Visualizing the "Greenhouse Effect" – Light and Heat

Guest Post by Ira Glickstein

Solar “light” radiation in = Earth “heat” radiation to Space out! That’s old news to those of us who understand all energy is fungible (may be converted to different forms of energy) and energy/mass is conserved (cannot be created nor destroyed).

My Visualizing series [Physical Analogy, Atmospheric Windows, Emission Spectra, and Molecules/Photons] has garnered almost 2000 comments, mostly positive. I’ve learned a lot from WUWT readers who know more than I do. However, some commenters seem to have been taken in by scientific-sounding objections to the basic science behind the Atmospheric “Greenhouse Effect”. Their objections seemed to add more heat than light to the discussion. This posting is designed to get back to basics and perhaps transform our heated arguments into more enlightened understanding :^)

Solar "light" energy in is equal to Earth "heat" energy out.
[Click on image for larger version]

As I’ve mentioned before, during my long career as a system engineer I’ve worked with many talented mathematical analysts who always provided precise results, mostly correct, but some precisely wrong, usually due to mistaken assumptions. I got into the habit of doing a “back of the envelope” calculation of my own as a “sanity check” on their results. If their results matched within reasonable limits, I accepted them. If not, I investigated further. In those days my analysis was really done using a slide rule and scrap paper, but I now use spreadsheets.

The graphic above is based on an excellent spreadsheet from http://serc.carleton.edu/files/introgeo/models/mathematical/examples/XLPlanck.xls. It uses Planck’s Law to calculate the black body radiation spectrum from the Sun, as observed at the top of the Earth’s Atmosphere. It also may be used to calculate the radiation spectrum from the Earth System (Atmosphere and Surface, see below for explanation) at any assumed temperature. (I will refer to this spreadsheet as “Carleton” in this posting.)

I modified the Carleton spreadsheet to compute the mean Solar radiation per square meter absorbed by the Earth System, which turns out to be 240 Watts/m^2. I then used the spreadsheet to determine the effective mean temperature of the Earth System that would emit an equal amount of energy to Space, and that turned out to be 255 Kelvins (-18ºC which is 1ºF).

Since the mean temperature at the surface of the Earth is 288 Kelvins (+15ºC which is 59ºF), that leaves 33 Kelvins (33ºC which is 58ºF) to be accounted for. Guess how we acount for it?

The yellow curve (above left) shows that Solar radiation is in a tall, narrow “shortwave” range, from about 0.1μm (microns, or millionths of a meter) to about 4μm, which we call ultra-violet, visual, and near-infrared. The vertical axis is Intensity of the radiation, measured in Watts/m^2/μm, and the horizontal axis is Wavelength, measured in μm. If you divide the area under the yellow curve into vertical strips, and add up the total area, you get 240 Watts/m^2.

Since we humans sense the visual portion of this radiation as “light”, that is the name we give it, and that has led to the false assumption that it contains no “heat” (or “thermal”) energy.

The violet curve (above right) shows that, assuming a mean temperature of 255 K, Earth System radiation to Space is in a squat, wide “longwave” range, from about 5μm to beyond 40μm, which we call mid- and far-infrared. If you divide the area under the violet curve into vertical strips, and add up the total area, you get the same 240 Watts/m^2 as is under the yellow curve.

DETAILED EXPLANATION

Left: Actual Solar radiation spectrum observed at top of Atmosphere, compared to black body model. Right: Black body Earth System radiation spectrum out to Space.

The graph on the left shows the actual observed Solar radiation spectrum (in red) as measured at the top of the Atmosphere. It is superimposed on a black body model (in blue) showing very good correlation. Thus, while the Sun is not exactly a black body, it is OK to assume it is for this type of “sanity check” exercise.

If you calculate the area under the curve you get about 1366 Watts/m^2. That means that a square meter of perfect black body material, held perpendicular to the Sun, would absorb 1366 Watts.

However, the Earth is not a perfect black body, neither is it a flat surface perpendicular to the Sun! So, to plot the yellow curve at the top of this posting, I had to adjust that value accordingly. There are two adjustments:

  • The Earth may be approximated as a sphere, with the Sun shining on only half of it at any given time. The adjustment factor for this correction is 0.25.
  • The albedo (reflectiveness) of the Earth system, primarily clouds and light-colored areas on the Surface such as ice, causes some of the Solar radiation to be reflected back out to Space without contributing any energy to the Earth System. The adjustment factor for this correction is 0.7.

After applying these adjustments, the net Solar energy absorbed by the Earth System is 240 Watts/m^2.

The graph on the right shows the black body model for an Earth System at a mean temperature of 255 K, a temperature that results in the same 240 Watts/m^2 being emitted out to Space.

Of course, the Earth System is not a perfect black body, as shown by the graph in the upper panel of the illustration below, which plots actual observations from 20 km looking down. (Adapted from Grant Petty, A First Course in Atmospheric Radiation, Figure 8.2, http://www.sundogpublishing.com/AtmosRad/Excerpts/index.html.)

The actual measured radiation is the dark squiggly curve. Note that it jigs and jags up and down between the topmost dashed curve, which is the black body spectrum for a temperature of 270 K and a lower dashed curve which is the black body spectrum for 230 K. This data was taken over the Arctic, most likely during the daytime. The Petty book also has a graph looking down from over the Tropical Pacific which ranges from 300 K down to 210 K. Observations will vary by tens of degrees from day to night, summer to winter, and Tropical to Polar.

However, it is clear that my result, based on matching 240 Watts/m^2, is within a reasonable range of the true mean temperature of the Earth System as viewed from Space.

NOTE ABOUT THE ABOVE ILLUSTRATION

WUWT readers will notice some apparent inconsistencies in the graphs above. The top and bottom panels, from Petty, peak at 15μm to 20μm, while the purple, blue, and black curves in the middle panel, and the Earth System curves from the Carleton spreadsheet I used (see above) peak in the 9μm to 11μm range. Also, the Petty black body curves peak at a “Radiance” around 100 mW/m^2/sr cm^-1 while the black body curves from Carleton peak at an “Intensity” of around 14 W/m^2/μm. Furthermore, if you look closely at the Petty curves, the labels on the black body curves are mirror image! What is going on?

Well, I know some of the reasons, but not all. (I hope commenters who are more fluent in this than I am will confirm my explanations and provide more information about the differences between “Radiance” and “Intensity”.) I have Googled and Wikied the Internet and am still somewhat confused. Here is what I know:

  • The horizontal axis in Petty’s plots are what he calls “Wavenumber”, increasing from left to right, which is the number of waves that fit into a cm (centimeter, one hundredth of a meter).
  • This is proportional to the frequency of the radiation, and the frequency is the inverse of the wavelength. Thus, his plots are the mirror image of plots based on wavelength increasing from left to right.
  • The spreadsheet I used, and my previous experience with visual, and near-, mid-, and far-IR as used in military systems, always uses wavelength increasing from left to right.
  • So, when I constructed the above illustration, I reversed Petty’s curves, which explains why the labels on the black body curves are mirror image.
  • Fortunately, Petty also included a wavelength legend, which I faithfully reproduced, in non-mirror image, at the top of each plot.

But, that still does not explain why the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet and other graphics on the Internet. I tried to reproduce Petty’s blackbody curves by multiplying the Carleton values by the wavelength (μm) and that did not move the peak to the right enough. So, I multiplied by the wavelength again (μm^2) and, voila, the peaks agreed! (I hope some WUWT reader will explain why the Petty graphs have this perverse effect. advTHANKSance!)

ANSWERING THE OBJECTIONS TO BASIC ATMOSPHERIC “GREENHOUSE EFFECT” SCIENCE

First of all, let me be clear where I am coming from. I’m a Lukewarmer-Skeptic who accepts that H2O, CO2 and other so-called “greenhouse gases” in the Atmosphere do cause the mean temperature of the Earth Surface and Atmosphere to be higher than they would be if everything was the same (Solar radiation, Earth System Albedo, …) but the Atmosphere was pure nitrogen. The main scientific question for me, is how much does the increase in human-caused CO2 and human-caused albedo reduction increase the mean temperature above what it would be with natural cycles and processes? My answer is “not much”, because perhaps 0.1ºC to 0.2ºC of the supposed 0.8ºC increase since 1880 is due to human activities. The rest is due to natural cycles and processes over which we humans have no control. The main public policy question for me, is how much should we (society) do about it? Again, my answer is “not much”, because the effect is small and a limited increase in temperatures and CO2 may turn out to have a net benefit.

So, my motivation for this Visualizing series is not to add to the Alarmist “the sky is falling” panic, but rather to help my fellow Skeptics avoid the natural temptation to fall into an “equal and opposite” falsehood, which some of those on my side, who I call “Disbelievers”, do when they fail to acknowledge the basic facts of the role of H2O and CO2 and other gases in helping to keep temperatures in a livable range.

Objection #1: Visual and near-visual radiation is merely “light” which lacks the “quality” or “oomph” to impart warmth to objects upon which it happens to fall.

Answer #1: A NASA webpage targeted at children is sometimes cited because they say the near-IR beam from a TV remote control is not warm to the touch. Of course, that is not because it is near-visual radiation, but rather because it is very low power. All energy is fungible, and can be changed from one form to another. Thus, the 240 Watts/m^2 of visible and near-visible Solar energy that reaches and is absorbed by the Earth System, has the effect of warming the Earth System exactly as much as an equal number of Watts/m^2 of “thermal” mid- and far-IR radiation.

Objection #2: The Atmosphere, which is cooler than the Earth Surface, cannot warm the Earth Surface.

Answer #2: The Second law of Thermodynamics is often cited as the source of this falsehood. The correct interpretation is that the Second Law refers to net warming, which can only pass from the warmer to the cooler object. The back-radiation from the Atmosphere to the Earth Surface has been measured (see lower panel in the above illustration). All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.

Objection #3: The Atmospheric “Greenhouse Effect” is fictional. A glass greenhouse works mainly by preventing or reducing convection and the Atmosphere does not work that way at all.

Answer #3: I always try to put “scare quotes” around the word “greenhouse” unless referring to the glass variety because the term is misleading. Yes, a glass greenhouse works by restricting convection, and the fact that glass passes shortwave radiation and not longwave makes only a minor contribution. Thus, I agree it is unfortunate that the established term for the Atmospheric warming effect is a bit of a misnomer. However, we are stuck with it. But, enough of semantics. Notice that the Earth System mean temperature I had to use to provide 240 Watts/m^2 of radiation to Space to balance the input absorbed from by the Earth System from the Sun was 255 K. However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.

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JT
May 7, 2011 8:34 pm

Ira, the basic bookkeeping of the simplest greenhouse calculations assumes that the re-emission of infra-red radiation by the greenhouse gas molecules is spatially symmetrical. I am well aware that spontaneous emission is equally likely to be in any direction so that approximately 1/2 will head downward to the surface. However, there is another kind of emission – stimulated emission – and it is massively biased in the same direction of travel as the direction of travel of the stimulating photon. Thats what creates laser light, but stimulated emission can occur without lasing. IF there were any significant amount of stimulated emission happening in the atmosphere then the basic 1/2 back down calculation would likely be wrong because most of the IR through the atmosphere comes from the ground up and, if it is stimulating atmospheric emission, would enhance outgoing IR. Has anyone ever looked for evidence of stimulated emission of outgoing IR in the atmosphere?

Cherry Pick
May 7, 2011 8:38 pm

Let’s add two more objections to this partial visualization of the Earth System.
1. Where are the storages? Oceans store and release huge amounts of energy. Elaborate your views of the energy/radiation balance.
2. Energy is not just radiation. Earth System is more complex than that. For example, absorbed radiation energy may result in increased cloudiness that has impact on who much radiation reaches the earth. Ignoring major part of the system does not help in believing your conclusions.

charles nelson
May 7, 2011 8:58 pm

CO2 vs Earth’s Atmosphere Screen Saver!
1. Find out how many pixels there are on your computer screen. Using basic math calculate the proportional size in pixels of an image that represents 380 and 400 ppm.
For example if your screen had exactly 1million pixels the image would be exactly 380 or 400.
2. (Sticking with the above example) Go to Google images and search for images that are…380…400..or 20 pixels. Download and place in the centre of an empty screen.
3. Show the images to people who Believe and ask them if they still think that CO2 is anything other than a trace gas and ask them if the eensy teensy baby image; 20 pixels (the increase in CO2) looks sufficient to destabilize the other 999,980 parts of the atmosphere!

May 7, 2011 9:15 pm

This posting is designed to get back to basics and perhaps transform our heated arguments into more enlightened understanding :^)
A little humour goes a long way… Nice, Ira!

martin mason
May 7, 2011 9:23 pm

Ira, the Second Law of thermodynamics works at every level and you cannot transfer heat from a colder body to a warmer body. A black body cannot also reabsorb and re-emit lower intensity radiation that it has already emited. I’d say that it is also possible to arrive at the temperature difference between the black body radiating temperature and the average surface temperature without any recourse to radiant heating (adiabatic compression by gravity) and same for the surface temperature on the surface of venus. Greenhouse heating by back radiation is surely wrong because it says that if you put a frozen steak inside a vacuum flask with reflective interior that the steak would cook, that if you stand in front of a mirror you will heat up from the reflected rays or that you can save on your heating bills by filling the loft with CO2?
I agree that we can shoot down the dire predictions of the AGW whilst acknowledging that the basic greenhouse theory is correct but there are many now refuting that there is any greenhouse effect.

