convection occurs and drives the lapse rate back down to the adiabatic lapse rate.

As a result they inhibit the upward radiation and conduction from the surface more than would have been achieved if the air molecules above the surface had simply been at the cooler average temperature for the whole atmospheric column.

An atmosphere that cannot absorb or emit radiation cannot in any way “inhibit” radiation. You are talking nonsense, not physics.

January 13, 2012 11:09 am

Robert Brown said @ January 13, 2012 at 7:13 am

The fact that you are an astrophysicist makes me smile. I’ve thought for some time that physical climate modelling is something that should be done by astrophysicists working with condensed matter theorists working with computational fluid dynamicists working with statistical mechanics, perhaps with a few quantum mechanics and complex systems humans thrown in for good measure. Any single physicist knows something about all of this, but can’t possibly be an expert in it all. The traditional “I work over here all by myself with a one or two collaborators and a grant” model just doesn’t look big enough to be able to crack this particular nut.
Crowd-sourced physical climatology. I like it.

Like it? I love it 🙂

Stephen Wilde
January 13, 2012 11:10 am

davidmhoffer says:
January 13, 2012 at 10:54 am
Well done David.
You have just set out in your own words the content of all my work over the past 4 years.
Whatever forcings operate to try to destabilise the global climate the system response is always firmly negative and works via a surface pressure redistribution that alters the relative sizes intensities and latitudinal positions of ALL the permanent climate zones.
That applies whether the forcing in question is oceanic, solar or anthropogenic.
The question then is what difference do human GHGs make in relation to the natural solar and oceanic forcings that shifted the climate zones by 1000 miles or so from MWP to LIA and LIA to date.
I’d be surprised if it were more than a mile or so.
My collected works can be found here:
http://climaterealists.com/index.php?tid=37&linkbox=true

George E. Smith;
January 13, 2012 11:12 am

“”””” Myrrh says:
January 12, 2012 at 2:33 pm
Robert Brown says:
January 12, 2012 at 9:00 am
I think it would be very educational to do this and would take the guesswork out of the question “what does water do” or “what does an atmosphere do” to the sphere (relative to superconducting or insulating static sphere).
Why not start from where water has been included?
………………………
And there’s also that forty years ago NASA had to junk Stefan-Boltzmann to get real moon temps estimated for the landings, somewhere on-line, as they needed three dimensions not the flat earth of SB and they had to include that thermal energy absorbed from the Sun penetrated and was released later. “””””
Why is it that many posters here at WUWT throw around “Stefan-Boltzmann” as if it were some universal snake oil elixir, that cures gout, the dts, toe-nail cancer or whatever else ails you ??
“”””” NASA had to junk Stefan-Boltzmann to get real moon temps “””
Total BS, absolute nonsense and moreover, quite false.
Anybody who chooses to “junk Stefan -Boltzmann”, is invited to step up, and get tarred and feathered in the public square; Well of course unless you can present a serious peer reviewed paper setting out your new theory to replace the junked science you chose to discard.
The Stefan-Boltzmann “law”, is nothing more nor less, than a total integral of the Planck formula for the Spectral Radiant Emittance of a BLACK BODY, at a uniform Temperature. Integration is a simple mathematical operation that produces a specific result when applied to a well behaved specific function; so a ceremonial junking of “Stefan-Boltzmann” is synonymous with a junking of the Planck formula for the Black Body Radiation spectrum.
In turn, a BLACK BODY is an entirely fictional theoretical object, that is defined to absorb any and all Electromagnetic Radiation that falls on it; that is radiation of any frequency or wavelength from zero to infinity arriving from any source in any direction. It is NOT defined as any cavity or other geometry; it’s only required property is the complete absorption of ANY incident Electro-magnertic radiation.
No such object exists anywhere in the universe; it is a completely theoretical contrivance.
And decades of very careful experimental measurement of experimental approximations to an ideal black body, over every conceivable wavelength and practical range of Temperatures, has produced measured results that agree with the Planck formula to within the limits of the best experimental practices.
No repeatable systematic deviation from the Planck formula has ever been reported in peer reviewed papers. It is one of the most thoroughly established theories of modern Physics; despite the fact that it relates ONLY to a completely fictional object.
Also, importantly, the Planck formula contains no arbitrary constants or parameters that have to be finagled to get a match with the theory; other than the new fundamental physical constant that it introduces, which is (h) ; Planck’s constant that simply sets the scale of everything in the formula.
So please stop talking nonsense about NASA “junking Stefan-Boltzmann”. They did no such thing; they simply observed, that it does not define the Temperature of the moon. Whoopee, it doesn’t define the Temperature of anything else that is real either.
It is a useful ideal model to use as a starting point in calculating the properties of actual real objects.

Stephen Wilde
January 13, 2012 11:13 am

“An atmosphere that cannot absorb or emit radiation cannot in any way “inhibit” radiation. You are talking nonsense, not physics.”
Warm air above a surface will reduce the net upward energy flux more than would cooler air above that surface.Temperature differentials are what dictate the rates of energy flux.

January 13, 2012 11:13 am

Joe Postma said @ January 13, 2012 at 9:40 am

So we have two scenarios in which the exact same physical processes are going on, but in one it is theorized to cause warming (alarmist GHE) and in the other it is theorized to cause cooling (modern astrophysical theory). It’s one thing to question colleagues on conformism in climate science; it would be another thing entirely if I asked them to completely re-write the standard theory of star formation: sorry, but your molecules now cause heating, not cooling, you need to find another way to explain how stars form out of gravitationally-collapsing gas clouds.

And the science is settled. Right…

gbaikie
January 13, 2012 11:27 am

Jim D says:
January 12, 2012 at 9:32 pm
“It is very simple. Net incoming solar radiation for a spherical earth with albedo 0.3 is 240 W/m2.
Black-body temperature required to radiate 240 W/m2 is 255 K. QED. Any questions?”
Yeah.
Use 1/2 the world.
Put up 100 mile high wall that stops atmosphere. Have earth always face the sun.
So half the world freezes and other half in constant day. Now we have doubled 240 W/m2.
As above we have 480 W/m2 entering and 480 W/m2 leaving.
Such arrangement would change weather and would change climate. Evening and morning parts of world would be cooler. Temperate regions in mid day would become warmer.
And tropics in mid day would remain about the same temperature as any day currently in mid day.. Or generally average temperature may increase. But oceans aren’t going boil. Hot deserts aren’t going to get much hotter.
You are going to disrupt normal weather- and have the severe climate change- fools fear.
One could get repeating patterns, tons rain someplace, and little in others. Trying to predict this seems rather difficult- what part world is constantly facing the sun maybe the biggest variable in terms these numerous possible changes.
But my point is I think average temperature on the sun lit side will not change much.
I am excluding the night side- it obviously will become very cold. And seems obvious that average global temperature would be lower- but that isn’t the issue.
The question is if one doubles the incoming energy [240 to 480 W/m2] for one side, how much effect does have on the sun lit side.
Or does the chance frying eggs on the sidewalk improve?
I imagine if you believe greenhouse effect theory, frying eggs or boiling water on the sidewalk would be how you cook breakfast.
The issue is how much energy does the earth absorb- how much joules of energy.
Obviously per year it’s not much. Compared to each day of warming and cooling- gain and loss,
the daily cycle is losing and gain far more energy than is gained or lost per year.
How many joules on average per day is gained then lost.
Or rotate the walled world, how many joules are gained and lost per day. On each earth half.
If rotating it obviously absorbs far more energy. But anyways how much? In either case.

January 13, 2012 11:31 am

Energy is moved from the tropics to the poles to satisfy the energy balance overall by multiple mechanisms. These include oceanic currents, atmospheric currents, and yes, “back radiation” from GHG’s. However, for net absorption to equal net emission, the exact transport mechanism actually doesn’t matter. The latitudes beyond 60 North and South are required by the laws of physics to warm up to a temperature that results in their net emission to space to exactly match the net absorption of the lower latitudes. The exact mechanism is immaterial because the amount that the high latitudes must increase in temperature such that their net loss to space balances the net absorption by the low latitudes is the same regardless of mechanism.
I’m having just a bit of difficulty with this, in the context of the extended discussion on radiative balance. Although (as I’ve said) I’m withholding judgement until I’ve done the integrals, it appears to be strictly true that radiative loss is always favored by higher radiation temperature inhomogeneity. In other words, moving heat from the tropics to the poles cools the tropics and heats the poles so it provides a more uniform temperature, but a more uniform temperature is always a warmer temperature on average because in fact you lose heat far faster from the hot tropics than you do from the poles.
In fact, “most” of the net heat (the gain) absorbed during the day in the tropics is lost during the night, at least in any sort of naive static BB model.
Having a bit of epiphany of my own, trying to think of how this could be otherwise, I can think of only one way. If we leave the surface out of it (and for that matter, leave the ocean out of it and everything else out of it) it can increase the mobilization of the heat only by thermalizing the upper troposphere, that is moving heat from the tropics to the poles in the troposphere. In the poles there is an inversion (one visible in the data, incidentally) which slightly warms the ground there but loses the transported heat without reducing the radiative rate at the tropics.
The point is that once one starts to consider the surface and the troposphere as being to independent channels for heat loss, making the tropospheric temperatures more homogeneous by warming them at the poles allows for more rapid heat loss without actually warming the surface commensurately. If you simply warmed the poles at the expense of the tropics homogeneously (at all levels) the more uniform temperature would be net (on average) warmer all things being equal.
Is this not correct? Presuming that there isn’t anything funny that emerges from doing the integrals, of course, matching hot vs cold areas vs their actual temperatures and radiation rates?
More comments later. Life intervenes once again.
rgb

January 13, 2012 11:37 am

– Re geothermal gradient and heat flow.
Here are two links relating to geothermal heat flow:
SMU Geothermal Lab, US heat flow maps from USGS
Univ. of N. Dakota, Int’l Heat Flow Commission with a valuable Marine data map. The vast majority of marine data points are listed as < 0.1 W/m^2, i.e. 0.5 W/m2 for the ocean in general would require huge heat flow from the rifts that I don’t think is there in the data.
The geothermal gradient is good for estimating temperatures any a given depth in a given area. But without a measure of thermal conductivity, it tells you nothing about heat flow. The Gulf of Mexico is quite cold, largely because the Mississippi River has been dumping huge amounts of sediment into the basin, pushing the isotherms deeper. It is hotter in the GOM where the sedimentation rates are low today and where salt (a good conductor of heat) is rooted.

January 13, 2012 11:44 am

Correction (darn greater than signs!) “The vast majority….” should read
The vast majority of marine data points are listed as < 0.1 W/m^2, i.e, less than 100 milliwats/m^2. The distribution of points is probably lognormal, so whether the high end tail is over or under sampled is a good question. But even if the mean Ocean Heat flow is driven by rifts, getting the mean above 0.5 W/m2 for the ocean in general would require huge heat flow from the rifts that I don’t think is there in the data.

eyesonu
January 13, 2012 11:50 am

davidmhoffer says:
January 13, 2012 at 10:54 am
Earth Shattering Aha! Moment To Follow
=============
Great comment. I highlighted “Joe’s” earlier post for a comment of my own but lost the inititive due to the fast moving pace of this thread.
Your summary is spot on!
I believe there is a big hammer falling and AGW will not survive its impact.

eyesonu
January 13, 2012 12:00 pm

Stephen Rasey says:
January 13, 2012 at 11:44 am
=============
Thanks for the info. I will be looking into it. I don’t know the answers, but I have a curious state of mind. Kind of like ” iNQUISITIVE / INQUIRING MINDS NEED TO KNOW”.

Joel Shore
January 13, 2012 12:15 pm

Stephen Wilde says:

“An atmosphere that cannot absorb or emit radiation cannot in any way “inhibit” radiation. You are talking nonsense, not physics.”
Warm air above a surface will reduce the net upward energy flux more than would cooler air above that surface.Temperature differentials are what dictate the rates of energy flux.

The way that temperature differences influence net radiation is that two objects at different temperatures and different emissivities radiate different amounts and hence the radiation from the colder object partially cancels the net heat flux from hotter to colder object.
However, you are proposing that this effect still occurs (i.e., the atmosphere reduces the net radiative emission) even if the atmosphere is transparent to IR radiation, i.e., it neither emits nor absorbs it. This is, once again, nonsense…not physics.

Tim Folkerts
January 13, 2012 12:17 pm

Joe says:
>Anything with a temperature radiates…in the case of non-spectral gases
>like N2 or O2, the radiation will arise from inter-molecular collisions.
OK — sounds good. But these effects are MUCH weaker than radiation from molecular vibrations.
Joe then starts grasping at straws
>Perhaps we haven’t explored the spectrum at far enough
>wavelengths to see this emission;
IR spectroscopy is very well studied, so if there were important emissions from N2, they would be known. Besides, for 250-350 K objects, the IR radiation will only be strong in wavelengths of ~ 2 – 40 um. There is no need to look “the spectrum at far enough wavelengths” because the emissions would be too weak to matter.
>perhaps this emission is what helps constitute the entire profile of the
>“black-body” output curve of the Earth as seen from space.
The “entire profile” from the ground is known to be very close to a blackbody curve — nothing must be added to turn it into BB curve. Whatever contributions N2 does make are indistinguishable from the what was already there — ie N2 adds or subtracts nothing that is measureable.

Phil.
January 13, 2012 12:19 pm

Joe Postma says:
January 13, 2012 at 9:40 am
Anything with a temperature radiates…in the case of non-spectral gases like N2 or O2, the radiation will arise from inter-molecular collisions. Perhaps we haven’t explored the spectrum at far enough wavelengths to see this emission; perhaps this emission is what helps constitute the entire profile of the “black-body” output curve of the Earth as seen from space in any case.

No, it’s been investigated and found to be negligible under the conditions of the Earth’s atmosphere, that’s why we don’t see it. The final speculation is nonsense, just look at the spectrum given above from above Antarctica.
However, there is another important point to consider, which Alan Siddon’s has discussed elsewhere: if the spectraly-neutral gases like O2 and N2 don’t radiate, that means that they collect heat-energy from the solar-heated surface by conduction, and then hang on to that heat: they can’t shed it, they can’t radiate it away spectraly, they just hold on to it. A “GHG”, on the other hand, once having absorbed heat energy from outside into its internal vibration, can then shed that energy by radiating it away. Not being able to lose and radiate the energy away, vs. being able to, should be the difference between a heat-trapping gas and a heat-shedding gas.
That’s the problem with astrophysicists like Postma when they look at the Earth’s atmosphere they think of it as a stellar atmosphere and therefore make elementary errors as Postma has here. Below ~3 scale heights in the Earth’s atmosphere CO2 predominantly loses its excess energy to the surrounding atmosphere by collisions not radiation and thereby warms that part of the atmosphere hence Greenhouse warming. Above that altitude CO2 radiational loss to space starts to win hence the ‘cooling’ of the stratosphere due to GHGs.
Postma, understand the way planetary atmospheres work before modelling them as a stellar atmosphere, the lecture notes recommended above would be an excellent start.
http://maths.ucd.ie/met/msc/PhysMet/PhysMetLectNotes.pdf
In fact, given my perspective from astrophysics, this is exactly the theory that we use to explain how interstellar gas-clouds are able to overcome the thermal response from gravitational collapse (potential energy converts to kinetic = temperature), and continue to collapse to form stars. The spectraly-emitting molecules in the gas (like CO2, but typically CO and others) provide a “vector” through which the thermal energy build-up of the collapsing cloud can escape the cloud. They absorb thermal energy via collision into internal degrees of freedom, then radiate that energy away, out of the cloud. This effectively “damps” the thermal response and then causes cooling. This allows the cloud to collapse into a star.
Yes but we aren’t concerned with a stellar atmosphere here and the ‘perspective from astrophysics’ is wrong! Perhaps you should stick to stars?

January 13, 2012 12:25 pm

Robert Brown;
In other words, moving heat from the tropics to the poles cools the tropics and heats the poles so it provides a more uniform temperature, but a more uniform temperature is always a warmer temperature on average because in fact you lose heat far faster from the hot tropics than you do from the poles.>>>>
EXACTLY!!!
1. Moving heat from the tropics to the poles results in a more uniform temperature.
2. A more uniform temperature results in a higher average T
3. But, a higher average T does NOT necessarily mean a higher average P!
4. Unless a change in any given transport (for sake of argument, an increase in GHG’s) affects the amount of P absorbed, then it doesn’t matter one wit as to energy balance. If it did, the laws of thermodynamics would be breached.
In other words, if average energy absorbed is 240 w/m2, then that’s it. Energy emitted must ALSO equal 240 w/m2. It doesn’t matter if you have 10 ppm of CO2, 100, 1000 or ten thousand ppm. UNLESS the CO2 changes the amount of energy absorbed in the first place, it can have no, none, zero, nada, zippidy doo da, did I mention zero affect on the amount of energy radiated (at equilibrium).
That being the case, the ONLY affect that CO2 can have in terms of GHE is to become part of the mechanism that redistributes energy from the tropics to the poles. CO2 suppresses emissions in the tropics, forcing more energy to be transferred from tropics to poles. The net energy emitted at equilibrium changes not one single fraction of a what. The only thing that changes is that less is emitted at tropics and more at poles.
That in turn results in an increased average T for the exact some P absorbed. No change in energy balance need occur to account for the increase in T.

Joel Shore
January 13, 2012 12:32 pm

davidmhoffer says:

The observed data supports exactly Joe’s premise, my premise, and that of N&Z.

There is no data whatsoever that supports your premise. If you actually want to support your premise, I suggest that you:
(1) Show that if one averages T^4 instead of T, the warming seen disappears. Good luck doing that, as Essex & McKitrick (& one other co-author) have already tried…but were only able to get rid of the warming trend seen for an artificially-small set of 12 stations by doing averages of T^n where n get really big. The difference in trend that they got between averaging T and T^4 was small and it will get even smaller if you consider more stations.
(2) Come up with an explanation of how you can take GHGs away and still have the current average radiative emission that the Earth’s surface has.

Thank you to N&Z for publishing the formula which I believe will ultimately prevail for many reasons but most importantly, that when you consider their formula within the big picture, it is N&Z that preserve the laws of thermodynamics, and it is the warming attributed to GHG’s that violates them.

Congratulations, David. You used to make fun of posters who claimed that the greenhouse effect violates the laws of thermodynamics (and the Second Law specificially). Now you have become one of them…and, like them, you make that claim on the basis of no evidence whatsoever.
I hope you find your full embrace of nonsense more fulfilling than your previous partial embrace.

January 13, 2012 12:37 pm

Stephen Wilde says:
January 13, 2012 at 11:10 am
davidmhoffer says:
January 13, 2012 at 10:54 am
Well done David.
You have just set out in your own words the content of all my work over the past 4 years.>>>
Well I did kinda cheat. I didn’t have to do any original research, it was all done for me and publsihed in various threads on WUWT. All I really did is say hey! all these pieces fit together.
It seems to incredibly obvious. I think the nay sayers are so caught up caught up in how every little detail fits together thah they’ve lost sight of the big picture.
If the Earth absorbes 240 w/m2, then it emmits 240 w/m2. Period. Unless CO2 changes the amount absorbed, then the amount emityteed is 240 w/m2, and it matters not how much CO2 there is or isn’t. the only thing that changes with CO2 concentration is the average of T. The average of T^4 changes not one bit, and hance neither does the energy balance.

Spector
January 13, 2012 12:50 pm

According to the Stefan-Boltzmann law, radiant energy is proportional to the fourth power of the temperature. Thermal heat balance equations are based on average radiated energy. The characteristic temperature associated with a given average radiant energy flow is *NOT* supposed to be an average temperature in any usual sense. It is just a handy reference number. In this case, an average temperature is actually less useful and more complex because it depends on factors unrelated to energy balance equations. This should be an obvious apples and oranges situation, but it appears this is one of those fine distinctions that has gotten lost in the communication noise.
I note that there is one exception sometimes used in climate science and that is to assume that such a narrow range of temperatures are involved that the Stefan-Boltzmann equation can be ‘linearized’ by treating the average third power of the temperature as a ‘constant.’ This is being done whenever you see a number involving degrees per W/m².

Stephen Wilde
January 13, 2012 1:02 pm

“It seems to incredibly obvious. I think the nay sayers are so caught up caught up in how every little detail fits together thah they’ve lost sight of the big picture.”
I agree, it seems incredibly obvious but only once the bulb lights up in the mind. There are so many out there that will fight tooth and nail against any such ideas that it is going to be an uphill struggle.

Myrrh
January 13, 2012 1:05 pm

George E. Smith; says:
January 13, 2012 at 11:12 am
“”””” NASA had to junk Stefan-Boltzmann to get real moon temps “””
Total BS, absolute nonsense and moreover, quite false.
Anybody who chooses to “junk Stefan -Boltzmann”, is invited to step up, and get tarred and feathered in the public square; Well of course unless you can present a serious peer reviewed paper setting out your new theory to replace the junked science you chose to discard.

Shrug. The thing I found confusing at first in all these arguments was that pro AGW’s are perfectly content to argue with whatever out of context or made up physics came to hand, actually as I later concluded because they don’t have real physics to argue from as their base premise is junk. I don’t know all the ways that SB and Planck is used, argued from, in these calculations, there were countless arguments and i found myself drawn rather to the other silly physics that wasn’t discussed and the history, so if you have a problem with that message I suggest you take it up with the source I got it from, John O’Sullivan. The link I had to the article is now defunct, he had some problems with suite101 as I recall, and I’d been trying to find it, but got distracted by a more recent piece on the debunking of the claim that a colder object can heat up a warmer which grabbed my interest because it was debunking something I had read a while back and which at the time finally convinced me that using crap physics was par for the course in pro AGW arguments, regardless for how apparently well educated in science and mathematically literate they appeared.. .
Ah, here it is:

http://johnosullivan.livejournal.com/16242.html
“Thus, the ‘blackbody approximations’ were proven to be as useful as a chocolate space helmet; the guesswork of using the Stefan-Boltzmann equations underpinning the man-made global warming theory was long ago debunked. If NASA had made known that Stefan-Boltzmann’s numbers were an irrelevant red-herring then the taxpayers of the world would have been spared the $50 billion wasted on global warming research; because it would have removed the only credible scientific basis to support the theory that human emissions of carbon dioxide changed Earth’s climate.”

I’ve lost count of the number of times I’ve been told to ‘educate myself in the real physics of SB and Planck’, and the number of times a given story suddenly changes and a different argument presented and even denial that anything to the contrary had been said…, so I’m not interested in arguing about it. Not that I’m suggesting you’d do such a thing, but this commonly has the effect of confusing these arguments even more and the subject isn’t one that grabs me enough to put in the effort. I gave it in for background.

Stephen Wilde
January 13, 2012 1:12 pm

“(2) Come up with an explanation of how you can take GHGs away and still have the current average radiative emission that the Earth’s surface has.”
You mean just taking away non condensing GHGs presumably.
Whether David has the equations exactly right or not doesn’t really matter. The fact is that even without non condensing GHGs the surface temperature would be set by solar input to the oceans which would keep them liquid and produce the necessary water vapour for a water cycle.
The water cycle might be a little less vigorous with no non condensing GHGs but that would just mean a slight adjustment in the positions of the permanent climate zones.
No need for any change in the equilibrium temperature of the system (primarily the oceans) at all.
All climate change is just a redistribution of surface energy as the rate of energy flow from surface to space varies over time. The equilibrium temperature for the system as a whole (as opposed to just the atmosphere) is set by solar shortwave into the oceans, atmospheric pressure and the phase changes of water.

Joe Postma
January 13, 2012 1:13 pm

======================================
Phil. says:
January 13, 2012 at 12:19 pm
No, it’s been investigated and found to be negligible under the conditions of the Earth’s atmosphere, that’s why we don’t see it. The final speculation is nonsense, just look at the spectrum given above from above Antarctica.
That’s the problem with astrophysicists like Postma when they look at the Earth’s atmosphere they think of it as a stellar atmosphere and therefore make elementary errors as Postma has here. Below ~3 scale heights in the Earth’s atmosphere CO2 predominantly loses its excess energy to the surrounding atmosphere by collisions not radiation and thereby warms that part of the atmosphere hence Greenhouse warming. Above that altitude CO2 radiational loss to space starts to win hence the ‘cooling’ of the stratosphere due to GHGs.
Postma, understand the way planetary atmospheres work before modelling them as a stellar atmosphere, the lecture notes recommended above would be an excellent start.
http://maths.ucd.ie/met/msc/PhysMet/PhysMetLectNotes.pdf
Yes but we aren’t concerned with a stellar atmosphere here and the ‘perspective from astrophysics’ is wrong! Perhaps you should stick to stars?”
====================================
Indeed it was speculation when I was discussing where we might find the collisional emission from the N2 and O2 in the atmosphere. It would be interesting to see it and acknowledge it rather than jump down its throat. But it doesn’t bear on my central theses in any case.
As Phil’s post was mostly antagonistic I will just fill in the blanks for others who are here willing to learn and have a scientific discussion with experts.
Phil made a fundamental error in his critique when he criticized my “talking about stellar atmospheres”. I wasn’t talking about stellar atmospheres.
The case of an interstellar gas cloud, and the physics we talk about with them, is entirely analogous to that of the terrestrial atmosphere in the functionality of the radiation. In both cases, you have reduced optical thickness “looking out”, and increasing optical thickness to complete opacity “looking in”. Of course, the idea that spectral absorption and line emission is different just because it’s found in a gas cloud, a star, or a planetary atmosphere, rule by different physics is absurd – it’s the exact same phenomenon, one set of physics describes it.
The point remains: in astrophysics, a gas cloud cools due to the presence of radiating molecules like CO2. They absorb heat energy collisionally and then radiate that heat away due to their ability to do so. The other molecules in the gas cloud, like N2, O2, etc, don’t radiate the heat away, just like we talk about in the terrestrial case. The heat that these molecules pick up due to collapse needs to be shed out of the system or else the gas cloud would heat up too fast and blow itself apart before it collapsed. The heat from gravitational collapse, that N2, H2, O2, etc, pick up, is transferred collisionally to radiating molecules like CO2, and others. Because the more complex molecules like CO2 have internal vibration which can be excited by collision, these thermal collisions are damped, and this is the first step in reducing the rate of heat build-up.
The next step comes when the CO2 molecules radiate the heat energy they’ve internally picked up from the previous collision. This radiation is emitted isotropically; the outward component can eventually escape due to the reducing optical thickness of the cloud, but the inward component can not escape because the gas cloud is optically thick & opaque inside.
So this describes, pretty much precisely, the exact same phenomenon in our terrestrial atmosphere.
Now, given the history of malfeasance by alarmist climate science, is it astrophysics or is it climate science which is more likely to be correct, on the question of cooling vs. heating…
It is perfectly logical from the astrophysical perspective: damped collisions reduce heat build-up, check, and radiating molecules move energy out of the system, check, rather than adding more energy and heat than was already there, major check.

Stephen Wilde
January 13, 2012 1:19 pm

“However, you are proposing that this effect still occurs (i.e., the atmosphere reduces the net radiative emission) even if the atmosphere is transparent to IR radiation, i.e., it neither emits nor absorbs it”
I am considering an atmosphere with water vapour and some naturally occurring non condensing GHGs. The real world in fact.
Conduction warms ALL the gases whether GHGs or not. The Nitrogen and Oxygen warm up via conduction from the surface and from collisions with GHGs.
Warmer air at the surface inhibits upward energy transfer of all types in that situation.

John West
January 13, 2012 1:42 pm

wayne says:
“Radiation therefore allows an atmosphere to equalize faster that without it.”
Agree.
Robert Clemenzi says:
“That ignores the energy added by conduction.”
True, I should’ve included that in my “list”.
Ian W says:
“This energy output is NOT shown by Stefan Boltzmann maths nor is it linked to ‘temperature’.”
Cool, thanks for the additional info.
So, do we all (at least us 4) agree that an increase in temperature will result in an increase in outgoing longwave radiation (IR); such that the premise espoused by certain pro-CAGW advocates (SKS, RC) that an increase GHG’s necessarily increases the representative emission height (TOA) that is cooler thereby reducing the outgoing longwave radiation is complete hogwash?

Bart
January 13, 2012 1:43 pm

Tim Folkerts says:
January 13, 2012 at 12:17 pm
“The “entire profile” from the ground is known to be very close to a blackbody curve…”
How? In what way, specifically? Here are my thoughts.
A blackbody with purely radiative outward energy flux will settle to a Planck energy distribution, and the integral of that distribution across frequency gives you the Stefan-Boltzmann relationship. But, when you have energy being dissipated by convection and conduction from that surface, will the dissipation be uniform across frequency of the blackbody emitters, and simply scale the energy distribution so that it still looks like a Planck distribution? Or, will it distort the distribution, picking off specific frequency bands, so that the distribution is now non-Planckian? If the latter, then the S-B relationship no longer holds.
How do we know? Proponents of the GHG hypothesis point to measures of the emissivity in laboratory and in-situ experiments conducted in calm conditions, e.g., the emissivity of ocean water near the coast in clear weather with particular wind speeds. But, can such behavior actually be extrapolated to the entire oceanic surface of the Earth? I do not think that is a given.
Moreover, the emissivity numbers are given as single bulk numbers. Do the sensors used measure the entire frequency spread, or do they sample a narrow band, and extrapolate emissivity based on assuming an underlying Planck distribution?
Why does the emission spectrum at TOA (a href=”http://www.john-daly.com/smoking.htm”>Figure 3) look like two separate distributions reflecting radiating bodies at two distinct temperatures, with a gap in between due mostly to water vapor? Is it possible the water vapor gap reflects energy removed by conduction and convection at the surface, rather than absorption within the higher atmosphere? Since convection occurs with moist air, is it not possible that these energy bands are precisely the ones which would be preferentially dissipated from the surface?
It is also noted that emissivity of ocean water very much depends on the observation angle, and this is hypothesized to be due to surface roughness or waves. Should not the emissions, then, be de-weighted by the ratio of the integral of the angle sensitivity to the integral of normal diffuse emission?

January 13, 2012 1:47 pm

Don’t forget the moon and tidal dragging… that really had some interesting effects earlier in the planet’s history….
Say, it has been an idea floating around in my head that the major difference between the K and the T, beyond that boloid, is the break of a pangea into sub-continental masses that also break up the single ocean. Plus once a continent got into the southern polar position we seem to get these little glacial periods… like it was acting as a hit sink for the planet.
Going from nice, warm, placid oceans covering continents and having polar regions closer to today’s WA State to getting a broken up oceanic system with a continent at the pole seems to be doing something to the entire climate deal.
If you ever get a model running, I’d like to check this out.

Bart
January 13, 2012 1:47 pm

“Should not the emissions, then, be de-weighted by the ratio of the integral of the angle sensitivity to the integral of normal diffuse emission?”
Rather, that would be the ratio of the integral of normal diffuse emission weighted by the angle sensitivity to the integral of normal diffuse emission. Thus, the result would always be less than unity.

Joel Shore
January 13, 2012 2:04 pm

Stephen Wilde says:

I am considering an atmosphere with water vapour and some naturally occurring non condensing GHGs. The real world in fact.

Well, when did this change occur? I thought your argument was that one could have the current Earth surface temperature even in the absence of a radiative greenhouse effect. Now, you seem to be saying that it is necessary to have the radiative greenhouse effect to explain the Earth’s surface temperature. That has been what Willis, I and the rest of the scientific community have been trying to tell you for the last couple of weeks!

Tim Folkerts
January 13, 2012 2:30 pm

Davidmhoffer says:
“That being the case, the ONLY affect that CO2 can have in terms of GHE is to become part of the mechanism that redistributes energy from the tropics to the poles. “
That is one part (although a very small one, since all gases can transport energy via large scale movements like Hadley cells).
The OTHER effect of CO2 is to move the location of the radiating surface from the ground level up higher in the atmosphere. Some of the radiation comes from the very cold upper atmosphere and some comes from the ground, with the net result being 240 W/m^2 on average. This has been discussed multiple times and until people understand the implications of this, they will continue to misunderstand the GHE.

Editor
January 13, 2012 3:08 pm

Bob Fernley-Jones says:
January 12, 2012 at 10:53 pm

… I’ve had some interesting intercourse with Willis starting in the link below, but after several exchanges he seems to have taken a premature withdrawal:
http://wattsupwiththat.com/2012/01/08/the-moon-is-a-cold-mistress/#comment-860094

Oh, piss off with your ugly sexual innuendo, it does not make me want to continue the conversation. I have a life. I have a day job. Sometimes I ignore people because they can’t see the obvious. If you want an answer, ask an intelligent question. Hang on, let me see which comment that was …
Ahhh … right, I remember you now. That was your opening salvo, your first contribution to the discussion. In it you accused me of a variety of unpleasant things, said I was of “ranting” and claimed I was somehow plagiarizing Richard Courtney. I went on past that, I ignored your rudeness and answered your question.
Then you went (from my perspective) drifting off into statements that had little to do with what I’d done in the head post. For example you said:

… what I meant was that to apply S-B over a supposed average temperature of an airless sphere based on average insolation spread over that sphere does not give a sensible result.

I know that, THAT WAS ONE POINT OF THE ARTICLE. So of course, I didn’t do that. I converted each individual temperature into radiation and then averaged the radiation. So you were accusing me of something I did not do.
Now, I could have tried to correct your lack of reading skills. But I’m constantly doing triage on the new posts—short answer, long answer, or no answer. And you had come in full of spite and unpleasantness. So at that point, since you hadn’t even read the head post, you went into the “not worth it” pile. I’m here to talk to people who are following the bouncing ball.
And now you are making nasty sexual innuendoes about how I didn’t answer you? Were you born that dumb, or do you have to work at it?
I can see I was foolish to answer such an offensive, venom-filled introductory message as yours. You walk in the door and you start slanging me like that, my friend, you are on a very short leash. I may drop the discussion with you at a moments notice.

Oh, and I think he also does not understand the warming effect of even a transparent atmosphere, but one thing at a time!

When will you learn to QUOTE MY WORDS, you unpleasant man? Where am I wrong about the atmosphere? Yes, it can reduce the cooling from the temperature fluctuations, but that’s all it can do.
w.

