People Living in Glass Planets

Guest Post by Willis Eschenbach

Dr. Judith Curry notes in a posting at her excellent blog Climate Etc. that there are folks out there that claim the poorly named planetary “greenhouse effect” doesn’t exist. And she is right, some folks do think that. I took a shot at explaining that the “greenhouse effect” is a real phenomenon, with my “Steel Greenhouse” post. I’d like to take another shot at clarifying how a planetary “greenhouse effect” works. This is another thought experiment.

Imagine a planet in space with no atmosphere. Surround it with a transparent shell a few kilometres above the surface, as shown in Figure 1.

Figure 1. An imaginary planet surrounded by a thin transparent shell a few kilometres above the surface (vertical scale exaggerated). The top of the transparent shell has been temporarily removed to clarify the physical layout. For our thought experiment, the transparent shell completely encloses the planet, with no holes. There is a vacuum both inside and outside the transparent shell.

To further the thought experiment, imagine that near the planet there is a sun, as bright and as distant from that planet as the Sun is from the Earth.

Next, we have a couple of simplifying assumptions. The first is that the surface areas of the planet and the shell (either the outside surface or the inside surface) are about equal. If the planet is the size of the earth and the transparent shell is say 1 kilometre above the surface, the difference in area is about a tenth of a percent. So the simplification is warranted.

The second simplifying assumption is that the planet is a blackbody for longwave (infra-red or “greenhouse”) radiation. In fact the longwave emissivity/absorptivity of the Earth’s surface is generally over 0.9, so the assumption is fine for a first-order understanding. You can include the two factors yourselves if you wish, it makes little difference.

Let’s look at several possibilities using different kinds of shells. First, Fig. 2 shows a section through the planet with a perfectly transparent shell. This shell passes both long and shortwave radiation straight through without absorbing anything:

Figure 2. Section of a planet with a shell which is perfectly transparent to shortwave (solar) and longwave (“greenhouse”) radiation.

With the transparent shell, the planet is at -18°C. Since the shell is transparent and absorbs no energy at all, it is at the temperature of outer space (actually slightly above 0K, usually taken as 0K for ease of calculation). The planet absorbs 240 W/m2 and emits 240 W/m2. The shell emits and absorbs zero W/m2. Thus both the shell and the planet are in equilibrium, with the energy absorbed equal to the energy radiated.

Next, Figure 3 shows what happens when the shell is perfectly opaque to both short and longwave radiation. In this case all radiation is absorbed by the shell.

Figure 3. Planet with a shell which is perfectly opaque to shortwave (solar) and longwave (“greenhouse”) radiation.

The planet stays at the same temperature in Figs. 2 and 3. In Fig. 3, this is because the planet is heated by the radiation from the shell. With the opaque shell in Fig. 3, the shell takes up the same temperature as the planet. Again, energy balance is maintained, with both shell and planet showing 240 W/m2 in and out. The important thing to note here is that the shell radiates both outward and inward.

Finally, Fig. 4 shows the energy balance when the shell is transparent to shortwave (solar) and is opaque to longwave (“greenhouse”) radiation. This, of course, is what the Earth’s atmosphere does.

Here we see a curious thing. At equilibrium, the planetary temperature is much higher than before:

Figure 4. Planet with a shell that is transparent to shortwave (solar) radiation, but is opaque to longwave (“greenhouse”) radiation.

In the situation shown in Fig. 4, the sun directly warms the planet. In addition, the planet is warmed (just as in Fig. 3) by the radiation from the inner surface of the shell. As a result, the planetary surface ends up absorbing (and radiating) 480 W/m2. As a result the temperature of the surface of the planet is much higher than in the previous Figures.

Note that all parts of the system are still in equilibrium. The surface both receives and emits 480 W/m2. The shell receives and emits 240 W/m2. The entire planetary system also emits the amount that it receives. So the system is in balance.

And that’s it. That’s how the “greenhouse effect” works. It doesn’t require CO2. It doesn’t need an atmosphere. It works because a shell has two sides, and it radiates energy from both the inside and the outside.

The “greenhouse effect” does not violate any known laws of physics. Energy is neither created nor destroyed. All that happens is that a bit of the outgoing energy is returned to the surface of the planet. This leaves the surface warmer than it would be without that extra energy.

So yes, dear friends, the “greenhouse effect” is real, whether it is created by a transparent shell or an atmosphere.

And now, for those that have followed the story this far, a bonus question:

Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?

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TomB
November 27, 2010 12:01 pm

Because we’re not in a vacuum inside the shell?

keith at hastings uk
November 27, 2010 12:13 pm

Well, how about day & night, water vapour (incl stratospheric) & clouds, heat transport by atmosphere & water currents to the poles, albedo changes, biosphere reactions, oceanic heat storage, solar variations (incl Milankovitch cycles), cosmic ray effects, aerosols, soots, volcanoes – and I guess a few more things.
I don’t pretend to be an expert, btw.

November 27, 2010 12:14 pm

It only deals with radiation and does not include the major energy storage and transfer mechanisms in our atmosphere water and how the maintenance of an equilibrium in the amounts of each state of water (solid – liquid – vapor) stores and releases energy independent of temperature at each phase change.

val majkus
November 27, 2010 12:22 pm

TomB thanks for going first; I’m no scientist but I would guess that’s the answer; from my understanding the earth’s atmosphere is not uniform, for example there are oceans over about 70%; there are variations in cloud cover, vegetation, terrain and all these things make a difference to the way radiation is absorbed and reflected. I know this is pretty simplistic and I’m looking forward to being educated by all you clever people out there; thanks Willis for such an interesting puzzle to start off my Sunday

XCapglider
November 27, 2010 12:23 pm

Thanks for clarifying the ‘greenhouse effect’ by means of the Steel Greenhouse and Glass Planet posts. Because the term is a misnomer, there is a lot of confusion about this topic.
This is the most simple and to-the-point explanation I have found:
http://www.ems.psu.edu/~fraser/Bad/BadGreenhouse.html
Simple and beautiful, indeed.
Regards,
Noud.

November 27, 2010 12:26 pm

So, if only those few molecules of water/CO2 are the reason why Earth night is warmer than night on Moon, why on Sahara, with its negligible humidity, today midnight temperature will be +19°C? Overall “greenhouse effect” there is by magnitude weaker than over equal, but humid area.
http://www.meteogroup.co.uk/meteo/gfs/MediterraneanSea/2010112712/MediterraneanSea_2010112712_tmin2m_12.png
http://www.meteogroup.co.uk/meteo/gfs/MediterraneanSea/2010112712/MediterraneanSea_2010112712_rh925_0.png
How long the simple heat-keeping capacity of remaining >99% of the atmosphere will be totally ignored?
Btw, real experiment was done by well known physician Woods.
http://neighbors.denverpost.com/blog.php/2009/02/04/greenhouse-theory-disproved-a-century-ago/

Edvin
November 27, 2010 12:29 pm

Convection (et. al.). Which is the main contributor to keeping real greenhouses warm.

pig farmer
November 27, 2010 12:30 pm

because it doesn’t include clouds and atmosphere/GHG’s?

Will Nelson
November 27, 2010 12:33 pm

Adiabatic lapse rate?

Thierry
November 27, 2010 12:36 pm

Here comes the -18°C again. In their excellent paper “Falsification Of The Atmospheric CO2 Greenhouse Effects Within The Frame Of Physics”, the two german physicists have clearly explained that -18°C comes from a wrong way of applying Stefan-Boltzmann law. On the non rotating and atmosphere free planet, the real average temperature would have been -129°C, which is more or less what we have on the moon. They have also explained why this glass approach is not to be used, since the so called greenhouse gazes do not behave as a physical or reflecting barrier. This post is unfortunately pointless.

pettyfog
November 27, 2010 12:36 pm

aarrggh!
Some time ago this site had a GREAT graphic showing the actuality of the ‘shield’. It aint a light filter over a petri dish!
It was a global shell AROUND a globe, not a flat shell with some translucence. I fully accept that the carbon based molecule absorbs long wave radiation and re-emits it. However that radiation emits in all directions, to be absorbed by the next molecule it happens to come across. Which then re-emits, and so on…
But here’s the kicker: Does, or does not, that radiation travel in a straight line?
And what happens to ANYTHING projected in a straight line from, or above, the surface of a globe?
Now forgive me, I’m just a poor dumb engineer {and electrical, at that}, son of a poor dumb farmer, but how the hell does it not occur to most science and physics wonks to take that MAJOR factor into account?
So, SURE, to extrapolate to water vapor.. a cloudy night after a cloudless day is gonna be warmer than a cloudless night after the same cloudless days. And maybe the day after too.. but EVENTUALLY an equilibrium is reached. Regardless of the amount of vapor on a given night.

Archonix
November 27, 2010 12:39 pm

Convection and heat transport from the tropics to the poles, I’d say. That completely changes the dynamics of the whole system.

Steeptown
November 27, 2010 12:42 pm

The real situation is far more complex, with a continuum of temperatures through a continuous medium, not through a shell, with conduction and convection, with a water cycle, with a spectrum of IR frequencies, with the SR and IR from the sun cycling down to zero every 24 hours, with etc, etc.

Anything is possible
November 27, 2010 12:44 pm

Because the shell is not opaque to all long-wave radiation, only that at certain wavelengths, nor is it transparent to all short-wave radition, a proportion of it is reflected, mainly by clouds.
This is where things start getting complicated…

EthicallyCivil
November 27, 2010 12:48 pm

Off the top of my head…
1) The “shell” is a gas column of varying temperature
2) Convection transports engery through “the shell”
3) The shell is is a multi bandgap filter, not opaque
4) The “shell” bandgap filters are a non-linear function of the temperature, given the that critical gas components (water vapor) undergo phase change within the range of

Stephen Brown
November 27, 2010 12:49 pm

The earth is not a black body insofar as radiation and absorption of heat is concerned. The equation-confusing factors of gases and vapours between the shell and the ‘not-black-body’ planet are conveniently missing as well.

okie333
November 27, 2010 12:49 pm

Because the atmosphere has many distinct layers, because CO2 is not the only gas in the atmosphere, because all of the gases have different properties that must all be considered, and because cloud cover is not being considered?

Vorlath
November 27, 2010 12:49 pm

Your third graph doesn’t work if the shell is opaque.

Sam Hall
November 27, 2010 12:51 pm

Your results are correct, but your diagrams are sure messed up. They imply that the earth is an energy source.

John S
November 27, 2010 12:56 pm

Where are the feedback mechanisms? No polar ice caps, no water vs. land specific heat capacity, no cloud formation.

pettyfog
November 27, 2010 1:00 pm

The Earth IS a long wave infrared source. Taking the ‘vacuum under the opaque shield’ instance mentioned above, the earth’s crust would not be at absolute zero because of magma temperatures bleeding off through the crust.

Slabadang
November 27, 2010 1:02 pm

Willis!
Well ive had this question in m head for a long time and I think you have to calculate the atmospfere as a part of the “blackbody`s surface” to get the energy balance right.
“The atmospfere is no shell its a part of the surface” thats my answer.
And I am a total layman in fysics so its just a guess. 🙂

Stephen Wilde
November 27, 2010 1:09 pm

The oceans are far more significant in terms of slowing the sun’s energy loss to space than anything the air can achieve:
See here:
http://climaterealists.com/index.php?id=1487
” The Hot Water Bottle Effect”
and for a more general analysis see here:
http://climaterealists.com/index.php?id=1562
“Greenhouse Confusion Resolved”

DocMartyn
November 27, 2010 1:09 pm

You cannot use the word ‘equilibrium’ to describe a steady state, the thermodynamics of steady states and equilibrium systems are quite different.
You models suck. The Earth rotates. At least 50% of the time the Earth is radiating into space and is not absorbing light. However, your greenhouse material is a superconductor and must be at the same uniform temperature all over.
(You will also note that we can establish the (internal) heat radiated from the actual planet from the mid-winter Antarctica, -89°C or 184K)

Oxbridge Prat
November 27, 2010 1:12 pm

Multiple concentric shells.

FrankK
November 27, 2010 1:13 pm

Variability of cloud cover thickness and distribution; atmospheric air flow from tropics to poles; transmission of internal core heat; there’s no “plastic shell” in reality

Noblesse Oblige
November 27, 2010 1:22 pm

The alititude of the effective radiating surface of the earth is a function of concentration. Below this altitude heat is transported mainly by convection/conduction. Above this altitude it is transported via radiation. Glass greenhouses don’t do this.

Golf Charley
November 27, 2010 1:29 pm

clouds?

Vorlath
November 27, 2010 1:30 pm

Willis Eschenbach says:
November 27, 2010 at 1:21 pm
Vorlath says:
November 27, 2010 at 12:49 pm
Your third graph doesn’t work if the shell is opaque.
Huh? Why not?
———————
Because if the shield is opaque to longwave, then the only way you can have longwave emitted from said shield is if it came from outer space. This would also mean that no longwave could escape from inside and your calculations are wrong. What your third graph is describing is something *like* a black hole with respect to shortwave radiation.
To fix it, you’d need something that allows half of the longwave radiation through.
Also, the planet acts as an energy source in all three graphs. Where’s that energy from?

Malaga View
November 27, 2010 1:30 pm

am not so sure about the left hand up arrow in Fig 4 being at 480 W/m2… I would have thought 240 W/m2…
However, for the earth, it all depends on how opaque (or transparent) the atmosphere is… and we know cloud cover is variable… and the sun might encounter snow, ice, water, plants, sands, rocks etc on the surface of the earth… we also know that the heating effect of the sun is not instantaneous – it take times… as does the cooling in the evening… and then we have the other half of the day when the sun has gone to bed… and we know it gets cooler at night.
Now my thought experiment is more of a real experiment… I park my sedan in the middle of the car lot at my local Wall Mart in the early morning… and I sit in the car… as the sun comes up I get hotter… and by midday it is hotter in the car than it is outside the car… and by midnight it is older inside the car than it is outside… it might even be frosted over while the parking lot is frost free. Now being tired and hungry I go home.
So the next day I decide to change my experiment a little bit… I take my girl friend’s convertible to the car lot… now when it gets hot I open up the roof to let the heat out…. and I don’t close it at night and the car stays frost free.
Therefore, my conclusions are:
1) The greenhouse effect only works when the air is trapped inside the car… after all hot air rises.
2) If the sun stayed up all night then it would be too hot all the time.
3) If the sun stayed in bed all the time then it would be too cold all the time.
4) It takes time for everything to warm up in the morning.
5) It takes time for everything to cool down in the evening.
So my guess is that the earth has a very clever mechanism called CLOUDS… they work a bit like a convertible… they form when it gets hot and stops the earth from getting too hot… and they tend to disappear when it gets very cold so the earth has its best chance of being heating up the sun… but as to pinning down the precise science of the heat flows… dream on… everything is variable and subject to time delays.

Owen
November 27, 2010 1:32 pm

The shell is best represented by a “venetian” blind, or slat blind – with the slats vertical to the surface & gaps between the slats – as the “greenhouse” is only absorbing a small portion of the radiation – the rest escapes directly to space. The slats are also “opaque” as the absorb all they can – and making the slats longer in the vertical direction (more greenhouse gases) has little additional effect.

Kev-in-UK
November 27, 2010 1:37 pm

Just looked into this post and comments and I am puzzled at the lack of mention at the fact there is no real equilibrium in the real earth. solar in and energy out are highly unlikely to ever be in equilibrium, the atmosphere and oceans, planetary surface, etc are all just giant storage heaters, storing energy at daytime and releasing at night. some days the charge is hotter, some days its less hot (due to albedo etc). but the time period (lag if you like) it takes for the thermal ‘mass’ to equalise around the planet and then emit out to space and this will always define the actual climate temperature as rising or falling. (I am ignoring the different ‘spheres’ upper trop, etc).

November 27, 2010 1:39 pm

Come on this is stupid. The greenhouse effect doesn’t exist for the simple reason that there is no “greenhouse” effect in a greenhouse – it would be like using the theory of the ether to explain radio waves and talking about the “ether effect” and not expecting most people who knew what was being talked about not to roll with laughter on the floor.
And … even putting it in quotes as in the “greenhouse” effect doesn’t explain why CO2 has to result in warming when it has a cooling effect due to the greater emissivity of CO2.
To summarise: the greenhouse effect is Noddy science and even when it is “Noddy” science it still doesn’t follow CO2 causes warming.

val majkus
November 27, 2010 1:42 pm

All right I went back and read your Steel Greenhouse post to get a clue, is the answer that upward radiation does not equal downward radiation?

Phil C
November 27, 2010 1:43 pm

well, the model system doesn’t have climate- no atmosphere, water, land, or biomass- all of which constitute climate, or perhaps more correctly, weather. So of course it doesn’t explain anything about climate or why or how the energy transfer through the atmosphere to the ground and back affects climate.
the model system is a good explanation of why the earth is in energy balance and perhaps the limits to the temperatures involved, as opposed to the current AGW hypothesis that CO2 in the atmosphere has a positive feed back that increases the response of temperature.

Malaga View
November 27, 2010 1:46 pm

PS
The other big problem with Fig 4 is the right hand down arrow… although you argue that 240 W/m2 is radiated back towards earth this does not mean that all this energy is received by the earth… there is a problem with distance… remember if you stand close to the fire you get hot… and as you step way from the fire you cool down… so size is important too 🙂

Tom in Florida
November 27, 2010 1:47 pm

Too many assumptions that are not real world. Kind of like thinking a sports team will always win because on paper they have the better players. Unfortunately, they still have to play a real game.

Kev-in-UK
November 27, 2010 1:51 pm

BTW – is there not a shift in emission spectrum due to higher BB temp? and of course, this would constantly change slightly, resulting in permanent non equilibrium condition?

Gary
November 27, 2010 1:58 pm

But there is a much more fundamental reason why that model is inadequate to represent the Earth’s climate system in even the most simplified way … and no one has mentioned it yet.

No thermostatic regulation in the model?

Malaga View
November 27, 2010 1:59 pm

PPS
Simple models simply don’t work for complex situations!
Just look at the accuracy of the AGW computer models if you need confirmation….