May 7, 2011 9:33 pm

It should be pointed out that the 33K difference referred to is the effect of the heat-trapping gases in the atmosphere. In the case of each gas, the first small amount has the greatest effect, with diminishing effect as more is added, as according to Beer’s Law. For CO2, it’s effect is 90-95% spent and, thus, the hypothetical doubling of atmospheric CO2 would have little effect, possibly 0.10–0.01 K.
It should also be pointed out that, as CO2 partitions 50 to 1 into the oceans, we would have to emit 50 times more CO2 that required to simply double atmospheric CO2. There is not enough carbon available to do this. If we really tried, we might be able to do 20%.
So, where is the majority of atmospheric CO2 coming from? Outgassing from the oceans with warming, as according to Henry’s Law. And, of course, with cooling, the oceans will soak up CO2. There is a lag period before increases reverse to decreases, but with 30–50 years of cooling ahead, we will be able to see it happen.

Jeff Norman
May 7, 2011 9:46 pm

Ira is just doing a gross energy balance, so it seems to me that JT’s and Cherry Pick’s objections all occur within the black box of the Earth-black-body emitter.
I don’t like that Ira treats albedo as a constant. That sounds like a fudge factor you would throw into a GCM.
“The adjustment factor for this correction is 0.7.”
What is that? 70%? I don’t think so but I wait to be illuminated after Ira reflects on this.
How much heat energy gets transferred to the Earth black box by the electromagnetic interactions between the Sun and the Earth?
240 Watts/m^2. What’s the error margin on that for the Sun (solar “constant” anyone?) and for the Earth? Seems you could hide a great deal of heat in there.

Henry
May 7, 2011 9:50 pm

3. Show the images to people who Believe and ask them if they still think that CO2 is anything other than a trace gas and ask them if the eensy teensy baby image; 20 pixels (the increase in CO2) looks sufficient to destabilize the other 999,980 parts of the atmosphere!
Yes! and then replace “CO2” with Cyanide and “atmosphere” with your body. If they don’t believe you, just ingest 2mg of cyanide for each kilogram of your body weight (1:500,000) and show those iteyucktuals what for!

May 7, 2011 10:02 pm

Ira; The surface air temperature is higher then theoretical because of the insulative mass of the air of the atmosphere. Just like the insulation in the wall of a cooler or a kiln.
Energy out equals energy in. The air is heated at the ground level and looses energy to space. So the air at ground level has more energy per unit then the theoretical black body temperature. The thicker and less dense the insulation the slower the energy flow out. The thinner more dense the insulation (atmosphere) the faster energy lose.
The solar radiation incoming is of wave lengths that are mostly transparent to the atmosphere and heat the oceans and land. Thermal energy as heat and that carried in water vapor is transported through the atmosphere towards space. The surface air temperature is just a single point of measurement in this energy flow. Solar radiation that energizes the atmospheric constituents change the density and thickness of the insulation and therefore the rate of energy flow to space. Ultraviolet and shorter wavelengths are more effective at energizing gas molecules to effect changes in the insulative value of the atmosphere. pg

Ted Dooley
May 7, 2011 10:06 pm

The basis for the adjustments to the total absorption due to geometric and reflectivity are based on???

Gnomish
May 7, 2011 10:10 pm

“The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
aren’t you overlooking the obvious? ever fry an egg on the sidewalk from back radiation of the atmosphere?

May 7, 2011 10:15 pm

One often meets the assertions that (a) the sea in polar and sub-polar regions is commonly several degrees warmer than the air above it; and (b) in these regions, incoming solar energy has heated the air which in turn heats the sea.
Please expand?

May 7, 2011 10:17 pm

I still have issues with down welling or back radiation in the troposphere. It needlessly complicates the simple fact that clouds, air have a temperature that varies. Increased CO2 can change that temperature but doesn’t beam back little phaser blasts of photons. The layering of temperatures in the atmosphere just complicates the radiative cooling like walls of differing layers of insulation thickness. Without the cutesy back radiation into a warmer surface invention, it might be easier to explain the potential radiative impact of CO2 and with a better model of the atmospheric layers, understand why it is not living up to its potential. Other than that, nice post.

Mooloo
May 7, 2011 10:27 pm

Show the images to people who Believe and ask them if they still think that CO2 is anything other than a trace gas
Remove that “trace” gas and all photosynthetic life of earth dies. And us with it.
Small concentrations can have powerful effects.

gbaikie
May 7, 2011 10:28 pm

“I modified the Carleton spreadsheet to compute the mean Solar radiation per square meter absorbed by the Earth System, which turns out to be 240 Watts/m^2. I then used the spreadsheet to determine the effective mean temperature of the Earth System that would emit an equal amount of energy to Space, and that turned out to be 255 Kelvins (-18ºC which is 1ºF).”
When you say absorbed by earth system, this mostly means amount the surface warms up and the atmosphere itself is only absorbing a small percentage of the incoming solar flux?
And assuming this is the case wouldn’t different material absorb different amounts. Or has this been simplified to mostly account for how much the oceans absorb energy, and since ocean absorb the vast majority of energy that reaches the earth surface, anything else can be mostly be ignored?

Günter Heß
May 7, 2011 10:37 pm

Ira,
Thanks for your visualization.
I’d like to comment on your sentence and adress some of the comments here:
“The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
To start, here is my view that I already posted on Climate etc awhile ago..
As I understand it: “backradiation” is the downwelling infrared radiation integrated over all directions that go into the surface.
I am an experimentalist. For me a physical cause is something that I can switch on and off, at least in a thought experiment, and get an intended effect.
If I switch off the sun in a thought experiment, I get cooling of the atmosphere and the surface. If I switch on the sun again, I get warming of the atmosphere and the surface.
If I switch off ”back radiation” in a thought experiment on the night side I get a higher cooling rate.
If I switch on “backradiation” on the night side I get a lower cooling rate.
If I switch on “backradiation” on the day side I get a higher heating rate. Combining my thought experiments I deduct that if I switch on “back radiation” on the day side I get a higher heating rate due to a lower cooling rate because of the “back radiation”
Moreover, I can deduct that with “back radiation” the earth system gets or stays warmer than a reference system without “back radiation”, because of a lower cooling rate.
One might also say “back radiation” leads to a warmer surface compared with a reference system.
However, I do consider sentences as incorrect with respect to physics that state only: “back radiation” is warming the surface.
Of course in any private conversation we state things like: my new coat is warming me nicely. But in a scientific discussion, we need to employ the correct wording, especially in lectures or papers.
For my opinion sentences like: “back radiation” is warming the surface
cause therefore a lot of misunderstanding in the blogosphere. The Gerlich and Tscheuschner discussion is based on it.
Thanks again for your article.
Best regards
Günter

dr.bill
May 7, 2011 10:38 pm

Ira, the radiance formula involves either a “df” or “dλ” at the end. Converting from frequency to wavelength thus involves replacing df by -cdλ/λ², which explains why multiplying the frequency data by λ² puts the peak back in the right place.
This same issue arises in Wien’s Law. The “peak color” is different, depending on whether you use wavelength or frequency as a variable. Intensity, however, is integrated over all colors, and gives the same result for either variable.
/dr.bill

Dishman
May 7, 2011 10:54 pm

In re Objection #2:
Gas molecules follow ballistic trajectories between collisions…
… unless all the work on Gravity since Galileo is wrong, and doesn’t apply at the molecular level, of course.

Mike McMillan
May 7, 2011 10:57 pm

Would it make any difference if we doubled the average insolation for half a day, then dropped it to zero for the other half (we’ll call that half night)? We’re radiating outward the entire time, day and night. Messes up the simplicity of the model, but since the temperature varies on a 24 hour clock, the outward radiation does, too.

Steeptown
May 7, 2011 11:14 pm

“As I’ve mentioned before, during my long career as a system engineer I’ve worked with many talented mathematical analysts who always provided precise results, mostly correct, but some precisely wrong, usually due to mistaken assumptions. I got into the habit of doing a “back of the envelope” calculation of my own as a “sanity check” on their results. If their results matched within reasonable limits, I accepted them. If not, I investigated further. In those days my analysis was really done using a slide rule and scrap paper, but I now use spreadsheets.”
During my long career as a nuclear engineer, I got into the habit of doing a “back of the envelope” calculation of my own as a “sanity check” on other’s results. If their results matched within reasonable limits, I accepted them. If not, I investigated further. I also encouraged new engineers to do the same.

RJ
May 7, 2011 11:51 pm

What happened to your previous thread. It was getting interesting until no more comments could be posted
http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/
This above thread convinced me even more that backradiation does not further heat the planet. Luke warmers are as scientifically wrong as alarmists if they believe this
The average temperature might slightly increase due to more CO2. But the highest temperature is due entirely to the Sun. In no way can CO2 and backradiation further heat the surface. It can only slow the cooling rate and by so doing might raise the average temperatures
But a colder object can never further heat a warmer one. And saying this time and time again will not make it true. So this objection is correct IMHO.
“Objection #2: The Atmosphere, which is cooler than the Earth Surface, cannot warm the Earth Surface”.

Gilles
May 7, 2011 11:51 pm

I’d like to comment the “However, it is clear that my result, based on matching 240 Watts/m^2, is within a reasonable range of the true mean temperature of the Earth System as viewed from Space.”
Actually for an inhomogeneous body, and even more if it is not a perfect black body, there is nothing like “the true mean temperature”. You can define different averages, but not a single one; the important thing (never clearly stated by climate scientists, although they probably know it very well), is that the average surface temperature (the integral of TdS divded by the surface), is different from the effective temperature Teff which is such that Prad = S.sigma.Teff^4 , where sigma is the Stefan constant. And these temperatures would also be different from a “best fit Black Body temperature”.
The important thing is that you can easily make one of the temperature vary , holding another one constant ! for instance, the average temperature can vary without varying the effective temperature, that is with the same energy budget – and conversely. This can be demonstrated very simply by noticing that the local temperature can be written as +∆T where is the average surface temperature and ∆T a surface (not temporal) “anomaly”. But the only condition is that = 0. Any repartition with a vanishing surface average gives the same average temperature, but NOT the same effective temperature, since on average the won’t vanish.
This lets a lot of room for spontaneous variation even with no change of forcings. This is carefully hidden under the name of “unforced variability”. After all during El Niño/La Niña cycles, the average temperature changes by several tenths of °C (= several decades of observed trend ! ) without any significant change in the energy budget.
The gospel of climate science is that the unforced variability is restricted to 30 years, which would mean that its power spectrum is cut-off for longer variations. But it can be easily seen that this cannot be true. Natural variability is necessary to explain that long term, secular variations, at century and millenium scales. If it were restricted to 30 years, for instance, the paleoclimate data should show that the average temperature should follow very closely the forcing (Milankovitch) curve, which is obviously not the case. And there are obvious physical reason why the Earth climate could oscillate naturally on millenium scale – for instance that’s the characteristic timescale of thermo-haline circulation and it can be expected to give cycles at this frequency. This is much overlooked by climate scientists, in my sense.

Gilles
May 7, 2011 11:52 pm

(sorry for the italic tags, can you fix it Anthony. Sorry also for the possible bad english, that’s not my mother language 🙂 ).

Ronaldo
May 7, 2011 11:54 pm

Ira
Congratulations, you have provided a good, visual explanation of the basic energy inputs and outputs for the earth. The devil is, of course, as I am sure you would agree, in the detail.
When calculating surface temperatures, variations in the assumed constants in the equations need to be considered. To me the most obvious ‘variable’ constant is that assumed for the earth’s albedo. Spatial and temporal variations due, amongst other things, to the ability of H2O to exist as all three of its phases within the surface and atmospheric regions of the earth, coupled with the large amounts of energy emitted or absorbed at the transition temperatures result in variations in cloud and ice cover, affecting both incoming and outgoing radiation and provide major feedbacks.
When these are coupled with the vast energy storage capabilities of the oceans and more subtle long-term variations of energy input imposed on the earth and solar system from ‘external’ sources known and unknown, the prediction of the earth’s surface temperature variation over an extended period is not within our present capabilities.
In my view the attempts by the powers that be to convince us that the science is settled are the only significant source of ‘hot air’ in the debate!

Alcheson
May 7, 2011 11:54 pm

“How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
Since the earth’s core is molten (very hot), there must also be convection of heat from the core to the surface. How much of the extra 33K heat can be attributed to earth’s own heat source? Is it none?… a little?…. or perhaps a significant amount of the 33K? Just wondering.

Gilles
May 7, 2011 11:54 pm

grrr I should’nt use > for average. The correct thing is that
T = Taverage + ∆T
with the condition ∆Taverage =0

wayne
May 7, 2011 11:56 pm

Mike McMillan, you seem to need something like these:
Here are a couple of energy budgets using KT97 and TFK09′s data for the overall averages of reflection (albedo) per the papers, irradiance at the surface, and measured TOA LW upwelling radiation. These are somewhat rough, a first stab at a cosine weighted across 24 hour view but they give you a closer idea of what actually occurs with Trenberth and Kiehls’s numbers. You will notice the totals all match what you see on their graphics.
http://i56.tinypic.com/avc5g.jpg : KT97
http://i53.tinypic.com/ir6lie.jpg : TFK09
They need some small adjustment I can see already:
• Need a better estimate of the actual evaporation rate at nighttime compared to evaporation in the daytime.
• The exact time that the minimum and maximum temperatures are reached, on the average globally, might need to be moved 30 minutes or one hour sooner.
• Window radiation is strictly in those spreads gauged by the surface temperature and I don’t know if this is strictly correct since the cloud layer over 62% of the earth needs to be considered, but, the average should be close.
• …
These are built upon what is actually happening in our atmosphere. You will notice the lack of the mention of back radiation for back radiation is a mirage. A figment of you imagination brought to life by the way the Stefan-Boltzmann equation is applied. In a layered gravity held atmosphere there is no real backward movement of net radiation or any other energy for that matter of fact. When the Stefan-Boltzmann equation is applied correctly to each thin layer you will find this to be true.
Also, radiation in the CO2 and H2O main bands which lie outside the radiative window frequencies cannot pass downward to any great distance to any great degree for the atmosphere is nearly completely “black” to these frequencies. Another way to put it is the mean free path (mean distance a radiation will travel before being absorbed) is simply to short in the lower troposphere to allow this imaginary backward propagation from high in the atmosphere to the surface to ever really occur. Both of these GHGs are simply fast and longer reaching conduction of parcels of energy just like conduction but conduction has to move molecule to molecule, this radiation can jump many meters. Above the tropopause these limits are no longer present enhancing the movement of radiation to space.
So, that is why the familiar 390 & 396 W/m^2 huge red arrows of upward LW radiation and the 324 & 333 W/m^2 of downward IR radiation are missing. They do not belong there. However, they are real and can be measured by radiometers at the surface but this radiation that is being measured there is in the bottom few hundred meters or so except within the window frequencies. It is local, very local as to compared to vastness of the atmosphere. They deserve to be mentioned on those famous budget diagrams but would have more honestly portrayed as a footnote.
Mike, fell free to use these to make you contention a bit stronger.