Tim Folkerts
January 13, 2012 3:08 pm

Joe Postma says January 13, 2012 at 1:13 pm
>Now, given the history of malfeasance by alarmist climate science,
>is it astrophysics or is it climate science which is more likely to be
>correct, on the question of cooling vs. heating…
We know science is not decided by consensus; it is also not decided by the politics or personalities of the scientists involved. So let the science speak for itself.
>It is perfectly logical from the astrophysical perspective: damped collisions
>reduce heat build-up, check,
>and radiating molecules move energy out of the system, check,
>rather than adding more energy and heat than was already there, major check.
But you are missing one major component — the warm object in the center (the earth). Lets start with the warm earth – suppose it is 300K. It will be radiating energy into space and will start to cool at some rate. Now consider a second earth (also at 300 K) with a nebula of your N2/CO2 mix around it. Earth2 will radiate as much energy as Earth1. But now the gas around it will also be radiating. Yes, some of that energy will head outward, but some will head inward. Earth2 will absorb some of that energy. No matter what the temperature of the surrounding nebula (as long as it is above 2.7 K), the net loss from Earth2 will be less, and Earth2 will cool slower that Earth1.

Kevin Kilty
January 13, 2012 3:33 pm

Tim Folkerts says:
January 13, 2012 at 2:30 pm

The OTHER effect of CO2 is to move the location of the radiating surface from the ground level up higher in the atmosphere. Some of the radiation comes from the very cold upper atmosphere and some comes from the ground, with the net result being 240 W/m^2 on average. This has been discussed multiple times and until people understand the implications of this, they will continue to misunderstand the GHE.

Tim, this is an excellent explanation of the greenhouse effect. That by moving the effective radiator partially up into cooler regions, the surface must warm, and in turn lead to a somewhat warmer atmosphere to achieve the same energy balance. This explanation avoids the non-pertinent argument that CO2 cannot be effective because one cannot transfer heat from a cooler place to a warmer one. Sol maintains input to the absorber end; and, CO2 affects the radiator end of the thermal circuit.

Kevin Kilty
January 13, 2012 3:56 pm

George E. Smith; says:
January 13, 2012 at 11:12 am
…Anybody who chooses to “junk Stefan -Boltzmann”, is invited to step up, and get tarred and feathered in the public square; Well of course unless you can present a serious peer reviewed paper setting out your new theory to replace the junked science you chose to discard.
The Stefan-Boltzmann “law”, is nothing more nor less, than a total integral of the Planck formula for the Spectral Radiant Emittance of a BLACK BODY, at a uniform Temperature…….

It is becoming very difficult to decide who says what in these exchanges; so, if these aren’t your words, George, then I apologize in advance. All you say is so, however, when the aperture to a cavity acts as a band-limiting filter, then S-B, if you wish to define it as a fourth power T function, does not describe the emitted power. If, on the other hand, you look at SB as an integral over all wavelengths including the band weights, then I suppose you can claim the universality of SB. I suppose the long and short of this is that once active gasses enter a radiation problem, it becomes quite difficult to describe in terms of the simple S-B , T to the fourth power, function.

Theo Goodwin
January 13, 2012 3:58 pm

Tim Folkerts says:
January 13, 2012 at 2:30 pm
“The OTHER effect of CO2 is to move the location of the radiating surface from the ground level up higher in the atmosphere. Some of the radiation comes from the very cold upper atmosphere and some comes from the ground, with the net result being 240 W/m^2 on average. This has been discussed multiple times and until people understand the implications of this, they will continue to misunderstand the GHE.”
I doubt that you mean what you wrote. If some of the radiation comes from the ground then it can contribute to UHI but nothing else. What percentage of CO2 molecules do you see as near the ground?

markus
January 13, 2012 4:08 pm

“Kevin Kilty says:
January 13, 2012 at 3:33 pm
Tim, this is an excellent explanation of the greenhouse effect. That by moving the effective radiator partially up into cooler regions, the surface must warm, and in turn lead to a somewhat warmer atmosphere to achieve the same energy balance. This explanation avoids the non-pertinent argument that CO2 cannot be effective because one cannot transfer heat from a cooler place to a warmer one. Sol maintains input to the absorber end; and, CO2 affects the radiator end of the thermal circuit”.
All you have said would be correct, except for the thermal dynamic entropy of pressure.

Kevin Kilty
January 13, 2012 4:09 pm

Tim Folkerts says:
January 13, 2012 at 12:17 pm

…The “entire profile” from the ground is known to be very close to a blackbody curve — nothing must be added to turn it into BB curve. Whatever contributions N2 does make are indistinguishable from the what was already there — ie N2 adds or subtracts nothing that is measureable….

This is hard to decipher, Tim, and maybe I ought to return to Joe P’s comment to figure it out, but the spectrum of the ground from space only looks like a BB at some fixed T if you are willing to interpolate over the reject bands. The spectrum is full of holes.

Kevin Kilty
January 13, 2012 4:12 pm

markus says:
January 13, 2012 at 4:08 pm
“Kevin Kilty says:
January 13, 2012 at 3:33 pm
Tim, this is an excellent explanation of the greenhouse effect. That by moving the effective radiator partially up into cooler regions, the surface must warm, and in turn lead to a somewhat warmer atmosphere to achieve the same energy balance. This explanation avoids the non-pertinent argument that CO2 cannot be effective because one cannot transfer heat from a cooler place to a warmer one. Sol maintains input to the absorber end; and, CO2 affects the radiator end of the thermal circuit”.
All you have said would be correct, except for the thermal dynamic entropy of pressure.

What’s that?

Bart
January 13, 2012 4:23 pm

Kevin Kilty says:
January 13, 2012 at 4:09 pm
“The spectrum is full of holes.”
And, where do the holes come from? Atmospheric filtering, or is it that way very close to the ground already? That is what I have been trying to nail down..
Moreover, it looks not like a single blackbody, but a piecewise paste together of two blackbodies, with a reference 275K blackbody at low wave number, and 300K blackbody at high (before getting to the apparent methane band, where it drops off precipitously (Figure 3).

Bart
January 13, 2012 4:24 pm

Figure 3. I seem to keep having trouble getting that link right.

Kevin Kilty
January 13, 2012 4:35 pm

Joel Shore says:
January 13, 2012 at 12:32 pm
davidmhoffer says:
The observed data supports exactly Joe’s premise, my premise, and that of N&Z.
There is no data whatsoever that supports your premise. If you actually want to support your premise, I suggest that you:
(1) Show that if one averages T^4 instead of T, the warming seen disappears. Good luck doing that, as Essex & McKitrick (& one other co-author) have already tried…but were only able to get rid of the warming trend seen for an artificially-small set of 12 stations by doing averages of T^n where n get really big. The difference in trend that they got between averaging T and T^4 was small and it will get even smaller if you consider more stations….

Mr. Shore, it appears you have labelled the suggestion that the true average radiation temperature would be done better by using the n-th root of the sum or integral of T to the n-th power, where n is not one as nonsense. Now maybe it is true that the difference with the arithmetical average is not material, but taken literally you are, in effect, saying that T to an exponent of one describes emitted power.

Bob Fernley-Jones
January 13, 2012 4:43 pm

Willis Eschenbach @ January 13 3:08 pm
Willis, it was partly in payback for your arrogance and rudeness to me some months ago, when amongst other things you described me as a member of the faunal order; Rodentia.

Joe Postma
January 13, 2012 5:41 pm

===============================
Tim Folkerts says:
January 13, 2012 at 3:08 pm
But you are missing one major component — the warm object in the center (the earth). Lets start with the warm earth – suppose it is 300K. It will be radiating energy into space and will start to cool at some rate. Now consider a second earth (also at 300 K) with a nebula of your N2/CO2 mix around it. Earth2 will radiate as much energy as Earth1. But now the gas around it will also be radiating. Yes, some of that energy will head outward, but some will head inward. Earth2 will absorb some of that energy. No matter what the temperature of the surrounding nebula (as long as it is above 2.7 K), the net loss from Earth2 will be less, and Earth2 will cool slower that Earth1.
===============================
However, the situation is still exactly the same, but you might have overlooked it not being familiar with stellar formation. At the center of the gas cloud we DO have a warm object – the primordial stellar core. So it is correct to say that the surrounding gas mixture will be heated by both conductive and radiative effects from the warmer thing in the center; however, in our gas-nebula case, this outward propagation of thermal energy, assisted by the radiating molecules, effects exactly just that: an outward propagation of energy. The heat/energy isn’t returned in such a way as to heat the core even more, the CO2 (etc) molecules KEEP the surrounding gas from getting too hot, thus allowing the cloud to continue collapse. Agreed that the surrounding gas will also be radiating bc of the CO2 (etc) molecules, and that represents a vector for energy loss out of the system. Otherwise, without that spectral radiation, the gas would just continue to heat because the energy wouldn’t be able to escape via radiation.

Kevin Kilty
January 13, 2012 5:50 pm

Bart says:
January 13, 2012 at 4:23 pm
Kevin Kilty says:
January 13, 2012 at 4:09 pm
“The spectrum is full of holes.”
And, where do the holes come from? Atmospheric filtering, or is it that way very close to the ground already? That is what I have been trying to nail down..
Moreover, it looks not like a single blackbody, but a piecewise paste together of two blackbodies, with a reference 275K blackbody at low wave number, and 300K blackbody at high (before getting to the apparent methane band, where it drops off precipitously (Figure 3).

I agree with you here. I agree that it does not look like a single temperature body, and this is why I have said here in a few places that it looks more like a T^4.6 function, which would produce the effect you describe. As to exactly where the holes come from… Some bands are so highly absorptive that you could see them in the spectrum from a couple of meters above the surface; whereas others are weaker and do not develop fully until a substantial thickness of atmosphere is passed. I don’t know enough IR spectroscopy to tell you myself which are which.

Bart
January 13, 2012 6:12 pm

Joe Postma says:
January 13, 2012 at 5:41 pm
“Otherwise, without that spectral radiation, the gas would just continue to heat because the energy wouldn’t be able to escape via radiation.”
This is a rather fascinating observation, and on the one hand, it would be just desserts if it turned out that the added CO2 were actually cooling the planet. However, in 30 years when we are at the bottom of the incipient cooling part of the natural ~60 year cycle, I foresee that it will be seized upon to argue that we are driving the Earth into a new deep freeze, and the only thing to do is hand unaccountable power over to the political class to deal with it by drastically lowering the standard of living for those not of the nomenklatura.

richard verney
January 13, 2012 6:42 pm

eyesonu says:
January 13, 2012 at 8:36 am
//////////////////////
eyesonu
You are right to point out the energy involved in techtonic plate movement and a by product of which must be heat going into the oceans.
Following the Japanese Tsunami (last March/April – I can’t recall the exact dates), I posted a comment on one of the threads (discussing the Tsunami/nuclear issues) raising this very point.
I firmly consider that oceanic geomechanical and geothermal issues have not been adequately considered.

jorgekafkazar
January 13, 2012 7:51 pm

What about atmospheric tides? These are mostly thermal, rather than gravitational, but represent another form of energy transfer from the Sun to Earth besides temperature. They also influence the upward transfer of energy back towards space.

wayne
January 13, 2012 8:00 pm

Here is a very preliminary integration test set. Apologies in advance if there are any mistakes within.
These numeric integration cases were created taking Dr. Brown’s suggestions in the top post above. I have highlighted a few average temperatures that most of you will recognize. The first is the figure you will find in the IPCC reports and seems what current climate science is built upon. The second you will find within Dr. Nikolov & Zeller’s poster summarizing their paper. They are all hypothetically smooth cases and have not been thoroughly tested, so if you question them, run your own integration to verify.
The cells are cosine weighted using normal Simpson rule interpolation to find the exact center point of the cells. Latitudinal, the cell spacing is every degree. Latitudinal, the spacing is every three degrees. These were chosen for efficiency, but, running a test at 0.1 degree spacing in both dimensions returned the same results up to the fifth digit of precision. So these use 10800 points on the globe and the finer resolution was using 3,240,000 points on the globe. Funny, very little difference but the latitude direction is much more sensitive to the cell spacing and that does make sense due the decreasing area of latitude bands. The solar temperature was set to 5774 K to give a TSI of 1362.3 W/m^2 which seems very close to today’s readings.
These can also be used to evaluate the moon, for none of the case have an atmosphere yet, so the 0.12 albedo cases are close for both the Earth and the moon at this point.
Dr. Brown’s words in quotes.
[A case]
“Start with a non-rotating superconducting sphere, zero albedo, unit emissivity, perfect blackbody radiation from each point on the sphere. What’s the mean temperature?”
This run leave all points on the entire globe at equal temperatures, both lit & unlit sides:
254.6 K, 0.30 albedo, 1 emissivity
272.8 K, 0.12 albedo, 0.955 emissivity
278.4 K, 0 albedo, 1 emissivity
[B case]
“Now make the non-rotating sphere perfectly non-conducting, so that every part of the surface has to be in radiative balance. What’s the average temperature now? This is a better model for the moon than the former, surely, although still not good enough. Let’s improve it.”
Rotating or non-rotating, non-conducting.
This run leaves all points on the globe at different temperatures:
154.3 K, 0.12 albedo, 0.955 emissivity
157.5 K, 0 albedo, 1 emissivity
[C case]
Rotating or non-rotating, superconducting only vertically or only horizontally.
This run leaves all points latitude-wise at the same temperatures:
165.8 K, 0.12 albedo, 0.955 emissivity
168.2, 0 albedo, 1 emissivity
[D case]
Rotating or non-rotating, superconducting only vertically or only horizontally.
This run leaves all points longitudinal-wise at the same temperatures:
179.5 K, 0.12 albedo, 0.955 emissivity
183.2 K, 0 albedo, 1 emissivity
[E case]
Rotating or non-rotating, superconducting on lit side only
This run leaves the temperatures of all points the same on 1) the lit side and 2) the dark side:
192.9 K, 0.12 albedo, 0.955 emissivity
196.8 K, 0 albedo, 1 emissivity
After looking at these results, to me is seems the case D with an actual albedo and emissivity would be the one showing the moon’s true average temperature. Just a hunch.
You will notice the emissivity and albedo have just a few degrees of variance while the differences between each case is much larger.
Now to add the capability for rotation and thermal inertia at the surface interface. First in the soil and oceans, later an atmosphere.
– Wayne Jackson

Tim Folkerts
January 13, 2012 8:00 pm

Kevin Kilty,
I should have been a little more specific. I meant “The ‘entire profile’ from the ground is known to be very close to a blackbody curve AT GROUND LEVEL WHERE IT IS EMITTED.” By the time you are a few km up, then the “bites” due to GHGs begin to show up. If N2 had any major effect, then you should see the absorption or emission from N2 affecting the spectrum more and more as you go up. But even above the entire atmosphere, there is no hint of any effect of N2 shown in the satellite images I have seen. There is no hint of any effect of N2 in calculations of spectra for the atmosphere. (people can play around with the calculations here: http://geoflop.uchicago.edu/forecast/docs/Projects/modtran.html)

Tim Folkerts
January 13, 2012 8:10 pm

Joe Postma says: January 13, 2012 at 5:41 pm …
Joe, I would prefer sticking to my planet-in-a-nebula case. Stars have many complicating factors (they DO have gravitational heating, they have nuclear energy, the plasma is good at absorbing/emitting radiation so they don;t need polyatomic molecules to radiate, … )
You never addressed my conclusion –> the planet-in-a-nebula will cool slower than the naked planet. Can you argue in this simple case that I am wrong? (BTW, I am pretty sure I can argue that even your protostar situation leads to a warmer star if there is a surrounding cocoon of GHG, so even that doesn’t support your conclusions.)

Tim Folkerts
January 13, 2012 8:20 pm

Wayne,
I like your work on calculating the various radiative balances. I haven’t confirmed them independently, but they seem reasonable.
One note: when Dr Brown said ‘Now make the surface have some thermalized heat capacity — make it heat superconducting, but only in the vertical direction and presume a mass shell of some thickness that has some reasonable specific heat. ” I am 99% sure he meant that it conducts heat in a truly vertical direction (ie into/out of the ground). Your C & D instead allow heat conduction north-south or east/west, both of which are “horizontal”. They are still interesting cases, but not the next step Dr. Brown was envisioning.

Bob Fernley-Jones
January 13, 2012 8:25 pm

richard verney @ January 13, 6:42 pm
Richard, concerning geothermal stuff, I’m sorry, I can’t remember the commenter’s name, (you maybe?), but there was mention on another thread that the oceanic crust is generally very much thinner than the continental, with uhm erh implications.
Furthermore, that at some of the depths spoken of for the oceans, it would be rather hot at the same depth on land. AND, sea water is a far better conductor of heat than rock, and being so cold at the water’s bottom, it would seem to be a HUGE heat sink via conduction and convection/advection. (T1 – T2). I don’t know what assumptions were involved in others suggesting that average geothermal heat loss is only ~0.08 W/m^2 (I recall ?), but seeing how hot it can be in deep mines, intuitively, given an apparently better conduction path, it don’t look right to me for over 70% of the surface.
Incidentally, I’m no geologist, but I suspect that the greater sedimentary layering on the continents, resulting in more conduction interfaces, and some rocks such as sandstones and limestones notionally being perhaps the worst conductors, there seems to be a big curly: ?.
Bob Carter, where are you? Help!

wayne
January 13, 2012 8:35 pm

Some may wonder why I was using Simpson’s rule to calculate the cosine weights.
Let’s say your cell spacing is 30 degrees, using huge cells to divide up a sphere. At first glance some might just chose 15 degrees for middle of the cell and take the cosine of that, but you will find your results are skewed to high, for cos(15) gives 0.966 which is really to large. You need to take (cos(0)+4*cos(15)+cos(30))/6 to give 0.955 or the cosine at 17.2626 degrees which is much, much closer to the actual average of the cosines of all points in one dimension within that cell, when squared it gives you the best “center”. (wonder if IPCC knows that trick?)
Further, (cos(0)+3*cos(10)+3*cos(20)+cos(30))/8 is Simpson’s 3/8th rule and is even closer to the actual mean of the cosines across each dimension in the cell, the cosine of 17.2652 degrees.
Most here know it well, but just in case someone here had never happened across that neat “trick”, just had to put it out, it can save you from being WAY off in many cases.

wayne
January 13, 2012 9:03 pm

A commenter has pointed out an error in my descriptions of the integration cases. Thank you.
These should have read: (somehow my changes got reverted)
[C case]
Rotating or non-rotating, superconducting only vertically. (very unrealistic)
[D case]
Rotating or non-rotating, superconducting only horizontally. (very realistic)

Septic Matthew
January 13, 2012 9:07 pm

davidmhoffer: If the Earth absorbes 240 w/m2, then it emmits 240 w/m2.
Unless it’s warming or cooling. Whether it’s warming or cooling, and whatever CO2 may be doing to (transiently) change the balance of incoming and outgoing, are the unknowns. You seem to assume what everyone else wants to test.

Septic Matthew
January 13, 2012 9:26 pm

davidmhoffer quoting Joe: “The primary mechanism responsible for maintaining the incredible 4-billion-year stability of our system must be the physical transfer of heat from the equator to the poles where it is radiated away.”
The variation within the bounds of that incredible 4-billion year stability is not negligible for human and other life, and is indeed the topic of the AGW debate. The arguments based on thermodynamic equilibria are inadequate to determine whether a doubling of CO2 will produce a transient change in the surface temperature distribution or rainfall distribution sufficiently to reduce crop production disastrously. The argument is that by reducing the rate of radiation of heat the CO2 will cause an increase in temperature until at some time, when the earth surface has heated, the eflux matches the influx.
All you have shown is that the CO2 does not change the equilibrium rate of eflux. How does this matter in a system that has never been in equilibrium?

Septic Matthew
January 13, 2012 9:38 pm

davidmhoffer: The laws of thermodynamics require this statement to be true. The amount of net loss of energy in high latitudes must balance, to the last photon, the net absorption in low latitudes.
Again you are assuming what has to be decided: whether there is or is not net accumulation of energy on the earth, either now, over the last 10 years, over the last 150 years, over the last 10,000 years, or whenever; and whether CO2 accumulation affects that net accumulation, if it occurs
You have shown that if the total heat content of the climate system can not change, then CO2 changes can not change it.

January 13, 2012 9:47 pm

Septic Matthew;
Unless it’s warming or cooling. >>>
Don’t be silly, I said “at equilibrium” half a dozen times in first comment. The debate revolves around what the actual equilibrium temperature because we can’t calculate any of the other numbers (like GHE) unless we know that first. I’m not assuming anything and actually stipulated otherwise.
Tim Folkerts;
The OTHER effect of CO2 is to move the location of the radiating surface from the ground level up higher in the atmosphere. Some of the radiation comes from the very cold upper atmosphere and some comes from the ground, with the net result being 240 W/m^2 on average. This has been discussed multiple times and until people understand the implications of this, they will continue to misunderstand the GHE.>>>
Agreed. My point was that CO2 only moves the location from which radiance is emitted to space, not the amount that is emitted to space. This applies to both geographical location (decreased emission in the tropics, increased at high latitudes) but the exact same thought process applies to altitude as well, thanks for bringing that up, I’d missed it.
Myrrh;
Shrug. >>>
Gee, you stole my response to you. Thread after thread you ask for proof of this thing or that thing and when it is offered to you, your response is, yeah but I want proof. 2+2=4 and Myrrh hollers, yeah, but I want proof. Well, here’s two pospsicle sticks Myrrh and here’s two more popsicsle sticks and if we put them together and count….one, two, three, four…to which the inevitable rebuttal from Myrrh arrives…. yeah, but I want proof. To which I reply with my new stock answer to your serial idiocy. Shrug.
Stephen Wilde;
I agree, it seems incredibly obvious but only once the bulb lights up in the mind. There are so many out there that will fight tooth and nail against any such ideas that it is going to be an uphill struggle.>>>
So…you’re saying the lights aren’t on over at Joel’s place…. 😉
Joel Shore;
Congratulations, David. You used to make fun of posters who claimed that the greenhouse effect violates the laws of thermodynamics (and the Second Law specificially). Now you have become one of them…and, like them, you make that claim on the basis of no evidence whatsoever.>>>
I’m sorry that you can’t follow the logic nor the physics and so understand neither my explanation of the matter currently being discussed, nor how it differs from isolated and specific instances of radiation balance. The GHE is of course a real change in energy transfer that can be measured and verified. The notion that the GHE breaks the laws of thermodynamics isn’t correct. What is ALSO not correct is that one can determine if the earth is gaining or losing energy by averaging T instead of T^4. what is further not correct is that a change in CO2 levels results in an energy imbalance rather than an energy redistribution that results in a higher average T without changing average T^4 and average P. If you cannot understand these dead simple issues, then I have little choice but to relegate you to the Myrrh bin. Shrug.
Joel Shore;
(2) Come up with an explanation of how you can take GHGs away and still have the current average radiative emission that the Earth’s surface has.>>>
But that is exactly what my previous comments explain. Shrug.
Joel Shore;
(1) Show that if one averages T^4 instead of T, the warming seen disappears. Good luck doing that, as Essex & McKitrick (& one other co-author) have already tried…>>>
The point was to understand what the actual theoretical average black body temperature of the earth is when properly calculated so that we can COMPARE to the observed temperatures and determine how much should be allocated to insolation and how much to GHE. REALLY JOEL? YOU DIDN’T GET THAT?
Shrug.

Editor
January 13, 2012 10:02 pm

wayne says:
January 13, 2012 at 8:00 pm

Here is a very preliminary integration test set. Apologies in advance if there are any mistakes within.
These numeric integration cases were created taking Dr. Brown’s suggestions in the top post above.

Interesting analysis, wayne. Props for running the numbers yourself.
All the best,
w.

January 13, 2012 10:38 pm

Septic Matthew;
The arguments based on thermodynamic equilibria are inadequate to determine whether a doubling of CO2 will produce a transient change in the surface temperature distribution or rainfall distribution sufficiently to reduce crop production disastrously. The argument is that by reducing the rate of radiation of heat the CO2 will cause an increase in temperature until at some time, when the earth surface has heated, the eflux matches the influx.>>>
Given that an increase in CO2 must result in an increase in P (w/m2) that in turn results in an increase in T (degrees K) at earth surface, let us put aside for a moment the question of what the equilibrium temperature of earth actually is or isn’t and simply demonstrate what an idiotic statement this is when the fact that P varies with T^4 is taken into account. Let us consider the tropics, temperate zones and arctic zones to be at “average” temperatures of +30C, +10C and -20C respectively. Let us further use 3.7 w/m2 as the accepted increase in P at earth surface for a supposed doubling of CO2 (direct effects only for this example).
For an additional 3.7 w/m2, the increase in “average” temperature would be:
Tropics +0.58 degrees
Temperates +.72 degrees
Arctics +1.0
Now let’s take the next step and see how that “average” temperature change is distributed. too many combinations and permutations to bother doing them all, particularly when we only need to do one to demonstrate the issue at hand. Let’s use the temperate zones with Winter, spring, summer and fall “average” temps of -20C, +10C, +30C and +10C which “average” to plus 10. The increase in each of those due to 3.7 w/m2 would be:
Winter +1.0
Spring +0.72
Summer +0.58
Fall +0.72
See the pattern emerging? The tropics heat up the least, the arctic zones the most. Agricultural ouptut doesn’t get adversely affected a whole lot by +.58 and the warming in the temperates actually moves the “average” temperature into a temperature range that his MORE beneficial to agricultural output in the temperates and arctics, not LESS. But wait! There’s more!
Those numbers are WRONG!!!!!!
I added 3.7 w/m2 to the “average” temperature in the temperate zones and calculated via SB Law that the “average” temperature went up by 0.72 degrees. But then I took four values for the different seasons that average to the exact same number, but this time, I calculated the temperature increase for each season due to 3.7 w/m2 on a season by season basis. The average of those numbers isn’t 0.72, it is 0.75! Uh oh, which one is right?
Neither. We haven’t accounted for daily fluctuations. Let’s just consider winter and summer with a daily temperature fluctuation of 16 degrees. That would mean that the “average” day in winter would range from -28C to -12C and the average day in summer would average from 22C to 38C. (OK, I may have picked an unrealistic range but I typed too much to go back and start over so just live with it, focus on the principle being illustrated). That would give us for an additional 3.7 w/m2:
Winter night time low +3.3
Winter day time high +2.7
Summer night time low +0.63
Summer day time high +.54
See the pattern more clearly now? Winter night time lows go up 3.3 but summer day time highs go up only 0.54. Do we grow crops in winter at -20C? No? Then plus 3.3 is immaterial to crop production, and an increase of day time highs of 0.54 is neglible.
Of course if you average those numbers you won’t get the same amount as adding the 3.7w/m2 to the seasonal average either. Which one is right?
Neither. The numbers are STILL wrong. Why? So glad you asked!
The “average” increase in P from a doubling of CO2 MIGHT be 3.7 w/m2, but the increase at any given point on earth surface will almost (almost) NEVER be 3.7 w/m2. The GHE is based on upward bound LW being absorbed by CO2 and then re-radiated back to earth surface. OK,let’s walk through a couple more easy SB Law calcs.
In the tropics at +30C, the upward bound LW is 478 w/m2
In the arctic at -20C, the upward bound LW is 232 w/m2
OK geniuses, can someone exlain to me how doubling of CO2 can, when exposed to upward bound LW of 478 w/m2 return 3.7 w/m2 and how the exact same concentration of CO2 when exposed to 232 w/m2 returns 3.7 w/m2 also? Of course it doesn’t.
So if you’ve followed all of this the take aways are:
1. The most warming for a given increase in w/m2 occurs at night time lows, in winter, in the arctic zones.
2. The least warming for a given increase in w/m2 occurs at day time highs, in summer, in the tropics.
3. In other words, the warming occurs the least where it could do potential harm and the most where it can only improve conditions.
4. Averaging T to understand a change in energy balance is hopelessly useless.
5. Averaging P from a doubling of Co2 is equally nonsensical.
There can be no usefull discussion about how much the earth is warming, if the earth is warming, how much GHE raises the surface temperature, what the actual equilibrium blackbody temperature is, or if we are in a positive or negative energy balance if we do not first understand the implications of P varying with T to the 4th power.
Hey. I forgot to include the variance introduced by altitude and terrain. T at the top of as mountain versus T in the valley below. Then I also left out the fact that the earth orbit is elliptical and so P varies a few percent over the course of the year from that. I also left out the Gore effect which has to be added in by tracking the location of Al Gore at any given time “on average” and since we have no clue how the Gore Effect works and what factors govern the magnitude at any given time or place of the Gore Effect,the task of coming up with any meaningful numbers to base any discussion on is that much more impossibler.

January 13, 2012 10:45 pm

Septic Matthew;
Again you are assuming what has to be decided: whether there is or is not net accumulation of energy on the earth, either now, over the last 10 years>>>
Please Matthew, try and keep up.
The whole point of the explanation is that we CANNOT draw any conclusions about the energy balance from trending of temperature data. The whole point of my explanation was to show that there can be a change in uniformity of the planetary temperature that would be interpreted from the trending and averaging of T to be an increase in accumulated energy when, in fact, no increase had occurred.
In order to determine how much of any given temperature trend is due to ACTUAL energy imbalance and how much is due to NO change in energy balance but a change in distribution of temperature, all we’ve got is a bunch of numbers and a bunch of trends that mean nothing. It is possible, as I have demonstrated through simple examples in other threads, to produce a scenario in which the “average” temperature is increasing while the earth is actually LOSING energy, not gaining it.
I haven’t assumed a single thing. I have shown why the assumptions regarding the trending of temperature data are false and misleading.

Editor
January 13, 2012 10:57 pm

wayne says:
January 13, 2012 at 8:35 pm

Some may wonder why I was using Simpson’s rule to calculate the cosine weights.
Let’s say your cell spacing is 30 degrees, using huge cells to divide up a sphere. …

That’s slick, and I like the use of Simpson’s rule, but it’s roundabout and at the end of the day still an estimate.
When I area weight, I do it by the actual area. As a percentage of the total sphere, the area between two latitudes Lat1 and Lat2 is given by ( sin(Lat1) – sin(Lat2) ) / 2. For your example, 0-30° N latitude is a quarter of the globe,
Additionally, if you are doing the whole globe, this method lets you use a simple sum of the weighted values to give the weighted average.
All the best,
w.

johndo9
January 13, 2012 11:50 pm

@ Wayne
Yes Thanks for trying the integrals.
I can’t help but think the non-conducting case is nearest to reality. Perhaps look at the post a few days ago from Willis with actual on Moon temperature observations and his first pass at estimating heating and cooling rates. These may suggest approximately the low rate of conduction and some sort of heat capacity for the near surface few metres.
However the rate of rotation will also make a significant difference.
Nikolov and Zeller also had 154 K, perhaps its time for some interaction with them? Or to suggest some revision or qualification to their poster?
I wonder if Anthony has given consideration to a specific page for
“Crowdsourcing” modelling and “sauce”.
Perhaps in “References” or “Resources”?
It may need to include sections such as
Blackbody or SB calculations
Albedo Models
Atmosphere models
GHGs and how they effect the atmosphere models
In relation to Albedo, P.R. Goode, E. Palle´ / Journal of Atmospheric and Solar-Terrestrial Physics 69 (2007) 1556–1568 suggest up to 10% change in albedo from clouds over two decades (up to 1998).
Also from remote observation, Tyler D. Robinson, et al. Astrobiology. June 2011, 11(5): 393-408. doi:10.1089/ast.2011.0642. say 4 different cloud types are needed to get a vaguely reasonable model of the albedo of Earth as measured from space.
The atmosphere and albedo model sensitivities will be complicated before getting to GHGs.
The “Team” have spent millions of manhours on the GHG part, perhaps when the basics have been defined, some of them may be prepared to contribute.