November 27, 2010 2:00 pm

A transparent shell would not be at 0K. If it were perfectly transparent (which is physically unreal, but never mind) it would retain whatever temperature it had when you first put it there, because it is neither absorbing nor losing any heat. If it were merely very nearly transparent (which is the physically realisable case) , then in this example it would be at a temperature ~ 255K. The temperature would be somewhat higher or lower than this, depending as the emissivity were higher or lower in the visible or thermal.
This model implicitly assumes uniform insolation, which is not valid for a sphere illuminated from one direction. Steady state requires that the planet does not rotate (relative to the sun), or has negligible heat capacity; the temperature would then range from a factor 4^1/4 higher (ie 361K) at the subsolar point, through ~73K around the terminator, to ~0K on the dark side. There would be no meaningful average or global temperature at all. The greenhouse effect of the shell would be the same factor of 2^1/4 (for the visibly transparent, thermally opaque case). Add diurnal rotation and finite heat capacity and thermal conductivity, and things become much more complicated. There is no steady state, and the temperature curve for a given latitude is messy to compute. The thermal capacity of the shell has to be considered too. The general effect of thermal capacity and rotation is to transfer (some) heat between the day and night sides.

INGSOC
November 27, 2010 2:12 pm

Doesn’t the atmosphere deflect some of the incoming radiation out to space without even penetrating?

Braddles
November 27, 2010 2:13 pm

The comments that dispute Willis’ excellent post highlight some of the problems with the AGW sceptic arguments. Those who argue that the greenhouse effect does not exist do the sceptics’ side a disservice, by arguing an untenable position. It creates a weakness in the sceptical position, which warmists focus on to try to discredit all sceptic arguments.

INGSOC
November 27, 2010 2:15 pm

In other words, the atmosphere “denies” entry of a lot of what the sun etc tosses our way, but not all, and that changes all the time for various reasons.

November 27, 2010 2:17 pm

Willis – I can’t be sure what model inadequacy you have in mind, but perhaps the answer is that climate involves much more than just temperature. Another thought is that climate (and weather) vary with time and location on the Earth’s surface, but your model seems incapable of representing either of these aspects.

Barry L.
November 27, 2010 2:21 pm

Willis
Thanks for the post
Where is Length of day? Doesn’t this have an affect on outgoing radiation?
Picture Earth with the day on one side, and night on the other. Just like a roast on a spit. mmm
http://farm1.static.flickr.com/1/369706_795fabc842.jpg?v=0

Jack Maloney
November 27, 2010 2:24 pm

DocMartyn says: The Earth rotates. At least 50% of the time the Earth is radiating into space and is not absorbing light.
Have I missed a response to this? The graphics appear to assume a Flat Earth, which reinforces what the RealClimate folks have been saying about climate skeptics… 😉

Schrodinger's Cat
November 27, 2010 2:31 pm

I guess that a couple of relevant factors would be the extinction coefficient or saturation of the long wave absorption capacity of the shell and the permanence or otherwise of the shell (residence time of GHG in our case).

James
November 27, 2010 2:33 pm

Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?
Too hot?
Third picture shows equilibrium of ~30 C at the surface. Average for Earth is lower than that.

Bill Illis
November 27, 2010 2:36 pm

Here is my short explanation of why it is wrong (this could be expanded into several books).
At the real Earth surface, the atmosphere has a molecule every 3.7e-26 square metres. The longwave photons, therefore, have to travel by/through 5.4e+30 (54 with 29 Zeros behind it) layers of atmospheric molecules (opaque shells) on its way to space.
About 20% of shortwave photons from the Sun hits one of those 5.4e+30 molecules on its way to the surface so a longwave photon only needs to hit a few more than 20% to cause just enough lag in the time between the two types of photons so that the Earth’s surface is warmer than it should be. Then factor in the fact that the Earth’s land and ocean surface then absorb 85% of the 85% of the shortwave photons that reach the surface and those land molecules hold onto them for a little while and the surface is a nice comfortable 288K (for some of us).
All physics equations should have “Time” included in them. Given that 1 watt/m2 = 1 joule/m2/second, the shortwave photons actually come in 960 joules/m2/second at the height of the day (which is 3e+21 individual sunlight photons per square metre per second) and 0 joules/m2/second at night (but don’t forget the Cosmic Background Radiation).
The surface temperature at 2 metres, however, only changes by +0.001 joules/m2/second during the day and loses 0.0017 joules/m2/second through the night despite the large difference in shortwave energy coming in during that 24 hour period. Given the longwave photons are at lower energy levels, there are still almost as many thermal photons emitted per second at night as the sunlight coming in at the height of the day. Lots of layers and molecules have to be navigated by that unbelievable number of photons even though they travel at the speed of light and could escape the atmopshere in 0.0003 seconds if they weren’t intercepted so often.
The greenhouse effect description needs to move to the level it occurs at, the quantum level. Technically, each shortwave photon from the Sun ends up in about 8 billion different molecules before it escapes from the Earth system – that is a lot of random walking for a lot of individual photons.

Dr A Burns
November 27, 2010 2:38 pm

Here’s a detailed explanation including the very important effect of gravity:
http://realplanet.eu/atmoseffect.htm

Baa Humbug
November 27, 2010 2:42 pm

Layer upon layer upon layer

James
November 27, 2010 2:45 pm

Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?
Earth is a sphere?
Your picture illustrates one point on the surface. But if you “zoom out” to show whole earth, you realize it’s receiving radiation on one hemisphere but radiating to space from a full sphere.

Vorlath
November 27, 2010 2:51 pm

Willis Eschenbach says:
November 27, 2010 at 2:36 pm
Vorlath, the shell (shield) absorbs shortwave and emits longwave … as does almost everything on the earth. We are warmed by the shortwave radiation of the sun, and re-radiate that energy as longwave (infra-red) radiation.
———————-
I’m talking about Figure 4, the third graph.

pax
November 27, 2010 2:53 pm

Your model is inadequate because the earth rotates and so the incoming radiation will oscillate. In other words, your model is static, while the real world i dynamic.

Baa Humbug
November 27, 2010 2:56 pm

I think it’s erroneous to add the Wm2 from the sun to the Wm2 from the shell (atmosphere) which effectively doubles the energy i,e, the original 240Wm2 becomes 480 Wm2.
Take a basic strip heater of say 1000W, the above fallacy suggests all we have to do is add a well conducting bar across the front of the element, say an inch away (unwired and not receiving any power, i.e. benign) and voila’ we now have a strip heater putting out twice the energy.
Can someone help me with patents please?

Iskandar
November 27, 2010 2:57 pm

The greenhouse effect is the most abused non-existing item on this planet.
Let me explain:
CO2 does have an absorption in the far IR. However, H2O does also. CO2 sits in a window of H2O vapour.
Both of them absorb the black body radiation of the earth. The fact is, earth is NOT a perfect black body. It is a reflecting, colourful body, which means that it reflects a lot of its incoming energy. When emitting, it will only emit radiation with a max temp of 300K. Which means that it does nothing to heat the atmosphere. All atmospheric gasses emit the same amount of radiation at their black body temperatures. Since there is no boundary layer, emissions cancel out.
Ther is no green house effect.

K
November 27, 2010 2:58 pm

The shortcoming in the model is that each molecule absorbs heat emitted from the earth, the re-radiates it out spherically, meaning that half of the energy is returned to the earth, and that this process repeats itself until vacuum is reached. So the surface will be warmer than the top, though the earth will always find equilibrium and emit its 240 W/m2. Or did I not get the right answer?

simpleseekeraftertruth
November 27, 2010 2:58 pm

My use of the greater than symbol has messed up the html! Again:
Day/night cycles are not considered. Day: outgoing is less than incoming. Night: outgoing is greater than incoming. The change (raising) of surface temperature is the difference between these two budgets.

Michael J. Dunn
November 27, 2010 3:04 pm

Dear Folks,
There is a simple way to understand this effect. It is a case of: Radiant Power in = Radiant Power out
The radiant power in is simply (S alpha piR^2), where S is the solar constant (watts/meter^2), alpha is the absorption coefficient, and piR^2 is the area of the Earth’s disk in terms of its radius R.
The radiant power out is given by the grey-body version of Stefan’s Law, equal to epsilon signma T^4 (4piR^2), where epsilon is the emission coefficient, sigma is the Stefan-Boltzmann constant, T is the surface temperature of the Earth, and the expression in parentheses is the surface area of the sphere of the Earth.
If we equate them, we get
S alpha piR^2 = epsilon sigma T^4 (4piR^2)
By cancellation and rearrangement of terms, the temperature is solved as
T = (S alpha / 4 epsilon sigma)^1/4
Now we are in a position to understand the influence of the various parameters. S and sigma are physical constants and are not subject to variation (in this analysis). The absorption coefficient, alpha, may vary from zero (perfect reflection) to unity (perfect absorbtion) and characterizes the Earth’s absorption of radiant power from the sun. If alpha = 0, the Earth cannot absorb any power and will cool to zero temperature. Likewise, the emission coefficient, epsilon, may vary from zero to unity and characterizes the Earth’s emission of thermal power from Stefan’s Law. If epsilon = 0, the Earth cannot emit any power and the temperature will increase infinitely.
The crux is this: if alpha = epsilon (both close to unity), then a certain equilibrium temperature will be reached that will be close to the “airless planet” temperature (for S = 1400 w/m2 and sigma = 5.67 x 10–8 w/m2K4 this will be T = 280.3 K or 7.15 C).
However, if alpha > epsilon, as is the case when there is imperfect emission (grey-body conditions, epsilon < 1), then the equilibrium temperature will be higher (if alpha/epsilon = 1.3 then T = 299.3 K or 26.15 C). This is what happens with a re-radiative atmosphere; it is simply equivalent to the Earth’s emission coefficient being reduced by the thermal radiation backscatter from the atmosphere.
Contrariwise, if alpha is reduced (as by higher albedo from increased clouds), the equilibrium temperature will be lower. As this equation makes clear, albedo has direct leverage over the result, whereas CO2 has very indirect leverage.
The mechanics of heat transfer within the atmosphere and to and from the oceans and land masses are details that affect the values of alpha and epsilon, but are not otherwise involved in the equilibrium radiative thermal balance among the Earth, Sun, and outer space.

jack morrow
November 27, 2010 3:05 pm

The earth is not in a vacuum.

Malaga View
November 27, 2010 3:07 pm

Willis Eschenbach says:
November 27, 2010 at 2:39 pm
In Fig. 4, the surface receives 240 W/m2 from the sun, and 240 W/m2 from the shell … how much should it radiate at equilibrium, if not 480W/m2?

Because a Thermos Flask helps keep things hot… but it doesn’t make them hotter… your only input is 240 W/m2 from the sun!

DocMartyn
November 27, 2010 3:10 pm

Willis, you have had the same answer twice; your diagram needs to show
Sun
————–Shield
EEEEEEEEE Earth
————–Shield
The shield radiates at the same temperature all over the sphere, but half the sphere is in darkness. This is why the heat flux budget diagrams are such bollocks, the Earth rotates. We do know that the Earth generates heat from internal radioactive decay; something missing from most heat budget.

Kev-in-UK
November 27, 2010 3:11 pm

Willis, I must be missing something too…
but I do not quite understand your query…the reason, in basic terms that the planetary surface is ‘warmer’ is because the rate of thermal energy (or rather just energy) transfer is reduced or ‘slowed down’ by the shell.
In terms of light transmission through a substance, the light is bent and deviated, resulting in slower (altered) transmission (i.e. refraction – which is different for different wavelengths too!) and this means the light travels a longer path. This could be analagous to the thermal ‘storage’ in that the atmosphere acts as a deviation for the thermal ‘light’ energy (ok – its radiation, but it is an analogy) – thereby lengthening its stay in the atmosphere, which is not a vacuum and its contents can be excited (vibrated) by the passing radiation, thereby allowing it to warm the atmosphere.
Mind you – it’s 11pm here in UK, and after a very nice red wine, with nothing on the TV – this may not be making much sense! LOL

Leonard Weinstein
November 27, 2010 3:11 pm

Willis,
Your model does show some essential features that determine planetary “greenhouse effects”. However, it does not tell the full (still simplified) story of what actually happens. The following is a somewhat more complete version. The point is that it is actually the combination of the fact that a lapse rate becomes established in combination with the movement of the outgoing radiation to a high altitude that is required.
Atmospheric greenhouse effect:
Once a sufficient amount of atmospheric long wavelength IR absorbing gases (called atmospheric greenhouse gases by convention) are present, they reduce the direct radiation transmission to space enough so that convection and relayed radiation becomes the dominant modes to transport absorbed energy from the surface to a high altitude where it is radiated to space. The only way the energy can be radiated to space from the high altitude is from the absorbing and radiating gases (or clouds, but that is another story). Once this situation is encountered, adding more absorbing and radiating gases (but not enough to significantly increase the total mass of the atmosphere) do not change the fact that convection is still the dominant heat transport mechanism within the atmosphere. In fact, this reduces the radiation conduction, so that convection carries even more of the total energy to the high altitude. However, adding more absorbing gas does raise the altitude of outgoing radiation somewhat (not because of the tiny added mass, but because the height where the radiating gas concentration is suitable to radiate to space is increased). It is this increase in altitude of the outgoing radiation that results in the slight temperature increase.
Once the dominant mode of heat transport is convection (buoyancy, winds, and turbulent mixing), the atmosphere will form and maintain (on the average) an adiabatic lapse rate. This lapse rate is a temperature gradient due to the cooling effect of rising gas in a dropping pressure (due to gravity). Evaporation from the surface and condensing water vapor in the atmosphere change the level of the lapse rate from the dry air value, and this is called the wet adiabatic lapse rate. The outgoing radiation has to equal the incoming absorbed radiation unless the temperature is changing, but here we consider the case where the temperature has leveled off for simplicity (as it has on Earth for the last decade or so). In that case, the match of radiation out to space, from a particular effective altitude, to absorbed input radiation, determines the effective temperature of the gas at that altitude. This temperature is then added to the lapse rate times the effective altitude of outgoing radiation, and this gives the ground effective temperature. The combination of moving the location of the effective level of the atmosphere, where the radiation to space occurs, to a higher altitude, and adding the effect of the lapse rate times increased altitude is the source of higher surface and low altitude temperatures.

Schrodinger's Cat
November 27, 2010 3:11 pm

Your model assumes a transparent shell and a vacuum within. These are not true for the earth since clouds can reflect or scatter the short wave radiation and the atmosphere is not a vacuum.

Lew Skannen
November 27, 2010 3:13 pm

I am intrigued by this thread.
Perhaps I have missed some glaring fundamental principle but the only answer to the bonus question that I can come up with is the scale of complexity of the model.
For any realistic calculation to be made the model would have to include all the goings on of the many layers of the shield as well as other factors like convection , cloud formation etc.
I think it is safe to say that once the radiation has arrived on Earth it can never again be emitted as short wave so the question is whether the long wave emissions can maintain the energy balance. It occurs to me that the long wave opacity of the atmosphere might have gaps in it which will affect the numerical calculations but I do not see any holes in the principle.
In summary: My guess is the inability of the basic model to cover all necessary parameters.

bubbagyro
November 27, 2010 3:17 pm

Molecules, like air, do not reflect energy. If they absorb, and then retransmit, it is not directional, but in all directions from the molecule, as a tensor, not a vector.

Steve Oregon
November 27, 2010 3:17 pm

It’s the impervious nature of a shell. Like the glass in a real greenhouse. It traps and elevates heat. Our atmosphere does not resemble glass or a shell.
If it did we would all be cooked.

bubbagyro
November 27, 2010 3:17 pm

Statistically in all directions, I meant.

Schrodinger's Cat
November 27, 2010 3:18 pm

Our atmosphere is multishell

Leonard Weinstein
November 27, 2010 3:23 pm

Willias,
I forgot to add: The higher value of the ground temperature results in a higher radiation out level from the ground than if the ground were not as hot. However, the atmospheric radiation from just above the ground (due to the local atmospheric temperature acting on a radiating gas) is omnidirectional, and the downward component, called back radiation, almost equals the upward radiation from the ground, resulting in a greatly reduced NET radiation heat transfer. It is convective heat transfer that carries most of the energy from absorbed solar radiation in the ground to the upper atmosphere. Thus the higher radiation level up and back radiation do not cause the heating, they are a result of it.

Edvin
November 27, 2010 3:23 pm

I think Paul Birch got it. The earth is a sphere, with a “point” source of energy in the sun a single 1D model can’t capture the system.

Rob M
November 27, 2010 3:27 pm

Earth’s ‘shell’ is not fixed.An atmosphere can expand/contract.

Vorlath
November 27, 2010 3:34 pm

I’ll just say it again as clear as I can since I and others have mentioned it several times. The last figure. Figure 4. The math is wrong. If the shell is opaque to longwave, then you have two possibilities for the 240W/m2 radiating out into space from the shell.
1. Longwave is coming from space or the Sun.
2. Shortwave is coming from the surface of the Earth and converting to longwave as it passes through the shell.
Neither of those are shown in your graph. You have all of 240W/m2 going to the surface of the Earth. So option #1 is no good. Second, you have blue lines indicating longwave radiation on the inside of the shell. So option #2 is no good. Option #2 has another issue that makes it wrong is that the line of radiation from the Sun is shown to go all the way through to the surface of the Earth without any conversion. So if there is no conversion going in, then there is no conversion possible going out either.
IOW, the math is critically wrong in figure 4. Please fix this before continuing.

Jimmi
November 27, 2010 3:35 pm

Well this looks as if it will be an interesting thread.
I am not sure what the most important missing factor is, but would vote for the fact that the earth rotates. This means that on average the incoming radiation only warms some of the time, but the outgoing is present all the time. However this only affects the average temperature, and the basic principle that warming occurs is still correct.

Lady Life Grows
November 27, 2010 3:44 pm

The Earth is vastly more complicated than the model, and has many known and unknown feedback mechanisms.
Most of these must be negative feedbacks, or we’d have had a runaway greenhouse or freezer effect eons ago and life would not have arisen.

AusieDan
November 27, 2010 3:51 pm

Willis – Roy Spencer claims that there is a net negative feedback which more than counters the CO2 effect.
He claimes that this is not just theory but that he has measured it from observations.
Please comment.