May 8, 2011 12:04 am

Alcheson;
Since the earth’s core is molten (very hot), there must also be convection of heat from the core to the surface. How much of the extra 33K heat can be attributed to earth’s own heat source? Is it none?… a little?…. or perhaps a significant amount of the 33K?>>>
When I first became interested in climate science that was one of my questions. After much reading and research and my own back of the envelope calculations, the answer came out “insignificant”. Same for radioactive decay of elements in the earth, and I even calculated the energy released by burning of fossil fuels… in the end, all insignificant. Solar rules.

Mike McMillan
May 8, 2011 12:16 am

wayne says:May 7, 2011 at 11:56 pm . . .
Thanks, Wayne. Seems to me the atmosphere is more of a blanket than greenhouse, just slowing the outward movement of energy. The temperature rises down here to the point where the increased outward movement once again roughly equals the insolation, not that they’re ever exactly equal.

dr no
May 8, 2011 12:19 am

“The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
Has anyone actually calculated how much energy can be absorbed by the GHG molecules in the air? Radiation is not the only heat transfer mechanism.
I have difficulty rationalizing how 2% of the molecules in the air (H2O and CO2) can heat themselves, the other 98% of the air molecules, and the earth surface, by 33K only through back radiation of IR. The GHGs absorb and emit IR but the other air molecules can’t so the heat transfer has to be through collisions. What is the probability of a non-GHG molecule to collide with an excited GHG molecule compared to the probability of colliding with the earth surface and have direct energy transfer from there? (Equally reasonable, we could have “back-collisions” as well, when an air molecule with higher energy than the earth surface, transfer its excess energy back when colliding). If the energy can be inhibited from instantly going out into space by being absorbed by 2% air molecules then it must also be inhibited by being absorbed by e.g. oceans, and released at a later moment. (Some of the energy is also stored in vegetation through the photosynthesis which uses energy from the sun to convert H2O and CO2 into oxygen and cellulose, but that is a longer time perspective.) As has been pointed out in previous postings, we can’t gain more energy than we receive from the sun, but we can delay its disappearance so the 33K energy that is missing has to originate from the sun anyway. Earth is not at instant equilibrium (never at equilibrium at all), and there are many variables that can affect the energy budget at any instant.

gbaikie
May 8, 2011 12:23 am

“However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
Suppose you cover all the ocean of the world with plywood floating “floor”.
You can paint the plywood any color- a nice blue, perhaps.
Would this cause any significant change of global temperature?
And would average global temperature increase or decrease?
This “plywood ocean” could be hotter [though depends on type of paint- not necessarily on simply what color pigment it has] during the day. Since we are not sealing the ocean just covering it, the ocean water would still have some evaporation, but still the nite should be cooler.
Of course such thing would cause a massive extinct event and have all kinds wild effects.
But the point is, would it warm the planet or cool it?
And my answer is it would cool it, significantly.
It could increase global average daytime temperatures- especially if you included the area of the plywood oceans and have somewhat lower global average nite time temperatures. But the northern hemisphere [where most earth land mass is] would get colder- mostly nite time temperature. And other land masses would get colder, and the ocean [beneath the plywood] would get colder.
And looking from space and measuring earth’s radiation would indicate that earth has lost a significant part of it’s “greenhouse affect”
[And if painted the “plywood ocean” black, or put asphalt on it, the earth would still cool.]

May 8, 2011 12:26 am

Ira,
Very detailed explanation. Hard to argue with the fact that when one breaks the data down into that much detail, one gets only one possible conclusion. Either that or all the physics we know that engineers use every day to make planes fly, internal combustion engines that work, air conditioners that cool, furnaces that heat, are wrong. Since the planes they design do fly, the engines work as expected, the air conditioners cool exactly as designed and furnaces heat right to specification, either the physics is right or a few billion calculations based on the physics got it right by pure coincidence. I’ll put my money on the physics!
That said, you can break this down to a much simpler calculation. The sun’s radiance doesn’t get absorbed 100% by earth surface. Some of it is absorbed by the atmosphere. Similarly, the earth’s radiance is not released 100% from the earth’s surface, some of it is released by the atmosphere itself. So the earth surface might be 288K, but as seen from space, any given outward bound photon could have come from the earth’s surface or from the atmosphere. Seen from space, earth is not a “surface” but a sort of fuzzy/murky/semi-transparent sphere. Take any “area” that you can “see” from space, and measure the w/m2 being emitted, and you get a total that is mostly originating from the surface, and some that originated from the atmosphere.
If we assume that your 240 w/m2 is correct, we can use Stefan-Boltzman Law to calculate the “effective” black body temperature of the earth, which if you accept “average” as an argument, would occur not at earth surface, but somewhere between earth surface and the top of the atmosphere.
SB Law being P=5.67*10^-8*T^4
If we use P of 240 watts/m2 we get T of 255.07 degrees K or -17.93 degrees C.
The same logic still applies. Since we can measure what’s going into the system at TOA (top of atmosphere) as well as what is coming out, and the various satellite measurements and experiments like ERBE and others all arrive at a MEASURED value of arounf 240 w/m2, then there is only one possible conclusion. Earth surface is 288K or +15C on average because the atmosphere retains heat that keeps the surface warmer than it otherwise would be. Call it “less cooling” or “back radiation” or “invisible pajamas” but the measurements and the math arrive at the same conclusion. 240 watts/m2 going in =240 watts/m2 going out = -18C. Since the earth surface is +15C, it can only be 33 degrees warmer because the invisible pajamas are making it warmer.

May 8, 2011 12:33 am

Cherry Pick says:
May 7, 2011 at 8:38 pm
“Energy is not just radiation. Earth System is more complex than that. For example, absorbed radiation energy may result in increased cloudiness that has impact on who much radiation reaches the earth. Ignoring major part of the system does not help in believing your conclusions.”

What you have just used as an example is one that would cause a cooling effect. I’m not sure if that was your intent, while trying to educate Dr. Glickstein.
Good nickname, by the way.

Alcheson
May 8, 2011 1:09 am

Thanks David. Seems the radioactive decay of elements is estimated to be ~30TW (thought to be ~80% of the earth’s internal heating based on Wikipedia) which would result in about 0.1K of the 33K difference. Wonder how well that value is actually known? Off hand, 30TW doesn’t seem like a lot to keep the earth’s core molten. I might have to spend some more time investigating how that 30TW was derived, and see if it makes sense given the thermal conductivity and temperature gradient of the earth to convince myself as well that it is totally insignificant.

May 8, 2011 1:09 am

Energy in = Energy out
Only if integrated over period spanning centuries !

May 8, 2011 1:13 am

Ira
You should read the article below.
It will answer all of your misconceptions about the “greenhouse theory” which in fact is pure fiction.
1. It is true that the radiative equilibrium temperature of the Earth is -18C.
2. This figure is confirmed by Satellite measurement from space.
3. It is NOT true that all the radiation emitted from the Earth surface.
4. The emission is mostly from the cloud level.
5. If we pick say around 5km as the average emission height to space we will be near enough.
6. Now use thermodynamics expressed in the form of the lapse rate.
7. An average moist lapse rate is -6.6K/km.
8. The lapse rate has nothing to do with back radiation as even scienceofdoom admits.
9. Work back down from -18C at 5km to the Earth surface and we get 5×6.6 = 33K.
So now you have an Earth surface temperature of 15C without any “greenhouse effect”.
http://www.ilovemycarbondioxide.com/pdf/Understanding_the_Atmosphere_Effect.pdf

Don K
May 8, 2011 1:48 am

Alcheson says:
May 7, 2011 at 11:54 pm
Since the earth’s core is molten (very hot), there must also be convection of heat from the core to the surface. How much of the extra 33K heat can be attributed to earth’s own heat source? Is it none?… a little?…. or perhaps a significant amount of the 33K? Just wondering.
=====
From memory … about a tenth of a watt per meter squared. Big enough that it should probably be explicitly counted in really serious models, but probably not enough to really affect anything. As I recall, the number is in Wikipedia somewhere.

Hector Pascal
May 8, 2011 1:50 am

I’d like to run a little thought experiment here, and ask a question. I’m not pushing an agenda. It’s a very long time since I had the maths to follow the physics above, and I’m too idle to re-learn.
Given a warm radiating surface, and cooler atmosphere of, say, 1 molecule. The surface radiates 10 units of energy out. Nine escape, one unit is trapped by the molecule, and re-radiated back to the surface. Net result is a loss of nine radiation units from the surface. The 2nd law of thermodynamics is satisfied: the cool body is not heating the warm body, merely slowing the cooling rate. Effectively “insulating” the surface (but not in the sense of a “blanket” preventing convection).
Isn’t this the effect of “back-radiation”?

Bob_FJ
May 8, 2011 1:56 am

At a quick flick through I don’t think there was much discussion on the fact that the earth is a rotating sphere.
The solar constant for Earth is commonly given at 1366 W/m^2, and to make a simple average of what is absorbed over the surface and atmosphere, and absorption rates etc, (and the outgoing emissions, spread over the entire sphere), when the wattage is proportional to the fourth power of T, introduces a few complications. In short, might I suggest that the effective radiative T of Earth as is popularly touted is uhm questionable.

charles nelson
May 8, 2011 1:57 am

Gotcha Henry!
He thinks CO2 is like a toxin in the atmosphere!
An amazing toxin which is NOT TOXIC at 380ppm but suddenly BECOMES TOXIC at 400ppm. There in a single post is the fuzzy thinking of the Warmists! Fantastic.
Textbook example of ‘Carbonaphobia’!

charles nelson
May 8, 2011 2:17 am

Oh and by the way Henry, do bear in mind that every time you breath out you are breathing out that toxic gas…be careful not to inhale too quick after you breath out!!
Intellectual…ha…like shooting fish in a barrel.

MartinGAtkins
May 8, 2011 2:22 am

JT says:
I am well aware that spontaneous emission is equally likely to be in any direction so that approximately 1/2 will head downward to the surface. However, there is another kind of emission – stimulated emission – and it is massively biased in the same direction of travel as the direction of travel of the stimulating photon.
This is I believe wrong in that you are assigning some kind of kinetic impact of the photon even though you call it stimulated emission. The photon has no mass, so when it is absorbed it is the electromagnetic properties of the photon that is acting on the molecular structure. In an instant the electromagnetic fields of its components are both attracted and repelled depending on the phase polarity of the photon at the time of absorption.
I find wave theory a better example of the action. At rest the molecule is in balance with its competing fields. The magnetic wave is an oscillation of both positive and negative fields. Dependant on it’s phase, the approaching wave will attract all components of the molecule that are the opposite polarity and repel all those the are the same polarity. As the wave progresses, the polarity is reversed and the components that were repelled are now attracted and those that were attracted are now repelled.
The energy of the wave has now been absorbed and for a brief time the molecule has kinetic energy as, notionally the competing fields struggle to reach equilibrium. In doing so the molecules components that are now vibrating disturb the electromagnetic balance of all the surrounding molecules electromagnetic fields which due to their state of equilibrium with the wider body of gas offer resistance but must acquiescent to the force of the field by propagating the energy though out medium.
The above thought bubble only applies to an ideal gas.

Martin Lewitt
May 8, 2011 2:24 am

The average global surface albedo is about 0.121 to 0.124. The .3 figure includes the cloud contribution. The AR4 model albedos averaged about 0.14 per Roesch (2006), which amounts to about 3W/m^2 globally and annually averaged. This 3W/m^2 of absorbed radiation that the models currently under represent, chiefly from earlier observed spring snow melts while they allegedly “match” the 20th century climate, will be added as the models project the next 100 years, catching up with the snow melt, an addition of heat comparable to the projected CO2 forcing.
Another way to think of the significance of the CO2 concentration is that it is about 2.66 grams per square inch, that is about the mass of a penny. Imagine that penny flattened into a square inch plate, and contiguous over the surface of the earth with all the other pennies. Instead of having the optical properties of copper, this contigous shell is opague to certain characteristic frequencies of CO2. Obviously this is thick enough for several absorption of photons in that range and to allow practically no chance of these photons escaping directly from the surface to space. I don’t see how one can just hand wave this away with no deeper analysis than a dismissive mention of the concentration.

May 8, 2011 2:44 am

You state that the back radiation has been measured. Well it might, or at least some long wave IR that is more probably short wave IR that has lost energy traveling through the atmosphere. It would be impossible to tell the difference from measurements at the surface. It is also true that if a molecule of any so called ‘greenhouse gas’ adsorbs energy it must, by 2nd law rules, share this energy with the surrounding non ‘greenhouse’ gasses, ie it loses heat through conduction and radiation.
I do not see in your calculations any adiabatic effect which is important. Taking this into account raises the temperature at the surface. Also, when any gas is heated it will convect. Not an easy effect to measure in the laboratory compared to the real atmosphere. Convecting air will cool.
Jupiter, with an atmosphere of mainly hydrogen, has warming at the 1 atmosphere level similar to Earth. So there it must be an adiabatic temperature rise not ‘greenhouse effect’. Solar input out there is much lower than Earth’s.
True, glass does not transmit IR energy but a greenhouse, transmits visible light and so doing allows energy losses inside, warming the interior by visible light. These losses mean that the lower wave light converts to IR which cannot be transmitted out through the glass. This, together with lack of mixing with cooler external air raises the internal temperature. At night the temperature falls rapidly. 2nd law dictates that the higher the temperature difference of two bodies the greater the heat flow between the two bodies. Entropy must increase.
Taking into account the adiabatic effect, which is real and measurable, there is no need for the GHG effect.
The greenhouse effect also has one fatal flaw. The warming in the mid troposphere in the tropics shown by all models has yet to be found in real observable data. I do not care what your calculations show (in reality another model), if this warm area shown by models is not there then the theory falls flat on its face!

