Visualizing the "Greenhouse Effect" – Molecules and Photons

Guest Post by Ira Glickstein

This series began with a mechanical analogy for the Atmospheric “Greenhouse Effect” and progressed a bit more deeply into Atmospheric Windows and Emission Spectra. In this posting, we consider the interaction between air molecules, including Nitrogen (N2), Oxygen (O2), Water Vapor (H2O) and Carbon Dioxide (CO2), with Photons of various wavelengths. This may help us visualize how energy, in the form of Photons radiated by the Sun and the Surface of the Earth, is absorbed and re-emited by Atmospheric molecules. DESCRIPTION OF THE GRAPHIC

The animated graphic has eight frames, as indicated by the counter in the lower right corner. Molecules are symbolized by letter pairs or triplets and Photons by ovals and arrows. The view is of a small portion of the cloud-free Atmosphere. (Thanks to WUWT commenter davidmhoffer for some of the ideas incorporated in this graphic.)

  1. During the daytime, Solar energy enters the Atmosphere in the form of Photons at wavelengths from about 0.1μ (micron – millionth of a meter) to 4μ, which is called “shortwave” radiation and is represented as ~1/2μ and symbolized as orange ovals. Most of this energy gets a free pass through the cloud-free Atmosphere. It continues down to the Surface of the Earth where some is reflected back by light areas (not shown in the animation) and where most is absorbed and warms the Surface.
  2. Since Earth’s temperature is well above absolute zero, both day and night, the Surface radiates Photons in all directions with the energy distributed approximately according to a “blackbody” at a given temperature. This energy is in the form of Photons at wavelengths from about 4μ to 50μ, which is called “longwave” radiation and is represented as ~7μ, ~10μ, and ~15μ and symbolized as violet, light blue, and purple ovals, respectively. The primary “greenhouse” gases (GHG) are Water Vapor (H2O) and Carbon Dioxide (CO2). The ~7μ Photon is absorbed by an H2O molecule because Water Vapor has an absorption peak in that region, the ~10μ Photon gets a free pass because neither H2O nor CO2 absorb strongly in that region, and one of the 15μ Photons gets absorbed by an H2O molecule while the other gets absorbed by a CO2 molecule because these gases have absorption peaks in that region.
  3. The absorbed Photons raise the energy level of their respective molecules (symbolized by red outlines).
  4. The energized molecules re-emit the Photons in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.
  5. This frame and the next two illustrate another way Photons are emitted, namely due to collisions between energized GHG molecules and other air molecules. As in frame (2) the Surface radiates Photons in all directions and various wavelengths.
  6. The Photons cause the GHG molecules to become energized and they speed up and collide with other gas molecules, energizing them. NOTE: In a gas, the molecules are in constant motion, moving in random directions at different speeds, colliding and bouncing off one another, etc. Indeed the “temperature” of a gas is something like the average speed of the molecules. In this animation, the gas molecules are fixed in position because it would be too confusing if they were all shown moving and because the speed of the Photons is so much greater than the speed of the molecules that they hardly move in the time indicated.
  7. The energized air molecules emit radiation at various wavelengths and in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.
  8. Having emitted the energy, the molecules cool down.

DISCUSSION

As in the other postings in this series, only radiation effects are considered because they are the key to understanding the Atmospheric “Greenhouse Effect”. I recognize that other effects are as important, and perhaps more so, in the overall heat balance of the Earth. These include clouds which reflect much of the Sun’s radiation back out to Space, and which, due to negative feedback, counteract Global Warming. Other effects include convection (wind, thunderstorms, …), precipitation (rain, snow) and conduction that are responsible for transferring energy from the Surface to the Atmosphere. It is also important to note that the Atmospheric “Greenhouse Effect” and a physical greenhouse are similar in that they both limit the rate of thermal energy flowing out of the system, but the mechanisms by which heat is retained are different. A greenhouse works primarily by preventing absorbed heat from leaving the structure through convection, i.e. sensible heat transport. The greenhouse effect heats the earth because greenhouse gases absorb outgoing radiative energy and re-emit some of it back towards earth.

That said, how does this visualization help us understand the issue of “CO2 sensitivity” which is the additional warming of the Earth Surface due to an increase in atmospheric CO2? Well, given a greater density of CO2 (and H2O) molecules in the air, there is a greater chance that a given photon will get absorbed. Stated differently, a given photon will travel a shorter distance, on average, before being absorbed by a GHG molecule and be re-emitted in a random direction, including downwards towards the Surface. That will result in more energy being recycled back to the Surface, increasing average temperatures a bit.

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March 29, 2011 12:11 am

I like that animation. I have never been very good at that. 🙂
It is worth noting that gases can gain energy from collisions as well as by absorbing a photon. A molecule doesn’t care where it got the energy from. Once a molecule has more energy than the molecules around it, it will dissipate through releasing a photon or it could also lose it by another collision. Each molecule is having billions of such collisions per second under normal atmospheric conditions.
You are correct in keeping it simple as the more complex the animation is, the harder it is to follow.

rusureuwant2know
March 29, 2011 12:34 am

Convection currents are the “great equalizer”…. I have a question I’m hoping someone can answer – I thought Al Gore got his graph backwards and the increase of CO2 followed warming – why are we still working with the assumption it creates it???

bananabender
March 29, 2011 12:41 am

The Ideal Gas Law PV=nRT quite accurately predicts the atmospheric temperature of Earth, Mars, Venus and Titan (at all altitudes) based on a simple pressure effect. Why add the unnecessary complication of the highly dubious Greenhouse Effect to an elegant and simple solution?

Jer0me
March 29, 2011 12:43 am

Can we have one with the number of different GHG molecules represented proportionally for reference? We might not be able to see the CO2 one among all the others, though….

mccall
March 29, 2011 1:09 am

Don’t know how I missed this, but here’s a catchy Harrington Bro’s tune for part of the soundtrack: http://www.youtube.com/watch?v=9Ts4WxMHw4E&feature=related

mccall
March 29, 2011 1:17 am

Much of Newfoundland (“The Rock”) is at higher Lat than Minnesota. Though not as biting as “Hide the Decline,” tune-wise the M4GW guys have some professional competition…

Hank
March 29, 2011 1:21 am

Let us use the back radiation according to the global warming science theory to warm our houses without burning fossil fuels.
It is simple: paint a black square on your ceiling.
Say the black paint covers 10% of the ceiling and let us compare the backradiation of the black square to the back radiation of the Co2 molecules in the air .
Let us assume that the Co2 molecules in the airabove the ceiling are all present next to each other on the ceiling and radiating all the energy back to the floor . The backradiation of the Co2 molecules on the ceiling would be much stronger than the the real situation in the air.
According to the global warming theory doubling the Co2 concentration would increase the temperature with one degree
Instead of doubling the Co2 concentration we have a painted a black square.
The surface of the black square is 250 times bigger than the Co2 surface
The black paint will absorb and emit the whole infrared band from the floor
whilst the Co2 surface will only backradiate from a narrow band in the spectrum.
Because of its size the black square we can compare this to a (at least) 250 times increase in Co2 concentration.
In other words we can expect an enormous increase in temperature in the room without using energy. Be carefull don’t paint the whole ceiling black !!!.

Steve in SC
March 29, 2011 1:31 am

Ira,
I think a review of classical thermodynamics is in order here.
You can not win
You must lose
You can not get out of the game.

John DeFayette
March 29, 2011 1:38 am

Very interesting discussion thread. I appreciate the simplification and the limitation of the argument, since this basic physical concept seems to confuse the waters in more cases than not.
In the meantime, I’ve been pondering the question of atmospheric absorption of energy lately, and maybe you could pull out some interesting figures for the audience. Do you happen to have in your files the energy dependent photon absorption curves for various important gases in the air (i.e. N2, O2, H2O vapor, CO2, and why not Ar, H2, He, CH4, CO, and N2O)? I imagine the non-asymmetric molecules will not be very opaque, but it’s still a nagging question in my mind.
Also, for the sake of education it my be instructive to show what the blackbody radiation spectra of a surface at, say 240K (Arctic), 285K (a normal summer night), and 310K (a hot day) look like.
Thanks for the good work.

Stephen Wilde
March 29, 2011 1:38 am

Although there clearly is a greenhouse effect it is infinitesimal compared to the energy retaining effect of the oceans.
Furthermore the non radiative processes seem to mostly or entirely negate its effect with the real reason for the Earth’s temperature being down to oceanic and not atmospheric energy retention.
Really, we should regard the oceans as part of the atmosphere for energy budget purposes.
The recent suggestion that removing CO2 would cause a snowball Earth cannot be right because there would still be solar shortwave into the oceans keeping them liquid and so a continuing water cycle.
As far as I can see a removal of CO2 would just mean a little less energy in the atmosphere, lower the height of the tropopause a little for a shallower troposphere and redistribute the pressure patterns for a slightly slower water cycle for no significant temperature effect.
The cooling effect of less CO2 would be all or mostly offset by the slowing of the water cycle just as the warming effect of more CO2 seems to be all or mostly offset by a speeding up of the water cycle.

John Marshall
March 29, 2011 1:52 am

Sounds all very well BUT.
This explanation completely ignores convection which is probably the most important heat distribution system. When this LW radiation adds energy to a GHG, ie warms up a CO2 molecule, it will immediately transfer this energy to molecules of lower energy levels, ie warm up the surrounding gasses. This is as per the 2nd law which states, through inference, that heat cannot be stored.
When a parcel of gas becomes warmer than that surrounding it will convect due to density difference and rising air cools adiabatically. This cooling air will be warmer than that surrounding to continue convecting but cooler than the surface so energy will not be radiated to the surface because the 2nd law of thermodynamics forbids this to happen, energy only able to flow from a high to lower level.
If it were possible to get energy to flow from cold to hot then we would have all our energy requirements solved. We would have a perpetual motion machine. This is impossible.
So the above explanation may work in the laboratory but the atmosphere is a different place. If convection were not a major player in heat transport then we would see far less cloud than we do, most cloud being due to convection.
When we look at geological history there were periods when atmospheric CO2 levels were many times today’s. There were ice ages and periods of warmth both independent of the atmospheric CO2 levels. Present day research shows that CO2 levels rise after rises in temperature which goes a long way to prove that CO2 does not drive temperature.

Thomas
March 29, 2011 1:57 am

bananabender. Go ahead, try to explain the temperature of the Earth using only the ideal gas law! It can’t be done, you see, if you actually start to think about what you do.

Massimo PORZIO
March 29, 2011 2:04 am

“The Photons cause the GHG molecules to become energized and they speed up and collide with other gas molecules”
I’m not really convinced about that, or better it could be not always true. In the particular case of the CO2 absorption band at 15um, as far I know, the energy of the photon is converted into molecular bending which just polarize the molecule. From my point of view, this shouldn’t increase the speed at all, but it should increase the probability of share energy with the surrounding molecules because the polarized molecule needs only two simultaneous collisions to share energy, while the unpolarized one needs three simultaneous collisions, one in one direction applied to the C atom and the other two applied in the opposite direction, or at least applied having angles on the O atoms which can bend the aligned O=C=O chain. If I’m right here, the CO2 molecules should be better as “photons absorbers” than “photon emitters”. So, it should have very little efficiency as “optical photon spreader”, but it should be a very efficient “thermodynamic photon sharer”.

stephen richards
March 29, 2011 2:05 am

Ira
Sorry to poop the party but over at climaterealists http://climaterealists.com/index.php?id=7457
there is an excellent proof using the physics of black body spectrum which disproves the greenhouse theory conclusively.

Cassandra King
March 29, 2011 2:06 am

So a single CO2 molecule can both absorb photon energy and then re emit this energy in random directions? Given the fact that atmospheric CO2 only amounts to 0.039% of the atmosphere the net amount of re re radiated photons hitting the earth must be truly tiny, so small in fact as to be barely measurable.
“The greenhouse effect heats the earth because greenhouse gases absorb outgoing radiative energy and re-emit some of it back towards earth.”
The problem I have with this explanation is that the amount of re radiated energy directed downwards by such a trace gas as CO2 would be infinitesimally small and it is not constant and because it is not constant and the amount is so small this heating would be lost almost immediately within the far greater ‘noise’ signal of the other atmospheric gases which exist in far far greater amounts.
The obvious question is one of proportions and percentages and given that atmospheric CO2 is present in such trace amounts just how much is CO2 heating the earth by? And given that global temperatures have not increased in the last decade and most of the 20th centurys small net warming occurred BEFORE the steep rise in atmospheric CO2 then I have strong reservations about CO2 being responsible for any measurable warming whatsoever.
Here is the problem as I see it, the rise in atmospheric CO2 and the rise in global temperatures should be linked together and yet they are not, we have seen a fall in global temperatures even as CO2 increased and if CO2 was the driver of increasing temperatures we would have seen a greater increase in global temperatures at the same time as increasing CO2 levels yet the opposite has happened. CO2 rose quickly post war while global temperatures rose much more slowly post war and even as CO2 has been rising since 1998 global temperatures have not measurably increased at all.

MattB
March 29, 2011 2:06 am

Careful – you’ll get Louis, Richard and the “Violation of the 2nd Law of Thermodynamics” crew on to you with outrageous statements like:
“The greenhouse effect heats the earth because greenhouse gases absorb outgoing radiative energy and re-emit some of it back towards earth.”

cal
March 29, 2011 2:07 am

I am happy with this description overall but I am not sure about this part:
“The Photons cause the GHG molecules to become energized and they speed up and collide with other gas molecules, energizing them”
It is a long time since I studied absorption and scattering but this is how I see it.
Photons interact with electrons in molecules. Photons with energies which excite any of the vibrational or rotational modes of the molecule will be absorbed, other wavelengths will generally not be absorbed. There are some inelastic scattering modes which result in photon absorbtion and re-emission at slightly different wavelengths but I do not think that this band broadening is what you are describing.
A CO2 molecule colliding with, say, an N2 molecule, will pass on energy as well as receive energy during that collision. However the CO2 molecule has vibrational and rotational modes in the long wave energy band but N2 does not so any energy passed to the N2 molecule will not be radiated away whilst energy absorbed by the CO2 molecule might well be. On balance however I do not see this as being significant overall since the nett effect of the collisions should be small.
Was it not you who showed the radiation plot looking up from the earth at night? This showed that there is virtually no atmospheric radiation outside of the absorption bands of the greenhouse gases. I do not understand why you feel a need to include these small scattering effects in your description while you exlude major effects like convection which change the altitude and therefore amount of radiation to space far more significantly.

Anders
March 29, 2011 2:25 am

If all that is going on is absorption and re-emitting radiation there would be no heating of the atmosphere. Heating happens because of the re-distribution of energy through collisions between molecules. The “simple” visualization is thus too simple. The temperature of a gas IS the average kinetic energy of the gas molecules, this is something Roger Pielke Sr. has pointed out a number of times and for many years.

MattB
March 29, 2011 2:27 am

Stephen Richards – that link that “disproves” theory of AGW – well for mine it is based on an incorrect understanding of the 2nd law and how it relates to warming. I see you beat me to my post at 2:06 though – damn moderation lag:)

wayne
March 29, 2011 2:29 am

Jer0me says:
March 29, 2011 at 12:43 am
Can we have one with the number of different GHG molecules represented proportionally for reference? We might not be able to see the CO2 one among all the others, though….

Like this one?
http://i55.tinypic.com/a76tw.png
( In proportion: N2-green; O2-gray; Ar-brown; H2O-aqua (2.5%); CO2-red (400 ppmv) )

lgl
March 29, 2011 2:51 am

What a lot of nonsense. Here is how it really works http://www.ems.psu.edu/~fraser/Bad/BadGreenhouse.html
“The atmosphere radiates because it has a finite temperature, not because it received radiation”

steveta_uk
March 29, 2011 2:56 am

Further to cal’s points, I don’t understand how CO2 and H2O are able to re-radiate in the way described.
We are told by the AGW crowd that much of the expected (and so far undetected) warming occurs in the middle atmosphere, where the air temps are pretty low (-10C to -30C, up where the planes live).
If a warm surface emits a photon that is captured at this height, and conduction results in almost instant passing on of the captured energy to surrounding gasses (primarily N2) then how do the H2O and CO2 molecules regain enough energy from the comparatively cold gasses to re-emit photons of similar energy?
Surely they can only re-emit much lower energy photons, of similar energy levels to the black-body curve at -20C.

1DandyTroll
March 29, 2011 3:06 am

So far the best write up.
However, there is no mention of how much the re-emitted photon helps to warm the earth, and it is only a conjecture that it really does. For does it not hold true that if the re-emitted direction is random above a sphere the, in this case outer, space around it is the most likely direction?

anopheles
March 29, 2011 3:08 am

How much of this is theory, and how much observed, in the real atmosphere? Can we see it happening? Can we devise an experiment or observation which would work in the field? Personally, I’ve been struggling with radiative theory, I’ve seen stuff from all sides, and plainly theorists cannot agree. So what do we KNOW?

mindert eiting
March 29, 2011 3:15 am

Hank: thanks for your example. Although I am not a physicist, I thought that at the end in step 7 by Glickstein a little miracle occurred. If we have a perpetuum mobile working a tiny tiny bit, we can use that for heating our homes at least. Perhaps the next winter I will do your painting trick.

steveta_uk
March 29, 2011 3:15 am

Stephen Richards (2:05)
the article in Climate Realists makes exactly the same mistakes as all those that think back-radiation somehow violates 2nd law thermodynamics. The last-but-one paragraph sums up his errors quite nicely – it’s basically completely wrong.
Oddly, he describes a thought experiment with a black body and mirrors that he could easily try out, and which would show him he is wrong, Clearly he knows he is right, so doesn’t bother to actually test it.
The other indicator that he is wrong is the massive introduction that is just waffle, before reaching any solid science. Why bother with all that if you are right?

Stephen Wilde
March 29, 2011 3:45 am

“where their energy is absorbed, further heating the Earth”
That phrase is the crux of the difficulty that some have with the greenhouse effect.
There is no ‘further’ heating.
What happens is that the downward longwave radiation reduces the net upward flow by partially offsetting it until the temperature rises and a new equilibrium is reached.
There is always a net energy flow from a warmer body to a cooler body in accordance with the Laws of Thermodynamics but in fact both bodies still radiate towards each other.
A cooler body doesn’t stop radiating just because it is in the presence of a warmer body.
It is the net rate of energy transfer between the two that changes with no need for the cooler body to effect any direct warming of the warmer body.
The suggestions that the greenhouse effect somehow offends the Laws of Thermodynamics is a non starter and a hindrance to scepticism of the theory of AGW.
Ira doesn’t seem to make that mistake but he has used a form of words that perpetuates the misunderstanding.

Michael Larkin
March 29, 2011 4:12 am

Ira,
Thanks for this, but for me the animation is way too fast to follow. Any chance you could slow it down?

Joe Lalonde
March 29, 2011 4:30 am

Ira,
Good presentation!!!
You’ve only touched partially on the tip of the iceberg of how complex this planets climate system is.
Thermodynamics is garbage in being too simple and too broad in coverage so that simple minded scientists can grasp some hope of understanding.
This planet is a globe and not a tube, that rotates and has gravity. From the equator to the poles, there are different factors going on besides the introduction of solar radiation, CO2, gases, water vapor, etc. Cold compression and superheated compression has very different areas in complexity as well as storing energy on a planet that is thrown through space at 300km/sec. We have yet to understand infused energy into the planet and solar system at creation and what that accomplishes in storing energy for slow release.
Our current problem is that science can only grasp one area at a time and fails in understanding a multi-understanding of many areas into a super complex system.
Following temperatures in a 150 year time period is ridiculous in the 4.5 billion year time frame and for what? To know what clothes to wear?

Ian W
March 29, 2011 4:31 am

Ira,
I know you have the caveat paragraph “As in the other postings in this series, only radiation effects are considered because they are the key to understanding the Atmospheric “Greenhouse Effect”……
But in comparison to the power of the hydrologic cycle your ‘radiation only’ approach is not unlike saying you are only looking at the retardation effects on the locomotive of hitting insects.
The AGW CO2 warms the world hypothesis depends on the hydrologic cycle. It only works if they can show the ‘water vapor feedback’ which all the GCMs show as a tropospheric hotspot that in the real world does not exist therefore all the models are falsified. You cannot disregard the locomotive of the hydrologic cycle whose effects nobody has been able to quantify but is accepted as the main transport of heat from the surface to the tropopause and the ‘iris’ that increases albedo reducing incoming energy and only worry about the insect collisions of outgoing IR with CO2 in three small radiation bands.

bananabender
March 29, 2011 4:38 am

@Thomas says:
March 29, 2011 at 1:57 am
bananabender. Go ahead, try to explain the temperature of the Earth using only the ideal gas law! It can’t be done, you see, if you actually start to think about what you do.

Try this:
Timothy Casey B.Sc.(Hons.)
http://greenhouse.geologist-1011.net/
The temperature of the earth’s surface is often explained using the “Greenhouse Effect”. However, having refuted the “Greenhouse Effect”, we may wonder if it was necessary in the first place. The earth orbits the sun in the vacuum of space. There is no aether as Fourier, Tyndall and Arrhenius believed. Moreover, there is no heat capacity or thermal conductivity in space. The only way for heat to escape the planet is by emission to space. That makes the temperature of the absorbing mass of the earth a question of radiative heat transfer. Hereafter, I will refer to the that portion of the earth’s mass which absorbs solar radiation as the “solarsphere” because the atmosphere does not include the surface layer warmed by the sun on a day to day basis and there is no other term to encompass both. The method of calculation is to treat the solarsphere as an absorbing body subject to incident radiation from the sun.
Given the solar constant of 1368 Wm-2 (Fröhlich & Brusa, 1981) and the fact that the cross-sectional area of solar radiation incident upon the earth is roughly one quarter of the earth’s surface area, it is unsurprising to observe that authors such as Kiehl & Trenberth (1997) arrive at 342 Wm-2 as the mean quantity of solar radiation that falls on the entire surface of the earth. Using this, we may calculate the expected geographical and altitudinal mean temperature of the earth’s solarsphere.
Wm = σT4
T4 = Wm/σ
T = {Wm/σ}0.25
Given Wm = 342:
T = {342/0.000000056704}0.25 = 278.7ºK = 5.5ºC
This figure, is an average or mean temperature for all times, latitudes, and altitudes of the the earth’s solarsphere. Just as the balance point or centre of gravity is found at the centre of mass, this average temperature may be found at the centre of heat capacity. In materials of similar heat capacity, this can be found near the centre of mass. Thus, in order to determine how well our 5.5ºC result -calculated above- corresponds to observed reality, we must first determine the average observed temperature at the barometric median in the part of the earth penetrated by solar energy.
From the diagrams supplied by Vallier-Talbot (2007, pp. 25-26), we may roughly determine the centre of mass for a one square metre column extending from two metres below the surface to 50 kilometres above the surface. Soils and clays amount to roughly 2 tons per cubic metre, with the atmospheric column having to weigh 10 tons in order to yield a mean barometric pressure of roughly 1000 hectopascals at the surface. The total column weighs 14 tons with the centre of gravity corresponding to the barometric median at 700 hPa. Referring once again to Vallier-Talbot (2007, p. 26) we may determine that on average, this pressure corresponds to an elevation of roughly a mile or 1600m above the surface. Given the observed average atmospheric thermal gradient of -7ºC with every 1000m of elevation above the surface (Vallier-Talbot, 2007, p. 25), we may calculate the average absorbing mass temperature as it occurs at the altitude of the barometric mean for our absorbing column. No doubt you’ve worked out that the temperature drop over a tropospheric ascent is 11ºC per mile, and we all know that the average surface temperature is 15ºC (Arrhenius, 1896, p. 239; Burroughs, 2007, p. 124). Notwithstanding 100 years of apparently constant mean temperature from Arrhenius to Burroughs, we may determine that the observed temperature at the altitude corresponding to the centre of absorbing mass is 4ºC or 277ºK. This, via the reasoning above, extends to an observed average absorbing mass temperature for planet earth of 4ºC or 277ºK. This is slightly cooler than the mean absorbing mass temperature calculated above from the solar constant (278.7ºK, 5.5ºC) even if we do allow for 0.5º warming over the last century. However, if we were to consider the impact of convective cooling, I think we can agree that the temperature we derive from the Stefan-Boltzmann equation is well within the tolerance we must allow for such tests.
Adding the tropospheric thermal gradient of 11ºC per mile we got from Vallier-Talbot (2007) above, our temperature (278.7ºK, 5.5ºC), calculated from the Stefan-Boltzmann Equation using the Solar Constant, yields a calculated surface temperature of around 16.5ºC. The fact that this is warmer than the observed mean surface temperatures of Arrhenius and Burroughs (15ºC) leaves no room for such dubious free energy mechanisms as Arrhenius’ “Greenhouse Effect”. The surface temperature of the earth can be much more simply explained without resorting to such complex and unverifiable entities as radiative amplification and power recycling via backradiation of the “Greenhouse Effect”. Absorptivity of any of the parts can vary, but that only alters the overall emissivity, which in turn leaves unchanged, the gross power flowing though the system. Once equilibrium is reached it is only the power flowing through a thermally isolated system that controls and maintains mean temperature. This is because power is required to offset the amount of heat that is lost spontaneously and continuously due to emission of radiation.
Our calculation of mean surface temperature without the “Greenhouse Effect” above (16.5±0.5ºC corresponding to 16-17ºC) is made without considering the effect of carbon dioxide. According to Arrhenius (1906a, translated by Gerlich & Tscheuschner, 2009, pp. 56-57) the observed temperature should be 20.9ºC higher than that yielded by a calculation such as this, owing to the carbon dioxide in the atmosphere. The observed surface temperature of 15ºC (Arrhenius, 1896; Burroughs, 2007) is actually 1-2ºC lower than the calculated mean surface temperature of 16-17ºC. The lower atmosphere will always be warmer than the upper atmosphere because higher material density in the lower atmosphere dictates a much higher thermal conductivity, absorption and density of heat. In contact with an opaque surface warmed by the bulk of the heat absorbed from the sun, it is not difficult to explain why the surface is so much warmer than the altitude corresponding to the centre of mass in the solarsphere. Moreover, the Ideal Gas Law (PV = nRT) dictates that the temperature of a gas containing a given amount of heat invariably increases with pressure. As the highest atmospheric pressure is at the surface, it makes sense that the higher temperature is there, especially if obstruction to radiative outflow decreases with altitude.
Turning our attention to the example of Langley’s greenhouse experiment on Pike’s Peak in Colorado (mentioned by Arrhenius, 1906b), we may be tempted to ask how it is that a greenhouse can reach such high temperatures. Qualitatively, we may attribute the difference between the 15ºC mean surface temperature and the 113ºC observed in Langley’s greenhouse to the fact that noon-time radiation at the surface is three to four times as intense as the mean radiation over the whole of the earth’s surface. Repeating our calculation method, this time for the midday conditions of a greenhouse:
T = {Wm/σ}0.25
Given Wm = 1368:
T = {1368/0.000000056704}0.25 = 394.1ºK = 121.0ºC
As you can see, our application of the Stefan-Boltzmann Equation predicts that incident Solar radiation at 1368 Wm-2 should produce a maximum daytime temperature of 394.1ºK or 121.0ºC in a greenhouse fully protected from heat losses to conduction. Although Langley’s temperature is lower by eight degrees, it is near enough and, allowing for conductive heat loss, remains a testament to the insulating effectiveness of double glazing.
What is demonstrated in the above examples, is the fact that surface temperature and the temperature in a greenhouse can be explained without resorting to the extraneous entity called the “Greenhouse Effect”. This is significant in light of Ockham’s Razor, which states:
Entia non sunt multiplicanda praeter necessitatem.
This reads in English as:
Entities are not to be multiplied beyond necessity.
Although the terminology may seem unfamiliar in light of 20th century usage, if we look at the words for what they mean, we can nonetheless understand this statement. In short, William of Ockham is urging us not to hypothesise beyond what is necessary to explain the material evidence we possess. A hypothesis that does go beyond the support of material evidence violates this principle in that the evidence is already explained by a simpler theory.