1DandyTroll
November 27, 2010 3:53 pm

The “greenhouse effect” is, if I remember, only a PR definition that was used to simplify stuff for the policy makers or something.
Secondly the greenhouses isn’t used to trap heat per se (if they were they’d be padded down with loads and loads of insolation) but to control the inside environment for maximum growth using sunlight, artificial light and heat (for night time and winter), humidity, CO2 and Nitrogen levels, pesticides, and of course water flow and nutrients. CO2 is of course not used to trap heat or to add to the heat because if it were that much of a problem plants getting heat stroke would be a real problem and you’d end up spending more money on cooling systems, so the trace gas is simply used as nutrient in a controlled environment.
So the analogy that gases acts with a “greenhouse effect” is just bad since the greenhouse is to control a balanced environment.

INGSOC
November 27, 2010 4:03 pm

I get the feeling I’ll be slapping my forehead and exclaiming; Doh! I shoulda thoughta that one!
Let me be the first (?) to state that I am in fact a blathering blowhard, and a stuffed shirt! I take what little I know and make up the rest! I pass wind at your silly little mystery question! Hahahahahahahahahahaha… I’m dressed like a turnip!

PJP
November 27, 2010 4:03 pm

Willis,
I see the following as a being a problem with your model:
The incoming energy (from the sun) you express in w/m^2, lets simplify it even more and say that energy is delivered in truckloads.
Lets say we get 2 truckloads per hour.
With your transparent shell, we end up with 2 truckloads in and two truckloads out, things are balanced.
But when we come to your semi-transparent shell, you are still getting two truckloads per hour, but you say that these two truckloads are delivered to both the earth and to the shell — that makes 4 truckloads/hr.
Where did the extra two truckloads come from?
You then make things even worse, by saying that the magical two extra truckloads that are delivered to the shell are again magically multiplied, with two going out to space, and two more going to the earth.
Where do these magical extra truckloads of energy come from?

u.k.(us)
November 27, 2010 4:08 pm

“There is a vacuum both inside and outside the transparent shell. ”
=======
A vacuum inside ??
Not my kind of planet.

Hannibal B.
November 27, 2010 4:10 pm

The sun does not hit the earth on all sides at once. There is no night and day in your model.

Baa Humbug
November 27, 2010 4:27 pm

Willis I’m having trouble with fig.3
If the shell is radiating 240Wm2 to space AND 240Wm2 to the planet, it must therefore be receiving a total of 480Wm2, which is what fig.3 shows.
IF it is receiving 480Wm2, how can it be the same temp as the planet surface i.e. 255K or -18C? The planet surface is receiving only 240Wm2.
Maybe I’m not comprhending this situation too well. To put it another way..
Take an energy source radiating 240Wm2 and point it at a black body object. This object will reach 255K.
Now, if we add a SECOND energy source of 240Wm2 from the opposite direction, according to fig.3 this object will stillbe at 255K???

Golf Charley
November 27, 2010 4:29 pm

If it is not the cloudy thingies, could it be that the basic theory is a load of dung beetles raison d’etre?

November 27, 2010 4:38 pm

“Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?”
Because the oceans also have their own “greenhouse effect”. They allow downward shortwave radiation (visible light) to warm the oceans to depths of 200 meters (with the warming diminishing with depth), but only release heat at the surface. John Daly made the argument that a planet that was all ocean would be warmer than a planet of all land. Refer to the following post inder the heading of “The oceans also behave this way”:
http://www.john-daly.com/deepsea.htm

Speed
November 27, 2010 4:38 pm

” … it does not give enough energy to allow for the known losses in the climate system.”

DirkH
November 27, 2010 4:43 pm

“Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?”
When your shell absorbs the blackbody radiation from the planet, it heats up to a lesser temperature than the planet’s surface, so it will emit a lower frequency blackbody radiation than it received. CO2 re-emits the same frequencies it receives.

DirkH
November 27, 2010 4:53 pm

Wait. Your figure 3 is wrong. 480 W go up to the shell, it must reach the same temperature as the planet’s surface, otherwise it can’t re-emit 480 W. (Again, this is not the case with CO2 in the atmosphere, as it re-emits the frequencies it receives due to absorption bands, not blackbody spectra)

DirkH
November 27, 2010 4:54 pm

DirkH says:
November 27, 2010 at 4:53 pm
“Wait. Your figure 3 is wrong. ”
Sorry, i mean figure 4

davidmhoffer
November 27, 2010 4:55 pm

Observed from a distance, Mr Eschenbach’s imaginary planet and shell would appear to be in equilibrium at 240 w/m2 in and 240 w/m2 out in the first part of his explanation. In the latter part of his explanation, with the shell opaque to LW, oberved from a distance the planet would appear to be in equilibrium at 240 w/m2 in and 240 w/m2 out. In other words, the distant observer would see both situations as the planet being the exact same temperature.
The only thing the observer might notice out of the ordinary is that if the imaginary shield went from “clear” to “opaque” to LW in an instant, there would be a perturbation (spelling not my strong point at best of times, so I’m pretty certain that’s wrong) to the outgoing energy until equilibrium was re-established. The distant observer would conclude that “something happened” but that the temperature of the planet as observed from space was the same before and after.

pwl
November 27, 2010 4:56 pm

The energy conversion between the different forms of energy is not perfect, there is loss, entropy. No one has mentioned that yet.
“In physics, the term energy describes the amount of work which may potentially be done by forces or velocities (kinetic energies) within a system, without regard to limitations in transformation imposed by entropy. Changes in total energy of systems can only be accomplished by adding or subtracting energy from them, as energy is a quantity which is conserved, according to the first law of thermodynamics. According to special relativity, changes in the energy of systems will also coincide with changes in the system’s mass, and the total amount of mass of a system is a measure of its energy.
Energy in a system may be transformed so that it resides in a different state. Energy in many states may be used to do many varieties of physical work. Energy may be used in natural processes or machines, or else to provide some service to society (such as heat, light, or motion). For example, an internal combustion engine converts the potential chemical energy in gasoline and oxygen into heat, which is then transformed into the propulsive energy (kinetic energy that moves a vehicle.) A solar cell converts solar radiation into electrical energy that can then be used to light a bulb or power a computer.
The generic name for a device which converts energy from one form to another is a transducer.
In general, most types of energy, save for thermal energy, may be converted to any other kind of energy, with a theoretical efficiency of 100%. Such efficiencies may even occur in practice, such as when potential energies are converted to kinetic energies, and vice versa. Conversion of other types of energies to heat also may occur with high or perfect efficiency.
Exceptions occur when energy has already been partly distributed among many available quantum states for a collection of particles, which are freely allowed to explore any state of momentum and position (phase space). In such circumstances, a measure called entropy, or evening-out of energy distribution in such states, dictates that future states of the system must be of at least equal evenness in energy distribution. (There is no way, taking the universe as a whole, to collect energy into fewer states, once it has spread to them).
A consequence of this requirement is that there are limitations to the efficiency with which thermal energy can be converted to other kinds of energy, since thermal energy in equilibrium at a given temperature already represents the maximal evening-out of energy between all possible states. Such energy is sometimes considered “degraded energy,” because it is not entirely usable. The second law of thermodynamics is a way of stating that, for this reason, thermal energy in a system may be converted to other kinds of energy with efficiencies approaching 100%, only if the entropy (even-ness or disorder) of the universe is increased by other means, to compensate for the decrease in entropy associated with the disappearance of the thermal energy and its entropy content. Otherwise, only a part of thermal energy may be converted to other kinds of energy (and thus, useful work), since the remainder of the heat must be reserved to be transferred to a thermal reservoir at a lower temperature, in such a way that the increase in entropy for this process more than compensates for the entropy decrease associated with transformation of the rest of the heat into other types of energy.”
http://en.wikipedia.org/wiki/Energy_transformation
How this applies to the alleged greenhouse effect is an interesting question.
How it applies to the actual real planet Earth is an even more interesting question.
Is the effect significant? If so by how much and where does it take place?

Matthew Sullivan
November 27, 2010 5:00 pm

I do believe that there are errors in figures 3 and 4. It is stated that the shells are at a temperature equivalent to 240 W/m^2, but the sum of the energy emissions is actually 480 W/m^2, because there is 240 W/m^2 towards earth and 240 W/m^2 out into space for both. This might be a source of confusion.
I’m interested in hearing where the failure is in the model. Nothing really sticks out to me.

November 27, 2010 5:01 pm

I’m not sure what deficiency Willis is thinking of, but the one that occurs to me is that the main transfer of energy between the surface and the ‘shell’ in the real planet is not radiation but convection. There is enough GH gas in the atmosphere to absorb all the photons in the absorption bands within about ten metres. The main radiation layer is where the H2O ceases (the upper atmos. is dry) and the H2O radiation band gets a clear line of sight to outer space – that is the height corresponding to Willis’ shell. Hot air rising from the surface layer to the upper atmosphere is the main energy transport mechanism, and so the extra 240 Wm-2 is not needed to get the energy up there.
Also, people keep saying “multiple shells”. The above is the reason why that is not the reason.

Kev-in-UK
November 27, 2010 5:04 pm

Am going to hit the sack now but I just wondered if Willis’ query is perhaps related to the enthalpy of the system?

Steve Schaper
November 27, 2010 5:05 pm

I’m going to have to agree with those who point out that the shell is not at thermal equilibrium. Our atmosphere is not made of a shell of Maxwell’s daimon’s radiating with a directional variance like that.

Baa Humbug
November 27, 2010 5:06 pm

A further thought experiment…
I take a hot rock and place it outside where it’s zeroDeGC. The hot rock will cool down in a given period of time.
If I repeat the experiment, but this time place a 2nd hot rock alongside the first one, the first hot rock will take some time longer to cool down compared to the first experiment, BUT IT’S TEMPERATURE WILL NOT RISE. This we know to be true.
Hence this is the problem facing the proponents of the Enhanced Greenhouse Effect (EGHE) hypothesis. The EGHE should manifest itself in higher minimum temps. The proponents need to explain why we should rearrange our energy use and economies just because of higher minimum temps.
If they believe higher min T’s will eventually lead to higher max T’s, they then need to first prove this, and second, demonstrate empirically how long it will take before we reach these higher max T’s. Is it 1yr? 5yrs? 20yrs?
So far as I know, this hasn’t been done

INGSOC
November 27, 2010 5:10 pm

The Earth itself is producing heat.

Jimmi
November 27, 2010 5:24 pm

Firstly, your link to Judith Curry’s site is not correct (wrong format)
Now let try this by the Sherlock Holmes principle of eliminating the unnecessary.
Things which only affect the actual magnitude of the effect, and not its existence, are not necessary for a proof-of-concept model. So, I earlier suggested that the earth’s rotation was needed, but it is not, as that only affects the actual temperature, not the existence of warming. Likewise clouds – they affect albedo, but that does not matter. Oceans likewise – who says this model planet has oceans. Is an atmosphere needed? Well this provides extra mechanisms for heat transport i.e.convection but that does not seem to be essential.
Objections to the name “greenhouse effect” on the grounds that it is not a real greenhouse are just silly, and objections to the energy balance in figure 4 are also silly (figure 4 is correct).
What else – the shell in the model has to be infinitesimally thin otherwise you need convection to get heat through – is that it? You need several concentric shells of finite thickness in a real system?

Battiwallah
November 27, 2010 5:33 pm

I think the clue is given away in the question and that the simplification of a thin shell is the weak point. If you allow for a thick shell with a temperature gradient you can find a heat balance that allows different temperatures for the planetary surface. For example a planetary surface temperature of 35C (510 W/m2) and a temperature of -10C for the lower surface of the shell (270W/m2) would balance. The key then would be to find how the heat gradient can be calculated through the shell.

INGSOC
November 27, 2010 5:39 pm

One thing for certain, is that there is no “barrier” separating the atmosphere from space as in the model presented.

Craig Moore
November 27, 2010 5:42 pm

I have observed that women like Dr. Curry have demonstrated that they can break through the gas ceiling. ;>)

Matthew Sullivan
November 27, 2010 5:50 pm

I take back my remarks about the temperatures of shells. I missed the fact that the shell has twice the area since it has a top and a bottom, and so the per-area radiation is still just 240 W/m^2 for the shell.

November 27, 2010 5:53 pm

Lets do a thought experiment on these shells of yours.
What happens if the shell is further away or closer. Does the temperature increase or decrease?
What happens if there are several shells? Does each lower level increase in temperature? Could we build a device that uses solar energy with say 500 shells that creates a temperature of say 25300 kelvin?
So if we look at what happens with the “greenhouse effect” we currently have, does moving the distance from say 10 feet off the ground to only 5 feet off the ground increase the temperature, because the bandwidth of the spectrum in question is fully absorbed within 10 of leaving the ground surface by CO2.
Does the fact that we are not in a vacuum change the temperature? My understanding is that as pressure increases, so too does the temperature. It has been shown that the density and weight of the Venusian atmosphere captures 100% the temperature one would expect without CO2 methane and other “greenhouse gasses” in the atmosphere.
To point, the 255 kelvin theoretical temperature of the Earths surface is as a black body with nothing surrounding it allowing the heat to escape back to space. But our Earth does not end at the surface, there is an ocean of air that extends this surface outward several miles. Thus, the temperature of 255 kelvin would be at the central point in the system from which the radiation is emitted, several miles above the surface of the Earth. Points below this position would increase in temperature and those beyond it decrease in temperature. So, is 30 degrees kelvin beyond the scope of this?
Basically, your whole idea is a joke original author, and while it can make sense, it is not compatible with what science shows.

INGSOC
November 27, 2010 6:00 pm

James Delingpole has an interesting article that is somewhat pertinent here.
http://blogs.telegraph.co.uk/news/jamesdelingpole/100065683/why-i-now-deeply-regret-my-last-post/

November 27, 2010 6:01 pm

@Eschenbach
> Finally, Fig. 4 shows the energy balance when the shell is transparent to
> shortwave (solar) and is opaque to longwave (“greenhouse”) radiation.
> This, of course, is what the Earth’s atmosphere does.
The Earth’s atmosphere is not completely opaque to longwave. There is a transparent transmission band extending from 8 to 13 microns, in the center of the terrestrial thermal black body curve. This allows up to 30% of the longware to escape.
http://clivebest.com/blog/wp-content/uploads/2010/01/595px-atmospheric_transmission.png
The rest of the longwave is absorbed by the atmosphere, mostly by water vapor.
Note that CO2 has 4 absorption bands, but the 2,3 and 5 micron bands are not in the main part of the thermal curve and don’t have much effect. The 15 micron band is the largest and is centered in the 310K thermal curve.
CO2 is a powerful absorber of longwave energy. Though present in only “trace” amount (380 ppm), it completely absorbs 100% of the 13-16 micron longwave emitted from the surface, all within the few hundred meters above the surface.
But adding more CO2 does not make this band “more opaque”. Water (in all of its states) is doing all of the “heavy lifting” here on Earth, in terms of warming and regulating Earth’s climate. CO2 has a relatively small “greenhouse effect”.
To see this, consider Mars, whose atmosphere is 95% CO2. Though much thinner, it contains almost 30 times as much CO2 per surface area unit, than Earth. Yet it has virtually no greenhouse warming effect: the mean surface temperature is very close to the theoretical black body temperature of 210 Kelvins.
Mars Facts http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html
Visual geometric albedo 0.170 (Earth 0.367)
Solar irradiance (W/m2) 589.2 (Earth 1367.6)
Black-body temperature 210.1 K (Earth 254.3 K)
Average temperature: ~210 K (Earth 287 K)

TGSG
November 27, 2010 6:02 pm

6 hrs worth of guesses…. BZZZZZZZZZ time’s up!

Editor
November 27, 2010 6:03 pm

Thierry says:
November 27, 2010 at 12:36 pm

Here comes the -18°C again. In their excellent paper “Falsification Of The Atmospheric CO2 Greenhouse Effects Within The Frame Of Physics”, the two german physicists have clearly explained that -18°C comes from a wrong way of applying Stefan-Boltzmann law. On the non rotating and atmosphere free planet, the real average temperature would have been -129°C, which is more or less what we have on the moon. They have also explained why this glass approach is not to be used, since the so called greenhouse gazes do not behave as a physical or reflecting barrier. This post is unfortunately pointless.

I’m glad no one else has commented on this, and I probably shouldn’t too. I thought this had been put to bed long ago.
Lunar conditions don’t apply due to many things beyond the lack of CO2. Lunar soil is a poor heat conductor, in part because its dry. That’s one reason why some deserts, e.g. the high deserts of Oregon or Arizona, cool off so quickly at night. (Basalt walled canyons in Oregon will bake you all night.) Worse, a lunar night is 14 days long – that’s a lot of time for radiational cooling to cool!
Also, the effect of real greenhouse is achieved by locking convection so that warmed air doesn’t rise and leave the greenhouse. There’s no convection in a real greenhouse. As Willis noted, he’s talking about the “Greenhouse Effect,” which is unfortunately very different from real greenhouse physics.

November 27, 2010 6:09 pm

The approximately 70% water does not absorb/reflect the same as the remaining 30% land.
Or was that said already?

Vorlath
November 27, 2010 6:09 pm

Sorry, but figure 4 is incorrect. The 240 going into space comes from nowhere. I don’t know what kind of games Willis Eschenbach is playing, but this flaw is really trivial. And the outgoing energy is not the only flaw. It’s not about energy balance. It’s about the description. It’s wrong. You cannot have outgoing energy without that energy coming from somewhere. But nowhere in the graph is it shown how that energy can pass through a shell that doesn’t allow that kind of energy to pass through it.

November 27, 2010 6:13 pm

> … I still have not seen anyone point to it.
Are you perhaps thinking of “convection”, which is a major transporter of heat in the Earth’s atmostphere?

David J. Ameling
November 27, 2010 6:13 pm

The atmosphere is the shell. For climate we meaurse the temperature of the shell not whats is encased in the shell. That is why the greehouse effect does not reflect the real world.

Green Sand
November 27, 2010 6:14 pm

Answer:- Chaos squared
PS. People in glass planets should not get as cold as I am.