May 8, 2011 3:02 am

Martin Lewitt says:
……Another way to think of the significance of the CO2 concentration is that it is about 2.66 grams per square inch………
Its an odd unit, but it seems much too high;
Ive worked it out in g/cubic inch as 0.000074!

May 8, 2011 3:05 am

Bob_FJ says:
The solar constant for Earth is commonly given at 1366 W/m^2, and to make a simple average of what is absorbed over the surface and atmosphere, and absorption rates etc, (and the outgoing emissions, spread over the entire sphere), when the wattage is proportional to the fourth power of T, introduces a few complications. In short, might I suggest that the effective radiative T of Earth as is popularly touted is uhm questionable.>>>>
One of… OK, my BIGGEST pet peave. There is no such thing as an average temperature or w/m2 that is all that meaningfull. The rough estimate commonly used in climate circles is 1366 x .5 (for day/night) = 683. 683 x .5 (cuz its a sphere) = 360. 360 x .7 (for albedo) = 250, pretty close to Ira’s 240.
But that of course would be around 500 w/m2 at high noon at the equator, and pretty much 0 for six months at a time at the poles. So despite being relatively cold, the poles actually beam way more energy into space than they recieve from the sun, and the equator, despite getting very high insolation year round (in daytime at least) is a net absorber (the excess being moved away by wind and water currents).
So trying to come up with an average insolation value and an average temperature value is meaningful in terms of the total of each. But since insolation varies with w/m2 and w/m2 varies with T^4… trying to track “warming” by following “temperature” is silly. 1 watt of extra insolation results in a much larger temperature anomaly at the south pole than it does at the equator. So when the IPCC says that the “average” temperature of the earth has increased about 1 degree “on average” since 1880, that is highly misleading.
But if they said about 0.3 degrees at the equator and about 1.5 degrees at the poles, it would sound a lot less alarming. And before anyone starts screaming about the ice caps melting, let’s keep in mind that T^4 applies by latitude as well as season. So to get an even more reasonable expression of that “one degree average” they would need to say something like:
Equator +0.3 degrees all year round.
Poles +0.5 degrees in summer
Poles +2 degrees in winter
In other words, that scary “one degree” is mostly confined to the dead of winter in high latitudes. Sorry “the ice caps are melting we’re all going to drown” crowd, but if the poles go from -40 in the dead of winter to -38, the extra melting would be…. between almost zero and zero. And neither the polar bears nor the seals will notice. Or mind.

John of Kent
May 8, 2011 3:09 am

I’m surprised that a “skeptical” site such as WUWT would print such nonsense:-
“Answer #2: The Second law of Thermodynamics is often cited as the source of this falsehood. The correct interpretation is that the Second Law refers to net warming, which can only pass from the warmer to the cooler object. The back-radiation from the Atmosphere to the Earth Surface has been measured (see lower panel in the above illustration). All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.”
[b]No![/b] The earths surface is already emitting radiation in the Infra Red. It’s atoms and molecules are ALREADY excited sufficiently to be emitting IR. Therefore that IR scattering back to the surface CANNOT warm the surface any further because the atoms and molecules of the ground are already kinetically excited to this level. The IR will merely be scattered back to the atmosphere again- and eventually espace to space. The radiation from a body can ONLY transfer heat to a cooler body. The second law applies also to radiation.
This sentence is true however:-
” Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.”
Yes, and temperature of that surface is one of those characteristics. The radiation, does not need to “know” anything! If the surface is ALREADY emitting radiation that is “hotter” , i.e. more energetic then the “back radiation ” will in effect just be reflected again.
Heat always passes from hot to cold, never the other way round, THERE ARE NO EXCEPTIONS!
Ira Glickstein does not seem to understand how matter interacts with radiation.

John of Kent
May 8, 2011 3:18 am

“I modified the Carleton spreadsheet to compute the mean Solar radiation per square meter absorbed by the Earth System, which turns out to be 240 Watts/m^2. I then used the spreadsheet to determine the effective mean temperature of the Earth System that would emit an equal amount of energy to Space, and that turned out to be 255 Kelvins (-18ºC which is 1ºF).
Since the mean temperature at the surface of the Earth is 288 Kelvins (+15ºC which is 59ºF), that leaves 33 Kelvins (33ºC which is 58ºF) to be accounted for. Guess how we acount for it?”
We can account for the 33C difference without any need for an imaginary greenhouse effect. I urge everyone at WUWT to read this brilliant paper by physicist Joe Postma ;
http://www.globalwarmingskeptics.info/forums/attachment.php?aid=327
In this paper, Joe accounts for the 33C. The truth is that the radiative temperature of the Earth is measured from space IS -18C. And it is true that the average air temperature near the surface of the earth is 15C. However, if one includes the whole of the atmosphere right up to the edge of space and the surface of the earth in calculating the average radiative temperature of the Earths surface [i]and[/i] atmosphere, then we get to the -18C figure.
There is no discrepancy, there is no need for an imaginary “greenhouse effect”.
Surely the Phlogiston of our times!

May 8, 2011 3:19 am

….and just for the “cold things can’t heat warm things” crowd, here is a link to Denmark’s DMI which publishes average temperature in the arctic going back to 1958. The green line is the average and the red line is “this year”:
http://ocean.dmi.dk/arctic/meant80n.uk.php
Now take a close look at days 1 to about 60. That’s roughly January 1 to the end of February. The average is flat at about 247 K or -26 degrees C. Think about that for a moment. ITS DARK THE WHOLE TIME! NO SUNSHINE! NO INSOLATION!
Despite which it drops down to around -26C by December… and then just stays there. Why doesn’t it keep cooling off? After all, the atmosphere above the arctic is even colder than the arctic itself… Heck, at 25,000 feet its colder than that even at the equator! So… with 0 watts/m2 from the sun, the arctic nonetheless sticks right around -26 degrees. Hmmmmm…..wonder what keeps it from cooling off more….

May 8, 2011 3:43 am

Alcheson;
Off hand, 30TW doesn’t seem like a lot to keep the earth’s core molten.>>>
You’re thinking about it wrong way round. The 30 TW (actually I think it is 44 TW but no matter) is what ESCAPES from the molten core to the surface. OK correction, the outer core is molten, the inner core is actually solid. Fact is that dirt and rocks are pretty good insulators. In high latitudes where winter temps might be as low as -40, houses have to be constructed on a foundation that goes below the “frost line”. That line is a few feet below the surface and in even the harshest of winters, the dirt only reaches the freezing temperature by no more than an inch or so deeper than average. Get about 10 feet down and the temperature varies by almost nothing all year round. So figure that molten outer core is insulated by kilometers of dirt and rock…the heat gets out…but slowly.
There’s a decent summary at wikipedia (not the best source for climate info but good enough in this case) http://en.wikipedia.org/wiki/Earth's_energy_budget
Geothermal about 0.08 w/m2
Tidal about 0.006 w/m2
Fossil Fuel consumption about 0.025 w/m2.
Compared to 250 w/m2 from solar… pretty minor.

Allan M
May 8, 2011 4:05 am

All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.
But will be reflected by a body of a higher energy level.
http://www.vermonttiger.com/content/2008/07/nasa-free-energ.html
In fig. 3 a box states:
CO2 & H2O ~15µ absorption; re-emit 7µ, 10µ, 15µ.
But re-emission at a higher frequency (shorter wavelength) than the original absorption is not possible.

Richard111
May 8, 2011 4:28 am

Notice that the Earth System mean temperature I had to use to provide 240 Watts/m^2 of radiation to Space to balance the input absorbed from by the Earth System from the Sun was 255 K. However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.
255K is the mean earth radiative temperature at around 4,000 metres altitude and above. Escaping IR radiation is from a volume of the atmosphere, not a flat surface.
The mean surface temperature of 288K is simply the compression factor of gravity on the air. Air at the surface has the highest density and returns highest temperature. Think deserts, polar regions, equitorial regions. All very different suface temperatures but always the same adiabatic lapse rate, water vapour content allowing.

Allan M
May 8, 2011 4:43 am

http://www.vermonttiger.com/content/2008/07/nasa-free-energ.html
………
Figure 3 has a box which states:
CO2 & H2O ~ 15µ absorption; re-emit 7µ, 10µ, 15µ.
a body cannot re-emit at a higher frequency (energy level) than it absorbs.

David
May 8, 2011 5:17 am

There is a bathtub which can hold energy. (the earth’s atmosphere/surface/oceans)
There is a energy tap flowing into it. (the sun)
There is a hole in the tub where energy flows out. (radiative emission into space)
Inflow = outflow, energy in = energy out.
The quantity of energy the black-box “bathtub” is able to hold (wants to hold) can go up and down and is indicated by the temperature. When the variables which alter the level stored in the tub are changed (knobs turned) there will be an imbalance between inflow and outflow until the new equalibrium is reached – a equilibrium at which a different level of energy is stored in the tub. We expect a change in the total level of energy stored to affect the temperature at the bottom of the tub.
I expect the IR interference to be one of those variables but I’m sure there are many others. The claim that 30 degrees celsius at the surface is all due to the greenhouse effect is rubbish.
3 scenarious.
A. Black body earth (-30 % albido).
B. Earth with atmosphere and no IR interference.
C. Earth with atmosphere and IR interference.
A is cooler than B is cooler than C.
C – B = greenhouse warming.

Fred Souder
May 8, 2011 5:35 am

Davidmhoffer,
The ground and ocean warm the lower levels of the atmosphere in the polar winters.
Does anyone else wonder what the rate of heat transfer to the surface is from conduction through the crust from the mantle?

RJ
May 8, 2011 5:46 am

We can account for the 33C difference without any need for an imaginary greenhouse effect. I urge everyone at WUWT to read this brilliant paper by physicist Joe Postma ;
http://www.globalwarmingskeptics.info/forums/attachment.php?aid=327
Agree it was a brilliant paper. But this Postma paper was posted on Ira’s previous thread and Ira and others criticised the paper or would not read it all.
http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/
Ira Glickstein, PhD says March 30, 2011 at 2:02 pm
This post by Ira has not really moving forward at all with Ira still posting this for example
“How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
This is just total nonsense. (The only rational explanation). It is one explanation and not a very good one. Unfortunately Ira is just ignoring comments on previous thread so this series is not moving forward. He could summarise the main areas of contention without giving a viewpoint rather than posting again his beliefs that he knows many disagree with

Cementafriend
May 8, 2011 5:55 am

Engineering covers a considerable range of displines such as chemical, civil, electrical, mechanical etc. Heat transfer, thermodynamics, fluid dynamics are basic subjects for some of these disciplines but are not taught in all the disciplines. It is not surprising that some engineers do not understand the fundamental theory and experimental knowledge in these subjects. It is surprising, however, that qualified professional engineers who supposedly comply with a code of ethics which includes the requirement of competence should express opinions about a subject which they do not understand.
Ira, you have been sucked in by climate scientists who have no understanding of heat transfer, fluid dynamics and thermodynamics. You make no mention of heat transfer by convection and phase change (evaporation and condensation). One can only assume that you have no knowledge of this subject. For a quick overview I suggest that you look at Perry’s Chemical Engineering Handbook. Learn about Nusselt and Reynolds numbers. Also, take note of the definition of a black body -it has a surface, and all absorbs energy of all wavelengths. The atmosphere is not a surface and the trace gas CO2 only absorbs radiant energy in very narrow wavelengths ie it is not a black body. The basic Stefan Boltzman equation is not applicable to the atmosphere. Engineers have carried out a huge amount of research on radiation involving gases in heat exchangers.
Then read the following http://climategate.nl/wp-content/uploads/2011/02/CO2_and_climate_v7.pdf

May 8, 2011 5:59 am

Dishman says:

In re Objection #2:
Gas molecules follow ballistic trajectories between collisions…
… unless all the work on Gravity since Galileo is wrong, and doesn’t apply at the molecular level, of course.

Ballistic trajectories are trajectories under the action of gravity. Note, he didn’t say they travel in straight lines.
However: the approximate speed of molecules in the atmosphere is about 1,700 km/h. The path isn’t very curved, in other words. And the mean free path of a molecule in air is about 65 nanometres. You’d be pretty amazing if you could measure the deviation from a straight line of a path segment with that length and curvature.

David
May 8, 2011 6:19 am

I’m reading the post by Joe Postma now. He appears to be saying my energy tub can only be filled to a maximum level and can’t hold more after that.
It already occurred to me that climate scientists didn’t really have any firm understanding of the mechanism of action of the greenhouse “theory”, hence this very subject has been at the front of my mind lately, hence my bathtub theorizing. I’m glad to see this particular issue is slowly but surely coming to a head.
If Joe is right they’ll just start saying that energy is being redistributed in a way to make the surface warmer.

Leonard Weinstein
May 8, 2011 6:22 am

Bryan,
I read the link you gave, and the guy got it partially wrong. He stated that a black body near a light bulb in equilibrium would not heat up if a reflector was then introduced to reflect back onto the black body some of the radiated heat. This is wrong. The black body would absorb the reflected radiation, and thus the equilibrium temperature would be increased. However, this is not related to the back radiation issue. The main remainder of yours and his points are correct, i.e., it is the effective average altitude of outgoing radiation and lapse rate that heats the surface, and backradiation is a consequence rather than cause of the heating. The reason is that buoyancy overcomes absorbed radiation heat transfer to prevent stagnant heating from the radiation.