Richard111
March 29, 2011 4:45 am

I agree with John Kehr says:
March 29, 2011 at 12:11 am
What he is describing is the Maxwell-Boltzmann kinnetic energy effect in the atmosphere.

Richard111
March 29, 2011 4:53 am

Hmm… from my second reading I note that no mention is made of energy absorbed being passed directly to nearby air molecules (collisions) and warming those before the absorbing molecule can fire off a photon and cool itself that way.

John Brookes
March 29, 2011 5:00 am

[taunting is not an attractive color on you. go ahead and reply, some other moderator will not know why I deleted your little attempt at a humorous humorless polemic. ~ ctm]

commieBob
March 29, 2011 5:04 am

A quibble:
Ira says:

The energized air molecules emit radiation at various wavelengths

I say:

The energized air molecules emit radiation at various longer wavelengths

The vast majority of downward radiation coming from the atmosphere is longer wavelength than 13 um. The area under the curve around 10 um looks like about 5 percent of the area under the curve greater than 13 um. In other words, the vast majority of the energy coming back at us is explainable as blackbody radiation. Note that the graphs I am attempting to display here are for arctic data.
If I interpret the graphs correctly, even if CO2 absorbed all the energy around 10 um, and that energy was immediately passed to other molecules to re-radiate around 15 um, the energy absorbed by the CO2 wouldn’t account for much of the back radiation.

Spartacusisfree
March 29, 2011 5:16 am

This ‘GHG physics’ is bunkum. Let’s take the 15 micron band. 95% is absorbed in 1 m air. That’s why mirages exist – the air above a hot surface is warmed by IR absorption.
OK, the air does re-radiate downwards but a lot of the hot air is convected upwards thus ensuring most of that original IR radiation from the ground never makes it back.
What GHGs do is to increase the adiabatic lapse rate a bit, also raise the tropopause. And because more latent heat means more efficient precipitation, the upper atmosphere dries and it’s easier for the heat to radiate to space.
So, GHG warming is controlled to a near constant level independent of [CO2]. it’s about time that ‘climate scientists’ and politicians realised it.

Thomas
March 29, 2011 5:39 am

bananabender. That calculation starts off with the simple mistake of forgetting to include albedo, that Earth reflect some of the sunlight, and thus gets a too high non-greenhouse temperature. (This temperature is measurable as the blackbody temperature of Earth as seen from space and is -19 C)
Then the author come up with an altitude of 1600 meters that he for some reason thinks is important. Multiplying this arbitrary altitude with the temperature gradient and adding the (faulty) base temperature does give the right answer, but only because he picks that altitude to get the right answer.
Now take a look at what the real atmosphere looks like:
http://apollo.lsc.vsc.edu/classes/met130/notes/chapter1/vert_temp_all.html
Still think you can explain it?

AdderW
March 29, 2011 5:40 am

…rubbish !

Bob Shapiro
March 29, 2011 5:46 am

John Kehr says:
March 29, 2011 at 12:11 am
“I like that animation. ”
I find Dr. Glickstein’s animations (this isn’t his first) to be annoyingly distracting. Since it interfered with my focusing on the text, I decided to skip the essay.
Did I miss an interesting and important post? Maybe. But it just wasn’t worth my effort.

Joe Lalonde
March 29, 2011 5:48 am

Ira,
There are a few reasons that I do not trust “averaged out global mathematical calculations”.
One time frames. Our current calculations fail due to the calculation do not include adjusting for planetary motion, slowdown or unknown surprising factors(such as solar flares, massive eruptions, salt changes, ocean current changes, impacts, etc.).
Next, planetary positioning. Since the suns diameter at it’s equator is the greatest mass and the poles are much smaller to the drifting of the planet between the two.
Thirdly is planetary shape of having a huge diameter equator and smaller diameter poles on a rotating planet.
So, where on this planet or atmosphere does make a difference to any other place taking a measurement.

Thomas
March 29, 2011 5:54 am

Spartacusisfree, the amount of energy the Earth has to radiate to remain in thermal equilibrium is constant. Since we have GHG:s in the atmosphere only a small part of the radiation emitted into space comes from the surface, the rest comes from the atmosphere, and we can assign an “effective altitude” from which the average radiation is emitted. If you add more GHG:s radiation will have a harder time escaping and thus this effective altitude will increase, but since the amount of energy has to be constant its temperature has to remain the same. Now take the lapse rate into account. If the layer of constant temperature rises, the temperature at the surface has to increase as the lapse rate multiplied by the increase in altitude. If more GHG:s raises the tropause by 100 meters the temperature at the surface rises 0.7 degrees as a first approximation.

Harold Pierce Jr
March 29, 2011 5:59 am

Ira
Carbon dioxide is a linear molecule. You show it with a bent structure like H2O
Fix that mistake. It makes you look stupid.
When a CO2 or H2O molecules absorbs an IR photon, its linear velocity does not increase. The absorbed photon causes an increase in the vibrational frequency of a bond or combination of bonds (e.g, bending). Fix that mistake.
A vibrationally- excited CO2 or H2O molecule in the troposphere will not re-emit the absorbed photon but will undergo immediate collision deactivation with N2, O2, Ar or H2O. This causes an increase in the speed of these, i.e., they become slightly warmer.
The collision frequency at 1 atm and room temp is about 100 billion collision per sec. Incidently, this is why nat gas in air explodes with great violence, i.e., the reaction goes at the collision frequency.
FYI: CO2 is a weak absorber of IR because it does not have a permanent electric dipole like H2O.

March 29, 2011 6:02 am

All of this is totally unnecessary. All we need is one valid graph of long-term temperature vs CO2 concentration, showing the lack of correlation. And we already have plenty of valid graphs like that.

Jose Suro
March 29, 2011 6:10 am

The greenhouse effect hypothesis, and specifically the warming caused by increasing CO2 in the atmosphere goes back to 1896, when Svante Arrhenius wrote an article and then a book that postulated the theory. All kinds of people ran with it afterwards although the theory was later found to be based on grossly erroneous spectra. These gross spectroscopy errors can be seen as conclusively proven wrong in MODTRAN(R) simulations. For a list of references to Arrhenius, including some by people that still believe it to be true look here:
http://www.lycos.com/info/svante-arrhenius.html
For the MODTRAN(R) simulations disproving the theory look here:
http://members.casema.nl/errenwijlens/co2/arrhrev.htm
Although I disagree with his conclusions (he is a warmist), Spencer Weart wrote a pretty thorough history of how this whole AGW thing evolved over the last century, ending with Charles (Dave) Keeling, before going into the modern research and model predictions that are so questionable.
Regardless of his conclusions, I give Weart credit for being quite thorough and his book is worth a read because it encapsulates all the early prominent research that went into building AGW theory as we know it in it’s present form. This book is also a great place to look at all this greenhouse research in one place and to see where the all the holes and leaps of faith are. A pretty complete condensation of the book can be found here:
http://www.aip.org/history/climate/co2.htm
Sadly, and contrary to the whole point of the book, which should have been citations and discussions of the history of AGW research, Weart himself takes the ultimate leap of faith and joins the warmist camp, tainting a great read with his own conclusions. You can see that in his personal notes:
http://www.aip.org/history/climate/SWnote.htm
Best,
Jose

MikeEE
March 29, 2011 6:11 am

rusureuwant2know
“I thought Al Gore got his graph backwards and the increase of CO2 followed warming – why are we still working with the assumption it creates it???”
Greenhouse gasses definitely do warm the planet, there is no question of that. And, increasing the amount of CO2 in the atmosphere will increase the amount of heat retained here on Earth. The real question is, how much will the change be as a result of a doubling of the CO2? First order effects seem to point to a change of 0.8 to 1.5 degrees. All of the other warming the alarmists are trumpeting is from second order effects that are definitely not known. Will the H2O in the atmosphere increase significantly, thereby significantly warming the planet? or perhaps there is negative feedback and the H2O is reduced and the planet cools. I don’t know.
As to your other point about the CO2 following temperature, there is definitely evidence of that too. Warmer oceans can’t hold as much CO2 so more of it ends up in the atmosphere. It’s a complicated system.
Stephen Wilde
Although there clearly is a greenhouse effect it is infinitesimal compared to the energy retaining effect of the oceans.
The greenhouse effect and the energy retaining effect of the oceans are different things. The atmosphere and the ocean may retain heat, but the atmospheric heat retention is infinitesimal compared to the heat retention of the oceans. The greenhouse effect changes the amount of heat that flows out of the system.
Perhaps you could think of the heat retention of the ocean as a bucket of water. Then, you have a stream of water flowing into the bucket and a hole in the bottom of the bucket that lets an equal amount of water flow out. The system is in equilibrium because the same amount of water is going out as is coming in, the water level remains constant. When you change the greenhouse gases you change the amount of water flowing out so the water level will raise of lower as a result.
MikeEE

Alan McIntire
March 29, 2011 6:12 am

To be thorough, there’s also an “anti-greenhouse” effect. See
http://en.wikipedia.org/wiki/Anti-greenhouse_effect
The atmosphere of Titan is transluctent to some wavelengths, and translucent to part of the surface radiation.
This results in a surface cooler than a surface with NO atmosphere
sun —–> 4 watts 2 watts 2 watts Titan surface 2 watts4 2 watts from atmosphere<——-
2 watts from surface 2 watts from atmosphere
4 watts 2 watts Earth
The earth heats up, reradiating the 2 watts to the atmosphere, which will also heat up. The final
balanc will be
Sun —> 4 watts to atmosphere 4 watts to earth from atmosphere
–>4 watts from sun, 4 watts from earth4 watts from atmosphere to earth
<—4 watts to earth from atmosphere
So the net effect of the surface warming from an infrared absorbing atmosphere is the same as the effect of NO atmosphere.
See Trenbeth's figures here:
http://stephenschneider.stanford.edu/Cl … lance.html
Note that the zero greenhouse effect would apply for those 67 watts absorbed directly by the atmosphere.
Caveats: Working out the temperature drop in a purely radiation
cooled atmosphere, it can be shown that the temperature drop would be
greater than the
moist pseudoadiabetic lapse rate, making a purely radiation
controlled atmosphere unstable. The adiabetic lapse
rate places an upper limit on drop in temperature with height.

March 29, 2011 6:18 am

Nice animation and explanation. But I think it is giving ony a passing thought to a very important component countering any heating effect of more CO2 — convection. As noted, a warmer molecule will be moving faster and hence have a faster Mean Free Velocity and it will rise. That warmer air will cool as it rises and condenses water pulled up with that rising mass, losing even more energy. Hence transporting ground air heat to the cooler higher atmosphere.
Also missing from this, is that this affect in the animation is only when the sun it at noon in the location where the sun is directly over head. Daytime everywhere else gets lower energy from the sun because of the angle. Nighttime, more of that stored daytime heat is lost into space. Also, the warmed portions of the planet intermix that warm air to the colder regions of the planet (causing storms which disapates even more heat).
In other words, the atmospheric system has a buffering affect and is why the planet have never “cooked” when CO2 was 20 times today.
The planet is not suseptable to over heating, it’s suseptable to significant cooling, which is bad.

Spartacusisfree
March 29, 2011 6:23 am

Reply to Thomas: I agree that as the adiabatic lapse rate and the tropopause rise, surface temperature will increase. However, the reduction of [H2O] in the upper atmosphere will allow easier radiative heat transport to space. The interaction of this radiative heat transport with the ALR is actually quite interesting.
The evidence of lower [H2O] is fairly conclusive. So, I expect GHG warming from CO2 to be very low.

MikeEE
March 29, 2011 6:23 am

Igl
What a lot of nonsense. Here is how it really works…
Don’t believe everything you read. He says it’s not reradiation because “When the atmosphere emits radiation, it is not the same radiation”.
What is ‘the same radiation’? that’s just nonsense. Any material (CO2 for example) that is not at absolute zero emits radiation to stay at equilibrium. If you raise the amount of energy that you put into that material you will raise it’s temperature and the amount of energy that it radiates. Whether it is the same of different energy is irrelevant.
MikeEE

Richard M
March 29, 2011 6:25 am

Although this article mentions CO2 absorbing energy through collisions and radiating that energy (half to space) it does not mention that this is a cooling effect. And, when you add more CO2 you increase this cooling effect. So, what is the trade-off here. How much warming vs. how much cooling. I’ve never seen this tackled anywhere.

MrCannuckistan
March 29, 2011 6:26 am

Looking at Ira’s animation and Wayne’s proportional image, is it not clearly evident that CO2 does not have the volume required to have any dramatic impact on the Greenhouse effect?
For example, the total Greenhouse effect represents approximately 33 degrees of warming for our planet. Simplifying the atmosphere as Wayne has done it would seem that CO2 can only add ~0.44 degrees (400ppmv CO2 / 30,000ppmv H2O x 33 degrees). This discounts other GHGs but I believe they are all in far less concentrations.
MrC

mindert eiting
March 29, 2011 6:36 am

Stephen Richards: thanks for the link. I did not have enough time for reading the complete article by Postma. Perhaps tonight. My first impression is like yours: it’s over with the greenhouse theory and everything based on it.

commieBob
March 29, 2011 6:40 am

Here’s a link to the graph I tried to display in my previous post: http://www.skepticalscience.com/images/infrared_spectrum.jpg
I tried: <img src = “http://www.skepticalscience.com/images/infrared_spectrum.jpg”> but that obviously didn’t work.
[Reply: WordPress doesn’t really like the img tag. Either cut ‘n’ paste the URL like you did here, or use the “href=” command. ~dbs, mod.]

bananabender
March 29, 2011 6:40 am

When you adjust for atmospheric pressure the absolute atmospheric temperatures of Earth, Mars and Venus only vary by 10-20%. This is despite Venus receiving about 8x as much solar radiation as Mars and 1.9x as much as Earth and the totally different atmospheric composition on all three planets.
Measured Temperature vs Atmospheric Pressure:
0.001Bar:
Mars ~210K, Earth ~220K, Venus ~190K.
1Bar:
Mars (n/a), Earth 288K, Venus 350K.
When you adjust for the different gas composition (relative molar mass) of each planet’s atmosphere the temperature differences are even smaller.
This fact is explained far better by the Ideal Gas Law than any mythical Greenhouse Effect.

March 29, 2011 6:41 am

It doesn´t matter how much Greens may insist on the so called “green house effect”, the only “effect” that exists is that of “confined heat”…and, still, it can´t heat sea water up, as heat transfer goes from the denser to the less dense:
http://es.scribd.com/doc/28018819/Greenhouse-Niels-Bohr

Richard E Smith
March 29, 2011 6:51 am

Ira Glickstein says:
“7 The energized air molecules emit radiation at various wavelengths and in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth …”
As Ira states, heated air molecules radiate. So why are we so concerned about the ‘backradiation’ from CO2? As most of the heating of the atmosphere is by conduction and convection and all gases radiate in close proportion to their temperature, then all gases are greenhouse gases. If it were otherwise, N and O could not radiate away to the vacuum of space. Consequently even if it is accepted that there is backradiation from CO2 ‘further heating the Earth’ (impossible as this means the surface is heating up itself as it is the surface that has warmed the atmosphere in the first place) this effect would be dwarfed by the radiation from the remaining 99% of the atmosphere.
Some commenters on this thread who deride those of us who keep on propounding the second law claim that we misunderstand the position. They say that the greenhouse gases ‘reduce cooling’ (rather than ‘further heating the Earth’). But reducing cooling cannot add heat. In the constant irradiance models that are used to demonstrate the greenhouse effect (e.g. Kiehl & Trenberth’s) the Earth is not cooling down. The most that can be said about the ‘reduced cooling’ theory of greenhouse gases (properly called an atmosphere effect) is that it will reduce heating during the day and cooling at night. Therefore the average temperature may be higher than it would be otherwise (e.g. on the Moon) as the possible minimum is 3K whereas the maximum is determined by the quantity of solar radiance. Those who think this is the greenhouse effect have not understood what real warmists (like K&T and Gavin Schmidt) really believe. Conventional greenhouse theory says that the downward radiation from greenhouse gases raises the Earth’s temperature higher than solar radiation can.

Dave Springer
March 29, 2011 6:52 am

4. The energized molecules re-emit the Photons in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.
This isn’t true for cold (below ionization energy threshhold) dense gases. They do not exhibit spectral emission lines they only have spectral absorption lines. The reason is that in a dense gas the molecules are packed so tightly that any increase in energy in an absorption line is immediately distributed to surrounding molecules through collisions. The emissions from the GHGs and all the other gases in the lower atmosphere are approximately continuous blackbody spectrum with a peak emisson frequency corresponding to the temperature of the air.
7. The energized air molecules emit radiation at various wavelengths and in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.
This is correct except for the part about “heating the earth”. That phrase justifiably causes people with a basic understanding of the laws of thermodynamics to balk. In the big picture this simple phrase (courtesy tallbloke) describes what’s happening: the sun warms the ocean and the ocean warms the air. So-called back radiation, which is what you’re describing, does not warm the surface except in limited and exceptional circumstances where a mass of air transported convectively (read winds) is warmer than the surface over which it blows and even then it is limited because the air has so little heat capacity compared to the surface – imagine trying to heat a cup of coffee by stirring it with a hot feather.
What actually happens is the back radiation slows down the rate of surface cooling. This is because radiative transfer between two objects of different temperature is a two-way street. Say the warmer object is emitting 10 watts per square meter. The cooler object, unless it is at absolute zero, is also emitting energy. Say it’s emitting 5 watts per square meter. So the warmer object is emitting 10w/m and absorbing 5w/m while the cooler object is absorbing 10w/m and emitting 5w/m. The net transfer of energy is 5w/m from warmer to cooler.
Absent greenhouse gases essentially all the radiation from the surface charges through the atmosphere at nearly the speed of light and is absorbed by the virtually infinite heat sink of the cosmic void. The cosmic void is 3 Kelvins – just a few degrees above absolute zero so the back-radiation from the cosmos is just about nil. Put some greenhouse gases in between the surface and the cosmic void and then you have some significant back radiation which slows down the surface cooling rate.
At the end of the day because the ocean doesn’t cool as quickly at night with GHGs in play it’s a little warmer in the morning than it would be absent the greenhouse gases. The sun does its thing during the day warming the ocean and because the ocean was a little warmer at sunrise it’s a little warmer at the end of the day too. The cosmic void doesn’t change temperature but now that the surface is a bit warmer than it would be absent GHGs it cools more quickly at night – the higher the difference in temperature between two objects the faster warmer object loses energy to the cooler object. Thus a new equilbrium temperature is obtained where the energy received by sun during the day is emitted at night in perfect balance.
In the real world equilibrium is moving target due to other factors like clouds (which reduce the amount of solar energy reaching the surface), convection (which mechanically transports energy from one place to another both horizontally and vertically), evaporation, and conduction. But the one thermodynamic fact is always operative in the real world – the farther out of equilibrium the system is the harder it tries to move back to the theoretical equlibrium temperate. So when everything is said & done and absent any feedbacks – GHGs raise the theoretical surface equilbrium temperature and for non-condensing (read everything except water vapor) GHGs which are well mixed and slow to change in concentration the change in theoretical surface equilibrium temperature is calculable and works out to about a 1C rise for every doubling of CO2.
A 1C rise in surface equilibrium temperature per CO2 doubling is a very good thing especially when one considers that the bulk of the warming actually occurs at night, at higher latitudes, and in the winter. This has the beneficial effect of extending growing seasons when and where longer growing seasons are most needed. Add to that that higher CO2 concentration means plants grow faster/larger (provided sunlight, nutrients, and water are not limiting factors) and they use less water per unit of growth in higher CO2 as well.
The whole brouhaha over scary global warming rests on the wholly fabricated idea that CO2 greenhouse warming puts more water vapor in the atmosphere and that, because water vapor is a powerful greenhouse gas, it constitutes a positive feedback that turns 1C per doubling into 3C per doubling. There isn’t a shred of empirical evidence to support that belief and IMO overwhelming evidence that the feedback is actually negative – the water cycle speeds up which serves to mechanically transport heat from the surface by evaporation and convection to the cloud deck where it then has an easier radiative path out to space. It also results in slightly greater cloud cover which reflects more sunlight back out into outer space before it can reach the surface to warm it. This prevents any chance of a runaway greenhouse and this is borne out by there never having been a runaway greenhouse in the earth’s history even though CO2 concentration in the past was as much as 20x greater than present. In fact the l paucity of CO2 today, relative to most of the earth’s history, is IMO undoubtedly a major factor in why the earth has been in an ice age for the past 3 million years – ice ages are rare but there is a tipping point where factors can combine that allow ice and snow to dominate the surface through positive feedback mechanisms – ice reflects a lot more sunlight than ocean or unfrozen land surface thus reducing the amount of solar surface heating (which makes conditions even better for glacier expansion) and because cold air holds less water vapor than warm air it also greatly reduces what’s normally our primary greenhouse gas in a viscious cycle of cold begetting even more cold. There is substantial but controversial evidence that glaciations in the remote past have frozen everything from pole to equator – appropriately called “snowball earth”. So global cooling, especially now with the Holocene interglacial overdue for its ending, should be our greatest concern and we should be welcoming whatever global warming we can get with wide open arms.

bananabender
March 29, 2011 6:55 am

re: Thomas says:
March 29, 2011 at 5:39 am
bananabender. That calculation starts off with the simple mistake of forgetting to include albedo, that Earth reflect some of the sunlight, and thus gets a too high non-greenhouse temperature. (This temperature is measurable as the blackbody temperature of Earth as seen from space and is -19 C)
If you totally ignoreevery factor including albedo, atmospheric composition and incident solar radiation there is still a <10% temperature variation between the atmospheres of Mars and Venus at the same pressure. This is despite Venus receiving 8x as much solar energy as Mars.
I prefer to accept an irrefutable Ideal Gas Law than the idea of "magical greenhouse gases" which defy virtually every basic principle of chemistry.
The calculated value of -19C for Earth without an atmosphere is based on absurd reasoning.

Jeff Carlson
March 29, 2011 7:21 am

if I am reading it correctly (probably not) #7 violates the 2nd Law … the cooler air cannot heat up a warmer body …
if you mean it slows down the rate of heat loss of the earth then I am with you but it cannot “heat” the earth …

Jan Oortwyn
March 29, 2011 7:29 am

My impression is that there is confusion between thermal transfer between molecules (rotation, vibration etc) and radiation emission/absorption. The first depends on the thermodynamics, the second on quantum physics.
As far as I remember, the absorption of a photon concerns only the electrons which can gain a higher (not stable) energy level. After some time the excited electron will re-emit the received excitation energy as a photon and fall back into the stable configuration. The radiation energy balance would be zero. But may be I have forgotten some essential laws of physics.

Steve Keohane
March 29, 2011 7:31 am

Anders says:
March 29, 2011 at 2:25 am
If all that is going on is absorption and re-emitting radiation there would be no heating of the atmosphere. Heating happens because of the re-distribution of energy through collisions between molecules. The “simple” visualization is thus too simple. The temperature of a gas IS the average kinetic energy of the gas molecules, this is something Roger Pielke Sr. has pointed out a number of times and for many years.
Your comments are as close to what makes me uncomfortable with this representation of the absorbtion and re-radiation of a photon. The kinetic energy seems like it could be represented as the mean free path and temperature. What I have a problem with is the absorbtion and re-radiation at the same wavelength. You can’t have that and warm the molecule as well. If warming occurs, then the re-radiation must be at a different wavelength, lower energy. Otherwise, energy is created from nothing.

March 29, 2011 7:42 am

This article is incompetent, in the face of definitive experimental data that simply invalidates its premise:
Venus: No Greenhouse Effect
The simple analysis I did, within hours of finding the data on the internet, should have been done nearly 20 years ago. Every scientist who promulgates the greenhouse effect is incompetent. You will find the simplest, clearest discussions about the greenhouse effect on my website. My presentation of the definitive evidence, the first and only such presentation to date as far as I know, should be properly confronted and accepted by the entire science community; I have submitted it to “Physics Today” to get it before that community, but have received no response, which shows a complete lack of integrity and professional commitment to scientific self-correction — a fundamental failing of scientific institutions today.
There is no longer any validity to arguments for the greenhouse effect. The science is settled, and what is needed is re-education of all those like Ira Glickstein who are promulgating false science.

Marc77
March 29, 2011 8:01 am

If the “greenhouse effect” had increased, a hot spot would be found over the tropics. This hot spot has not been measured.
Now, let’s see where warming has been measured: mostly where you can find snow or ice. Maybe it could be good to evaluate the “greenhouse effect” inside of snow. First, it is wrong to think that snow reflects light like a giant mirror. Most of the light does enter inside of the snow where it is reflected in all directions by pieces of ice, and after a while, most of the light is reflected out of the snow. So it is similar to a “greenhouse effect”.
Now, the impurities inside of the snow and the spectrum of light will probably have an effect on how much light is absorbed. Up to now, I have not seen any study about this.

dp
March 29, 2011 8:06 am

I think a more accurate animation will look like tsunami open sea wave models like this (where the islands represent GHG molecules):
http://www.amath.washington.edu/~dgeorge/talks/movies/TsunamiMovies/IndianOceanAmrMV.gif
It certainly is not like a bunch of projectiles flying around.

March 29, 2011 8:07 am

Thermodynamics gives us an idea of the direction of energy transfer. The transfer mechanism controls the rate. The rates of convection, evaporation/condensation, freezing/thawing are much slower than radiative energy transfer that can be considered “line of site, speed of light”. Model the energy input/output of a column of air in winter, over Arctic sea ice considering all these processes. Test your model by analyzing the reanalysis data. Take a guess as to the relative contributions of each process to slowing down the loss of energy to space. In your model design, don’t forget to include the insulating qualities of sea ice (thermal conductivity and thickness change). Energy is being transferred from water below the ice to the upper surface, as it freezes.

Thomas
March 29, 2011 8:21 am

Spartacusisfree, if you have any evidence for your claims I suggest you publish them. The general view is that H2O increases as temperature go up, and that seems much more reasonable to me. If you look at different areas of the Earth with different temperature the trend is definitely that warmer areas have more water vapor, so why should the trend be in the opposite direction if the warming comes from AGW? (But at least you don’t violate any obvious law of nature unlike most of the geniuses here)

BigWaveDave
March 29, 2011 8:25 am

Thomas says:
“bananabender. Go ahead, try to explain the temperature of the Earth using only the ideal gas law! It can’t be done, you see, if you actually start to think about what you do.”
Ideal gas law PV=nRT can be rewritten as P/R=rho T . Ascending in altitude through the troposphere, the density decreases at a lesser rate than pressure

stephen richards
March 29, 2011 8:30 am

steveta_uk says:
March 29, 2011 at 3:15 am
You need to explain in the same terms and manner as to why back radiation doesn’t violate black body theory not wave your arms around. I have read the entire doc twice and compared it to the BB theory and can find no problem with what he says. So, explain yourself using the known laws of physics and perhaps we can come to consensus 🙂
Stephen Richards BSc Physic MSc Solid state Physics.

Thomas
March 29, 2011 8:34 am

bananbender, again I ask you to look at that figure of the real atmosphere. Where is the 0.001 bar altitude on Earth? It’s around the stratopause, and according to the standard atmosphere the temperature at that altitude is 270 K, not 220 K as you claim. You just make up your numbers to make them fit! I think I prefer the regular kind of science.

Tenuc
March 29, 2011 8:54 am

steveta_uk says:
March 29, 2011 at 2:56 am
“…Surely they can only re-emit much lower energy photons, of similar energy levels to the black-body curve at -20C.
Yes, however the temperature of a gas is actually defined as the average kinetic energy of the gas molecules, for example at -20C and not the amount of radiant energy it is emitting. In fact the laws of black body radiation only have a good fit to solid blackbodies which are in thermal equilibrium – something that seldom happens in our chaotic weather systems. The tiny effect the odd molecule of CO2 has is swamped by other processes which drive our turbulent, non-linear system.