Ben Peterson
November 27, 2010 6:15 pm

Could it be that the model results in a equilibrium temp of 303 k? Way too hot for me.
Ben

Grumbler
November 27, 2010 6:17 pm

I’m going with spherical chickens in a vacuum…?

November 27, 2010 6:18 pm

Our average radiation is based on our average temperature. I don’t believe 390PPM of CO2 contributes anything measurable to our average temperature. I think the difference between the calculated black body temperature and our actual temperature is because we don’t live on a black body disk, we live on a gray, spinning sphere with lots of integrating thermal mass and a homogenizing atmosphere.
Want to see some negative feedback? Try increasing radiation by the fourth power of a temperature difference. Mother Nature does not want us to be warm!

Eric (skeptic)
November 27, 2010 6:30 pm

There is a “greenhouse” effect in a greenhouse, although it is less important than trapping the warmer air inside. Like CO2, glass is transparent to the sun’s shortwave energy. Also like CO2, the glass absorbs IR and then reemits it in all directions. The half that goes back into the greenhouse helps heat the greenhouse. Having two panes of glass means, in a very crude sense, that only 1/4 of the IR makes it out.

Matthew Sullivan
November 27, 2010 6:35 pm

Vorlath: The 240 W/m^2 outwards is thermal radiation from the shell, not the radiation from the earth somehow passing through the shell.

davidmhoffer
November 27, 2010 6:37 pm

Willis Eschenbach;
Even with that, there is still something that prevents this single-shell planetary “greenhouse” in our thought experiment from representing our real planet in even the simplest way. I still have not seen anyone point to it.>>
There are so many defficiencies in this model Willis that you’re probably going to have to throw out a hint. You’ve so oversimplified things that there are dozens of simple things wrong with it. What perspective are you looking at this from?
o An outside observer would see the temperature of the planet as identical if the shell is opaque or clear to LW. Both radiate 240 w/m2.
o For the shell to radiate 240 w/m2 it must have some sort of mass that heats up to emitt 240 w/m2. Depending on what material you presume the shell to be made of, that could be a very large amount of energy, or very small.
o For the model to be vaguely accurate, it cannot presume a shell of 0 thickness. Per above, the shell must have mass to function, and therefore some sort of thickness. If we were to change the model to presume the shell surface is one kilometer above the earth surface, and that the shell thickness is 1 km, then the model breaks down entirely. Both the clear and opaque to LW shells would then have a temperature gradiant between earth surface and TOS (Top of Shell). The distant observer would still see 240 w/m2, but it would be EFFECTIVE black body temperature that exists somewhere between planetary surface and TOS. The difference between the clear and opaque models would then be that the effective black body temperature is identical, but the altitude at which it occurs is different. Further, the earth surface in this case would NOT double its radiance as it must in a o thickness 0 mass shell model.
o The sun emitts considerable LW though mostly SW. An opaque shell would absorb and re-emitt LW from the Sun, and I expect this would be measurable.
o Though you postulate the Sun as some sort of cloud emitting constant radiance to earth surface at all times, the earth surface is not constant. Even assuming constant 240 w/m2 constant over entire surface, land and water will react differently. In the case of a presumed vacuum between shell and earth surface, I presume also that we have oceans that don’t boil, but otherwise have identical properties. In that case, LW only penetrating a few microns of water, the water surface would heat up enormously with the hot water rising to the top and becoming a barrier to mixing. Land would heat up less as it does not flow and conductance would carry some heat to some depth. This is the reverse of the real world where land would heat up faster than ocean surface. If we presume instead that the shell is again 1 km thick and touches the earth surface, then the top few microns of ocean would heat up and transfer the bulk of the energy back into the shell without appreciable changing the heat content of the ocean at all.
I could go on, but I think you’ve got to narrow the scope somewhat.

Morris Minor
November 27, 2010 6:38 pm

Sorry Willis… but you have just added to the rather large chunk of climate science junk.
You said “In the situation shown in Fig. 4, the sun directly warms the planet. In addition, the planet is warmed … by the radiation from the inner surface of the shell. … As a result the temperature of the surface of the planet is much higher than in the previous Figures.”
NO, No and No.!!! The colder inner surface of the shell can not warm the warmer surface of the Earth (Second Law of Thermodynamics). However, if the colder shell surface was slightly less cold (due to the absorbed infra-red radiation from the surface of the Earth) than the loss of heat from the surface of the Earth would be less due to the lower temperature differential. This in effect would lead to increase in the minimum temperature readings at the Earth’s surface (as stated by Baa Humbug above). I think there is evidence that there is an increase in minimum temperatures leading to an increase in the average surface temperatures ( Can someone verify this?). If so then this may be evidence of an increased “greenhouse” effect?

TH
November 27, 2010 6:45 pm

Water, clouds and wind is missing from this simple model…??

val majkus
November 27, 2010 6:46 pm

Willis can we have your answer please? I’m exhausted trying to keep up

Michael D Smith
November 27, 2010 6:47 pm

Willis,
Add one more variable to your thought experiment. Add a layer of 100% IR transparent atmosphere to the model. Assume very high gravity so that even with a thin layer, the atmosphere is extremely dense at the surface, say, 100x earth’s, yet near zero at the outer diameter that you considered, like ours. Now some energy will be transferred by conduction and diffusion from the surface to the atmosphere at the lowest levels so that the lowest layer will be the same temperature as the surface. The temperature profile will decrease via the usual adiabatic lapse rate for whatever gravity we choose (this might take some time as conduction and diffusion are both slow, but we will wait for equilibrium). The highest level will be near absolute zero, due to near zero pressure. What happens? Now what happens when the density at the surface is 1000x? 1 million x? By failing to consider temperature simply due to pressure, and the atmosphere whose molecules are also emitting IR (even though they are IR transparent), the greenhouse effect by greenhouse gases is overstated, while the effect simply due to pressure rise with depth is understated or ignored. Forget convection for this small addition to your model.

Peter Milford
November 27, 2010 6:47 pm

Willis,
I normally find your posts very interesting, but this ‘bonus question’ lark is a bit too juvenile. There are a whole host of reasons why your model is inadequate to represent the climate system of the earth – earth has an atmosphere, no hard shell, with convection, water that changes state, and with oceans, and with a hot core, etc, etc, etc, etc, etc.
This strikes me as a bit like a weird game of ‘one of these things is not like the others’ where the question-poser eventually reveals the fact that the odd one out is obviously the one which has a different penultimate letter, and no other differences count.

November 27, 2010 6:53 pm

I get accused of believing some odd things, but I’m not the one denying that N2 and O2 molecules have temperatures. 781,000PPM of N2 has no thermal capacity or thermal mass? I don’t care how many degrees of freedom your fantasies have, the temperature of 781,000PPM N2 has nearly nothing to do with anything 390PPM of CO2 can do. Stefan-Boltzmann says things radiate based on their temperature, not whether they’re made of CO2 or water vapor.

old construction worker
November 27, 2010 6:54 pm

The earth is round and it’s not a black body.
NASA Abandoned Flawed Climate Calculations in 1960’s
“Lunar Temperatures Cast Doubt on Climate Theory
NASA had found that daytime temperatures on the lunar surface were lower than expected because planetary bodies also conduct heat to their inside rather than radiating it all into space – an empirical fact that challenges the GHG theory. Computer models supporting GHG theory had predicted that such heat energy would be ‘blanketed’ above a planet’s surface.”

Billy Liar
November 27, 2010 7:06 pm

The m^2 of the shell is greater than the m^2 of the earth’s surface.

Matthew Sullivan
November 27, 2010 7:12 pm

Morris Minor:
The second law of thermodynamics is about the net flow of heat. In Figure 4, the net heat flow between the warm planet and cooler shell is still towards the shell, so there is no violation.

tim maguire
November 27, 2010 7:16 pm

You’re killin’ me Willis. People say, it’s missing this or it’s missing that and you say, well, all models miss stuff. The obvious rejoinder being, if it’s missing something important, than it’s not much of a model, is it? And you say, even if it included everything, it still wouldn’t provide a workable model of the “greenhouse effect” on our planet.
I was hoping by the time I got to the end of the comments, you’d spill the beans, but not yet. So out with it! Put an end to this infernal guessing!

Graham Green
November 27, 2010 7:25 pm

In our world the atmosphere has a temperature, that’s why it radiates some photons to earth. The model has but a vacuum and a shell of presumable infinite thinness so there is nothing there to have a temperature and thereby radiate energy.
Do I get some sort of prize?

November 27, 2010 7:28 pm

I think it’s the fact that you are showing us a simplified model of a “perfect” system, sort of in the spirit of the ‘ol college physics 101 “assume there is no friction”.
In your model, the shells operate at 100% efficiency and the “earth” is a perfect blackbody for exactly what you are modeling.
Nothing is perfect or works at 100% efficiency. Iskandar hit on this a little earlier up in the posts.

Dave Springer
November 27, 2010 7:32 pm

Did anyone mention the bottom line? Two additional watts of forcing at the surface from all anthropogenic GHGs together for a total of 242 watts per square meter temporarily until equilibrium is reestablished. New equilbrium temperature will be close to 2 degrees higher if nothing else changes.
There’s barely a third of that surface warming actually observed if we can trust the instrumental record. Therefore Trenberth’s “missing heat”.
I suspect the missing heat is being rejected by higher speed water cycle which means more convection and higher albedo from more clouds. Humans still emit aerosols along with their CO2 and methane and aerosols have a surface cooling effect so some of the 2 watt GHG forcing is negated by those. Nothing is missing. Trenberth should relax about it.

Steve R
November 27, 2010 7:32 pm

Earth does not receive ALL of it’s energy from the sun. Their is an additional source from within the earth radiating heat away from it’s active core.

davidmhoffer
November 27, 2010 7:35 pm

Morris Minor;
NO, No and No.!!! The colder inner surface of the shell can not warm the warmer surface of the Earth (Second Law of Thermodynamics).>>
No violation of 2nd Law occurs. Consider a similar thought experiment. A house is heated by a 10,000 watt heater and exists in a -40 C atmosphere. At equilibrium, the temperature inside the house is higher than -40, and the house radiates 10,000 watts to the oustide. Double the thickness of the walls from say R2o to R40. At equilibrium, the house radiates exactly 10,000 watts to the outside. But the temperature in the house is higher. Why? Because the layer of extra insulation added must heat up until the house as a whole is radiating 10,000 watts to the outside. But since it radiates in two directions, not one, the interior temperature of the house must increase until the amount of extra energy it radiates balances the energy absorbed by the extra insulation and radiated back into the house.
I’m a hardcore skeptic and I don’t have a single problem with this. My problem is with all the other factors in a chaotic system that physics requires to be a mitigating factor that the climate alarmists conveniently ignore. Well I have two problems. The second is trying to stay awake long enough to find out what the heck Willis is getting at.

Geoff Sherrington
November 27, 2010 7:36 pm

The blue down arrow on your figure 4 disappears in the case of a vanishingly thin transparent sphere. This is because the mass of the transparent sphere affects its capacity to capture and reradiate energy.

PhilinCalifornia
November 27, 2010 7:41 pm

The shell as proposed is a constant, and does not allow for any feedbacks due to, for example, that pesky water vapor.
Interestingly, I think that both Miscolskzi and Hansen would agree with me on this.

Dave Wendt
November 27, 2010 7:45 pm

Your opaque shell example seems nonsensical. If the shell is opaque at all wavelengths there is no albedo and the shell absorbs near total TSI and would radiate half that to the surface, But if the shell is opaque to LW, the energy reradiated by the planet couldn’t escape and would build up in the interstice between shell and planet.

Vorlath
November 27, 2010 7:45 pm

Matthew Sullivan says:
November 27, 2010 at 6:35 pm
Vorlath: The 240 W/m^2 outwards is thermal radiation from the shell, not the radiation from the earth somehow passing through the shell.
————-
Thermal radiation from what energy source? And why doesn’t fig 3 show this if what you say is true? Also, Willis Eschenbach disagrees with you. He’s saying it does come from the Earth and it’s passing through the shell.
I’m seeing a lot of goalpost movement here. The math is flawed. Plain and simple.
—————
Willis Eschenbach says:
November 27, 2010 at 6:55 pm
Baa Humbug says:
November 27, 2010 at 4:27 pm
Willis I’m having trouble with fig.3
If the shell is radiating 240Wm2 to space AND 240Wm2 to the planet, it must therefore be receiving a total of 480Wm2, which is what fig.3 shows.
IF it is receiving 480Wm2, how can it be the same temp as the planet surface i.e. 255K or -18C? The planet surface is receiving only 240Wm2.
The key to understanding the effect is to remember that the shell has two sides. It radiates energy from both sides equally.
—————————–
If so, then it’s no longer opaque. Here’s the definition of opaque as it relates to energy.
2. not transmitting radiation, sound, heat, etc.
So please stop changing the goalpost, or at the very least stop using better terms to describe your models.

F. Ross
November 27, 2010 7:47 pm

Willis. My answer [guess] to your question.
The posited globe receives much less energy at the higher latitudes than at the equator.
This would make for an energy imbalance and [consequently] an energy flow between the low latitudes and the poles.

RoyFOMR
November 27, 2010 8:01 pm

As always Willis, a great, thought-provoking post.
Judging by the responses, so far, you’ve hit the odd raw nerve!
Nice demonstration of just how shaky the application of the adjective “Greenhouse” is to the”consensus” that “underpins” CAGW.

Richard Patton
November 27, 2010 8:02 pm

I would say the missing component is convection from the surface to TOA.

Vorlath
November 27, 2010 8:04 pm

typo in my last message: stop = start
Here’s the corrected version:
So please stop changing the goalpost, or at the very least *start* using better terms to describe your models.
Also, if you’re letting the shell transmit energy, is this happening in figure 3 as well? So it’s not truly opaque. But even so, you’ve got a delay that doesn’t show up even if the totals do add up eventually with a continuous supply of energy. If half the energy from the beam goes back into space, then that’s 120W/m2 leaving 120W/m2 going to the planet surface and back to the shell where half again goes to space. So we’re at 60W/m2. On and on until all of it is sent back into space. It will add up to 240W/m2. But the description of opaque is completely off.

old construction worker
November 27, 2010 8:07 pm

‘Willis Eschenbach says:
November 27, 2010 at 7:50 pm
Why The Thought Experiment Can’t Represent Earth’
Well throw another log on the fire, I’ll get the beer and popcorn. This sholud get interesting.

Martin C
November 27, 2010 8:13 pm

Well, I’ll take a stab at the ‘bonus’ question: I would say in short because the earth’s climate system of predominately the amosphere and the oceans move heat around the earth (convection, latent heat of vaporization from the oceans, and ocean currents) to locations they can more easily escape, in essence ‘negating the shell to some extent.
The oceans absorb heat, and ocean currents move the heat towards the poles, where it can radiate the heat into space, without being ‘reheated’ the same as at the equator because of lower angle of incidence of the sun (thus the poles are acting as a ‘heat outlet’ of the earth.
The atmosphere transports a lot of heat in water vapor from the oceans (the latent heat of vaporization) higher in the atmosphere, where when the water vapor condenses, it gives up its heat higher into the atmosphere, where it is radiated into space quicker (due to the thinner atmosphere) than heat radiated at a lower level in the atmosphere.
. . is this explanation on the right track . . ?

davidmhoffer
November 27, 2010 8:20 pm

Willis Eschenbach;
For radiation/convection/evapotranspiration models of the earth, the absolute minimum configuration is two shells.>>
So when I said one shell of o mass and o thickness was unrealistic, that much better would be a shell of 1 km thickness resulting in a temperature gradient from earth surface to TOS, that would be different how? You proposed two shells as being the minimum and I proposed (in effect) a near infinite number of shells of near 0 thickness stacked to a depth of one kilometer. Your two shell minimum is better than one shell, but an infinite number of shells is far superior to both, easily described by calculus, and much much more accurate than two shells. Or three. Or 10.
Great thread by the way, provoked a very interesting discussion. But your two shell minimum isn’t much better than your original one shell.

Barry Moore
November 27, 2010 8:22 pm

Here goes Willis again wandering off into the never never land of naively simplistic science. This post if I can understand it correctly deals with Greenhouse effect which most people will interpret as the greenhouse GAS effect. It has already been pointed out correctly that there is no shell but a continuum of atmosphere from the surface to the TOA.
Now the facts are that CO2 can only interact with about 7% of the outgoing radiation and H2O somewhat more than that but the majority of the incoming and outgoing radiation interacts with the particulate in the atmosphere which is not resonant frequency sensitive as are the gasses. Particulate is defined as water droplets and solid particles such as volcanic ash carbon and fly ash etc.
Clearly the vast majority of the particulate is water droplets or clouds. Therefore the so called greenhouse gas effect is a very small contributor to the general warming produced by our atmosphere.
The debate over whether clouds have a net positive or net negative forcing effect has raged for years IPCC claim net positive thus the effect of an increase in temperature is an increase in humidity which produces more cloud therefore warms the earth Drs Lindzen, Christie and Spencer feel the reverse is true and that clouds have a net negative forcing effect which creates an equilibrium.
The cloud cover according to satellite data has varied from 62% to 69% over the last 30 years which has had a significant effect on temperatures.
To get back to Willis’s childish illustrations there is never a 100% transparent or 100% opaque atmosphere and to cite a “global average” radiation is even more ridiculous than trying to compute a “global average temperature” from a bunch of computer generated statistics.
Basically the clouds do tend to smooth out the average radiation budget although they will reduce the average as they increase and increase the average as they decrease.

phlogiston
November 27, 2010 8:27 pm

Paul Birch says:
November 27, 2010 at 2:00 pm
A transparent shell would not be at 0K. If it were perfectly transparent (which is physically unreal, but never mind) it would retain whatever temperature it had when you first put it there, because it is neither absorbing nor losing any heat. If it were merely very nearly transparent (which is the physically realisable case) , then in this example it would be at a temperature ~ 255K. The temperature would be somewhat higher or lower than this, depending as the emissivity were higher or lower in the visible or thermal.
This model implicitly assumes uniform insolation, which is not valid for a sphere illuminated from one direction. Steady state requires that the planet does not rotate (relative to the sun), or has negligible heat capacity; the temperature would then range from a factor 4^1/4 higher (ie 361K) at the subsolar point, through ~73K around the terminator, to ~0K on the dark side. There would be no meaningful average or global temperature at all. The greenhouse effect of the shell would be the same factor of 2^1/4 (for the visibly transparent, thermally opaque case). Add diurnal rotation and finite heat capacity and thermal conductivity, and things become much more complicated. There is no steady state, and the temperature curve for a given latitude is messy to compute. The thermal capacity of the shell has to be considered too. The general effect of thermal capacity and rotation is to transfer (some) heat between the day and night sides.