May 8, 2011 6:36 am

John of Kent says:

I’m surprised that a “skeptical” site such as WUWT would print such nonsense:-
“Answer #2: The Second law of Thermodynamics is often cited as the source of this falsehood. The correct interpretation is that the Second Law refers to net warming, which can only pass from the warmer to the cooler object. The back-radiation from the Atmosphere to the Earth Surface has been measured (see lower panel in the above illustration). All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.”
[b]No![/b] The earths surface is already emitting radiation in the Infra Red. It’s atoms and molecules are ALREADY excited sufficiently to be emitting IR. Therefore that IR scattering back to the surface CANNOT warm the surface any further because the atoms and molecules of the ground are already kinetically excited to this level. The IR will merely be scattered back to the atmosphere again- and eventually espace to space. The radiation from a body can ONLY transfer heat to a cooler body. The second law applies also to radiation.

Firstly, I do wish that people who think they know something would just be polite about it; this is not a peer-reviewed journal (thank goodness). Anthony is offering a wonderful service here. The way I look at it is: the entire article, including comments, is the publication. If an article contains fallacies, we will learn from the more knowledgeable commenters.
Having said that, you are not one of those more knowledgeable commenters. All radiators are absorbers. Contrary to your assertion, anything capable of emitting a certain wavelength at a certain temperature can absorb ditto. So yes, a photon radiated back towards the surface can indeed be absorbed by the surface. What this absorption will do, however, is raise the temperature slightly so that the surface now emits at a higher rate. Heat still travels on average from hot to cold. The temperature increase of the hotter surface, whilst happening only if the back radiation is happening, still comes about because of the impinging heat from the sun.
Think of it this way: focus a flame upon a single cubic inch of steel that happens to be part and parcel of a massive steel girder. The flame’s heat will flow into the steel all around and you will find it very hard to increase the temperature of the cubic inch by very much. Now focus your flame on a cubic inch embedded in another surrounding cubic inch, embedded in a very poor conductor. Your heat will heat the first cubic inch, which will heat the second. The poor conductor will prevent escape of much heat and the molecules of steel will bump upon each other. Molecules in your first cubic inch will bump (heat) those in the second, those in the second will bump those in the first. The entire two cubic inches will quickly increase in temperature. The outer cubic inch will never become hotter than the cubic inch being directly heated, and yet the directly heated part will rise in temperature. So there is such a thing as back flow of heat and it is by no means contrary to the 2nd law.
But having said that, the idea of back radiation as formulated by some warmists doesn’t hold much water. The most likely thing to happen to a CO2 molecule that has absorbed a photon and risen to a higher energy state is that it will collide with another molecule and transfer the energy into general heat of the whole gas. When that heats the gas enough, then the number of CO2 molecules undergoing the reverse phenomenon (getting knocked into a higher state by an impinging molecule – remember, these processes are always reversible) will rise enough to provide enough radiation that back radiation becomes significant. But there is a huge heat capacity to fill before that point is reached, since the CO2 is a trace gas heating a vastly huger bulk of N2, O2 etc. In the meantime, the heating of the gas starts convection, limiting the rise in lower air temperature to the adiabatic lapse rate. So once again, the back radiation has its effect auto-limited. Since all planets we see (Earth, Venus, Jupiter) are at close to maximum lapse rate, this is clearly a phenomenom that is quickly reached by almost any trace amounts of greenhouse gas in the atmosphere, and so the mechanism required for the AGW theory to work is already exhausted to within a few percent, and thus the phenomenon is basically harmless.

Leonard Weinstein
May 8, 2011 6:36 am

Ira,
I enjoyed your clear presentation of detail until I saw the kicker “How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
There have been several good presentations of the correct reason. The presence of IR absorbing gases acts as a partial radiation insulator. However, the atmosphere is unbounded, so if there were even a small amount of local heating occurs due to absorbing IR, the warmer air would rises and then cool by adiabatic expansion. In fact, the atmosphere mixes and rises so strongly, that noticeable radiation warming does not occur. The absorbed radiation passes energy into the surrounding air, but the more energetic air molecules (from the velocity distribution) impacting the absorbing molecules cause radiation, and the average is so in balance that the air is considered to be in LTE. However, the fact of the partial radiation insulation (i.e., prevention of direct radiation to space) results in the radiation to space occurring at higher altitude (average about 5 km), and it is the adiabatic lapse rate combined with the equilibrium temperature at 5 km that results in the ground temperature. The back radiation and ground up radiation are a result of the increased temperature, not the cause!

David
May 8, 2011 6:48 am

I didn’t think he was right with the mirror example. If you shine a bright torch and a dim torch at a object it will get hotter than if you shone only the bright torch. If you shine the bright torch and reflect a mirror at the object it’s like adding another dimmer torch.

gbaikie
May 8, 2011 6:53 am

“Heat always passes from hot to cold, never the other way round, THERE ARE NO EXCEPTIONS!”
What about the sun corona? It’s a million degree and sun surface is cooler.
Perhaps you referring to solids. Such as warm air can not heat up a warmer surface via radiant energy.
Say, you had an object in space and it’s temperature was 100 K. This is a cold object, -173 C but compared to background temperature of Space 2.7 K it’s fairly warm. And it is certainly radiating energy into space. No part of earth is as cold as 100 K, if colder can not pass to warmer this means that with infrared telescope on Earth you could never detect such a cold object in space. It also means that your telescope mirrors would need to as cold or colder than any cool object you wanted to observe.

Dave in Delaware
May 8, 2011 7:00 am

* Blackbody – As you have stated, the inbound radiation spectrum from the Sun can be approximated as black body (at it’s temperature) and from the surface (at it’s temperature). The implication being, a blackbody emits in a radiant distribution dependant on it’s temperature. But you incorrectly indicate that Atmosphere is also a black body and it is NOT.
“It also may be used to calculate the radiation spectrum from the Earth System (Atmosphere and Surface, see below for explanation) at any assumed temperature. ”
If CO2 absorbs at 15 micron, it also emits at 15 micron (not a blackbody distribution of all wavelengths based on atmospheric temperature). So where your chart says CO2 ‘Re-Emit 7, 10, 15 micron’ – that is not correct. Similarly for H2O. The purple, blue and black curves in your atmosphere graphic have no meaning.
The ‘looking down’ emission spectrum tells you where the energy was emitted if you look at the temperature curves. The atmospheric window wavelengths are like the surface temperature, cold from the antarctic, hot from the Sahara, with no emission from the intermediate atmosphere. The CO2 wavelengths always look like the tropopause temperatures no matter if they are measured over the desert or the poles, which tells us we are seeing emissions from high in the atmosphere, not re-emitted near the surface. And the H2O wavelengths seem to be in between, more like upper atmosphere and perhaps indicative of cloud tops or ice crystals.
But ‘looking down’ distribution doesn’t look like a one-temperature blackbody distribution. Using a one size fits all blackbody emission from the atmosphere at any one temperature is probably not a good assumption.

Dave in Delaware
May 8, 2011 7:03 am

Radiant energy accounts for all of the outbound energy at the top of atmosphere, but only a fraction of the energy transfer at the surface. Understanding how the energy moves up through the atmosphere via non-radiative pathways is key in estimating potential GHG effects.
If we use values from Kiehl & Trenberth 1977, they show 60% of the surface energy transfer to the atmosphere is due to Thermals (convection) plus Evapo-transpiration (water evaporation and release as latent heat). Leaving no more than 40% as Radiant Energy transfer, with 24% of that 40% as direct loss to Atmospheric Window. Whatever the GHG affects may be, they are happening in the 16% of the total that is left. (And I read somewhere that Trenberth used 40 W / m-2, knowing that the actual ‘window’ value was more like 66 W/m-2. That would increase the ‘atmospheric window’ from 24% to about 39%, leaving only about 1% of the total for the net of all other radiant emissions).
How can that be, since we know the surface emits a blackbody distribution dependent on the surface temperature? One possible explanation is to look at what the energy ‘pathway’ might be. Sort of by definition, the radiant energy that is NOT in the atmospheric window wavelengths, is susceptible to being absorbed by one of the GHG, mostly CO2 and H2O. So the surface emits radiant energy, and the GHG susceptible wavelengths are absorbed fairly close to the surface. Next, we think that the absorbed photons are more likely to be ‘thermalized’ rather than re-emitted as photons (10,000 : 1 more likely thermalized). Thermalized means the GHG molecule collides with a non-GHG molecule in the air, converting the photon to kinetic energy, which can now be seen on a thermometer. Now all of the non-radiative pathways take over. In essence, the radiant emissions from the surface have ‘speeded up’ the energy transfer to get the energy into the other pathways. Hot air rises, clouds and thunderstorms happen when conditions are right, and generally energy moves up through the atmosphere at something like the wet or dry Lapse rate, as appropriate to conditions. When the energy gets high enough, the radiant pathways again become significant, and eventually take over, sending the energy out to space.
Of course the devil is in the details, but that description seems to fit with what others have said earlier. It is not all about just the radiant component.

Nullius in Verba
May 8, 2011 7:06 am

“The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
There’s an objection you haven’t addressed yet – that if extra heat tried to build up near the surface, convection would immediately carry it away again because warm air rises.
In an atmosphere without convection, it would indeed work as you suggest, and the Earth’s surface temperature would be about 60 C on average. When we stop convection artificially – with greenhouses and solar ponds – we do get much higher temperatures. But the mechanism doesn’t apply in a convective fluid, and the 60 C prediction (and other things like an exponential relationship of temperature with altitude) conflicting with observation falsify it.
The back-radiation explanation (or the “shells model” as I sometimes call it) was the original one developed by De Saussure, Fourier, etc. and was overturned in the 1960s by the work of Manabe, Moller, Strickler, and others. The actual explanation is that extra greenhouses gases raise the average altitude of emission to space, raising the altitude of the level that equilibrates to -20 C, and the warmth of the surface relative to this level is due primarily to the adiabatic lapse rate (adjusted for water vapour latent heat).
The surface temperature is the effective radiative temperature (-18 C) plus the average height of emission (5.5 km) times the (moist) adiabatic lapse rate (6 C/km).
T_surf = T_eff + z_emiss * MALR.
The CO2 greenhouse works by increasing z_emiss.
The tropical hotspot, incidentally, is supposed to be caused by also decreasing MALR, through water vapour feedback.
I’m very pleased to see a lot of people now picking up on the convection/lapse rate connection. Things have much improved over a few years ago.

May 8, 2011 7:35 am

Leonard Weinstein
……”The black body would absorb the reflected radiation, and thus the equilibrium temperature would be increased.”……
Yes I agree, he could have been clearer on that point.
This aspect causes a great deal of confusion and some intellectual mischief.
See if you would agree with this formulation.
The reflected radiation is the reflective insulation component.
If increased it will slow down the heat loss from the black body.
If the black body has an unchanged energy input its equilibrium temperature would be increased.
However if the energy input to the black body was cut off the temperature of the black body would drop despite the backradiation.

DirkH
May 8, 2011 7:40 am

Ira, explaining the 33K temp difference between surface and top of atmosphere in terms of LWIR backradiation AKA greenhouse gas effect is misleading; as it is already entirely explained by the adiabatic lapse rate.
(see for instance William C. Gilbert here
http://bit.ly/fIhWPS
)
This leaves us with the question – if the lapse rate explains the temp difference, why does the purported greenhouse effect contribute nothing – to which my answer would be: As Absorption and re-emission of LWIR is an extremely fast process the re-radiation process can only delay heat transport via LWIR radiation to space by minutes at best – it’s a ping pong game of photons, but no heat is “trapped”, no matter how often Dessler et. al. use the term “heat-trapping gases” ( see here:
http://www.chron.com/disp/story.mpl/editorial/outlook/6900556.html
)
If this delay rises by a few percent due to increased CO2 levels then it is still only a short term delay.
And that is why we see no warming related to CO2 (if there were such warming, it would correlate to the Keeling curve, which it doesn’t).

DirkH
May 8, 2011 7:43 am

And i see that others have mentioned Postma. A very good explanation of the lapse rate.

Alistair
May 8, 2011 7:56 am

The 33K is the result of the warming by gravitational potential energy of the atmosphere below the -18°C isotherm determined by equality of radiation out to radiation in. It’s about 5.5 km. Because it would be the same for an atmosphere of the same average Cp but without GHGs, it is unconnected to greenhouse warming.
All that does is to provide extra heat in the lower atmosphere and it’s convected to the top where it can radiate away. In the absence of convection, we’d get another 44K warming so the convection/negative feedback is extremely efficient.
If you increase [CO2], [H2O] ust decrease in the upper atmosphere, as observed in 61 years’ radiosonde data. The mechanism is whatever is consistent with the 2nd Law’s requirement of maximum rate of entropy production.
Climate science’s confusion over greenhouse warming comes from the mistake made by Arthur Milne in 1922 when he solved a differential equation for IR absorption in the atmosphere using as boundary condition infinite thickness. The term it generated, extra ‘back-radiation’ is imaginary.
This is not to say there is no back-radiation: that is essential to match exactly the IR radiation from the atmosphere below that point in specific spectral ranges consistent with the local deviation from the lapse rate, e.g from clouds. So, the spectral curve will vary depending on whether there are clouds for example.
The problem is that most scientists haven’t a clue about practical conductive, convective, radiative heat transfer so are easy meat for the charlatans who have made a good career out of pretending there’s an effect of CO2. As an exercise they need to understand the UHI effect is caused by convective heat transfer. You get higher temperature because to get the required thermal equilibrium, you need more radiative flux from the Earth’s surface. Yet the sum of all the heat transfer components, equal to the short wave heating, remains the same once you have exhausted thermal storage effects.
Similarly, at night, you can freeze water by digging a hole in the desert to restrict convection even though air temperatures can be 20°C+.
Before anyone else contributes, they need to know some chemical engineering: http://rpaulsingh.com/teaching/LectureHandouts/Convection%20Heat%20Transfer_handout.pdf

DirkH
May 8, 2011 7:56 am

davidmhoffer says:
May 8, 2011 at 3:19 am
“Why doesn’t it keep cooling off? After all, the atmosphere above the arctic is even colder than the arctic itself… Heck, at 25,000 feet its colder than that even at the equator! So… with 0 watts/m2 from the sun, the arctic nonetheless sticks right around -26 degrees. Hmmmmm…..wonder what keeps it from cooling off more….”
David, if the atmosphere above the arctic is colder than the arctic surface, do you want to imply here that this cold atmosphere is nevertheless capable of WARMING the surface? Are you sure you wanted to say that?