Arfur Bryant
March 29, 2011 9:18 am

I think Cassandra King nailed it. To me, this is a problem of scale which, of course, is not represented in the original graphic.
I would use the following analogy to visualise the scale:
.
Let us take a large sports stadium such as the Melbourne Cricket Ground. Capacity 100,000 seats. Although the concentrations of ghgs are by ppm by volume, let us approximate to 10.000 molecules per percentage volume (1 million divided by 100). This means that of the 100,000 spectators in the MCG, only 40 (wearing red for GHGs) are able to both hear the tannoy (LW radiation) and re-transmit what it says. Another 250 spectators (wearing blue for water vapour) can hear the tannoy but cannot re-transmit the message (because they are mute). The rest (wearing white) are deaf-mute. Let us now pretend the tannoy message was a very funny joke. The only way that the 99,710 spectators in white can even realise something funny has been said is by ‘conduction’ withthe 290 people, who are shaking with laughter (vibration). If they are touching the laughing people, they may well laugh in sympathy – but maybe with less conviction. As most of the ‘water’ spectators are in the lower stand, that is where most of the shaking is done. Back in 1850, the number of emitting spectators was only 28 (probably 29 if you include all the other dry ghgs). Therefore, according to the radiative forcing theory, the addition of 12 spectators (of any type of ghg) has increased the degree of overall shaking by 2.5% (0.8 C). According to the IPCC, doubling the number of emitters to 56 will lead to a ‘best estimate’ increase of 10% (3 C).
Does anyone think the original 28 red people can have a significant effect, and will the addition of another 12, or even doubling the original figure, be likely to incur catastrophic warming (laughter)?
.
A lighthearted analogy, I admit…

RJ
March 29, 2011 9:33 am

“The greenhouse effect heats the earth because greenhouse gases absorb outgoing radiative energy and re-emit some of it back towards earth”.
To be true doesn’t this mean that CO2 must act like a mini sun with an energy multiplying effect.
As an example if the earth contains 1 trillion units of energy. If 100,000 leave the earth’s surface as radiation. Somehow even though half or move does not return to the surface the energy that does return must be greater than 100,000 units. Otherwise there can not be a heating of the earth.
But even this example is not correct as the radiation returning has a cooler energy impact than the earth and a cooler body can not heat a warmer one.

mkelly
March 29, 2011 9:37 am

Ira, you do keep pluggin’ away.
1. If you are going to have H2O in the mix then you also need to recognize that it has a much higher specific heat than CO2. water vapor about 1.9 CO2 about .84. Water vapor can hold on to the heat. CO2 cannot.
2. You should show the area under the curve. (this was mentioned before) It is about 8% for the 15 micro. People are only a small protion (3%) of the 8%. So our effect if any is small.
3. You should define “surface”. If a proton of IR strikes me it disappears. If it hits a leaf or a blade of grass it is gone. So there is a huge portion of the downward IR that never gets to the surface. Again, the oceans cover 2/3 of planet. So your claim is that 1/3 of the surface area of the planet heats the other 2/3’s of the atmospher over water due to backradiation. I have my doubts.
4. Heat transfer is totally and only dependent on temperature difference. q/A=(emissivity * sigma*(T^4-T^4)) If both e and sigma are the same no heat transfer but that does not cause either to stop radiating at its temperature. If I take a blow torch at 3000 C and add a second torch so the flame is exactly opposite I don’t get 0 at any point. If I combine the flames tips in the same direction I don’t get 6000 C. 3000 C is it.
5. What is the emissivity of CO2 at partial pressure and the temperatures you list? Without that knowledge you cannot figure heat transfer and I bet it is very tiny. My heat transfer book does not go that low in temperature so I have to extrapolate to less than .1.
6. PV=nRT for the air temperature at or near the surface is valid since the critical temperture of N2 is so far away. (Paul please note I said at or near. ;)) If this was not valid why do we have STP, standard temperture and pressure. Using standard pressure and 273 K the equation is in balance.
7. Because of the equation in my #4 and the atmosphere under almost all conditions being at a lower temperature than the ground no heating of the surface by air can take place. Not to mention the mass diparity of both and the heat capacity difference.
Please try again.

George E. Smith
March 29, 2011 9:42 am

Where Argon ?

George E. Smith
March 29, 2011 9:55 am

“”””” Jan Oortwyn says:
March 29, 2011 at 7:29 am
My impression is that there is confusion between thermal transfer between molecules (rotation, vibration etc) and radiation emission/absorption. The first depends on the thermodynamics, the second on quantum physics.
As far as I remember, the absorption of a photon concerns only the electrons which can gain a higher (not stable) energy level. After some time the excited electron will re-emit the received excitation energy as a photon and fall back into the stable configuration. The radiation energy balance would be zero. But may be I have forgotten some essential laws of physics. “””””
In the lower regions of the atmosphere where weather happens, the mean time between intermolecular collisions, is much shorter than the mean lifetime of the molecular excitted states; so before an excited molecule gets to spontaneously decay to some lower state, it collides wiht another molecule or atom (of Argon) and energy is redistributed between the collision participants. That is what temperature is all about, so that original photon energy, becomes a melange of thermal energy of kinetic motion of the gas molecules. As a result of those collisions eventually some IR active species will receive enough kinetic energy from a collision, to put it into soe excited state from which it can radiate (before the next collision).
Only in the rarified stratospheric regions, is the lifetime of the excited states shorter than the mean collision time.

Cassandra King
March 29, 2011 9:59 am

Arfur Bryant says:
March 29, 2011 at 9:18 am
I think Cassandra King nailed it. To me, this is a problem of scale which, of course, is not represented in the original graphic.
I would use the following analogy to visualise the scale:
Now that is a perfect and descriptive analogy and explanation of the facts, I really hope Ira reads your post and comments on it because it goes right to the heart of my concerns and I am not a scientist nor do I hold any scientific qualifications so your post really appealed to me and was a joy to read and understand and visualise. I say visualise because that is the key to understanding. IMHO you have created a masterpiece that should have a wider airing in order for us mere mortals to gain a better understanding of the issues.

Spartacusisfree
March 29, 2011 10:10 am

Reply to Thomas: 8.21 Stratospheric drying: http://www.npr.org/templates/story/story.php?storyId=123075836
Solomon thinks it’s fewer thunderstorms: I think it’s more efficient precipitation in thunderstorms, part of the GHG control system that maintains near constant GHG effect in the atmosphere. Miskolczi has found from radiosonde data that the reduction of [H2O] in 61 years’ radiosonde data maintains constant IR optical depth/ 1.87 average IR photon absorptions in passing through the atmosphere [within 0,.1% of the theoretical value]: http://miskolczi.webs.com/
My own work concerns the aerosol optical physics in the climate models. All the two-stream approximations are wrong because they fail to take into account direct backscattering you get from applying Mie theory to the first few scatterings as the light wave enters a cloud. What this means is that what is claimed to be ‘cloud albedo effect’ cooling is either neutral or heating, quite substantial in the palaeo-climate.
Hence you don’t need CO2-AGW: it could well be net zero. The quid pro quo is that none of the models can predict climate.

cal
March 29, 2011 10:12 am

I am getting a bit depressed by the level of physics being discussed here and in some of the other web sites that we are being directed to. I am beginning to wonder if there is a strategy amongst the warmers to refute the greenhouse effect using bogus physics in order to quote these arguments, say how stupid they are, and thus justify the dismissal of all sceptic arguments as unscientific. Or I might just be paranoid!
There are serious problems with the quantification of the greenhouse effect. There is a serious doubt about feedbacks. There is a serious problem that the “fingerprint” of an enhanced greenhouse effect cannot be identified.
But there is no doubt that the atmosphere absorbs radiation emitted by the earth and eventually radiates it to space at a higher altitude which, being at a lower temperature, means a smaller energy loss. That reduction has to be compensated for by increased radiation losses elsewhere and that means a warmer surface.
This is the overall energy balance scenario and Ira has tried to explain it at a mechanistic level to show what is happening within the climate system. I have problems with some details, as I have already commented, but overall I think it is a good effort.
Some of the objections however are bizarre.
For example some say the sky cannot radiate to the earth because it is warmer yet you can point an infrared detector at the sky at night and measure it! Anyway when a CO2 molecule radiates a photon downwards how does it know the earth is colder and it should not do it!? Of course if the earth is colder the nett energy flow would be upwards (from warm to cold) and that is the only necessary requirement thermodynamically.
Then there are the arguments like CO2 is only a “trace gas” and H2O is a far more powerful absorber. If we a want to make the warmists accept the evidence of measurement we must accept the same rigour. One can measure the absorbtion by CO2 at say 1Km and do the same for H2O. By this altitude both have already trapped all the outgoing radiation at their characteristic absorption frequencies and re-radiated at least once. You can measure the radiation and see the big gaps in the black body spectrum associated with this absorption. You may think that CO2 (or H2O) at such low densities cannot do this but it does and there is already more than enough to do it totally. It is pointless coming up with simplistic arguments to deny these facts when we have data to confirm them.
The greenhouse effect is very badly named but it is a real effect. It would do the sceptic cause a great deal of good for everyone to accept this and then focus on the real issue – what is the effect of further increments of CO2? This question is by no means obvious. When the warmists tackle this question they tend to use the same simplistic arm waving physics that I have been reading above. We need to do a better job not a worse one.

Thomas
March 29, 2011 10:18 am

Spartacusisfree, you brought up water vapor with this statement: “And because more latent heat means more efficient precipitation, the upper atmosphere dries and it’s easier for the heat to radiate to space.”
Now you start to talk about the stratosphere which is clearly inconsistent with a statement about precipitation since there is far too little water vapor in the stratosphere to cause precipitation. Anyone brining up Miskolczi obviously have rather shallow understanding of the subject, so I think I pass on further discussion.

Dave Springer
March 29, 2011 10:20 am

Marc77 says:
March 29, 2011 at 8:01 am
If the “greenhouse effect” had increased, a hot spot would be found over the tropics. This hot spot has not been measured.
Now, let’s see where warming has been measured: mostly where you can find snow or ice. Maybe it could be good to evaluate the “greenhouse effect” inside of snow. First, it is wrong to think that snow reflects light like a giant mirror. Most of the light does enter inside of the snow where it is reflected in all directions by pieces of ice, and after a while, most of the light is reflected out of the snow. So it is similar to a “greenhouse effect”.
Now, the impurities inside of the snow and the spectrum of light will probably have an effect on how much light is absorbed. Up to now, I have not seen any study about this.

Recommend you start here with snow absorption:
http://www.google.com/search?sourceid=navclient&ie=UTF-8&rlz=1T4GGLL_enUS382US382&q=%22black+carbon%22+snow+albedo
The so-called tropical upper troposphere hotspot I understand is the mark of the mythical water vapor amplification that turns 1C per CO2 doubling into as much as 3C per doubling. Supposedly the extra CO2 increases water vapor content of the upper troposphere more than lower troposphere resulting in upper troposphere (which is always far colder than the surface) warming more than the surface. “Hotspot” is an inaccurate name since it’s really differential warming between upper and lower troposphere.
The increasing magnitude of the differential warming, if any, is difficult to measure and controversial. I believe notoriously poor GCM modeling of clouds is why the models are wrong. If the water vapor being delivered to the upper troposphere increases so does the amount of heat carried away from the surface by evaporation and convection and it condenses at an increased rate rather than giving the upper troposphere a higher absolute humidity. Thus evaporation, convection, and cloud formation is a negative feedback that negates the effect of more water vapor due to warmer surface temperature. There are several well developed hypotheses on this effect. The best IMO is
http://www.scribd.com/doc/25071132/The-Saturated-Greenhouse-Effect-Theory-of-Ferenc-Miskolczi
which is basically saying that the atmosphere, with regard to water vapor greenhouse warming, is saturated and we can’t get any more greenhouse warming out of water vapor – we can get additional surface warming only from increases in non-condensing greenhouse gases like CO2 and methane. I am in fundamental agreement with the conclusion simply by considering that the earth has never experienced a runaway greenhouse effect which water vapor amplification must certainly entail. So regardless of whether Miskolczi is right something is putting a ceiling on the maximum surface temperature and this appears to me to explain why. CO2 levels in the remote past have been several doublings greater than present and the global average temperature was several degrees higher fitting perfectly with there being no water vapor amplification.
My own pet theory is that CO2’s primary role is increasing the earth’s average temperature from below freezing to above freezing and this is accomplished by the first 100ppm in the atmosphere. CO2’s LWIR absorption curve is linear for low concentrations of the gas and becomes logarithmic in greater concentration (which has been known for over 150 years since John Tyndall experimentally measured it). It serves as “kindling” to get the water cycle going. Once the water cycle is going it takes over as the primary greenhouse gas and unlike CO2 is self-limiting because it’s a condensing gas. After that CO2’s role is limited to about 1C per doubling. There isn’t enough recoverable fossil fuel in the world to do more than two or three doublings which might possibly mean we could end the ice age and in a matter of thousands of years see the earth get green from pole to pole (which is its usual state – no polar ice caps at all). The downside of having a vastly increased amount of arable land is that ocean levels will return to their normal non-ice age level which is about 100 meters higher. That will take thousands of years to happen. No plants or animals will have any trouble migrating to higher ground. Humans have the bigger problem as we built a lot of immobile crap very close to the ocean under the mistaken impression that ocean level rises and falls only with the tides and is constant otherwise.

March 29, 2011 10:29 am

Harry Dale Huffman,
I agree with your analysis up to a point. In particular, the temperature ratio of the “effective radiating surface” of Venus should be 1.176 times the temperature of the “effective radiating surface” of earth (ignoring albedo effects). (I have also heard this called the earth’s “photosphere” since that is the regions from which photons are emitted from earth.)
That “effective radiating surface” is a combination of the surface (warm), the clouds (cooler), and the GHGs (cooler yet since they radiate from close to the top of the troposphere). On Venus, almost no radiation comes from the surface, since it is blocked by continuous thick clouds and GHGs. On earth the surface DOES radiate a significant amount.
To a zero-order approximation, you could say all the “effective radiating surface” is around the 500 mBar location on either, so this layer should have similar (within a factor of 1.176) temperatures on either planet.
However, any more accurate estimate would need to include the actual locations of clouds and amount of cloud cover, as well as the location and emission spectrum of GHGs, since these all affect the “effective radiating surface”.
(Once you get below this effective radiating surface, then the lapse rate is the key factor in determining the temperature for lower layers, so Venus with a much deeper atmosphere will have a much higher surface temperature.)

RJ
March 29, 2011 10:31 am

“For example some say the sky cannot radiate to the earth because it is colder yet you can point an infrared detector at the sky at night and measure it! Anyway when a CO2 molecule radiates a photon downwards how does it know the earth is colder and it should not do it!? Of course if the earth is colder the nett energy flow would be upwards (from warm to cold) and that is the only necessary requirement thermodynamically”.
Measure what though. See below that might be relevant. From the slaying blog
And this has been discussed before but the issue is not whether radiation travels downwards. It is whether it has a warming impact if it returns to the surface.
From slayingtheskydragoncom/en/blog/102
Lord: “Back radiation can be simply demonstrated by pointing a simple infrared detector at the underside of a cloud. Try it.”
Chorus: “My IR detector only cost $60! Simple! Agreed! Agreed!”
Slayer: “Clouds do not absorb and re-radiate heat back to Earth. Clouds add THERMAL MASS which takes longer to heat and cool. Warmists ‘support’ this false hypothesis with IR thermometer readings, but the IR readings of a hot Barbie is the same from any distance; ENERGY is not. Your $60 REMOTE thermometer is not measuring the radiant energy you are receiving, it is measuring the resonance of the Barbie.”

Matt
March 29, 2011 10:40 am

Yeah, yeah… that seems to be a very British concern to me. The country where they still have single glasing all over the place, and anyway, doors and windows never seal. I can totally see how they could save 90% heating by simply arriving in the 3rd millenium to that end. After many years, and by sheer luck, I finally ended up in a London house with double glasing for a change. And you can tell the difference – when friends come around for a visit, everybody goes like: ” Hey wow, you got double glasing…”
It is also a national habit here that people simply won’t shut their car engines down in the street and let them run for ever, etc, etc, etc The truth is that the British are wastrels and being a bit more sensible alone would cut their energy consumption in half over night – no policy needed at all.

Matt Sullivan
March 29, 2011 10:55 am

@RJ: If the earth radiates 100,000 units of energy and then receives even 1,000 unit through back-radiation, then it’s still warmer than if there were no back-radiation at all. At the same time, the sun is radiating energy to the earth. The net effect is that the earth is warmer thanks to the back-radiation.

Spartacusisfree
March 29, 2011 11:02 am

I did not say precipitation occurs in the stratosphere, it’s much lower down. And it’s easy to prove the optical physics in the climate models, which predicts the cloud part of ‘global dimming’, is wrong by looking at clouds as droplets coarsen prior to raining: the base gets darker, exactly the opposite of what the models predict.
After NASA discovered there was no experimental proof of this cooling, they published this: http://geo.arc.nasa.gov/sgg/singh/winners4.html
Twomey’s physics, which he warned could not be extrapolated to thicker clouds, is based on Mie theory. There is no ‘surface reflection’ but it apparently fooled climate science, hence AR4 contained the imaginary cooling.
So the high feedback hypothesis was always the artefact of a scientific mistake. Until it’s corrected, no model can predict climate. Correct it and out pops another AGW, ‘cloud albedo effect’ heating that explains palaeo-climate [probably via phytoplankton bio-feedback] far better than CO2-GW, which if not net zero, is much lower than claimed.

Reed Coray
March 29, 2011 11:20 am

The argument “backradiation warms the surface of the Earth” is often used when discussing the greenhouse effect. Is it fair to infer from this argument that an object’s temperature will always be higher in the presence of backradiation than it would be in the absence of backradiation? If so, I disagree.
Consider two disjoint blackbody objects (a spherical active object, and an inert object) in the vacuum of cold space. Assume the internal thermal conduction properties of each object are such that the surface temperature is uniform. By blackbody I mean the surface of an object (a) absorbs all electromagnetic energy incident on it, and (b) radiates electromagnetic energy in accordance with Planck’s blackbody radiation law. By “active object” I mean an object with a constant internal source of thermal energy–e.g., radioactive decay. By “inert object” I mean an object that has no internal source of thermal energy.
The active sphere in isolation will experience no back radiation–i.e., all radiation leaving the sphere will escape to space. The active sphere in the presence of the inert object will experience back radiation–i.e., some of the radiation emitted from the active object will be absorbed by the inert object–causing the inert object’s temperature to rise–causing the inert object to radiate energy, some of which will be directed “back” toward the active object. Provided there is no object-to-object thermal conduction or convection, I believe the steady-state temperature (the temperature when the rate of energy leaving an object is equal to the rate of energy entering the object) of the active object will be higher in the presence of the inert object than it will be in isolation. However, if conduction and/or convection between objects is allowed, then depending on the rate of energy transfer via conduction / convection, in the presence of the inert object and hence in the presence of backradiation the steady-state temperature of the active object can be either higher or lower than in the absence of the inert object. If true, then the argument that “backradiation warms (i.e., increases the steady-state temperature of) the active object” is not true.
Ira explicitly stated that his arguments apply only in the absence of conduction and convection; and with that caveat, I agree. However, since conduction and convection are both present in the Earth/Earth atmosphere system, I am not convinced that “greenhouse gas backradiation” produces a net increase in the surface temperature of the Earth. Any attempt to discuss the “greenhouse effect” in the Earth’s atmosphere without including a discussion of conduction/convection is in some sense like trying to get your car to the top of a steep hill by pushing–you’ll waste a lot of time; and when you’re done pushing, your call will likely be farther down the hill than when you started.

mkelly
March 29, 2011 11:20 am

Ira this from Jennifer Marohasy blog. The calculation for the emissivity of CO2 is found in this thread.
Comment from: Nasif Nahle March 30th, 2011 at 1:01 am
” The algorithm was developed (derived) from experiments by Hottel, Leckner, Lapp, etc., etc., etc.”
Additionally, you can see the insignificant total emissivities of the carbon dioxide on the tables that you can find in any book on heat transfer by radiation. I’m not alone on this THEORY and LAWS. Mine is a humble calculation based on experimentation and observations made by many physicists.
…” The total emissivity of the atmospheric carbon dioxide is 0.0017, whether you like it or not. …”
:)Hottel found also that the total emissivity of the carbon dioxide in a saturated state was very low (Ɛcd = 0.23 at 1.524 atm-m and Tcd = 1,116 °C). [6]
As Hottel diminished the partial pressure of the carbon dioxide, its total emissivity also decreased in such form that, below a partial pressure of 0.006096 atm-m and a temperature of 33 °C, the total emissivity of the carbon dioxide was not quantifiable because it was almost zero. [6] [7] [8]
If has shown in ( http://jennifermarohasy.com/blog/2011/03/total-emissivity-of-the-earth-and-atmospheric-carbon-dioxide/#more-7739) that at STP the emissivity off CO2 is almost zero then no back radiation can take place.

RJ
March 29, 2011 11:28 am

Matt
All this might do is slow the rate of cooling. It will not warm the planet. And as radiation travels at the speed of light\(and the quantity of CO2 is small) the slower cooling if it occurs would be very minimal.
But there is still the problem of a cooler body heating a warmer one. If radiation leaves earth’s surface at say 10 degrees and collides with CO2. Won’t this lower the potential energy in the radiation. So when it returns (if it does) its vibrational energy will be lower than the surface of earth. So can not warm the planet unless the planet has in the mean time cooled by more than the radiations potential energy.

cal
March 29, 2011 11:41 am

RJ says:
in reply to March 29, 2011 at 10:31 am
“For example some say the sky cannot radiate to the earth because it is colder yet you can point an infrared detector at the sky at night and measure it! Anyway when a CO2 molecule radiates a photon downwards how does it know the earth is colder and it should not do it!? Of course if the earth is warmer the nett energy flow would be upwards (from warm to cold) and that is the only necessary requirement thermodynamically”.
Measure what though. See below that might be relevant. From the slaying blog
And this has been discussed before but the issue is not whether radiation travels downwards. It is whether it has a warming impact if it returns to the surface.
—————————–
If it returns to the surface and does not warm it must be totally reflected. If it is totally reflected then the surface cannot emit at that wavelength either (basic law of thermodynamics). Therefore it is not possible for radiation to be emitted from the surface, absorbed by the atmosphere and re-radiated downwards without increasing the temperature of the surface.

RJ
March 29, 2011 11:49 am

Ira
I always had a problem with the logic of the GWG theory. And thankfully all the queries I had were answered and explained in the brilliant slayers book.
I now debate with warmist in various ways. One is to dispute the GWG theory as per the slayers book. Another to dispute the extent of the warming assuming the GHG theory is correct. Another to point out the pointlessness of CO2 taxes and cap and trade etc.
Assuming the flawed GHG theory is almost a fact (as you are doing) without understand how weak this theory is, and its flaws, was a hindrance not a help to me.

Brian W
March 29, 2011 12:00 pm

Dave Springer (march 29@2:11pm)
“So-called back radiation, which is what you’re describing, does not warm the surface except in limited and exceptional circumstances where a mass of air transported convectively (read winds) is warmer than the surface over which it blows and even then it is limited because the air has so little heat capacity compared to the surface – imagine trying to heat a cup of coffee by stirring it with a hot feather.” So let me get this straight you think backradiation is caused by a warm wind blowing across the surface, and that this is a limited and exceptional circumstance? You’re kidding, right?
“What actually happens is the back radiation slows down the rate of surface cooling.”
Does it now, and where are the experiments that quantify this? Objects above absolute temperature emit electromagnetic radiation sure, but HOW MUCH? How much IR does a block of iron at freezing emit. Is it sensible. Backradiation is a myth, go measure it and tell me how many btu’s or degrees kelvin it’s good for. I’d really like to know as it’s 3 weeks into spring, it’s freezing, and we have been below normal temps by 8-10 degs. How’s this, you or anyone else can’t measure or quantify it. To see what you’re up against see Dr. Roys Spencer’s amusing but failed “cavity experiment”.
” But the one thermodynamic fact is always operative in the real world – the farther out of equilibrium the system is the harder it tries to move back to the theoretical equlibrium temperate.” Does it now. You already invalidated any equilibrium by calling it a moving target. Any evidence that it “tries harder”.
The bottom line is that there is no greenhouse effect. It is an atmosphere effect due soley to water vapor. Both Feedbacks and forcing are nothing but plausible fantasies.

old engineer
March 29, 2011 12:07 pm

Ira, I really enjoy your contributions and the discussions they create. With so may ideas, one really has to think and ponder what is the most realistic explanation.
After reading through the various contributions to this thread, it seems to me that the dualistic nature of electromagnetic radiation is not taken into account sufficiently. We talk of wavelength (the wave nature of electromagnetic radiation) and photons (the particle nature). Your diagrams are of molecules being hit by photons. So it is like a billiard ball situation.
Someone said it was more like a tsunami wave. While I agree with that somewhat, radiation is not a wave front. It is more like a constant field. The billiard ball type diagram makes it seem as though the chance of a CO2 molecule being hit is small. And for one ray of radiation it is. But, with a constant field of radiation, the CO2 molecule is always being hit by some ray of radiation.
Nevertheless, it is hard to understand how this could cause enough increase in the water vapor in the air for the water vapor to produce the increase in atmospheric temperature that the models show.

cal
March 29, 2011 12:12 pm

Reed Coray says:
March 29, 2011 at 11:20 am
The argument “backradiation warms the surface of the Earth” is often used when discussing the greenhouse effect. Is it fair to infer from this argument that an object’s temperature will always be higher in the presence of backradiation than it would be in the absence of backradiation? If so, I disagree.
Consider two disjoint blackbody objects (a spherical active object, and an inert object) in the vacuum of cold space. Assume the internal thermal conduction properties of each object are such that the surface temperature is uniform. By blackbody I mean the surface of an object (a) absorbs all electromagnetic energy incident on it, and (b) radiates electromagnetic energy in accordance with Planck’s blackbody radiation law. By “active object” I mean an object with a constant internal source of thermal energy–e.g., radioactive decay. By “inert object” I mean an object that has no internal source of thermal energy.
The active sphere in isolation will experience no back radiation–i.e., all radiation leaving the sphere will escape to space. The active sphere in the presence of the inert object will experience back radiation–i.e., some of the radiation emitted from the active object will be absorbed by the inert object–causing the inert object’s temperature to rise–causing the inert object to radiate energy, some of which will be directed “back” toward the active object. Provided there is no object-to-object thermal conduction or convection, I believe the steady-state temperature (the temperature when the rate of energy leaving an object is equal to the rate of energy entering the object) of the active object will be higher in the presence of the inert object than it will be in isolation. However, if conduction and/or convection between objects is allowed, then depending on the rate of energy transfer via conduction / convection, in the presence of the inert object and hence in the presence of backradiation the steady-state temperature of the active object can be either higher or lower than in the absence of the inert object. If true, then the argument that “backradiation warms (i.e., increases the steady-state temperature of) the active object” is not true.
If I follow the logic here you are suggesting that the loss of heat via convection and conduction to the inert body can be greater than the heat gained by back radiation. That means that the inert body is inside the atmosphere of the radiating body (or vice versa). In either case the inert body is in dynamic equilibrium with this atmosphere. In which case it would be the properties of the inert body relative to the atmosphere it replaces which would determine whether the surface warmed or not. If the inert body was thermally more conductive the surface would be cooler and if it were a better insulator the surface would be warmer. However I do not see how this is relevant to the situation we have where, in order to simulate the actual atmosphere, your “inert body” would have to have exactly the same composition and properties as the atmosphere itself.