Joe
January 14, 2012 12:49 am

David M Hoffer,
Thank you for finding the link to the NASA satellite data showing net in-flow and out-flow of energy spatially across Earth’s globe. I had seen it so long ago I did not know where to find it again. NASA’s own data proves beyond a shadow of a doubt that local radiative balance does not occur across most of the earth’s surface. In the lower latitudes the average temperature is below that which it should be in local radiative balance. The opposite is true at the higher latitudes. The excess incoming energy in the lower latitudes must go somewhere if the temperatures there are to stay stable at values below that required for local radiative balance. That somewhere can only be in one direction: orthogonal to the incoming radiation, i.e. parallel with the earth’s surface. The very fact that the poles are hundreds of degrees above the temperature at which they should be if driven only by incoming solar radiation means that some other place on the face of the earth must be below the temperature at which it should be for the local incoming solar radiation flux. That is the true meaning of the parameter “average global temperature”. If heat is really flowing from the equator to the poles, mathematics states beyond a shadow of a doubt that these deviations from local radiative balance must occur. NASA’s data say it does.
That last sentence leads me to an interesting yet disturbing aside. Ever notice how Alaska and the Arctic are the showcase for global warming because their temperatures are increasing two or three times faster than the average global increase? For that to be true, and I do not doubt that the measured temperature rise is true, someplace else on the earth’s surface of equal area must be decreasing in temperature in order for the average world-wide temperature increase to be only 1 degree. Where is that place? Another way to reach the average of a 1 degree global increase is for the temperature to decrease slightly over the rest of the planet to average out the large increase in the north. Is that global warming? You could also achieve the small average global increase with a large increase in the north even if no other place in the world is increasing its temperature. Would that constitute global warming? The only variable that changes to create these different scenarios is the relative surface area of the warming and non-warming or cooling regions. And yet I do not hear questions in the community about the non-warming or cooling regions. Where are they? What is the temperature doing in those regions? Do these regions move around constantly or do they remain in the same geographical region all the time? I ask this question because the above-average warming is sticking to one spot, the Arctic. When global maps of temperature change over the previous 30 years are released by researchers, does the average change across the entire map equal the average change in the temperature record? This is middle school mathematics and any valid model must withstand this type of evaluation.
The spatial distribution of the heating gives an indication of what is happening to the earth’s stored heat. If the heat flow to the poles slows down, they will cool while the lower latitudes will heat up. If the poles heat up while the lower latitudes remain the same, more energy must be entering into the surface system. If the lower latitudes retain more heat due to carbon dioxide in the atmosphere, the poles will heat up as more energy is transferred in that direction at a higher rate. If the poles heat up and the lower latitudes cool slightly, the earth is losing energy. If the poles heat up enough to melt the ice caps, the earth will lose energy fast and the lower latitudes will cool more quickly.
When writing my first post last night, I realized an answer to question of whether the greenhouse effect is a real phenomenon. This answer is a resounding YES! It can be proved using the same trigonometry I used last night. In my previous post, I approximated values for NASA’s energy budget numbers, which can be found at
http://earthobservatory.nasa.gov/Features/EnergyBalance/page4.php
I simplified that budget to 50% absorbed at the surface, 25% reflected by the atmosphere and surface, and 25% absorbed by the atmosphere. Keep in mind that this absorbed energy is from the energy in-flux from the sun and is not from the earth’s surface as heat. Therefore, that special 1 square meter column of atmosphere at the equator at high noon sees 25% x 1366 or 341W/m2. 341W/m2 is the average value across the entire globe so we can assume that if that square meter column of atmosphere at the equator at high noon was in radiative balance only with the incoming energy, it would be at 288k or there abouts. As that square meter rotated away from high noon, its temperature would fall. At any latitude north or south from the equator, the incoming energy absorbed by the local atmosphere will be less so, everywhere else on the face of the earth besides the equator at high noon, the atmosphere should be colder than the average global temperature of the earth as a black body, a grey body, or a greenhouse body! We know that this is not true, especially those folks in New York City that see 32C (90F) in the summer at their latitude. If we ignore magnetic field coupling, the solar wind, cosmic rays, or other yet-to-be-explored incoming energy sources, the only other source of energy to enable the atmosphere to remain at its elevated temperature is the warmer ground. Energy transfer from the ground to the atmosphere will take place by conduction, convection, and radiative transfer. I am not an expert in the area of fluid thermodynamics so I cannot begin to construct the equations to describe how the atmosphere is warmed but it is beyond doubt that it is warmer than it should be in radiative balance with the incoming flux. Hence, the greenhouse effect.
How much does carbon dioxide contribute to that warmth is an excellent question. It most likely does since the physics are relatively easy to prove. Does it balance the local radiative transfer in and out of the system? No because the local temperatures in the lower latitudes remain lower than they should be in balance with the sun during daylight. How will it affect the transfer of energy to the poles? It helps because it increases the heat content of the air which is then moved by weather and the Coriolis effect to the poles to be released. It probably also reduces heat loss at night when the energy flow for the surface goes only one way -> out. This is good because if the temperature dropped so far every night that ice formed everywhere every night, things would not be good for us. It is startling that the rate of Earth’s rate of rotation is exactly right so water does not freeze everywhere every night! That could happen if our day was twice or three times as long.

gbaikie
January 14, 2012 12:50 am

“I’m having just a bit of difficulty with this, in the context of the extended discussion on radiative balance. Although (as I’ve said) I’m withholding judgement until I’ve done the integrals, it appears to be strictly true that radiative loss is always favored by higher radiation temperature inhomogeneity. In other words, moving heat from the tropics to the poles cools the tropics and heats the poles so it provides a more uniform temperature, but a more uniform temperature is always a warmer temperature on average because in fact you lose heat far faster from the hot tropics than you do from the poles.
In fact, “most” of the net heat (the gain) absorbed during the day in the tropics is lost during the night, at least in any sort of naive static BB model.”
It seems that most energy loss occurs in the tropics. And tropic ocean doesn’t change in much in terms of temperature. But the tropics have a lot of heat to lose.
The tropics should be absorbing more joules of energy as compared to anywhere else, and dumping more joules of energy into space- in terms of joules of energy per sq meter. The tropics also is 40% of the surface area of the planet, to poles have.. well antarctic is “The total surface area is about 14.2 million sq km” earth is 510 million. So say 30 million for both pole 5 % of surface. And the poles are cold so can’t radiate much energy.
I would say it this way, the pole allow the tropics to absorb more energy.
So if no poles to give heat to, then tropic region absorbs less energy.
So Poles extracting a low tax rate, which doesn’t affect the tropics heat wealth.
If government were so wise as poles, we would have a better world.
Hmm, so what does that do with theory that antarctic is major cause of the last tens of million of years of cooler global climatic?
Let’s see, how energy could be transported to Antarctic from tropics per second?
Well first when tropics are near the poles, summer or winter [either pole] the distance is less and and if wind going same speed- no that will not make any difference.
How about this the atmosphere is equal to 10 meters height of water- only some portion of this can the transporting medium- and it high atmosphere so air density is much less. I am sure it’s less, but most it could equal the equivalent of 1 meter or water. Whereas you an ocean of water will probably transporting tens of sea water [per square meter surface area]- per mass water carries more energy than water. Though if tropics is transporting to pole as water vapor this can a lot of energy per mass, one very water vapor in high altitudes.
The other factor in term of the Antarctic the rotation air mass [entire atmosphere] which circles the Antarctic land mass. It seems the entire atmosphere and oceans have to be transporting more energy and the roaring nineties are going affecting the temperate regions, and ocean current going to tropics.
So what’s all this mean?
Well the tropics aren’t really involved much with this last tens of million of years of cooling- tropics are unchanging- very stubborn area.
I would say, it’s the roaring nineties are bigger effect. Hmm, no I give up. Wait- it seems if earth isn’t closest when southern hemispere is in summer- would get bigger sea ice around the Antarctic. You have deep water with ice floating on top of it exposing huge surface area to ocean water. Maybe this is lowering tropical sea surface temperature.
That’s seems unlikely.
It seems difficult to see how south arctic is affecting northern hemisphere where one getting all the glaciation.
So, got to go with arctic ocean sea ice melting – and that through various means cools northern hemisphere.
More ice around antarctic, more rain in Africa, and snow all over northern hemisphere.
Northern hemisphere cover ice and cold, is much large area than arctic. Arctic cold front, become North America and European cold fronts.
In current situation American and Europe are buffer states, when ice covered and cold, there are disrupting the tropics. And tropics are dumping moisture at them.
Let see, what is significant about Antarctic is it is cold and it has [with all it’s ice] a very high average elevation.
Oh here something:
“The winter sea-ice cover at the LGM [Last Glacial Maximum] was, therefore, twice the modern surface. Similarly, the LGM summer sea-ice edge was projected northward of its modern position, overlying the modern winter sea-ice margin (Fig. 2). The summer sea-ice cover was, therefore, 6-7 times greater than modern extent.”
http://pages-142.unibe.ch/products/newsletters/2007_2/Special%20section/Crosta_2007-2%2813-14%29.pdf
According graph in ref, the Antarctic sea ice it gets close to reaching the 45 parallel- or if compared to north hemisphere that reaches down to US. [49th parallel is most US/Canadian- west of Ontario. Or all of UK and the tip northern part of Japan].
Maybe with large area having sea ice and large amount cold water dropping into ocean something like Gulf Stream is drive huge amount warm surface ocean towards the south pole- having cooler surface in tropics would have global affects.

Tim Folkerts
January 14, 2012 6:05 am

Joe ponders”

That last sentence leads me to an interesting yet disturbing aside. Ever notice how Alaska and the Arctic are the showcase for global warming because their temperatures are increasing two or three times faster than the average global increase? For that to be true, and I do not doubt that the measured temperature rise is true, someplace else on the earth’s surface of equal area must be decreasing in temperature in order for the average world-wide temperature increase to be only 1 degree. Where is that place?

The short answer is that that place is the “top of the atmosphere”.
Adding more CO2 raises the average altitude for emission of IR. Higher altitudes are colder, so the CO2 will emit less IR.

Bill Illis
January 14, 2012 6:12 am

I have a bunch of the basic climate numbers separated into each 10 degree latitude band.
Surface area, Albedo, Solar Forcing, Average Temp, Resulting Greenhouse Effect (which would include the temperature re-distribution from the equator to the poles through atmospheric circulation).
http://img24.imageshack.us/img24/9751/climatefiguresbylatitud.png
I can probably put this up in a spreadsheet if someone wants.

Bryan
Reply to  Bill Illis
January 14, 2012 6:42 am

Bill Illis or anyone.
Does anyone have a link to a satellite obtained IR spectrum looking down at night?
I would like to compare it with one obtained during daytime .

Bill Illis
January 14, 2012 6:14 am

I should have said that the numbers are off a little as a total but one would need more accurate numbers for solar forcing etc. (four decimal points really) than I have been able to find.

Myrrh
January 14, 2012 6:19 am

davidmhoffer says:
January 13, 2012 at 9:47 pm
Myrrh;
Shrug. >>>
Gee, you stole my response to you. Thread after thread you ask for proof of this thing or that thing and when it is offered to you, your response is, yeah but I want proof. 2+2=4 and Myrrh hollers, yeah, but I want proof. Well, here’s two pospsicle sticks Myrrh and here’s two more popsicsle sticks and if we put them together and count….one, two, three, four…to which the inevitable rebuttal from Myrrh arrives…. yeah, but I want proof. To which I reply with my new stock answer to your serial idiocy. Shrug.
But… you’ve never come up with any proof! Most of the time what I get is the arrogant response from y’all to go read books on physics! Hello, is anybody in? I’m arguing about the physics in those books..
Prove that ‘the heat we feel from the Sun comes from visible light’ of the nonsense Ira gave and yet you say you’ve given proof? I’m not interested in the blackbody arguments, far too much maths for me find time following, what does interest me is the sleight of hand produced in support of AGW which has very subtley created an impossible, imaginary, world, which you all then argue about as if it is real..!
And to do this I’ve noticed a pattern of taking laws out of context, of giving the properties of one thing to another, of missing out rebuttals from the past to continue emphasising errors as if they’re not – Arrehnius the classic example here.* My call, if I was allowed to make it here, is for you all to go back to basics.. Until you get that right, have some grip on the reality of the world around us, you’ll continue arguing about this fantasy world created by AGW pushers and ever so often you’ll come out with ‘gosh, we found that carbon dioxide wasn’t well-mixed at all, but lumpy’, or similar surprise when you finally get some real data which you could have extrapolated if knowing that CO2 is heavier than air and the atmosphere around us wasn’t empty space with molecules zipping around at gazillion miles an hour bouncing off each other in elastic collisions and so throughly mixing, and even, adding Brownian motion to that to prove carbon dioxide will sponaneously diffuse into empty space oblivious that Brownian motion presupposes a fluid medium.., etc. Do you understand what I’m being so sarky about here? Nice to see that conduction and convection is finally making an appearance…
And, I’m fed up with being told that ‘experiments’ prove that greenhouse gas warming is a reality, but when I ask for these to be produced they never are. Where does Tyndall, who was experimenter of high reknown, prove that? I haven’t been able to find it in his writings. Maybe I missed it, I’m willing to stand corrected if so.
I’ll say it as clearly as I can, what I’ve been finding is very subtle tweaks to the physics of the real world which give an erroneous impression of the real world, added together they have created a world which doesn’t exist, and that’s what you’re arguing about.
As the man says here: “Latour insists that the apparent errors in atmospheric physics made by climatologists are because they work in a ‘generalist’ field of science, unlike most ‘hard’ sciences such as physics, chemistry, biology, engineering and medicine where detailed and in-depth specialization is essential so that products and services actually work.” http://johnosullivan.livejournal.com/43659.html
Which is how I came to see the tweaks, through following both sides arguing and the applied scientists won hands down on logical physics. My conclusion though is that certain key erroneous memes about the physical world have been deliberately introduced to create this climate of confusion about the physics, and it takes someone with a very thorough grounding in real science to do this and come up with ‘experiments’ that apparently ‘prove’ these memes, see Anthony’s deconstruction of Gore’s experiment. Attention! Magicians at work.
Typical is that introduced into classrooms some time ago, to show that carbon dioxide was well mixed and diffused through the atmosphere rapidly to be that, opening a bottle of scent. That this says nothing about carbon dioxide is not the message got by the children, but now they argue pro AGW by saying carbon dioxide can do this and can’t separate out from the atmosphere in layers, etc. .. Do you understand the point I’m making here? They can no longer understand the basics of the world around them.
This has now been so thoroughly confused that I think the easiest way would be to go back to basics – perhaps taking each of those memes and examining the tweaks to physics creating them.
We need more applied scientists..
Where does the meme come from that radiation in the atmosphere is in all directions? There’s a distinct difference between that travelling direct from the Sun in concentrated force to reach us, the great heat from the Sun reaching us in straight lines in eight minutes, and that given off by the much colder Earth from the variety of parts into the atmosphere.
As Tyndall said in: On the Absorption and Radiation of Heat by Gases and Vapours, and on the Physical Connexion of Radiation, Absorption and Conduction
“The solar heat possesses… the power of crossing an atmosphere; but, when the heat is absorbed by the planet, it is so changed in quality that the rays emanating from the planet cannot get with the same freedom back into space. Thus, the atmosphere admits of the entrance of the solar heat, but checks its exit; and the result is a tendency to accumulate heat at the surface of the planet.” -Tyndall
What’s he mean by “so changed in quality”? No, he didn’t mean that shortwave becomes longwave .., he’s talking about heat from the Sun, thermal infrared, and not light, visible, he knew the difference even if y’all arguing AGW KT meme that shortwave heats land and oceans, don’t.
……….
Pierre R Latour says:
January 13, 2012 at 7:18 am
GHG Theory 33C Effect Whatchamacallit
GHG Theory was invented to explain a so-called 33C atmospheric greenhouse gas global warming effect. In 1981 James Hanson stated the average thermal T at Earth’s surface is 15C (ok) and Earth radiates to space at -18C (ok). Then he declared the difference 15 – (-18) = 33C (arithmetic ok) is the famous greenhouse gas effect. This is not ok because there is no physics to connect these two dissimilar numbers. The 33C are whatchamacallits. This greenhouse gas effect does not exist.
I see this from a different slant, there is a physics to connect them.. These figures were sleight of handed to give the impression that there was such a thing as ‘greenhouse gas global warming’. There is no Water Cycle in this, in the KT97 and ilk.
These come from:
Earth with atmosphere as we have it, 15°C
Earth without atmosphere, -18°C
Earth with atmosphere but without water, 67°C
Hidden by simply removing it from the AGW energy budget which now is ubiquitously taught to represent the real world, let me present to you, drum roll, the Water Cycle.
This cools the Earth by 52°C, to get it down to the 15°C
The main greenhouse gas cools the Earth, ergo, greenhouse gases cool the Earth. There is no greenhouse gas warming, it’s a trick, in the slip between cup and lip the water has disappeared.
Carbon dioxide is fully part of that cycle – all pure clean rain is carbonic acid. Water vapour and carbon dioxide have irresistible attraction for each other, what carbon dioxide doesn’t come down to Earth by displacing the molecules of oxygen and nitrogen in the fluid gaseous ocean around us because heavier than air, will come down in the rain, water vapour hoovering in whatever is around – how then can carbon dioxide ‘accumulate for hundreds and thousands of years in the atmosphere’?
Getting rid of the Water Cycle they can pretend that convection etc. doesn’t exist, with no convection no atmosphere necessary, all can now be reduced to the one dimensional imaginary empty space of the basic imaginary ideal gas volumeless world where dot molecules travel unimpeded at great speeds bouncing off each other with no attraction and so becoming thoroughly mixed, and only radiation necessary as the means of transferring heat. So trace CO2 the greenhouse culprit who raises the temp of the Earth from -18°C to 15°C.
Supermolecule defying gravity! He wears his knickers on the outside.
And I’m continually told that the ‘greenhouse gases warm the Earth’ is real and proved by experiments ‘somewhere over there’ but never fetched for perusal and the argument should be only how much it warms the Earth..
So, David, effin’ prove it.
* I’ll fetch something I found interesting about this.

January 14, 2012 7:33 am

That in turn results in an increased average T for the exact some P absorbed. No change in energy balance need occur to account for some of the the increase in T.
(Boldface my addition.) Sure, with the addition, since we now have multiple mechanisms in place. This doesn’t mean that differential radiation atmospheric effect (not really “greenhouse”) warming or cooling doesn’t also take place, only that there are multiple channels and one cannot attribute all warming observed to differential radiation rate local warming in the naive picture, one has to mentally split it up into warming (and cooling) caused by other processes. Even though balance is still purely radiative in the end, transport matters, and (like albedo) might well matter more than small changes in local differential forcing. Regardless, the mere existence of the channel implies that “greenhouse warming” per se is a smaller fraction of the total observed warming than unity.
The really interesting thing is the modulation of mixing. One has to assume that the big oscillations all mix, but they also drive extremes, or at least modulate the mixing. If mixing leads to relative warming, and oscillations modulate mixing, oscillations modulate relative warming and relative cooling. Q.E.D. All that remains is to consider magnitudes.
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January 14, 2012 7:40 am

When writing my first post last night, I realized an answer to question of whether the greenhouse effect is a real phenomenon. This answer is a resounding YES!
Absolutely, which is why I try not to let it pass when people assert otherwise. The radiation data are absolute proof of it. The question isn’t “is there a GHG effect that contributes to the warming”, it is “what are the feedbacks, and what are the other mechanisms that also contribute to warming among which total observed warming (or cooling) must be split up”.
Short version, what are the feedbacks. A difficult question to answer, either empirically or theoretically — theories depend on assumptions, most of the untestable or difficult to test in the short run. Empirically one has to understand all of the dynamics, not just simplified idealized pictures of the dynamics, in order to “measure” it. “Catastrophe” is based on assumptions of large feedback. “Reality” seems to be empirically lowering the upper bound on the feedback quite systematically. It also strongly suggests that the models are leaving out important physics and feedbacks, notably solar modulation of albedo and possibly mixing.
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Joel Shore
January 14, 2012 8:14 am

Joe says:

That last sentence leads me to an interesting yet disturbing aside. Ever notice how Alaska and the Arctic are the showcase for global warming because their temperatures are increasing two or three times faster than the average global increase? For that to be true, and I do not doubt that the measured temperature rise is true, someplace else on the earth’s surface of equal area must be decreasing in temperature in order for the average world-wide temperature increase to be only 1 degree. Where is that place?

No…There can just be places that are rising somewhat less rapidly than average. In particular, one important thing to realize is that the arctic is a fairly small part of the global area: In particular, only 13.4% of the Earth’s surface lies north of 60 deg N latitude (and, of course, another 13.4% lies south of 60 deg S) and fully half of the Earth’s surface lies in what one might call the tropics between 30deg N and 30deg S latitude.
To show you a concrete example: Let us say (to make the numbers simple) that the Earth has warmed an average of 1 deg. Let us also assume that the arctic region north of 60 deg N has warmed at 2.5 times this rate, and the rest of the world outside the tropics have warmed at exactly the average rate. Then, how fast would the tropics (30deg S to 30 deg N) have had to warm? They would have had to warm ~0.6 deg.

January 14, 2012 8:15 am

Myrrh;
I’m not interested in the blackbody arguments, far too much maths for me >>>
Shrug.

Joel Shore
January 14, 2012 8:50 am

davidmhoffer says:

I’m sorry that you can’t follow the logic nor the physics and so understand neither my explanation of the matter currently being discussed, nor how it differs from isolated and specific instances of radiation balance. The GHE is of course a real change in energy transfer that can be measured and verified. The notion that the GHE breaks the laws of thermodynamics isn’t correct. What is ALSO not correct is that one can determine if the earth is gaining or losing energy by averaging T instead of T^4. what is further not correct is that a change in CO2 levels results in an energy imbalance rather than an energy redistribution that results in a higher average T without changing average T^4 and average P. If you cannot understand these dead simple issues, then I have little choice but to relegate you to the Myrrh bin. Shrug.

The point is that I understand this well enough to know that it is a fairly small effect. Let’s take a concrete example: Suppose we have a world that is half at 268 K and half at 308 K…That’s a distribution that likely has a larger standard deviation than temperatures on the Earth (particularly when you consider, as I have noted, that half the world is in the tropics, defined by 30 deg N to 30 deg S and 70% is ocean where temperature variability is much lower than on land).
Let us further suppose that the entire world warms by an average of 1 K but that the warming is twice as large (1.33 K) for the colder half as for the warmer half (0.67 K). Then, if we look at taking the fourth root of the average of T^4 rather than just averaging T, how much will we conclude the world has warmed? The answer is that we would conclude the world has warmed 0.93 K rather than 1 K.
Heck, even if we assume that ALL of the warming occurs in the colder half (so it warms 2 K) and none in the warmer half, then the method of averaging T^4 and taking the fourth root gives us the result that the world has warmed 0.80 K.

The point was to understand what the actual theoretical average black body temperature of the earth is when properly calculated so that we can COMPARE to the observed temperatures and determine how much should be allocated to insolation and how much to GHE. REALLY JOEL? YOU DIDN’T GET THAT?

And, the answer, as I have demonstrated above, will be that most of the rise is due to the increase in the amount of power being radiated and only a fairly small fraction is due to the temperature becoming more uniform on the Earth.
So, at the end of the day, what you have shown is what we already know: That a single number like average temperature is an imperfect metric. We already knew that the average is only an average and that actual warming will generally be greater in cold regions than warm regions, considerably greater over the continents (where people live) and less over the oceans, etc., etc. Characterizing a whole distribution by one number always necessarily results in a large loss of information.
And, as for the main focus of what we have been discussing here, the magnitude of the natural greenhouse effect, we know the following: The Earth would be about 33 K colder without the greenhouse effect because the surface would be emitting ~240 W/m^2 rather than ~390 W/m^2. To the extent that such a world without a greenhouse effect has a broader temperature distribution, this could make the 33 K number even larger although I would argue that this is somewhat of a “false effect”, i.e., it is due just to a redistribution of the energy that the surface would be emitting without a change in the actual amount emitted.
In general, I think that you are making a mountain out of a molehill, although less so in your latest post than in your previous one ( http://wattsupwiththat.com/2012/01/12/earths-baseline-black-body-model-a-damn-hard-problem/#comment-863072 ) where you were just totally off the rails.

Joe Postma
January 14, 2012 9:24 am

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Tim Folkerts says:
January 13, 2012 at 8:10 pm
Joe Postma says: January 13, 2012 at 5:41 pm …
Joe, I would prefer sticking to my planet-in-a-nebula case. Stars have many complicating factors (they DO have gravitational heating, they have nuclear energy, the plasma is good at absorbing/emitting radiation so they don;t need polyatomic molecules to radiate, … )
You never addressed my conclusion –> the planet-in-a-nebula will cool slower than the naked planet. Can you argue in this simple case that I am wrong? (BTW, I am pretty sure I can argue that even your protostar situation leads to a warmer star if there is a surrounding cocoon of GHG, so even that doesn’t support your conclusions.)
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I can understand the desire to turn the argument around back into the standard GHE paradigm…that the molecules in the cloud cause heating. But that’s not what happens or what we address. I wanted to present the astrophysical theory because it had been presented to me by a colleague who IS an expert on that particular issue; he thought I would be interested in the information.
The center of the gas cloud is where the heat is being generated; this is just like the Earth where the heat is generated by the action of sunlight upon the ground, at a temperature potential of up to +121C, at the bottom of the atmosphere.
In astrophysics, that heat finds a vector to leave the gas via the mechanism I have explained above – if it behaved the opposite way, that of increasing the internal temperature, it would do the exact opposite thing it is needed to do, which is assist the cloud in collapsing by damping the thermal build-up. Otherwise the cloud would be prevented from collapsing.
The heat is generated internally, by collapse, or by sunlight, and this is what sets the internal temperature and amount of energy; THAT energy is then modulated and distributed by the gases, and there is no mechanism (in astrophysics) by which the gases would amplify said preexisting energy; if they did, then stars shouldn’t be able to form.
Anyway I don’t think you and I can get much further on this issue, as it is quite a new and unique angle to look at the situation from. I will discuss at length this physics more with the colleague who does work in this field, and will write about the findings in an upcoming paper.

Phil.
January 14, 2012 9:27 am

Robert Brown says:
January 13, 2012 at 10:08 am
“Anything with a temperature radiates…in the case of non-spectral gases like N2 or O2, the radiation will arise from inter-molecular collisions. Perhaps we haven’t explored the spectrum at far enough wavelengths to see this emission; perhaps this emission is what helps constitute the entire profile of the “black-body” output curve of the Earth as seen from space in any case.”

These emissions are so weak by orders of magnitude in the Earth’s atmosphere that they are negligible. We do see O2 emissions from O2 at ‘far enough wavelengths’, they’re in the microwave range and they’re what MSU uses to measure the atmospheric temperature but their contribution to the energy balance is minuscule.
To clarify this a bit, if one looks at the actual spectrum associated with the cold top of atmosphere, one doesn’t see “CO_2 lines”, one sees pretty much a BB curve, but at a colder temperature in the general IR window. There is clearly a reduction of ground level IR at T_s and its replacement by atmospheric IR at T_a. That’s the physical reality of actual measurements. This is why I have very carefully avoided giving any impression that I “deny” that the “greenhouse effect” or “atmospheric warming effect” in general terms exists.
However, the data tells us something else as well. It is not CO_2 emissions lines. It is well-thermalized emissions from a colder radiator. It clearly does not come from CO_2, which is 0.03% of the atmosphere, recall. It comes from all of the air, well mixed.

It’s the CO2 emission band with the lines thermally and pressure broadened and the resolution of the spectrometer causing the lines to merge. It does not come from well mixed air!

Joe Postma
January 14, 2012 9:29 am

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davidmhoffer says:
January 13, 2012 at 10:38 pm
3. In other words, the warming occurs the least where it could do potential harm and the most where it can only improve conditions.
=====================================
Wow!

January 14, 2012 10:09 am

But… you’ve never come up with any proof! Most of the time what I get is the arrogant response from y’all to go read books on physics! Hello, is anybody in? I’m arguing about the physics in those books..
The proof isn’t theoretical, although you do have to understand at least the idea that as far as the Earth as a whole is concerned, the fundamental issue is how hot the Earth has to be in order for radiation in to equal radiation out.
At that point, one goes up into space, measures the radiation coming out as a function of wavelength (or frequency, if you prefer) and goes “wow, the fact that the atmosphere is experimentally observed to take up some of the total heat delivered and radiate it at cooler temperatures than the surface means that the surface temperature is somewhat higher than it would be if it were an ideal black body.
That’s all. Forget any discussion of mechanism. To understand that, you’d have to actually work through and learn to understand all of the physics, and I’m a physics Ph.D. and it is still a daunting proposition for me. But the energy balance you can understand without knowing a damn thing about “upwelling”, “lapse rate”, convection, conduction, thermodynamics and so on. All you need to know is that ins have to equal outs in (dynamical) equilibrium, and the higher the average temperature of the whole thing, the greater the outs. Lower the effective temperature of part of the outs, you raise the surface temperature to compensate and keep ins (fixed) equal to outs.
Done. End of story. The proof is in the measured spectrum of the Earth and elementary knowledge, it isn’t complex or difficult to understand. The details are difficult, the feedback, convection, radiation balance inside and all that are difficult, but the proof that the process occurs isn’t theoretical, it is purely observational. We might have the mechanism(s) completely wrong. Invisible fairies could be responsible for channelling heat from the surface into the atmosphere to radiate out in the IR at cooler temperatures. It wouldn’t matter. The net effect would be surface warming.
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Phil.
January 14, 2012 10:16 am

Joe Postma says:
January 13, 2012 at 1:13 pm
======================================
Phil. says:
January 13, 2012 at 12:19 pm
Indeed it was speculation when I was discussing where we might find the collisional emission from the N2 and O2 in the atmosphere. It would be interesting to see it and acknowledge it rather than jump down its throat. But it doesn’t bear on my central theses in any case.
Around 50-200cm^-1 and very weak, and it doesn’t have any bearing on anything.
As Phil’s post was mostly antagonistic I will just fill in the blanks for others who are here willing to learn and have a scientific discussion with experts.
Interesting that you regard your errors being pointed out as ‘antagonistic’, also that you are unwilling to learn and have a scientific discussion with experts.
Phil made a fundamental error in his critique when he criticized my “talking about stellar atmospheres”. I wasn’t talking about stellar atmospheres.
You certainly weren’t talking about the Earth’s atmosphere.
The case of an interstellar gas cloud, and the physics we talk about with them, is entirely analogous to that of the terrestrial atmosphere in the functionality of the radiation. In both cases, you have reduced optical thickness “looking out”, and increasing optical thickness to complete opacity “looking in”. Of course, the idea that spectral absorption and line emission is different just because it’s found in a gas cloud, a star, or a planetary atmosphere, rule by different physics is absurd – it’s the exact same phenomenon, one set of physics describes it.
Indeed, but under different situations different approximations are used, in your case you appear to use the one where collisional deactivation of the excited states is negligible, this is not the case in the lower earth’s atmosphere!
The next step comes when the CO2 molecules radiate the heat energy they’ve internally picked up from the previous collision. This radiation is emitted isotropically; the outward component can eventually escape due to the reducing optical thickness of the cloud, but the inward component can not escape because the gas cloud is optically thick & opaque inside.
So this describes, pretty much precisely, the exact same phenomenon in our terrestrial atmosphere.

No it does not for the reasons given above.
Now, given the history of malfeasance by alarmist climate science, is it astrophysics or is it climate science which is more likely to be correct, on the question of cooling vs. heating…
It is perfectly logical from the astrophysical perspective: damped collisions reduce heat build-up, check, and radiating molecules move energy out of the system, check, rather than adding more energy and heat than was already there, major check.

I have no problem with astrophysics, my problem is with an astrophysics student who’s done some classes of interstellar gases and thinks that he can apply the same equations to the earth’s atmosphere and assumes that those scientists who specialize in the area of atmospheric physics and chemistry are too stupid to understand what’s going on. Perish the thought that he should read some of the science on the subject and realize why the equations he uses are an approximation which doesn’t understand that the relationship between the radiative life time and the mean time between collisions is important!

Joe Postma
January 14, 2012 11:02 am

Hi Phil,
Well, astrophysics is generally recognized as a super-set of all other physics because it has to incorporate molecular, atomic, quantum, gravitational, atmospheric, electromagnetic, relativistic, Newtonian, strong & weak, physics in the practice of its research to explain phenomena in which these universal principles do (or do not) apply, and in situations that, by definition and observation, bound the conditions found on Earth. I think it’s a very good perspective to approach things from…the best, in fact, I can’t imagine a better field of physics to come into this from, given the breadth astrophysics covers. The Earth is a small sliver somewhere in the middle of the bound of conditions which astrophysics characterizes.
If you look at the type of “science” that AGW alarmists have peddled, it is clear that the poorer scientists who don’t understand the basic moral method of it are ending up in a particular field. Not understanding the difference between correlation vs. causation, as a simple example in climate science, is telling.
Anyway, I will report on further findings from this angle at a later date, after discussing it with other astrophysical experts who study it.

January 14, 2012 11:08 am

Joel Shore;
blah blah blah>>>
Shrug.
Robert Brown;
(Boldface my addition.) Sure, with the addition, since we now have multiple mechanisms in place. This doesn’t mean that differential radiation atmospheric effect (not really “greenhouse”) warming or cooling doesn’t also take place>>>
Agreed. My point is that until (and unless) we accurately quantify how much of a given temperature trend is due to a change in distribution rather than an increase in net energy, we have no possible way of calculating how much of the trend is due to any given mechanism. Calculating GHE without knowing what the baseline is, is impossible.
gbaikie;
It seems that most energy loss occurs in the tropics.>>>
Of course. The warmest regions radiate the most energy to space. BUT, the tropics still absorb more energy than they lose, so the tropics are a net absorber over all. Here is ERBE outbound LW showing that the tropics lose WAY more heat to space than any other regions:
http://eos.atmos.washington.edu/cgi-bin/erbe/disp.pl?olr.ann.
and here is ERBE net LW showing that the tropics still absorb more than they lose to space:
http://eos.atmos.washington.edu/cgi-bin/erbe/disp.pl?net.ann.
If you scroll back upthread to Joe’s first comment on the issue, you’ll see that he pretty much nailed exactly where the “break even” latitude is just from theoretical physics.
Joe;
Ever notice how Alaska and the Arctic are the showcase for global warming because their temperatures are increasing two or three times faster than the average global increase? For that to be true, and I do not doubt that the measured temperature rise is true, someplace else on the earth’s surface of equal area must be decreasing in temperature in order for the average world-wide temperature increase to be only 1 degree.>>>
I’ve been harping on that exact issue for months and months, though one could still have the effect you speak of but still with a warming trend across the board, one doesn’t actually need to have one section of the earth to be cooling to show that there is a balance that does not meet the eye if all one looks at is raw temperature data because 1 degree in the Arctic and 1 degree in the tropics are two differen things. By SB Law:
+30C + 1 degree = +6.3 watts
-40C + 2.15 degree = 6.3 watts
Every time I hear the “itz happening in the arctic already!” I think…uh huh. I’d expect warming from ANY given source to be more pronounced in the artic! More to your point however, it would be very possible for an increase in temperatures in the arctic to be balanced by cooling in the tropics, and this would be hard to spot because the temperature decrease in the tropics required to establish a balance versus warming in the arctic would be a much smaller change in temperature than would be exhibited in the arctic warming. In fact, I’ve demonstrated via some very simple models in other threads that it is in fact possible to have a set of conditions in which average temperature is increasing, but net energy is actually negative, so temps can actually say “warming” when earth is actually “cooling”.

Bart
January 14, 2012 11:19 am

wayne says:
January 13, 2012 at 8:00 pm
” Latitudinal, the spacing is every three degrees. “
To get equal area coverage of a sphere, you should not space uniformly by latitude, but by vertical height. That is why you had to do the workaround you describe at January 13, 2012 at 8:35 pm. Remember, the area element is cos(theta)*d_phi*d_theta, where theta is the latitude. Transform to a new set of coordinates with z = sin(theta) and the area element becomes dz*d_phi.
Tim Folkerts says:
January 13, 2012 at 8:00 pm
“There is no hint of any effect of N2 in calculations of spectra for the atmosphere.”
Why would there be any hint in calculations which do not assign it any particular weight? You appear to be using a model to confirm… the model.
Robert Brown says:
January 14, 2012 at 7:40 am
“Empirically one has to understand all of the dynamics, not just simplified idealized pictures of the dynamics, in order to “measure” it.”
So true. Nature is complex. Climate science is in a panic due to back-of-the-envelope calculations. Nature is not cooperating.