Dave Springer
May 8, 2011 8:08 am

Ira:
As a sanity check you should do this calculation for the moon which receives exactly the same amount of solar radiation as the earth system and there’s no atmosphere or water to complicate the situation. Further, the average surface temperature of the moon was experimentally measured by two separate Apollo experiments where a borehole 3 meters deep was made and a probe with thermocouples at various depths was put into the boreholes. These experiments transmitted temperature data back for several years. The average temperature of the surface is that depth at which there is no daily, seasonal, or annual variation i.e. a constant year-round temperature. At two mid-latitude locations that temperature is -23C or -9F at all depths of 1 meter or more. The experiments were actually designed to measure the thermal conductivity of lunar regolith.
So it would appear on the face of it that your theoretical figure for the earth of 33C warmer than it would be for two approximate black bodies is somehow in disagreement with experimental reality by about 5C.
So before talking about the earth as an approximate black body, which we can’t measure because it ain’t one with a dynamic ocean/atmosphere wrapping it, you have to do a sanity check and see if your black body calculations agree with the measured temperature of the moon which actually is an approximate black body.
I can’t possibly take seriously any discussion which hinges on this commonly number that the earth is 33C warmer than it would be sans ocean and atmosphere when it appears to be 5C smaller than experimental observations. Near as I can tell from the facts that number should be 38C. Normally when two numbers like that differ by something near 10% I shrug it off because it’s still pretty close to agreement. But in this case 5C of disagreement five times greater than the commonly claimed 1C of anthropogenic warming to date.
This goes to the heart of the problem. Anthropogenic effects are so tiny as to easily buried in the noise of the real world and in the error margins of all these toy models of the real world.

Retired Engineer
May 8, 2011 8:22 am

This makes sense. Absent CO2 and water vapor, we would freeze. How much warming does each contribute? If CO2 did it all, and mankind producses about 3% of the CO2, then we are responsible for 3% of the 33C warming or about 1C. Thus AGW is real. Except for water vapor which may cause 90% of the 33C, in which case we cause 0.1C warming. Big deal. Given the daily and seasonal variations in temperature, I find it very hard to believe that 0.1C will have any measureable effect. Or be measureable at all.

May 8, 2011 8:22 am

“How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
Are we suggesting that the earth’s surface is one black body and the atmosphere another? And thus the atmosphere radiates to the surface? However, the atmosphere and the surface are part of the same black body at TOA. Treating them separately below TOA may lead to nonsensical results.
A more reasonable explantion is that the missing 33 K is due to the weight of the atmosphere which compresses the gas nearer the surface, which we see as an increase in temperature.
We know this by comparing the atmospheres of Venus (CO2), Earth(N2O2) and Mars (CO2). All three planets show that temperature varies as the pressure of the atmosphere and the distance from the sun, independent of the composition of the atmosphere.
Venus is totally covered in clouds. If albedo has a 30% blocking effect on earth, then it should have a near 100% blocking effect on venus and the surface should be cold. But it is not.
If CO2 has a warming effect, the mars, which has a greater partial pressure of CO2 at the surface than does earth, should be warm, but it is not.
Climate science continues to build models. Why? Because the argument goes that their is only one earth, so we need models to tell us the effects of CO2 on climate.
This is a rubbish argument. Our two nearest neighbors have CO2 rich atmosphere. These are perfect models for us to judge if the GHG theory is correct, but climate science continues to ignore this more accurate alternative.
We hear about “run-away” greenhouse effect on Venus, without any explanation of how this is possible with 100% cloud cover if albedo and aerosols works as we think they do.
The facts are that the surface pressure of Venus is 90 times. Yes 90 times greater than on earth. This is why the surface temperature is high, even though according to climate theories about clouds and albedo, it should be low.
Why are the temperatures at the bottom of the oceans on earth not higher than the surface? After all this happens in the atmosphere, should it not happen in the oceans.
The reasons is compression. The oceans do not compress, so a cubic meter of surface water has the same number of molecules as a cubic meter of water from the deep oceans. However, the atmosphere does compress. So, a cubic meter of atmosphere at the surface has more molecules than a cubic meter of atmosphere higher up.
It is these molecules that carry the heat. If the energy of the molecules is unchanged, then the more you have per cubic meter, the greater the energy per cubic meter, which we will see as increased temperature.

Theo Goodwin
May 8, 2011 8:31 am

What is so bothersome about this topic is that people who take it up do nothing but rehearse the Great Assumptions, such as the claim that Earth should be treated as a black body. It’s like listening to discussions among Ptolemy and his colleagues. The solution is always the same: more epicycles and maybe an eccentric or two. Yet in setting forth the Great Assumptions, you mention known facts which contradict them. Earth has albedo. Suppose for a moment that Earth’s albedo became 100%, would that motivate you to change the Great Assumptions. Is an object with 100% albedo properly treated as a black body? Can anyone state a plausible account of Earth’s radiation budget that does not treat Earth as a black body? I don’t believe anyone can and I believe that fact reveals a limitation of imagination, not the firmness of accepted science. I guess we will just have to wait for the science to advance. Anyone know somebody named Kepler?

Dave Springer
May 8, 2011 8:36 am

Ira (con’t):
The first pass you need to make in accounting for this discrepancy is to take albedo into account. The moon’s albedo is constant and accurately measured at some 16% so that amount of insolation doesn’t get absorbed the surface. That will account for the theoretical difference between black-body absorption and reality. The moon would be about 5C warmer if the surface albedo was close to 0%.
The problem doesn’t go away however because the earth has a non-zero albedo too and it isn’t known nearly as well as the moon’s with, depending on who you ask, an average albedo in the range of 32% which is primarily the result of some 70% being shrouded by clouds of some sort at any given instant. The problem here is there is no satisfactory agreement between experiments attempting to measure the earth’s average albedo and estimate range from as little as 30% to as much as 40%. All the experimental attempts to measure the earth’s average albedo do agree on one thing – it isn’t constant and varies by as much 1% year over year.
One percent doesn’t seem like much variation but once again it’s a very large number when compared to anthropogenic warming. A 1% change in the amount of insolation actually reaching the surface and not being reflected directly back out into space is some 2.5 watts per square meter. The IPCC third assessment is “95%” confident that anthropogenic forcing lies in the range of 0.6 to 2.4 w/m2. Yet measured variations in earth’s albedo over just several years has it changing by more watts/m2 than the highest estimate of alleged anthropogenic warming. Worse yet, the actual albedo can’t be pinned down and estimates vary by about 7% which is 7 times greater than anthropogenic forcings.
So how can we possibly start talking about anthropogenic forcings and surface temperature changes wrought by same when we don’t even know to +-5C what the average temperature of the earth should be due to our albedo measurements being so imprecise and having no bloody idea how, when, and why the earth’s average albedo varies.
The earth being a dynamic water world makes all attempts to model it exercises in futility. The models are toys and should be regarded as precisely that – toys with zero credibility.

May 8, 2011 8:53 am

6. Now use thermodynamics expressed in the form of the lapse rate.
7. An average moist lapse rate is -6.6K/km.
8. The lapse rate has nothing to do with back radiation as even scienceofdoom admits.
9. Work back down from -18C at 5km to the Earth surface and we get 5×6.6 = 33K.
So now you have an Earth surface temperature of 15C without any “greenhouse effect”.
Exactly, the earth and atmosphere are part of the same black body. Considering them as two different black bodies, based on observation of their combined effect at the TOA without consideration of lapse rate, convection and condensation seems unlikely to provide an accurate answer.

Alan D McIntire
May 8, 2011 8:56 am

Bryan stated that using clouds, you get a 33 C temperature difference between clouds and surface “without any greenhouse effect”. I’d say inseead, that a significant fraction of the greeenhouse effect is due to clouds.
Ira Glickstein stated that the warming due to CO2 would only be about 0.2C. Here’s a way to get a rough estimate of the figure:
We get an average of 342 watts/M^2 from the sun.
As others have pointed out, excluding clouds, the average reflectivity of earth’s surface is about 0.124
Earth is not a blackbody, most of the earth is covered by ocean, which has an emissivity between 0.92 and 0.96- I’ll use a 0.94 average. This non- blackbody emissivity means earth actually radiates away less than a blackbody at the same temperature.
A non-greenhouse earth surface would receive an average of
342 *(1-.124reflectivity)/0.94 emissivity would give an average surface flux of
342*0.876/.94 = 318.7 watts. All of the positive and negative feedbacks to the
greenhouse effect give an average temperature of 288K, or effectively 390.7 watts/m^2, for an effective magnification of 1.226
From Trenbeth’s figures,
http://content.imamu.edu.sa/Scholars/it/net/trenbert.pdf
161 watts /m^2 hit the earth’s surface, with back radiation we get a total of 492 watts, not all in sensible heat, for a magnification of
493/161 = 3.06,
Putting the two formulas together, a theoretical additional factor of 2.06 multiplier due to the greenhouse effect becomes an actual multiplier of 0.226, thanks to negative factors like increases in albedo due to clouds, decreases in lapse rate, etc.
A doubling of CO2 would supposedly increase the surface flux by something less than 3.7 watts/m^2, for a total of 496.7 watts, an increase to 335.7 from 332 over the original 161 watt/m^2 hitting the surface.
Multiply that 335.7/332 * 0.226 and we get 1.2285* 318.7 for an effective
wattage increase from 390.7 to 391.52 watts, or a temperature increase to
(391.52/390.7)^0.25 = 288.15K from 288 K – Close to Ira Glickstein’s estimate of 0.2K- a drop in the bucket.

jae
May 8, 2011 8:57 am

Ira:
“The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
This is the crux of the issue. It is the point of contention. You and others keep repeating the same GHE theory over and over again, with slightly different words. By now, I doubt that there are many people with any scientific background who do not understand what you are saying. The problem is that it has not been demonstrated empirically AND there are other ways to explain why the surface of the Earth is higher than -18C!
The -18 C represents the average radiation temperature which comes from an altitude of about 5 km, NOT the surface. Due to the lapse rate, it is much warmer on the surface (viola, about 33C higher!).
Now, that 33C “increase” may well be due to the GHE that you describe. However, it could also be explained simply by heat storage and the ideal gas law. So I’m still not sure.
BTW, I hope you are not still trying to say that the surface is HEATED by the backradiation from the atmosphere. The atmosphere may manage to keep the surface from cooling so fast (due to the backradiation), but it is certainly NOT heating it. A colder body cannot heat a warmer one. Period.

Dave Springer
May 8, 2011 8:59 am

Ira Glickstein, PhD says:
May 8, 2011 at 8:35 am
You say the earth’s average is pretty close to 70%.
First of all that’s the reciprocal of albedo. Albedo is a number ranging from 0-1 with 0 being dead black and 1 being a perfect mirror. Multiply by 100 for a percentage.
Now I know wikipedia is frowned on as a reference but when it comes to global warming their distinct bias is always on the warmist side so if you find something in wikipedia that is contrary to warmist dogma you can bet your bottom dollar it’s a painful admission.
I quote:
http://en.wikipedia.org/wiki/Black_body

Estimates of the Earth’s average albedo vary in the range 0.3–0.4, resulting in different estimated effective temperatures. Estimates are often based on the solar constant (total insolation power density) rather than the temperature, size, and distance of the sun. For example, using 0.4 for albedo, and an insolation of 1400 W m−2, one obtains an effective temperature of about 245 K.[25] Similarly using albedo 0.3 and solar constant of 1372 W m−2, one obtains an effective temperature of 255 K.[26][27]

This is exactly the figure I gave you of 30% to 40% for estimated average albedo. However I didn’t get it from wiki. I got it from much more in depth reading of the experimental attempts to measure it. Wikipedia in this instance is accurate and for the warmista it is a fatal blow with regard to the credibility of their propaganda. We don’t know the average insolation that reaches the surface of earth any closer than +-25w/m2. To start talking about anthropogenic forcings in the range of -0.6 to 2.5w/m2 is ludicrous when the error margin in experimental attempts to measure average insolation is 25 watts.
Before any toy model of the earth can begin to be given any credibility it must first have a accurate number for the earth’s average albedo down to a fraction of percent and then it must further model the observed variation in albedo down to a fraction of a percent. If it can’t do that it can’t even begin to seperate anthropogenic from natural forcing.

DirkH
May 8, 2011 9:11 am

Ira Glickstein, PhD says:
May 8, 2011 at 8:35 am
“The point of my simplified calculation was to show that, absent back radiation from the Atmosphere, there is no good explanation for the approximately 33ºC (58ºF) temperature difference between the “light” input from the Sun and the “heat” output from an Earth that would have to be cooler than it is for the heat balance to work.”
Again: The Earth radiates from the top of the atmosphere and that’s where you find exactly the temperature needed to radiate enough. See also
Harry Dale Huffman : No Greenhouse effect on Venus
http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html
where Mr. Huffman does the same computations for Venus.

jae
May 8, 2011 9:27 am

Probably should have read more comments before posting. fred burple already said about the same thing that I did. I think he is correct.