Dave Springer
March 29, 2011 12:30 pm

mkelly says:
March 29, 2011 at 11:20 am
If has shown in ( http://jennifermarohasy.com/blog/2011/03/total-emissivity-of-the-earth-and-atmospheric-carbon-dioxide/#more-7739) that at STP the emissivity off CO2 is almost zero then no back radiation can take place.

You were doing fine until you said no back radiation takes place. CO2 has no line emission in a cold dense gas. It has an absorption line. After absorption it loses the energy kinetically via collisions primarily with oxygen and nitrogen. This thermalizes the gas mixture i.e. sensibly warmer. The air is ALWAYS emitting a continuous blackbody spectrum! Everything with a temperature above absolute zero emits a continuous blackbody spectrum with a peak frequency that rises with rising temperature. The higher the peak frequency the more energy in the emission. This is where the downwelling radiation comes from and its existence is without doubt.

Dave Springer
March 29, 2011 12:33 pm

Brian W says:
March 29, 2011 at 12:00 pm
Dave Springer (march 29@2:11pm)
“So-called back radiation, which is what you’re describing, does not warm the surface except in limited and exceptional circumstances where a mass of air transported convectively (read winds) is warmer than the surface over which it blows and even then it is limited because the air has so little heat capacity compared to the surface – imagine trying to heat a cup of coffee by stirring it with a hot feather.” So let me get this straight you think backradiation is caused by a warm wind blowing across the surface, and that this is a limited and exceptional circumstance? You’re kidding, right?
“What actually happens is the back radiation slows down the rate of surface cooling.”

Does it now, and where are the experiments that quantify this?

“Heat: A Mode of Motion” by John Tyndall, 1859. This is old stuff. If you don’t know this you don’t know basic experimental physics.

Brian W
March 29, 2011 1:06 pm

Dave Springer
” This is old stuff.” OH IS IT NOW! That’s funny NOWHERE in the 512 pages of Tyndall’s great experimental work does he EVER mention backradiation, nor does he mention the term greenhouse EVEN ONCE. You are blowing a lot of hot MOIST air and I’m calling you on it. Bring out some of that old stuff and let’s find out!

stephen richards
March 29, 2011 1:22 pm

Dave Springer says: Everything with a temperature above absolute zero emits a continuous blackbody spectrum with a peak frequency that rises with rising temperature. The higher the peak frequency the more energy in the
It’s not a peak frequency. It’s a median energy. Blackbody spectrum is a spectrum with a central energy and energys higher and lower thus forming the BB spectrum.

RJ
March 29, 2011 1:25 pm

“If it returns to the surface and does not warm it must be totally reflected. If it is totally reflected then the surface cannot emit at that wavelength either (basic law of thermodynamics). Therefore it is not possible for radiation to be emitted from the surface, absorbed by the atmosphere and re-radiated downwards without increasing the temperature of the surface”.
My understanding is that this is not correct. As a cooler body can not warm a warmer body. Otherwise if for example a human was in a container of CO2 they would heat up rather than just cool at a slower rate. Or a human in a vacuum with a perfect radiation reflector would eventually cook to death. This surely is not possible.
So my understanding is that radiation from a colder body can not heat a warmer body. As a teacher can pass and increase knowledge of a student. And a student to a younger student. But not increase knowledge (heat) by passing it back
.

stephen richards
March 29, 2011 1:27 pm

Gerge E Smith says: My impression is that there is confusion between thermal transfer between molecules (rotation, vibration etc) and radiation emission/absorption. The first depends on the thermodynamics, the second on quantum physics
I would lkie an euro for every time I’ve said this on this site and stil people continue.

stephen richards
March 29, 2011 1:30 pm

Surely they can only re-emit much lower energy photons, of similar energy levels to the black-body curve at -20C.
This is a confusion between kinetic heat energy and quantum radiative energy.

cal
March 29, 2011 1:36 pm

Dave Springer says:
March 29, 2011 at 12:30 pm
The air is ALWAYS emitting a continuous blackbody spectrum! Everything with a temperature above absolute zero emits a continuous blackbody spectrum with a peak frequency that rises with rising temperature. The higher the peak frequency the more energy in the emission. This is where the downwelling radiation comes from and its existence is without doubt.
Of course it does not. Only a black body emits a black body spectrum. The atmosphere is not a black body. If it were a black body we would not be able to see the sun! Indeed most things are not even close to black bodies – that is why the world is colourful. Did you not see the spectral distribution looking up at the night sky that was posted about a week ago? It looked nothing like the black body curve. It was completely dominated by the greenhouse gas spectral lines as one would expect. The atmosphere is pretty transparent in the visible region and has an atmospheric window around 10 micron where there is not much absorption, so these wavelengths are absent in the downward radiation as well.

björn
March 29, 2011 1:44 pm

This was a great post!
Brilliant.
Very very nice presentation.
Thank you, now I will consider this in depth.

stephen richards
March 29, 2011 1:47 pm

The greenhouse effect and the energy retaining effect of the oceans are different things. The atmosphere and the ocean may retain heat, but the atmospheric heat retention is infinitesimal compared to the heat retention of the oceans. The greenhouse effect changes the amount of heat that flows out of the system.
The oceans absorb the vast majority of the RADIATIVE ENERGY stiking the planet from the sun. This RADIATIVE energy leaves the planets surface by a route proportional to the density of the atmosphere (not just GHGs) and the speed of light. That’s why Stephen Wilde said what he said. ALL the energy on this planet comes from the sun ~ 99.9%. The Earth approximates to a blackbody in as much as it’s surface albedo modifys the amount of RADIATION that is reflected back into space. A blackbody cannot re-absorb radiation it has given up after equilibrium is reach. After equilibrium all radiation striking the BB is reflected away that’s why the earth and the moon cannot heat the sun by radiative reflection. After bombarding the earth with radiation for 5 billion years they are as near as damn it at BB equilibrium unless the sun’s radiative output changes.

mkelly
March 29, 2011 1:58 pm

Dave Springer says:
March 29, 2011 at 12:30 pm
You were doing fine until you said no back radiation takes place.
I was speaking very narrowly about CO2 per the drawing of Ira’s. If CO2 has at STP no emissivity then by Kirchoff’s Law it did not absorb.

wayne
March 29, 2011 2:07 pm

Inter-radiation of energy between molecules in all planetary atmospheres are random walks with an upward bias due to the curvature. The only place this does not occur is in the window where radiation can move from the surface to space directly.
Cassandra King, you should be able to see this effect. It is what further reduces the effect of that 0.039% factor you mentioned for we live on a round Earth.
Try reviewing this comment and the contained image: http://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/#comment-629487

Reed Coray
March 29, 2011 2:44 pm

cal says:
March 29, 2011 at 12:12 pm

Cal, thank you for your response.
My argument has little if any relevance to the computation of the surface temperature of the Earth in the presence of greenhouse gases. However, it does have relevance to Ira’s statement: “That said, how does this visualization help us understand the issue of “CO2 sensitivity” which is the additional warming of the Earth Surface due to an increase in atmospheric CO2? Well, given a greater density of CO2 (and H2O) molecules in the air, there is a greater chance that a given photon will get absorbed. Stated differently, a given photon will travel a shorter distance, on average, before being absorbed by a GHG molecule and be re-emitted in a random direction, including downwards towards the Surface. That will result in more energy being recycled back to the Surface, increasing average temperatures a bit.
In particular, his statement implies backradiation results in (a) more energy being returned to the Earth surface, and (b) an increase in average temperatures. If any situation can be constructed where (a) backradiation does NOT result in more energy being recycled back to the active object’s surface, and/or (b) backradiation does NOT increase the active object’s surface temperature, then in my opinion Ira’s statement requires additional justification.
By applying Planck’s blackbody radiation law to the surfaces of (1) an active, uniform-surface-temperature sphere and (b) a thin, co-centered, disjoint, inert, planar, uniform-surface-temperature, circular annulus, it can be shown that in the presence of object-to-object conduction, the rate of inert-object to active-object backradiation, although present, can be made less than the rate of conductive energy transfer from the active object to the inert object. This will result in a lower active-object temperature. Furthermore, even in the absence of object-to-object conduction, increasing the outer radius of the inert annulus beyond a certain point, which provides a larger surface area from which backradiation can originate, results in less backradiation, not more. It may be true that greenhouse gas backradiation warms the surface temperature of the Earth, but Ira’s statements that (a) adding more greenhouse gas will increase the amount of backradiation, and (b) that the additional backradiation will warm the surface of the Earth do not convince me. I want to see an analysis of how and to what degree conduction and convection affect the amount of backradiation received by the Earth’s surface relative to the rate of energy transfer from the Earth’s surface by these thermal transfer mechanisms. For example, wouldn’t the increase in the amount of greenhouse gas increase the rate of conductive thermal transfer and possibly convective thermal transfer from the Earth’s surface to the atmosphere? If so, which effect is dominant, and under what conditions?

March 29, 2011 3:13 pm

cal says: March 29, 2011 at 10:12 am
“I am getting a bit depressed by the level of physics being discussed here and in some of the other web sites that we are being directed to…. It is pointless coming up with simplistic arguments to deny these facts when we have data to confirm them.”
AMEN!
As to the original post
* in (7) you say “The energized air molecules emit radiation at various wavelengths”
The emission is still by the GHGs since other molecules are notoriously poor at emitting any sort of IR.
* along with the previous point, the diagram in step 7 seems to show an O2 molecule emitting IR. This would not happen (to any significant degree).
* Most importantly, you miss one major way that molecules can get the energy to emit IR — collisions with other molecules. Just as GHG molecules that have absorbed extra energy from IR photons can pass that energy on to other molecules, other molecules can collide with GHG molecules and give them extra energy from the collision that can then be emitted as IR photons.

Reed Coray
March 29, 2011 3:16 pm

cal says:
March 29, 2011 at 11:41 am
If it returns to the surface and does not warm it must be totally reflected. If it is totally reflected then the surface cannot emit at that wavelength either (basic law of thermodynamics). Therefore it is not possible for radiation to be emitted from the surface, absorbed by the atmosphere and re-radiated downwards without increasing the temperature of the surface.
Cal, I have to disagree. Consider a spherical active Earth surrounded by an “atmosphere” comprised of co-centered, thin, blackbody-surface (both sides), spherical shell with a vacuum filling the space between the Earth’s surface and inner surface of the shell. I know this isn’t a gaseous atmosphere, but in the sense that the shell surrounds the Earth, I will take license and call it an atmosphere. In isolation, the Earth’s surface temperature will be (a) directly proportional to the fourth root of the Earth’s internal rate of energy generation and (b) inversely proportional to the square root of the Earth radius. No backradiation exists for the Earth in isolation.
When the shell is added, because in steady state the shell must radiate to space the energy generated internal to the Earth, the steady-state temperature of the shell (for a thin shell, the inner and outer shell surface temperatures will be very near the same) will be directly proportional to (a) the fourth root of the Earth’s internal rate of energy generation and (b) inversely proportional to the square root of the shell radius. Backradiation (radiation from the inner shell surface to the Earth) will exist. Since the radius of the shell is greater than the radius of the Earth, the shell’s surface temperature will be lower than the Earth surface temperature in isolation. If a thermal conduction path is provided from the Earth to the shell and if that path supports a rate of energy transfer greater than the radiative rate of energy transfer from the shell’s inner surface to the Earth, the Earth’s temperature in the presence of the shell with conduction will be lower than (a) the Earth’s temperature in the presence of the shell without conduction, and (b) the Earth’s temperature in absence of the shell. Since radiation is being emitted from the active object (the Earth) and backradiation from the inert object (the shell/atmosphere) exists whenever the shell is present, how can you claim backradiation from the shell/atmosphere must warm the surface temperature of the active object?

BigWaveDave
March 29, 2011 4:15 pm

Gravity is a force that is constantly pressurizing the atmosphere. Consider a warmed parcel of air at the surface. It will rise, no? When it rises, it eventually cools. When it cools and is pulled to the surface by gravity, its temperature increases because it is compressed.
The constant compression of atmosphere by gravity results in higher pressure, density and temperature at the surface. No GHG’s are necessary for this..

cal
March 29, 2011 4:22 pm

Reed Coray says: (in reply to my reply)
March 29, 2011 at 2:44 pm
…………. If any situation can be constructed where (a) backradiation does NOT result in more energy being recycled back to the active object’s surface, and/or (b) backradiation does NOT increase the active object’s surface temperature, then in my opinion Ira’s statement requires additional justification. ………It may be true that greenhouse gas backradiation warms the surface temperature of the Earth, but Ira’s statements that (a) adding more greenhouse gas will increase the amount of backradiation, and (b) that the additional backradiation will warm the surface of the Earth do not convince me. I want to see an analysis of how and to what degree conduction and convection affect the amount of backradiation received by the Earth’s surface relative to the rate of energy transfer from the Earth’s surface by these thermal transfer mechanisms. For example, wouldn’t the increase in the amount of greenhouse gas increase the rate of conductive thermal transfer and possibly convective thermal transfer from the Earth’s surface to the atmosphere? If so, which effect is dominant, and under what conditions?
————————————–
Reed, I think it is totally reasonable to question whether there is a limit to the impact any change can have and whether there are compensating effects. Almost all natural effects have a limit and negative feedbacks. That is why the earth is ruled by cycles and, for example, does not get permanently hot or permanently cold. The greenhouse effect is logarithmic so we know that without positive feedback one has to double the concentration to get a 0.7C temperature rise and double again to get another 0.7C. Nothing to get concerned about. That is why positive feedback has to be postulated in order to scare us. With such small effects one has a duty to look at all other small effects that might create negative feedbacks which might partially compensate or even overcompensate and drive the temperature in the opposite direction as CO2 increases. You suggest this might be possible and while I do not follow your reasoning as to the specific mechanism I entirely agree that this is possible. So while I am convinced that back radiation warms the surface I am not convinced that further increases in CO2 will warm the surface further. I see these as two entirely different questions and is the reason why I believe sceptics should focus on this second question and not the first.
You might be interested in one possibility that I raised (in person) directly with the UK met office. I have never had a satisfactory answer even though it was repeatedly promised. As you will probably be aware CO2 is the most significant molecule involved in the COOLING of the atmosphere. It radiates most of the energy from the thermosphere and stratosphere and together with H2O radiates almost all the energy from the top of the troposphere and tropopause. H2O is the dominant source of radiation from the mid troposphere and of course the surface radiates directly to space, mainly through the atmospheric window around 10 micron.
So clearly there is the potential for a negative feedback through increased cooling as the concentration of CO2 increases. Which effect dominates? I believe the physics is too complicate to work this out. To find out the answer we have to do something outrageous. We have to measure what is actually happening!
We need answers to questions like: what is happening to the position of the tropopause? what is happening to the temperature in the stratosphere immediately above the tropopause? If the tropopause and lower stratosphere are cooling then we might expect the surface to continue to warm but if there is no change to the tropopause or there is warming then (perhaps non intuitively) this means more radiation to space and a lower surface temperature. A couple of years ago there was a report of a measured increase in the altitude of the tropopause and implied cooling but since then it has gone very quiet, so I suspect the measurements are not yet providing the “right” answer!!

Joel Shore
March 29, 2011 4:24 pm

Ian W says:

The AGW CO2 warms the world hypothesis depends on the hydrologic cycle. It only works if they can show the ‘water vapor feedback’ which all the GCMs show as a tropospheric hotspot that in the real world does not exist therefore all the models are falsified.

The water vapor feedback is now extremely well-verified, especially over timescales of months to a few years. See http://www.sciencemag.org/content/310/5749/841.abstract and http://www.sciencemag.org/content/323/5917/1020.summary
Amplification in the tropical troposphere (the so-called “hot spot”) is also well-verified over these timescales. Over the multidecadal timescales, the data is simply too uncertainty to reach any definitive conclusions but it is difficult to imagine mechanisms by which the physics responsible for this would break down on these timescales without also breaking down on shorter timescales.
Dave Springer says:

The whole brouhaha over scary global warming rests on the wholly fabricated idea that CO2 greenhouse warming puts more water vapor in the atmosphere and that, because water vapor is a powerful greenhouse gas, it constitutes a positive feedback that turns 1C per doubling into 3C per doubling. There isn’t a shred of empirical evidence to support that belief and IMO overwhelming evidence that the feedback is actually negative – the water cycle speeds up which serves to mechanically transport heat from the surface by evaporation and convection to the cloud deck where it then has an easier radiative path out to space.

The only way that radiation escapes to space is by stuff radiating and the amount of radiation that stuff emits is determined by the temperature. Hence, your statement amounts to the hypothesis that the “hot spot” ought to be more pronounced, not less pronounced, than the models predict.
The models already take into account the negative feedback due to the lapse rate, which basically seems to be what you’re describing. But a literal interpretation of the claim that the hot spot is missing would be that the models are including a negative feedback that does not exist.

If the water vapor being delivered to the upper troposphere increases so does the amount of heat carried away from the surface by evaporation and convection and it condenses at an increased rate rather than giving the upper troposphere a higher absolute humidity.

Except that the satellite data seem to show otherwise, as I noted above.

Thus evaporation, convection, and cloud formation is a negative feedback that negates the effect of more water vapor due to warmer surface temperature. There are several well developed hypotheses on this effect. The best IMO is
http://www.scribd.com/doc/25071132/The-Saturated-Greenhouse-Effect-Theory-of-Ferenc-Miskolczi
which is basically saying that the atmosphere, with regard to water vapor greenhouse warming, is saturated and we can’t get any more greenhouse warming out of water vapor – we can get additional surface warming only from increases in non-condensing greenhouse gases like CO2 and methane.

If that’s the best, I can’t imagine what the worst looks like! Miskolczi’s paper is basically nonsense with lots of extremely strange things like misapplying the virial theorem.

Max Potter
March 29, 2011 4:45 pm

I agree with Michael Larkin that the animation was much too fast. If you right-click on the picture and save it and open it in Irfanview you can save it as eight separate pictures and then move forward and backwards at leisure (Options – Extract all frames).

Joel Shore
March 29, 2011 4:51 pm

BigWaveDave says:

The constant compression of atmosphere by gravity results in higher pressure, density and temperature at the surface. No GHG’s are necessary for this..

Your notion doesn’t even satisfy the 1st Law of Thermodynamics! In the absence of a greenhouse effect, the surface temperature of the earth is set by energy balance considerations. So, unless you are proposing a large source of energy…e.g., that the earth’s atmosphere is undergoing continual gravitational collapse, which would certainly be a novel hypothesis…your notion is completely in violation of known laws of physics.

eadler
March 29, 2011 5:27 pm

bananabender says:
March 29, 2011 at 12:41 am
The Ideal Gas Law PV=nRT quite accurately predicts the atmospheric temperature of Earth, Mars, Venus and Titan (at all altitudes) based on a simple pressure effect. Why add the unnecessary complication of the highly dubious Greenhouse Effect to an elegant and simple solution?
You are wrong. John Tyndall showed that you are wrong in 1859, and his explanation of why the earth is as warm as it is has been accepted by scientists ever since. Even Fourier in 1829 knew you were wrong, but it was Tyndall who showed it conclusively.

Jim D
March 29, 2011 5:33 pm

Two main mistakes as pointed out above.
1. Absorbing by CO2 at 15 microns does not speed up the molecule, it raises its vibration state, and may change its rotation state (15-micron side-bands).
2. Photons are not emitted by molecules other than GHGs and not at wavelengths other than those where they have spectral bands.
Other than that, fine.

BigWaveDave
March 29, 2011 5:36 pm

Joel Shore says:
March 29, 2011 at 4:51 pm
Your notion doesn’t even satisfy the 1st Law of Thermodynamics! In the absence of a greenhouse effect, the surface temperature of the earth is set by energy balance considerations. So, unless you are proposing a large source of energy…e.g., that the earth’s atmosphere is undergoing continual gravitational collapse, which would certainly be a novel hypothesis…your notion is completely in violation of known laws of physics.
You are mistaken. You can’t ignore that gravity is a force, and that a gas is compressible, and will increase in temperature as it is compressed. Unlike the GHG hypothesis, these are physical phenomena that are measurable and reproducible.

George E. Smith
March 29, 2011 5:38 pm

“”””” cal says:
March 29, 2011 at 10:12 am
I am getting a bit depressed by the level of physics being discussed here and in some of the other web sites that we are being directed to. I am beginning to wonder if there is a strategy amongst the warmers to refute the greenhouse effect using bogus physics in order to quote these arguments, say how stupid they are, and thus justify the dismissal of all sceptic arguments as unscientific. Or I might just be paranoid!
There are serious problems with the quantification of the greenhouse effect. There is a serious doubt about feedbacks. There is a serious problem that the “fingerprint” of an enhanced greenhouse effect cannot be identified.
But there is no doubt that the atmosphere absorbs radiation emitted by the earth and eventually radiates it to space at a higher altitude which, being at a lower temperature, means a smaller energy loss. That reduction has to be compensated for by increased radiation losses elsewhere and that means a warmer surface. “””””
Well there is one thing seriously wrong with your thesis, cal.
The earth’s surface is mostly hotter than the air is at altitude. Therefore the surface emits at a higher emission level than the cooler atmosphere, and moreover it does so in a generally higher frequency (lower wavelength range; per wien’s Displacement Law.
So the very hottest desert surfaces, are emitting plenty of radiation that goes right on by the CO2 15 micron absorption band right out into space unimpeded by CO2. And as you get higher, and the atmosphere gets colder and less dense, the absorption bandwidth of the 15 micorn CO2 band keeps getting narrower; each and every one of the high resolution lines in the CO2 band keeps get less broadened by doppler, and collisions, as you go up in the atmosphere, so it keeps on absorbing an even lower fraction of the surface emitted radiation, and escape gets easier, as you go higher. So you are not dependent on energy being transported to the highest least dense, and coldest upper reach4es of the atmospehre, and then expecting that material to radiate the entire heat load of the planet.
And if you go in the opposite direction (atmospheric emissions are isotropic, so only half of it goes down, the downward path (that “back radiation thing”) sees a continuously warming, and denser atmosphere so the CO2 absorption spectrum gets more and more opaque for the downward radiation, instead of more and more transparent for the upward escape path.
So the earth des not sit at some BB Temperature of 255 K due to some high altitude evanescent layer of colder rarified atmosphere. Certainly the energy borne aloft as heat by conduction, convection, and evaporation (latent heat) must eventually be radiated by higher altitude gases; but the very ground itself, is making a significant direct contribution to the radiative cooling that augments the other (thermal) processes.
The earth is anything but Isothermal, and we should not expect it to radiate some isothermal radiation spectrum.

Joel Shore
March 29, 2011 6:12 pm

BigWaveDave says:

You are mistaken. You can’t ignore that gravity is a force, and that a gas is compressible, and will increase in temperature as it is compressed. Unlike the GHG hypothesis, these are physical phenomena that are measurable and reproducible.

I am well-aware that gravity is a force and that gases are compressible. I, and other atmospheric and climate physicists are also aware of the constraints on lapse rates set by the decrease of pressure with height. However, that does not allow you to violate the Laws of Thermodynamics at will. And, these Laws tell us that the surface temperature of an IR-transparent atmosphere will go toward a steady-state value determined by the condition that the radiative emission balance the radiation absorbed from the sun (in the absence of any other significant energy source).

Max Hugoson
March 29, 2011 6:26 pm

Harold Pierce Jr says – – –
Classic error in thinking and the modeling. The concept that once the CO2 molecule “absorbs” the IR, and becomes vibrationally excited, it “losses” that energy to the O2 and N2, and thus “traps” the energy in the atmosphere.
NOT TRUE! There is a distribution of energies of the ensemble of molecules. There is EXCHANGE. Therefore, collisions between O2 and N2 molecules on the “high end” of the distributed energy, LOSE that translational kinetic energy to CO2 and H2O molecules.
WHICH, can…with significant probability, re-emit an IR photon.
That’s why the REAL question here is what is the “optical transparency” of the ensemble of the atmosphere, and has the increased CO2 since WWII changed that number significantly.
Studying the work of Miskolczi, Ferenc M. 2010. “THE STABLE STATIONARY VALUE OF THE EARTH’S GLOBAL AVERAGE ATMOSPHERIC PLANCK-WEIGHTED GREENHOUSE-GAS OPTICAL THICKNESS”, Energy and Environment, 21, 243-262 , indicates that this is the case. I’m well aware of the critics of Miskolczi’s work. I’m also aware that based on “empirical observations” up to 1942, Dr. Elsasser came to the same conclusion as Miskolczi, and neglects CO2 in his “general radiation chart”, which has been used successfully to model day to day heat up and cool downs for “weather forecasting” since publication of his paper, “On the IR Heat Balance of the Atmosphere”, (Harvard Met. Series, 1942)
Dr. Glickstein: I hope you are retired, and have the time to examine these matters in depth. I, alas, still have to work for a living.
Max 🙂

Editor
March 29, 2011 6:33 pm

As soon as you put in convection, which completely dominates IR radiation in the troposphere, then I’ll be interested. Until then, it’s just a matter of “Given these conclusions what assumptions can a draw?” followed by “Assume there is no convection”…