January 14, 2012 11:57 am

Bill Illis;
I can probably put this up in a spreadsheet if someone wants.>>>>
You gave me an idea.
Seems to me what we need to get this discussion back from thrashing around theoretical physics to understanding the meaning of the observational data we have, is not lots of calculus, but a means of manipulating the grid data we have but with the relationship of T^4 understood as part of the manipulation. We don’t have data that would mimic dx/dt so why bother? Can we not get a better approximation of the data that we do have, granular though it may be, by just using the grid data properly?
To do that, we could start with the NASA/GISS temperature data. What they do is average all the grid data from the globe. Each grid cell is in turn averaged from the annual data which is in turn averaged from whatever increments (hourly?) the weather stations in that grid cell use. By the time one gets done with the average of the average of the average of the who knows what, we’ve got a cute graph showing a trend in T that is meaningless.
What we need (I think) is to go back to as granular a copy of the data as possible, daily at a minimum if not hourly, and convert each and every value for T in each and every grid cell to P via SB Law.
Once that is done, THEN average P over the course of a day and year and the globe and graph THAT trend. We could then look at the change in ACTUAL DIRECTLY MEASURED w/m2 and see what change has occurred over the course of the temperature record. I will bet that if there is a significant increase in w/m2 at all, the trend will be well under what we see by trending T calculated from a global average of an annual average of a daily average which tells us NOTHING about the change in energy balance of the planet over the course of time because the only way to calculate that is from the trend in average P, the trend in T is USELESS and misleading.
That’s quite beyond my spreadsheet skills however. Probably beyind the memory capacity of my aging notebook too.

January 14, 2012 12:48 pm

That’s quite beyond my spreadsheet skills however. Probably beyind the memory capacity of my aging notebook too.
Almost all of the things we are discussing aren’t spreadsheet computations. Nor are they particularly trivial numerical computations. Just covering a sphere with “cells” is nontrivial (think about the patches of leather that make up a soccer ball). The simplest coordinates are spherical polar, but they have a horrible “bias” in the Jacobian because of azimuthal compression at the poles. Integrating over a sphere is a pain in the ass, in other words (been there, done that).
rgb

Joel Shore
January 14, 2012 1:56 pm

davidmhoffer says:

Joel Shore;
blah blah blah>>>
Shrug.

What you are saying here is essentially, “Don’t bother me with details, such as the fact that the magnitude of the effect that I am making such a big deal about is demonstrably small. I will just continue to believe in the overwhelming importance of my result, because…Well, just because it is what I want to believe.”

Myrrh
January 14, 2012 2:04 pm

davidmhoffer says:
January 14, 2012 at 8:15 am
Myrrh;
I’m not interested in the blackbody arguments, far too much maths for me >>>
Shrug.
Chuckle. Par the the course, you’ve missed out “time” from my sentence and changed the meaning..
Robert Brown says:
January 14, 2012 at 10:09 am
Re Myrrh to davidmhoffer; “But… you’ve never come up with any proof! Most of the time what I get is the arrogant response from y’all to go read books on physics! Hello, is anybody in? I’m arguing about the physics in those books.. ”
The proof isn’t theoretical, although you do have to understand at least the idea that as far as the Earth as a whole is concerned, the fundamental issue is how hot the Earth has to be in order for radiation in to equal radiation out.
At that point, one goes up into space, measures the radiation coming out as a function of wavelength (or frequency, if you prefer) and goes “wow, the fact that the atmosphere is experimentally observed to take up some of the total heat delivered and radiate it at cooler temperatures than the surface means that etc.

Not what I asked you for, but interesting. Oh, did you miss my p.s. to you on your other thread re text books on this?
http://wattsupwiththat.com/2012/01/09/strange-new-attractors-strong-evidence-against-both-positive-feedback-and-catastrophe/#comment-862221
I asked you for proof of the claim you keep making that the ‘greenhouse gas warming effect is real’ and it’s therefore only the detail of that which should be argued about. As you’ve just written again here in this discussion:
Robert Brown says:
January 14, 2012 at 7:40 am

Re Joe says:
January 14, 2012 at 12:49 am
“When writing my first post last night, I realized an answer to question of whether the greenhouse effect is a real phenomenon. This answer is a resounding YES!”
Absolutely, which is why I try not to let it pass when people assert otherwise. The radiation data are absolute proof of it. The question isn’t “is there a GHG effect that contributes to the warming”, it is “what are the feedbacks, and what are the other mechanisms that also contribute to warming among which total observed warming (or cooling) must be split up”.

I asked you for proof of this. You haven’t actually come back with any proof. You don’t appear to have noticed that you haven’t come back with any proof or any detail of the experiments you say there’s an abundance of. Can’t you find any?
But meanwhile as I’ve given above, in the real world of convection and gravity in the fluid volume of gaseous ocean pressing down a ton on your shoulders,
YOU’VE MISSED OUT THE MAIN GREENHOUSE GAS WATER VAPOUR,
WHICH:
COOLS THE ATMOSPHERE BY 52°C.
COOLS IT DOWN TO 15°C.
Carbon dioxide is fully part of the Water Cycle, therefore, greenhouse gases COOL the Earth.
therefore, greenhouse gases COOL the Earth
The proof isn’t theoretical – just walk out into a desert.
Now imagine the whole Earth without the unique (to our present knowledge) WATER CYCLE.
Now, just how significant is any warming that the trace gas carbon dioxide can manage against the daily 52°C cooling of the Earth by the greenhouse gas water vapour? Of which it’s any way integrated into the cooling of that cycle as carbonic acid?
“Hello, is anybody in? I’m arguing about the physics in those books.. ”
Neither shall I let this pass – “The radiation data are absolute proof of it.”
It’s irrelevant against the great greenhouse gas cooling by water vapour in the Water Cycle.
Greenhouse gas warming is fiction not real world physics.
Now, back to your “The proof isn’t theoretical, although you do have to understand at least the idea that as far as the Earth as a whole is concerned, the fundamental issue is how hot the Earth has to be in order for radiation in to equal radiation out.
At that point, one goes up into space, measures the radiation coming out as a function of wavelength (or frequency, if you prefer) and goes “wow, the fact that the atmosphere is experimentally observed to take up some of the total heat delivered and radiate it at cooler temperatures than the surface means that the surface temperature is somewhat higher than it would be if it were an ideal black body.”
Now you have where it comes from – an Earth heated to 67°C directly by the straight lines of thermal energy, heat, from the Sun which is the invisible thermal infrared, of which the greenhouse gas water vapour removes 52°C from the surface by evaporation and release of heat in condensing out in the colder heights.
The “greenhouse effect” is not a real phenomenon – it is fiction not fact.
Greenhouse gases in the “greenhouse effect” do not warm the Earth from -18°C to 15°C
Greenhouse gases in the real world and real physics cool the Earth by 52°C to 15°C

January 14, 2012 2:05 pm

Robert Brown;
Almost all of the things we are discussing aren’t spreadsheet computations. Nor are they particularly trivial numerical computations. Just covering a sphere with “cells” is nontrivial >>>
Agreed again, but I think you missed the point.
What NASA/GISS and Hadcrut use, is gridded temperature data. They average that gridded temperature data spatially and over time and then trend it. They attribute the rising trend that results not just to warming in general, but to increased GHE from CO2 in particular. They even take their average of an average of an average to calculate what the surface radiance is, and they also take an average of an average of an average for insolation to calculate the “non GHE” temperature of the earth, and then subtract the latter number from the first to arrive at a conclusion as regards what the total GHE of the atmosphere is. This calculation is invalid.
The “average” of 240 w/m2 (P) cannot be used to arrive at an “average” T anymore than the average of T can be used to arrive at an average P. The relationship between T and P doesn’t exist. Only the relationship between P and T^4 exists. 240 w/m2 can arrive at a surface temperature of 255K if, AND ONLY IF, P is uniform in both time and space. It is not. The planet is exposed to insolation which varies from 0 to 1000 watts every day and does based on both latitude and time. The “real” effective temperature of earth calculated via blackbody would not only be less than 255K, it would be a LOT less. Similarly, one cannot average T across the temperature record in both time and space to calculate average T to determine how many degrees above 255K should be attributed to GHE. That results in an average of 288K and 390 w/m2 that is valid if, AND ONLY IF, T is uniform in both time and space. It is not. 288K is the lower bound of T for a surface emitting 390 w/m2 in a uniform temperature distribution, the actual T is higher.
So would my spread sheet idea provide the correct answers? No. But what it would provide is proper mathematical evaluation of the temperature record as maintained by HadCrut and GISS which would show conclusively the fallacy of two things. The first is the fallacy that the total GHE is 33 degrees. My back of the envelope estimate is about 150 degrees, commensurate with N&Z. The other thing it would show is that the supposed warming trend is due in part to the temperature of the earth becoming more uniform over the course of the temperature record (it has, tropics up 0.2 degrees arctic zones up 1 degrees) and that much of what is being attributed to CO2 is actually just a consequence of math, not a change in energy balance.
Is there NO change in energy balance? Probably not! The point is to take the data we have and see if we can come up with a way to extract the warming trend in T introduced by increasing uniformity in the earth’s temperature profile and then see how much is left over to attribute to other factors such as changing GHE.
Would the results be accurate? H*ll no! The data isn’t granular enough, it doesn’t even provide sufficient coverage of large tracts of the global surface. BUT, itz the only data we have, and we should deal with the only data we have via proper mathematical analysis rather than the total hocum that NASA/GISS and HadCrut are using to make conclusions about GHE and earth’s energy balance that they cannot possible make on the basis of the math they have used

Agile Aspect
January 14, 2012 2:34 pm

Responding to the title, the goal is estimate an upper bound on the surface temperature of the Earth – which is an exercise in thermodynamics.
If the only thermodynamic tool you have at your disposal is the Stefan-Boltzmann Law, then you’re screwed.

George E. Smith;
January 14, 2012 3:02 pm

“”””” Kevin Kilty says:
January 13, 2012 at 3:56 pm
George E. Smith; says:
January 13, 2012 at 11:12 am
…Anybody who chooses to “junk Stefan -Boltzmann”, is invited to step up, and get tarred and feathered in the public square; Well of course unless you can present a serious peer reviewed paper setting out your new theory to replace the junked science you chose to discard.
The Stefan-Boltzmann “law”, is nothing more nor less, than a total integral of the Planck formula for the Spectral Radiant Emittance of a BLACK BODY, at a uniform Temperature…….
It is becoming very difficult to decide who says what in these exchanges; so, if these aren’t your words, George, then I apologize in advance. All you say is so, however, when the aperture to a cavity acts as a band-limiting filter, then S-B, if you wish to define it as a fourth power T function, does not describe the emitted power. If, on the other hand, you look at SB as an integral over all wavelengths including the band weights, then I suppose you can claim the universality of SB. I suppose the long and short of this is that once active gasses enter a radiation problem, it becomes quite difficult to describe in terms of the simple S-B , T to the fourth power, function “””””
Kevin, essentially everything you excerpted above is what I wrote; but the quoted part:
“junk Stefan -Boltzmann”, was a verbatim extract from some total nonsense that Myrrh posted, so those originally were his words, that prompted me to post what I did. I have no idea why I bothered.
And I thought I explained the issue quite thoroughly; I don’t see how there can be any misunderstanding. Now anybody who chooses to challenge the truth or accuracy of what I wrote is of ourse free to do so; my attitude simply is: believe it or not, it’s ok by me.
So the essence of my message was this. Earlier researchers, notably Wien, Lord Raleigh, and James Jeans investigated the “expected” electromagnetic radiation spectrum from a COMPLETELY FICTIONAL IDEAL, NON EXISTING OBJECT called a BLACK BODY. No such object exists anywhere in the universe, so it is inherently impossible; no matter what, to experimentally observe the spectrum of the radiation of a BLACK BODY, since there is no such thing.
BUT, those researchers still considered it instructive to use the discipline of Statistical Mechanics; a well known branch of CLASSICAL PHYSICS, to investigat the expected radiant emission properties from such an object it such existed.
The BLACK BODY fictitious object they investigated, is defined by one simple property; well two actually. The first requirment, is that the fictitious body was comprised of a very large number of individual particles, which could be atoms or molecules or any other extremely large collection of particles, comprised of REAL MATERIALS HAVING A FINITE MASS. The other requirement was that the fictitious object should completely absorb any and all electromagnetic radiation that falls on it; period; without restriction or conditions.
Other than that, the theory placed no restrictions, and assigned no special properties to any of the particles in the collection; they are of undefined composition, unrelated to any known element, or chemical or any other known material; so no atomic or molecular structure or “energy ;evels”, or any other trappings of atomic or molecular physics is germane to their investigation.
Kirchoff’s law for the state of a closed system where the material is in equlibrium with the electromagnet radiation in the system, requires that the emission of EM radiation from such material must be at every wavelength or frequency identical to the amount of such radiation absorbed by the material. The required state of thermal equilibrium of course requires the whole system is at a uniform Temperature.
So the researchers sought to explain the OBSERVED emission spectra of carefully constructed systems, that approximated the conditions required of the mythical black body. Cavities with small apertures, were systems with such properties, and they showed observed spectra with constantly occurring similarities.
So they sought to describe the expected EMISSION SPECTRUM from their ideal BLACK BODY, AT A FIXED TEMPERATURE.
Wien discovered a mathematical form for the spectrum, that was not a bad fit to the observations from experimental approximations to a black body, particularly at short wavelengths; but it deviated at long wavelengths.
Jeans, and Raleigh on the other hand, using an analysis based on the equipartition principle, obtained a fomula, that gave good agreement at long wavelengths, but predicted an ever increasing amount of radiation at ever shorter wavelengths and consequently an infinite total energy emission; the so-called ultraviolet catastrophe.
now recall, this is a purely mathematical derivation, without reference to the known properties of any material; statistical mechanics,based on the equipartition principle, in this case assigning equal amounts of energy to each and every possible frequency or wavelength. So it was equipartition of radiant energy, rather than kinetic energy such as was used successfully in studies of the thermal capacities and specific heats of gases and solids.
So the failed spectral formulae for the black body spectrum, were purely mathematical exercises in statistical mechanics, and CLASSICAL PHYSICS. In particular, they did not even contemplate what the basic physics that caused the emission of ANY em radiation was, and it takes no part in the derivation.
Max Planck introduced a completely arbitrary restriction sans any precedent for such a rule. Namely that although there was to be equipartition of energy among all the possible radiating frequencies (which were infinite in number). He simply decreed thatthe energy aqssigned to any frequency must be a quantized function of that frequency; that is the energy had to be n.h.f where f was the frequency of the radiation, (h) was a new constant (Planck’s constant) and (n) is just some integer.
With that restriction, a higher frequency component, could only have a smaller set of values for (n) which prevented the total energy from taking off to infinity.
The result of that restriction was to lead to the Planck formula for the expected radiation from a fictitious BLACK BODY, that didn’t actually exist. The calculated spectrum was found to match with ever increasing accuracy, the observed of real approaches to making a body that absorbed all radiation that fell on them.
So the Planck radiation formula describes a fictitious object; not ANY real object. The formula is difficult to integrate in closed form over all frequencies, but when it is so integrated, the result it the Stefan-Boltzmann equation for the TOTAL RADIANT EMITTANCE OF A BLACK BODY
It isn’t the emittance for ANY other object.
So the Planck formula, and the S-B result are useful starting points to investigate emission from real objects. No real object obeys either the Planck radiation law, or the S-B law. In particular, the earth’s moon, is not even approximately close to being able to absorb ALL EM radiation that falls on it so the moon doesn’t obey either Planck or S-B, but despite Myrhr’s declaration; it doesn’t “junk” S-B, nor does NASA
The Planck Radiation spectrum, and the Stefan-Boltmann law, are results of classical physics; statistical mechanics; it isn’t quantum mechanics.

January 14, 2012 3:28 pm

Joel Shore;
What you are saying here is essentially, “Don’t bother me with details, such as the fact that the magnitude of the effect that I am making such a big deal about is demonstrably small>>>
I will ask you sir, to not put words in my mouth.
I am giving up because you raised the exact same issue using a nearly identical example in another thread and then completely ignored my response showing you where your error was and wht the magnitude must be considered from BOTH sides of the equation which you have not done either because you don’t want to or you don’t understand in the first place. There’s no point repeatedly going over known ground, hence:
Shrug.

Gary wilson
January 14, 2012 3:29 pm

Robert,
Good luck. I have been working on it in excel in my spare time with some success but there is only so far you can go.
One comment I would like to make is the idea that you can calculate average temperature of a body from the average radiant flux. This is the basis of the 255K and many other wrong conclusions. Consider 2 1m2 surfaces varying in radiative flux such that the average flux is constant . For emissivities of 1 and an average flux of 340 w/m2 this is what you get:
Q1 w/m2 Q2 w/m2 Q avge T avge C T1 C T2 C Actual T avge
680 0 340 5.275 57.926 -273.000 -107.54
612 68 340 5.275 49.324 -86.906 -18.79
544 136 340 5.275 39.971 -51.696 -5.86
476 204 340 5.275 29.696 -28.087 0.80
408 272 340 5.275 18.252 -9.824 4.21
340 340 340 5.275 5.275 5.275 5.27
As pointed out by Willis the larger the difference in temperatures or range the larger the difference between true average temperature and calculated radiant temperature.
These cases can represent two opposite sides of a revolving sphere or even two areas on the same side such as equator and poles.
Another important point is that the atmosphere and the oceans does more than the greenhouse effect in that it squashes the temperature range by mixing. By the above analogy this action alone raises the average temperature without greenhouse. Each case above can be viewed as heat being transferred from surface 1 to surface 2.
This begs the question whether the warming of the Arctic is due to more efficient mixing hence raising average temperature. If it is purely CO2 then why not the Antarctic as well. As ocean circulations change and mixing reduces will the Artic start cooling again and we will see global temperatures falling. Are most natural variations in GAT just variations in mixing and heat transfer from one area to another.
Robert, I also suggest you enrol yourself in Chem Eng 102 Radiant Heat Transfer (too complex for 101) and get yourself a copy of what was the bible some 40 years ago “Radiative Transfer” by HC Hottel and A F Sarofim

January 14, 2012 3:36 pm

Robert Brown;
Re; Myrrh
Myrrh doesn’t believe that visible light is capable of transferring energy to earth surface. He doesn’t believe that the physics equations in modern text books are the same as, and/or applied in the same manner, as the equations in physics text books from last centurey. Myrrh is of the belief that there is a giant conspiracy to hide the “real” physics and seems to think that there is some sort of “fake” physics that has been substituded in the text books for “real” physics just to support the GHG theory. How it is that these “fake” physics are used every day to design everything from bridges to hair dryers to night vision goggles to internal combustion engines to nuclear reactors with success he hasn’t been able to explain to me.
But he wants proof. Obviously itz your choice to waste time on him, or not. But I’ve found that reason, logic and facts are about as effective as:
Shrug.

Myrrh
January 14, 2012 3:47 pm

There’s lot’s of interest in this piece, but the something of interest I found about Arrehnius was specificially this:

http://greenhouse.geologist-1011.net/
The Shattered Greenhouse: How Simple Physics Demolishes the “Greenhouse Effect”.
“While Arrhenius credits Tyndall with the thermal buffer idea expressed in Plimer (2001) and Wishart (2009), he then goes on to express the more complicated idea described in Press & Siever (1982) and Whitaker (2007). The “atmospheric re-emission” that “helps heat the surface of the earth” of Whitaker (2007, pp. 17-18) is the key to Arrhenius’ original proposition, which revolves around the backradiation notion first proposed by Pouillet (1838, p. 42; translated by Taylor, 1846, p. 61). However, Pouillet used this idea to explain rather than add to the thermal gradient measured in transparent envelopes while, as we shall see, Arrhenius treated backradiation as an addition to the conductive (i.e. net) heat flow indicated by the thermal gradient.
…….
Moreover, when Arrhenius (1896, p. 255) added the radiative transfer between the earth’s surface and the atmosphere to the conductive transfer between the earth’s surface and the atmosphere, he effectively duplicated the radiative transfer quantity because it was already included in the conductive transfer quantity (“M”). This quantity is representative of net heat flow in accordance with Fourier’s Law which, further, does not distinguish between kinetic and radiative modes of heat transfer across a thermal contact

My bold.
Compositional variation can change the distribution of heat within a body in accordance with Fourier’s Law, but it cannot change the overall temperature of the body. Arrhenius’ Backradiation mechanism did, in fact, duplicate the radiative heat transfer component by adding this component to the conductive heat flow between the earth’s surface and the atmosphere, when thermal conduction includes both contact and radiative modes of heat transfer between bodies in thermal contact. Moreover, the temperature of the earth’s surface and the temperature in a greenhouse are adequately explained by elementary physics. Consequently, the dubious explanation presented by the “Greenhouse Effect” hypothesis is an unnecessary complication. Furthermore, this hypothesis has neither direct experimental confirmation nor direct empirical evidence of a material nature. Thus the notion of “Anthropogenic Global Warming”, which rests on the “Greenhouse Effect”, also has no real foundation.

wayne
January 14, 2012 3:50 pm

Bart says:
January 14, 2012 at 11:19 am
wayne says:
January 13, 2012 at 8:00 pm
” Latitudinal, the spacing is every three degrees. “
To get equal area coverage of a sphere, you should not space uniformly by latitude, but by vertical height. That is why you had to do the workaround you describe at January 13, 2012 at 8:35 pm. Remember, the area element is cos(theta)*d_phi*d_theta, where theta is the latitude. Transform to a new set of coordinates with z = sin(theta) and the area element becomes dz*d_phi.
— — —
Hi Bart. Thanks. I did realize that slicing even heights for the latitudes, as computing areas for spherical caps or latitude bands, is much more elegant and really more proper mathematically. I was just not sure at the moment that there might not be some math later to give me problems, like finding the proper cell centers and areas. I just stayed with what I could visualize and knew I could get there correctly though it was the long way around. Heh, one of my past supervisors said I always did look at all problems upside down and backwards, always wondered if that is not why I ended up with all of the very hard programming tasks that none of the other forty didn’t even want to think of touching. But I will look into switching to that form now that I have a set of numbers to compare against for correctness. Thanks again.
Bart says:
January 14, 2012 at 11:19 am
wayne says:
January 13, 2012 at 8:00 pm
” Latitudinal, the spacing is every three degrees. “
To get equal area coverage of a sphere, you should not space uniformly by latitude, but by vertical height. That is why you had to do the workaround you describe at January 13, 2012 at 8:35 pm. Remember, the area element is cos(theta)*d_phi*d_theta, where theta is the latitude. Transform to a new set of coordinates with z = sin(theta) and the area element becomes dz*d_phi.
— — —
Hi Bart. Thanks. I did realize that slicing even heights for the latitudes, as computing areas for spherical caps or latitude bands, is much more elegant and really more proper mathematically. I was just not sure at the moment that there might not be some math later to give me problems, like finding the proper cell centers and areas. I just stayed with what I could visualize and knew I could get there correctly though it was the long way around. Heh, one of my past supervisors said I always did look at all problems upside down and backwards, always wondered if that is not why I ended up with all of the very hard programming tasks that none of the other forty didn’t even want to think of touching. But I will look into switching to that form now that I have a set of numbers to compare against for correctness. Thanks again.

Myrrh
January 14, 2012 4:07 pm

Really very sorry Mods, it looks like there’s an italic problem in my post,
Myrrh says:
January 14, 2012 at 2:04 pm
It looks like I’ve put in an open italic where there shouldn’t be one, in front of the following :
“I asked you for proof of this. You haven’t actually come back with any proof. You don’t appear to have noticed that you haven’t come back with any proof or any detail of the experiments you say there’s an abundance of. ”
which comes after the blockquote beginning:

Re Joe says:
January 14, 2012 at 12:49 am
“When writing my first post last night, I realized an answer to question of whether the greenhouse effect is a real phenomenon. This answer is a resounding YES!”

apologies, please would you fix.
[Reply: this is a common WordPress glitch. It’s not the fault of any commenters. Fixed now. ~dbs, mod.]

Myrrh
January 14, 2012 4:20 pm

Brilliant, you worked it out before I’d even seen it. Many thanks.

Bill Illis
January 14, 2012 5:02 pm

davidmhoffer says:
January 14, 2012 at 11:57 am
What we need (I think) is to go back to as granular a copy of the data as possible, daily at a minimum if not hourly, and convert each and every value for T in each and every grid cell to P via SB Law.
Once that is done, THEN average P over the course of a day and year and the globe and graph THAT trend. We could then look at the change in ACTUAL DIRECTLY MEASURED w/m2 and see what change has occurred over the course of the temperature record. I will bet that if there is a significant increase in w/m2 at all, the trend will be well under what we see by trending T calculated from a global average of an annual average of a daily average which tells us NOTHING about the change in energy balance of the planet over the course of time because the only way to calculate that is from the trend in average P, the trend in T is USELESS and misleading.
———————-
I did something like this for one of the global SURFRAD stations monitoring solar radiation, downwelling IR, etc. over one 24 hour period.
http://www.srrb.noaa.gov/surfrad/index.html
The radiation flows do not match what the actual surface temperature is doing. The 2 metre air temperature changes by absolutely miniscule amounts on a per second basis while the radiation flows indicate temperatures should change by a wide margin.
http://img140.imageshack.us/img140/4109/tablemountainall.png
http://img12.imageshack.us/img12/3225/tablemountainnets.png
So, I did some further calculations and realized this is the general condition all over the world and there is even a greater out-of-balance for water surfaces, the tropopause etc. The changes in energy content are so small per second, that the radiation model does not work. We have to move down to another level to make sense of it.

wayne
January 14, 2012 5:03 pm

Robert: “Integrating over a sphere is a pain in the ass, in other words (been there, done that).”
My first was just about two month’s ago, monte carlo, and the next a week ago, cellular, and I have to agree, many choices and tradeoffs and hard to change tracks once the train is moving. Definitely not excel territory. 🙂
Did you see the results from a first run per your laid out cases of hypothetical planets?
http://wattsupwiththat.com/2012/01/12/earths-baseline-black-body-model-a-damn-hard-problem/#comment-863484
(C&D has wrong descripts, skip on down a few comments for details)

Richard G
January 14, 2012 5:47 pm

LazyTeenager says:
January 12, 2012 at 6:32 pm
******************
Hey Lazy, try your demonstration at 3 AM and tell me how it works out. For both cases. Inquiring minds want to know the out come of the real world test.
P.S Are you really a checkout clerk?
[REPLY: No he’s not. Don’t you think, maybe, that is a snark too far? -REP]

Myrrh
January 14, 2012 5:59 pm

George E. Smith; says:
January 14, 2012 at 3:02 pm
So the Planck formula, and the S-B result are useful starting points to investigate emission from real objects. No real object obeys either the Planck radiation law, or the S-B law. In particular, the earth’s moon, is not even approximately close to being able to absorb ALL EM radiation that falls on it so the moon doesn’t obey either Planck or S-B, but despite Myrhr’s declaration; it doesn’t “junk” S-B, nor does NASA
🙂 Well, whatever else it did, my choice of wording spurred you into an excellent, but what do I know?, explanation…
From which I can now see that NASA indeed did not junk SB and Planck, but used these imaginary constructs as a base for their calculations of the moon, much as those working with gases will begin with the ideal gas law which likewise is purely imaginary and which no real gas obeys and by stages add back all that is missing.
I think now this is of like ilk as I have previously pointed out elsewhere, that NASA has changed its story to fit in with AGW. NASA used to teach that the heat we feel from the Sun was the invisible thermal infrared, now it teaches that this doesn’t even reach us and has given the properties of heat to visible light, in the ‘shortwave in, thermal out’ meme of the energy budget that heats their imaginary land and oceans, which it can’t do in the real world.
So, what this is, is another example of the AGWScience Fiction meme producing department’s fiddling with physics to promote their truly fictitious ‘greenhouse gas warming’ and generally dumb down the education system, as it follows the same pattern I found in discovering that ideal gas law was used as proof that carbon dioxide diffused spontaneously into the atmosphere and could accumulate for hundreds and thousands of years regardless that its real gas weight, attraction etc. precluded such a thing happening.
So, it seems, that AGWSF uses SB&Planck in raw form without any tweaking for reality around us just as carbon dioxide, oxygen and nitrogen are designated the fictitious ideal gas by them and used without tweaking to explain such things as ‘well-mixed’. It’s not that NASA junked it then, but that now NASA teaches something different, hence all the confusion.
From the confusion by using the basic ideal gas, we now have taught that the atmosphere around us is empty space and these molecules zip around at vast speeds thoroughly mixing, that is, ideal gas without volume, gravity, attraction etc. And a whole generation of children have been educated into thinking that it is the atmosphere around us, because it is taught so, as the PhD in physics teaching the subject explained to me. Which is why I call it an imaginary world, because based on the imaginary ideal gas law without any tweaking.
But it has to be that to get rid of the Water Cycle..
[This teacher was first adamant that carbon dioxide could not separate out from the atmosphere in which it had spontaneously thoroughly mixed as per ideal gas law. I couldn’t quite believe that he thought this, let alone taught it, (he said he’d fail me for my views if taking the exams he set and marked), so I proposed a thought experiment to make sure I really understood what he was saying.
At first he adamantly denied that carbon dioxide could separate out of the atmosphere to pool at the ground after becoming thoroughly mixed as per AGWSF, but after I had shown him real world examples of carbon dioxide displacing air to pool on the ground, breweries etc., he conceded that it did ‘come down’ but created a novel explanation for it, that somehow it brought down the whole package of air it was in with it…
Anyway, the thought experiment was simply to imagine a room in which an amount of carbon dioxide had pooled on the floor and what happened next without any change to the conditions in which the carbon dioxide had pooled, no work done such as opening windows or putting on a fan.
He said that the carbon dioxide would rise spontaneously as per ideal gas and quickly diffuse into the atmosphere so thoroughly mixing that it would take a great amount of work to separate it out again.
I said it would remain pooled on the floor because it was heavier than air and so would not spontaneously rise into the atmosphere of the room without any work done.]
So, OK, how does AGWSF use the imaginary basic SB&Planck and what kind of world does that create if based without tweaking on this imaginary blackbody…?

January 14, 2012 7:31 pm

Bill Illis;
The radiation flows do not match what the actual surface temperature is doing. >>>
Thanks for those graphs! Actually, that’s pretty much what I would expect to see. In fact, if you sift through N&Z, that’s what they say to expect as well.
The surface temperature can’t possibly correlate well to insolation and net radition in the short term. The surface temperature is modified by net radiation, conductance, convection, and most importantly, heat capacity.
Your graphs show exactly why applying “average P” to arrive at “average T” for the earth makes no sense what so ever. The theoretical black body temperature of the earth is governed by P which varies by hundreds of w/m2 every day. If earth was a perfect black body that would mean that T would also vary by in excess of 300 degrees K every day. Instead, it varies by only a few degrees because heat capacity establishes a central range that the fluctuation can never stray very far from, and that value is further mitigated by convection and conduction which must speed up due to higher insolation and fall of during periods of low insolation (hence night time cooling slows down to a crawl even though insolation is zero and the theoretical black body of earth would be approaching absolute zero at night).
Earth surface as a whole must redistribute energy via ALL possible mechanisms until thermal equilibrium is reached. If the GHG’s were not present, the redistribution must still end up in the same place, just more gets done by other means. It doesn’t matter to the end state if the GHG’s are there or not, the redistribution stabilizes at equilibrium, it just perhaps doesn’t get there as quickly.
But changing the mass of the atmosphere DOES change things. It changes the heat capacity of the atmosphere and it changes conductance and convection. These factors in turn govern the distribution curve that results in equilibrium, and hence the observed average temperature canbe calculated from simply surface pressure and insolation.

Paul Vaughan
January 15, 2012 6:03 am

Dr. Brown,
The solid Earth does some of the integrals:
ftp://ftp.iers.org/products/eop/long-term/c04_08/iau2000/eopc04_08_IAU2000.62-now
ftp://ftp.iers.org/products/geofluids/atmosphere/aam/GGFC2010/AER/
“Apart from all other reasons, the parameters of the geoid depend on the distribution of water over the planetary surface.” — N.S. Sidorenkov
Sidorenkov, N.S. (2005). Physics of the Earth’s rotation instabilities. Astronomical and Astrophysical Transactions 24(5), 425-439.
http://images.astronet.ru/pubd/2008/09/28/0001230882/425-439.pdf
Gross, R.S. (2007). Earth rotation variations – long period. In: Herring, T.A. (ed.), Treatise on Geophysics vol. 11 (Physical Geodesy), Elsevier, Amsterdam, in press, 2007.
http://geodesy.eng.ohio-state.edu/course/refpapers/Gross_Geodesy_LpER07.pdf
http://geodesy.geology.ohio-state.edu/course/refpapers/Gross_Geodesy_LpER07.pdf

You mention the coupling of evaporation & wind.
Visualizing Shared Patterns:
1. Column-integrated Water Vapor Flux with their Convergence:
http://i51.tinypic.com/126fc77.png
Compare 1 with 2 & 3:
2. Near-Surface (850hPa) Wind:
http://i52.tinypic.com/nlo3dw.png
3. Near-Surface (850hPa) Wind — Polar View:
http://i54.tinypic.com/29vlc0x.png
Compare with wind-driven ocean gyres:
4. Wind-Driven Ocean Currents:
http://upload.wikimedia.org/wikipedia/commons/6/67/Ocean_currents_1943_%28borderless%293.png
Note the place of 2 & 3 (near-surface wind) in 5:
5. Zonal Wind Vertical Profile:
http://i51.tinypic.com/34xouhx.png
For another perspective on 5’s westerlies (mid-latitude ~200hPa spots), see 6 & 7:
6. 200hPa Wind:
http://i52.tinypic.com/zoamog.png
7. 200hPa Wind — Polar View:
http://i52.tinypic.com/cuqyt.png
Credit: Climatology animations have been assembled using JRA-25 Atlas [ http://ds.data.jma.go.jp/gmd/jra/atlas/eng/atlas-tope.htm ] images. JRA-25 long-term reanalysis is a collaboration of Japan Meteorological Agency (JMA) & Central Research Institute of Electric Power Industry (CRIEPI).
Regards.

Septic Matthew
January 15, 2012 5:19 pm

davidmhoffer: Don’t be silly, I said “at equilibrium” half a dozen times in first comment.
Ah. All of your subsequent comments apply to equilibria. I missed that.
Since the earth has never been in equilibrium and isn’t now, and isn’t projected to be (that is, no one has supplied any kind of proof that the dynamic system of the earth climate is capable of equilibrium), then you are not arguing about anything that actually exists. Dr. Brown’s post summarized the reasons why the climate system is not known accurately enough to predict what effects CO2 accumulation will have, and your comments have all been beside the point.