Useful analysis. So even this (supposedly) extremely simplified scenario results in thermal complexity and a non-equilibrium thermal state, with all the non-equilibrium implications of nonlinear dynamics etc.
Does this not make it even more absurd to make any attempt to analyse earth’s real climate in terms of linear, equilibrium physics? OK there is always a heat in – heat out equation. But a complex chaotic system can respond in complex including adaptive ways…

Chico sajovic
November 27, 2010 8:33 pm

There is no greenhouse effect because the greenhouse effect assumes all energy is transfered radiatively. Your model shows only radiative energy transfer.
I highly doubt that it is even possible to reliably model all the seperate modes of energy transfer in the earth-atmosphere system, conduction, convection, radiation, evaporation, condensation and mass transfer(i.e. wind and precipitation)
Here’s a couple thought experiments on heat transfer:
1. You have a really hot cup of hot chocolate and you want to cool it down do you A. Drop an ice cube in it (conduction), B. blow on it (convection) or C. hold it up to the clear night sky (radiate)
2. Thermoses use a vacuum for their insulation, because purely radiative energy transfer is slower then conduction and convection when a gas is present.
3. Argon gas is used in insulated windows becuase of its low thermal conductivity and co2 isn’t used because the greenhouse effect isn’t real.

Matthew Sullivan
November 27, 2010 8:39 pm

Vorlath: The shells aren’t transmitting the radiation. All of the radiation leaving the shell is thermal radiation originating from the shell, the same way the energy going up from the modeled planet is thermal radiation originating from there.

phlogiston
November 27, 2010 8:39 pm

Bill Illis says:
November 27, 2010 at 2:36 pm
Bill, can you give a quantum explanation of why the CO2 “greenhouse” or IR radiative heat trapping effect does not saturate in a few hundred meters?
The mean free path and radiative balance scenario that you describe could surely be turned on its head by missing out just one subtle feedback somewhere.

Sera
November 27, 2010 8:54 pm

1) The earth has multiple shells of insulation.
2) The earth is not a black body- it is a grey body.

Barry Moore
November 27, 2010 9:01 pm

Just another comment on Willis’s article.
Go to Climate4you.com select global temperature and go down to the top of atmosphere radiation section.
You will see a wild variation in TOA radiation from 216 W/m2 to 238 W/m2 with a mean around 230 W/m2, not exactly what Willis was promoting, The CO2 curve v’s TOA radiation is also very instructive.

Baa Humbug
November 27, 2010 9:35 pm

Willis Eschenbach says:
November 27, 2010 at 6:55 pm
Baa Humbug says:
November 27, 2010 at 4:27 pm
Willis I’m having trouble with fig.3
Yes thnx Willis.

Roger A
November 27, 2010 9:37 pm

The model does not work in that if you make the ‘shield’ distance to the earth surface thinner and thinner until it is actually AT the surface you are in fact at the same condition as either figure 2 or 3. This cannot be if the model is to be consistent.
Roger

don penman
November 27, 2010 9:46 pm

The glass sphere is uniform but greenhouse gasses(water vapour,clouds and co2) are not uniform around the earth,they are variable over time.

Robert Weber
November 27, 2010 10:25 pm

Willis,
You still haven’t replied to several posts that say you can’t add the two 240 W/m2 downward radiation element and get 480 W/m2 upward.

Michael D Smith
November 27, 2010 10:29 pm

Geoff Sherrington says:
November 27, 2010 at 7:36 pm
The blue down arrow on your figure 4 disappears in the case of a vanishingly thin transparent sphere. This is because the mass of the transparent sphere affects its capacity to capture and reradiate energy.

Thickness will only affect the response time for a change in input. Near 0 = fast response, thick = slow response. As small at 1 atom thickness would still be in equilibrium.

Sera
November 27, 2010 10:32 pm

It would be difficult to have an earth model with only two shells. You would need one for the gasses (which are mostly uniform), one for the water vapour (which is not uniform, although can be persistant in some areas), and one for the clouds (which are not uniform, although can be persistant in some areas). These ‘shells’ overlap, for the most part, but are distinctive in their responses. You could also include the ozone layer and thermosphere. But I would not calculate the earth as a black body.

JimboW
November 27, 2010 10:33 pm

Vorlath,
I don’t see what’s wrong with the third diagram. Shortwave radiation comes through shield (transparent to shortwave) hits planet (opaque to SW) and heats it, where upon planet emits LW. LW radiation hits shield (opaque to LW) heating it, where upon it begins to emit LW in both directions, further heating the planet surface, and losing LW to space. Just because a material is opaque (non transparent) to a wavelength doesn’t mean it can’t emit that wavelength. I would suggest that white light won’t pass from a source behind the tungsten filament in an incandescent globe to the other side, even when the incandescent light bulb is on.

Dave F
November 27, 2010 10:54 pm

Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?
Without reading the comments, I would venture to guess that the energy in the system you are postulating doesn’t do any work.

Richard111
November 27, 2010 10:57 pm

Okay, my guess. A lot of incoming shortwave radiation is absorbed by LIFE on the planet, it is not immediatly reradiated as longwave radiation?

Soren F
November 27, 2010 11:06 pm

Braddles says:
November 27, 2010 at 2:13 pm
The comments that dispute Willis’ excellent post highlight some of the problems with the AGW sceptic arguments. Those who argue that the greenhouse effect does not exist do the sceptics’ side a disservice, by arguing an untenable position. It creates a weakness in the sceptical position, which warmists focus on to try to discredit all sceptic arguments.
True, then on the other hand, warmists occupation with this self-created strawman does a devastating disservice to its side. It exposes its rage as being unfocused, not centering on the strong skeptic gems in there. Informed minorities on both sides should ignore both the untenable and the rage.

Grey Lensman
November 27, 2010 11:17 pm

In for a penny, in for a pound. Had to read the lot and hope I did not miss any points. Several made great store on their observation that the Earth was not an emitter.
Really
Its a humungus ball of molten rock at about 2,000 degrees with a very thin layer of colder rock on top. Space is about zero degrees and if i am correct hot moves to cold, I.E. the earth emits. Volcanoes, hot rocks, hot springs, geothermal vents, it emits all the time and is a very effective means of warming the oceans and thus the air.
It is not a black body like the moon

G.L. Alston
November 27, 2010 11:26 pm

Ok, Willis, so I have a really stupid question relating to the thought experiment.
You posited a shell and the implied assumption was that the shell was unchanging over time. Obviously this was to restrict the experiment to simply defining the upper boundry of radiation.
The real shell (atmosphere) is created by life, which is constantly replenishing it. We’re now at 390 ppm of CO2. An increase in CO2 ought to result in an increase in O2, although the climate community doesn’t seem to have accounted for this. As such the shell ought to be somewhat thicker now than 1890. Maybe not by much, but thicker all the same. Moreover, the assumption of CO2 being (n)% of the atmosphere and growing seems off. If plants are also emitting more O2 then the (n)% isn’t growing as much as it may seem at first blush. As such this would imply that much of the apparent temp increases are more related to land use etc than CO2 (or natural recovery from the LIA, etc.) since it appears that the AGW argument is that CO2 is rising whereas all else is purely static.
So… my question. Am I missing something really basic here? Thanks…

cohenite
November 28, 2010 12:27 am

Hi Willis; these types of thought experiments are interesting and serve a purpose. If I may point out an aspect to do with your figure 4; in the first instance only 240 w/m2 reaches the surface in the form of SW so only 240 w/m2 is reemitted to the disc in the form of LW; of that 240 w/m2 isotropic effect causes 120 w/m2 to be reemitted from the disc back to the surface and 120 w/m2 outwards to space; so in the next instance the surface has 240 w/m2 SW + 120 w/m2 LW for a total of 360 w/m2 and so on; in short this is, of course, a limiting sum geometric series:
Sn=a/1-r where a=1 and r=0.5; or 1/n^2 for n=1 to infinity is 2.
In otherwords the surface will only receive 480 w/m2 at infinity.

Malaga View
November 28, 2010 12:31 am

Willis Eschenbach says:
November 27, 2010 at 2:51 pm
but my thought experiment is in a vacuum.

So standing your Glass Thermos Flask in the sun at mid-day will slow down the cooling during the daytime… but by mid-night the contents will be cool… and probably cold just before dawn…
On a more constructive note it would be wonderful if you could create, based upon the very intelligent comments you have received so far, a revised version of your diagram that is technically correct and isn’t based upon a vacuum… because I remember being taught: Nature abhors a vacuum

Richard S Courtney
November 28, 2010 12:37 am

Willis:
Please forgive the abruptness of this post which O am fitting in between duties I have today.
At November 27, 2010 at 8:20 pm davidmhoffer said to you:
“Your two shell minimum is better than one shell, but an infinite number of shells is far superior to both, easily described by calculus, and much much more accurate than two shells. Or three. Or 10.”
You have replied to that at November 27, 2010 at 10:12 pm saying:
“I’m looking for the simplest model that will give me earthlike results. That model contains two shells. Yes, you can model it with four, or four hundred. I’m looking to simplify, not complicate.”
OK. I understand that, but it can be misleading to simplify beyond the physical process being modelled: an analogue is not a model of a mechanism but may be an accurate model of the behaviour of that mechanism.
GHG molecules in the atmosphere absorb IR photons from the Earth’s surface then (ignoring collisional energy exchange) release that energy by emitting photons in all directions. So, the molecules emit half the energy in upward directions and half in downward directions.
Thus, the GHG molecules act like half-silvered mirrors transmitting visible light. They accept radiation from below, then emit half of that energy upwards and half downwards.
Ignoring effects of collisional energy exchange, the number of ‘half-silvered mirrors’ to model the observed GHG effect is 7.
So, I suggest that your “simplest model that will give me earthlike results”is 7 spheres each with 50% transmittance and 50% reflectance. And a more realistic model would include energy extraction by molecular collisions from each shell.
Richard

Malaga View
November 28, 2010 12:58 am

I also found the following passage very interesting…
see http://www.newton.dep.anl.gov/askasci/env99/env223.htm

The flow of heat through the sandstone and limestone overlying Wind Cave’s passages is extremely slow. Temperature fluctuations of over 60°F between day and night are not uncommon on the surface, but if we were to monitor the temperature only two feet below the surface on such a day, the fluctuation in temperature would be only about 1°F. Therefore, it does not stay warm long enough during the day, nor does it stay cool long enough during the night, to significantly change the temperature
of the rock only two feet underground. The same principle holds true for seasonal temperature fluctuations as well, although the depth at which temperatures begin to stabilize is greater. Seasonal temperature fluctuations of 80°F are reduced to only l°F at a depth of about 50 feet.

So does the temperature gradient look like from the earth’s core up to ground level?
Is the earth internally generating heat or just simply cooling down?
Sorry to be throwing another spanner into the works….

Cold Lynx
November 28, 2010 1:03 am

You can solve this with one shell. No problem.
But You must add little bit of reality.
The energy balance between surface and your shell is NOT a radiative balance.
In real world is it a heat balance driven by convection. Then You get the height of the real world “shell”. 33K/6 Kk/km = 5,5 km
6 K/km are average lapse rate.

Lu
November 28, 2010 1:05 am

I’m no scientist, but how about uh… barbecues, campfires, air-conditioners, billions of cars, motorcycles, ships, airplanes, heaters, stoves, buildings on fire, industries, power stations, nuclear stations, Borneo, Russia & Sumatra forest fires, geysers, electricity, volcanoes, lava streams, E=MC2, energy and heat produced by nature, earth and man? Doesn’t that add substantially and explain a fraction of a degree rise of temperature perhaps? Do I get at least a bonus?

Sera
November 28, 2010 1:05 am

Willis,
Are you calculating the earth as a black body, because it behaves like a black body?

FrankK
November 28, 2010 1:07 am

I am puzzled by the long wave radiation out to space of 240 W/m2 (top RHS of figure 4)
If the boundary does not allow long-wave radiation out then where does this radiation come from?
Under steady state conditions (equilibrium)
Outflow must equal Inflow. Outflow=Inflow (1)
Under these conditions storage has no effect on the result
i.e. S dT/Dt is zero where S is storage and T temp and t time.
As far as I can see the components don’t obey equation 1 in Figure 4.

Dr A Burns
November 28, 2010 1:19 am

Willis,
Please take a look at Fig 3b here: http://realplanet.eu/atmoseffect.htm
A surface temperature of 15 degrees is achieved with a single shell, simply by adding gravity and a circulating atmosphere. This seems more realistic than your 2 shell model.

simpleseekeraftertruth
November 28, 2010 1:32 am

“Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?”
================================
Assuming you wanted the simplest answer and noting that you said ‘planetary’, I posted at 2:58 pm;
“Day/night cycles are not considered. Day: outgoing is less than incoming. Night: outgoing is greater than incoming. The change (raising) of surface temperature is the difference between these two budgets.”
You first need two conditions, not two shells, for your construct to work. When I was at school (1960’s), a greenhouse was explained to work because glass is semi-transparent to IR. Therefore, when the sun is shining, its internal temperature increases as less IR escapes than is received. When the sun is not shining, its internal temperature decays to ambient more slowly (than without glass) because of the same phenomenon – the semi-transparency. Convection losses are present at all times but vary as delta T changes – applicable to a greenhouse but not a planet.
I claim my $5!-)

tallbloke
November 28, 2010 1:33 am

Willis Eschenbach says:
November 27, 2010 at 7:50 pm
For radiation/convection/evapotranspiration models of the earth, the absolute minimum configuration is two shells. My iterative Excel model of the two-shell system is here (I hope, server hassles lately). All models are wrong. Some models are useful.

Damn. I started to reply that the atmosphere is made of an almost infinite number of ‘shells’ but didn’t submit because you said it was a simple model.
I’d like you to expand though. How do you think your insight relates to the real world? The path lengths of longwave radiation go from about 10^-27m near the surface (due to the density of the atmosphere) to around 1m at the top of the ionosphere. Radiation becomes more important than convection as a transporter of heat somewhere near the tropopause.
Humidity near the tropopause has been falling since 1948 on the average. I think this will have negated any rise in co2. Specific humidity near the tropopause mysteriously mimics solar activity levels: http://tallbloke.files.wordpress.com/2010/08/shumidity-ssn96.png
So what caused the warming? Reduced cloud cover and high solar activity. ISCCP data shows a drop in tropical cloud cover from 1980-1998: http://tallbloke.files.wordpress.com/2010/11/isccp-temp.jpg My calcs show this was causing an extra 4W/m^2 forcing on the surface in the 1993-2003 decade.
Simples. 🙂

kuhnkat
November 28, 2010 1:53 am

More junk science.
The problem with thought experiments is that there are always factors left out to simplify. Climate Science has already left out too much science. Leaving out more simply proves the thought experiment is worthless for this application.
Try your thought experiment again with a reasonable figure for the energy flux within the earth. Then keep adding the other factors left out and viola, no NET greenhouse effect!!!
Capiche??

Baa Humbug
November 28, 2010 2:12 am

Between Willises post here and J Currys “Skeptics make your best case”, my brain is frazzled.
If I’ve got it right, Willis says at 7:50pm “It’s not powerful enough…..This means that at a minimum a system to model the earth must have two separate and distinct shells. One shell is not enough to concentrate enough energy to satisfy the known constraints.”
At 2:42 pm I gave a simple short answer to the simple question thus:
Baa Humbug says:
November 27, 2010 at 2:42 pm
Layer upon layer upon layer
So, do I get a cyber prize or a cyber kick up the backside?

Oxbridge Prat
November 28, 2010 2:27 am

Willis, see

Oxbridge Prat says:
November 27, 2010 at 1:12 pm
Multiple concentric shells.

Can I have a virtual banana please?

John Marshall
November 28, 2010 2:28 am

The Greenhouse Effect is a ridiculous name for an effect that has been used as blame for climate. Greenhouses do not act like the atmosphere. Greenhouses are heated by visible light and the energy loss of that light to become IR energy. Glass does not transmit IR, or UV for that matter, but greenhouses still get warmer than ambient. Incoming visible light is transmitted and this warms the interior and suffers energy loss so that some if that light becomes IR which remains inside the greenhouse because the glass traps it inside. The temperature will build and build until someone opens a vent to let in cooler air and expel the hot air.
The atmosphere is transparent to IR, and some UV light, as well as visible light. Yes water vapour and CO2, as well as other misnamed GHG’s, do warm up by adsorbing this energy, but this heat is not stored to be re-radiated later because the 2nd law of thermodynamics does not let this happen. The adsorbed heat is immediately conveyed to the other atmospheric gasses by convection, conduction or radiation to cooler areas. (Heat will only be transferred from hot to cold never in the other direction). So some of this heat will be transferred back to space because the lower atmospheric layers will be warmer!
The whole idea that something can ‘store’ heat violates the laws of thermodynamics and so makes the GHG warming hypothesis a failure.