March 29, 2011 6:46 pm

If backradiation was a phenomena that would heat up earth surface it would also heat up the sun up to the core.
By using Stefan-Boltzmann the non-GHG surface temperature is found by using E = sigma T^4 (SB) where E can only originate from the sun. If now T would go up further because of GHG, this would mean more energy input would be needed. So where does this come from, are GHG capable of generating energy? Many people seem to think so, but what about the following?
If we look at a CO2 layer as a plate with T2 above earth, you have something like the classic example of radiation between two plates here, where the sun would heat the lower one which is T1.
So yes radiation will go both ways, there is ‘backradiation’ and I go with that. But read that in the state (equation 19.2) where also plate number two would be T1, Qnet will be zero. Surface 1 will never get any warmer by the (back)radiation, despite the accumulated (heat)energy in plate 2 and radiation going everywhere. Read from this that radiation is energy but never heat.
So you can make calculations with multiple layers (like CO2 layers) and the result will be the same. T1 will not get any warmer with Qnet getting zero, and all the layers acting as insulation (like some people think CO2 does) and all plates actually permanently getting radiation from both sides. (Earth atmospheer would also be forced to become T1 from the surface up if it was’nt for gravity).
Now think about multiple plates where the sun’s surface is the first plate and earth the second etc. So the greenhouse effect says that the second plate is supposed to get warmer by backradiation from the plates next in line? Then also the first one, the sun’s surface will get warmer (no matter how tiny) by this backwards phenomena. The sun has many more layers, but this process will go on up to the core! The source of the energy!
Well, what is wrong here. The problem is that in our real universe the temperature of the sun in space is the driving force, telling all other layers down the line what temperature they must take on, in its path to the 3 Kelvin of space and not the other way around.
With new layers somewhere in the line, the temperature of the layers behind the new one will simply go down to establishing a larger dT maintaining the same energy flux.
Yes, it’s that pesky Second law, that in fact is constantly trying to lower the temperature of the sun.
So what are the tricks with radiation that many people don’t see? For starters I want to say that the photons of EM radiation can be regarde as cold, only interaction with matter gives any thermal energy (to get a temperature) which is taken away (partly) when a(nother) photon leaves.
Most greenhouse effect statements/sites claim an energy balance on earth surface because of the First law, the greenhouse theory makes fame by using energy in-out balances etc. convincing everyone who has no clue about thermodynamics. But this is dead wrong, the Second law rules here and this means simple energy conservation is not what determines radiative equilibrium temperature at the surface.
The earth could conserve it’s energy by accumulating it until it would be as hot as the sun, there is nothing against the First law with this.
So in GHG physics T of earth surface is supposed to go up with while receiving the same amount of energy from the sun. Now we know the Second law wants to establish a dynamic radiative equilibrium at the surface with a certain temperature but also that T = dE/dS, so this means that for T to be able to go up with backradiation Entropy must decrease during the absorption-emission process?? That will never happen.
Looking closer it works like this:
Why does the sun heat the surface to T? Because HQ radiation from the sun can leave earth as LQ IR radiation (to the cold space) in the process gaining entropy while raising kinetic energy as potential WASTE HEAT on the surface (and they don’t call this irreversible for nothing, the downgraded (photon)energy won’t be able to do the same trick at the surface again).
The radiation Quality is expressed as theRadiation Energy Density
Some LQ IR returns (backradiation), but it can never ever create more heat here. Because in that case it would have to leave with less energy so as even lower quality IR, and hey …… this can only happen at a LOWER surface temperature (think of the BB-spectra of earth next to the sun, all the energy flows from the high spectrum to the low one, anyone thinks some energy goes the other way here?).
See the contradiction, that’s the mighty Second law of thermodynamics in action that says: it’s all wrong with this heating by backradiation philosophy.
So backradiation will simply leave in a reversible proces (like reflection) without leaving waste heat, and with the same frequency as it had before and bounce whatever way it wants with no effect on the surface. It’s like in the Qnet = zero situation between the two plates (lotsa radiation, no more heating by the cold photons).
So what is ‘the heat’ that can’t go from low to high temperature, warming the earth surface and why the confusion?
Maxwell’s classic Theory of Heat states: heat is something which may be transferred from to another, according to the Second law of thermodynamics.
In any case, heat transfer needs matter which has a temperature (as kinetic energy) and another piece of matter at lower T. Radiation has no temperature and can never ever represent heat. And so the photons of radiation can be regarded as cold.
Heat is the waste from the irreversible radiation phenomena happening due to what the Second law dictates and is equivalent to the entropy gained and what is called dissipation of energy.
Heat can be released when a photon interacts with matter. Many think ‘the heat’ has to do with all the energy (E= h x v) related to the cold travelling photon. But we have seen that this is not the case, this is imaginary heat (which is not heat but energy) that is never released as long as matter is’nt involved. Energy is only released when the SWR photon disappears in the matter and after that heat from this matter may flow if other colder matter is present, but this heat is not equivalent with the photon energy but is a function of the temperature difference between surface temperature and colder matter above it (like Q = k*A*dT). If there is no colder matter, there is no heat (dT=0).
If earth had no atmospheer, the photons would warm earth surface to SB temperature but no heat would ever be exchanged. Then if there was a layer of CO2 around it at say 10 meters or 10 km, this would create backradiation, but the vacuum inbetween would be as cold as the 3K space, and the backradiation would not make the surface warmer, and there would be no heat despite all the radiation.
Now because heat is flowing from earth surface we know we have a second irreversible process (waste heat, making entropy rise) we also know for sure the new lower energy photon can never release heat at this temperature where it came from ever again, so to think backradiation can go beyond that and even create a higher temperature is impossible by the Second law of thermodynamics.
So the heat coming from the surface is the part of the solar radiation that did’nt get radiated away (directly) as IR (and the total energy of this IR is thus lower than the LWR that hits the surface). Heat is the part that has nothing to do with radiation, and the IR that left has nothing to do with heat.
And so we get the surface balance: IR = LWR – HEAT. So the IR that leaves the surface does’nt only have a lower frequency, but the total energy of it is also lower than the incoming LWR. This is the second blow for the imaginary effect of backradiation.
It is this HEAT only that warms the first layers of the atmospheer and gives the ground temperature’s measured (in air, so does not even represent the surface), and it can only take place by conduction from surface to air initially, it must be transported from the surface at T1 to matter at T2 and it finds its way up (conduction/convection – radiation) to space separated from the IR photons that left the surface.
Papers about the sun that are not infected with the greenhouse virus go along the lines of my view. Photons jiggle around for tenthousands of years from layer to layer, and they do not heat the core. This degradation of high quality X-ray photons to relatively low quality optical photons, is only to be expected from the Second law of thermodynamics.
Further: a photon can only travel a tiny distance before running into another hydrogen nucleus. It gets absorbed by that nucleus and the re-emitted in a random direction. If that direction is back towards the center of the Sun, the photon has LOST GROUND! It will get re-absorbed, and then re-emitted, over and over, trillions of times. The path it follows is called a “random walk”

March 29, 2011 7:06 pm

Max gets a good part of it, but there are a couple of things missing, the higher you go in the troposphere, the colder it gets, and the slower the emission from collisionally excited CO2, water vapor or any other greenhouse gas. Second, the distance a photon emitted from CO2 can go is not very far, so effectively, the height at which emission in the CO2 bands can reach space is about 10 km, where it is very cold. You can see this in the modtran spectra (just hit submit the calculation) which show what would be observed from say the space station. The emission between 600 and 700 cm-1 is much lower than it would be without CO2, because it comes from a layer 10km up where the temperature is 220 K.

BigWaveDave
March 29, 2011 7:08 pm

Joel Shore says:
“I am well-aware that gravity is a force and that gases are compressible.”
So, what happens to cooled air that descends?
“I, and other atmospheric and climate physicists are also aware of the constraints on lapse rates set by the decrease of pressure with height.”
If this is supposed to be some sort of appeal to authority, I’m not at all impressed. First, you have to say something that demonstrates you understand what you are talking about.
You say: “However, that does not allow you to violate the Laws of Thermodynamics at will.”
Nor does it allow you to ignore the system conditions with statements like this : “And, these Laws tell us that the surface temperature of an IR-transparent atmosphere will go toward a steady-state value determined by the condition that the radiative emission balance the radiation absorbed from the sun (in the absence of any other significant energy source).”; which totally ignores the influence of gravity on the initial state of the compressible gas that makes up your hypothetical IR-transparent atmosphere.
But, where have I violated any laws of thermodynamics?

enough
March 29, 2011 7:12 pm

When I got to the part where you demonstrated your complete lack of understanding the thermalization process I quit reading. The idea of an IR photon being absorbed and then quickly remitting with half going up and have down is absolute BS. Go back and study how the atmospheric thermal reservoir works.

dp
March 29, 2011 8:47 pm

I’m stunned that there is debate about the ability of one radiating object being able to heat another radiating object. It is so numbingly obvious that this happens that it begs the question: What the hell would keep it from happening?!?
As for which direction molecules radiate energy, try thinking of each of them as isotropic radiators (they’re not, but the principle applies). Except for geometry limitations of the molecule (asymmetric construction), they will behave like an isotropic radiator and radiate equally in all directions. In the atmosphere, where molecular orientation is assuredly random, even asymmetric molecules on average will not favor up vs down vs over there when it comes time to shed their packet of new energy.
Here’s another mental widget to help get your head around this. We’re talking about radiation here, remember. Picture a flash bulb sitting in free space. Imagine it is lit off. Imaging a sphere of light radiating away from the bulb. Imagine that light hits other things. Imagine some of those things are other lightbulbs. Some lit, some not. Now try to understand – there is no intelligence delivered with that sphere of light. It simply illuminates any surface it touches. That includes other lit surfaces. The bulbs don’t know this is going on and don’t care. They are just following the rules of thermodynamics.
Now the test – you put a very sensitive light sensor in the lab space and it has lenses so it sees but one single bulb. The entire field of vision of the sensor it filled with that bulb. Turn on the bulb and notice the sensor response. Now turn on another bulb that is conveniently located behind the sensor and so out of it’s sensing view. Note that the sensor detects that new light coming from the only thing in its view and it registers a higher light level than before. Some call it BS. In fact that just happened in this thread, but it is true that radiating objects light up other radiating objects in all parts of the spectrum including IR.
Ever wonder why three logs stacked in a pyramid will burn to ash but one log alone will quickly stop burning and turn cold? It is because the logs radiate heat onto each other while they burn and heat is one of the three essentials for combustion.
Nuther example because I sense a lot of ignorance here. You are sitting in your kitchen typing on your laptop, filling the WUWT blog with ignorance. It is dark out, and very cold – snow everywhere. You look at the outside thermometer and it is 5ºF – damn cold. You pull aside the drapes on your double paned sliding glass door so you can look out at the beautiful snowy wonderland. And after a few minutes you start to feel cold where before you were quite comfortable. How can that be? It is 72º in your kitchen – how can you feel cold? You granny comes over and pulls the drapes closed, calls you a dummy and that you’re going to freeze if you leave those drapes open. What does granny know? Did you notice too you were cold only on side of you facing the snowy winter wonderland?
Granny knows your body is radiating the maximum possible heat out of the house and into that winter wonderland. If you pull the drapes closed that heat loss stops. Why? Think about that then send your granny a card. A real card, not a lame iCard, and let her know how smart she is.

Reed Coray
March 29, 2011 8:59 pm

Ira Glickstein, PhD says:
March 29, 2011 at 4:12 pm
…there are at least three issues: 1) The small difference in surface area, 2) The large temperature increase due to the Atmospheric “Greenhouse Effect”, and 3) Convection/conduction/precipitation are not anywhere near 100% effective at transfering thermal energy from the Surface to the Atmosphere. Let us address each in turn:
1 – If the shell is at an altitude similar to the effective altitude of the Atmosphere, which is a teeny tiny fraction (less that 1/1000) of the radius of the Earth, the shell’s surface area will be almost the same as the Surface area of the Earth, so the supposed temperature difference will be miniscule.
2 – The effect of “Greenhouse” warming of the Earth is over 30ºC, which is many, many times the maximum possible temperature difference due to (1).
3- Even if (1) was as large as 30ºC, the combined sum of convection/conduction/precipitation would fail to transfer anywhere near 100% from the Surface to your shell.
Nice try though and THANKS for getting me thinking.

I wish to thank you Ira. I too enjoy and benefit from these discussions. They often make me consider something I had completely missed. In that vein, I would like to retract my statement that it’s a “waste of time” to discuss Earth surface temperature without including a discussion of conduction and convection. Discussions of incorrect analyses of physical phenomena often lead to better understanding and to viable formulations. BTW–the immediately above statement applies to my analysis. If my analysis is incorrect, I will likely learn from my errors; and in the process maybe give the person who corrects my errors insight into an issue he didn’t have before the discussion.
Now to address your three points.
First, “The small difference in surface area,
It’s true a shell that has the diameter equal to the atmosphere diameter will have at most a slightly larger area than the Earth. However, we’re discussing the simultaneous existence of backradiation and a cooler active object surface temperature. As such, there is no restriction on the radius of the shell. It’s also true that the larger the radius of the shell, the smaller will be the backradiation from the shell to the active sphere. The reason a larger shell radius results in less backradiation is as follows. For a thin shell, (a) the area of the inside shell surface is close to the area of the outside shell surface, and (b) for a highly conductive shell, the temperature of the shell inner surface will be close to the temperature of the shell outer surface. In a steady-state condition, the area of the shell outer surface must radiate energy away from the shell at a rate equal to that produced by the active-object internal thermal source. Since the inner shell surface area and temperature are close to the outer shell surface area and temperature, the rate energy is radiated from the inner shell surface will be close to the rate energy is radiated from the outer shell surface, and hence close to the rate of internal energy supplied by the active object. However, as the radius of the shell increases, more and more of the energy radiated from the inner surface of the shell will be directed toward another part of the shell inner surface, not toward the active sphere. As such, a corresponding drop in the backradiation to the active object will take place. Backradiation will still exist, but the rate of backradiation energy flow decreases as the sphere radius increases.
Second, The effect of “Greenhouse” warming of the Earth is over 30ºC
The approximate 30ºC number is obtained by assigning to the Earth a reception albedo of approximately 0.3 and an emissivity of approximately 1 [i.e., an albedo of approximately 0). If you treat the surface of the Earth as a graybody (i.e., you use a modified Planck blackbody radiation law with an emissivity (1 minus albedo) somewhere between 0 and 1], then you must use the same albedo for both reception and emission. For example, if you completely enclose an inert graybody object (non-zero albedo) within a surface that is maintained at a uniform fixed temperature, the steady-state temperature of the enclosed graybody object will equal that temperature. However, if to that scenario you apply Planck’s graybody radiation law with an emission albedo different from its reception albedo, the result will be that in radiation balance (rate of energy absorbed equals rate of energy emitted) the inert enclosed object’s steady-state temperature will be either higher or lower than the temperature of the surrounding material. The steady-state temperature of the inert object cannot both be “equal to” and “not equal to” the temperature of the surrounding (enclosing) material. I believe of these two choices (assuming one of them is correct), the correct answer is “equal temperature”. If true, then the use of different reception/emission albedos and Planck’s graybody radiation law are incompatible. As others have pointed out, I question the use of Planck’s graybody radiation law for a gas/surface interface. It may be valid for minute amounts of gas, but as the gas density increases, I have a feeling the law starts to break down. [As an aside, for a gas there is no clearly defined “surface”. Since one term in Planck’s blackbody radiation law is a differential area at a known temperature, any use of Planck’s blackbody radiation law or a graybody version of that law for gases is inherently impossible.] If the same albedo is used for both reception and emission, the difference between the Earth’s actual temperature and it’s graybody temperature without an atmosphere is more like 7ºC, not 30ºC.
Third, “Even if (1) was[n’t] as large as 30ºC, the combined sum of convection/conduction/precipitation would fail to transfer anywhere near 100% from the Surface to your shell.” The rate of energy transfer via mechanisms other than radiation doesn’t have to be anywhere near 100%. The active object temperature will be decreased if the rate of energy transfer via non-radiative mechanisms is greater than the rate of backradiation. Since the rate of backradiation decreases with increasing shell radius, for a sufficiently large shell, it won’t take much conduction to produce a lower active object temperature. I believe experiments using eight or nine silver struts connecting the active object to the shell can be constructed that will likely transfer a sufficient rate of conductive thermal energy. If you don’t believe so, turn on your stove, place the blade of a looooong sterling silver knife on the element, and see how long you can hold onto the knife handle.
Bottom line, I don’t claim greenhouse gases won’t increase the surface temperature of the Earth. However, I do argue that the backradiation argument is insufficient to make such a claim. This gets us back to my statement to the effect that any meaningful discussion of the Earth’s surface temperature must include a discussion of all thermal transfer phenomena, not just radiation.

Tom T
March 29, 2011 9:30 pm

Well ok, I guess, although many here don’t seem to think it is ok. But why is the sensitivity of atmosphere to an additional molecule of CO2 logarithmic? Your model implies that it wouldn’t be.

Cherry Pick
March 29, 2011 9:30 pm

In this essay you have qualitatively described a small part of a complex nonlinear system which has huge amounts positive and negative feedbacks. Without numbers coming from theory and/or measurements and without describing connections of your radiation physics to the other parts of the system it is not possible to assess its validity and relevance. It is good that you did not have any conclusions or advice to the policymakers.
Nice graphics, anyhow 🙂

bananabender
March 29, 2011 10:30 pm

@DP
It’s a shame none of your examples support your argument.
The first experiment is simply completely wrong.
The three logs burn better because a pyramid allows much more air into the fire and increases the surface area exposed to the fire promoting combustion. Boy Scouts 101. Nothing to do with radiation.
The reason it is cold near the window is because:
a) The window is a much poorer insulator than the walls. This means the air near the window is cooled by conduction and radiation to the outside and convection on the inside surface of the window.
b) The window is also further from the fire and receives much less thermal energy according to the inverse square rule of radiation.
c) The window has a very low albedo so it doesn’t reflect heat back into the room.
If you repeated the experiment with triple glazed windows it wouldn’t work.

dp
March 29, 2011 11:18 pm

@DP
It’s a shame none of your examples support your argument.
The first experiment is simply completely wrong.
The three logs burn better because a pyramid allows much more air into the fire and increases the surface area exposed to the fire promoting combustion. Boy Scouts 101. Nothing to do with radiation.

How is it possible for a pyramid of logs to allow more air to the burning part of the log that a log standing alone without competition for that air from other logs? It really is the mutual heating that sustains the burn.
FAIL

dp
March 29, 2011 11:28 pm

If you repeated the experiment with triple glazed windows it wouldn’t work.

Multiple panes of glass work against conduction. They are useless against radiated heat unless they are designed (coated) to radiate IR back into the room in which case it is exactly the same as drawing the drapes closed but much more expensive. Pick up a good science book and don’t put it down until you understand how radiated energy works. Then teach your kids.

bananabender
March 30, 2011 1:50 am

It’s more than 20 years since I did any spectroscopy so I’m pretty rusty. However we certainly weren’t told about any (alleged) heat trapping abilities of molecules by our PhD qualified chemistry lecturers.
The author seems to have confused spectrophotometry which only involves the outer electron orbitals and IR spectroscopy which involves changes in the vibrational modes of molecular bonds.
IR spectroscopy is primarily used for non-destructive testing of carbon rich solids such as coal, plastics and pharmaceuticals.
It should be remembered that spectroscopy is an empirical tool for measuring structures, quantities or concentrations of a substance. It isn’t designed for determining the existence of a greenhouse effect.

March 30, 2011 2:11 am

DP;
How is it possible for a pyramid of logs to allow more air to the burning part of the log that a log standing alone without competition for that air from other logs? It really is the mutual heating that sustains the burn.>>>>
Thank you thank you thank you thank you! What a wonderful, obvious, easy to understand example! Mind if I borrow it from time to time? I’ve been relying on my igloo example of cold things radiating heat to warm things and have become exasperated with the nonsensical objections.
As for bananabender….
If you’d actually taken Scouts 101 pretty much the first thing you learn is that any logs of a decent size simply will not burn well unless the burning faces are used to heat each other up. In your own example, if the pyramid of logs promoted surface area exposed to air, why wouldn’t the logs burn from the outside in instead of the inside out? If you took Scouts 201 you’d also know that arranging the logs in layers, each layer a set of parallel logs at right angles to the layer below, makes a fire that puts the pyramid to shame. It creates a whole series of surfaces radiating at each other to achieve much higher temps.

cal
March 30, 2011 2:55 am

George E. Smith says:
March 29, 2011 at 5:38 pm
I did not say that all the earth’s energy was absorbed by the atmosphere and re-radiated. It is only true for a small range of wavelengths associated with the absorbtion bands of the greenhouse gases. I have said this many times in different ways (including other comments of mine on this thread) but Eli Rabbet covers it pretty succinctly a few comments back. This is not a theory. It has been measured to be the way the world is.

RJ
March 30, 2011 3:13 am

DavidM
I’m just reading this article. This section below seems at variance to your very dubious Igloo example.
tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf
“Thermodynamics says that no object in the universe can heat itself by its own radiation, nor can heat flow from cold to hot”.

Joel Shore
March 30, 2011 5:06 am

BigWaveDave says:

Nor does it allow you to ignore the system conditions with statements like this : “And, these Laws tell us that the surface temperature of an IR-transparent atmosphere will go toward a steady-state value determined by the condition that the radiative emission balance the radiation absorbed from the sun (in the absence of any other significant energy source).”; which totally ignores the influence of gravity on the initial state of the compressible gas that makes up your hypothetical IR-transparent atmosphere.

I’m not ignoring gravity. I am just accepting the fact that gravity doesn’t magically create energy out of nothing. You write P/R=rho T and say that “density decreases at a lesser rate than pressure” in the troposphere and conclude that temperature must rise. However you fail to note that:
(1) This is just an observed feature of the troposphere. It is not a necessary feature and, in fact, in other levels of the atmosphere (like the stratosphere) it is not the case that the temperature decreases with height.
(2) The fact that temperature decreases with height does not imply that gravity somehow sets a constraint on the surface temperature. And, indeed, it does not…In the absence of an IR-absorbing atmosphere, the constraint on surface temperature is set by simple radiative balance considerations. (In the presence of an IR-absorbing atmosphere…i.e., the greenhouse effect, it is more complex but is still set by energy balance considerations, although one has to consider energy balance within the atmosphere too and convection is also involved.)

But, where have I violated any laws of thermodynamics?

You are proposing that the earth’s surface can be at a temperature where the earth would continuously emit much more energy than it absorbs from the sun, which is not possible for it to do without rapidly cooling down unless there is a significant other energy source that comes into play.

March 30, 2011 5:56 am

RJ says:
March 30, 2011 at 3:13 am
DavidM
I’m just reading this article. This section below seems at variance to your very dubious Igloo example.
tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf
“Thermodynamics says that no object in the universe can heat itself by its own radiation, nor can heat flow from cold to hot”.>>>
NET heat cannot flow from cold to hot.
If it didn’t then you couldn’t stay warm inside of an igloo anymore than you could standing outside the igloo. Igloos work. Period.
Pour boiling hot coffee into a thermos at room temp. Which is hotter? The thermos of the coffee? The coffee obviously. Put a bowl out on the counter and fill it with an equal amount of the same coffee, cover it to prevent heat loss via convection. Wait six hours. Coffee in the thermos is still hot. Which is hotter? The coffee or the thermos? The coffee. What about the coffee in the bowl? Room temp. Huh.
So what heat source was it that kept the coffee hot? Given that it was in a thermos that was at a lower temperature than the coffee and there were no other heat sources being put into the thermos, why did it stay hot?
The answer is that it got some of its own heat back. It radiated heat to the thermos, which, as a consequence of warming up, radiated some back into the coffee.

Richard E Smith
March 30, 2011 6:08 am

According to davidmhoffer an igloo is an example of ‘a cold thing radiating to a warm thing’.
Snow is an excellent insulator as it contains pockets of air and, like a greenhouse, prevents convection. This means that a warm object (a human body, for example) will warm the interior of the igloo by conduction and radiation. The snow cannot add energy to a warmer body. The molecules of the warmer body are vibrating and this is the signal of its temperature. The radiation from a cooler body cannot vibrate the molecules more than they are already vibrating as it has less energy. The snow is not an extra radiator adding to the heat created by the warm body. If it were otherwise a an infra-red reflector (a mirror) would be a heater.

bananabender
March 30, 2011 6:37 am

davidmhoffer says:
March 30, 2011 at 2:11 am
As for bananabender….
If you’d actually taken Scouts 101 pretty much the first thing you learn is that any logs of a decent size simply will not burn well unless the burning faces are used to heat each other up. In your own example, if the pyramid of logs promoted surface area exposed to air, why wouldn’t the logs burn from the outside in instead of the inside out? If you took Scouts 201 you’d also know that arranging the logs in layers, each layer a set of parallel logs at right angles to the layer below, makes a fire that puts the pyramid to shame. It creates a whole series of surfaces radiating at each other to achieve much higher temps.

I was never a boy scout. That means I didn’t learn any of their incorrect explanations for how fires work.
A pyramid or a “log cabin” fire forms a venturi drawing fresh air in from outside the base like a chimney. Hot air rises via convection up the the inside of the structure. This causes the evaporation of volatiles which then ignite. The outside doesn’t burn well because wood is a very poor conductor of heat and there is very little direct heating from the flames.
In the case of a horizontal log the heat is carried upwards by convection so the wood below is not well heated. Charcoal is an exceptionally effective insulator so the wood below the flame doesn’t reach combustion temperature and the fire exhausts the fuel supply.
Radiation is a negligible contributor to flame propagation in small fires. Heat flow is always from hot to cold.

Stephen Wilde
March 30, 2011 6:49 am

A number of burning logs in contact with one another will certainly aid combustion by aggregating their heat.
However the jumble of logs will also obstruct and redirect and locally accelerate the flow of air between them so it is also true that accelerating the flow of air between the logs will fan the flames further.

bananabender
March 30, 2011 7:06 am

dp says:
March 29, 2011 at 11:28 pm
Multiple panes of glass work against conduction. They are useless against radiated heat unless they are designed (coated) to radiate IR back into the room in which case it is exactly the same as drawing the drapes closed but much more expensive.
Curtains reduce heat loss primarily by preventing convection currents near the window surface. This in turn reduces the effectiveness of the glass as a conductive heat exchanger. Curtains have very little effect on radiative heat loss. Triple glazing is far more effective at preventing heat loss than curtains in a cold climate
Triple glazed windows are effective primarily because they prevent conductive heat loss. Radiative heat loss across triple-glazed windows is low because glass is only partly transparent to IR radiation.
Pick up a good science book and don’t put it down until you understand how radiated energy works. Then teach your kids.
I’ve got three science degrees and have studied physics and chemistry at university. Unlike you I actually understand how radiation works.

bananabender
March 30, 2011 7:14 am

Igloos are temporary structures that work primarily by preventing wind chill and heat loss by convection. Snow is also a fairly good insulator reducing conductive heat loss to the outside air. The cold igloo doesn’t have any warming effect at all on the inhabitants.

mkelly
March 30, 2011 7:23 am

Ira says “2 – The effect of “Greenhouse” warming of the Earth is over 30ºC, which is many, many times the maximum possible temperature difference due to (1). ”
This is only true if you disregard that STP is 0 C. If taken into account then GHG is 15 C max.

Ian B
March 30, 2011 7:27 am

A few things that stick out in this discussion:
1 – The concept that the Earth surface warms by back radiation is a mis-understanding. The surface warms because of a rather large object 93 million miles away. The issue is that the OLR radiation leaving the Earth’s surface is slowed in its escape through the atmosphere because of absorption of certain wavelengths including the one about 14 microns that corresponds to an absorption band in CO2. This has the effect of increasing the energy and temperature of the near surface atmosphere, so slows the heat loss from the surface. If this is a violation of the 2nd law of thermodynamics, so are all forms of insulation…
2 – There is a confusion between attempting to explain things at an atomic or quantum scale and in the macro world of classical physics.
3 – the logarithmic effect of absorption (or at least that it is a reasonable approximation at the current CO2 concentrations) is easily demonstratable in absorption spectroscopy (i.e. can be verified by experimental observation). Absorption of a photon is probablistic, and dependent on how close the photon’s energy is to the absorption band of the CO2 molecule – this is why the argument that the CO2 wavelength is ‘saturated’ with regard to IR is a bit over-simplified, those CO2 molecules that have not absorbed a photon ideally suited to its absorption band can still absorb a photon at a markedly lower or higher energy, although the likelihood of absorbing this photon reduces the further from the ‘ideal’ wavelength the photon is.
4 – Despite the above, all of this discussion is rather theoretical until adequate consideration is given to the dominance of convection as a bulk heat transfer mechanism in the lower atmosphere, and to the moderating effects of the oceans as a heat sink and source, remembering that the issue from a human experience perspective really is how much is the temperature increasing in the biosphere.

RJ
March 30, 2011 8:15 am

“Ira says “2 – The effect of “Greenhouse” warming of the Earth is over 30ºC, which is many, many times the maximum possible temperature difference due to (1). ”
Here’s a section from the document noted above
It explains a reason why GHGs do not cause 30+ degrees of warming. This GHG theory is looking more and more shaky. (Is the GHG theory built on such poor science).
tech-know.eu/upload/Understanding_the_Atmosphere_Effect.pdf
[This link did not work for me. It brought up a plea for a donation, and I could not see any way to view the supposed .pdf document. It appears to be a scam of some sort. – Ira]
The greenhouse theory states that the reason the ground is warmer than -18 C is because the atmosphere, via greenhouse gases like CO2, re-emits thermal radiation towards the ground and therefore amplifies the heat at the ground from -18 C up to +15 C.
This is the same point as discussed just above, in using a projection factor of p=4. But to make the point more clearly: the entire surface of the Earth is not simultaneously illuminated by the incoming solar energy around all sides, but only half of the Earth is ever actually illuminated. When the solar input heating is incorrectly averaged over the entire Earth at once, there isn‟t a high enough radiative energy flux density to explain why the temperature ever gets above -18 C. Therefore a greenhouse theory must be proposed in order to explain why the ground temperature is +15 C.
However, if the solar energy is correctly averaged over only the single hemisphere that actually physically receives sunlight, the heating temperature equals +30 C for that hemisphere (and much higher directly under the solar zenith, as we have seen). The +15 C average over both day and night, which is less than +30 C, is then easily understood as simply being due to the fact that the night-side is cooler. Of course, the night-side has to be cooler because it receives no solar energy, but it doesn‟t cool very fast because of the thermal capacity of the atmosphere and the ground. The average temperature of both the day and night hemispheres then comes out to +15 C, which is less than the input solar heating, not more.
It is not surprising therefore that in the first case, we need to theorize a greenhouse effect which ends up violating the laws of physics,

March 30, 2011 8:22 am

Ira Glickstein says
…..” The effect of “Greenhouse” warming of the Earth is over 30ºC, which is many, many times the maximum possible temperature difference “…….
Looking at Ira’s CV perhaps he comes from an electrical engineering background.
Perhaps the degree options he picked did not include much thermodynamics.
However he has to be congratulated for his attempts to make a moving pictorial representation of the atmospheric radiative effects.
However these effects are greatly exaggerated such as the claimed 33K magnitude.
Ira consider this analogy from electrical engineering.
The suns effect on the earth surface as it spins resembles a half wave rectified AC signal (say after passing through a diode represents day/night) .
The earth has a great capacity to store the energy it receives from the sun.
This can be represented as three large capacitors connected in parallel.
The largest by far represents the oceans.
Another represents the atmosphere and a third represents the land surface say to a depth of two metres.
Completing the circuit is a resistor in which represents radiation to space.
Such a system would smooth out the day/night effect and would account for most of the 33K falsely attributed to the so called “greenhouse effect”.