Septic Matthew
January 15, 2012 5:31 pm

davidmhoffer: 4. Averaging T to understand a change in energy balance is hopelessly useless.
I wrote that as well. However, computing the spatio-temporally averaged mean T is useful for estimating the net warming or cooling over a given time span, such as since 1850. Estimating the effect of CO2 on past or future warming isn’t possible right now. Claiming that you know that CO2 has no effect in the recent past, and can’t have an effect in the near future, if that’s what you have been trying to establish, is not substantiable.

January 15, 2012 9:24 pm

Setpic Matthew;
I wrote that as well. However, computing the spatio-temporally averaged mean T is useful for estimating the net warming or cooling over a given time span, such as since 1850. >>>
No! it does not! The energy balance can be completely unchanges down to the last joule but a change in the distribution (cooler tropics warmer arctic for example) will show up as a positive temperature trend if you average T, but the total watts in and the total watts out haven’t changed.
If the earth is “warming up” and by “warming up” we mean retaining more energy than it is losing, averaging T may STILL give you the wrong answer. It is possible to have an average of T that is increasing while the energy retained is decreasing and vice versa, which I’ve shown via example calculations on several occassions.
You CANNOT average T and calculate P, nor may you average P and calculate T! Doing so arrives at meaningless numbers from which only incorrect conclusions may be drawn.

January 15, 2012 9:33 pm

Septic Matthew;
Dr. Brown’s post summarized the reasons why the climate system is not known accurately enough to predict what effects CO2 accumulation will have, and your comments have all been beside the point.>>>
If you can say that, then I’m sorry to say that you’ve completely missed the point.

Mark T
January 15, 2012 10:26 pm

Point him to the definition of an intensive variable.
Mark

Joe
January 15, 2012 10:50 pm

This has been a very instructive interchange but it appears to be petering out -> finally! I can rarely spend much time reading blogs and comments except late at night and I could not spend the time on this one to get more involved. However, the experience has sharpened my understanding of the problems with the debate itself, not just the science. There are too many terms, labels, pre-accepted conclusions, incomplete physics, improperly applied mathematical relationships, shortcuts, and limited models upon which people on both sides, both amateur and professional, base their arguments. Even more noticeable is that some of these things apparently mean different things to different people participating in the same discussion. Nevertheless, if the foundation of a model is weak, no amount of arguing will make the result true, even if the argument is won. In seems that winning the argument is becoming more important than the truth of understanding. This “debate club” mentality has completely overwhelmed political discussion in the US and is now entering the realm of science as well. Let’s get it right, not win arguments. I was pleasantly surprised at the mostly calm interchange of comments to this post but that may have been due to the moderator! That said, I am no saint myself. I can sometimes get it embarrassingly wrong and I have in the past publicly crashed and burned!
I am not being critical of the commenters to this blog, which no doubt range from interested non-technical people to highly educated scientists. Obviously Dr. Brown in his posting and everyone who commented are passionate and want to get it right. I am critical more of the climate community itself which supports many “facts” which would not pass muster in a 2nd year undergraduate engineering class (Yes, I am an engineer!) or directly contradict 100 years of research by geologists, a similarly well developed scientific discipline. For instance, pick up a geology guide at the Yellowstone National Park Visitors’ Center and read what the Grand Teton glaciers did during the Little Ice Age that did not occur.
Below are my responses to those who thought enough of my posts to comment on them for better or worse:
Tom Folkert:
In answer to my question about where there is surface cooling to counter the Arctic heating in order to make the average rate of change, Tom said:
>>The short answer is that that place is the “top of the atmosphere”.
>>Adding more CO2 raises the average altitude for emission of IR. Higher
>>altitudes are colder, so the CO2 will emit less IR.
Thanks for pointing out that I was not thinking 3-dimensionally about the temperature problem. I am a pilot and I should have known better. Nevertheless, I do not think you can claim that surface heating can be countered by top-of-the-atmosphere cooling to make the average. First, the average is calculated only from surface temperatures and is not integrated over the entire height of the atmosphere. It is correct to compare top-of-the-atmosphere photonic emission with surface temperature in a balance equation? They are two different things. Secondly, since the entire argument about radiative balance means that energy reaching the surface must also leave the earth’s emissions sphere at the same location it entered, you cannot make the argument that the top of the atmosphere cools as the surface heats up. That implies, from an engineering point of view, that the thermal conductance of the atmosphere decreases as its temperature goes up. That would form an unstable system resulting in the surface temperature running away while the top of the atmosphere freezes, clearly something that is not physically realizable. Third, if the equal-intensity emission surface rises in altitude, it by definition still emits the same amount of energy so the energy transfer has not changed. You actually can make the argument that the atmosphere got hotter. If you mark a coordinate on the emission surface above my favorite spot over the equator, after that surface rises to a higher altitude because of heat absorption by carbon dioxide in the lower atmosphere, the temperature at my fixed coordinate, at the original altitude but now below the emission surface, most likely increased. The upper atmosphere got hotter, not cooler. Finally, the photonic energy emitted by the carbon dioxide comes from the change in a vibrational state of the molecule, not from a change in its temperature, which as you know is the linear velocity of the molecule. That the carbon dioxide is cooler (lower linear velocity) when the iso-emission surface rises should not change the emission statistics for that vibration state. Am I wrong about that?
The iso-emission surface is a mathematical concept that I suspect was created to be a summary of a variety of mechanisms, a summary that could then be used to make communication more efficient. It is my observation that it has created confusion. I prefer to reduce things to the absolute. For instance, the iso-emission surface is often described as a sphere representing the altitude at which 240W/m2 is emitted from the earth. It is a sphere only if you use the average emission value. In reality it should touch the earth’s surface at the poles (assuming that the earth’s axis is not tilted and energy balance is achieved only by local outbound radiation) and reaches very high in the atmosphere over the equator where the incoming sunlight reaches 1366W/m2. If you do construct a spherical emitting surface at some fixed altitude above the earth, it will have a different temperature at each point and will emit at a different intensity at each point. That sphere will form a potential surface. The derivative of a potential surface, that is its local slope, defines the “force” between two points on that surface that transfers energy from one point to the other. In non-technical terms, the same thing happens that I described in my first post: at a specified altitude, the potential surface will have slope from the equator to the poles, forcing energy to flow to the poles to be radiated .
Tom, your observation has made clear to me a very important point I had not considered before and have never seen mentioned: Energy comes into the system photonically and can only leave the system photonically because hot gas molecules cannot leave Earth’s gravitational field like steam leaves a boiling pot of water. No matter how the sloar energy bounces around inside our system, to leave our system it must excite a state in a molecule, a liquid, or a solid that can convert it back to a photon.
David M Hoffer:
>>If you scroll back upthread to Joe’s first comment on the issue,
>>you’ll see that he pretty much nailed exactly where the “break even”
>>latitude is just from theoretical physics.
Thanks!
Robert Brown of Duke:
My model did not get more complex than a non-rotating black body plus the differential heat load on the globe due to the local slope of the globe. I did not begin to add the levels of complexity you mentioned in your post but already the boundary limits of my world differ from the climate models. You were very right!
Tom Folkert
>>So if you had a container with a mixture of hot N2 & CO2,
>>the N2 could lose some energy via radiation, but it would lose
>>more by transferring energy to CO2 via collisions, and then
>>having the CO2 radiate the energy to space.
I think you are right about nitrogen’s rate of radiation. I ran across information about carbon dioxide lasers when researching another topic and found that they are like a fluorescent light bulb but with nitrogen and carbon dioxide inside the tube. The electron current hits the nitrogen molecules and excites them. The nitrogen molecules do not readily emit but do lose their energy to the carbon dioxide molecules by collision. The carbon dioxide emits the energy as photons, albeit at the famous 10 micron wavelength!
This description raises the following question: if the atmosphere is populated with excited carbon molecules and IR energy of the same wavelength travels only upwards from the surface, is there any stimulated emission from the excited carbon dioxide? I think the yes or no of this question is determined solely by how long the atmospheric carbon dioxide stays excited before colliding with another molecule and de-energizes. If the answer is yes, it would mean that a cubic meter of atmosphere does not emit its IR radiation in all directions equally but instead preferentially in the direction of space. It may be only a very small percentage or even a fraction of a percent but it would change the outcome of the energy balance equations for the green house effect.
Good night everyone! I enjoyed the discussions!

Septic Matthew
January 15, 2012 11:24 pm

davidmhoffer: No! it does not! The energy balance can be completely unchanges down to the last joule but a change in the distribution (cooler tropics warmer arctic for example) will show up as a positive temperature trend if you average T, but the total watts in and the total watts out haven’t changed.
If the earth is “warming up” and by “warming up” we mean retaining more energy than it is losing, averaging T may STILL give you the wrong answer. It is possible to have an average of T that is increasing while the energy retained is decreasing and vice versa, which I’ve shown via example calculations on several occassions.

Ah yes, I have written myself that the distribution can change. If that was the point that I missed, I do agree that I missed it.

Septic Matthew
January 15, 2012 11:35 pm

dmh, it looks like I shall have to reread everything that you wrote.

Septic Matthew
January 15, 2012 11:46 pm

dmh: From Joe’s comment upthread:
“The primary mechanism responsible for maintaining the incredible 4-billion-year stability of our system must be the physical transfer of heat from the equator to the poles where it is radiated away.”
Precisely. The laws of thermodynamics require this statement to be true. The amount of net loss of energy in high latitudes must balance, to the last photon, the net absorption in low latitudes. Not only does this satisfy the laws of thermodynamics, it also explains why the concentration of GHG’s in the atmosphere is, in fact, immaterial.

The “stability” is not “equilibrium”, and includes sufficient variation to include climates that are hostile to human civilization. The laws of thermodynamics do not require the balance that you claim, as they only require that all the heat be accounted for; you can have transient net influxes and net efluxes.

January 16, 2012 1:08 am

Septic Matthew;
The “stability” is not “equilibrium”, and includes sufficient variation to include climates that are hostile to human civilization. The laws of thermodynamics do not require the balance that you claim, as they only require that all the heat be accounted for; you can have transient net influxes and net efluxes.>>>>
For the second time, please note the number of time I said “at equilibrium” in my comments.
Further, please note the title of the thread which begs the question as to how to calculate the blackbody temperature of the earth. Calculating blackbody is, by definition, a calculation of the equilibrium state. Further, the laws of thermodynamics must absolutely be satisfied under ALL circumstances, they don’t get an exemption for transient states. These include transient states in which some of the energy is being asorbed (for example) by heat capacity during the transient state. The numbers must balance to satisfy the laws of thermodynamics even when some of the numbers come from processes that do not exist in equilibrium states.

January 16, 2012 9:37 am

I guess it depends what you are setting out to prove as to whether all the effort is worthwhile. perhaps Hans Jelbring made a smart move with his model atmosphere and isometrically heated planet surface in his 2003 paper:
http://tallbloke.wordpress.com/2012/01/01/hans-jelbring-the-greenhouse-effect-as-a-function-of-atmospheric-mass/
At least if he is right you won’t have to worry about including radiative effects of GHG’s

OK, here’s my first referee comment to this paper (which I would not have accepted). He creates an artificial adiabatic planet. He asserts that the atmosphere is in thermal equilibrium. He explicitly states indeed that the average kinetic energy of the molecules of the atmosphere is a constant throughout. Then he proceed to derive a lapse rate for this atmosphere.
No. Equipartition states that T is proportional to the internal energy per molecule. If the atmosphere is in equilibrium, it is at a uniform temperature T. Period. In fact, this is the zeroth law of thermodynamics, the definition of thermal equilibrium.
http://en.wikipedia.org/wiki/Zeroth_law_of_thermodynamics
See especially the discussion on thermal equilibrium wherein it is proven that if two systems i and j (or two parts of one system) are in thermal equilibrium, T_i = T_j. End of story. So there can be no adiabatic lapse rate for a gas in thermal equilibrium, that is, a gas in a container, no matter how you apply external forces or accelerate the system to create a pressure gradient. It is entirely possible to have a gas in a vertical column where the pressure drops off with height all at the same temperature, and if the gas is in equilibrium it is at the same temperature.
Thus I don’t understand Jelbring’s argument from the beginning. He asserts equilibrium, but then imposes an adiabatic lapse rate in temperature that contradicts equilibrium, at least thermal equilibrium. To put it another way, the molecules of gas at the cooler temperature have to have a completely different maxwell-boltzmann distribution of temperatures, do they not, with completely different average speeds? They are not in thermal equilibrium.
If you have a thermal gradient in any system with a mechanism for the transfer of energy, you have a heat transfer from hotter to colder in any steady state solution to the problem. This too is a law of thermodynamics (or nearly so) — the second law. Heat is irreverisbly transferred only from hot (higher T) reservoirs to colder (lower T) reservoirs, not the other way around. If it ever goes the other way around, you can create heat engines that violate the second law. In the case of your adiabatic lapse rate atmosphere, one could just put an electrical thermocouple between the upper and lower atmosphere and generate electricity forever, right? Gravity somehow causes the temperature to re-equilibrate with a gradient, but the gradient will drive a current that makes a resistive wire hot in one place.
So Jelbring’s paper egregiously violates the laws of thermodynamics on page 2 or 3. Not just a little, mind you — a lot. Enough so that I can build a perpetual motion machine quite easily (and, mentally, just did).
The only way you can have an adiabatic lapse rate is if you have a differential delivery of heat to the upper and lower surfaces that maintains them at different temperatures. You have to have a constant flow of heat, and the system is never in global thermal equilibrium, quite the contrary.
This is where any static model for a thermal gradient to the atmosphere fails before it gets off of the ground. The thermal gradient is not due to statics. It is caused by heat flow, and if the lapse rate is a vertical decrease in temperature, that means that heat is always flowing from the hot lower to the colder upper.
“Thermodynamics” is tricky for open systems. You can talk about local thermal equilibrium — a sufficiently large volume of well mixed air or water might have a “temperature” relative to a thermometer placed in contact only with that limited volume — but if there is heat flow and thermal gradients, it isn’t in thermodynamic equilibrium globally. Theoretical computations of e.g. adiabatic lapse rate do not assume global thermal equilibrium, they assume local thermal equilibrium and adiabatic expansion consistent with the pressure. I’m not certain that they are completely consistent, but perhaps they are sufficiently so, given the relative timescales in question.
I have much the same issues with N&Z, BTW. I’d have to see how the temperature differential is maintained that doesn’t involve differential radiative cooling rates (greenhouse) somewhere in the system and that doesn’t allow me to build a perpetual motion machine across the thermal gradient. Gravitation is reversible and can never replenish energy in a steady state system, provide additional free energy to drive an engine. You need heat input from the Sun, and you need direct atmospheric cooling to keep the top of the atmosphere cooler than the surface.
I think that both Jelbring and N&Z have things upside down, especially with regard to venus. Venus’s atmosphere is optically thick in pretty much all relevant wavelengths. The atmosphere itself reaches thermal equilibrium with a thermal gradient.
rgb

Editor
January 16, 2012 11:38 am

Robert Brown says:
January 16, 2012 at 9:37 am

OK, here’s my first referee comment to this paper (which I would not have accepted). He creates an artificial adiabatic planet. He asserts that the atmosphere is in thermal equilibrium. He explicitly states indeed that the average kinetic energy of the molecules of the atmosphere is a constant throughout. Then he proceed to derive a lapse rate for this atmosphere.
No. Equipartition states that T is proportional to the internal energy per molecule. If the atmosphere is in equilibrium, it is at a uniform temperature T. Period. In fact, this is the zeroth law of thermodynamics, the definition of thermal equilibrium.
http://en.wikipedia.org/wiki/Zeroth_law_of_thermodynamics
See especially the discussion on thermal equilibrium wherein it is proven that if two systems i and j (or two parts of one system) are in thermal equilibrium, T_i = T_j. End of story. So there can be no adiabatic lapse rate for a gas in thermal equilibrium, that is, a gas in a container, no matter how you apply external forces or accelerate the system to create a pressure gradient.
… [much good stuff snipped]

Thanks, Robert, With great trepidation, I must disagree with you.
Consider a gas in a kilometre-tall sealed container. You say it will have no lapse rate, so suppose (per your assumption) that it starts out at an even temperature top to bottom.
Now, consider a collision between two of the gas molecules that knocks one molecule straight upwards, and the other straight downwards. The molecule going downwards will accelerate due to gravity, while the one going upwards will slow due to gravity. So the upper one will have less kinetic energy, and the lower one will have more kinetic energy.
After a million such collisions, are you really claiming that the average kinetic energy of the molecules at the top and the bottom of the tall container are going to be the same?
I say no. I say after a million collisions the molecules will sort themselves so that the TOTAL energy at the top and bottom of the container will be the same. In other words, it is the action of gravity on the molecules themselves that creates the lapse rate.
What am I missing here?
Many thanks for all your contributions to the discussion,
w.

Editor
January 16, 2012 11:44 am

Robert, one further thought. You might think that this would be a violation of the conservation of energy, because you can use that temperature difference to do work.
But if you do so, you remove energy from the tall container, so it is not a perpetual motion machine. You still have to add energy to the system to get continuous work out of it.
w.

Septic Matthew
January 16, 2012 1:00 pm

davidmhoffer: Further, the laws of thermodynamics must absolutely be satisfied under ALL circumstances, they don’t get an exemption for transient states.
I didn’t say that the laws of thermodynamics were ever not satisfied. I said that Earth may transiently receive more energy from the sun than it radiates to the rest of space.
No matter how many times you used the word “equilibrium” it is clear from your writing that you address non-equilibrium cases — unless you are asserting that the earth has actually been in equilibrium.
Further, please note the title of the thread which begs the question as to how to calculate the blackbody temperature of the earth. Calculating blackbody is, by definition, a calculation of the equilibrium state.
This is one of the ways in which the calculations can not be assumed to be accurate for the earth; that is, inaccuracies of at least a few percent are introduced by using equilibrium approximations to model processes on earth. The temperature of the earth, its mean and its particular temperature in each place, fluctuates. This indicates that the earth system is not in thermal equilibrium. Sometimes it is unclear whether you really mean “equilibrium” in place of “steady state” or “within the basin of attraction”. In equilibrium there is no heat flow; in steady state the inflows equal the outflows, for the whole earth as well as for every 3D region of it; within the basin of attraction you can have heat flows (and temperature changes) within a range (like claimed for earth above) without exact periodicity, and without steady-state for any duration of time. Strictly speaking, we can not know for sure that we are within a basin of attraction, but the earth climate system is surely never at equilibrium or in steady state.

Pierre R Latour
January 16, 2012 2:38 pm

Robert Brown repeated a common misunderstanding at 1008, 13Jan2012 with: Anything with a temperature radiates.
The better way to write it is: The Stefan-Boltzmann equation assigns a radiation temperature vector equivalent and proportional to the radiation intensity vector of any black body that radiates.
Horses come before carts. S-F simply assigns a measurable radiating temperature vector to a mathematically defined black body radiating vector.
George E Smith got it right at 1502, 14Jan12. He does not confuse radiation temperature vectors with thermal temperature scalars.
The language in the GHG theory debate is appallingly confusing to laymen. As Prof Richard Lindzen, MIT Meteorologist, is so fond of saying, the science of meteorology is indeed in its infancy.
This professional chemical process control system engineer says the quality of the multivariable dynamic matrix model relating independent inputs like fossil fuel combustion rate and solar incidence to atmospheric CO2 content and both thermal and radiant temperatures of Earth’s surface and atmosphere are utterly inadequate for designing Earth’s thermostat. Control engineers developed the commercial art of rigorous and empirical modeling to suit an engineering purpose in the 1970’s; GHG theorists are apparently oblivious to its existence. I proved this chemical process system is not measurable, unobservable and uncontrollable in 1997.
I have not found a rebuttal to my WUWT essay of 0718, 13Jan12 explaining why the 33C GHG effect is a whatchamacallit.

dr.bill
January 16, 2012 3:14 pm

@Willis

If you gentlemen will pardon me for intruding between you, I don’t think there’s a conflict between the two notions, and I’m referring here to the real world, not the one that Willis has going on the other thread. ☺
Anyway, whether by conduction or convection or intelligent design (!), if the column of air has arrived at a stable state with hotter stuff at the bottom and cooler stuff at the top, it will only stay in that condition if continually heated from the bottom, thus satisfying Robert’s need for a perpetual energy flow in order to maintain the gradient. If the column didn’t have that energy input, it would gradually lose energy and settle to the ground until limited by increasing density. I don’t quarrel about the system not being in TD equilibrium, it won’t be, but it will be ‘stable’ as long as the energy keeps coming.
By the way, I have greatly enjoyed Robert’s many contributions of late. I am about to start back to classes a week from today, and as luck would have it (well, careful planning actually), I’m teaching a section of Thermodynamics. I may “poach” some of the turns of phrase that I’ve seen you use. ☺
/dr.bill

George E. Smith;
January 16, 2012 3:32 pm

“”””” Myrrh says:
January 14, 2012 at 5:59 pm
George E. Smith; says:
January 14, 2012 at 3:02 pm
So the Planck formula, and the S-B result are useful starting points to investigate emission from real objects. No real object obeys either the Planck radiation law, or the S-B law. In particular, the earth’s moon, is not even approximately close to being able to absorb ALL EM radiation that falls on it so the moon doesn’t obey either Planck or S-B, but despite Myrhr’s declaration; it doesn’t “junk” S-B, nor does NASA
🙂 Well, whatever else it did, my choice of wording spurred you into an excellent, but what do I know?, explanation… “””””
You know Myrrh; you just done gone and earned my respect.
It takes a real man to admit that maybe he has it wrong; and that IS the learning process.
I don’t know it all not even a tiny bit of it; and I have put my foot in it so many times; just ask Phil about some of my bobby dazzlers.
But I am willing to make the effort, to try and reduce parts of this stuff; the small parts I understand, to a level where ANYBODY can understand it.
And I think you have grasped it. The Planck and S-B constant (it’s a fundamental Physical constant) ONLY WORKS for a theoretical gizmo, that doesn’t really exist.
Fortunately we can and do make some very close approximations to a black body at some specific Temperature. You can actually buy a “Freezing Copper” black body cavity, that is set up to heat copper until it melts, and then let it refreeze as it slowly loses energy (radiation (NOT HEAT)), and then when it reaches the freezing point of copper, the Temperature fall stops, while the copper gives up it’s “latent heat” of melting, and while it holds at the copper freezing Temperature, the cavity aperture will emit a nearly perfect black body radiation spectrum centered on that Temperature’s peak wavelength. When the copper has all frozen, then the Temperature will continue to drop. Hopefully you got your measurements made during the freeze.
Filthy rich people can have a platinum freeze black body source. I dunno how filthy they are but they have to be rich.
So although the Planck and S_B relations don’t apply to anything real, it is a very good starting point as MANY real objects follow it fairly well OVER USEFUL WAVELENGTH RANGES
And that point is important.
The black body radiation (Planck) curve has 25% of its energy at wavelenghts shorter than the spectum peak, and 75% at longer wavelengths. It has only 1% at wavelengths shorter than 1/2 of the peak wavelength, and only 1% at wavelengths longer than 8 times the peak. So 98% of the eergy is between 0.5 and 8.0 times the peak wavelength, adn for most purposes that’s plenty good enough.
So for the solar spectrum at 6,000 K which peaks at about 0.5 microns, the 98% range is from 250 nm to 4.0 microns. For the earth surface emitted long wavelength infra-red for say 300 Kelvin, the peak is at 10.0 microns, so the useful range is 5.0 to 80.0 microns, which covers the important 15 micron CO2 band, and the 9.6 micron Ozone band.
You see it really doesn’t matter if a 300 K body goes out of whack (invites junking) at wavelengths less than 5 microns, or longer than 80 microns, because we aren’t expecting to have any interest out there for the 300 k earth surface, so if it is fairly close within that 98% spectrum range, we feel justified in using Planck and S-B to describe it; at least until we learn it is further out of whack.
So I have hopes for you Myrrh, congratulations.

January 16, 2012 5:07 pm

What am I missing here?
The definition of equilibrium.
You keep talking about things falling. Nothing is falling. The air column is static. If it is truly adiabatic (isolated) and you wait a long time, it will reach equilibrium. The definition of equilibrium is uniform temperature. I’m tempted to say “end of story” (because it is) but I will relent and give you a very, very simple example.
Suppose you have two containers of oxygen. The left hand container has a pressure of P. The right hand container has a pressure of 2P. We’ll assume that we’re far away from any sort of critical point so that the oxygen is an “ideal gas” or near enough, although that doesn’t matter. Make it a van der Walls gas, who cares?
Put the two in “thermal contact” and otherwise wrap them in insulation. Wait. When you come back you find a) The first container has half the temperature of the second; or b) The two have the same temperature; or c) something else (your choice) independent of the initial temperature of the gasses in the containers?
When you understand the correct answer to this question, you will understand that a pressure gradient per se has nothing to do with being in thermal equilibrium.
You can also consider Mr. Ocean. Below the thermcline, the Ocean has a temperature that is constant to within a hair (within a degree C). Of course the
pressure increases by an atmosphere every ten meters or so.
Note well that temperature has nothing to do with energy density in the example above. It has to do with the energy distributed per degree of freedom in the system. The number of degrees of freedom for oxygen at normal roomish temperatures is 5 — 3 translation and two rotation — per molecule. Heat it up enough and it goes up first to six, then to seven (or more) as one excites additional modes (there is one more rotational mode and vibrational modes but there is a quantum barrier that prevents them from participating in the sharing of energy close to room temperature — not enough molecules that can provide a full quantum of energy). Read:
http://en.wikipedia.org/wiki/Heat_capacity
especially remarks on monoatomic and diatomic gas, as well as:
http://en.wikipedia.org/wiki/Equipartition_theorem
see especially figure 4.
Again — if gravity created a spontaneous, permanent temperature gradient, our energy problems would be over. We would just generate electricity by putting a heat engine between the hot side and the cold side and wait for gravity to re-separate out the molecules like a little “Maxwell’s Demon”. But sadly (well, really it is rather fortunate, as violating energy and entropy rules would be “bad” as far the overall consistency of the Universe is concerned) reversible laws of nature like gravity do not.
This isn’t terribly easy to understand, I agree. Introductory thermodynamics is the undergraduate class that is a strong competitor to quantum mechanics in terms of level of difficulty. Statistical mechanics (the sound theoretical basis for thermodynamics) is more difficult than quantum theory. It is easy even for physicists to state things that are wrong, that make no real sense, that violate the laws of thermodynamics. This particular argument seems to be at the heart of one big, long, extended mistake: the idea that gravitational compression creates a static, stable, temperature gradient in a fluid. It does not. Period. To get a temperature gradient in anything, you have to have heat flow (or work being done) and gravity does no net work on a gravitationally confined fluid in static equilibrium.
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January 16, 2012 5:11 pm

But if you do so, you remove energy from the tall container, so it is not a perpetual motion machine. You still have to add energy to the system to get continuous work out of it.
Not at all. I use the energy to drive a fan — in the gas. All the heat I turn into work remains, as energy, in the gas. The fan just runs forever.
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January 16, 2012 9:34 pm

Anyway, whether by conduction or convection or intelligent design (!), if the column of air has arrived at a stable state with hotter stuff at the bottom and cooler stuff at the top, it will only stay in that condition if continually heated from the bottom, thus satisfying Robert’s need for a perpetual energy flow in order to maintain the gradient. If the column didn’t have that energy input, it would gradually lose energy and settle to the ground until limited by increasing density. I don’t quarrel about the system not being in TD equilibrium, it won’t be, but it will be ‘stable’ as long as the energy keeps coming.
Agreed. The reason the atmosphere takes on the thermal profile that it does is, to put it mildly (having just read through Caballero’s chapters on fluid thermodynamics again) “complicated”, but Caballero actually has as an exercise showing that thermodynamic equilibrium in a static fluid is uniform temperature both horizontally and vertically.
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January 16, 2012 9:58 pm

Robert Brown repeated a common misunderstanding at 1008, 13Jan2012 with: Anything with a temperature radiates.
Well, there were a few assumptions in there. The matter has to be charged matter, for example. The emissivity may be very low for quantum mechanical reasons. The matter has to be present in sufficient density and with sufficient interaction coupling so that the concept of “temperature” is itself relevant.
But beyond that, yes, anything with a temperature radiates. I didn’t say that it follows a blackbody radiation curve, mind you, only that as charged particles accelerate, they radiate, and as charged matter not in its ground state interacts (even quantum mechanically) there have to be very special, unusual circumstances for there to be no electromagnetic coupling at all connecting the excited states to the ground state, no channels at all for the system to “cool” by emitting photons.
As was correctly pointed out, for transparent gases (low absorptivity and emissivity) the thermally generated radiation may be very weak, but it is not zero.
OTOH, “dark matter” and “dark energy” may not radiate. No charge (apparently) and/or no electromagnetic coupling — hence it is “dark”. But I await definitive confirmation, some way of seeing dark matter before I completely believe in it or worry about it further, and it is irrelevant to the current discussion.
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January 16, 2012 10:21 pm

Joe says:
January 15, 2012 at 10:50 pm
This has been a very instructive interchange but it appears to be petering out -> finally! I can rarely spend much time reading blogs and comments except late at night and I could not spend the time on this one to get more involved…

Joe, you would find reading, and working through, Caballero to be very instructive, because many of your remarks indicate a misunderstanding of how radiative balance is managed. It isn’t local immediate equilibrium, because the temperature of everything is constantly changing — either warming or cooling, and doing so differently at different heights. All that matters is whether total energy absorbed on average over years equals total energy radiated away on average over years. If not decades. It doesn’t matter at all where the energy is incident on the Earth — usually “somewhere sunlight is illuminating” since the Sun is the big source of external energy — or where it is emitted as radiation — from the surface of the ground, from the top surface of a cloud, reflected from the surface of the ocean, from water vapor at some height, from CO_2 at a different height, from O_3 at still a different height, or whether the heat was absorbed in the tropics but radiated away near the north pole. All that matters is that on average ins equal outs to maintain (on average) a constant temperature. To the extent that they never precisely balance minute to minute, hour to hour, day to day, the temperature everywhere is in flux, warming here, cooling there, and the Earth’s outgoing radiation balance is similarly fluctuating with respect to time of day, albedo, latitude, time of year, emissivity, what the major oscillations and weather are doing to the jet stream, sea surface temperatures.
So it is important to understand that radiating some of the long wavelength outgoing radiation from the top of the troposphere instead of directly from the ground does lead to higher average temperatures on the ground. However, it’s not at all clear that adding more CO_2 will substantially alter the effective radiation height of the upper atmosphere — the atmosphere is optically thick already as far as the CO_2 band is concerned and the radiation is more or less emitted (within an optical path of) the top of the troposphere already. There isn’t any room for it to move higher — it runs into the stratosphere! In fact, if CO_2 concentrations did indeed push it much higher, one might actually see the CO_2 band emission temperature increase, and faster cooling. Unless the addition of CO_2 is going to somehow increase the actual thickness of the troposphere while maintaining the same lapse rate, I suspect that we are well into a regime where additional CO_2 has almost no effect on the outgoing radiation profile, which may be why it has been very difficult to directly observe in e.g. NASA IR spectroscopy of the surface. At the very least the signal is buried in noise that is orders of magnitude greater.
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Editor
January 17, 2012 1:57 am

Robert Brown says:
January 16, 2012 at 5:07 pm

What am I missing here?

The definition of equilibrium.
You keep talking about things falling. Nothing is falling. The air column is static. If it is truly adiabatic (isolated) and you wait a long time, it will reach equilibrium. The definition of equilibrium is uniform temperature. I’m tempted to say “end of story” (because it is) but I will relent and give you a very, very simple example.

Thank you for the explanation, Robert. I fear it didn’t respond to my question:

Consider a gas in a kilometre-tall sealed container. You say it will have no lapse rate, so suppose (per your assumption) that it starts out at an even temperature top to bottom.
Now, consider a collision between two of the gas molecules that knocks one molecule straight upwards, and the other straight downwards. The molecule going downwards will accelerate due to gravity, while the one going upwards will slow due to gravity. So the upper one will have less kinetic energy, and the lower one will have more kinetic energy.
After a million such collisions, are you really claiming that the average kinetic energy of the molecules at the top and the bottom of the tall container are going to be the same?

I don’t see anything in there about “things falling”. I asked about whether after a million collisions where one atom goes up and decelerates and one goes down and accelerates the KE is the same top and bottom. I don’t see why that should be. And unfortunately, although your explanation was clear, it didn’t answer my question.
What is wrong with what I describe here? I’m starting to think you may be right, but I don’t understand why my description above is inaccurate. I thought that the relation between potential and kinetic energy was the standard explanation for the reason that there is a DALR. It equalizes the energy in the gas, and it is the reason for the “g” term in the equation for the DALR, g / Cp.
So why would that not work in a sealed container? What is different in a sealed container than out here on the surface of the planet, that out here we have a DALR, and inside the sealed container there is none?
Many thanks for your answers to my questions,
w.

January 17, 2012 2:04 am

With regard to gas at equilibrium in a gravitational field, it appears that the Velasco et al. paper discussed here http://tallbloke.wordpress.com/2012/01/04/the-loschmidt-gravito-thermal-effect-old-controversy-new-relevance/#comment-13608 says that there is indeed a lapse rate but that it’s negligible for gas-molecule ensembles the size of our atmosphere.

wayne
January 17, 2012 3:28 am

Dr. Brown, there is a paper http://arxiv.org/PS_cache/arxiv/pdf/1003/1003.1508v2.pdf by Gerlich and Tscheuschner and they end up with exactly the opposite of your stance on what you and Willis were discussing above, whether a lapse would naturally form under gravity in an idealized column of air. Have you read it and have any thoughts? I’m leery (and a bit confused after this last week).
They summarize:

Results
By combining hydrodynamics, thermodynamics, and imposing the above listed assumptions
for planetary atmospheres one can compute the temperature profiles of idealized atmospheres.
In case of the adiabatic atmosphere the decrease of the temperature with height is described
by a linear function with slope −g/Cp, where Cp depends weakly on the molecular mass. …

Not trying to put you on the hot seat but I cannot seem to find the error in their derivative and equations but also I am rather new to thermodynamics. Two weeks ago I firmly fell as you do, isothermic, the I waivered. The courses I just went through seemed to not delve to such depths. It may just be they end up on the last page combining in a manner that ends up breaking the zero’th law… that is according to your explanation.
This seems to be one of the most misunderstood areas that should be well known but brother, such diverse views!