May 8, 2011 9:27 am

jae –
Someone who would benefit an understanding of ‘traveling waves’; incident and reflected waves (energy), S-Parameters (Scattering Matrix Parameters).
Falling back on the simple ‘laws of thermodynamics’ does you a disservice …

May 8, 2011 9:37 am

Ira Glickstein, PhD says:
“The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
“absent back radiation from the Atmosphere, there is no good explanation for the approximately 33ºC (58ºF) temperature difference”

As many comments above have pointed out, the adiabatic lapse rate alone can completely explain this. Thus, Ira you need to explain why in effect you consider this to be an irrational or poor explanation.

May 8, 2011 9:44 am

I disagree with answer three. As a fellow systems engineer I appreciate and support your back of the envelope checks. But remember, we also have to explore all feasible answers, and the difference between black body and the actual measured radiation back to space does NOT require a ‘greenhouse’ effect.
It simply requires a long time constant between heating and radiation, which is where heat sinks and non-solar heat sources come into play.
The 33° K difference could simply be a result of a time delay between solar heating and thermal cooling. The heat capacity of the Earth could require the system to bank 33°K of heat before the there is sufficient flow of energy to the upper atmosphere to radiate.
Moreover, the inner core and the active underwater volcanoes could easily be dumping a extra wedge of gravity-friction generated heat (akin to what drives the Io system). What our massive oceans provide is a heat store, not just a heat sink. Gravity-friction (which produce wave action, which transfers kinetic energy into heat) is a source of energy transfer at the surface, as is any flowing water and its passage to the sea.
How much kinetic energy is in a river flowing to the sea? How was that energy stored? It was stored as water vapor evaporating from ground and bodies of water from solar heat, which rise to be COOLED, where the kinetic energy from gravity is then brought back to Earth with the rain. Friction supplies the transfer.
You can identify numerous transitions where heat is captured, converted, released, transferred and held for days, weeks, months and years.
What is wrong with ALL these models is they lack a time constant from which a joule of solar energy can travel through our system – over any one of hundreds of energy transformation paths – before it is made available for radiation. And we never remember to include the yet still unquantified heat source that is at the core of our planet. There is NO dissipation path for that heat except through radiation. It is a massive heat source which could EASILY account for the 33°K difference.
We know so little, yet act as if we have all the answers. The minute that happens we stop being scientists and become zealots.

May 8, 2011 9:49 am

DirkH says:
May 8, 2011 at 7:56 am
davidmhoffer says:
May 8, 2011 at 3:19 am
“Why doesn’t it keep cooling off? After all, the atmosphere above the arctic is even colder than the arctic itself… Heck, at 25,000 feet its colder than that even at the equator! So… with 0 watts/m2 from the sun, the arctic nonetheless sticks right around -26 degrees. Hmmmmm…..wonder what keeps it from cooling off more….”
DirkH;
David, if the atmosphere above the arctic is colder than the arctic surface, do you want to imply here that this cold atmosphere is nevertheless capable of WARMING the surface? Are you sure you wanted to say that?>>>>
Dirk, I didn’t say the atmosphere made the surface warmer. I said, what keeps it from cooling off more? Why does the temp drop from 275K in July, to 240K in December, a drop of 35 degrees over 5 months, and then over the next 2-3 months, despite having 0 insolation, it drops by perhaps two degrees more. Since it isn’t getting any energy from the sun, where is the energy to keep it from cooling any further come from?
There are several possibilities. The arctic is mostly ocean covered by ice, so warm water currents could be bringing in heat. Wind from warmer latitudes could be bringing in warm air. Reflection from clouds that are high enought in altitude to catch some sunlight cold contribute. All of these have been measured in various ways, all are part of the picture, but combined they are insufficient to keep the surface at -26.
So why doesn’t it cool off more? Answer is (in part) that some of the outward bound radiance is in fact returned to the surface. In fact, the atmospheric window starts to close pretty fast right around that temperature, and by -40 it is shut tight. That’s why the very coldest nights at 80 N latitude are about -40 and the very coldest nights at 50 N latitude, where the “average” temp is MUCH higher, are around -40.
Once the temp hits -25 not only is the “colder” atmosphere above the artic backradiating, reradiating, simulating pajamas, what ever term you want, it get’s more efficient at it. So it can’t EVER cause an INCREASING temperature trend, but it can certainly recycle, reradiate, pajamas, the surface radiance to the point where continued decreases in temp slow to a crawl.

Dave Springer
May 8, 2011 9:54 am

ferd berple says:
May 8, 2011 at 8:22 am
I don’t want to rain on your parade but the temperature at the top of the atmosphere is in the thousands of degrees. The TOA boundary for energy budget calculations is the top of the thermosphere. So the temperature is indeed much higher at TOA than it is at BOA.
Your argument based on lapse rate is only valid to the top of the middle atmosphere and falls apart in the upper atmosphere where temperature increases with altitude.
It’s very thin atmosphere but it’s still atmosphere and its temperature can be measured. So the temperature at the bottom of the atmosphere is indeed much colder than at the top.

Gilles
May 8, 2011 10:02 am

In my view, the explanation of the 33°C difference is somewhat tricky.
If the atmosphere were totally transparent to radiation, the effective temperature of 280 K would be that of the ground (although, as I mentioned above, this effective temperature does not coincide with the average surface temperature). So the atmosphere must be opaque to thermal IR radiation to produce an effective temperature at the TOA different from the ground one. But absorption of IR radiation IS greenhouse effect. So “some ” greenhouse effect must be present to make the atmosphere partly opaque.
BUT the open question is to compute the average difference between ground and TOA and this is much less obvious. In the absorption lines, the outgoing LWR intensity is determined by the local temperature at the last diffusion surface – so the upper atmosphere temperature. But the heat transfer in the troposphere is a complicated mixture of convection and radiation, and is indeed dominated by convection. So the difference temperature is mainly controlled by the adiabatic law between ground and and tropopause – and this can NOT be computed only with radiative physics. So in some sense, those claiming that the difference between the ground and the TOA is mainly due to gas thermodynamics are also right – this is a very complicated mixture of convective transport and radiative opacity.
This can only be solve through detailed modeling – and it is quite possible that this modeling is inaccurate.
A further complication is that as I mentioned , average ground temperature is NOT the same as effective temperature, so there is no clear relationship between the two: The Earth is NOT an isothermal copper sphere ! actually the average temperature depends strongly on meridional circulation that transports a lot of heat towards the high latitudes – especially with oceans that are, to say the least, not very well described and understood. Any variability in oceanic circulation could have strong effects on local, and hence average temperature, even with a fixed energy budget. This is again totally overlooked in “simple” isothermal , radiative arguments. I would say IMHO that this is actually the less known feature of current GCM – and the main reason for doubting about their accuracy and predictive power. And I am not really surprised that they struggle so much with the “lack of warming” .. 🙂

Dave Springer
May 8, 2011 10:04 am

ferd (con’t)
The reason the middle and lower atmosphere get colder with height isn’t because of compression. It’s because in the middle and lower atmosphere the sun doesn’t do any appreciable heating of it.
One must always begin with the big picture and work down to the smaller details not try to add up the details into a big picture.
In the big picture the sun heats the ocean, the ocean heats the atmosphere, and the frigid cold of the cosmic background (3K) cools the atmosphere.
The lower atmosphere is warmer than the middle atmosphere because the lower atmosphere is closer to the source of the ocean that heats it.

Massimo PORZIO
May 8, 2011 10:15 am

Ira said:
“Notice that the Earth System mean temperature I had to use to provide 240 Watts/m^2 of radiation to Space to balance the input absorbed from by the Earth System from the Sun was 255 K. However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
No, by my point o f view there is another explanation. The incoming radiation from the Sun is truly measured from the radiometers because of the distance from the source we get all the parallel energy rays in the SW, while the emission of the Earth in the LW range is scattered in a spheric space around the planet, so the satellites’ radiometers can’t really measure all that energy because those rays are not parallel each other and many of them will never be perpendicular to the radiometer input slit.

Dave Springer
May 8, 2011 10:23 am

I suppose this is my segue into talking about the temperature of the ocean since it is responsible for the temperature of the lower atmosphere where we live and breathe.
The average temperature of the ocean is 4C. 90% of the volume of the global ocean is at a near constant temperature of 3C. Only the surface layer comprising 10% of its volume gets any warmer (or colder) than that. It’s not a temperature/density thing either because seawater, due to its saline content, increases in density all the way down to its freezing point of about -2C.
The only possible explanation for why the average temperature of the ocean is 4C is because that is the average surface temperature of the earth taken over a period of time long enough for convection and conduction to equilibrate the entire volume.
Ostensibly I should think that period of time is about 120,000 years or once complete glacial/interglacial cycle.
Any sane person armed with the facts should find their source of angst to be that huge volume of icewater lying just below the warm thin surface layer of the ocean. The current interglacial period is getting long in tooth so again by any sane measure we should be worried about an impending ice age and if there is any merit whatsoever in anthropogenic global warming we should be glad for it and try to get as much of it as we can in the hope that it might delay the inevitable return of glaciers a mile thick covering everything north of Washington, D.C.
I wonder how many of the people I share the planet with are sane in this regard. The facts on the ground are incontrovertable. The only thing we have to fear is more ice not less ice. For this reason I call the warmista’s “ice huggers” in fond remembrance of their predecessors the “tree huggers”. The tree huggers morphed into ice huggers. Ain’t that a hoot? Trees don’t grow well in ice. They are lucky to survive where ice prevails much of the year much less thrive in such conditions. A warmer earth is a greener earth and a greener earth is what we all want, innit?

jae
May 8, 2011 10:25 am

Dave Springer:
Please “debunk” the article at the link given by DirkH: http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html
You folks are, in essence, saying that the lapse rate is CAUSED by the GHE. That is nonsense. The lapse rate is dependent upon only two variables, gravity (g) and the heat capacity of the air (Cp) and can easily be calculated from these two variables. GHE has no “place” in this formula. The lapse rate for ANY atmosphere (even pure nitrogen or oxygen) would be calculated the same way–without “allowance” for any GHE.

Anthony Zeeman
May 8, 2011 10:58 am

No mention is made of ocean and atmospheric heat transport. With one side of the Earth exposed to near absolute zero temperatures and the other exposed to the Sun; any theory has to take into account the movement of vast amounts of heat energy from hot to cold and vice versa. Although the Earth may lose heat at night, there is still substantial heat transfer from the day side to the night side keeping the night side warmer than it would be otherwise. The true “greenhouse effect” is a perfect example. Without convective heat transport a greenhouse becomes much hotter than its unenclosed environment. Simplistic models with lots of fudge factors to make the numbers fit are insufficient to describe the climate.

Dave Springer
May 8, 2011 11:02 am

More on heating and cooling of gases due to compression and expansion.
Heat and cooling only takes place as the gas pressure is changing. When the pressure stops changing so does the temperature. There is very little dynamic variation of atmospheric pressure at any constant altitude so there is very little heating or cooling taking place due to pressure variation.
In a real world example I have an air compressor in my shop. If the air in the tank is not compressed and I run the compressor to bring up to 10 atmospheres the tank will get much warmer than the air in my shop and if I quickly bleed it back to 1 atmosphere the tank will get colder than the air in my shop.
However (this is the key) if I leave the tank full of compressed air it won’t stay warmer than the air in my shop for long. That’s because the pressure is constant.
Ferd’s assertion that atmospheres get warmer as you go deeper into them because of compression is quite wrong. If there were some mechanism that were taking volumes of air high in the atmosphere and transporting them lower in the atmosphere his thesis would have some merit but there is no such mechanism at work on Venus, Earth, or Mars. The pressure gradient is nearly static and if an actual volume of gas is neither expanding nor contracting there is no associated change in temperature.

May 8, 2011 11:07 am

“A warmer earth is a greener earth and a greener earth is what we all want, innit?”
It certainly is surprising that people fear the earth getting warmer, yet when it comes to spending their hard earned money to go on vacation, few it any head towards the poles. If warming is something to be feared, where did the term “tropical paradise” come from? Why isn’t it “polar paradise”?

DirkH
May 8, 2011 11:10 am

davidmhoffer says:
May 8, 2011 at 9:49 am
“Once the temp hits -25 not only is the “colder” atmosphere above the artic backradiating, reradiating, simulating pajamas, what ever term you want, it get’s more efficient at it. So it can’t EVER cause an INCREASING temperature trend, but it can certainly recycle, reradiate, pajamas, the surface radiance to the point where continued decreases in temp slow to a crawl.”
David, this means that at -25 deg the upper atmosphere is no more colder than the ground; an equilibrium is reached. Of course, nothing happens from there; but your premise was to show that the people who say a cold object cannot warm a warm object are somehow wrong – your example shows that two equally cold objects are in equilibrium and does not contradict those people at all. In your example, the upper atmosphere is no more colder than the ground.

May 8, 2011 11:13 am

I don’t want to rain on your parade but the temperature at the top of the atmosphere is in the thousands of degrees.
Then it must be back radiating a fantastic amount of energy to the earth. At thousands of degrees this radiation would be in the visible or even UV.

Gnomish
May 8, 2011 11:14 am

” there is no good explanation for the approximately 33ºC (58ºF) temperature difference between the “light” input from the Sun and the “heat” output from an Earth”
temperature is not heat – you can’t measure heat with a thermometer. degrees can not be converted to watts.
the sidewalk doesn’t get hot because the air above it warmed it.
the sidewalk stores heat and warms the air above it.
incidentally, phase change from liquid to gas has no temperature change – but loads of heat is involved.
incidentally, avogadro’s ‘laws’ trump co2 fetish fantasy – co2 molecules don’t get to be hotter than everything else around them – all molecules in a local region are basically at the same temperature via kinetic transfer.
/me sick of co2 shrimpers. it’s like an old ed wood flick with fat ladies in a playpen and some old fart claiming to be the walrus. or maybe it’s like divine following that pooch.