Max
November 28, 2010 2:28 am

To all those who think that the Figure 4 is wrong:
If we say we only absorb long-wave radiation in this example, then we get only 240 W/m^2 to the earth. In the first time step, we emitt 240 W/m^2 to the shielding shell (because Heat Capacity is 0). This shell then emits 120 W/m^2 into space and the same back to the planet. In the next step. The planet emits 360 W/m^2 and half of that gets remitted to the planet: 180 W/m2. In the next time step we emit 420 W/m^2 to the shell, the shell divides it and reemitts 210 W/m^2 back to earth. Then we get 450 W/m^2 that is emitted by the planet and thus 225 W/m2 back to earth. This continues until we reach the equilibrium.
E(t) = Solar rate + S'(t)
S'(t) = E(t) /2 (half of it into space)
S(0) = 0
Solar rate = 240 W/m^2
Shell can’t emitt in the initial condition, because it has no energy.
You can easily reiterate this in Excel and you will see that E(t) converges to 480 W/m^2.

cohenite
November 28, 2010 3:12 am

Max says:
November 28, 2010 at 2:28 am
Max, as I said earlier, there cannot be an equilibrium, defined as per Willis’s Figure 4 where the surface receives 480 w/m2, until infinity:
cohenite says:
November 28, 2010 at 12:27 am
Hi Willis; these types of thought experiments are interesting and serve a purpose. If I may point out an aspect to do with your figure 4; in the first instance only 240 w/m2 reaches the surface in the form of SW so only 240 w/m2 is reemitted to the disc in the form of LW; of that 240 w/m2 isotropic effect causes 120 w/m2 to be reemitted from the disc back to the surface and 120 w/m2 outwards to space; so in the next instance the surface has 240 w/m2 SW + 120 w/m2 LW for a total of 360 w/m2 and so on; in short this is, of course, a limiting sum geometric series:
Sn=a/1-r where a=1 and r=0.5; or 1/n^2 for n=1 to infinity is 2.
In otherwords the surface will only receive 480 w/m2 at infinity.

TimM
November 28, 2010 3:24 am

I assume that the amount of heat humans create is known to some degree and accounted for in climate modelling (unless it is considered trivial)?

Geoff Sherrington
November 28, 2010 3:41 am

Max,
Then how do you answer Baa Humbug who says:
November 27, 2010 at 2:56 pm Take a basic strip heater of say 1000W, the above fallacy suggests all we have to do is add a well conducting bar across the front of the element, say an inch away (unwired and not receiving any power, i.e. benign) and voila’ we now have a strip heater putting out twice the energy. Can someone help me with patents please?
……………………………
I try to make the same point, that the shell has to have some rather special properties to do anything. Also, there is a big mix-up (as usual) between static analysis and dynamic analysis. This shows with what cohenite says:
November 28, 2010 at 12:27 am – In otherwords the surface will only receive 480 w/m2 at infinity.
It’s actually worse than that because less than 50 % of isotropic emission will return to Earth from the shell because of geometry.
Similarly where Michael D Smith says:
November 27, 2010 at 10:29 pm – Thickness will only affect the response time for a change in input. Near 0 = fast response, thick = slow response. As small at 1 atom thickness would still be in equilibrium.
That’s dynamic analysis, not static. You have to wonder how a thickness of 1 atom interacts with a wavelenth of IR light of say 10 microns. It would take a rather long time to show. Logically, a tickness of 1 atom would imply a separation between atoms, so the sphere is not merely transparent to IR, it has holes in it. Ad absurdum, if the mechanism works with holes in the sphere , it will work with one atom. (Don’t think so).

MostlyHarmless
November 28, 2010 3:44 am

The issue is surely not the existence of a “greenhouse effect”, but its magnitude. Errors abound in calculations for the “non-greenhouse” earth, for example the usual assumption that the surface acts as a perfect “black body”. If it did, radiation scientists wouldn’t need to struggle to devise perfect”black body” reference sources for measurements, they’d just need a handful of dirt or a bucket of sea-water.
The relationship between temperature and radiation is non-linear. Averaging a wide range of one element to calculate the other introduces an error – the wider the range, the larger the error. Kiehl & Trenberth show that for small increments in either radiation or temperature, the relationship can be considered to be linear. However, they do this AFTER averaging wide ranges in radiation to calculate average temperature and vice-versa, and assuming that surface and atmosphere behave as perfect “black bodies”. Also water-vapour is NOT uniformly distributed over the earth’s surface, therefore neither are clouds, and the heat transfer involved in evaporation and condensation is very large, and not properly included in the internal atmospheric “budget” These factors mean that the averaging process used invalidates the “energy-balance” diagrams by as much as several degrees K. For example, averaging the radiation deriving from temperatures of 0°C and 40°C results in a figure which gives 22.1°C rather than 20°C. That’s not insignificant when temperatures are calculated to a tenth of a degree.

November 28, 2010 3:56 am

Willis Eschenbach says:
November 27, 2010 at 7:50 pm
“Why The Thought Experiment Can’t Represent Earth
It is not powerful enough.
What I have illustrated above is the energy balance in a theoretically perfect single shell planetary “greenhouse”. It shows that the maximum amount that such a system can produce is a doubling of the input.”
This is silly.
1) Plenty of people upthread have pointed out that there isn’t just a single shell; you poo-pooed them for adding complexity.
2) Doubling the input is ample for a simple model of Earth’s climate; you have only made it apparently inadequate by adding additional shells or absorption mechanisms that cut down the energy reaching the surface in places beyond what you assumed, which is an unnecessary complication.
3) Even a single shell can in principle produce any degree of greenhouse warming required. Off hand, I can think of at least three mechanisms:
(i) if the ratio of the thermal emissivities of the lower and upper surfaces is E then the temperature of the shell is enhanced by a factor E^1/4, and the flux re-radiated to the planet is increased by the same factor of E. One could easily engineer a factor of up to ~100 (using, eg, aluminium film), which would push the Earth’s temperature to almost 1000K! In the real atmosphere, with its thick shell, those emissivities do not differ by anything like so much, but they are not identical.
(ii) if the shell is reflective (as you have implicitly assumed by setting the albedo greater than zero) then the amount of energy radiated up and down from the shell can be much less than the amount of thermal energy simply reflected or scattered back down. Again, an aluminium film could boost the heat reaching the surface a hundredfold!
(iii) if the shell is not perfectly conductive, there is a temperature difference between top and bottom, and there is no fundamental limit on how large that temperature difference can be. In the real atmosphere, there is indeed such a temperature gradient.
You could argue, I suppose, that (i) and (iii), in using a shell with different properties for the top and bottom surface, are really two shells; but that pedantic objection would not apply to (ii).
Note that in principle one is not even limited to the 6000K temperature of the Sun; if one were to allow through only the blue or “hot” end of the spectrum, the planetary temperature could theoretically exceed the source temperature. This does not contravene the second law of thermodynamics, because there is still a net entropy gain (more photons leave the planet+shell than reach it).

tallbloke
November 28, 2010 3:57 am

Willis Eschenbach says:
November 28, 2010 at 2:26 am
in my Excel model referred to above. I still find a net warming effect. Here is a more complete view of a two-shell system, with the other factors included. The two shells are shown as horizontal gray bands:

Willis, that’s an interesting reworking of the Kiehl-Trenberth energy budget sketch. Have you been able to produce an estimated sensitivity using it?

bessokeks
November 28, 2010 4:13 am

The so called “greenhouse-effect” does not exist.
Radiation of heat through a body only depends on the temperature of that body and its coefficient for its emissivity. Back radiation cannot influence this. It only could transfer heat back to that body.
But transfering heat energy from “cold” to “warm” violates the second law of thermodynamics.
There is no way around that.

RC Saumarez
November 28, 2010 4:34 am

The thermodynamics of this model neglects entropy.
The Earth is not at equilibrium and it while the energy balance may be that the net energy flux integrated over a surface surrounding the Earth may be zero, the system is not at equilibrium, in which the entropy production is zero, but is in an approximate steady state. There are gradients of pressure, temperature, velocities of air and water, i.e.: decreased entropy, which are driven by an energy flux through the system. These would decay if there was an opaque shell surrounding the earth. A simple radiative transfer model striles me as somewhat simplistic.

November 28, 2010 4:53 am

Here’s the problem that I see. In any of your “fully opaque” scenarios, the 240 units of energy from the sun should be REDUCED because the energy is the sum total of all wavelengths. So if none of the longwave energy can pass from the sun to the earth, then it’s not 240 that hits the earth, it’s 170 (or whatever… just an example.)

G.L. Alston
November 28, 2010 5:05 am

Willis — I don’t follow that. If anything, it seems like making more CO2 should decrease the amount of O2 in the atmosphere.
Sorry I was not more clear. Your shell is the atmosphere. The atmosphere is always replenished in that there’s leakage into space. The atmosphere is (re)generated by life, so it’s not a steady state where increase of A necessarily means B decreases.
Specifically I was thinking of photosynthesis. If the number of plants is a steady state, then more CO2 = decrease in O2. On the other hand more CO2 = more plant growth (more plants?), hence more O2 released. More O2 = slightly more atmosphere. All things being equal there ought to be more plant activity with more CO2 meaning there ought to be more O2 emitted as well. In real life the volume of the atmosphere, which is a living system of sorts, ought to change slightly as parts are lost and subsequently replenished, so we know it’s being created. But enough to be measureable? I don’t know.
(Mental model: CO2 concentration goes up, plant volume up. More O2 created. Atmospheric thickness goes up. )
I also don’t know how the ppm # is created. If machinery counts up a million molecules and derives ppm, then OK, I’m dead wrong. OTOH if the machinery is calibrated against a steady state and ppm = derivation from this state, then what’s being measured is an absolute CO2 molecule count vs pctg of atmosphere, hence this count may not mean what it ought to mean.
Is it far off enough to make a difference? No idea. I’m simply wondering if we’re measuring what we think we’re measuring.
No I’m not questioning the greenhouse effect or offering alt theory here. Mostly it’s a nagging question regarding the reality of the hard numbers, and I’d thought of your list of known losses etc in your thought experiment.
Thanks for listening.

lgl
November 28, 2010 5:21 am

Willis
“For example, you can say that 390 W/m2 is emitted by the surface as radiation, and 100 W/m2 is lost from the surface through convection and evapotranspiration. So somewhere around a fifth of the energy is not going into warming the surface.”
“there is not much chance that the net feedback is positive”
So there must be 390+100 watts absorbed by the surface. There is only about 164 watts solar absorbed + lets say 36 watts solar re-emitted from the atmosphere, i.e there is an 490/(164+36) amplification. Why wouldn’t an additional 4 W/m2 (~1% change) see about the same 2.45 amplification?

Bryan
November 28, 2010 5:29 am

The greenhouse effect was shown to be bogus by R W Wood.
He showed that;
1. A glasshouse only worked by stopping convection.
2. The remaining real residual radiative effect was so small that it could be almost ignored.
See page 17
http://www.friendsofginandtonic.org/assets/hutton%20-%20climate%20change.pdf#page=17
Here is a paper that more people should read.
Especially as it comes from a source with no “spin” on the AGW debate.
The way I read the paper is it gives massive support for the conclusions of the famous Woods experiment.
Basically the project was to find if it made any sense to add Infra Red absorbers to polyethylene plastic for use in agricultural plastic greenhouses.
Polyethylene is IR transparent like the Rocksalt used in Woods Experiment.
The addition of IR absorbers to the plastic made it equivalent to “glass”
The results of the study show that( Page2 )
…”IR blocking films may occasionally raise night temperatures” (by less than 1.5C) “the trend does not seem to be consistent over time”
http://www.hort.cornell.edu/hightunnel/about/research/general/penn_state_plastic_study.pdf

Nylo
November 28, 2010 5:31 am

OMG, I really got desperate while reading some of the comments. Perhaps 30%. Is the general commenter here THAT clueless?
The next time I see a post which explains something very basic and/or proposes readers to solve a problem, I promise not to read any of the comments. It puts my morale down. Twice was I tempted to answer with expletives… It is incredible the authority with which absolutely clueless people can talk about things that they cannot understand.
Willis, really, just give up. Whoever cannot understand the GHE, after the several posts that have appeared here in the last months about the subject, will never do. It’s not worth the effort.

simpleseekeraftertruth
November 28, 2010 5:37 am

Following on from my post at 1:32 am.
Received average W/m^2 for half a cycle = 240 = Wr
Transmitted average W/m^2 for a full cycle = Wt
For 1 cycle: Wr+2Wt = 0 (steady state). Therefore Wt = Wr/2 = 120
This is true with and without a ‘greenhouse’ effect. So what lifts average temps from -18 deg? The answer is in the semi-transparent nature of the atmosphere to solar radiation, particularly IR, which creates a lag or hysteresis. With hysteresis, the work done on the system is the area under the hysteresis graph. In this instance, the work done is the raising of average temperature from -18 deg, an increase in potential energy. To understand lag, imagine yourself as a photon travelling into the atmosphere along-side a group of others: as it is semi-transparent, some, including yourself, pass through but on the way out some get out but you don’t. Your eventual departure has been delayed: you are lagging behind the others.
How you would show that in a diagram, I don’t know but that diagram will have to show 1/2 cycle input and time lag.

John Whitman
November 28, 2010 5:45 am

Willis asks, “Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?”
Willis answers, “It is not powerful enough.”

——-
Willis E.,
Your answer is not the most fundamental answer to your own question.
So, let’s play the game with you instead of your game with us!
MY COUNTER QUESTION: “What is the most fundamental reason why the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?”
John

BSM
November 28, 2010 5:55 am

People,
lots of interesting stufff here, but we are all overcomplicating and over critising this simple thought experiment to explain the fundamental mechanics of the green house effect. Not the earths atmosphere, BUT the greenhouse effect.
My GUESS is that the thin shell is described only in its characteristics of being transparent or opaque (as to whether or not it passes or blocks shortwave or long wave radiation.) It does not take into account its mass (in this example not stated so taken as zero) and as such does not account for the heat that it itself would store, how long it would heat up and also cool down….
depending on the temparature differential from one side to the other of the shell.

Max
November 28, 2010 5:55 am

@Geoff:
Well, Willis posted a good reply for the heater system up there, so I won’t repeat it here a second time.
But when it comes to static between dynamic systems, we have a steady-state system as long as all boundary constraints stay the same, as Willis showed in his figures. He only displayed the steady-state, because (and I only give my opinion here) of time constraints and simplification.
What I added was merely to show how you arrive from the initial condition of only solar radiation influx to the steady-state over a certain amount of time (read: iterations). As you can see the pure mathematical model NEVER actually arrives at 480 W/m^2, which is the convergation limitation of the dynamic model AND not a fault in the thinking. From there on, the model is in a steady-state until the influx from the solar radiation changes.
There is no violation of any kind of Energy law. The Earth in this model is only an intermitting energy source, BUT not the ultimate energy source, that is still the sun. In reality, the Earth heat capacity would further delay the arrival of a steady-state (along with several other factors which would help avoid any steady-state for a longer period). If you want to check the energy balance, you have to cut the whole dynamic system free at the shell. Then you have as much energy going away from the shell as is coming in (480 comes in, 240 goes into space and 240 is reemitted back to earth). The same you can do for the Earth side of the equation: 240 + 240 is coming in (from the sun and from the shell) and 480 is emitted to the shell. There is no “new” energy or energy creation on earth, it is just blind power, which you can also find in hybrid drive trains (as any engineer can tell you).
In those drive trains you have to ways to transmitt power and thus torque from the engine to the wheels. First you might have a mechanic transmission and a hydraulic transmission (for the sake of simplicity). The power is divided at a planetary gear just behind the combustion engine and then part of it is transmitted via the hydraulic transmission and the rest through the mechanic transmission. It is then added to a single shaft by three gears. We there have an addition and/or subtraction of torque and it might be that we actually have power that is only “circulating” between the two transmission and IS NOT leaving it due to their relative rotation speeds. I could go more into detail, but I don’t have the time to design some accompanying drawings that would help illustrating the point.
Also, similar effects can be observed in electrical systems (alternate current): http://en.wikipedia.org/wiki/Reactive_power#Real.2C_reactive.2C_and_apparent_power
It is sometimes a difficult to grasp concept, but nonetheless valid and proven.
quote:
Max,
Then how do you answer Baa Humbug who says:
November 27, 2010 at 2:56 pm Take a basic strip heater of say 1000W, the above fallacy suggests all we have to do is add a well conducting bar across the front of the element, say an inch away (unwired and not receiving any power, i.e. benign) and voila’ we now have a strip heater putting out twice the energy. Can someone help me with patents please?
……………………………
I try to make the same point, that the shell has to have some rather special properties to do anything. Also, there is a big mix-up (as usual) between static analysis and dynamic analysis. This shows with what cohenite says:

John Bowman
November 28, 2010 6:02 am

Greenhouses work by containing convection and not by preventing the re-transmission of radiant energy.
Whatever the “effect” on the Earth’s climate it has little to do with what happens in greenhouses.
It should more properly be called ” the insulation effect” as the atmosphere limits the transmission of heat energy by its attenuation within a material.

Michael D Smith
November 28, 2010 6:12 am

Willis, I’m curious about the effect of a high mass, 100% IR transparent atmosphere as posed in my first question above… Not sure if you saw it. Thanks, Mike S.

November 28, 2010 6:14 am

Willis
How does the glass shell radiate twice as much energy as it receives? Inquiring minds would really like to know.

Dave Springer
November 28, 2010 6:16 am

Ken Coffman says:
November 27, 2010 at 6:53 pm
“I get accused of believing some odd things, but I’m not the one denying that N2 and O2 molecules have temperatures.”
Individual molecules will have wildly varying temperatures if you could somehow find a way to measure them individually.
“781,000PPM of N2 has no thermal capacity or thermal mass?”
Of course it has thermal mass and as a group of molecules will have a measurable temperature which is the average of the group.
“I don’t care how many degrees of freedom your fantasies have, the temperature of 781,000PPM N2 has nearly nothing to do with anything 390PPM of CO2 can do.”
This is where you’re wrong. Imagine two thick sheets of copper sitting out in the sun. On one we put a put a very thin coat of black paint and on the other we put a very thin coat of white paint. The paint weighs next to nothing compared to the weight of the metal. Will the paint cause a difference in the temperature of metal as it sits out in the sun? I think you know the answer to that.
“Stefan-Boltzmann says things radiate based on their temperature, not whether they’re made of CO2 or water vapor.”
That is correct. But what governs absorption? The key is that CO2 absorbs infrared radiation at characteristic infrared frequencies while N2 is completely transparent in infrared. In a dense mixed gas (air is “dense” in this context) of N2 and CO2 when a CO2 molecule absorbs infrared radiation it translates to kinetic energy – the molecule is moving and vibrating faster as a result. The motion is what registers on a thermometer. Since it’s in a dense gas it almost immediately bumps into a neighboring molecule and transfers some of its kinetic energy to the molecule it bumps into. In that manner CO2 acts to absorb infrared energy emitted from the earth’s surface and raises the temperature of all the molecules surrounding it. Without the CO2 the radiation would have passed straight though the N2 and the N2 would not have been made warmer.