Bryan
January 17, 2012 4:13 am

wayne
G&T say
” In case of the adiabatic atmosphere the decrease of the temperature with height is described
by a linear function with slope −g/Cp, where Cp depends weakly on the molecular mass. …”
However earlier they say that the in a mixture of gases such as air the radiative contributions are included in the bulk thermodynamic quantities such as Cp.
Certainly looking up the tables and comparing say N2 and CO2 over the atmospheric temperature range 250K to 350K we find that
N2 changes by 0.04% i.e. almost constant
CO2 changes by 13% !!!!!!!
Why does CO2 change so much?
In the case of CO2 extra degrees of freedom over and above translational become available corresponding for instance to wavelengths 15um and 4um.
G&T warn here is an obvious danger here that if an attempt is made to do a separate purely radiative calculation then some energy could be counted twice.

dr.bill
January 17, 2012 5:16 am

re wayne, January 17, 2012 at 3:28 am :
Hi Wayne,
Pardon me for responding to a question directed to someone else, but I think what’s missing in the interpretation of “eventual equilibrium” is the final physical location of the gas molecules. If you keep heating at the bottom, you can make the top of the gas column stay where it is, or rise, or fall by some amount, depending on the energy input, and you can get a stable lapse rate if you persist with the same amount of heating for sufficent time.
If, however, the column receives NO heating, it will eventually end up as a “dense as possible” thin layer on the ground, and THAT layer will eventually attain the same temperature as the surface it’s sitting on (assumed to have a never-varying temperature), and the gas will have a uniform temperature throughout its now “very short” height.
/dr.bill

January 17, 2012 5:56 am

Willis:
Pardon me for butting into your colloquy with Dr. Brown, particularly because this is a non-rigorous answer. For the time being, I believe the rigorous answer is given by the Velasco et al. paper I mentioned above.
But a non-rigorous but more intuitively appealing answer may be that a lower number of higher-altitude molecules are being knocked upward by a larger number of lower-altitude molecules being knocked downward by fewer higher-altitude molecules, and the difference maintains goes into maintaining the greater potential energy.
Not rigorous, but maybe it helps?

Joe Postma
January 17, 2012 8:09 am

===================================
Willis Eschenbach says:
January 16, 2012 at 11:38 am
Consider a gas in a kilometre-tall sealed container. You say it will have no lapse rate, so suppose (per your assumption) that it starts out at an even temperature top to bottom.
Now, consider a collision between two of the gas molecules that knocks one molecule straight upwards, and the other straight downwards. The molecule going downwards will accelerate due to gravity, while the one going upwards will slow due to gravity. So the upper one will have less kinetic energy, and the lower one will have more kinetic energy.
After a million such collisions, are you really claiming that the average kinetic energy of the molecules at the top and the bottom of the tall container are going to be the same?
I say no. I say after a million collisions the molecules will sort themselves so that the TOTAL energy at the top and bottom of the container will be the same. In other words, it is the action of gravity on the molecules themselves that creates the lapse rate.
==================================
Willis this is a very nice description of the physics. But consider not a million collisions, but ~10^30 particles undergoing ~10^9 collisions per second.
I would like to point out that the atmosphere should not be referred (as some people have) to a as a static fluid – it is not, it is a compressible gas. Fluids don’t compress, gases do.
Also, the equation for balance of energy is very simple and is discussed at great length in these two papers, which I will re-link to again:
http://www.tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf
http://principia-scientific.org/publications/The_Model_Atmosphere.pdf
For a gas in a “general” state, or near-state, of thermal equilibrium (the total energy in the gas column 1m^2 is about ~10^7 Joules and the daily variation about this value is some very small percentage – this will be shown in my upcoming paper on the differential heat-flow equation which characterizes this air column), there is a balance between kinetic and potential energy. That very simply, as Willis very beautifully described it, leads to more kinetic energy being found at the bottom of the air column and less at the top, and hence the gas temperature tracks that. Combine that with the definition of what an average means, and you must conclude that the bottom of the atmosphere HAS to be warmer than the average of the entire thermodynamic ensemble average. Again, please refer to the linked papers for an in-depth discussion of that. The definition of an average temperature means that ~half of the particles WILL be hotter than the average. That the half are hotter doesn’t mean conservation of energy has been violated! Such is the meaning of “average”, but when you apply the meaningless conservation of power in place of conservation of total energy, it really confuses the thermodynamics.
Lastly, here is a video simulation of an ideal gas column in a gravitational field:

This simulation can be scaled up to a very large number of particles, at the expense of CPU time, but you actually don’t need to. The nice thing about ideal gases is that they scale – i.e., the behavior of a 3-particle ensemble might be quite unique, but as soon as you start getting up to ~a few dozen particles in a confined space the general behavior of the ensemble will be essentially identical to any other larger set of particles in the same space. This is the basis of statistical thermodynamics, of course, if I may not have described it very well.
In any case, the simulation lets you SEE this balance between kinetic and potential energy taking place, and also density. There is a higher density of particles near the ground and they are also bouncing around faster (the data can be extracted from the sim to show this), thus, more dense and more hot near the surface. And then cooler and less dense at altitude.
So, the thermal equilibrium profile of a compressible gas column in a gravitational field IS one of the temperature distribution described. That IS its thermal equilibrium state. Start with U = CpT + gh : that simple. Thermal equilibrium as equating to a constant, uniform temperature throughout is only one possible state for a given type of ensemble to which that end-state would apply – a metal bar heated on both ends, perhaps. But given that pressure changes non-linearly with altitude when in thermal equilibrium, and there’s no problem with that, there’s no reason to assume temperature can’t also have a distribution. There IS energy coming in to the system (the Sun) and energy leaving (LW radiation), and it is roughly equal, and that keeps the column “afloat”.
Also, given the fact that U = CpT + gh leads to the exactly-measured DALR in reality, I would say it’s a successful application of basic physics. G&T are correct in stating that the radiative effects are already included in the Cp, and that counting them over again is a double-counting. This was described and explained very nicely by Timothy Casey here:
The Shattered Greenhouse: http://greenhouse.geologist-1011.net/

January 17, 2012 8:58 am

The Postma animation above is not inconsistent with my (non-physicist’s) interpretation of the above-mentioned Velasco et al. paper, which does conclude that there’s a lapse rate at equilibrium. But (again, if my calculations are correct) it says that the lapse rate is negligible for as many molecules as are in our atmosphere.
I’m just a layman, but I have elsewhere implored others to show me where this (non-intuitive) conclusion is wrong, and I have so far received no convincing response.

January 17, 2012 9:05 am

I don’t see anything in there about “things falling”. I asked about whether after a million collisions where one atom goes up and decelerates and one goes down and accelerates the KE is the same top and bottom. I don’t see why that should be. And unfortunately, although your explanation was clear, it didn’t answer my question.
Imagine a plane surface in the gas. In a thin slice of the gas right above the surface, the molecules have some temperature. Right below it, they have some other temperature. Let’s imagine the gas to be monoatomic (no loss of generality) and ideal (ditto). In each layer, the gravitational potential energy is constant. Bear in mind that only changes in potential energy are associated with changes in kinetic energy (work energy theorem), and that temperature only describes the average internal kinetic energy in the gas.
Here’s the tricky part. In equilibrium, the density of the upper and lower layers, while not equal, cannot vary. Right? Which means that however many molecules move from the lower slice to the upper slice, exactly the same number of molecules must move from the upper slice to the lower slice. They have to have exactly the same velocity distribution moving in either direction. If the molecules below had a higher temperature, they’d have a different MB distribution, with more molecules moving faster. Some of those faster moving molecules would have the right trajectory to rise to the interface (slowing, sure) and carry energy from the lower slice to the upper. The upper slice (lower temperature) has fewer molecules moving faster — the entire MB distribution is shifted to the left a bit. There are therefore fewer molecules that move the other way at the speeds that the molecules from the lower slice deliver (allowing for gravity). This increases the number of fast moving molecules in the upper slice and decreases it in the lower slice until the MB distributions are the same in the two slices and one accomplishes detailed balance across the interface. On average, just as many molecules move up, with exactly the same velocity/kinetic energy profile, as move down, with zero energy transport, zero mass transport, and zero alteration of the MB profiles above and below, only when the two slices have the same temperature. Otherwise heat will flow from the hotter (right-shifted MB distribution) to the colder (left-shifted MB distribution) slice until the temperatures are equal.
The only way for this not to be true is to have a “Maxwell’s Demon” at the interface. Gravity is reversible and cannot act as a Maxwell’s Demon. Basically what I am saying is that a state where the two slices are at different temperatures is unlikely compared to a state where they are the same — there are many, many more ways to arrange all of the total energy of the system where the internal kinetic energy per molecules is on average the same in the two slices than there there are ways to arrange it so that they are separated, as you can understand from the simple description of the process describing transfer of heat above. The entropy of the system is greater when the temperatures are the same. The two-color (otherwise identical) salt grains are mixed, not separated, as you bounce them around, even in gravity.
This argument works for any partitioned gas, and is how one proves (with a bit more work) that thermal equilibrium is equivalent to “the same temperature throughout” very nearly independent of anything. I mean that this statement is true for a staggering array of cases where it can be shown explicitly or measured, so much so that it is the Zeroth Law of thermodynamics at this point, not really an option. Thermal equilibrium is isothermal as long as there is any pathway, reversible or not, between two reservoirs and quite independent of whether or not the reservoirs are stacked vertically or horizontally or have the same or different pressures or have the same or different gases in them in the same or different initial states — after all of the mixing and sharing of energy is done, the equilibrium system is isothermal.
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Joe Postma
January 17, 2012 9:16 am

Sorry for double-posting this video but I want to make another point with it:

I don’t know if everyone here has an intuitive-mind for physics, but when I look at that simulation (and having wrote it) I can see in my head exactly what would happen if you introduced additional parameters.
Imagine you take this simple gas, and add a molecule which has various internal degrees of freedom that can be excited by collision. When these internal degrees of freedom become excited, they DO so because they have absorbed kinetic energy from the collision. Kinetic energy is therefore taken out of the aggregate ensemble, to be converted into individual internal energy states of the additional molecule(s). A new equilibrium state of the aggregate ensemble will thus be established, and that state MUST have lower kinetic energy, because a small parcel of energy was “lost” to an internal state of the molecule(s).
Now, as long as that internal energy isn’t lost, this new equilibrium state will be constant, though cooler than before. However, now consider that the internal energy of the molecule can be exited from the aggregate system, by radiating the energy away. That parcel of energy then leaves the ensemble completely. Soon after, the internal state will again be re-activated by another collision, and thus another parcel of energy is taken out of the system. Now you keep on running that sequence, and that is how radiative emission causes kinetic cooling. Even for a simple gas, some small bit of radiation is lost just in the collisions themselves, though others have pointed out it is a very small rate. Molecular emission greatly amplifies the radiative loss, then, of course, and thus causes cooling.
So this is another project I want to finish: to write that ideal-gas simulation including molecules with internal states. It seems obvious to conclude initially that having some (non-radiating) molecules with internal states (initially inactive) will cause a “relaxation” of the kinetic thermal profile due to the energy being taken out of such, into the internal states activated by collision, compared what said profile would be otherwise.
If you then turned on radiative emission for those molecules you would obviously have eventual complete collapse of the whole gas column due to cooling.
So therefore you’d have to, and could, provide additional input kinetic energy to the gas simply via its collisions with the ground surface, which would be akin to the sunlight-heated surface transferring heat to the gas. I’ve approximated some of that already and seen the results, just by making the collisions damped with inelasticity.
Anyway, it’s pretty clear that having molecules with internal states causes damping of the kinetic, i.e. thermal, collisions, and that should relax the entire thermal profile.

Joe Postma
January 17, 2012 9:41 am

That’s a very good description Robert, just above in your last post. Painted a very nice mental picture of what’s going on, which I really like!
So, agreed that the same number of particles moving up, as moving down, between two slices.
The upward moving particles, from below, are initially warmer, say. The downward moving particles from above are initially cooler. Is this a stable equilibrium?
Or, lets change that and assume they are at the SAME temperature, above and below.
The upward moving particles have to lose kinetic energy, however, to potential energy. At some point, if they’re unimpeded, they actually pass though a complete zero of kinetic energy (vertically), and that has to be a “cooler” aggregate state than the maximum of kinetic energy they have at the surface.
So, the up-moving particles lose energy, the down-moving ones gain energy. The average of the whole ensemble is somewhere in the middle.
I am not sure there is a demon. Yes the number of particles passing through an infinitesimal slice is equal; however, they enter into a volume with less kinetic energy than they just previously had…so they have to be cooler.
So perhaps, on average, the particles would have equal kinetic energy when passing through each particular infinitesimal slice. But, the number of particles passing through each slice decreases with altitude (re: pressure, and not all of them bounce that high). And because the up-moving ones must be losing kinetic energy, and the kinetic energy becomes zero at the highest altitude, there must also be a decrease in the kinetic energy of the particles with altitude, i.e. temperature.
So the number of particles passing through each infinitesimal slice decreases with altitude, and they have to have a lower kinetic energy at each slice, with altitude, but each slice has on average an equal number of particles passing through it up and down.

Joe Postma
January 17, 2012 9:50 am

===========================================
Joe Postma says:
January 17, 2012 at 9:41 am
So the number of particles passing through each infinitesimal slice decreases with altitude, and they have to have a lower kinetic energy at each slice, with altitude, but each slice has on average an equal number of particles passing through it up and down.
===========================================
Actually, maybe the simpler thing to think about, would be: it is not quite true that each infinitesimal slice has a perfectly equal number of particles passing THROUGH it up and down: some of the upward moving particles, following a parabolic path due to gravity, just perfectly climax directly on the slice,and never actually pass completely through it and move above it. Hence, pressure gradient, fewer particles with altitude, and a thermal gradient. There’s a demon called gravity pushing some of the particles back down below the slice, never letting some though it.

January 17, 2012 10:02 am

I am not sure there is a demon. Yes the number of particles passing through an infinitesimal slice is equal; however, they enter into a volume with less kinetic energy than they just previously had…so they have to be cooler.
No, they don’t, because the mean free path is smaller than the thickness of my slices. Otherwise one cannot assign the slices a “temperature”.
Nothing is changed if you put a thin perfectly conducting barrier in between the slices and insist on the number of collisions above and below being equal. Again, the thermal distribution of speeds within the mean free path of the surface has to be identical.
You’re on the right track trying to build an applet. I’m looking for one that is already built, myself — I’m lazy. A simple MD simulation that sums and averages molecular speeds as a function of height whenever the mean free path is small compared to secular coarse grained distances will make the point. Wait a long time and you’ll observe equipartition of energy, precisely as you would if you inserted a whole series of aluminum foil “pistons” in between layers of gas.
The point is that you will violate the zeroth law and detailed balance of energy transport unless each degree of freedom has 1/2 kT energy, quite independent of whether or not gravity acts. This is independent of the pressure, the density, the number of molecules, the kinds of molecules. It is true for a mixture of gases, gases at different pressures and densities of different kinds in different containers. It is true for solids in contact with gases, liquids in contact with solids, everything in or not in gravitational fields. Equilibrium is isothermal.
How many times do I need to assert this? Do you really think that every thermodynamics textbook in the Universe is wrong and we’re just now discovering it?
rgb

January 17, 2012 11:36 am

Robert Brown: “Do you really think that every thermodynamics textbook in the Universe is wrong and we’re just now discovering it?”
It does seem arrogant, doesn’t it? Still, I don’t think that fairly describes what some of us think. What I for one think is that the Velasco et al. physics-teachers-journal comment I mentioned above says something different from what you do: it says gravity does indeed impose a lapse rate at equilibrium. If so, might not other physics folks say something similar?
Velasco et al. may be an outlier. More probably, I just don’t understand the (to this layman, daunting) math it contains. But as citizens we can’t simply throw up our hands and accept the word of whoever claims to be authoritative. My experience at least is that episodes in which that has been done tend to end badly.

Joe Postma
January 17, 2012 12:37 pm

Well there’s no question about equipartition of energy. But the energy available to equipart changes with altitude…it decreases. The available energy is not only specified by the aggregate sum divided equally to each slab, it is also specified by the altitude of the slab. U = CpT + gh. At each slice the energy is equiparted, surely, but when a particle falls it gains kinetic energy, when it rises it loses kinetic energy.
The relative difference between the slice thickness and MFP changes with altitude due to the change in pressure. I think it’s better to consider a true slice, an infinitesimally thin slice, because that’s simpler. Then, there’s a slab above and below the slice. The pressure of these slabs is certainly different. The total number of particles in each is different. And there’s some small number of particles that they’re constantly exchanging. Go to the next slice above the slab, and there’s fewer particles being exchanged, etc.
How do you explain, then, that the DALR is perfectly described and derived by U = CpT + gh ? The total energy of a slab of air is given by its thermal capacity, its temperature, and its potential energy i.e., by its kinetic and potential energy. If the total energy of the slab is not changing because it is in equilibrium with the inputs and outputs, then DT/Dh = -g/Cp. And that’s what is observed…we observe a temperature decrease with altitude, of exactly this value.
Maybe, then, we need not to be so bounded by the restriction of thinking of the gas column as being in thermal equilibrium. There’s obviously a great thermal load emplaced in the day time at the bottom of the column.
It may be correct that in an ideal-gas thermal equilibrium in a gravity field, the T should be uniform. But if that’s true then I don’t know why U = CpT + gh still gives a perfect description of the atmosphere, seemingly quite accidentally. I can run that sim and extract the speed distribution from slabs at various heights…something tickles my brain that I may have done that once already and found the result you insist..not sure. But I can also run with with damping included to see if that changes things, because damping certainly is occurring and the gas is not perfectly ideal.
We observe a temperature decrease with altitude and it is described nicely by the equation U = CpT + gh : the sum of kinetic and potential energy. If it’s a radiative effect that’s causing this and it is outside the bounds of the standard physics of kinetic & potential energy, then the DALR is still incorporated via the Cp parameter to make that equation still work. It means interpreting that equation as we intended to set it up, as a balance of kinetic and potential energy, is in error, and that what the equation represents, bc of the Cp parameter, is the resulting distribution due to the radiative effects included in the Cp parameter. But then you would think that CO2 should have a gigantic Cp so that it can dominate the DALR since the rest of the atmosphere is said to have little radiative significance. But we all know how the average Cp of the atmosphere is calculated and CO2 actually has a negligible contribution to it!
UPDATE: I went away and ran my sim to collect some data.
I have ran my ideal-gas simulation and extracted the speed distribution for two different sets of particles: at a snapshot occurring every 200 particle collisions I collect the data, and keep re-collecting it over a long period, letting it average out, building up the two speed-distributions. One set of particles is all of those between the arbitrary height of 0.5 & 1.0 in the sim, and the other set of particles is those corresponding to a height of 4.0 or above. The average height is about 2.5. MFP is close to that as found at STP, ~10 molecular diameters.
After a ridiculous number of collisions/data collections, two speed-distributions start to come up. As far as I can tell by looking at the graph, they peak at the same place, which means they have the same temperature. This is what Robert is saying should happen, and thus is why it is confusing that U = CpT + gh -> dT/dh = -g/Cp works.
So all this is very strange. -g/Cp works but Cp is dominated by radiatively inert gases. We observe -g/Cp in reality and it seems U = CpT + gh is a perfectly good way of characterizing the energy, and we would have thought this in the first place, before we even looked, that there should be a thermal gradient in the atmosphere when energy input = energy output, due to this very simple physics.
Maybe we just can’t think of the situation in terms of thermal equilibrium with the restrictions as described by Robert. There is equilibrium in regards to the basic energy coming in = going out, but thermal equilibrium as strictly defined, with isothermal end-state, it is not, by simple observation of course. There’s “new” energy always coming in (daytime) pumping up the system, and there’s “old” energy always being lost due to radiation.
The analogy would be to a metal bar heated on one end, with a cold-sink on the other, with either end held fixed in temperature by the source/sink. There’d be a thermal energy flow down the bar and that coming in would be equal to that leaving, but there would certainly be a temperature distribution down the length of the bar and it would eventually become static – unchanging but not uniform. You might call this “equilibrium” but it isn’t isothermal thermodynamic equilibrium as described by Robert.

Joe Postma
January 17, 2012 1:28 pm

Perhaps this is relevant:
http://en.wikipedia.org/wiki/Non-equilibrium_thermodynamics
“Temperature for bodies in a steady state but not in thermodynamic equilibrium:
While for bodies in their own thermodynamic equilibrium states, the notion of temperature safely requires that all empirical thermometers must agree as to which of two bodies is the hotter or that they are at the same temperature, this requirement is not safe for bodies that are in steady states though not in thermodynamic equilibrium. It can then well be that different empirical thermometers disagree about which is the hotter, and if this is so, then at least one of the bodies does not have a well defined absolute thermodynamic temperature.”
So, steady state but not thermodynamic equilibrium, sounds like the atmosphere.
“Temperature for bodies not in a steady state:
When a body is not in a steady state, then the notion of temperature becomes even less safe than for a body in a steady state not in thermodynamic equilibrium. This is also a matter for study in non-equilibrium thermodynamics.”
That sounds even more like the atmosphere, because it isn’t actually perfectly steady-state.
So the atmosphere is basically the worst-case study for applied thermodynamics. It’s not in steady state and not in equilibrium; it is close to being steady-state but still not in equilibrium. So the requirement of iso-temperature is in fact not there in the theory.
But we do still have the extensive/intensive properties of parcels of gas, i.e. thermal kinetic and potential energy, and somehow the equation for this results in the observed thermal profile of the atmosphere. I guess the “somehow” is due to the requirement of iso-temeprature not applying to the thermodynamic state of this system, since the system doesn’t fit the required parameters for such a state to be achieved in any case.

Joel Shore
January 17, 2012 2:28 pm

The lapse rate in the Earth’s atmosphere is well-understood: If convection were not possible, an atmosphere could have any lapse rate it wanted depending on where energy is absorbed and emitted. Once one considers the possibility of convection, it is found that lapse rates larger than the appropriate adiabatic lapse rate are unstable to convection and hence convection occurs and lowers the lapse rate back down to the appropriate adiabatic lapse rate.
This is why the troposphere, strongly heated from below and cooled from above, is approximately at the adiabatic lapse: It would be at an even higher lapse rate if convection could not occur but since convection is sparked, the lapse rate is lowered back down to the adiabatic lapse rate.
And, it also explains why other parts of the atmosphere, like the stratosphere are not at the adiabatic lapse rate. For these regions, the heating is such that the lapse rate is less steep than the adiabatic lapse rate (or even has temperature INCREASING with height) and hence it is stable and convection is suppressed.
The important thing to recognize is that the lapse rate alone does not determine the surface temperature. To determine the surface temperature, you must also know the temperature at some height. The constraint actually ends up being that the temperature at the effective radiating level has to be the Earth’s blackbody temperature of 255 K. Where the effective radiating level is depends on the opacity of the atmosphere to terrestrial radiation, i.e., on the greenhouse effect. For the Earth with its present constituents, the effective radiating level is about 5 km and the environmental lapse rate is about 6.5 K per km yielding a surface temperature of about 255 K + (6.5 K/km)*(5 km) = 287.5 K. As the levels of greenhouse gases in the atmosphere increases, the effective radiating level goes up and (to first order) the lapse rate doesn’t change. As a result, the surface temperature increases. For example, if the effective radiating level moved up to 6 km, then the surface temperature would rise to 255 K + (6.5 K/km)*(6 km) = 294 K.

Bryan
Reply to  Joel Shore
January 17, 2012 3:13 pm

Joel Shore says
.
“The lapse rate in the Earth’s atmosphere is well-understood: If convection were not possible, an atmosphere could have any lapse rate it wanted depending on where energy is absorbed and emitted.”
This is an odd way to express an idea as well as being wrong.
What is meant by “atmosphere could have any lapse rate it wanted “?
It seems to allow the atmosphere a consciousness to want something.
Its a bit like another statement often heard but is still nonsense
“nature abhors a vacuum”
The atmosphere at times has a still air condition(no convection) know as the neutral atmosphere.
It then follows rigidly the DALR of g/Cp = – 9.8 /Km in the Earths tropopause

Editor
January 17, 2012 3:56 pm

Dr. Brown, thank you so much. After following your suggestion and after much beating of my head against Caballero, I finally got it.
At equilibrium, as you stated, the temperature is indeed uniform. I was totally wrong to state it followed the dry adiabatic lapse rate.
I had asked the following question:

Now, consider a collision between two of the gas molecules that knocks one molecule straight upwards, and the other straight downwards. The molecule going downwards will accelerate due to gravity, while the one going upwards will slow due to gravity. So the upper one will have less kinetic energy, and the lower one will have more kinetic energy.
After a million such collisions, are you really claiming that the average kinetic energy of the molecules at the top and the bottom of the tall container are going to be the same?

What I failed to consider is that there are fewer molecules at altitude because the pressure is lower. When the temperature is uniform from top to bottom, the individual molecules at the top have more total energy (KE + PE) than those at the bottom. I said that led to an uneven distribution in the total energy.
But by exactly the same measure, there are fewer molecules at the top than at the bottom. As a result, the isothermal situation does in fact have the energy evenly distributed. More total energy per molecules times fewer molecules at the top exactly equals less energy per molecule times more molecules at the bottom. Very neat.
Many thanks,
w.
Cross-posted to the “Matter of Some Gravity” thread.

Myrrh
January 17, 2012 6:45 pm

George E. Smith; says:
January 16, 2012 at 3:32 pm
……………
Thank you, George. Much appreciated.
Pierre R Latour says:
January 13, 2012 at 7:18 am
GHG Theory 33C Effect Whatchamacallit
Further to my comment on this to you here and to an exchange I had about it here: http://wattsupwiththat.com/2012/01/15/sense-and-sensitivity-ii-the-sequel/#comment-866656
I have expanded on it here: http://wattsupwiththat.com/2012/01/13/a-matter-of-some-gravity/#comment-867720
Just to keep you in the loop because I’ve quoted you, and thank you, your post helped concentrate my attention on an aspect that’s been increasingly bothering me..

willb
January 17, 2012 7:35 pm

Robert Brown says:
January 17, 2012 at 10:02 am

No, they don’t, because the mean free path is smaller than the thickness of my slices. Otherwise one cannot assign the slices a “temperature”.

This seems to me to be a dicey premise to be using as input to your argument for isothermal equilibrium. The premise implies that, over a tiny altitude extent (something greater than the mean free path), the temperature does not change. In essence, you are assuming dT/dz = 0 over your slice. This assumption might be what your reasoning is extropolating over all z, resulting in a lapse rate of 0 everywhere.

January 18, 2012 9:53 am

Robert Brown:
What willb said.
Not that I think there’d be a detectable temperature lapse rate at equilibrium. But I do find persuasive the argument, advanced in the Velasco et al. paper I mentioned above, that it would be non-zero (but negligible) in theory. Yes, it’s something of a how-many-angels-can-dance-on-the-head-of-a-pin type of distinction, but I would have thought a physics professor would find that important pedagogically.

Robert Clemenzi
January 18, 2012 1:01 pm

Joe Postma says:
January 17, 2012 at 12:37 pm

If the total energy of the slab is not changing because it is in equilibrium with the inputs and outputs, then DT/Dh = -g/Cp. And that’s what is observed…we observe a temperature decrease with altitude, of exactly this value.

Thanks. I think exactly explains the problem. The DALR is just that – adiabatic. The explains the change in temperature with a change in altitude when no additional energy is added to the system.
However, as long as the maximum surface temperature is larger than the temperature of the air above it, additional energy will enter the atmosphere. The same argument applies as long as any layer of air is warmer than the layer above it – energy will move from warmer to colder and not follow the adiabatic (energy not changing) equations.

We observe a temperature decrease with altitude and it is described nicely by the equation U = CpT + gh

That does not describe the measured lapse rate. It does describe the temperature change of a parcel of air that rises because it is less dense than the surrounding air.

Robert Clemenzi
January 18, 2012 1:09 pm

Willis Eschenbach says:
January 17, 2012 at 3:56 pm

At equilibrium, as you stated, the temperature is indeed uniform. I was totally wrong to state it followed the dry adiabatic lapse rate.

And from this it follows that greenhouse gases cool the atmosphere. To be clear, without greenhouse gases the troposphere would not get cooler with increasing height.
From that, it immediately follows that increasing the amount of any greenhouse gas in the atmosphere will make the atmosphere cooler, not warmer as some people claim, as described in my 2009 paper – How Greenhouse Gases Work.

January 19, 2012 4:19 am

It does seem arrogant, doesn’t it? Still, I don’t think that fairly describes what some of us think. What I for one think is that the Velasco et al. physics-teachers-journal comment I mentioned above says something different from what you do: it says gravity does indeed impose a lapse rate at equilibrium. If so, might not other physics folks say something similar?
I don’t care what other physics folk say, not in this case. The number of people who have proposed violations of the laws of thermodynamics are legion, but they are never observed to be violated, are they?
However, I do teach this, so permit me to once again try to explain it in such a way that everybody understands it. This is a part of an answer I wrote to this same issue on Tallbloke’s blog:
Suppose you have a thermometer. A thermometer, recall, measures temperature. It does so on the basis of the zeroth law. You put the thermometer into thermal contact with (say) a fluid at some temperature and heat flows in or out of that fluid (presumed to have a lot more heat content than the variation so the thermometer itself isn’t changing the temperature much) until the two are in thermal equilibrium If it is a mercury thermometer, the mercury expands or contracts thermally out of a reservoir and you can read the temperature off of a little scale on the side that basically transforms the volume of the mercury into a temperature. Understand? Suppose your thermometer reads 20C.
Now you move the thermometer somewhere else. You put it into a glass of water, or take it with you on a road trip and drive a few hours. You wait for it to come to equilibrium with its environment and check it and again you see it real 20C. What does this mean?
Specifically, it means that the two systems (with well-defined local equilibrium temperatures) have the same temperature. If you put the thermometer into system A and it reads T_A, and put it into system B and it reads T_B, if T_A = T_B they have the same temperature. But what, exactly, does “have the same temperature” mean?
It means “if A and B are placed in thermal contact, they will be in mutual thermal equilibrium, specifically no net heat will flow from A to B or B to A.”
That’s the zeroth law. It defines the thermal equilibrium of two systems as the condition where no heat flows in between them, and establishes the transitivity of equilibrium:
Zeroth Law: If system A is in thermal equilibrium with system C, and system B is in thermal equilibrium with system C, then system A is in thermal equilibrium with system B.
This law is the basis of the thermometer, system C in my example. If we know the thermal properties of system C so that we can transform its equilibrium state into a linear scale of temperature (I will skip the historical process that lead to not only managing this but establishing the absolute or kelvin temperature scale) we can make it into a thermometer, and use it to predict whether system A and B, brought into thermal contact, would exchange heat.
The first and second law are also important in this process. In order for the systems in question to be in equilibrium, we have to manage energy flow into and out of them. First law states that we can’t have heat flow in or out (adiabatic) or work coming in or out, although it allows for a quasi-static progression through a set of states with the same temperature that takes in heat and transforms it directly into work (isothermal expansion). This kind of process can occur, but cannot be made into a cyclic process that “just” makes heat into work. The Second law establishes (among other things) the direction of heat flow. It always flows from the hotter to the colder system when they are placed in thermal contact so that they can exchange heat. Heat can never flow from the colder to the hotter system, nor can a single system evolve in time into a thermal equilibrium with different temperatures at different places as long as heat exchange is possible between those places.
So now imagine a supposed column of air that has spontaneously separated into a low temperature at the top and a higher temperature at the bottom because gravity has compressed the fluid until it is in static mechanical equilibrium. From the zeroth law, this means that if you put your thermometer in the top it will read T_t, and if you put your thermometer in at the bottom it will read T_b, where T_b > T_t.
Now imagine taking a thermally insulated silver wire that is exposed at the top and the bottom so that it is in thermal contact with the fluid there and only there. Place it into the fluid vertically, so that the top of the wire is in contact with gas at T_t, the bottom of the wire is in contact with the fluid at T_b. What will happen? Well, now you’ve got a piece of silver (an excellent conductor of heat) with a temperature gradient across its ends. Obviously, heat will flow out of the fluid at the bottom and up to the fluid at the top, cooling the bottom, warming the top. You must now use your common sense and experience of the world to predict which of the following will be true:
a) Heat will flow forever — as fast as it is delivered to the top, gravity will somehow re-sort the energy so that it flows back to the bottom and keeps it warmer than the top, to be picked up by the silver and conducted back up to the top, to fall to the bottom, to be conducted to the top, to fall to the bottom…
b) Heat will flow until the top and bottom are, in fact, in thermal equilibrium. They were not in equilibrium before. Only when the temperature of the top and the temperature of the bottom are the same will heat stop flowing in the wire, and once that is established the system is truly in equilibrium.
In the latter case, of course, you don’t need the wire. The gas itself conducts heat. In general, you never need “a wire” within a system. If you imagine using your thermometer to measure the temperature of any two neighboring coarse grained chunks of fluid (big enough to internally be in “thermal equilibrium”, small enough to be considered differential chunks as far as calculus and secular changes in gravitational potential and so on are concerned), the only way that heat will not flow between the two chunks (that are in thermal contact) is if the thermometer reads the same thing when it is in contact with each chunk separately. Otherwise if you connect them with an imaginary silver wire (that really only represents the process of heat conduction), heat would flow.
That’s the importance of the zeroth law in this discussion. Thermal equilibrium is isothermal, period. Otherwise it literally contradicts the simplest and most ubiquitous of our experiences of heat — that it flows from hot to cold, that it only flows if things aren’t at the same temperature, that thermal equilibrium is transitive (so we can build devices that measure equilibrium and quantify it as a temperature), and — by direct implication — if any isolated system ever spontaneously separates into hot and cold sides (spontaneous implying that it is a stable process with a well-defined direction to the thermal gradient, e.g. bottom to top) one can build a perpetual motion machine of the second kind, one that converts thermal energy directly to work in a cyclic engine.
Yes, physicists are known to make mistakes in their thermodynamic reasoning that unwittingly lead them to conclusions that violate the laws of thermodynamics, especially physicists that don’t do thermo or stat mech in their research and physicists for whom their one real thermo course and prelims were long ago and far away who work in something else entirely. It’s easily done and I am not innocent of error in this sort of thing myself. Physics is not the easiest subject in the world to master and we are all mere mortals with ageing, beer-damaged brains.
However, some mistakes are more egregious than others, and any discussion that contains “thermal equilibrium” and “thermal gradient” in the same sentence describing a single system is not only wrong, it is so wrong that the utterer of the statement really does need to go back to basics and work on passing prelims again. Or worse, go back, pick up an introductory textbook on physics (such as Tipler and Mosca, or Halliday, Resnick and Walker) and work through thermo again.
Thermal equilibrium is isothermal. Thermal equilibrium is both defined and consistently observed to be isothermal. There is a law of thermodynamics that pretty much says “being in thermal equilibrium means having the same temperature throughout” in any system permitting internal thermal transport. It is easy to show that other laws of thermodynamics, in particular the law forbidding energy conservative magic (perpetual motion machines of the second kind) would be broken if thermal equilibrium was not isothermal. There is an entire heuristic argument (Maxwell’s Demon) and conceptual algebraic argument (detailed balance) and detailed derivation and proof (statistical mechanics and the increase of entropy/Third Law) that all physics majors should have gone through explaining why spontaneous thermal separation does not and can not occur (fundamental answer: it can, it’s just almost infinitely improbable, as likely as all of the molecules of air in the room you are sitting in suddenly deciding to bounce just the right way to end up as a blob of liquid air off in a corner and leave you gasping in a vacuum).
Now, can we please, please stop asserting egregious violations of laws of physics as explanations for the warming of the globe? That isn’t being skeptical of CAGW, that is just being stupid, especially after the error is pointed out and carefully, rigorously explained.
Note well that this argument says nothing at all about non-equilibrium thermal separation. In fact, thermal separation is by definition only possible in non-equilibrium thermodynamics, when accompanied by heat flow or external work. The Earth is an open system, so sure, it can and does maintain a steady state (average) thermal gradient in its atmosphere but not because gravity is providing steady state work or energy. To explain a bottom to top thermal gradient that is maintained over time, one must be able to describe the sources and transport of energy that maintains it.
I continue to await a reasonable description of energy transport — absorption of actual heat input from somewhere (where “somewhere” must ultimately be “the Sun”) and the processes that lead to it being distributed in such a way as to produce net “warming” in N&Z. Simply invoking density and PV = NkT as an “explanation” does not do it, not at all.
rgb

January 19, 2012 5:01 am

And from this it follows that greenhouse gases cool the atmosphere. To be clear, without greenhouse gases the troposphere would not get cooler with increasing height.
From that, it immediately follows that increasing the amount of any greenhouse gas in the atmosphere will make the atmosphere cooler, not warmer as some people claim, as described in my 2009 paper – How Greenhouse Gases Work.