Sam Parsons
March 30, 2011 8:24 am

Thomas says:
March 29, 2011 at 1:57 am
“bananabender. Go ahead, try to explain the temperature of the Earth using only the ideal gas law! It can’t be done, you see, if you actually start to think about what you do.”
Earth does not have a temperature, except in the fantasies of Climategaters. It also does not have a Santa Claus, an Easter Bunny, or a Thanksgiving.

RJ
March 30, 2011 9:05 am

Ira
Here’s one link
http://www.tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf
The section posted above is on pages 25 and 26
Or try this one to get to it
http://climaterealists.com/index.php?id=7457

March 30, 2011 9:22 am

bananabender says:
March 30, 2011 at 7:14 am
Igloos are temporary structures that work primarily by preventing wind chill and heat loss by convection.>>>
The Inuit lived in igloos for months at a time and if you knew the first thing about them you’d know they have a surprisingly large hole dead centre a the top. This is to promote fresh air exchange otherwise the inhabitants would suffer oxygen deprivation within days, let alone months. The design is promotes heat loss by convection and is a net cooling effect on the temp inside the igloo. As for wind chill, sorry, you’ve not done Scouts 101 nor any winter camping obviously. At -40 degrees, in the best sleeping bag you can buy, and with a wind chill of zero, you will most likely freeze to death sleeping outside beside the igloo. Don’t be daft, crawl into the igloo, deal with the -1 wind chill caused convection by you being a heat source inside the igloo and you will very likely be just fine.

March 30, 2011 9:39 am

bananabender;
I was never a boy scout. That means I didn’t learn any of their incorrect explanations for how fires work.>>>
You claimed the authority of Scouting 101 and now admit the credential was false. You also don’t appear to have much experience with building fires.
“A pyramid or a “log cabin” fire forms a venturi drawing fresh air in from outside the base like a chimney. Hot air rises via convection up the the inside of the structure. This causes the evaporation of volatiles which then ignite.”>>>
So….the fresh cool air streaming into the centre of the fire would cool it….convection through the “venturi” is exhausting hot air OUT of the fire…this process sounds like net heating to you? The volatiles evaporate DESPITE the net cooling of the convection process because the combined heat of the various surfaces radiating at each other is high enough to overcome the cooling via convection. Don’t believe me? Build yourself a decent sized fire and turn a fan or leaf blower on it to put 10 or 20 times the air through and around the hole thing than convection could. Watch what happens and report back.
“I’ve got three science degrees and have studied physics and chemistry at university. Unlike you I actually understand how radiation works.”>>>
Really? What are these science degrees you possess? Clearly not physics or chemistry since you only “studied” them, and you no more understand how radiation works than you do Scouting 101. Oh yeah, you didn’t have that credential either.

March 30, 2011 10:23 am

Stephen Wilde
However the jumble of logs will also obstruct and redirect and locally accelerate the flow of air between them so it is also true that accelerating the flow of air between the logs will fan the flames further.>>>
The air flow has a net cooling effect as cold air is pulled in while the hot air escapes out the top. Build yourself a small fire either teepee style or log cabin. Pull out your 3.5 Hrspwr leaf blower (mine’s hanging on the garage wall if you need to borrow it) and play it at idle on the flames. The illussion is that they get bigger. The volatile compounds burning at high enough temps to become incandescent get pushed farther and faster by the air, resulting in them travelling further before going out. The flames get longer, but the fire isn’t as hot. Now whip the leaf blower to max. Poof, fire went out.
In fact you can prove these things with matches and candles. Bend ther wicks horizontal to the rest is easy to do. Strike a match and try and light a candle wick with the very tip only of the match flame. Now try with just the base of the candle flame. Now try with the side of the flame about 1/3 up from the bottom. You’ll find that is by far the hottest part of the flame. A long flame is a cool flame, its the short fat ones that are hot. A jumble of charcoal gets even hotter and it has no flame at all! And the purpose of closing your BBQ lid is to minimize heat loss by convection. Despite which the ones in the middle always turn to ash faster than the ones at the edge, because they heat each other up.
For banana and stephen and richard, please be aware that I am a raging skeptic. CAGW is a total farce. But you can’t win the battle by arguing physics with warmists because on this issue they are correct and all you do is discredit yourself and any arguments that you might have that are correct. If any of you could falsify there being a positive flow of heat from cold to warm despite there being a NET negative flow of heat in total you will most likely win a Nobel prize. The laws of physics verified thousands of times through experimentation are now centuries old. They make it plain that any body radiates the exact same amount of heat at a given temperature no matter WHAT the temperature of the surrounding objects is. That heat impacts cooler things and warmer things and get absorbed by all of them. The warmer things radiate more heat than the cooler, and so the cooler things have a net heat gain relative to the warm things.
If that is wrong, then the Stefan-Boltzmann Law http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law (1879!), Planck’s Constant, Wien Displacement Law, Raleigh-Jeans Law, and Kirchoff’s Law have all been disproven. Things like combustion engines, turbines, ramjets, and a whole long list of other things were therefore designed using falsified physics despite which they worked anyway.
There is SO much the CAGW crowd has gotten wrong. Why attack them on a nitpicking little detail like this one where they just happen to be right? Order of magnitude, effect of feedbacks, THOSE are the places where their arguments fall apart.

Joel Shore
March 30, 2011 10:23 am

RJ says:

The average temperature of both the day and night hemispheres then comes out to +15 C, which is less than the input solar heating, not more.
It is not surprising therefore that in the first case, we need to theorize a greenhouse effect which ends up violating the laws of physics,

Bryan says:

Such a system would smooth out the day/night effect and would account for most of the 33K falsely attributed to the so called “greenhouse effect”.

Both of these comments are simply incorrect. The energy balance constraint on the surface temperature of the earth in the absence of an IR-absorbing atmosphere is that the fourth root of average of T^4 over the earth’s surface must be equal to ~255 K. If the temperature on the earth was uniform, this would mean that the average of T would also be 255 K. However, to the extent that temperatures are non-uniform over the earth’s surface, the average of T is actually less, not more, since there is an inequality where will always be less than the fourth root of (where denotes average). This inequality is one of the few things that Gerlich and Tscheuschner get right in their paper (although they misinterpret what it means).

sky
March 30, 2011 10:34 am

In arguing (March 29, 2011 at 10:50 am) for the primacy of sensible radiative transport of heat based on its presumed near-light speed, Glickstein once again overlooks a crucial physical fact: the RATE of thermal energy transport by MOIST convection. At ~590 calories per gram of evaporated water, the latent heat transport is enormous. In fact, the Bowen ratio of sensible-to-latent heat transport rates has been shown in numerous experiments world-wide to be below unity in general, and invariably well below in marine environments that account for the great majority of the globe’s surface. Only in ultra-dry environments (Sahara, Antarctica) does it much exceed unity. Even Kiel & Trenberth’s “global energy-budget” cartoon reflects that fact–once the nearly null-net radiative exchange between surface and atmosphere is properly accounted for.
Despite all the traffic generated by this pop-science series of posts on the ethereal notion of a “radiative greenhouse,” the lack of deeper comprehension of real-world physics does a disservice to WUWT readers.

Joel Shore
March 30, 2011 10:43 am

RJ:

Here’s one link
http://www.tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf
The section posted above is on pages 25 and 26

So now the link works but we are left wondering what your point is. If it is that one can find a lot of nonsense out on the internet, I agree with you. If it is something else, then I would not use a paper that is so clearly nonsense to try to make your point!
Here’s a hint to you on where that paper goes wrong in all of its claims regarding thermodynamics and the greenhouse effect: The greenhouse effect does not say that (net) heat flows from hot to cold. (Some would say the word “net” is redundant since heat is usually a macroscopic concept describing the net energy flow.) In all models of the greenhouse effect, whether they are toy models or full-blown line-by-line convective-radiative transfer models, the heat flows from hot to cold.
The greenhouse effect is simply stating this: In the absence of an IR-absorbing atmosphere, all of the terrestrial radiation would be emitted directly out into space. However, in the presence of an IR-absorbing atmosphere, some of the terrestrial radiation is absorbed and when the atmosphere subsequently emits radiation, some of this radiation comes back to the earth. This keeps the earth warmer than it would be in the absence of the IR-absorbing atmosphere.
Also note that the Second Law of Thermodynamics is not magic: It does not state that colder objects magically detect the presence of a hotter object and refuse to emit any radiation towards said object. Rather, it is a statement derived from statistical physics that says that the amount of radiation that the colder object absorbs from the hotter will always be greater than the amount of radiation that the hotter object absorbs from the colder.

Steve Keohane
March 30, 2011 11:32 am

davidmhoffer says:
March 30, 2011 at 10:23 am
Stephen Wilde
However the jumble of logs will also obstruct and redirect and locally accelerate the flow of air between them so it is also true that accelerating the flow of air between the logs will fan the flames further.>>>
The air flow has a net cooling effect as cold air is pulled in while the hot air escapes out the top. Build yourself a small fire either teepee style or log cabin. Pull out your 3.5 Hrspwr leaf blower (mine’s hanging on the garage wall if you need to borrow it) and play it at idle on the flames. The illussion is that they get bigger. The volatile compounds burning at high enough temps to become incandescent get pushed farther and faster by the air, resulting in them travelling further before going out. The flames get longer, but the fire isn’t as hot.

In my experience with bellows at a forge, very little, and at a fireplace, a lot, adding a stream of air increases the temperature of the flame. This can be seen as the fire without the air cannot melt a given metal, but with the stream of air it can. Varying the amount of airflow controls the temperature lower for forging, higher for melting.

Joel Shore
March 30, 2011 12:13 pm

Joel Shore says:

However, to the extent that temperatures are non-uniform over the earth’s surface, the average of T is actually less, not more, since there is an inequality where will always be less than the fourth root of (where denotes average).

Whoops…I guess the HTML markup doesn’t do well with the brackets that I used to indicate averaging. That sentence should read:

However, to the extent that temperatures are non-uniform over the earth’s surface, the average of T is actually less, not more, since there is an inequality where the average of T will always be less than the fourth root of the average of T^4.

Joel Shore
March 30, 2011 12:19 pm

davidmhoffer says:

But you can’t win the battle by arguing physics with warmists because on this issue they are correct and all you do is discredit yourself and any arguments that you might have that are correct.

David and I completely disagree on most things regarding AGW, but on this we completely agree. Trying to argue that the greenhouse effect doesn’t exist just discredits you in the eyes of any serious scientist. It is like arguments that the earth is only 7000 years old…It is just like wearing a sign around your neck that says, “I am an anti-science crackpot.” Better to argue about things like feedbacks and climate sensitivity, where there are at least some significant uncertainties in the actual science.

mkelly
March 30, 2011 12:33 pm

Ira says:”However, since the Earth reflects about 30% (or 28%) of the incoming sunlight, the planet’s effective temperature (the temperature of a blackbody that would emit the same amount of radiation) is about −18 or −19 °C, about 33°C below the actual surface temperature of about 14 °C or 15 °C. “” and same for mkelly regarding your “15 C max”.”
STP temperature is 0 C so -18 up to 0, the 15 C as actual surface temperature per your quote above. The difference between 0 C and 15 C would be GHG. You answered your own question.
However, all this is based on the -18 C assumption.
240 w/m^2 = SB X T^4
240/SB = T^4
T= 255 or -18 C
1360/4=340
340 times .7 = 240 rounded
The 340 average may infact be incorrect for TOA as my heat transfer book shows 1063 w/m^2 at 90 deg sun angle (over head at equator) for a surface W/m^2.

cal
March 30, 2011 12:43 pm

I have just realised that according to some of you I have transgressed terribly. I was tuning in my radio when I picked up some of the cosmic microwave background. Since this has a spectrum of a black body at 3K (that is 3 degrees above absolute zero) according to you I have clearly broken the second law of thermodynamics. I am currently 290 degrees warmer than 3K. I am sorry but my radio would only have warmed a little bit since the energy is very low.
On the other hand maybe you do not have it quite right!

Joel Shore
March 30, 2011 12:51 pm

mkelly says:

The 340 average may infact be incorrect for TOA as my heat transfer book shows 1063 w/m^2 at 90 deg sun angle (over head at equator) for a surface W/m^2.

On average, the sun is not at a 90deg angle overhead. The average power in W/m^2 at the top of the atmosphere is obtained by considering the solar constant and the ratio of the area of a disc to the surface area of a sphere. Needless to say, these geometric formulas are on a pretty solid footing!

Joel Shore
March 30, 2011 1:01 pm

mkelly: Okay…My last post may have been in haste, as I now see what you might be trying to say. However, the 1063 W/m^2 is probably a value for a sunny day, i.e., it presumably neglects cloud albedo. (But, it also includes some things that lower the value from the solar constant, such as scattering from aerosols and other components in the atmosphere and absorption by the atmosphere.) [At any rate, even if we took 1063 W/m^2, divided it by 4 and used it in place of the 240 W/m^2, it would only raise the 255 K value to ~261.5 K.]

Steve
March 30, 2011 1:07 pm

Is “back radiation” really required to explain the greenhouse effect? It seems to me that it just confuses the issue.
With a pure nitrogen atmosphere, almost all infrared radiation would pass through. The molecules of this atmosphere are primarily heated by direct contact with the surface of the sphere, and convection of this thin layer with the upper layers.
Now replace a mass of that nitrogen with a like mass of CO2. Now you have layers of the atmosphere (top to bottom) heated by infrared radiation that was previously not heating any portion of the atmosphere. The atmosphere is warmer – period. No need to talk about whether or not back radiation will heat up the surface or not, because it is obvious that the atmospheric blanket itself is warmer than it was before. (If you really want to make it warmer, exchange H2O for either of the other two gases.)
Do you feel warmer in a room with 15 degree air versus -15 degree air. Uh, yeah. But 15 degree air is below body temperature, so how could it possible be keeping you warm?! Well, neither is keeping you warm – that’s why you are cooling off. But you will cool at a slower rate in one versus the other.

RJ
March 30, 2011 1:10 pm

“This keeps the earth warmer than it would be in the absence of the IR-absorbing atmosphere.”
But its a 30 degrees increase because of backradiation. Not keeps the earth warmer / reduces the amount of cooling.
Some other points
The article was very good. But then it does not challenge my strongly held beliefs.
How much of the energy /heat leaves the surface by conduction. If the majority then the likelihood of the extra 30 degrees is surely even further reduced even if back radiation can add extra energy and heat the surface like the sun does.
We are playing into the alarmists hands if we accept the GHG theory as a fact. It clearly is not so why not acknowledge this. Explain the theory but also explain the possible problems with this theory

mkelly
March 30, 2011 1:31 pm

Joel Shore says:
March 30, 2011 at 12:51 pm
You can see by the earlier post I know where it came from and why. I could have said the chart in my heat transfer book shows from 90-5 deg and the range is 1063 to 41 w/m^2. I am starting to get an inkeling that the 1360/4 may not be correct for TOA that is all I said. I need to look into it further.

Joel Shore
March 30, 2011 1:57 pm

RJ says:

How much of the energy /heat leaves the surface by conduction. If the majority then the likelihood of the extra 30 degrees is surely even further reduced even if back radiation can add extra energy and heat the surface like the sun does.

This diagram shows the relative transfers: http://www.cgd.ucar.edu/cas/abstracts/files/kevin1997_1.html And, yes, convection and evaporation / condensation play a very important role in heat transfer in the troposphere. However, that doesn’t affect the conclusion that the role of the greenhouse effect is to raise the average surface temperature by ~33 deg above what they would be in the absence of the effect. That is based on experimentally-observed surface temperatures. In fact, what I understand is that if one uses a purely radiative model to calculate the greenhouse effect, one predicts around twice that amount of warming of the surface…and it is the addition of convective effects to the model that then reduces that to something much closer to what is actually observed.
By the way, although this back-radiation is a useful concept for explaining things, the technically-better explanation is to focus on the effect of greenhouse gases on the top-of-the-atmosphere radiation budget (because you know that the only significant heat transfer that occurs between the entire earth-atmosphere system and space is via radiation) and then to figure out what will happen on the surface in a way that accounts for the importance of convective transport in the troposphere. I believe that to the first approximation, what is expected to occur is that convective and radiative terms between the surface and the atmosphere will be such that the lapse rate is expected to remain about the same in an atmosphere with more greenhouse gases.

March 30, 2011 1:59 pm

Steve Keohane;
In my experience with bellows at a forge, very little, and at a fireplace, a lot, adding a stream of air increases the temperature of the flame. This can be seen as the fire without the air cannot melt a given metal, but with the stream of air it can. Varying the amount of airflow controls the temperature lower for forging, higher for melting.>>>
As with all real world examples, there’s no yes/no. In every situation, at every air flow, you get a curve and somewhere along that curve, yes, you can within a range on that curve control the heat of the fire. But at the far end of the curve, a large enough gust of air has the same result – fire goes out. Mini example – most people extinguish a match with a quick puff of air.
But getting back to your forge….thanks for yet another fine example. Could you bring metal to the melting point with the same amount of fuel in an open fire? No, of course not, if you could there’d be no need to build a forge. Lots of heat going out by convection alone, and then along comes you pumping that bellows to get the maximum temperature. So with all that heat being pumped by convection PLUS the bellows, which is WAY more than an open fire would lose, it is still WAY HOTTER inside the forge…
’cause the walls of the forge absorb heat from the fire, and as they heat up they radiate heat. Some away to the outside…and some right back into the fire. Which is why it is so much hotter in there than in an open fire.

Joel Shore
March 30, 2011 2:06 pm

RJ says:

We are playing into the alarmists hands if we accept the GHG theory as a fact.

It depends what your goal is. If your goal is just to sow confusion in the general public and hope that this helps keep policymakers from implementing the steps that you oppose, then you may be right…I don’t know. However, if your goal is to be scientifically-correct and to have an impact on the real scientific debate, then you are surely incorrect.
When I see people challenging the basic facts of the greenhouse effect, what it tells me is that they are either misguided or (if they are knowledgeable enough to know better) that they are trying to actively deceive people and confuse them on the science, i.e., their real goal has nothing to do with science and everything to do with pursuing their policy objectives at the expense of science.

Joel Shore
March 30, 2011 2:18 pm

Ira Glickstein says:

When you make a false assumption, you can “prove” false conclusions.

My favorite example of this, attributed to Bertrand Russell is showing that if 0 = 1 then I’m the pope: If 0 = 1, add one to each side to get 1 = 2. Then consider a room containing 2 people, myself and the pope. However, since 1 = 2, it can also be said that the room contains 1 person…and yet it must still be true that it contains me and the pope. Therefore, clearly I must be the pope!

wayne
March 30, 2011 2:44 pm

Steve says:
March 30, 2011 at 1:07 pm
Is “back radiation” really required to explain the greenhouse effect? It seems to me that it just confuses the issue.
With a pure nitrogen atmosphere, almost all infrared radiation would pass through. The molecules of this atmosphere are primarily heated by direct contact with the surface of the sphere, and convection of this thin layer with the upper layers.
Now replace a mass of that nitrogen with a like mass of CO2. Now you have layers of the atmosphere (top to bottom) heated by infrared radiation that was previously not heating any portion of the atmosphere. The atmosphere is warmer – period. No need to talk about whether or not back radiation will heat up the surface or not, because it is obvious that the atmospheric blanket itself is warmer than it was before. (If you really want to make it warmer, exchange H2O for either of the other two gases.)

That’s a great point you raise.
But Steve, sorry, you are wrong. If ignoring the tiny radiation bands in nitrogen, the pure nitrogen atmosphere would over time be very, very hot. During the daytime with no clouds and oceans the atmosphere would heat greatly at the surface and disperse upward by conduction and convection as you said. As the Earth turns and night comes, all convection would ceases and the atmosphere cannot lose any heat by radiation as you said. Day by day the atmosphere would get hotter and hotter until the entire atmosphere would reach almost the same temperature as the daytime soil. The only cooling occurring would be by strictly conduction near the surface at night in contact with a frosty soil for it would be radiating unimpeded.
CO2 and all GHGs are a great fast conductors of heat. It is the over-half portion of radiation upward to space that keeps us from being literally cooked.

March 30, 2011 2:47 pm

Joel Shores simple world consists of an earth without oceans and a land surface that does not retain heat.
Clouds, rain and snow (which give a much more continuous absorption/emission possibilities than gas molecules) apparently do not exist.
The moon comes closer to Joel’s model but even there his simplistic assumptions do not give a realistic account of actual temperatures.
According to Joel the so called 33K “greenhouse effect” is entirely due to CO2 and H2O gaseous molecules.
Perhaps we should leave Joel with his simple fables.
It seems pointless to explain how absurd his conclusions are as he is religiously attached to them.

Ben
March 30, 2011 3:01 pm

Probably already requested, but how about a page that has each graphic and each explanation below it? That way you can focus on the information in each graphic and not be distracted by the constant changing. Thank you.

DocMartyn
March 30, 2011 3:24 pm

where is the low level boundary layer? Pretty much all IR radiation coming from the surface that can be absorbed by CO2 is absorbed by CO2 at ground level. The actual photons that CO2 can absorb beyond this must be generated from the heat in the atmosphere itself.

Reed Coray
March 30, 2011 3:40 pm

Ira Glickstein, PhD says:
March 30, 2011 at 8:13 am
from Wikipedia
If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 °C. However, since the Earth reflects about 30% (or 28%) of the incoming sunlight, the planet’s effective temperature (the temperature of a blackbody that would emit the same amount of radiation) is about −18 or −19 °C, about 33°C below the actual surface temperature of about 14 °C or 15 °C. The mechanism that produces this difference between the actual surface temperature and the effective temperature is due to the atmosphere and is known as the greenhouse effect.
While I certainly do not trust Wikipedia completely, since anyone may post almost anything to obscure topics, I know from personal experience that important scientific topics are monitored by domain experts who protect them from spurious information. Reed Coray, please cite a respected source for your “7ºC” estimate, and same for mkelly regarding your “15 C max”.

I believe the Wikipedia argument is correct. However, it compares apples to oranges. In particular, I’ll assume Wikipedia’s 30% number is correct. However, I then ask: “What part of the Earth reflects 30% of the sun’s energy?” From the papers I’ve read, it’s primarily the clouds and/or particulate matter in the atmosphere, not the surface of the Earth. When quantifying atmospheric effects on Earth surface temperature, shouldn’t you compare the measured Earth surface temperature (i.e., the temperature with an atmosphere) …to… an Earth model that is devoid of an atmosphere? Wikipedia’s model for the Earth surface temperature includes an atmosphere in that it’s the atmosphere that is reflecting 30% of the solar energy. So Wikipedia is comparing measured Earth surface temperatures to a model of Earth surface temperature that includes an atmosphere; and using the temperature difference to claim that the atmosphere raises the Earth surface temperature 33 °C . I call foul. It’s kind of like John Kerry saying he voted against it before he voted for it. (Sorry, I couldn’t resist).
An Earth without an atmosphere will have an albedo much nearer 0 than 0.3. However, assume for a moment that the albedo of an atmosphereless Earth is “x”. Then as I discussed previously, if you’re going to use Planck’s graybody radiation law to determine the rate energy is emitted from that surface, the surface emissivity must be the same as one minus the surface albedo. If true, for an atmosphereless Earth albedo of “x”, the emissivity of that atmosphereless Earth will be “(1-x).” The energy absorbed by the Earth from the sun is equal to the product of (a) the solar energy density (Watts per square meter) at the Earth’s surface, (b) the surface area of the Earth perpendicular to the direction of radiation (pi times the radius of the Earth squared), and (c) one minus the albedo–i.e., (1-x). The rate energy is radiated from the surface of the atmosphereless Earth will be the product of (a) the surface area of the Earth (4 times pi times the radius of the Earth squared), (b) the Stefan-Boltzmann constant, (c) the Earth surface temperature to the fourth power, and (d) the emissivity of the Earth–i.e., (1-x). For steady-state conditions, these two rates are equal. Since the factor “(1-x)” appears on both sides of the Equation, the temperature of an “atmosphereless” Earth is independent of the Earth’s surface albedo (one minus emissivity). Thus, the temperature of an atmosphereless Earth is the same as the temperature of a blackbody Earth–using Wikipedia’s value: 5.3 °C, not -18 ºC or -19 ºC.

March 30, 2011 4:09 pm

“Then as I discussed previously, if you’re going to use Planck’s graybody radiation law to determine the rate energy is emitted from that surface, the surface emissivity must be the same as one minus the surface albedo. ”
Not quite! The surface emissivity FOR VISIBLE LIGHT must be 1-albedo.
The surface albedo FOR IR must be 1.
There is no a priori reason the reflection for visible and IR must be the same.

Steve
March 30, 2011 4:47 pm

wayne says:
March 30, 2011 at 2:44 pm
“But Steve, sorry, you are wrong. If ignoring the tiny radiation bands in nitrogen, the pure nitrogen atmosphere would over time be very, very hot. During the daytime with no clouds and oceans the atmosphere would heat greatly at the surface and disperse upward by conduction and convection as you said. As the Earth turns and night comes, all convection would ceases and the atmosphere cannot lose any heat by radiation as you said.”
I didn’t say that a nitrogen atmosphere cannot lose heat by radiation – is that what you are saying? I said that a nitrogen atmosphere would warm less than a CO2 atmosphere via the radiation emitted from a cooling earth surface. A molecule will radiate at wavelengths according to it’s emission spectrum, which is roughly equivalent to it’s absorption spectrum. Warming via conduction does not prevent cooling via radiation – the nitrogen in our atmosphere (most of the gas in it) cools every night. Are you proposing that the nitrogen gas in our atmosphere cannot cool off unless it collides with another type of atmospheric gas?

Joel Shore
March 30, 2011 5:47 pm

Bryan says:

Joel Shores simple world consists of an earth without oceans and a land surface that does not retain heat.

No…Retention of heat is not relevant to the considerations that I mentioned. The way that it comes into it is that the ocean helps to keep the temperature more uniform, which means that the fourth-root of the average of T^4 is closer to the average of T, so the greenhouse effect is close to 33 K. If oceans were not there, then the temperatures would be more variable (between day and night, seasons, etc.) and the average surface temperature in the absence of the greenhouse effect would tend to be even lower.

Clouds, rain and snow (which give a much more continuous absorption/emission possibilities than gas molecules) apparently do not exist.

According to Joel the so called 33K “greenhouse effect” is entirely due to CO2 and H2O gaseous molecules.

I never said that. Clouds are part of the reason why the atmosphere is not transparent to IR radiation and hence contribute to the greenhouse effect.
Reed Coray says:

Thus, the temperature of an atmosphereless Earth is the same as the temperature of a blackbody Earth–using Wikipedia’s value: 5.3 °C, not -18 ºC or -19 ºC.