Thierry
November 28, 2010 6:40 am

Hello all,
To Ric Werme’reply, November 27, 2010 at 6:03 pm
1/ Everytime you see a reference to -18°C in the climate debate, you have to remember this value comes from the use the other way around of the Stefan-Boltzmann’s law as related in the peer-reviewed paper “Falsification Of The Atmospheric CO2 Greenhouse Effects Within The Frame Of Physics” (Gerlich & Tscheuchner). Therefore, this -18°C value means absolutely nothing and should never be used in the climate debate.
2/ Since the wrong calculation is usually done for a non rotating and atmosphere free planet, the comparison with the moon do apply here since in the same situation, earth would end up with the same desert-like surface anyway. The moon low rotation motion (14 days) is slow enough to reach a surface temperature steady state.
3/ The whole point is that a pure radiative approach of earth climate is meaningless as pointed out by G&T.

Dave Springer
November 28, 2010 6:41 am

@Ken (con’t)
The “greenhouse effect” (a poor term but we have to live with it) is physics that was known theoretically 200 years ago. About 150 years ago John Tyndall demonstrated it expermentally in literally thousands of different experiments with various gases under various pressures irradiated at various frequencies. He published his results in a seminal tome titled “Heat: A Mode of Motion”. You can read the book in its entirety on Google Books:
http://books.google.com/books?id=eZEAAAAAMAAJ&dq=john%20tyndall%22%20heat%20motion&pg=PP1#v=onepage&q&f=false
For ease of reading I suggest downloading the PDF file.
The bottom line is the greenhouse effect was demonstrated experimentally over 150 years ago. Anyone claiming it isn’t real is sorely lacking in understanding of introductory level physics.

Nullius in Verba
November 28, 2010 6:44 am

Others may be interested in the related discussion starting here and particularly the following comment on the shell model, and also here. The general idea was covered further at greater length here.

Bill Yarber
November 28, 2010 7:25 am

Willis
A simple question for you. Recognizing that your analogy is extremely simplistic to prove the existence of the “greenhouse affect”, what happens if you make the shell only 80% opaque to long wave radiation and then increase that gradually to 80.1%?
My guess, it results in an unmeasureably small change in the planets temperature!
Please feel free to enlighten me at 2billyarber@Gmail.com
Bill Yarber

Pamela Gray
November 28, 2010 7:35 am

It stands to reason that heating a globe who’s core has been cooling overall since it’s formation would be an intensive matter. And because of our size and shape, we don’t capture much of the Sun’s infrared. If the Earth fails to capture and hold onto longwave, we roast during the day and freeze at night.
The premise here is to layer up. One blanket does not trap heat nearly as well as multiple blankets. And a single opaque blanket may lead to cold damp conditions. So it’s a good idea to have a bit of flow-through in each of these layers. Which equates to the idea of the shells being not absolutely opaque but nearly opaque to longwave.

Dave Springer
November 28, 2010 7:45 am

TimM says:
November 28, 2010 at 3:24 am
“I assume that the amount of heat humans create is known to some degree and accounted for in climate modelling (unless it is considered trivial)?”
Yes and no. It isn’t trivial when a thermometer is located in close proximity to an artificial heat source! Averaged over the entire surface of the earth it is indeed a trivial amount of heat. Also trivial is the amount of heat from the molten hot core of the earth that is emitted from the surface due to the crust being a very good insulator.

Jordan
November 28, 2010 7:49 am

Willis: “OK, let’s assume that the sun is some form of luminous cloud that illuminates the planet evenly with 240 W/m2 of solar radiation.”
Then the planet heats up to the same temperature as the luminous cloud. Luminous cloud is a simplifcation too far, principles need to be argued from point source.
Flow of energy is then crucially determined by geometry. A disc (facing into the point source) has much the same issue as luminous cloud. I’d say it is essential to stick to a sphere.
Even a highly idealised spherical model needs to be detailed enough to integrate incoming radiation distributed over the sphere (or one side of a sphere at any time), conduction over the surface, and differentials of OLR at different coordinates.
I’m not going to try to do that because it relies on too many assumptions (that I’m not in the best position to deal with). However I supect that the combination of assumptions, integrals and application of S-B will introduce some significant numerical factors not evident in the simple “vertical line” diagrams above.
(BTW, before you answered the question set at the top of the thread, my answer was going to be “no angles”.)

HankHenry
November 28, 2010 7:52 am

It seems there should be some time considerations that should play into a “greenhouse gas” model. I imagine that on the gasless moon radiation strikes the surface, heat is created in the atoms at the surface and some short time later the heat is available to be re-radiated (at a longer wavelength), but if there were no “short time later” you would have a reflective surface and presumably there would be no heating. Likewise the time needed for radiation to be absorbed and re-radiated within the atmosphere should have a bearing on what equilibrium temp is achieved. In other words the atmosphere can be thought of as hindering or slowing the passage of radiation from the surface, and how quickly that happens helps determine how much warming occurs – with less warming occurring when radiant energy passes more quickly.
Does the atmosphere have an R-factor? Shouldn’t a blanket or insulation be part of this awkward, inapt greenhouse metaphor? Maybe an electric blanket where the electricity coming in on a wire in an electric blanket is akin to the shortwave sunlight passing freely through the atmosphere to warm the surface?

bessokeks
November 28, 2010 8:02 am

@Dave Springer says:
“The bottom line is the greenhouse effect was demonstrated experimentally over 150 years ago. Anyone claiming it isn’t real is sorely lacking in understanding of introductory level physics”
Nobody is saying that there is no backradiation. The point is that backradiation emitted by a layer with -55 Celsius cannot transfer heat to a surface much warmer that -55 Celsius. This is the second law of thermodynamics. This second law is true for all of us: Being realists, physicist or oekofascist…

davidmhoffer
November 28, 2010 8:09 am

Willis,
If it is the simplest possible model you want that is USEFUL then I think I already suggested it. A shell one km thick being in contact with earth surface and a resulting temperature gradient from earth surface to Top Of Shell. Still no where close to reality, but sufficient to model re-radiance of energy based on varying amounts of CO2.
As for those who are jumping on Willis because they don’t get how the shell winds up doing what it does, I’m sad to say that this hard core skeptic agrees 100%, that is EXACTLY what happens. In the real world the “effect” is distributed across the thickness of the atmosphere, and it seems counter intuitive when you look at the drawing, but that is what happens. As upward bound photons are absorbed by CO2 molecules and then re-emitted in a random direction, the amount of energy they absorb gets re-emitted 1/2 upward and 1/2 downward. For conservation of energy to take place, the earth surface mut heat up enough to create upward bound photons that balance the re-emitted downward bound photons.
Which is why a “shell” of a given thickness with a temperature gradient from bottom to top is usefull. It isn’t the surface temp per se we are interested in, it is the change in the gradient. If you do it that way you will find that doubling co2=3.7watts=+1 degree AT SOME POINT ON THE GRADIENT and that surface temp of earth only goes up about 0.6 degrees and the observed temperature of earth from space goes up…. O.

Rienk
November 28, 2010 8:10 am

Sorry, model doesn’t work. In figure 2 the temperature of the shell is given as 0 Kelvin. Since energy is either transmitted perfectly or reflected perfectly, that temperature doesn’t change. As a result in figure 4 it has to radiate at 0 Kelvin.
Another way of going about it, after going opaque we change the suns output from 240W/m^2 to, say, 500W/m^2. What happens on the inside? The answer must be that nothing changes. The shell is opaque and all energy is reflected. Inside the numbers stay at 240W/m^2 and on the outside the numbers are now 500W/m^2. Does the shell now have two temperatures?

John Whitman
November 28, 2010 8:12 am

Dave Springer says:
November 28, 2010 at 6:41 am
The “greenhouse effect” (a poor term but we have to live with it) is physics that was known theoretically 200 years ago. . . .[edit] . . .
The bottom line is the greenhouse effect was demonstrated experimentally over 150 years ago. Anyone claiming it isn’t real is sorely lacking in understanding of introductory level physics.
. . . [edit] . . . The bottom line is the greenhouse effect was demonstrated experimentally over 150 years ago. Anyone claiming it isn’t real is sorely lacking in understanding of introductory level physics.

– – – – – –
Dave Springer,
I agree. The name ‘GHG effect’ is a terribly misleading and very unscientific name. So let us forthwith not use it. I suggest something like Atmospheric Effects of Molecular Dipole Moment Gases is much much better.
Also, regarding the GHG effect being proven, yes, but a simplified demonstration in laboratory does not mean the earth atmosphere effect when mixed with actual atmospheric processes has always the same effect as the laboratory at all altitudes and in all regions. It is simple in lab principle is all we can conclude.
John

Bill Illis
November 28, 2010 8:22 am

The actual surface temperature does not follow the Radiation Theory Explanation.
The Earth should get much, much hotter during the day and much cooler at night. Basically, the Earth absorbs just an extremely tiny amount of the solar energy received during the day and releases back that tiny amount at night.
It absorbs only 0.00026% of the energy received by the Sun during the 12 hours of sunshine and then releases back that 0.00026% at night (give or take some small daily seasonal change).
In other words, 99.99974% of the solar energy received during the day is sent back up to the atmosphere and eventually to space every second that the Sun is up.
A given square metre of Earth Surface receives 20,736,000 joules of solar energy during the day but the surface temperature only rises by 54 joules during that period.
You could think of it as, over time, the Earth has accumulated a tiny differential in outgoing versus incoming radiation and if the Sun didn’t come up tomorrow, it would all be gone in 86 hours (give or take a lag from the ocean and land – it might take a few decades to fall to -270C but it would fall to -75C in just a few days).
This is a fairly typical 24 hours of measured Radiation Flows for a given location. The explanation does not match reality. Table Mountain Colorado on November 17th, 2010.
http://img12.imageshack.us/img12/3225/tablemountainnets.png
http://img140.imageshack.us/img140/4109/tablemountainall.png

Alan McIntire
November 28, 2010 8:28 am

Good post! You addressed both greenhouse effect and zero greenhouse effect. You overlooked a third possibility. On Titan, the atmosphere absorbs some frequencies directly from the sun, and is transparent to some of the longwave frequenices from the planet, resulting in a possible ” anti- greenhouse effect.

Paul Bahlin
November 28, 2010 8:30 am

How about this fundamental problem…
In figure 4 the blue arrow is radiating energy from a low energy black body (the glass) towards a higher energy black body (the earth). Isn’t this a violation of entropy?

Paul Bahlin
November 28, 2010 8:32 am

Furthermore, this can’t happen unless work is somehow applied to the model and there isn’t any in the model is there?

Alan McIntire
November 28, 2010 8:34 am

Thierry’s G&T objection is not completely off the wall. The simplistic global warming models assume an instantaneous adjustment in temperatures. In practice, you’ve got to apply Newton’s law of cooling.
I plugged in Newton’s cooling law on my July 24, 2010 post here:
http://scienceofdoom.com/2010/07/24/the-amazing-case-of-back-radiation-part-two/

Vince Causey
November 28, 2010 8:55 am

bessokeks,
“But transfering heat energy from “cold” to “warm” violates the second law of thermodynamics.
There is no way around that.”
There is a lot of misunderstanding surrounding the second law of thermodynamics. The original formulation of this law says approximately what you attribute to it. This was applied to heat transfers in conducting bodies. The modern formulation is in entropy and states that in a closed system, the total entropy must always increase. The application to heat transfer has been changed to refer to net heat transfers. The implications to conduction of heat have not changed. However, we know that radiative transfers do indeed take place in both directions.
In order for the second law to be obeyed, the net transfer must always be from warmer to cooler. This should be intuitive – even a cool body emits photons of energy which must travel at the speed of light regardless of the fact that they may impinge on surfaces that are warmer. The net transfer argument should make it plain that more photons will be emitted from the warmer surface.
A modern version of Maxwells demon thought experiment is to imagine a demon sitting between the cool and warm surfaces. The demon is supposed to allow the photons from the cooler surface to reach the warmer surface, but prevent the flow in the opposite direction. The action of the demon would cause the cool surface to become cooler and the warm surface to become warmer. This is clearly in violation of the second law. The Maxwell demon paradox was solved when it was realised that the demon needs information to sort the photons and this prevents the system from being closed.
In practice, although photons from the cool surface carry energy to the warm surface, a higher flow occurs in the opposite direction. The result is that the warm surface cools and the cool surface warms. Ie, they equilibriate and their entropy increases. Thus, back radiation does not violate the second law.

Rienk
November 28, 2010 8:56 am

Thinking a bit further. in figure 4 the temperature of the planet has to rise until it radiates 240W/m^2 at and above the short wavelength where the shell is transparent. The radiated spectrum will look a bit unnatural but that’s to be expected when working with miracle matter.

Dave Springer
November 28, 2010 8:56 am

John Bowman says:
November 28, 2010 at 6:02 am
“It should more properly be called ” the insulation effect” as the atmosphere limits the transmission of heat energy by its attenuation within a material.”
Exactamundo!
That should be enough said and it’s a shame when it isn’t. Most people at least intuitively understand the effect of insulation if not the physical mechanism by which it works.

puzzled
November 28, 2010 9:15 am

willis,
can you explain why the temperature of the shell in figures (3) and (4) is only 255K rather than 303K? It seems to be absorbing and radiating at 480W per m2 – the same as the planet in figure (4).
thanks
nige

Barry Moore
November 28, 2010 9:16 am

I agree that thought provoking hypothesis can generate discussion amongst discerning individuals but they must be accompanied with caveats which clearly state that the thought presented is a special case and on its own has no practical implications in the real world except it is one of a long list of interdependent variables which make up the final result.
My concern is that some will take such a presentation too literally and form a wrong impression.
The assumption that 100% of the absorbed radiation is reemitted back from the earths surface is incorrect since the plant life of the world absorbs radiation energy to generate carbohydrate from CO2 and H2O.
The grossly oversimplified diagrams showing radiation from the gasses or particulate in the atmosphere has a free ride back to the earths surface is nonsense since radiation downward will have the same degree of absorption as radiation upwards. Further to this point with regard to gasses they absorb resonant frequencies very quickly ( try doing a Beer’s law calculation) however the close neighbors to the resonant frequencies are absorbed much slower so reach higher into the atmosphere, when a molecule reradiates however it does so at exactly its resonant frequency so its return path is much shorter. Thus none of this energy can return to the surface. I note that in one comment an exception was made to the transfer of energy by collision this is not a viable argument since the time taken by a molecule to reemit IR energy is about 10 000 times longer than the time taken to collide with another molecule thus there is virtually no reemission.
One last point, to assume the atmosphere of the entire world contains some mix of components which forms a 100% opaque screen to all the outgoing radiation is a worthless hypothesis I think a number of shells representing all the different components with their contribution to the warming effect of our atmosphere would be far more instructive.

old construction worker
November 28, 2010 9:48 am

Willis, I believe you figure 3 and 4 are misleading. Here why. Instead of using energy, use dollars with a checking account (the shield) and a saving account (black body). You must do this in cycles.
We have $240 going into the checking account half of which going out, $120. The other half, $120, goes into the saving account then comes out into the checking account. Half is goes out $60 the other half goes back into the saving account and so on. The most you are working with is $360 ($120 in the saving account plus a new cycle of money, $240.)
Figure 4 It doesn’t make any difference if the money hits the saving account first then is transferred to the checking account. Half, $120, goes out, half, $120, is going back into the saving account before starting a new cycle and another $240 is added. You are never working with $480. The most you are working with is $360.

November 28, 2010 9:54 am

the “greenhouse effect”
Maybe this would better be explained by the use of “S-Parms” as they are used in the characterization of uWave (Microwave) components and devices; S-Parms more formally known as Scattering Matrix Parameters.
A system by which the vector sums of the forward (incident) and reflected (back scattered) waves or energy are characterized and allows for the characteristics of the ‘system’ to have different behavior at different wavelengths (as uWave components characterization often uses frequency swept measurements).
Now, if one considers that incoming solar insolation and outgoing earth LW IR encounter elements in three-space (including an optional 3-dimensional atmosphere of some thickness) that have reflective and scattering properties at IR wavelengths then a measurable ‘flux’ exists between the earth, the sun and the infinite ‘sink known as deep space, and each element will have some associated temperature above absolute zero given ‘forward’ and ‘reflected’/re-radiated IR/thermal energy … and the earth will have _with_ an atmosphere some temperature above that what it would have without an atmosphere … especially true if some of the elements encountered exhibit different wavelength characteristics to the spectra of the incoming and outgoing energies (the EM waves, the IR etc.)
I don’t expect anyone exc a couple of uWave RF engineers or physicists to comprehend this … MODTRAN et al (as they account for CO2 and WV) are kinda examples of this.
S-Parms: http://www.google.com/search?hl=en&client=opera&hs=yf6&rls=en&q=s-param+scattering+matrix&aq=0&aqi=m1&aql=&oq=s-parmsscattering+matrix&gs_rfai=
.

November 28, 2010 10:33 am

Jordan says:
November 28, 2010 at 7:49 am
Willis: “OK, let’s assume that the sun is some form of luminous cloud that illuminates the planet evenly with 240 W/m2 of solar radiation.”
“Then the planet heats up to the same temperature as the luminous cloud.”
Only if the cloud is optically thick. If you make the cloud of a hot (6000K) but highly rarefied plasma, it could provide the same amount of “sunlight” as the sun, but stop only 1/160,000 of the planet’s thermal radiation. More simply, if you want a model with effectively uniform insolation (which may indeed be a “simplification too far”), then you can surround the Earth with a cloud or constellation of miniature suns, allowing the thermal radiation to escape through the gaps.