This is far from clear. First of all, cooler than what? What is your initial state from which you measure relative cooling vs warming? Second, how do you compute the average temperature? By volume? By volume weighted by mass density? The upper atmosphere ends up much cooler than it would be if the atmosphere were perfectly transparent, to be sure, but the bulk of the atmosphere is actually underneath all of that cold, and the bottom of the atmosphere — where we live, and snow falls, and summer rages — is in moderately good thermal contact with the surface and follows surface temperatures.
What greenhouse gases do is alter the pattern of the outgoing radiation that ultimately has to balance incoming radiation from following an approximate blackbody curve at the temperature of the surface to following a blackbody curve at the temperature of the surface in only part of the spectrum — the “water window” — while being modulated down to emission from the colder atmospheric greenhouse gases at a different, lower, temperature. In order to maintain overall thermal balance, the surface temperature has to be higher than it would otherwise be, because the cooler upper atmosphere emits less energy per unit time per unit “area” (above a given surface area of the Earth) than would have been emitted otherwise.
That’s all. Nothing complicated. True completely independent of any particular mechanism for heat transport from the surface to the upper atmosphere. It depends only on the atmosphere being optically thick in some bands, so that emission in those bands that actually escapes from the Earth occurs only from the cooler upper troposphere. No mention of “upwelling” or “downwelling” IR, no mention of convective or conductive heat transfer or mechanisms that establish the thermal gradient in the atmosphere in the first place.
The surface, therefore, will always be warmer, and the air near the surface (in good thermal contact with the surface) is also likely to be warmer. How much air is warmer vs the amount that is cooler, how much heat is involved in the heat capacity of the atmosphere vs the heat that is involved in the heat capacity of the relevant part of the ground or oceans? I don’t know, but I do know that computing it or even estimating it from measurements is a nontrivial process, not something one is likely to be able to pronounce upon without a lot of quantitative work based on actual global data. It is a solution to a non-equilibrium heat flow problem with lots of complexity, and the answer might well depend in some detail on the heat capacity of the various components involved. If the Earth is modelled as a thin layer of blackened aluminum foil (very low heat capacity) covered with a thick atmosphere, you’ll very likely get a different answer than you would get if the Earth is modelled as a layer of water sitting on top of the blackened aluminum foil and covered with a thin atmosphere, where by thick and thin I’m referring to their relative heat capacity, not their optical thickness.
That’s why I think it is very important to state your premises when you talk about heating or cooling. Heating or cooling relative to what state of which model system? The ideal superconducting blackbody? A rotating Earth with no lateral heat transport or atmosphere (and with what surface heat capacity?) A rotating tipped Earth where one has to average over many annual cycles to get an “average temperature” of some sort to use as the basis? With what heat capacity? With an atmosphere or not? With lateral transport?
These are all of the reasons I’m a skeptic. The GHE is real, and contributes to the net warming of the Earth’s surface, but when one talks about responses to changes in greenhouse driving, one has to understand all of the other things that lead to heating or cooling, and define the reference model from which heating or cooling are measured. One also has to worry about whether one cares if the overall heat content of the system is lower if the temperature of the surface is higher, or vice versa!
rgb

January 19, 2012 5:08 am

This seems to me to be a dicey premise to be using as input to your argument for isothermal equilibrium. The premise implies that, over a tiny altitude extent (something greater than the mean free path), the temperature does not change. In essence, you are assuming dT/dz = 0 over your slice. This assumption might be what your reasoning is extropolating over all z, resulting in a lapse rate of 0 everywhere.
Well then, feel free to build the perpetual motion machine of the second kind that is enabled by a steady state temperature differential that is not maintained by any external source of heat energy or work. Or, learn the laws of thermodynamics that make this not only impossible, but absurd.
rgb

January 19, 2012 5:17 am

Many thanks,
w.

You are quite welcome. Now, if only a few other people would take the time to read Caballero and work through it, or the time to go (back) to their intro physics textbooks and look at the thermodynamics section to understand what temperature is (Zeroth Law).
This whole discussion has inspired me to go ahead and set up my very first (very preliminary) draft of part III of my online physics textbook, with the first part devoted to thermodynamics. It still isn’t sufficient to help people answer all of these questions, but the arguments I’m putting into this and the other threads are — they are classic arguments used in all of the intro texts.
Big hint — any time your answer permits you to build a perpetual motion machine, it is wrong, and you can build a heat engine that runs between any temperature differential. If any process spontaneously creates a temperature differential (without external work or energy being exchanged, it permits a perpetual motion machine of the second kind to be built, and incidentally violates the direction of entropy (third law) as hot reservoirs gain \Delta Q/T_H of entropy which is strictly less than the \Delta Q/T_C that the cold reservoir loses so that the entropy of the (isolated) system decreases.
Hmmm, no it doesn’t. Never happens.
rgb

January 19, 2012 5:22 am

I am not sure there is a demon. Yes the number of particles passing through an infinitesimal slice is equal; however, they enter into a volume with less kinetic energy than they just previously had…so they have to be cooler.
Dear Joe,
I know you took thermo if you are an astrophysicist. I know that you know the laws of thermodynamics. I know that you can understand that one can build a perpetual motion machine of the second kind if your assertion is true. We both know that — no we can’t. We can both compute the fact that the entropy of your hypothesized system decreases as thermal separation occurs.
Can we stop now? Go to your shelf, pick up a textbook with thermodynamics in it, and read about the zeroth law. Thermal equilibrium is the very definition of isothermal. Or do the actual textbook exercise and compute the detailed balance to convince yourself that there can be no thermal gradient at equilibrium.
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Joe Postma
January 19, 2012 7:39 am

=====================================
Robert Brown says:
January 19, 2012 at 5:22 am
Dear Joe,
I know you took thermo if you are an astrophysicist. I know that you know the laws of thermodynamics. I know that you can understand that one can build a perpetual motion machine of the second kind if your assertion is true. We both know that — no we can’t. We can both compute the fact that the entropy of your hypothesized system decreases as thermal separation occurs.
Can we stop now? Go to your shelf, pick up a textbook with thermodynamics in it, and read about the zeroth law. Thermal equilibrium is the very definition of isothermal. Or do the actual textbook exercise and compute the detailed balance to convince yourself that there can be no thermal gradient at equilibrium.
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Robert, this is beside the point. I tried explaining above that we’re not talking about a system in isothermal equilibrium. The atmosphere IS OBSERVED to have a lapse rate. If YOU think that the OBSERVED atmosphere means we can violate the 2nd Law, then go ahead and try it. I never said that, you did; so stop putting words in my mouth. We also have the fact that -g/Cp describes the observed temperature profile for dry air; that’s a fact, and we know the various reasons for why the exact quantitative value of -g/Cp is not always realized.
The atmosphere IS OBSERVED to change temperature with altitude. That is all I have been talking about, and connecting it to -g/Cp. I don’t know where this stuff about the observed atmosphere violating the laws of thermo is coming in from. It is also mostly static in its distribution – only the lower atmosphere changes temperature but between 3 & 20km the gradient is static.
So, let’s stop introducing some ideal system which has nothing to do with what we’re talking about and prevents us from actually understanding the first thing about it. If the ideal definitions of thermal equilibrium means that the atmosphere must be isothermal, then those definitions are not the physics we must refer to since they do not match the observed system.
It sounds as if you are insisting that if there’s non-isothermicity in the atmosphere with altitude we could build a perpetual motion machine. Well, the atmosphere IS non-isothermetic with altitude, so you’re going to have to figure that one out since it was you, not I, who came up with the idea you build such a machine in this case. Besides, maybe one could work as you’re trying to build it, who knows, but if it did I am sure the energy would be coming from somewhere for it. It is over-unity efficiency that would be a problem, not that it could run forever given a system which has a continuous supply of new energy coming in from the Sun. Aren’t there rumours that Tesla figured out a “free=energy” device of some sort using static charge in the atmosphere? Anyway, this is all not relevant.

Joe Postma
January 19, 2012 7:42 am

I’d like to point out yet another oddity of standard GHE theory. There’s supposed to be back-radiation from IR-emitting molecules, predominantly CO2, that cause 33C of additional heating. That’s the basic GHE theory.
So tell me then: on a spectrometer plot taken from the surface of the Earth and pointing upwards, or even from one taken from above the Earth and looking down – where is the emission line?
If CO2 is radiating all this energy and by definition this emission has to be spectral, then where is the huge & incredibly bright emission line introducing an additional 150 W/m2 into the surface, and that which should also be exiting the TOA? The ENTIRE output spectrum of the Earth only comes out to 240 W/m2, so this additional 150 W/m2 from spectral emission must be huge! Where is it?
It doesn’t exist.
Instead, where it could exist, is a huge hole in the spectrum, a LACK of energy power. It HAS TO EMIT to be said to cause heating by radiative emission in the first place. Yet where it could emit, it doesn’t exist. And at the bottom of the notch where CO2 should be emitting to add all this extra power, is a smooth blackbody curve corresponding to something like -80C. There is a very small emission peak right at 15um, but it’s barely worth noting.
So fine, let’s pretend to go with the standard theory of GHG back-emission. And so I’ll ask: where is the emission?
Pointing an IR sensor at the sky and it telling you the temperature, converted to some power units, is simply taking a temperature reading! It has NO relevance to the huge emission line we should see from CO2 & GHG’s causing all this heating. That’s a fraudulent interpretation of the measurement! The IR sensor is measuring a rough black-body that has LACK of emission flux at GHG wavelengths.
It might just be that simple. Fine: radiation causes heating, we all know that. So show me the radiation from CO2 then. Oh it doesn’t exist? Well then what the heck…non-existent, non-observable spectral emission from GHG’s causes heating. Wonderful.
Now someone might try to back-track and say that the downward IR gets all absorbed by the time it reaches the ground, and because it is all absorbed this is why it is causing heating. But wait, it can’t very well be said to be heating the surface then, can it, if it never GETS to the surface. And additionally, you can’t hide at the bottom of the atmosphere anyway – the TOA is free to space and there’s NO reason we shouldn’t see the 150 W/m2 of GHG spectral emission there. But it’s not there either, is it; except for a tiny little pip at 15um with maybe a couple of Watts in it.
Someone might also try to back-track and say that because all the 15um radiation is absorbed and you don’t see it, that’s why it causes heating. But that’s still inconsistent with the spectral reading at the TOA – it should still be seen at the TOA – and it also implies that LACK of emitted radiation is what causes heating. But that’s what non-GHG’s implicitly do in the first place – not radiate and therefore trap heat – and the whole GHG theory is based on the idea that GHG’s spectrally radiate.
This whole theory is shot; full of holes. With this OP and the comments here in it, and the other one by Robert Brown and the comments in that one, it is clear to anyone reading that GHG Theory is dead. I still think my treatises give a good explanation of why it is dead.
http://www.tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf
http://principia-scientific.org/publications/The_Model_Atmosphere.pdf
http://principia-scientific.org/publications/Copernicus_Meets_the_Greenhouse_Effect.pdf
Read only the last of those links if you want – it is a very short paper with a succinct summary of the paradigm shift.
Of course, I give thanks to the book that started it all.

Bryan
January 19, 2012 8:00 am

The isothermal/adiabatic distribution for an isolated ideal gas in a gravitational field has long been debated.
For the isothermal distribution we have Maxwell, Boltzmann and Clausius.
For the adiabatic distribution we have Loschmidt, Laplace and Lagrange.
The smart money must be with the isothermal advocates but I would not regard this as a debate of which was settled and of historical interest only.
Clausius clincher argument of the perpetual motion machine being possible for the adiabatic distribution turns out to be very hard to prove with real components given 9.8K/km scale.
There has been no experiment to settle the matter!
Here for instance is a member of the physics department of the University of California making a very up to date case for the adiabatic distribution.
http://arxiv.org/PS_cache/arxiv/pdf/0812/0812.4990v3.pdf

Joe Postma
January 19, 2012 8:03 am

=======================
This whole theory is shot; full of holes. With this OP and the comments here in it, and the other one by Robert Brown and the comments in that one, it is clear to anyone reading that GHG Theory is dead. I still think my treatises give a good explanation of why it is dead.
========================
I meant this OP by Robert and the other one by Willis Eschenbach, and the comments therein.

Joe Postma
January 19, 2012 8:29 am

====================================
Robert Brown says:
January 19, 2012 at 4:19 am
The Earth is an open system, so sure, it can and does maintain a steady state (average) thermal gradient in its atmosphere but not because gravity is providing steady state work or energy. To explain a bottom to top thermal gradient that is maintained over time, one must be able to describe the sources and transport of energy that maintains it.
rgb
====================================
Alright yes, I see. My point is that the thermal gradient exists and is described for the most part by -g/Cp. That equation does not actually require thermal equilibrium to work, it merely requires that the net change of energy in the system is zero – that definition implies nothing about the thermal distribution. So that’s an important distinction. Just like the metal bar heated/cooled from two ends example: the net energy coming into the bar is equal to that going out, so there is “equilibrium” in THAT sense, but that is not actually thermal equilibrium in the strict sense of isothermicity. There will be a temperature distribution down the length of the bar. Static, no change in total internal energy, but not in thermal equilibrium within itself.
Now, the next part:
==================================
Robert Brown says:
January 19, 2012 at 4:19 am
I continue to await a reasonable description of energy transport — absorption of actual heat input from somewhere (where “somewhere” must ultimately be “the Sun”) and the processes that lead to it being distributed in such a way as to produce net “warming” in N&Z. Simply invoking density and PV = NkT as an “explanation” does not do it, not at all.
==================================
Yes I do agree with that. However, I would throw away the assumption everyone makes that there NEEDS to be “net additional warming”. As I explained in my papers I’ve linked to several times, the real-time power flux into the Earth-system actually has a temperature value of +121C at maximum, or +30C on average given a hemisphere & albedo. N&Z use an averaging approach that leads to an input temperature even smaller than the mistaken P/4 value of -18C.
I do agree that is doesn’t make sense to have gravity acting as a free-energy source. I understand why you’re bringing that up now and I struggled solving that myself early on. You (Robert) have my obedience in that regard.
And so, what you request:
“I continue to await a reasonable description of energy transport — absorption of actual heat input from somewhere and the processes that lead to it being distributed”
is exactly what I am attempting with the heat-flow differential equation that I am developing that models the input and outputs in real-time with, real-time – non-diluted by averaging – values. As I said, you don’t need to model the intricate workings of a capacitor to be able to model the voltage in an R-C circuit – you just need the basic gross values for the parameters describing the capacitor and the resistor. This is analogous to temperature (voltage) with mass (resistance) and thermal capacity (capacitance). The equation is analogous but it doesn’t “depend” on the analogy, as it were.
And what I have found with such an equation is that we have to appreciate the average ground temperature is stable and at around +5C all over the planet. You have to include that. You can’t just say that the ground temperature is zero Kelvin (or 2.3K) and as such has no effect on the surface. Changing the level of the plateau upon which the solar insolation varies has a HUGE effect on the resulting temperature balances at the surface.

January 19, 2012 8:51 am

Joe Postma says:
January 19, 2012 at 7:42 am
I’d like to point out yet another oddity of standard GHE theory.
Mr. Postma, here is another oddity that I find. Standard radiative heat transfer equation says that if T1=T2 then Q/A = ZERO. But some how in GHE theory if T1>T2 then T2 will transfer energy back to T1.

Joe Postma
January 19, 2012 8:52 am

===========================================
Bryan says:
January 19, 2012 at 8:00 am
The isothermal/adiabatic distribution for an isolated ideal gas in a gravitational field has long been debated.
For the isothermal distribution we have Maxwell, Boltzmann and Clausius.
For the adiabatic distribution we have Loschmidt, Laplace and Lagrange.
The smart money must be with the isothermal advocates but I would not regard this as a debate of which was settled and of historical interest only.
Clausius clincher argument of the perpetual motion machine being possible for the adiabatic distribution turns out to be very hard to prove with real components given 9.8K/km scale.
There has been no experiment to settle the matter!
Here for instance is a member of the physics department of the University of California making a very up to date case for the adiabatic distribution.
http://arxiv.org/PS_cache/arxiv/pdf/0812/0812.4990v3.pdf
=============================================
Yes well the problem we’ve just identified, between Robert and I going back and forth, is realizing that the usual statement leading into the derivation of -g/Cp of
“At thermal equilibrium the change in energy of the system is zero”
is incorrect. Functionally what we are seeking, and defining in accordance with reality, is that the net change of energy in the system is zero. There’s nothing wrong with that part of it. What is wrong in the leading statement is that this corresponds to thermal equilibrium.
The condition of dU = 0 does not actually have to correspond to thermal equilibrium as defined to the strict thermodynamic sense of isothermicity. Robert is quite correct in having pointed out what he has been. The condition of dU = 0 simply states, at most, that you might expect a static temperature distribution in the system. In that regard, then the equation (resulting in -g/Cp) does work quite nicely, doesn’t it.
So, this is actually a MAJOR development, in my opinion. We have all just participated in correcting and improving a standard definition and starting point of a common analysis. We now have a bridge that the isothermal people and the adiabatic people can meet upon. And to be fair, for the purpose of understanding, not for blame, we understand that the initial starting point of the adiabatic “group” was incorrect: it was incorrect to say that dU = 0 corresponds to thermal equilibrium. That condition actually only implies “stasis” within the system, but the system is free to have a thermal distribution since energy is continuously moving through it. This latter part is what the isothermal “group” didn’t catch on to.

don penman
January 19, 2012 8:59 am

It does not matter if the the non ghg atmosphere is isothermic in the hypothetical example you give because the actual data for the moon which has no atmosphere is 100k below what s-b predict it should be.The atmospheric pressure must raise the near surface temperature by 100k before we even consider the greenhouse effect.It is not therefore radiating more energy to space then it receives from the Sun.The surface can radiate all the short wave radiaton back into space as long wave radation then.

Joe Postma
January 19, 2012 9:11 am


Indeed!

Joel Shore
January 19, 2012 9:19 am

mkelly says:

Mr. Postma, here is another oddity that I find. Standard radiative heat transfer equation says that if T1=T2 then Q/A = ZERO. But some how in GHE theory if T1>T2 then T2 will transfer energy back to T1.

You have had this explained to you many, many times in many, many threads and yet you persist in spreading such falsehoods! The heat flow, that is the net flow of energy, is from the warmer planet (T1) to the colder atmosphere (T2). However, the amount of heat flow that occurs depends on T2 as well as T1. The way this comes about is because radiative energy is transferred in both directions. However, the radiative transfer from the planet to the atmosphere is larger than the transfer from the atmosphere to the planet…and hence, the net radiative energy flow, which we call the heat flow, is from the planet to the atmosphere.
This field is the only field where it is at all controversial that the rate of heat flow between objects at a temperature T1 and T2 with T1 > T2 depends on T2 as well as T1. And, the reason it is controversial is that some peope like you would apparently prefer to believe pseudoscientific nonsense over real science.

Joel Shore
January 19, 2012 9:21 am
don penman
January 19, 2012 9:27 am

It could be that the two sources of heat the surface temperature and the heating caused by atmospheric pressure are in equilibrium at the S-B predicted temperature therefore no temperature change takes place between the surface and the atmosphere.

Joe Postma
January 19, 2012 9:33 am

Yes I saw that Joel. I saw that several of you agreed that there is no spectral radiation from GHG’s. Thanks for the back-link.

Bryan
January 19, 2012 9:41 am

The isothermal/adiabatic distribution for an isolated ideal gas (no heat enters or leaves the gas) in a gravitational field has long been debated.
The outcome either way, though interesting, has no relevance to the greenhouse theory as far as I can see.

Joel Shore
January 19, 2012 10:03 am

Joe Postma says:

Yes I saw that Joel. I saw that several of you agreed that there is no spectral radiation from GHG’s. Thanks for the back-link.

That is not what was agreed.

Brian H
January 19, 2012 10:04 am

Robert;
That 2009 paper has given me a solid long-term awareness of the potent interactions that keep the tropopause paused ever since about early 2010. The mechanism of “handing off” between CO2 and H20 at the stratosphere boundary is a powerful insight. As is the heat-pipe-like functioning of the water cycle, using latent heat energy transfers to get said energy up to the cloud tops, whence it is “dumped” upwards. Thank you again.
I’d recommended it many times to others, though its impact hasn’t been all it should have been. Perhaps with this backing, its time has come!

Joel Shore
January 19, 2012 10:12 am

Robert Brown says:

Now, can we please, please stop asserting egregious violations of laws of physics as explanations for the warming of the globe? That isn’t being skeptical of CAGW, that is just being stupid, especially after the error is pointed out and carefully, rigorously explained.

This fellow physicist gives a big “Amen” to that! Thanks for explaining this so carefully and well!

Joe Postma
January 19, 2012 10:37 am

==================================
Joel Shore says:
January 19, 2012 at 10:03 am
That is not what was agreed.
==================================
So you CAN see 150 W/m2 of spectral emission from GHG’s then? Great, show it.

January 19, 2012 11:46 am

Joel Shore says:
January 19, 2012 at 9:19 am
Joel perhaps you missed the part about if T1=T2 then Q/A is ZERO. T1 supplies no energy to T2 and T2 supplies no energy to T1. But , according to GHE theory if they are different (T1>T2)then some energy does flow from T2 to T1. Joel you also assume that a higher energy state object absorbs lower energy state radiation. I have found no evidence to support that.
Also Joel which is more efficient pure reflection or absorption/emission? If the object (CO2) just reflected the IR back to the ground would you say reflected photons can heat the object that emitted the photon?
By the way Joel you never did tell us what emissivity you use for CO2 at 1 atm and 288K for a heat transfer equation.

Robert Clemenzi
January 19, 2012 1:04 pm

Robert Brown says:
January 19, 2012 at 5:01 am

From that, it immediately follows that increasing the amount of any greenhouse gas in the atmosphere will make the atmosphere cooler, not warmer

This is far from clear. First of all, cooler than what?

Cooler than it would be without adding the gas?
I really don’t understand the question.
BTW, if a greenhouse gas makes the atmosphere cooler, it has to be by emitting radiation. Some of that radiation goes to space and some of it is absorbed at the surface.

Robert Clemenzi
January 19, 2012 1:17 pm

Brian H says:
January 19, 2012 at 10:04 am
Thanks for the support. I have seen your posts scattered all over the place. I don’t understand why this is so hard to grasp.
If you haven’t tried it already, I suggest trying my lapse rate animation program. It uses real data and proves beyond all doubt that the atmosphere is IR opaque in the bands that matter. No math required, you can just look at the plots. It also disproves the idea above about the real lapse rate being determined by -g/Cp (9.8 K/km). I can’t believe that anyone supports that nonsense when actual data completely disproves it.

Editor
January 19, 2012 1:41 pm

Joe Postma says:
January 19, 2012 at 7:42 am

I’d like to point out yet another oddity of standard GHE theory. There’s supposed to be back-radiation from IR-emitting molecules, predominantly CO2, that cause 33C of additional heating. That’s the basic GHE theory.
So tell me then: on a spectrometer plot taken from the surface of the Earth and pointing upwards, or even from one taken from above the Earth and looking down – where is the emission line?

There’s not one emission line for the combination of H20, CO2, and methane, there are a multitude of lines. Here are the relevant emission lines, from MODTRAN rather than a spectrophotometer.

And here is spectrophotometer data, this one happens to be from the South Pole:

I’m sure a few minute search on Google will turn up many more. It’s not a mystery as you seem to think, the spectrum of downwelling radiation is routinely measured by scientists all over the world.
All the best,
w.

Editor
January 19, 2012 1:47 pm

Robert Brown says:
January 19, 2012 at 5:17 am

Many thanks,
w.
You are quite welcome. Now, if only a few other people would take the time to read Caballero and work through it, or the time to go (back) to their intro physics textbooks and look at the thermodynamics section to understand what temperature is (Zeroth Law).

When my science conflicts with established science, I research it until I find the error. When I’m wrong, I admit it. So let me join you in your recommendation that others do the same.
w.

Editor
January 19, 2012 2:06 pm

Robert Clemenzi says:
January 19, 2012 at 1:17 pm

… It also disproves the idea above about the real lapse rate being determined by -g/Cp (9.8 K/km).

Robert, there are several “lapse rates”, the dry adiabatic lapse rate (DALR), the moist adiabatic lapse rate, and the actual observed lapse rate at a given place and time. It is not clear which one you are calling the “real” lapse rate.
The dry adiabatic lapse rate is 9.8°C per kilometre. This is modified by the actual atmospheric conditions. If the actual lapse rate is less than the DALR, the atmosphere overturns until the DALR is restored.
Looking at your citation, the actual lapse rate (red line) for that particular place and time is slightly different from the standard atmosphere (blue line). This is no surprise, as at any given location and instant the actual lapse rate will be determined by local conditions, and will not exactly follow the DALR. You will notice, however, that it closely follows the DALR, although the tropopause is somewhat turbulent and at a different height than the standard atmosphere. This is quite typical of real soundings.
The DALR does not “determine” the actual lapse rate as you seem to think. All the DALR does is put a limit on the lapse rate such that when the lapse rate is less than that limit, the atmosphere overturns until the DALR is restored. In our dynamic atmosphere, of course, the readjustments and changes in the lapse rate are continually occurring, so the atmosphere never exactly follows the DALR.
Hope this helps,
w.

Editor
January 19, 2012 2:12 pm

Bryan says:
January 19, 2012 at 9:41 am

The isothermal/adiabatic distribution for an isolated ideal gas (no heat enters or leaves the gas) in a gravitational field has long been debated.
The outcome either way, though interesting, has no relevance to the greenhouse theory as far as I can see.

It has huge relevance to the “gravito-thermal” theories of Jelbring and N&Z, which have been proposed as alternates to the greenhouse theory. The relevance is that it shows them to be false.
Also, a more correct statement would be that the isothermal distribution for an isolated ideal gas (no heat enters or leaves the gas) in a gravitational field has long been established, although “gravito-thermal” adherents (and misguided folks like myself) occasionally try to debate it.
w.

Joel Shore
January 19, 2012 2:19 pm

mkelly says:

Joel perhaps you missed the part about if T1=T2 then Q/A is ZERO. T1 supplies no energy to T2 and T2 supplies no energy to T1. But , according to GHE theory if they are different (T1>T2)then some energy does flow from T2 to T1.

You seem to have no ability to distinguish between heat flow and energy flow. Heat is the net energy flow. If two objects are at the same temperature, they don’t magically stop emitting radiation at each other or absorbing radiation from each other. It is simply that the amount object A absorbs from object B is the same as the amount object B absorbs from object A and the net flow of energy, what we call heat, is zero.

Joel you also assume that a higher energy state object absorbs lower energy state radiation. I have found no evidence to support that.

Every physics textbook in the world also assumes it…and every physicist and engineer who uses the radiative transfer equations assumes it. And, the assumption has been proven correct time and time again by the fact that these equations work.

Also Joel which is more efficient pure reflection or absorption/emission?

I don’t know what “more efficient” means. Since absorption/emission can result in the radiation going either up or down, I would say the pure reflection is more efficient in returning the energy to the surface. However, that is not what happens to any appreciable degree in the Earth’s atmosphere, although it is an important component of Venus’s greenhouse effect.

If the object (CO2) just reflected the IR back to the ground would you say reflected photons can heat the object that emitted the photon?

In the sense that they enter into the energy balance as incoming energy, yes.

By the way Joel you never did tell us what emissivity you use for CO2 at 1 atm and 288K for a heat transfer equation.

That is because the question does not make sense. CO2’s emission is a strong function of wavelength; furthermore, you have to define a pathlength for a gas in order to talk about its absorptivity or emissivity even at a particular wavelength.
But, all of this is irrelevant since we haven’t even been talking about emissions of CO2 here. We have been talking about emissions from the Earth’s surface. If one wants to do detailed quantitative calculations of the radiative transfer in the atmosphere, one has to use a full-fledged line-by-line radiative transfer code.
If you never seem to learn anything from what we explain to you, why should we waste our time even responding to you? Are you trying to compete with Myrrh?

Joel Shore
January 19, 2012 2:23 pm

Willis Eschenbach says:

All the DALR does is put a limit on the lapse rate such that when the lapse rate is less than that limit, the atmosphere overturns until the DALR is restored.

Just to be clear, when Willis says “less than that limit”, he means a lapse rate that is more negative…or steeper…than the DALR (i.e., where the temperature drops with height faster than the DALR.)
[It is challenging to find words to describe this that don’t cause confusion!]

don penman
January 19, 2012 3:07 pm

N&Z still have the greenhouse effect but they say that it less than the heating by compression of the atmosphere.They compare the moons average temperature with the earths average temperature and show that the moon is much less then 33k colder then the earths temperature .Could the moon be losing heat to the space in contact with its surface by other ways then radiation.

Editor
January 19, 2012 3:12 pm

Joel Shore says:
January 19, 2012 at 2:23 pm

Willis Eschenbach says:

All the DALR does is put a limit on the lapse rate such that when the lapse rate is less than that limit, the atmosphere overturns until the DALR is restored.

Just to be clear, when Willis says “less than that limit”, he means a lapse rate that is more negative…or steeper…than the DALR (i.e., where the temperature drops with height faster than the DALR.)
[It is challenging to find words to describe this that don’t cause confusion!]

Thanks for the clarification, Joel. My point was that the DALR establishes a limit on the actual environmental lapse rate, it does not determine that rate.
w.

January 19, 2012 3:42 pm

The atmosphere IS OBSERVED to change temperature with altitude. That is all I have been talking about, and connecting it to -g/Cp. I don’t know where this stuff about the observed atmosphere violating the laws of thermo is coming in from. It is also mostly static in its distribution – only the lower atmosphere changes temperature but between 3 & 20km the gradient is static.
Its change in temperature is observed to closely match the adiabatic lapse rate, which is nicely described and derived in Caballero. The point is that it is not in thermal equilibrium. If you turned off solar input and surrounded the Earth with a perfect insulator, you would not get the same temperature distribution.
What I’m been trying to establish is that thermal equilibrium in a closed system is isothermal, largely because lots of people have been asserting otherwise. It sounded like you were among them, with your arguments about gravity slowing things down. In an open system, the distribution of temperatures has nothing to do with gravity per se and everything to do with where you add and remove heat. If you heat it at the bottom and cool it at the top under circumstances where the warmed air can adiabatically expand and rise (and cool, because adiabats cross isotherms), you end up with something close to the adiabatic lapse rate. If you heat it at the top and cool at the bottom, you will not have higher temperatures at the bottom. The temperature has nothing to do with the pressure, and nothing to do with the density, by which I mean that one can take a gas with any pressure or any density that you like and heat or cool it to any temperature that you like (within reason) and it will be in thermal equilibrium with any other gas, or liquid, or solid at the same temperature independent of it’s density or pressure.
I await a non-thermodynamics violating explanation of higher temperature at the bottom of an atmospheric column that doesn’t begin with delivering heat directly to the bottom of the atmospheric column and/or involve bulk transport (convection). Fourier’s Law and the Heat Equation are very general results. The entropic restrictions on the flow of heat and entropy are very stringent restrictions. Imagine a sphere around the Earth 100 meters off of the ground. Heat will flow from the warmer ground to the cooler overhead, but that heat flow can only be maintained by energy input inside the sphere and gravity cannot provide it!
rgb

willb
January 19, 2012 3:54 pm

Robert Brown says:
January 19, 2012 at 5:08 am

(So what?)