Sort of true…although really you are forgetting a couple of things. One is that I believe the albedo of the earth in the absence of clouds is still about 8% or so. The second is that such a colder earth would presumably have more ice on it and would therefore have even a higher albedo than this.
At any rate, the full statement is that the atmosphere (including clouds) produces two effects: One is about 33 K of warming due to the greenhouse effect and the other is some cooling due to the albedo of the clouds. So, yes, the net effect of the atmosphere is not a full 33 K of greenhouse warming. However, again, the sort of calculations you discuss are under the assumption of no change in surface albedo [actually, you wrongly assumed 0 albedo]…In reality, if you removed CO2 from the atmosphere, not only would you remove a lot of the water vapor too, you would also increase the surface albedo of the earth.

Joel Shore
March 30, 2011 5:55 pm

Reed –
On re-reading what you wrote, I realize that you didn’t assume a zero albedo for the earth as I said in my last post but you did assume, as Tim Folkerts points out, that the emissivity is the same in the visible as it is in the far IR…which is indeed not a very good assumption. Emissivities of the earth’s surface in the visible vary widely with the type of surface, whereas emissivities in the far IR tend to be very close to 1 for nearly all surfaces. (I believe I’ve read that the only place that there can be any significant deviations from 1 is in some desert regions.)

wayne
March 30, 2011 6:09 pm

“”””
Steve says:
March 30, 2011 at 4:47 pm
I didn’t say that a nitrogen atmosphere cannot lose heat by radiation – is that what you are saying? “”””
Hi Steve. Yes, I was portraying that on what Phil. said in the prior post. I would agree with that. Nitrogen, except for some small lines near 3 µm, cannot radiate at all. I tend to agree with him though. There are many claiming that all matter radiates when above 0 K as you just said. If all matter radiates, and an atmosphere of pure N2 radiates, then this is no different than what a GHG is portrayed to radiate in all directions and all atmospheres and all molecules always cause a “greenhouse effect”.

Reed Coray
March 30, 2011 6:41 pm

Tim Folkerts says:
March 30, 2011 at 4:09 pm
“Then as I discussed previously, if you’re going to use Planck’s graybody radiation law to determine the rate energy is emitted from that surface, the surface emissivity must be the same as one minus the surface albedo. ”
Not quite! The surface emissivity FOR VISIBLE LIGHT must be 1-albedo.
The surface albedo FOR IR must be 1.
There is no a priori reason the reflection for visible and IR must be the same

I agree. Real-world materials can exhibit emissivities (albedos) that are a function of frequency. For blackbodies, Planck’s radiation law contains a frequency dependent term of the form (f^3)*(df)/[e^(hf/kT) – 1]. When integrated over all frequencies (0 to infinity), the result is radiated power that is proportional to the fourth power of the temperature T. If to account for a different “visible light” emissivity and “IR” emissivity an additional frequency dependent factor g(f) is included in the law, then the integral over all frequency will only be proportional to the temperature to the fourth power if g(f) is constant. The T^4 rule may be a good approximation, but it won’t be theoretically correct. As such, if you’re going to compute the radiated energy using a temperature dependence of T^4, aren’t you implicitly assuming an emissivity (albedo) that is the same at all frequencies? If you’re going to use an emissivity that is a function of frequency, then shouldn’t you specify that dependence?
I’m curious. What do you believe is the primary source of solar reflection: (a) the surface of the Earth, or (b) the Earth’s atmosphere and matter in the atmosphere, or (c) something else? Do you agree or disagree that if you’re going to determine the effect of the atmosphere on Earth surface temperature by (a) treating the measured Earth surface temperature as the “Earth with atmosphere temperature”, and (b) using a model to compute the atmosphereless Earth surface temperature, the model should be entirely devoid of atmospheric effects?
If someone wanted to (a) model the position/frequency dependence of Earth surface absorption (no atmosphere) and the position/frequency dependence of Earth surface emission (no atmosphere), (b) compute the “average” Earth surface temperature using that model, and (c) argue that the effect of the Earth’s atmosphere is the difference between the average measured Earth surface temperature and the model average Earth surface temperature, I would have no objection. But that is not how the 33 degree number is computed. The 33 degree number is based on a reception albedo that, I believe, represents an “atmospheric albedo”, not an Earth surface albedo.

March 30, 2011 7:01 pm

Ira Glickstein, PhD says:
March 30, 2011 at 2:02 pm
“Having made that error, the author continues:
One might think that, because the blackbody is now absorbing more light, even if it is its own infrared light, then it should warm up. …
Yep, it should warm up and (if it is a true blackbody in the longwave region) it will absorb that reflected longwave energy and, all else being equal, will warm up.”
Ira, so when will this cycle of selfheating stop?
Lets slice a blackbody cavity which is at a equilibrium temperature of 288 Kelvin in half and replace the other half with a mirror. How much warmer will this cavity become?

bananabender
March 30, 2011 7:08 pm

Nitrogen can’t radiate heat?
Since when? Everything above absolute zero radiates energy.
Atmospheric nitrogen has kinetic energy and therefore must radiate heat.
The atmosphere is heated by molecular-kinetic effects i.e. molecular collisions. Atmospheric temperature is a function of pressure and density not radiative effects.

Phil.
March 30, 2011 7:21 pm

bananabender says:
March 30, 2011 at 7:08 pm
Nitrogen can’t radiate heat?
Since when? Everything above absolute zero radiates energy.
Atmospheric nitrogen has kinetic energy and therefore must radiate heat.

No that isn’t true for gases, N2 doesn’t have a dipole and therefore doesn’t emit in the IR.

bananabender
March 30, 2011 7:51 pm

davidmhoffer says:
March 30, 2011 at 9:39 am
bananabender;
So….the fresh cool air streaming into the centre of the fire would cool it….convection through the “venturi” is exhausting hot air OUT of the fire…this process sounds like net heating to you? The volatiles evaporate DESPITE the net cooling of the convection process because the combined heat of the various surfaces radiating at each other is high enough to overcome the cooling via convection. Don’t believe me? Build yourself a decent sized fire and turn a fan or leaf blower on it to put 10 or 20 times the air through and around the hole thing than convection could. Watch what happens and report back.
For every phenomenon there is a simple and elegant explanation that is totally and utterly wrong.
Your anecdotes are amusing folklore but they show absolutely no understanding of basic scientific principles.
Have you ever used a bunsen burner? You vary the flame temperature of this burner by altering the amount of air mixing with the gas. The more air you add the more completely the gas burns and the hotter the flames. Fanning a fire is exactly the same principle.
Air is an extremely poor heat conductor. The cooling effect of increased air flow into a fire is far less than the increased heat produced by more efficient combustion.
You can hold your finger 1cm from the side of a bunsen flame and not get burnt. Gases have almost no mass so they radiate very little energy. A bunsen transfers energy directly from the flame by conduction not radiation.
Really? What are these science degrees you possess? Clearly not physics or chemistry since you only “studied” them, and you no more understand how radiation works than you do Scouting 101. Oh yeah, you didn’t have that credential either.
I was employed as chemist in the food and beverage industries after I graduated from university. I regularly used various analytical spectroscopy methods including AAS, GC-MS, spectrophotometry, IR and near IR spectroscopy . Spectroscopy is based entirely on radiative physics.

March 30, 2011 8:14 pm

During the daytime with no clouds and oceans the atmosphere would heat greatly…
No need to talk about whether or not back radiation will heat up the surface or not…
and therefore amplifies the heat at the ground from -18 C up to +15 C….

Apparently, heat = “temperature”
but the mechanisms by which heat is retained are different….
that the temperature of a gas containing a given amount of heat…
The atmosphere and the ocean may retain heat, but the atmospheric heat retention is infinitesimal compared to the heat retention of the oceans….

Apparently, heat = “total thermal energy”
Some would say the word “net” is redundant since heat is usually a macroscopic concept describing the net energy flow….
As most of the heating of the atmosphere is by conduction and convection …
ice reflects a lot more sunlight than ocean or unfrozen land surface thus reducing the amount of solar surface heating…

Apparently, heat = “a flow of energy”
As long as the discussion contains such different ideas about what is perhaps the single most important word in this discussion, I can’t see and hope of resolving anything. We will all be talking past each other as we each try to make out own points using our own understanding of the words.
P.S. In thermodynamics, the 3rd definition (a flow of energy) is the accepted definition. If you see “Q” in an equation, it means how much energy was TRANSFERRED, not how much energy is CONTAINTED. If you see “Q” on a diagram, it is always associated with an arrow, not an object. Heat (and similarly work) are not “state variables” and do not describe the properties of any system.

Steve
March 30, 2011 8:50 pm

wayne says:
March 30, 2011 at 6:09 pm
“Nitrogen, except for some small lines near 3 µm, cannot radiate at all.”
That determines the wavelengths at which it cools, it doesn’t limit the ability to cool off. It will still radiate according to it’s temperature. Nitrogen will radiate a lot of energy in a narrow band, instead of a moderate amount of energy across multiple bands.
“If all matter radiates, and an atmosphere of pure N2 radiates, then this is no different than what a GHG is portrayed to radiate in all directions and all atmospheres and all molecules always cause a “greenhouse effect”.”
The greenhouse effect is not a matter of the gases radiating, but absorbing. The surface of the earth radiates at a much larger variety of wavelengths than some gases can absorb, so some radiation goes directly to space without warming the atmosphere. If surface radiation is absorbed by an atmospheric molecule, it’s going to radiate that energy back out (within it’s emission spectrum), but not until it’s gained some vibrational/stretching/rotational energy in it’s bonds – it has to heat up to cool down.
bananabender says:
March 30, 2011 at 7:08 pm
“The atmosphere is heated by molecular-kinetic effects i.e. molecular collisions. Atmospheric temperature is a function of pressure and density not radiative effects.”
For convection to occur there must be a collision of masses. The vacuum of space doesn’t have enough mass for the upper atmosphere to collide with. The sun’s energy is radiated to the earth, and the earth radiates this energy back out into space as it cools.
Atmospheric pressure is a function of temperature. Of course our atmosphere (as do most) behaves as a near ideal gas. There is no fixed volume – it can only expand against the force of gravity. The outward force of temperature balanced by the downward force of gravity determines the pressure and volume. There is no volume constraint, so changes in temperature generate changes in pressure. Do you think that the night side of earth is exposed to a lower pressure from outer space, so the temperature goes down?

wayne
March 30, 2011 9:07 pm

Steve says:
March 30, 2011 at 4:47 pm
Warming via conduction does not prevent cooling via radiation – the nitrogen in our atmosphere (most of the gas in it) cools every night. Are you proposing that the nitrogen gas in our atmosphere cannot cool off unless it collides with another type of atmospheric gas?

I must admit I have been on both sides of this subject; does our atmosphere radiate as a whole, all molecules included, or only IR active molecules, and both according to temperature. My interests of decades has been in astrophysics but I have never seen a definitive spectrum of purely nitrogen or argon or for pure hydrogen to indicate that those symmetric gases in our galaxy are opaque to IR. Neither have I searched for them. I am like many and have only seen the lab spectrums of nitrogen that show no interaction in those frequencies but the one bump in then near IR.
That is why I was careful to answer back and explain my assumption on that earlier comment.
I have been to countless university sites on the subject and have found totally conflicting reports, usually because the word ‘gases’ is not separated out. So, you could flip me back so easily if you could just show me a spectrum with lines or continuum ranges that are identified to be nitrogen or argon for that matter. I just can’t find any so I stay in line with the apparent evidence, that whatever our atmosphere is radiating, it is only via IR active molecules.
And surely you are not assuming Kirchhoff’s law does not apply in this fine point on a frequency by frequency basis. For if it radiates it must absorb equally. We all know for certain that solids and liquids radiate close to a gray body manner but I’ll ask you, why do think nitrogen gas is always radiating energy away in all directions?

March 30, 2011 9:36 pm

The whole “burning logs” thread seems to have too many variables. Let’s make it more controlled.
Three separate identical vertical logs. The logs are each surrounded by identically-sized tubes a few inches bigger in diameter open at the top & bottom to allow convection.
* Tube 1 is glass, which will not melt, but is mostly transparent to Visible and near IR
* Tube 2 is oxidized metal, with an emissivity near 1.
* Tube 3 is shiny metal, with an emissivity near 0 (ie almost a perfect mirror).
I predict that Log 3 will burn fastest, due to all the reflected photons (which will look like a black body at the temperature of the glowing log) . These reflected photons will help raise the temperature of the log.
I predict that Log 2 will burn next fastest, due to all the emitted photons at the temperature of the tube (which will be cooler than the log, but warmer than the room).
I predict that Log 1 will burn slowest, since there will be very few photons striking it, so it will not get any boost in temperature.

BigWaveDave
March 30, 2011 10:28 pm

BigWaveDave says:
The constant compression of atmosphere by gravity results in higher pressure, density and temperature at the surface. No GHG’s are necessary for this..

Your notion doesn’t even satisfy the 1st Law of Thermodynamics! In the absence of a greenhouse effect, the surface temperature of the earth is set by energy balance considerations. So, unless you are proposing a large source of energy…e.g., that the earth’s atmosphere is undergoing continual gravitational collapse, which would certainly be a novel hypothesis…your notion is completely in violation of known laws of physics.

Gravity is a continual force and gas molecules are in motion, In any process where you pressurize a sufficient amount of gas, its temperature will increase. In processes where the gas is pressurized from one side, the temperature will be highest at the surface where where pressure is highest. This can be demonstrated with a centrifuge, or the leading edges of an aircraft. There’s nothing novel, or in violation of the laws of Physics here.

Trying to argue that the greenhouse effect doesn’t exist just discredits you in the eyes of any serious scientist.

This would be true if you replace “scientist” with “greenhouse hypothesis enthusiast”.

However, in the presence of an IR-absorbing atmosphere, some of the terrestrial radiation is absorbed and when the atmosphere subsequently emits radiation, some of this radiation comes back to the earth. This keeps the earth warmer than it would be in the absence of the IR-absorbing atmosphere.

This violates the second law of Thermodynamics.

When I see people challenging the basic facts of the greenhouse effect, what it tells me is that they are either misguided or (if they are knowledgeable enough to know better) that they are trying to actively deceive people and confuse them on the science, i.e., their real goal has nothing to do with science and everything to do with pursuing their policy objectives at the expense of science.

What you should realize, is that you have no physical explanation of what you believe to be “basic facts”. and that it is highly probable that the misguided one is yourself.

bananabender
March 30, 2011 11:11 pm

From this thread I have discovered that:
– I can heat my house for free by filling the roof space with CO2.
– Pure N2 has a temperature of absolute zero because it can’t radiate energy.
– Chimneys make fires cooler.
– Igloos heat the occupants by radiating “cold” energy.
– Two 50W light bulbs placed together are brighter than one 100W bulb.
– CO2 always absorbs energy at ground level and radiates energy at the edge of space.
– Chemists are all morons who think intramolecular bond stretching and orbital electron excitation are different things.
Sarcasm off.

Steve
March 30, 2011 11:12 pm

wayne says:
March 30, 2011 at 9:07 pm
“My interests of decades has been in astrophysics but I have never seen a definitive spectrum of purely nitrogen or argon or for pure hydrogen to indicate that those symmetric gases in our galaxy are opaque to IR.”
I think you misspoke there. Nitrogen gas is primarily transparent to IR, not opaque, just like it is primarily transparent to visible light. If nitrogen gas were primarily opaque to IR, imagine how difficult it would be to take a thermographic image in an atmosphere that is almost 80% nitrogen gas. It would be like trying to take a photograph in the fog.
“So, you could flip me back so easily if you could just show me a spectrum with lines or continuum ranges that are identified to be nitrogen or argon for that matter.”
In the infrared ranges? I can’t find those either, which is understandable. Typical infrared spectrometers use nitrogen gas as a purge for a very good reason – it doesn’t interfere with the readings. I have found a few articles, unfortunately behind paywalls, that reference it’s absorption/emission spectrum in the far infrared:
http://link.aip.org/link/doi/10.1063/1.431648
http://iopscience.iop.org/0022-3700/10/3/018
http://www.informaworld.com/smpp/content~db=all?content=10.1080/00268978400102221
“Why do think nitrogen gas is always radiating energy away in all directions?”
Because outer space is much colder than the temperature of the atmosphere, and nitrogen gas makes up almost 80% of the atmosphere. I feel it get cold at night, so the heat is going somewhere. Since it doesn’t get continuously hotter every day, the heat must be leaving the atmosphere, right?
You may be interested to read how satellite measurements pertaining to our temperature records are primarily readings in the microwave band, which correlates to heat radiated by oxygen gases (including ozone):
http://en.wikipedia.org/wiki/MSU_temperature_measurements

March 30, 2011 11:16 pm

bananabender says:
March 30, 2011 at 7:51 pm
Your anecdotes are amusing folklore but they show absolutely no understanding of basic scientific principles. >>>
My anecdotes are an attempt to show via real world example how the laws of physics work. The equation to describe the amount of power in watts per square meter (P) that a black body will radiate at a given temperature (T) in degrees Kelvin is as follows:
P = 5.67 * 10^-8 * T^4
If you will take note Mr Banana, there is no variable in that equation pertaining to the temperature of the environment the black body exists in. In other words, the amount of energy being radiated by the black body is 100% governed by its temperature in degrees K. It matters not one whit if it is warmer or colder than its immediate environment, that is the amount of energy that it will radiate. The energy it radiates is in the form of photons, which travel at the speed of light (for purposes of this discussion they are light) and have a mass approaching a limit of zero. They exhibit the characteristics of both particles and waves, and are actually neither and both at the same time. Light exhibits a number of properties that demonstrate it exists physically outside of certain parameters we normally take for granted. Consider as an example a person on a high speed train reading a newspaper by the dome light above. If they were to calculate the speed with which photons emerge from the bulb and travel to the newspaper, the observer would arrive at a speed of exactly 1 times the speed of light. A stationary observer however would not see the light as going “straight down”. The stationary observer would see the light going down and sideways at the same time. If the stationary observer calculates the speed of the photons they will get… 1 times the speed of light. Before you start screaming that this violates simple pythagorean geometry and the laws of motion, please be advised that the speed of light is in fact immutable, the traveller on the high speed train does not pass through time as fast as the stationary observer, and hence they both see the light travelling at the same speed. If you want to argue with me as to this being correct or not, I shall cite one reference only:
Mr Albert Einstein
His theory of relativity provided for the quantification of precisely how much time a body in motion at a given velocity would lose in comparison to a stationary body, and this has been confirmed through experimentation multiple times. In brief, photons ain’t what you seem to think, they radiate from a surface according to the Stefan-Boltzman equation (above) and they don’t give a tinkers damn how fast you think they are going, they only have one speed, and that’s the speed they go until they collide with something, and they will collide with that something no matter what the temperature of that something is. Would you like me to explain Plank’s constant next?
“Have you ever used a bunsen burner? You vary the flame temperature of this burner by altering the amount of air mixing with the gas. The more air you add the more completely the gas burns and the hotter the flames. Fanning a fire is exactly the same principle. “>>>
Well, having discredited yourself as a physicist, chemist, and Scout, you now want to move on to Bunsen burners? Yes, I’ve used one. Many times. If you open the vent at the bottom all the way to let in as much air as possible, the flame will most likely blow out. At best you will get a crackling sound with the bottom of the flame starting about a quarter of an inch above the lip of the burner. Close down the vent until you get a tall silent flame instead. This results in the oxidizing molecules travelling as straight upward in a narrow column as possible in order to concentrate the heat best on the bottom of your flask or other object you happen to be heating. They’re very handy for heating up and bending glass tubing in this manner, but I recommend you keep the first rule of chemistry labs in mind when you try. Hot glass looks exactly the same as cold glass.
“I was employed as chemist in the food and beverage industries after I graduated from university. I regularly used various analytical spectroscopy methods including AAS, GC-MS, spectrophotometry, IR and near IR spectroscopy . Spectroscopy is based entirely on radiative physics.>>>
Congrats. I see you neatly side stepped the question, which I will repeat. You claim to have three degrees. What are they? Not physics or chemistry obviously or you wouldn’t be trying to build your credibility by quoting all the cool instruments you used at a job after university. Guess what? I can drive a car. That doesn’t make me an automotive engineer. I can use a computer, that doesn’t mean I can design one. Operating lab equipment doesn’t provide you with any understanding of the physics the equipment is based on. Unless you actually understand the physics, you’re just following the instructions in the manual and for all you know its just a big, fancy, thermometer.
If you still think that operating a thermometer makes you a physicist, I may be willing to go another round. But at least come up with a tough one that doesn’t make you look foolish in the end, Mr not-a-Scout/chemist/physicist but I operated a spectrometer at work once and have three degrees.

bananabender
March 30, 2011 11:22 pm

BigWaveDave says:
March 30, 2011 at 10:28 pm
Atmospheric pressure is a function of temperature. Of course our atmosphere (as do most) behaves as a near ideal gas. There is no fixed volume – it can only expand against the force of gravity. The outward force of temperature balanced by the downward force of gravity determines the pressure and volume. There is no volume constraint, so changes in temperature generate changes in pressure. Do you think that the night side of earth is exposed to a lower pressure from outer space, so the temperature goes down?
I will rephrase my comment. The atmosphere can be sufficiently heated by molecular kinetic effects without needy any radiation-absorbing “GHGs”. Radiative energy transfer only becomes more important than conduction and convection at the outer edge of the atmosphere.

martin mason
March 31, 2011 2:10 am

I have just read through Nasif’s thread on Jennifer Marohasy’s blog and I haven’t had such an interesting and informative read or learned more about AGW issues in years. It will prompt me to read more too as there it would seem to show a possibility that not only is AGW not significant but that the whole greenhouse effect is not like it is viewed by the majority. The responses from the warmists sound like even they don’t believe it any more.
I’m genuinely sorry that Nasif doesn’t get better coverage, he’s a bit eccentric but an excellent and thought provoking read

March 31, 2011 3:17 am

Ira Glickstein,
Thank you for your efforts to bring out the science of one isolated aspect, which I call the earth’s ‘atmospheric effect theory’. It is one out of many interacting aspects of our earth’s atmosphere. I understand you are doing so when there are many aspects (including the one you are discussing) of our atmospheres which do interact and which are not isolated in the actual complex atmospheric dynamic.
The single isolated aspect (out of many interacting aspects) which you are presenting for discussion was purposely selected and isolated by you to aid in a clear presentation of just the one aspect that I call the earth’s ‘atmospheric effect theory’. I support your efforts to that extent with the understanding that what is not being presented by you is the total net effect (negative or positive) of the single isolated aspect you are discussing on the total complex earth dynamic system.
This latest post is much better, in my view, than the previous posts in your series that used analogy. It is best to just present the science rather than use analogy. Analogies are always limited and have some misleading attributes.
Things I find not at issue wrt the earth’s ‘atmospheric effect theory’ from the so-called consensus focused IPCC science:
a) The gases in our atmosphere interact by collision with each other and in doing so transfer energy.
b) Some of the gases in our atmosphere have the ability to absorb electromagnetic radiations of certain wavelengths. And in certain conditions re-emit it.
c) Some of the gases in our atmosphere have varying interaction with the incoming solar electromagnetic radiation and in doing some energy is transferred to some of the gases.
d) Some of the gases have varying interaction with the blackbody (or near blackbody type) radiation generated by all the surfaces of the earth and in doing so some energy is transferred to some of the gases.
e) Some of the gases have varying interaction with the blackbody (or near blackbody type) radiation generated by the atmospheric gases themselves and in doing so energy is transferred to some of the gases and also sometimes to the earth’s surface.
f) The electromagnetic radiation from the sun and also from the gases in our atmosphere has varying interaction with each of the earth’s various types of surfaces and in doing so some energy is transferred to the surfaces.
g) Some of the electromagnetic radiation from the earth’s surfaces go directly to space.
h) The electromagnetic radiation from all the gases in our atmosphere can, under certain conditions, go directly to space.
Things I find at issue wrt the earth’s ‘atmospheric effect theory’ from the so-called consensus focused IPCC science:
a) I would like to see the formal theory for what I call earth’s ‘atmospheric effect’. Show me the integrated formal theory. Has it been only premised that there is one, not shown that there is one? It cannot be uncoupled from other aspects of the climate system and remain meaningful.
b) I would like to see the actual comprehensive /unbiased /multiple /corroborating observations that demonstrate that a theory of earth’s ‘atmospheric effect’ has a significant net effect on meaningful climate parameters. OK, on average GST too.
Ira Glickstein,
John

wayne
March 31, 2011 3:24 am

@ Steve
I’ll read those links if I can get to them. Thanks.
One thing you might consider is that just because nitrogen by itself does not radiate it doesn’t mean nitrogen can not cool by radiation in the real atmosphere. It will do it through greenhouse gases by re-exiting them by a high enough collision and that greenhouse gas molecule will then radiate the energy away. I have also found that even though most references will show pure nitrogen with only a small band in the near IR and oxygen having bands in the microwave region, they both do radiate a small amount in the mid-infrared just above and below 20 µm due to N2-O2 collision electron interactions together. So, just because a pure nitrogen atmosphere might actually not be able to cool alone, it can once other species of molecules are put in the mix. Just found the paper last week describing that very combo with spectrums included. Well, that’s about where I stand at this point. And that is why I still think a pure nitrogen atmosphere with the same sea level pressure would be hot for sure, you have then taken away all of the ways that nitrogen can possibly lose IR heat. Two month’s ago I would have said exactly the same thing you said.

RJ
March 31, 2011 3:34 am

Martin
Thanks for posting this.
What this GHG theory shows is how people will hold onto their beliefs no matter what. They are not acting as scientists should but hold onto their theory based on faith not science.
The killer for me is that logically the GHG theory does not stand up. It leads to absurd possible outcomes. Like for example heating the house with CO2 in the roof space. Or ice in a Igloo heating the body by backradiation.
Or an oven that heats a chicken without any outside power source.
http://www.slayingtheskydragon.com/en/blog/111-a-pictures-worth-a-1000-words
The GHG theory is flawed IMHO and its time to either move on or at least acknowledge the issues with this theory.

March 31, 2011 4:47 am

Hans asks
…..”Lets slice a blackbody cavity which is at a equilibrium temperature of 288 Kelvin in half and replace the other half with a mirror. How much warmer will this cavity become?”…..
Ira Glickstein replies
…..”With longwave IR it would not be a mirror, but an absorbing/re-emiting thin, black-painted metal shell surface that would return about half the energy it absorbed”…..
My comment to Ira;
This sounds as if you think that IR radiation cannot be reflected!
When I was at school we did experiments to show that all EM radiation including Infra Red radiation exhibit the full range of wave properties of;
Reflection, Refraction, Interference and Diffraction.
Surely you did not mean to imply the opposite to what every school level physics pupil should know?

March 31, 2011 5:20 am

I said
Joel Shore’s simple world consists of an earth without oceans and a land surface that does not retain heat.
Joel replies
No…Retention of heat is not relevant to the considerations that I mentioned. The way that it comes into it is that the ocean helps to keep the temperature more uniform.
My reply;
The Oceans, covering 70% of planet surface, absorb solar radiation.
Some shorter wavelengths penetrate the water to a considerable depth.
In other words the effect is to increase the temperature of the water.
Ocean currents(and tides) distribute the hotter water.
E.g. the Gulf Stream effect keeps the west coast of UK much warmer than without it.
Glasgow is at the same latitude as Moscow.
The atmosphere, in particular clouds, help insulate the Earth surface.
The gross error of IPCC dogma is to attribute a 33K increase effect as due solely to the radiative properties of CO2 and H2O in gaseous form.
The early climate models did not even include clouds although some attempts are now being made to rectify this gross defect.