Robert of Texas
November 28, 2010 10:42 am

I read through this post and realized – this post exemplifies exactly what is wrong with serious AGW arguments (note – serious does not include the alrmist fringe – there is no explaining their arguments).
Serious AGW arguments always seem to start off with a “pretend Earth is a Terrarium” tangent and then try to build off of it. Well, Earth is not a terrarium. Most (all?) serious AGW skeptics understand that the Earth will not behave like a simple system. It isn’t a black body, and it doesn’t have a glass shell. It has a very dynamic atmosphre and multiple non-linear response patterns to incoming energy.
Even if we tried to start fixing your “pretend a planet is surrounded by a shell” thought experiment – it soon becomes so complex as to lose its power to explain anything. For example – lets first give your planet 3 concentric shells representing high, medium, and low atmosphere. Each responds differently to temperature becoming more or less reflective (representing clouds) over part of their surface in a very dynamic way. Oh, and the planet has a range of colors and refelctivity, and two main materials that store and reflect heat differently, and also can change refelctivity due to temperature (snow). Now we have a simple (!) model that accounts for reflectivity and absorbtion. We are merely missing heat transport (air convection, ocean currents), pollution effects (not all man-made), and variable light input (solar changes, orbital patterns, tilted rotating and orbiting planet). Does this model help explain anything? No, it makes people bug-eyed.
If atmosperic science could be studied in a controlled environment, we could tease apart the various interactions. Unfortunately, all that can be studied in a controlled environment are tiny bits and pieces (like what happens if we add CO2 to the terrarium?). People with science degrees in specific areas, but no understanding of large complex (often chaotic) systems then run off with the bits and pieces and make grand predictions pointing at their “facts”. (I am also ignoring the people that make up their facts – I am talking about serious AGW supporters here)
This is where the two sides can never seem to find common ground. AGW supporters want to over-simply to the point its “easy to understand”, and AGW skeptics can’t help but wonder how many of the gazillion detials left out are actually important.
My experience tells me that natural systems are complex and dynamic. Climate science is not going to “fit” into a simple realtionship between CO2 and temperature response. If it did, we would have been warming for the last 20 years at a minimum of a linear response, and more likely an increasing response. It hasn’t happened – therefore AGW supporters needs to re-think their side of the argument.

Moritz Petersen
November 28, 2010 10:52 am

Answering your Test-Question:
-The wavelengths which water and CO2 absorb are (mostly) saturated. So this model would have to include multiple shells for certain wavelengths. The number of these shells depends on the average distance between radiation and absorption within the atmosphere. This distance is correlated to the concentration of the greenhouse gases.
– Heat transport through convection between these shells and between the shells and the the surface.
– Varying opacity due to clouds.

November 28, 2010 10:59 am

The Greenhouse Effect is not ill-named. The principle is quite simple and quite general; whenever you make it easy for sunlight to get in, and hard for heat to get out, you get greenhouse warming. This is true no matter how the outwards heat flow is impeded, whether as convection, conduction, radiation, or something more exotic. The Earth’s atmosphere impedes convection, conduction and radiation. The glass of a garden greenhouse impedes convection, conduction and radiation. They are not fundamentally different, though of course the relative importance of each transfer mechanism differs.
I should also point out that garden greenhouses do impede thermal radiation, more than is sometimes thought, because the water condensing on the glass at night is a strong absorber.

November 28, 2010 11:09 am

Bryan.
The woods experiment has nothing to do with radiative transfer, especially in the column of gases, miles high, that we call the atmosphere. The properties of transfer change as you go higher in the atmosphere and are very different in the dry stratosphere. So, just very simply, woods test fails to test the real issue. His experimental set up does not replicate the conditions in the atmosphere, and thus it says nothing about the radiative properties of C02 in the atmosphere. We know from measurements of back radiation that outgoing longwave is reflected back. We know from physics which molecules are responsible and why. we know the same from TOA measurements. The question, the only open question worthy of debate (worth our time) is what sort of feedbacks are present. every doubling of C02 results in a warming of say ~1C. First order estimation based on known working physics. everything else beyond this or below this can only be estimated based on models and by looking at transient responses of the system. Someone like Willis would argue that the system is damped ( naturally regulated) and climate science argues that it is undamped ( in the short term at least).

November 28, 2010 11:27 am

With all due respect, Mr. Springer, you mix up a thin varnish with 781,000 microdots of clear epoxy and 390 molecules of opaque microdots and tell me you can see the difference in visible light, particularly when I’m doing everything I can think of to distract you, like reflecting and slopping on the varnish unevenly and modulating the incoming white light and waving banners between you and the surface you’re trying to study. I accept the resonance on CO2 (and water vapor molecules), but I don’t accept that you can measure the effect of it. If you can’t measure it, then I want you to stop it with the AGW nonsense.

stephen richards
November 28, 2010 11:36 am

Willis, I like it. It is what it is. A thought provoking model that does not and was not meant to mimic reality.
My physics mentor gave me something similar. A 1Tonne Bull charges a lump (very large) of putty. Calculate the energy absorbed by the deformation, converted to heat and it’s location (the energy’s that is).
It’s a very simplistic model that when iterated to infinity will give you the answer to everything climatic. Just one thing, I might have missed it, but won’t the shell reradiate in all directions. 240w absorped, ideally, 240 radiated over 4/3 pi r²? 120 w up and 120w down from both sides of the opaque shell. In simplistic terms.

Nullius in Verba
November 28, 2010 11:39 am

“So, just very simply, woods test fails to test the real issue. […] We know from measurements of back radiation that outgoing longwave is reflected back.”
But the reflection of outgoing longwave radiation isn’t what controls the temperature, because of convection.
Take the analogous case of a pan of boiling water on a stove. The temperature of the water is 100 C. We turn up the heat. Double the power is entering the system at the base of the pan. But the temperature that results from all this extra heat is… 100 C.
The reason it doesn’t get hotter, even though more heat is entering the system, is that convection and evaporation speeds up the rate at which the heat rises and escapes.
Logically, you cannot leap from “more heat” to “higher temperature” without talking about how convection responds, any more than you can say the pan of water gets hotter when you put more heat in. And in a convective atmosphere below the tropopause, convection dominates in determining the temperature profile. The downwelling longwave radiation is still larger in magnitude, but the radiative effects discussed here are ‘short-circuited’ by the non-linearity of convection, and forced to match the adiabatic lapse rate. The adiabatic lapse rate is entirely unaffected by the radiative properties of ‘greenhouse’ gases.
There is a greenhouse effect, and the Willis’s description is how it would work in a non-convective atmosphere, but with convection present it’s effectively an entirely different mechanism and radiative balance is not what determines surface temperature.

tallbloke
November 28, 2010 11:45 am

Willis Eschenbach says:
November 28, 2010 at 4:13 am
you can say that 390 W/m2 is emitted by the surface as radiation, and 100 W/m2 is lost from the surface through convection and evapotranspiration. So somewhere around a fifth of the energy is not going into warming the surface.

Hi Willis, thanks for your reply. Although the figures for downwelling and upward radiation are big in comparison to other forms of energy transport, I think we need to step back and consider their relative importance from a wider perspective. The LW flux results in a net cooling of the surface of around 71W/m^2 according to your modified K-T diagram. This is less than the combined convection and evapotranspirationof 100W/m^2 also cooling the surface.
I thin we need to be careful with terminology to avoid confusion. You talked about downwelling LW ‘heating’ the surface. Since the net flow of down and up LW results in a cooling of the surface, perhaps it’s better to say that downwelling LW slows the rate of heat loss from the surface rather than saying it warms or heats it?
But then it’s clear that if there wasn’t as much downwelling LW there wouldn’t be as much upward flow either, and the other point to make here is that downwelling LW doesn’t penetrate water (70% of the surface) so contributes to evaporation by superheating surface water molecules, a cooling effect.
So it is apparent that the only way the greenhouse effect operates is by raising the altitude of the outward radiating shell. The additional co2 will have raised it around 1-1.5% or about 100 – 150m.

Ira
November 28, 2010 11:48 am

PJP says:
“… lets simplify it even more and say that energy is delivered in truckloads. Lets say we get 2 truckloads per hour. … when we come to your semi-transparent shell, you are still getting two truckloads per hour, but you say that these two truckloads are delivered to both the earth and to the shell — that makes 4 truckloads/hr. … Where did the extra two truckloads come from?”
OK, say the Sun is shipping two truckloads of orange juice to the Earth each day through a Shell that is transparent to orange juice – all the orange juice gets through unchanged and the two truckloads of the stuff are absorbed by the Earth.
Whatever incoming juice the Earth absorbs is converted to blueberry juice and put on trucks that head back to the Shell. So, two truckloads of blueberry juice head to the Shell. The Shell absorbs the two truckloads of blueberry juice and then (pay attention here!) sends one truckload of blueberry juice out to Space and one truckload back to the Earth where it is absorbed.
The next day, the Sun has sent another two truckloads of orange juice to Earth and Earth has absorbed and transformed it to blueberry juice. So, on the second day, the Earth has THREE truckloads of blueberry juice to send up to the shell (the truckload sent back by the Shell yesterday and two new truckloads from today). OK, those THREE truckloads get to the Shell and half is returned to Earth (1.5 truckloads) while the other half goes to Space.
On the third day, the Sun has sent yet another two truckloads of orange juice to Earth, which is absorbed and converted to blueberry juice. That, along with the 1.5 truckloads returned from the Shell yesterday, is sent back to the Shell, for a total of THREE AND A HALF truckloads today.
As you should be able to see, as the days pass, the traffic in blueberry juice from the Shell to the Earth and back will continue to increase asymptoticaly until there are FOUR truckloads of blueberry juice a day going from Earth to the Shell and TWO coming back.
So, in the steady state, each day there are TWO truckloads of orange juice and TWO truckloads of blueberry juice going from the Shell to the Earth, FOUR truckloads of blueberry juice from the Earth to the Shell, and TWO truckloads of blueberry juice going from the Shell to Space. Exactly as Willis shows in his Figure 4. QED!

Bryan
November 28, 2010 11:56 am

Steven Mosher
…..”The woods experiment has nothing to do with radiative transfer, especially in the column of gases, miles high, that we call the atmosphere.”…….
The part of the Woods experiment that most people pass by is that for the remaining real radiative effect the result is almost negligible.
Some people have argued that this is explained because the Woods experiment box had very little height.
Hence my post on the much larger polyethylene tunnels which bear out the conclusions of R W Wood.
The radiative effect of CO2 and H2O seems to be of little consequence in large greenhouses even though the IR blocking was 100%.
If this concentrated “greenhouse effect” is of such little consequence that the recommendation of the study group is not to bother adding it to polyethylene what does that tell us about the atmospheric greenhouse effect?
The only effect that the “greenhouse gases” seems to engender is speculative posts and thought experiments.
For bulk effects of the atmosphere such as;
The polytunnel study in post above.
Another bulk experiment is obtained by pointing parabolic mirrors at the dark night sky.
This is capable of freezing water at the mirrors focus even when the planet surface temperature is above zero.
I realise of course that the focus is not perfect for an extended source such as the atmosphere however a concentration effect seems to be evident.
On the other hand is there any hard physical evidence of a bulk effect to show that the large claimed magnitude of the “greenhouse effect” is real?
I discount the claimed assertion that it amounts to a 33C surface temperature increase unless backed with a plausible cause.

Owen
November 28, 2010 11:56 am

H2O, CO2, O3, CH4, N2O absorb outgoing infrared radiation, promoting these molecules to higher-energy excited states. The excited states collide inelastically and repeatedly with N2 and O2 molecules, increasing the kinetic energies of those N2 and O2, raising the average temperature of the upper atmosphere and returning the excited greenhouse molecules to their ground states. Outgoing radiation is therefore prevented from radiating into space but instead is absorbed and transferred to the atmosphere.
The warming of the atmosphere reduces the thermal gradient between the surface and upper atmosphere and therefore acts to insulate.
Is this explanation correct?

Jordan
November 28, 2010 12:04 pm

Paul Birch says: “hot rarified plasma” around the planet.
To be relevant to AGW discussion, this thought experiment needs to relate to our situation, or it should move in that direction when elaborated. You are running in the wrong direction.

Roy
November 28, 2010 12:21 pm

@ Mike Haseler:
“Come on this is stupid. The greenhouse effect doesn’t exist for the simple reason that there is no ‘greenhouse’ effect in a greenhouse – it would be like using the theory of the ether to explain radio waves and talking about the “ether effect” and not expecting most people who knew what was being talked about not to roll with laughter on the floor.”
Are all those gardeners who have greenhouses stupid? Have you never warmed yourself on a cold but sunny day by standing with your back to a window through which the sun is shining?

Nullius in Verba
November 28, 2010 12:29 pm

“Are all those gardeners who have greenhouses stupid?”
Nope. Greenhouses work by preventing heat loss via convection, not by the “greenhouse effect”.
~And how does CO2 in the atmosphere prevent convection?

Editor
November 28, 2010 12:33 pm

Ok for those of you who keep claiming fig 4 is wrong, here is why its not:
Solar energy to the planet: 240 w/m^2
Energy reradiated by the atmosphere back to the planet: 240 w/m^2
Add the two together: 480 w/m^2, which is what the planet radiates back.
The atmosphere intercepts half of the 480 and reradiates it back to the planet in a continuous cycle that is the equilibrium, while the atmosphere radiates half the 480, or 240 w/m^2, to space, which is the same amount the sun sends to the planet.
For those of you who keep saying convection: that does eliminate the interior vacuum, so you will get, in addition to reemission of heat from one molecule to another from the surface to the edge of space, you will get some degree of convection as molecules rise when they are warm, and drop when they are cool, but this is limited to certain layers of the atmosphere, there is no actual convection from the surface to the edge of space.
For those of you who point to the earth being round with one side in darkness, as Willis says, the model is time averaged. The peak solar insolation is far higher than 240 w/m^2, 240 is a 24 hour average, which includes the time the sun isn’t shining on the surface. Peak solar insolation, with no clouds, is about 1000 w/m^2 at the surface at the equator, and zero at the poles, while about 400 w/m^2 is blocked/reflected by the atmosphere from the start (more when there are clouds involved).
For those of you who keep saying the molecules in the atmosphere radiate in all directions: Yes, this is true, which is why the model shows 240w/m^2 going up, and 24ow/m^2 going down, but not quite exactly…..
The flaw in Willis’ model, however is that you wont have an equal division of 240 up and 240 down, because the earth is round, the surface curves away convexly from the molecule’s perspective, so there will always be slightly more heat radiated to space than reabsorbed by the ground.
There is also a factor of variance that involves how the solar cycle’s variation in solar wind and UV and other factors causes the earth’s atmosphere to vary in diameter. When it is swollen up during solar maximum, it has a larger diameter so it will intercept more radiation than during solar minimum, when it is a smaller diameter.

Nullius in Verba
November 28, 2010 12:55 pm

“For those of you who keep saying convection: […] but this is limited to certain layers of the atmosphere, there is no actual convection from the surface to the edge of space.”
Good answer. But there is convection throughout all the layers that actually radiate to space. Most ‘greenhouse’ radiation to space is from the lower 10 km of atmosphere, the troposphere, and it’s all convective.
The exception is the ozone layer in the stratosphere, which absorbs UV and is non-convective.
Incidentally, since everybody knows that hot air rises, why are the tops of mountains so cold? Why is the top of the troposphere 11 km up at a temperature of -54 C? What maintains the temperature gradient?

November 28, 2010 12:57 pm

Jordan says:
November 28, 2010 at 12:04 pm
Paul Birch says: “hot rarified plasma” around the planet.
“To be relevant to AGW discussion, this thought experiment needs to relate to our situation, or it should move in that direction when elaborated. You are running in the wrong direction.”
I was not running in this direction. On the contrary, it was I who pointed out that the Earth is not subject to a uniform insolation. However, your objection to Willis’s fix was erroneous. The uniform insolation simplification is workable. How useful it is may be debatable; but those who cannot or will not understand even this simple model are unlikely to profit from more realistic ones.

Mike Blackadder
November 28, 2010 1:00 pm

Willis,
Sorry, but I liked some of the other answers (like Leonard Weinstein’s) better than yours. There are all kinds of reasons why a single shell model separated by a vacuum would not be very representative of real Earth temperature, such that if it was anywhere close it would have been a fluke. The question is whether a given model is informative about the impact of higher greenhouse gas concentrations. I don’t think the two shell model is much more informative than the one shell model. Sorry to be critical, but you’re the one who asked a fuzzy question and then held out for a particular answer that isn’t particularly interesting.
I think that the rest of the post is fine, I just think the final question and answer is a bit silly. Next time maybe you should drop a couple of hints to move the conversation in the desired direction.

davidmhoffer
November 28, 2010 1:18 pm

Ira,
That was a brilliant explanation. I’ve tried explaining the physics a dozen different ways and never got to a concise explanation that was easily understood and still reasonably accurate. thanks, and with your permission, soon to be repeated to many.

November 28, 2010 1:34 pm

Willis, your model is what I like to call the “Planet Dewar” model of the GHG effect. Of course, the air really is in contact with the surface, and it really has an exponential temperature gradient as altitude rises caused by gravity, at least up to the tropopause, but it sounds like these are not the problems you have in mind.
One paradox that arises from this model is when it is extended to what I call the “Planet Dante” model, with more multiple crystalline heavens. The same reasoning that leads 2E to reach the surface when only E is incoming with one heaven leads 3E to the the sruface with 2 heavens, etc. With multiple heavens, the surface can be made arbitrarily hot by this reasoning, at least until the surface begins to incandesce and emit visible light and short-wave IR which can escape through the assumedly transparent shells.
So other than this, what is wrong with the good old Planet Dewar model?

Dr A Burns
November 28, 2010 2:24 pm

>>Willis Eschenbach says:
>>November 27, 2010 at 6:05 pm
>>Some folks upthread have said that what is missing is rotation. … I still have not seen anyone point to it.
You assume a shell with 2 sides but a one sided earth. The rotation of the earth carries heat from the hot to the shaded side, where it radiates back to the shell and out to space. The shell itself cannot be assumed to be a constant temperature because it receives less solar heating at the poles.
Your model needs gravity and a circulating atmosphere to create a lapse rate, giving different temperatures between the shell and the earth’s surface.