I’m not claiming your conclusion is wrong. I’m pointing out a curiosity that makes me think your proof is somewhat less than rigorous.

Bryan
January 19, 2012 4:07 pm

I said
The isothermal/adiabatic distribution for an isolated ideal gas (no heat enters or leaves the gas) in a gravitational field has long been debated.
The outcome either way, though interesting, has no relevance to the greenhouse theory as far as I can see.
Willis said
It has huge relevance to the “gravito-thermal” theories of Jelbring and N&Z, which have been proposed as alternates to the greenhouse theory. The relevance is that it shows them to be false.
My reply
I don’t take that view as they realise that thermodynamics is involved through solar heating and TOA cooling.
Certainly they realise that the atmosphere does not resemble an isolated sample of gas which no heat enters or leaves.
I have not had time to study the “gravito-thermal” theories of Jelbring and N&Z,” and a full explanation of their ideas is to be published on Tallbloke.
What seems true to say is that the diehard believers in a 33K greenhouse effect have no logical basis for their conjecture.

Myrrh
January 19, 2012 4:22 pm

Joel Shore says:
January 19, 2012 at 2:19 pm
mkelly says:
Joel perhaps you missed the part about if T1=T2 then Q/A is ZERO. T1 supplies no energy to T2 and T2 supplies no energy to T1. But , according to GHE theory if they are different (T1>T2)then some energy does flow from T2 to T1.
You seem to have no ability to distinguish between heat flow and energy flow. Heat is the net energy flow. If two objects are at the same temperature, they don’t magically stop emitting radiation at each other or absorbing radiation from each other. It is simply that the amount object A absorbs from object B is the same as the amount object B absorbs from object A and the net flow of energy, what we call heat, is zero.
============
I’ll ask you again. What is the mechanism that makes the net come out as flow from hotter to colder?
Until you have that you have no way of stopping an ice cube warming your hotter house.

dr.bill
January 19, 2012 4:24 pm

(if you’re still monitoring this thread):
I’ve been re-reading a number of comments on this thread, and came across something you said earlier that I’d like to ask you about. What you said was this:

The number of degrees of freedom for oxygen at normal roomish temperatures is 5 — 3 translation and two rotation — per molecule. Heat it up enough and it goes up first to six, then to seven (or more) as one excites additional modes…

This doesn’t correspond to what I have always taught my students (and obviously believe myself). My counting system is that there are 3n degrees of freedom in total, with n being the number of atoms in the molecule. For anything beyond monatomics, these are distributed as 3 to translation, 2 to rotation for linear molecules (or 3 for non-linear ones), and all the rest to vibrations. There is, however, a caveat, namely that while the first 5 (or 6) modes have ½kT each, the vibrational modes have a full kT each due to possessing two quadratic terms in their energy expression. Thus for something like O2, there would never be a total energy of 3kT for an individual molecule, it would be either 2½ or 3½. There could be an average of 3kT, but only because the system was at a temperature where half the molecules had their vibrational mode activated and the other half didn’t.
I’d like your comments on this.
Thanks in advance,
/dr.bill

willb
January 19, 2012 5:20 pm

Robert Brown says:
January 19, 2012 at 4:19 am

Now, can we please, please stop asserting egregious violations of laws of physics …?

I can see you’ve got your hands full responding to comments here, so I’m just going to make this one last comment riddled with egregious violations. I hope you can stomach it.
You say:

Thermal equilibrium is isothermal, period. Otherwise it literally contradicts the simplest and most ubiquitous of our experiences of heat — that it flows from hot to cold, that it only flows if things aren’t at the same temperature, …

Call me dense but isn’t it also our experience that the lapse rate on Earth is ubiquitous? It exists day and night, winter and summer, at the equator and in the Arctic. Are you so absolutely convinced that a non-zero lapse rate is not an equilibrium condition directly resulting from Earth’s gravity?
What about this seeming paradox?:
If the atmosphere were to become isothermal, the average kinetic energy per molecule would be the same at all altitudes. Since the force of gravity decreases with altitude, at high enough altitudes the upward moving molecules would no longer be tied to the planet by gravity. As long as they avoided collisions their kinetic energy would cause them to shoot off into deep space, leaving the atmosphere. The remaining molecules in the atmosphere would expand to fill up the void left by these deserters. Since the atmosphere is isothermal and no work is being done, the remaining molecules would organize themselves in a new density gradient but they would retain their kinetic energy. The higher altitude upward-moving molecules, which always have the same average kinetic energy, would continually escape to deep space as long as they avoided collisions. Given an infinite amount of time the entire atmosphere would be lost. The odd thing about this is that, if this happened, the escaped atmospheric gas would have the same kinetic energy as it had when it was tied to the planet by gravity.
So how does the atmosphere manage to climb the potential well of gravity without losing any energy?
Regarding your thermometer example of the Zeroth Law, you say:

Zeroth Law: If system A is in thermal equilibrium with system C, and system B is in thermal equilibrium with system C, then system A is in thermal equilibrium with system B.
This law is the basis of the thermometer, system C in my example.

Question: How do you factor in the work that has to be done on the thermometer to move it from a lower-altitude system A to a higher-altitude system B when measuring temperature? If we assume that no energy can enter or leave the thermometer in the process of measuring the two temperatures, then it seems to me the thermometer must lose heat (and temperature) as it moves up from system A to system B, and gain heat as it moves down from system B to system A. In either case the thermometer would not be in thermal equilibrium with the new system it is about to measure.
And btw I call bullsh*t on your perpetual motion machine. As I said previously, We live in a world with a ubiquitous lapse rate. If it were possible to exploit this temperature differential to produce boundless energy, it would have been done by now.

Robert Clemenzi
January 19, 2012 10:16 pm

willb says:
January 19, 2012 at 5:20 pm

The higher altitude upward-moving molecules, which always have the same average kinetic energy, would continually escape to deep space as long as they avoided collisions.

That is true and it would cause the top of the atmosphere to cool. However, the atmosphere is also loosing mass. As pointed out elsewhere, there are many competing mechanisms that will cool greenhouse gas free atmospheres, and this is just one. However, the point of investigating a greenhouse gas free atmosphere is to understand the function of greenhouse gases, not to explore all the details describing a real atmosphere.

So how does the atmosphere manage to climb the potential well of gravity without losing any energy?

Energy is not lost by moving against gravity, it merely converts from kinetic energy to potential energy. However, the associated loss in temperature is replaced by the Sun heating the surface.

Joel Shore
January 20, 2012 3:44 am

Myrrh says:

I’ll ask you again. What is the mechanism that makes the net come out as flow from hotter to colder?
Until you have that you have no way of stopping an ice cube warming your hotter house.

Myrrh: I’ve already explained to you a few months ago the modern understanding of the Second Law in terms of the statistics of large numbers of particles and how that leads from microscopic reversibility to macroscopic irreversibility. (Robert Brown has also explained it in one of the threads here.) This understanding is one of the triumphs of physics in the last century or so. You however refused to believe it .To deny all of modern physics that you don’t like is your prerogative but I am not going to waste my time with you.

Joel Shore
January 20, 2012 3:52 am

willb says:

Call me dense but isn’t it also our experience that the lapse rate on Earth is ubiquitous? It exists day and night, winter and summer, at the equator and in the Arctic. Are you so absolutely convinced that a non-zero lapse rate is not an equilibrium condition directly resulting from Earth’s gravity?

It is not ubiquitous in the stratosphere. The reason it is fairly ubiquitous in the troposphere (although you might to google the term “temperature inversion”) is that the troposphere is strongly warmed from below and cooled from above. This is what creates a large lapse rate…and the lapse rate would even exceed the adiabatic lapse rate were it not for the fact that lapse rates steeper than the adiabatic one are unstable to convection, which transfers heat upwards in the atmosphere until the lapse rate is driven down to the adiabatic lapse rate.

wayne
January 20, 2012 5:24 am

Respectably Dr. Robert Brown, hope you at sometime read this and comment.
when you said:
“Thus I don’t understand Jelbring’s argument from the beginning. He asserts equilibrium, but then imposes an adiabatic lapse rate in temperature that contradicts equilibrium, at least thermal equilibrium. To put it another way, the molecules of gas at the cooler temperature have to have a completely different maxwell-boltzmann distribution of temperatures, do they not, with completely different average speeds? They are not in thermal equilibrium.”
… I had to take a week or so to digest that statement, at the surface it seems true.
But I can’t believe I am commenting here to you, on this subject. After thinking through your conclusions, especially using Caballero as your source, I have found it lacking and if you will give me a moment I will try to explain why as simply as possible. I think your purely kinetic view of equilibrium is at the base.
You were saying the strictly kinetic energy at each altitude level vertically must always stay constant due to the molecular mean velocity and its adherence to the Boltzmann distribution. I find that lacking only in the vertical axis in a gravitational field, for horizontally, I totally agree.
A great amount of time in physics has been concentrated in gravitational effects at the Newtonian level, leaving any relativity effects out of this discussion. Also, you explicitly mention of the total energy density at each level not being constant vertically, I of course agree. So how do I put this simply?
Take a molecule’s position vertically at point ‘z’. To me it is imperative that that any molecule must naturally be able to be displaced to either point ‘z+dh’ or ‘z-dh’ and after that displacement has occurred, must exactly be the same member of the Boltzmann distribution as before at that new point.
If the strictly mean kinetic energy were to be view vertically, this membership would never occur exactly. The molecule moving upward would have shifted its ‘exact’ placement in the horizontal Boltzmann distribution at the new ‘z’ level to the right with a slightly lower velocity, violating the Boltzmann distribution at that level. The opposite if moving downward.
The base of my contention is that it is not the kinetic energy which must always be constant vertically but due to the gravity it is the total energy, kinetic plus potential, which must always be constant.
I hate to say that I see a slight defect in the way the Boltzmann distribution is derived but, I do see it vertically against the gravitational acceleration. If I were to be correct, it is that factor which gives the DALR it’s real physical ‘existence’ instead of it merely being a non-real descriptive ratio -g/Cp.
Hope that gives you a new and different way to look at that same relationship.
If you disagree or I haven’t made myself clear with so few words, please let me clarify. I sure hate to disagree with you, I have agreed on all other points I have read as you eloquently laid each out for the readers here… but this one point, Caballero or not, I must point out my disagreement.
(Loved you article on beer, used to be a brewer back in college myself… still have the five gallon earth crock pot,
cases long neck Bud bottles and a capper in the attic! Maybe at some later point a post on the Brownian suspension of particles seen in a freshly brewed glass over time would interest all.)

wayne
January 20, 2012 5:52 am

Robert, one explicit description I never mentioned in my above contention: this is about multiple horizontal Boltzmann distributions at each and every level upward a tall gravitationally held column with an actual DALR. Without that thought in mind my description above may be a bit foggy.

January 20, 2012 6:33 am

When my science conflicts with established science, I research it until I find the error. When I’m wrong, I admit it. So let me join you in your recommendation that others do the same.
…and Well Done, sir!
Others including myself. I’m still learning atmospheric science myself, and I expect that learning process to continue for some time. I just start with a better understanding of basic thermo and stat mech (I’ve published a number of papers on dynamical critical phenomena and regular critical phenomena, much of it based on simulations so that I’m quite familiar with modelling, idealization vs reality, thermalization/relaxation processes and so on). “Established Science” is nearly always incomplete and overidealized, and is often overtly wrong — with the exception, in context of established basic science, the fundamentals of physics such as classical mechanics, classical thermodynamics, classical electrodynamics (including optics).
There we both know the science itself very well indeed, and we also know very well indeed at least how the idealizations that underlie them transition to and break down at the next level of our understanding, quantum mechanics, quantum thermodynamics, quantum electrodynamics and more general field theory, relativistic theories classical and quantum. Somewhere out there our knowledge of fundamentals starts to fray at the seams, so that I have no good idea of how gravity, general relativity, and quantum theory consistently merge (although there are proposed mergers, none of them quite “work” and it is almost impossible to find experimental evidence of e.g. gravity waves or what’s really happening at the Planck scale).
The other place our knowledge is pretty seriously bounded is in complex systems, open systems, self-organizing systems, nonlinear systems, chaotic systems. There we may know in principle all of the underlying fundamentals, but still lack computational or algebraic tools to be able to work out the math to compare a theory of how it is all supposedly put together to nature, often complemented by a lack of convenient laboratories where we can just go in an measure things. Parts of physics end up being competing, nearly incomputable theoretical models for things that we can’t directly observe and that we certainly cannot confirm or directly measure in laboratory experiments.
Black Holes, for example. As a consequence things like Black Hole physics have a long and checkered history in physics, where most of the “conflict” has been associated with constructing physically consistent theories. Susskind’s book Black Hole Wars is well worth reading in this context (and is written to be accessible to a lay person) as it shows how even some of the most brilliant physicists on the planet can be misled, seduced by their own arguments into making grave pronouncements that later proved to be inconsistent with established physics and that slowly gave way to far more “interesting” theories (that are far more difficult to compute) that are not, so far as we can yet tell, inconsistent. Or at least, they are inconsistent with different things.
Atmospheric physics and climate modelling is in this latter category. It is easy to assert that it is all based on “established science” — in one sense that is true. Most of the dynamics are Newtonian, straightforward. Most of the quantum theory is semiclassical and semi-empirical, not requiring the complexities and computational uncertainties of serious quantum electrodynamics. There are plenty of uncertainties even in the fundamentals — witness the ongoing argument concerning the empirically observed modulation of GCR flux and the role of GCRs as a supplement to aerosols that can affect the dynamics of cloud formation to shift things like the Earth’s albedo significantly away from any sort of “mean value”, turning it into a function of solar state (with numerous climate feedbacks). There appears now to be direct evidence that the Earth’s albedo is indeed increasing, although there is far too short a baseline so far to reliably relate this to anything at all, and there are multiple phenomena that might explain it, not just the solar model.
Between the physics and physical chemistry that is still not, actually, established science even at the level of the fundamentals, the extreme multivariability of the models themselves, the nonlinearity of the models, the fact that the models are highly idealized in order to make them simple enough to compute, the fact that the inputs into the multivariable models are often poorly known or derived from idealized averages, the computational difficulty of the problem that puts strong limits on adding more detail, the fact that the models are already chaotic in the extreme and consequently nearly useless as actual predictors of behavior, and the evidence that the models fail to capture or spontaneously generate observed long time scale semi-stable dynamical structures of great importance in the transport of heat and the fed-back modulation of the parameters that are being entered as if there is no such modulation of feedback (because it makes the models even more difficult to compute and more chaotic and hence even less reliable and predictive), the safest conclusion to make is — wait for it — it’s a damn hard problem.
Just this morning I was facing one very important piece of it when I got up to drive my son to swim practice at 5. Measured air temperature outside is just above freezing — clear skies with a bit of haze, no wind. The ground is not frozen. The roadways are not frozen or icy. But my car is covered in a thin layer of frost — it is at least 1-2C degrees colder than either the ground or the air.
So much physics on display. The car, metal and glass, has very little heat capacity and is moderately isolated/insulated from the ground , so it radiatively cools faster than the ground in spite of the fact that its emissivity is probably not as great as that of the blacktop pavement. The pavement is part of an extended body with a very high heat capacity and sufficiently good conductivity that the ground beneath the surface warms the surface just enough to keep it from freezing as it loses heat in the same amount of time my car freezes. And what is the air doing while all of this is going on? Interestingly, it is cooling at almost the same rate as the ground, in spite of the fact that there is no wind and that heat-forced convection should have turned off as the ground cooled faster than the air. Why don’t we wake up with the

wayne
January 20, 2012 7:34 am

“I’m still learning atmospheric science myself, …”
Me too Robert. Had to top and write my agreement with that statement. I have already been wrong multiple times learning this climate science, but, that is how we learn, isn’t it. ☺

January 20, 2012 7:55 am

(Arrgh, stupid interface… continued from the previous patch)
… air at (say) 10C (the previous day’s high temperature) and the ground at (say) 0C, locked there by the latent heat of fusion on windless days? The ground quickly cools below the temperature of the air, creating an instant inversion that should turn convection nearly off. The air itself is a poor greenhouse emitter, is it not? It should cool much more slowly than the ground, Yet the ground and the car cool at about the same rate as the air around them (with the car cooling much faster than the ground), at least at the levels where my thermometer can reach, where by “cools faster” I mean “changes temperature” faster, not “loses more heat”. The ground almost certainly loses a lot more heat than the car or the surrounding air in the same amount of time. Maybe that’s the explanation — the air loses heat much more slowly, but it has much less heat to lose and in the end it pretty much keeps up.
The nearby lakes are even worse. They hardly change their temperature at all as the sun sets, and don’t warm much during the day either. They have a surface temperature that is almost never the same as either the nearby ground surface or the air. Then there is the effect of clouds and haze. On a cloudy day this time of year we often observe no variation of temperature anywhere near the ground. We’ll have a whole day — sometimes a couple in a row — where my thermometer reads e.g. 8C within a degree. Daytime, nighttime, doesn’t matter. 8C. Or 12C. Or 2C. Other days, as night falls it gets colder, when the sun rises, it gets warmer, with the clouds in place.
Mostly, our local temperature depends a lot more on the direction of the prevailing wind of the day than it does on overt radiation or cooling. When the wind and weather come up from the south, carrying Gulf moisture (or not), it tends to be warm. When we get an arctic high pressure center that blows down from the Northwest, it gets cold. When we get just the right combination of the two — moisture (which is usually warming) and cold air, we get snow. It would never snow on its own I don’t think, in NC the way it snows on its own in upstate New York or Maine or Michigan, but if we borrow some cold air from somewhere else we can make it. As long as you don’t live near the coast — the warm Gulf Stream just offshore makes ice and snow pretty rare indeed (even at points at exactly the same latitude) just a couple of hundred kilometers away. Although I have, in the last five years, sat on Ocracoke on Easter in April — Ocracoke is an outer banks island that sticks way out and is surrounded by water on both sides — being bombarded by ice and snow pellets instead of enjoying the more “usual” mild and temperate climate one expects by late April in NC. Hell, last frost in Durham is usually April 1, although I’ve seen May in the last twelve years (it killed all of my azaleas that year).
It’s this last part, the transport of heat laterally by the atmosphere (and the oceans) that I’m still working on. If I built a radiative climate model for Durham based on the assumption that Durham is in some sort of local radiative balance, the resulting temperatures would look nothing at all like the actual temperatures, with or without a greenhouse effect. Not even on average. A lot of the heat we lose from our patch of ground to radiation wasn’t absorbed here, it has moved up here from somewhere else. We are radiating more heat than one would expect on the basis of local equilibrium — the “ideal” heat associated with time of year, insolation, GHE and so on — plus all the heat that was actually absorbed somewhere semi-tropical and moved here.
Here my brain explodes — I can’t figure it out. I want to say that moving heat from the tropics to here is net cooling as far as the Earth is concerned. We’ve moved a packet of heat from a hot location to a colder location, doing work on it, and releasing it to space faster than it would have been released if it had stayed home. Everything in my intuition says that by providing an additional pathway to the 3K “reservoir” of outer space that is driven by free energy changes, the system cools faster, not more slowly. And yet the simple blackbody model considers lateral heat transport to be net warming.
My intuition — which could be wrong, mind you — tells me that it should almost be a physical principle. In a self-organizing open thermally driven system, spontaneously emergent structures should always increase the rate of energy transfer between the hot reservoir and the cold reservoir. Convective rolls are a case in point — they occur because they speed up the cooling of the hot side and decrease the average temperature of the fluid compared to a stratified conduction-only arrangement of the fluid in layers. If you have a pot of water with a fixed input of energy at the top surface and a fixed (equal) output of energy at the bottom surface, you will build up a certain temperature at the top and the bottom that suffice to drive the heat input conductively through the water to where it is removed. You will (or should) end up with a linear temperature profile, warm on top to cool at the bottom.
Put the same amount of heat in at the bottom and remove it at the top, and the problem becomes a lot more complicated. Convective rolls amost instantly appear, and carry water from the bottom (where it is warmed) directly to the top where it is cooled. The conduction rate that (along with the heat capacity of the water and rate of heat input at the surfaces) determined the final temperature gradient and hence average temperature in the first case is almost irrelevant now — only small packets of water are warmed only a little bit, and then they are zipped straight up to the top and the heat is removed. It feels like the average water temperature in this latter case should be much colder, just like my house is colder (for the same heat input) in steady state when I have walls with convection interrupting glass wool in them rather than just an air cavity large enough to support convection. The glass doesn’t really modulate the conductivity — it blocks convection.
So what I feel like I’m lacking in my mental model so far is a convective cooling effect for the Earth, one where more convection equals faster cooling and hence lower temperatures. Here is where the GHE is counterintuitive — it predicts net warming from the more uniform temperature distribution that (all things otherwise equal) should result from convection. But I don’t see that. It doesn’t really matter how the outer wall of my house cools — turning on convection should make the house itself lose heat faster than it would for the same amount of heat input from my furnace.
The only complicating factor in this is the T^4, which has to interact with r^2 d\Omega and the emission spectrum to determine outgoing flux. The house does have complicated walls. But my intuition suggests that this doesn’t matter — from the time of Prigogene on, we have understood that self-organized structures, structures that reduce the entropy of the system, almost invariably do so at the expense of increasing the entropy of their surrounding environment. The more structured the Earth’s atmosphere as the result of convection, nonlinear flow, and so on, the faster it loses heat to and thereby increases the entropy of the Universe proper.
Hurricanes are a perfect, if rather temporally local, example. Hurricanes are net cooling events, I think, from a climatological perspective. They take a hell of a lot of oceanic heat that would have taken a long time to cool out of the ocean radiatively in a stratified no-lateral-transport model and quickly transport it over a very large area where it is ultimately removed from the Earth far faster. It doesn’t materially affect the temperatures of the places where it carries the heat for very long, because it pushes them well above their equilibrium temperatures where the local rate of cooling will be much faster than it would have been out in the stratified ocean. Even in terms of the GHE, they blast oceanic heat up throgh most of the GHGs to where it can be lost quickly.
If things like convection, especially long term convective oscillations that move heat long distances both laterally and vertically, have a differential cooling effect, that is a strong negative feedback in the climate system. It also suggests that the climate will be very sensitive to even small changes in those oscillations. El Nino, for example, might blast heat around, locally warming the air and “climate” in lots of places, but still have a net cooling effect in the long run compared to what would happen if the heat gathered in the La Nina years just continued to build up. An increase in the violence of the swings might well be the visible signature of negative feedback in action, the system spontaneously adapting to lose heat faster and hence cool overall compared to the rate it would have lost heat otherwise.
I think that dynamical heat transport, in other words, is very likely as important as the GHE in both establishing and modulating the baseline equilibrium heat flow from the Sun “through” the Earth and out to space in ways that effect both the local distribution of surface temperature and overall “average” temperature of the surface from somewhere inside the crust out to the edge of the atmosphere.
Hard problem.
rgb

Joe
Reply to  Robert Brown
January 20, 2012 10:26 am

Robert,
The comments to your post just will not stop, which I think is great. I feel that your explanation of dynamic heat transport is correct with two small exceptions:
1) By radiation alone, the ground should not cool faster than the air even with differences of radiation efficiency for the simple reason that they are both at roughly the same temperature but the ground has 1000 times the heat capacity per unit volume than the air. Throw in convection and winds and you have a different story.
2) You feel that dynamic heat transport is at least as important as radiation effects in dissipating surface heat. I disagree just a little bit. I think it dominates! If you look back 550 million years or even longer, you see periods where the sun is 15% weaker, carbon dioxide concentration is 20 times higher, and the surface temperature is a lot colder or a lot hotter than today totally out of sync with each other. Something else sets the thermostat.

January 20, 2012 8:02 am

Robert Brown:
I appreciate the time you took with your response. Being just a retired lawyer, I appreciate why you would consider it futile to dispel ignorance of the magnitude that must afflict one such as me.
However, it has happened more than once that a seemingly naive question from me caused a Ph.D physicist to abandon a theory in which he had theretofore been sure enough to base a new enterprise on. In this case my naive question is whether your certainty about what the assumptions are for the above-invoked thermodynamics laws’ derivation is so great that you can be confident that their applicability extends to the issue I raised, namely, whether the theoretical lapse rate for the air column at equilibrium is actually zero rather than just small.
Velasco et al. derived their molecular-kinetic-energy-vs.-altitude e relationship—which, as I understand it, says the equilibrium temperature lapse rate is small rather than zero–from state densities dictated by the Hamiltonian equations. (I hasten to add at this point that I’m just parroting buzzwords; I found their derivations’ calculus challenging, and I have yet to master why theoretical mechanics dictates equal probabilities for equal state densities as so defined.) At least to this layman, that sounds like a more-fundamental approach than simply invoking a result, like “equilibrium is isothermal,” without showing that the assumptions on which such a result is based apply to the situation in which you intend to use it–and without showing that the result is actually intended to be exact rather than, as I understand Velasco et al. to contend, approximate, .
I’m a layman rather than a physics professor, so I fully expect that you are able to demonstrate where my misapprehension lies. I just think you have yet to do it.

January 20, 2012 11:53 am

“Robert Brown says:
Here my brain explodes — Everything in my intuition says that by providing an additional pathway to the 3K “reservoir” of outer space that is driven by free energy changes, the system cools faster, not more slowly. And yet the simple blackbody model considers lateral heat transport to be net warming.
My intuition — which could be wrong, mind you — tells me that it should almost be a physical principle. In a self-organizing open thermally driven system, spontaneously emergent structures should always increase the rate of energy transfer between the hot reservoir and the cold reservoir.”
Robert, some info on self-organisation out of chaos can be found from the Maximum Entropy Principle (MEP).
Papers about MEP in our climate system:
The
second law of thermodynamics and the global climate system: a review of the Maximum
Entropy Principle

Nonequilibrium
thermodynamics and maximum entropy production in the Earth system

Entropy
Production by Earth System Processes

Robert Clemenzi
January 20, 2012 11:55 am

Joe says:
January 20, 2012 at 10:26 am

1) By radiation alone, the ground should not cool faster than the air

This happens every night in the current real atmosphere. The surface cools very fast. Heat then flows from the much warmer atmosphere toward the surface, causing a temperature inversion. Just look at a few lapse rate plots. The thickness of the inversion layer is directly related to the IR opacity of the atmosphere. Adding more greenhouse gases will not “trap more heat” but instead makes the inversion layer thinner which, in turn, will make the surface colder.
At any rate, by looking at the data, it is obvious that the ground cools much faster than the atmosphere even though it has a greater heat capacity.

Myrrh
January 20, 2012 4:36 pm

Joel Shore says:
January 20, 2012 at 3:44 am
Myrrh says:
I’ll ask you again. What is the mechanism that makes the net come out as flow from hotter to colder?
Until you have that you have no way of stopping an ice cube warming your hotter house.
Myrrh: I’ve already explained to you a few months ago the modern understanding of the Second Law in terms of the statistics of large numbers of particles and how that leads from microscopic reversibility to macroscopic irreversibility. (Robert Brown has also explained it in one of the threads here.) This understanding is one of the triumphs of physics in the last century or so. You however refused to believe it .To deny all of modern physics that you don’t like is your prerogative but I am not going to waste my time with you.
===========
Perhaps I didn’t explain myself very well, or well enough. I’ll have another go.
But that “statistically” is based only on the known law that heat flows from hotter to colder, putting in a ‘and that includes that maybe there will be one event when it doesn’t’ might sound terribly sophisticated, but doesn’t add anything, but importantly, it isn’t any proof that there is such a thing as net flow from an exchange from hotter to colder and from colder to hotter that always ends in hotter to colder. That’s just semantics.
You must include a mechanism for that to happen in the real world.
That’s a hypothesis at best. This is science we’re talking about, prove it.
Put in flow of water.
Sure, there might be some infinitesimal chance that Niagra Falls will suddenly reverse direction.. (/sark).
What is happening at each stage of water flowing downhill? Is each molecule of water at every point in an exchange to get a net ‘always flows downhill’? Nope. All the molecules are flowing downhill. If there’s a reverse at any point it is because some outside energy/work is being done on that molecule to reverse its direction.
Just because I can create a ‘statistical explanation’ that there is an exchange of direction at every point doesn’t make that explanation mean anything or be saying anything, when I have to bring in something else to explain it irrelevant to it, with water flow it’s more obvious to see that ‘statistically’ is nonsense because you have to bring in an obviously impossible scenario for it.
That’s all I’m saying. Simply creating the idea that energy flows from colder to hotter ‘because photons travel in all directions’ isn’t good enough. Or “microscopic reversibility to macroscopic irreversibility” of large numbers. You have to prove it to make that real ‘statistical’ maths meaningful.
And, take the Sun for example, the photons from the Sun aren’t travelling in all directions, they are travelling in straight lines. So, your ‘statistical photon exchange’ has the same problem as with water flowing downhill.
So, you are using an unproven law ‘the net exchange is always from hotter to colder in the exchange of photons from hotter to colder and colder to hotter’- which you can’t prove exists – to prove the statistics of molecular large number possible reversal at micro to macro proves the law, and demanding that we bow before it in awe at its cleverness. Fail.

Joel Shore
January 21, 2012 6:53 am

Myrrh: I am not going to explain modern physics to you when you have shown that any intelligent explanations to you are a complete waste of time. If someone wants to remain ignorant of modern science, there is nothing I can do to stop them.
You will continue to be a source of pure amusement here, even among most fellow AGW skeptics.

Myrrh
January 21, 2012 4:27 pm

Joel Shore says:
January 21, 2012 at 6:53 am
Myrrh: I am not going to explain modern physics to you when you have shown that any intelligent explanations to you are a complete waste of time. If someone wants to remain ignorant of modern science, there is nothing I can do to stop them.
You will continue to be a source of pure amusement here, even among most fellow AGW skeptics.

Oh right, modern science = ‘there’s no empirical objective reality, reality is whatever the individual’s subjective interpretation decides it is’
Well, I’m still stuck in old fashioned science, and if you have no proof of your subjective interpretation of the world around you it remains at best unproven, hypothesis, or junk. Until you can provide proof your claim is valid in heat transmission you’re just pissing in the wind, and you don’t need any ‘modern statistical science’ to grasp what that result will do to you..
You can rewrite the second law by putting in extra processes, but that doesn’t prove that these processes exist. If you can’t see that, if that simple real science paradigm is beyond your modern science understanding, then of course you miss the point I’m making.
What I find terribly sad here, is that so many can’t see that the world you’re describing is not the world we see around us.

Brian H
January 22, 2012 5:04 am

Just to take the radiation/temperature calcs to the extreme, if the 255K average were ½ 0K and ½ 510K hemispheres, the OR would rise by a factor of 8. Same average temp, though!

Brian H
January 22, 2012 5:43 am

Davidmhoffer says:
“That being the case, the ONLY affect that CO2 can have in terms of GHE is to become part of the mechanism that redistributes energy from the tropics to the poles. “

I think that applies a fortiori to Venus. My speculation about Venus is that, at those high temps and densities, CO2–>CO2 radiative transfer is very swift and efficient indeed, so that the day and nightside temps are almost identical, despite a very slow rotation rate.

Baa Humbug
February 5, 2012 3:41 am

@Dr Brown, Mr Eschenbach or other interested parties.
I recently came across a published paper (relevant to this topic) that made intuitive sense to me, but the technical aspects are way beyond my capabilities.
I would very much appreciate a critique from those who may be kind enough to put in the time and effort.
Scrutinizing the atmospheric greenhouse effect and its
climatic impact
Gerhard Kramm, Ralph Dlugi

Abstract:
In this paper, we scrutinize two completely different explanations of the so-called atmospheric greenhouse effect: First, the explanation of the American Meteorological Society (AMS) and the World Meteorological Organization (W?MO) quan- tifying this effect by two characteristic temperatures, secondly, the explanation of Ramanathan et al. [1] that is mainly based on an energy-flux budget for the Earth-atmosphere system. Both explanations are related to the global scale. In addition, we debate the meaning of climate, climate change, climate variability and climate variation to outline in which way the atmospheric greenhouse effect might be responsible for climate change and climate variability, respectively. In doing so, we distinguish between two different branches of climatology, namely 1) physical climatology in which the boundary conditions of the Earth-atmosphere system play the dominant role and 2) statistical climatology that is dealing with the statistical description of fortuitous weather events which had been happening in climate periods; each of them usually comprises 30 years. Based on our findings, we argue that 1) the so-called atmospheric greenhouse effect cannot be proved by the statistical description of fortuitous weather events that took place in a climate period, 2) the description by AMS and W?MO has to be discarded because of physical reasons, 3) energy-flux budgets for the Earth-atmosphere system do not provide tangible evidence that the atmospheric greenhouse effect does exist. Because of this lack of tangible evidence it is time to acknowledge that the atmospheric greenhouse effect and especially its climatic impact are based on meritless conjectures.
http://www.scirp.org/journal/PaperInformation.aspx?paperID=9233
This is an open access paper in pdf format 8mb
Thank you in advance

Brian H
February 5, 2012 5:35 am

Baa H.;
Wow, that’s one thorough piece of work! I notice they demolish Halpern’s critique of G&T early on:
“If it is possible to publish such a physically
inadequate comment, we have to acknowledge that the
discipline of climatology has lost its rational basis.”

Baa Humbug
February 5, 2012 7:05 am


That’s what I thought (thorough) but I can’t verify the math.
Assuming the math is correct, this paper warrants scrutiny don’t you think Brian?

Brian H
February 5, 2012 6:26 pm

Yes, one of the most important papers for a while, I’d guess.

Robert Clemenzi
February 11, 2012 10:07 pm

Brian H says:

Just to take the radiation/temperature calcs to the extreme, if the 255K average were ½ 0K and ½ 510K hemispheres, the OR would rise by a factor of 8. Same average temp, though!

Which is why you have to use T**4 for the average. In that case, the warm side would be 303K, not 2*255=510K.

Brian H
February 14, 2012 10:52 am

Robert Clemenzi says:
February 11, 2012 at 10:07 pm
Brian H says:
Just to take the radiation/temperature calcs to the extreme, if the 255K average were ½ 0K and ½ 510K hemispheres, the OR would rise by a factor of 8. Same average temp, though!
Which is why you have to use T**4 for the average. In that case, the warm side would be 303K, not 2*255=510K.

Uh, yeah. That was, I believe, my point. Averaging temps gets dumber and dumber the wider the range of temps in the sample.