Brian H
March 31, 2011 5:21 am

Nah. You’ve all got it “wrong on the internet”!
Here’s the Real Deal:
Way back in Hypothetical History, the Earth had no CO2 and was in thermal and radiative equilibrium with Sol. But then one day the Three-Handed God of Gasses wiggled his middle pinkie and ~~BAZOWIE~~ CO2 joined the fun.
This interrupted the normal escape routes and timing of surface radiation in a few LW slots, delaying it by hundreds, or even thousands, of microseconds, and it consequently warmed the air and surface slightly. But then the leading edge of the delayed LW escapee photons made it to the TOA, and fled the trap. Ever since, the surface and air temps have been nice and stable at the new minutely higher temperature.
But, someday, the above-mentioned God of Gasses will wiggle his middle big toe, and ~~EIWOZAB~~ the CO2 will vanish! Then the “excess” will directly blow the joint and temperatures will return to pre-Pinkie levels.
And now you know!
~~~~~~~~~~
We Believers in the Gospel of the God of Gasses are called “The Laggards”! Join us now, even though it may already be too late.
Or not.
😉
Exegesis;
The sequence above is actually slightly more complex and interesting. When the CO2 first appears, a drop in OLW occurs until the lagged photons hit TOA and the exit doors there. The energy they represent, as a function of flux times the lag duration, is the measure of the “excess” which hangs around and keeps temps a smidge above normal. When the CO2 vanishes, that drains away, and so does the temperature anomaly vis-a-vis the Pinky Event.

March 31, 2011 5:38 am

Hans asks
…..”Lets slice a blackbody cavity which is at a equilibrium temperature of 288 Kelvin in half and replace the other half with a mirror. How much warmer will this cavity become?”…..
The cavity will remain at the same temperature! The phrase “at equilibrium” means “the photons in the cavity are at equilibrium with the walls of the cavity, all of which are at 288 K..” Putting a mirror down the center won’t change the temperature of either the photons or the walls. The left half of the cavity and the right half of the cavity will still be at 288 K.

cal
March 31, 2011 5:54 am

Hans says:
March 30, 2011 at 7:01 pm
Ira Glickstein, PhD says:
March 30, 2011 at 2:02 pm
“Having made that error, the author continues:
One might think that, because the blackbody is now absorbing more light, even if it is its own infrared light, then it should warm up. …
Yep, it should warm up and (if it is a true blackbody in the longwave region) it will absorb that reflected longwave energy and, all else being equal, will warm up.”
Ira, so when will this cycle of selfheating stop?
Lets slice a blackbody cavity which is at a equilibrium temperature of 288 Kelvin in half and replace the other half with a mirror. How much warmer will this cavity become?
Good question. But I do not think the answer is what you think it is.
If the total system is at equilibrium with its surroundings ( there is no other source of heat ) then it is at equilibrium and so the nature of the surfaces will have no effect on temperature.
But let us take the more interesting example where the cavity has a power source. It does not matter how big but say it is 100 watts. Assume also that it is in vacuum so that it has no conduction losses and has a highly polished external surface so that it has no external radiation losses. If this is in dynamic equilibrium (that is it is losing heat at a constant rate but has a constant temperature) then the 100 watts will be dissipated through the opening of the cavity with a near black body spectrum characteristic of that constant temperature. Lets say for argument’s sake that this is 800K. If you silver the inside of part of the cavity it will be a slightly poorer black body so its spectrum will not be perfect and and therefore a small rise in temperature will have to occur in order to dissipate the same 100 watts. However the effect will be very small and there is of course no conflict with the conservation of energy.
However an even more interesting experiment would be to take a sphere blackened on the outside with lamp black which, although not a perfect black body, is very close to perfectly black. Now heat this with 100 watts. Let us say that the sphere has a surface area of 100 units. The dissipation is therefore 1 watt per unit area (still in a vacuum so no conduction losses). If the surface area of this sphere was the same as the area of the apperture in the original black body cavity then the temperature would be close to but a little more than 800K because it is not quite perfectly black. Now place a semispherical shell a small distance from one half of the sphere. This shell has a black internal surface and a perfectly silvered outside. Energy dissipation on this side of the sphere is therefore absorbed by the shell but can only be radiated back to the sphere since the silvered external surface cannot radiate. Therefore, if we take the sphere and shell system as a whole, all the 100 watts has to be radiated from the 50 units of surface that is not covered. For this to happen the temperature has to rise to a level where the heat loss per unit area is double. If this temperature is T the fourth power of T is equal to 2 times the fourth power of 800. Thus T is 800 times the fourth root of 2. The shell will also be at this new temperature since the two facing surfaces must be in equilibrium if there is no nett heat flow (no dissiptation from the shell’s external surface).

March 31, 2011 6:03 am

“The killer for me is that logically the GHG theory does not stand up. It leads to absurd possible outcomes. Like for example heating the house with CO2 in the roof space. Or ice in a Igloo heating the body by backradiation. ”
The killer for me is that people could actually concluded that the GH effect predicts that you could heat a house with CO2 in the roof! No competent scientist would say that nor support that – the relative size and relative temperatures are wrong! Only poorly informed skeptics ever seem to come to this flawed conclusion (although poorly informed warmists can make it seem like a plausible conclusion based on their own faulty understanding of the GH effect).
Since you are claiming the GH theory predicts that a house will warm up from C02 in the roof, perhaps you could explain the logic behind that prediction. How specifically would the GH theory lead to the conclusion that you could heat your house (or even limit cooling) simply from a layer of GHGs inside the roof space?

martin mason
March 31, 2011 6:33 am

Tim
Why don’t you go to Jennifer’s blog and read Nasif’s work and the responses. It may surprise you and answer your question. It didn’t actually say you can heat your house with an attic full of CO2 but go and read and be ready to be surprised even.

RJ
March 31, 2011 6:53 am

Tim
I know no competent scientists would say this. But its what scientists seem to be saying with the GHG theory.
CO2 supposedly adds heat to the earth due to backradiation. Why would the same not apply in a house (or if a person was enclosed in a container of CO2).
Or as per the chicken in the oven example above.
Or do you believe CO2 and backradiation can operate at higher levels but not at ground level. If so why not. If CO2 reflects back radiation to increase the temperature of the surface why can this not also apply at ground level.

Steve
March 31, 2011 7:39 am

wayne says:
March 31, 2011 at 3:24 am
“And that is why I still think a pure nitrogen atmosphere with the same sea level pressure would be hot for sure, you have then taken away all of the ways that nitrogen can possibly lose IR heat.”
There is still adiabatic cooling. The nitrogen gas will rise and cool, and then radiate it’s heat away at whatever wavelengths that cooler nitrogen gas radiates at, be it far infrared or microwave.

cal
March 31, 2011 8:17 am

RJ says:
March 31, 2011 at 6:53 am
Tim
I know no competent scientists would say this. But its what scientists seem to be saying with the GHG theory.
CO2 supposedly adds heat to the earth due to backradiation. Why would the same not apply in a house (or if a person was enclosed in a container of CO2).
Or as per the chicken in the oven example above.
Or do you believe CO2 and backradiation can operate at higher levels but not at ground level. If so why not. If CO2 reflects back radiation to increase the temperature of the surface why can this not also apply at ground level.
CO2 in your loft space would be warming but the effect would be tiny since your ceiling and your roof already do a very good job of absorbing radiated heat and re radiating it back. Any downward radiation from CO2 would warm the floor of the loft and upward radiation would warm the roof. Since the CO2 would be at a slightly higher temperature than the roof my guess is that this warming of the floor would be slightly greater than it would have been if it were left to the roof alone to backradiate to the floor. Therefore there would be a slight drop in roof temperature and so a slight drop in total heat lost through the roof but it would be tiny. Silver foil would be more effective!
Furthermore the scale is different. Energy radiated from the earth’s surface in the 14-18 micron band is initially radiated at around 290K and if there were no greenhouse gases this energy would be directly lost to space (just like the energy around the 10 micron band). The absorption by CO2 means that radiation to space can only take place at an altitude where there is little chance of further absorption. This is just below or within the tropopause ( around 8km at the poles and 16km at the equator). The temperature here is about 220K. At this temperature the energy radiated is about 10 times less than the original energy at the surface. This difference has to be made up because the incident radiation from the sun is fixed. There is no option but for the surface to warm in order to increase the radiation at the other wavelengths to compensate.
In the case of your loft there is no radiation into space and even if there were the height of your loft would induce a temperature difference of about 1/50th of a degree.

March 31, 2011 9:10 am

cal says
…”The absorption by CO2 means that radiation to space can only take place at an altitude where there is little chance of further absorption. This is just below or within the tropopause ( around 8km at the poles and 16km at the equator). The temperature here is about 220K. At this temperature the energy radiated is about 10 times less than the original energy at the surface. This difference has to be made up because the incident radiation from the sun is fixed. There is no option but for the surface to warm in order to increase the radiation at the other wavelengths to compensate.”……
Cal has given a good outline of the more sophisticated version of the Greenhouse Theory.(GT)
This is quite a contrast to the cruder; …. atmosphere “toasts” the Earth Surface version favoured by Joel and others.
In this version the famous experiment by R W Wood is not contradicted as the radiative effect of CO2 and H2O can still be very small for that particular volume.
Ira likes analogies and the best one here is of a bath with an small open drain hole being filled with a bigger water supply from the tap.
I’m not saying I agree with this version of GT, but at least its more plausable.

Joel Shore
March 31, 2011 10:33 am

martin mason says:

I have just read through Nasif’s thread on Jennifer Marohasy’s blog and I haven’t had such an interesting and informative read or learned more about AGW issues in years.

You mean the thread that starts out with this statement?

Central to the theory of Anthropogenic Global Warming (AGW) is the assumption that the Earth and every one of its subsystems behaviors as if they were blackbodies, that is their “emissivity” potential is calculated as 1.0. [1]

Since that statement is utterly wrong, I don’t really see why you would get anything out of his post. He then goes on to do a calculation for “total emissivity” of CO2 without even defining what he means by “total emissivity” (of an entire atmosphere’s pathlength?) … or to what wavelength his calculations apply.
If you want to read about the science of calculating radiative transfer in the atmosphere, pick up a book on the subject or at least find a blog post by someone who has a clue what he is talking about.

cal
March 31, 2011 10:41 am

Bryan says:
March 31, 2011 at 9:10 am
cal says
…”The absorption by CO2 means that radiation to space can only take place at an altitude where there is little chance of further absorption. This is just below or within the tropopause ( around 8km at the poles and 16km at the equator). The temperature here is about 220K. At this temperature the energy radiated is about 10 times less than the original energy at the surface. This difference has to be made up because the incident radiation from the sun is fixed. There is no option but for the surface to warm in order to increase the radiation at the other wavelengths to compensate.”……
Cal has given a good outline of the more sophisticated version of the Greenhouse Theory.(GT)
This is quite a contrast to the cruder; …. atmosphere “toasts” the Earth Surface version favoured by Joel and others.
In this version the famous experiment by R W Wood is not contradicted as the radiative effect of CO2 and H2O can still be very small for that particular volume.
Ira likes analogies and the best one here is of a bath with an small open drain hole being filled with a bigger water supply from the tap.
I’m not saying I agree with this version of GT, but at least its more plausable.
————
Thanks Bryan I have never been called sophisticated before!
However I would point out that there is nothing contradictory about the two ways to describe the effect. I prefer this one (but have used both on this thread) because systems with many positive and negative feedbacks are difficult to predict. One can describe the greenhouse effect from bottom up, so to speak, as Ira has done, but it is hard to calculate the nett effect of small changes. However the top down approach (puns definitely intended) allows one to look holistically at the system and apply solid rules like energy balance equations to predict change on the basis of actual measurements. These are available at the macro level such as the temperature and position of the tropopause and radiation at different wavelengths leaving the atmosphere. It is because these measurements are readily available that I am sceptical of the warmists’ claims. The satellite data does not seem to provide evidence of the macro changes theory would predict. I can’t help feeling that if they had such data it would be spread all across the media. My conclusion is that rather than adapt their theory they are desperately trying to come up with an explanation. The issue is not whether the previous 10 doublings have had an effect. The issue is will this one. It is not obvious that it will because the effective level at which CO2 radiates into space is already at the coldest point in the atmosphere. For further reductions in radiation to occur the whole upper atmosphere has to cool. I will wait and see. I am sceptical, but everything is possible. I just think we would have seen at least a trend by now.

Joel Shore
March 31, 2011 10:44 am

Bryan says:

Cal has given a good outline of the more sophisticated version of the Greenhouse Theory.(GT)
This is quite a contrast to the cruder; …. atmosphere “toasts” the Earth Surface version favoured by Joel and others.

I am not in favor of the cruder one. I have often pointed out that the picture described by cal is necessary for a fuller understanding. See for example, this post: http://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/#comment-625057 and in general the discussion we had when I tried to explain to you why Wood’s experiment did not disprove the atmospheric greenhouse effect. I am glad to see you belatedly coming around to me point-of-view.
However, the cruder picture is not without merit for getting some of the basic ideas down. Yes, it neglects some things like convection that eventually should be included in a more refined picture. But, simple models are always limited.

March 31, 2011 11:51 am

“martin mason says: March 31, 2011 at 6:33 am
Tim
Why don’t you go to Jennifer’s blog and read Nasif’s work and the responses. It may surprise you and answer your question. It didn’t actually say you can heat your house with an attic full of CO2 but go and read and be ready to be surprised even.”
Well, the very first paragraph is off to a bad start

Introduction
Central to the theory of Anthropogenic Global Warming (AGW) is the assumption that the Earth and every one of its subsystems behaviors as if they were blackbodies, that is their “emissivity” potential is calculated as 1.0. [1]

http://jennifermarohasy.com/blog/2011/03/total-emissivity-of-the-earth-and-atmospheric-carbon-dioxide/#more-7739
There is no such “central assumption” that “every one of its subsystems behaviors as if they were blackbodies.”
The bulk of the paper deals with a rather involved equation that appears to be some empirical fit to CO2 absorption data. Without access to the original paper, it is very difficult to comment on the accuracy of the equation or Nasif’s application of of the equation.
However, the conclusion states “the potential of the carbon dioxide to absorb and emit radiant energy is negligible”. His calculations clearly contradict various experimental results (http://www.skepticalscience.com/images/infrared_spectrum.jpg or http://brneurosci.org/spectra.png for example) which clearly show CO2 having a clear and significant impact on IR in the atmosphere. I have to go with experimental evidence and reject his calculations.
Furthermore, Nasif’s comments in the discussion following the paper do not particularly impress me.

March 31, 2011 11:56 am

RJ;
The killer for me is that logically the GHG theory does not stand up. It leads to absurd possible outcomes. Like for example heating the house with CO2 in the roof space. Or ice in a Igloo heating the body by backradiation. >>>
What is absurd is taking the explanations, which are based on 200 years of documented physics theory verified by thousands of experiments, and instead of learning how to understand the equations and the physics, instead make wild accusations that are completely out of proportion. Instead of responding with the ludicrous “so why can’t I heat my house by putting CO2 in the attic” as a proper question.
Q; If I put CO2 in my attic, according to the laws of physics, would it heat my house?
A; It would provide an amount of insulation so small that you likely could not measure it without some very expensive and accurate instruments, but it would have some effect compared to a vacuum for example. However it would still be positive compared to vacuum. A foot of fibre glass insulation would have a measurable effect, and by the way, works on the same physics. Does it generate “new” heat? No. It just slows down the escape of IR resulting in a warmer house for a given amount of heating input from the furnace, stove, people, etc. And BTW, if you take the temperature of the fibre glass insulation on a very cold day, you’ll find that it is indeed colder than the ceiling of the house it rests on. Despite which, it keeps the house warmer.
Stop extrapolating actual physics to orders of magnitude higher claims than the physics actually makes, and the absurdity will go away with it.

March 31, 2011 12:04 pm

Joel
I certainly do not believe you ever advanced a version of the greenhouse theory that explained it as being almost totally confined to above the tropopause.
R W Wood and G&T are correct in saying that the radiative effects of CO2 are very small at atmospheric temperatures and pressures.

Richard E Smith
March 31, 2011 12:25 pm

In response to Hans, Ira Glickstein claimed that a radiating body would heat itself up with its own reflection.
“Yep, it should warm up and (if it is a true blackbody in the longwave region) it will absorb that reflected longwave energy and, all else being equal, will warm up.”
Bang go the laws of energy conservation then.
If you believe in back radiation, then try this suggestion of Alan Siddons. Shine a beam of light on a surface which reflects some of the light, such as a wall. Shine the light at 45 degrees so that the part which is reflected bounces off at the same angle. Now take a mirror and position it so that it reflects back to the illuminated part of the surface the light that is reflecting off the surface. You will see that it makes absolutely no difference to the brightness of the surface. (It will, of course, illuminate the wall where it is not already illuminated by the beam.)
Ira claims that the laws of thermodynamics are not broken because the energy exiting from inside his reflective shell is the same as that entering. But inside the shell (which in theory could be as large as a room, or a furnace) you are producing two units of energy out of one. This is energy creation, not conservation.

wayne
March 31, 2011 1:01 pm

@Steve
If the entire pure nitrogen atmosphere could radiate sufficiently to overcome the cloudless sky daytime irradiance transferred by conduction, then you are, or course, correct. Just wish I could get some definitive proof and the answer to why professors in physics and spectrometry at various universities disagree with each other on that very point. Some would say that the ones saying “all matter always radiates if over zero K” are only referring to liquids and solids.
I guess you do know what that would mean if you could prove that. That would mean that not only does nitrogen absorb and radiate adequately by itself but also oxygen and argon at 101325 Pa and there is no special attribute of GHGs at all over the atmospere as a whole, for the entire atmosphere then would be acting in exactly the same manner as they do. It’s just that the spectrums don’t seem to show that, or, possibly that their radiance is just so incredibly tiny and the hot atmosphere I mentioned above is still probably true though a bit cooler than with zero radiance.
But I still would ask you on that hypothetical situation, if I could: If there was zero radiation from a pure nitrogen atmosphere, heck, lets make it pure argon as dense as ours, no oceans, no clouds, would it be cold or hot? What do you think?

Joel Shore
March 31, 2011 1:45 pm

BigWaveDave says:

Gravity is a continual force and gas molecules are in motion, In any process where you pressurize a sufficient amount of gas, its temperature will increase. In processes where the gas is pressurized from one side, the temperature will be highest at the surface where where pressure is highest. This can be demonstrated with a centrifuge, or the leading edges of an aircraft. There’s nothing novel, or in violation of the laws of Physics here.

Yes, when you pressurize gas, it warms. However, if you pressurized some gas and then let it sit in equilibrium with its surroundings for billions of years, it will equilibrate with its surroundings. The earth’s atmosphere is not undergoing continual gravitational collapse.
It is also true that the lapse rate, i.e., the decrease in temperature with height in the troposphere is determined by understanding what would happen if a parcel of gas rises up in the atmosphere and undergoes adiabatic expansion or sinks down and undergoes adiabatic compression. So, yes, the fact that the temperature decreases with height in the troposphere is due to this fact (coupled with the fact that the atmosphere is primarily heated from the bottom…since the adiabatic lapse rate only sets a stability limit on the lapse rate, which is why the temperature actually increases with height in the stratosphere).
However, in the absence of an IR-absorbing atmosphere, the temperature at the surface would be set by the condition that the earth system must be in radiative balance with its surroundings. And, the temperature of the rest of the atmosphere would have to adjust accordingly.

Trying to argue that the greenhouse effect doesn’t exist just discredits you in the eyes of any serious scientist.

This would be true if you replace “scientist” with “greenhouse hypothesis enthusiast”.

That is like me saying that trying to argue that the Earth is only 7000 years old discredits you in the eyes of any serious scientist and you responding that it would be true if I replaced “scientist” with “evolutionist”. All serious scientists understand that the greenhouse effect exists; you’ll notice that Roy Spencer or Richard Lindzen don’t dispute it, for example. It is based on well-understood laws of physics correctly applied.
You may operate under the delusion that there is serious doubt about this…and, by all means, I very much encourage you to raise your lack of belief in the greenhouse effect with any serious scientist you might interact with; it will certainly help that scientist to quickly decide whether to take anything you say seriously.

This violates the second law of Thermodynamics.

No…You just clearly don’t have a clue what the Second Law actually says, as I have explained in posts above. The Second Law is not magic; it is a law deriving from statistical physics.

What you should realize, is that you have no physical explanation of what you believe to be “basic facts”. and that it is highly probable that the misguided one is yourself.

I and many others have endeavored to explain it to you. If you still cannot understand it, I don’t think it is all of our faults.

Joel Shore
March 31, 2011 2:00 pm

Bryan says:

I certainly do not believe you ever advanced a version of the greenhouse theory that explained it as being almost totally confined to above the tropopause.

That is not what cal said. What he said is that the radiation that successfully escapes to space is mainly emitted from close to or a little below the tropopause (because radiation emitted from a lower layer will likely be absorbed again before escaping and higher layers have less emission). The greenhouse effect is not confined above a certain layer; it comes about from a holistic understanding of radiative transfer within an atmosphere where temperature decreases with height.

R W Wood and G&T are correct in saying that the radiative effects of CO2 are very small at atmospheric temperatures and pressures.

I don’t even know what that sentence means. In one sense it is exactly backwards…It is because CO2 absorbs so strongly at atmospheric temperatures and pressures that radiation emitted from near the surface is unlikely to escape to space without being absorbed again. However, if you mean that the fact that the temperatures in the troposphere decrease with height is vital in understanding the greenhouse effect, then this is correct. And, that is where jumping from Wood’s experiment to any conclusion about the atmosphere turns out to be wrong.
Within the simple spherical shell models (which consider only radiative effects), one naturally finds that the temperatures of the shells decrease as one goes outward…i.e., the radiative equilibrium condition itself (coupled with the solar radiation mainly being absorbed at the surface) produces such a result. In the real atmosphere, convection prevents the lapse rate from being too steep…but you still get the temperature decreasing as you go up in the troposphere.

Myrrh
March 31, 2011 2:40 pm

And still the examples of short-wave Light energy heating oceans..

Myrrh
March 31, 2011 4:00 pm

AGW can never understand the objections re the Laws of Thermodynamics because AGWScience claims there is a “net” word added to the 2nd Law is because they have to pretend that Solar energy in the short-waves are absorbed by the Earth and so heat it, that the only Thermal IR in the energy budget comes from the thus heated Earth radiating into the atmosphere. They exclude the Thermal IR coming from the Sun that actually, really, heats the Earth, and us, that we can feel, to teach instead that non-thermal photons of light are in effect thermal, are actually felt as thermal. AGWScience has created another physics.
I bring to your attention this old NASA page for children of the classic physics of Light and Heat energies, that they are different. This page is soon to disappear and the clear, straightforward descriptions will go the way of all AGWBSSscience mangling, so that AGWScience can continue to teach that cold molecules radiating back from the atmosphere heat the earth and that a cold interior wall of an igloo can cook a lump of meat if left there for a few hours, and the rest. So for the educational record, (the exclamation marks are in the original, this is for children):
http://science.hq.nasa.gov/kids/imagers/ems/infrared.html
Infrared light lies between the visible and microwave portions of the electromagnetic spectrum. Infrared light has a range of wavelengths, just like visible light has wavelengths that range from red light to violet. “Near infrared” light is closest in wavelength to visible light and “far infrared” is closer to the microwave region of the electromagnetic spectrum. The longer, far infrared wavelengths are about the size of a pin head and the shorter, near infrared ones are the size of cells, or are microscopic.
Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared. The temperature-sensitive nerve endings in our skin can detect the difference between inside body temperature and outside skin temperature.
Infrared light is even used to heat food sometimes – special lamps that emit thermal infrared waves are often used in fast food restaurants!
Shorter, near infrared waves are not hot at all – in fact you cannot even feel them. These shorter wavelengths are the ones used by your TV’s remote control.”
Because AGW doesn’t teach this, is why there are so many here who can’t understand convection or explanations by others here who do understand the real Thermodynamics Laws, for example:
“If it were possible to get energy to flow from cold to hot then we would have all our energy requirements solved. We would have a perpetual motion machine. This is impossible.” (John Marshall March 29 @1:52 am)
Contrast example: “There is always a net energy flow from a warmer body to a cooler body in accordance with the Laws of Thermodynamics but in fact both bodies still radiate towards each other.
A cooler body doesn’t stop radiating just because it is in the presence of a warmer body.
It is the net rate of energy transfer between the two that changes with no need for the cooler body to effect any direct warming of the warmer body.
the suggestions that the greenhouse effect somehow offends the Laws of Thermodynamics is a non starter and a hindrance to scepticism of the theory of AGW.” (Stephen Wilde March 29 @ 3:45 am)
The AGWScience energy budget, as Ira et al teach it, says that it is Solar energies which heat the Earth, that is, Visible Light and the two short waves either side of UV and Near Infrared. These are not hot. We cannot feel them. No matter how far Blue Light, for example, penetrates into the ocean, it will not heat it. UV may burn surfaces, as artifically intensified Blue light can also burn, but these do not raise the temperature of matter the way that Thermal IR does.
Thermal IR heating the Earth is excluded from the AGW Energy Budget.
Why?
Where is the proof that Solar Light energies can heat the Earth, raise the temperature of the Earth?
It’s Thermal IR that can cook food, Near Infrared is not hot.

Steve
March 31, 2011 4:17 pm

wayne says:
March 31, 2011 at 1:01 pm
“If there was zero radiation from a pure nitrogen atmosphere, heck, lets make it pure argon as dense as ours, no oceans, no clouds, would it be cold or hot? What do you think?”
I’m still trying to understand what you mean. If there is zero radiation across an entire atmosphere, top to bottom, then it is extremely cold – absolute zero. So it wouldn’t even be an atmosphere, but a solid. I don’t think that is what you mean.
If there is zero radiation in a particular wavelength, there are multiple explanations. The most obvious is that your atmospheric molecules don’t emit at the target wavelength. Or the atmosphere might be the wrong temperature, and it would radiate the target wavelength if it was hotter/colder.
You also specify “as dense as ours”, and in a previous comment you specified “same pressure at sea level”. Since the atmosphere will approximately follow ideal gas laws, you have defined a priori the temperature of the atmosphere. Instead of asking a question, you have provided the answer via the assumptions.
The question is “Given an atmosphere of equal mass to the current atmosphere, but consisting entirely of nitrogen (or argon), what would the density and pressure be at sea level? (and thus, the temperature)” Note that in this question “at sea level” refers to a height of a meter or two above the ground. The average pressure exerted by the atmosphere directly upon the ground’s surface at sea level will be unchanged, assuming the total mass of the atmosphere is the same.
My hypothesis is that, given an equal mass of just nitrogen (or argon), the atmosphere will be denser at sea level because it will be colder. The atmospheric pressure at sea level (let’s say 2 meters above surface) will be lower, because the percentage mass of the atmosphere located below knee level will have gone up. The height of the top-of-atmosphere will also be lower.
What I can’t hypothesize is how much denser/colder a pure nitrogen or argon gas atmosphere would be (at sea level) compared to the one we’ve got. And note that the radiation experienced by a person, such as us, standing in direct sunlight would skyrocket during the day and plummet during the night. The surface of the moon, with practically no atmosphere, hits over 115 Celsius at midday and drops below -170 Celsius at night (at it’s equator). Our atmosphere both reflects/absorbs harmful radiation during the day and radiates helpful energy down to the surface at night.

wayne
March 31, 2011 5:24 pm

Thanks Steve, interesting viewpoint. Seems almost every parameter of that is known to a close estimate. That just might deserve a quick simulation of such a scenario to see what that would show.
You and I agree perfectly on the soaring temperatures during the day and frigid nights but I was including much weight on conduction and convection as the prime heater, so much that night would never be able to catch up with limited convection.
The reason I perked up on that is it’s a perfect example to see what a lapse rate would be in such a scenario. Thanks for your viewpoint again. Very interesting.

Myrrh
March 31, 2011 5:50 pm

The Earth’s atmosphere is greenhouse regulated by the water cycle, it would be 67°C without water.