May 8, 2011 11:16 am

“It was stored as water vapor evaporating from ground and bodies of water from solar heat, which rise to be COOLED, where the kinetic energy from gravity is then brought back to Earth with the rain.”
I recall years ago seeing a calculation that the amount of heat energy delivered to the earth by rainfall significantly exceed that delivered by bright sunlight.

Dave Springer
May 8, 2011 11:17 am

Ira:
We seem to be out of synchronization. In a series of comments I left the second in the series explained why the moon is not at its theoretical black-body temperature i.e.. because it isn’t a black-body. It’s a gray body with an albedo around 12%. I found the missing heat for you in that comment. I then went on to explain that the earth is also a gray body with an albedo of 30% to 40% and pointed out the crux of the issue – the earth’s average albedo isn’t know to any better certainty than +-5% and +-5 uncertainty in albedo translates into 25 watts per square meter of uncertainty in how much insolation is actually absorbed at the surface. Moreover all actual attempts to measure albedo, while disagreeing by 10% (I actually only found 7% disagreement in actual studies but still…) all do agree that the average in any given year is not static and varies from year to year. One experiment using earthshine as a measure that went on for about five years found a 1.5% difference between year one and year five. Their methodology in that one was measuring the brightness of the new moon which is illuminated solely by light reflected from the earth. It’s a bit surprising how difficult it is to get an accurate measure of the earth’s average albedo in any one year but on the other hand it isn’t difficult to measure change from year to year. The earthshine measure might not be accurate but it is precise and consistent. Precision and consistency is all you need to identify trends.
[Thanks Dave Springer and I appologize for not remembering that detail of your earlier attempts to clarify the issues. Please continue to share your special knowledge of climate science with me and others at WUWT. Your comments are most welcome here in my threads, even when they are corrections to my mistakes. In fact, especially when they help me improve my understanding and learn more about this topic. THANKS! – Ira 9:25PM EST]

May 8, 2011 11:19 am

jae says:

You folks are, in essence, saying that the lapse rate is CAUSED by the GHE. That is nonsense. The lapse rate is dependent upon only two variables, gravity (g) and the heat capacity of the air (Cp) and can easily be calculated from these two variables. GHE has no “place” in this formula. The lapse rate for ANY atmosphere (even pure nitrogen or oxygen) would be calculated the same way–without “allowance” for any GHE.

The lapse rate is not caused by GHE’s per se. Your statement about the lapse rate is more or less correct…except that this sets a maximum lapse rate. That is, it is possible for an atmosphere to have a smaller lapse rate than you get from the stability calculation that you allude to. However, it won’ t have a steeper lapse rate because such an atmosphere becomes unstable to convection, with then brings the lapse rate down to the marginal stability value.
So, the best way to put it is that the lapse rate is caused by the fact that the atmosphere is heated from below in concert with stability arguments based on buoyancy, adiabatic expansion, and the existence of convection.
What the GHEs determine is where in the atmosphere the emitted radiation is able to escape to space. Radiative balance of the earth system then sets the temperature at this level in the atmosphere and the temperature at the surface basically follows from the lapse rate.

DirkH
May 8, 2011 11:29 am

Dave Springer says:
May 8, 2011 at 10:23 am
“For this reason I call the warmista’s “ice huggers” in fond remembrance of their predecessors the “tree huggers”.”
Good idea. They really do love their Greenland ice and all that.

Anton Eagle
May 8, 2011 11:32 am

Ira,
I have been reading many posts here that discuss the blackbody radiation of the earth. You have claimed that the 33K difference is due to GHG back radiation. I don’t want to argue that with you… but I think it would be worthwhile to be very very clear on exactly what you are claiming.
So… I can boil this down to a simple yes or no question…
Are you claiming that if the atmosphere were replaced with a different set of gasses that do not contain GHG molecules, but still had all the other macro effects such as clouds etc in the exact same amount as our current atmosphere… are you claiming that this atmosphere would NOT cause (directly or indirectly) the surface of the earth to warm at all?
You see… it does seem as if the answer to the above question is yes. It does seem as if you are making that claim. Are you? Or, are we just misunderstanding your argument?
-Anton Eagle

May 8, 2011 11:33 am

John of Kent says:

We can account for the 33C difference without any need for an imaginary greenhouse effect. I urge everyone at WUWT to read this brilliant paper by physicist Joe Postma ;
http://www.globalwarmingskeptics.info/forums/attachment.php?aid=327

Bryan says:

You should read the article below.
It will answer all of your misconceptions about the “greenhouse theory” which in fact is pure fiction.
1. It is true that the radiative equilibrium temperature of the Earth is -18C.
2. This figure is confirmed by Satellite measurement from space.
3. It is NOT true that all the radiation emitted from the Earth surface.
4. The emission is mostly from the cloud level.
5. If we pick say around 5km as the average emission height to space we will be near enough.
6. Now use thermodynamics expressed in the form of the lapse rate.
7. An average moist lapse rate is -6.6K/km.
8. The lapse rate has nothing to do with back radiation as even scienceofdoom admits.
9. Work back down from -18C at 5km to the Earth surface and we get 5×6.6 = 33K.
So now you have an Earth surface temperature of 15C without any “greenhouse effect”.
http://www.ilovemycarbondioxide.com/pdf/Understanding_the_Atmosphere_Effect.pdf

Okay…So it is time to debunk the Postma nonsense again, even though we already did it in the last thread that Ira wrote.
The problem with the Postma argument is that the “average emission height” depends on…you guessed it…the IR-absorbing substances, namely greenhouse gases and clouds, in the atmosphere. So, the Postma argument does not get rid of the greenhouse effect…He merely sweeps it under the rug by obfuscating the fact that this average emission height being above the earth’s surface is due to the greenhouse effect.
Think about it. If the atmosphere contained no IR-absorbing substances, then all the IR emitted by the earth’s surface would escape into space and radiative balance would dictate that the earth’s average surface temperature (or really the average of emissivity*T^4 where T is the absolute temperature and the emissivity of most terrestrial materials in the wavelength range of interest is very close to 1) is set by the condition that the earth must radiate as much energy as it absorbs from the sun.
Now…Slowly add IR-absorbing substances. What happens is that some of the radiation from the earth’s surface is absorbed in the atmosphere, which subsequently radiates it again. This absorption can even occur multiple times but once the radiation occurs high enough up in the atmosphere that the remaining IR-absorbing substances above that layer are unlikely to absorb the radiation, then it can successfully escape to space. Hence, the average emission height is what is determined by the IR-absorbing substances in the atmosphere.
As the greenhouse gases in the atmosphere increase, this average emission height rises…and because of the lapse rate…this means less radiation is emitted back out into space. As a result, the earth system heats up until radiative balance is restored.

Anton Eagle
May 8, 2011 11:33 am

Ira,
I have a second question regarding the two emission diagrams you show.
The upper graph, showing down-looking readings at the top of the atmosphere shows that approximately 1/2 of the 15 um radiation is absorbed. It looks as if it would be about 90 mW/m^2 without absorption, and instead is about 45 mW/m^2 or so with absorption.
Then you show the lower graph showing the up-looking readings at the surface… showing about 90-100 mW/m^2 of radiation (at 15 um) down to the surface from the atmosphere.
So, my question is… if the GH gasses are absorbing about 45 mW/m^2… how are the re-radiating almost 100 mW/m^2 back to the surface?
It doesn’t look like the sun can be the source, since according to these curves the sun doesn’t radiate hardly anything at those wavelengths. So… where is the extra energy coming from?
Thanks,
-Anton Eagle

May 8, 2011 11:36 am

Ira Glickstein
Do you know what the adiabatic lapse rate is?.
On page 21 of the Postma paper it is derived.
This is not a controversial point, serious IPCC advocates include this derivation as part of their narrative.
If you like I can give you a couple of such references.
The clouds at 5Km are far better radiators than gaseous CO2 or H2O .
Yet strangely enough it is the Latent Heat and high specific heat capacity of water in clouds that accounts for cloudy warmer nights rather than radiative effects.
The convection mechanism is reduced if there is a smaller temperature difference between Earth Surface and clouds.

MartinGAtkins
May 8, 2011 11:40 am

John of Kent says:
No! The earths surface is already emitting radiation in the Infra Red. It’s atoms and molecules are ALREADY excited sufficiently to be emitting IR.
Yes..The molecules of the earth surface have kinetic energy because they are excited or are being exited with the absorption of radiation.
This excitement distorts the magnetic field of the molecules components in a way that they are not in balance as they would be if the molecule was at rest. This leads the electromagnetic field of the molecule to send out waves of electromagnetic energy into space at decreasing rates if the source of the energy is turned of.
When the source energy is turned off the emitting waves from the molecule don’t cease immediately because the molecules components have mass and it is the movement of these parts that cause the decreasing waves of electromagnetic disturbance as energy is lost.
Similarly when the molecule is hit by a magnetic wave it will not reach its full kinetic and radiative potential until all its parts are moving in such a way as to propagate a field of disturbance equal to the constant of the energy input over time. At this point it is at radiative equilibrium with the incoming energy and no more work can be done.
Let’s say we are on the moon so there is no overlaying gas and pick an arbitrary incoming value of 100Watts/m^2. The mass of the particles at the surface will eventually increase in temperature until the moving parts of the molecules are so excited they cause a electromagnetic disturbance that is equal to the incoming electromagnetic energy from the sun. At this point the waves of energy from the surface propagate outward and are lost to open space. 100Watts/m^2 in, 100Watts/m^2 out.
This however is not the case with our planet. The over laying gas doesn’t impede the inflow of energy to the surface but the principle remains the same as the moons surface in that in its self the surface will move toward radiative equilibrium with the incoming energy.
This time however the outgoing radiation is now blocked by a molecule of gas which of course also has mass and a state of rest.
The propagating electromagnetic wave from the earths surface is now absorbed by the molecule which in turn adopts kinetic and thereby electromagnetic properties as the state of rest is now disrupted and its components vibrate, sending out electromagnetic waves as it struggles to balance its competing internal forces.
Although free floating the rules for the gas molecule are the same as the molecules at the earths surface. It will absorb all the incoming energy from the earths surface until it radiates away the same amount of radiation it is receiving. It will do this in all directions, so as it moves to reach radiative equilibrium with the surface it will radiate half toward the surface and half away. The energy being radiated back to the surface is not kinetic, it is electromagnetic. It would be 50 Watts/m^2 in the first instance of time with the other 50 Watts radiating away. The surface doesn’t know where the 100 Watts or the 50 Watts are coming from it only knows it is receiving 150 Watts in the second instance of time.
Think of this. If we had a closed system where no energy is lost and we were to put 100 watts a second into the system we would have a system with 100 joules after the first second. Now in the next second we only put 50 Watts a second into the system we get 150 joules in the system. The system doesn’t care that the that the last second had a lower value than the first second it only knows that over two seconds it has received 150 Watts of energy.

May 8, 2011 11:49 am

John Marshall says:

Taking into account the adiabatic effect, which is real and measurable, there is no need for the GHG effect.

No…It does not as I have explained above.

The greenhouse effect also has one fatal flaw. The warming in the mid troposphere in the tropics shown by all models has yet to be found in real observable data. I do not care what your calculations show (in reality another model), if this warm area shown by models is not there then the theory falls flat on its face!

The prediction of tropical tropospheric amplification has nothing to do with the greenhouse effect. It has to do with the fact that in the tropics the temperature decrease with height is expected to closely follow the moist adiabatic lapse rate. Hence, this amplification is predicted for warming due to any mechanism.
It is also worth noting that tropical tropospheric amplification is well-verified by the satellite and balloon data for the temperature fluctuations that occur on monthly to yearly timescales, e.g., due to ENSO. Where the data is more ambiguous is with the long-term multidecadal trends, where such data is easily contaminated by artifacts due to changes in instrumentation over time, switches from one satellite to another, etc. The different data sets…and various analyses of the data sets tend to differ not only from the model predictions (depending on which data set you look at) but also from one another.
Even Richard Lindzen agrees that tropical tropospheric amplification follows from basic physics and has nothing to do with the greenhouse effect mechanism…and that the fact that is a problem with the observational data, not the models (and, at any rate, nothing to do with the mechanism causing the warming). His only point of departure is that he thinks the data more likely to be wrong is the surface data in the tropics whereas most others think it is the data at altitude.
The specific fingerprint of warming due to an increased greenhouse effect (at least relative to, for example, the mechanism of heating by increased solar radiation) is that the greenhouse effect is predicted to cause cooling in the stratosphere. And, this is indeed what is seen. (Some of the cooling is due also to the decrease in stratospheric ozone but the amount and altitude distribution of the cooling apparently cannot be explained solely by this mechanism.)

Dave Springer
May 8, 2011 11:53 am

Ira Glickstein, PhD says:
May 8, 2011 at 10:41 am
“I hasten to add that engineers are not as picky about precision of estimates as scientists and mathematical analysts are about their data. If we get estimates within 15% (33ºC vs 28ºC) that is usually close enough, since other uncertainties are likely to be in the mix. ”
Acceptable margins of error are established in context. Sometimes being within an order of magnitude is good enough and other times the proverbial nine nines isn’t good enough. But I understand your point. It’s illustrated by a trite expression that +-10% is good enough for government work. Left unsaid is that the government screws up a lot because what’s considered “good enough” by them often isn’t.

Dave Springer
May 8, 2011 12:06 pm

Ira:
A chain is no stronger than its weakest link. I think we can agree on that.
When it comes to trying to sort out natural forcings from unnatural forcings we run into a link in the chain where we have a 25w/m2 range of uncertainty in surface forcing from natural factors. We might reliably know that unnatural forcings fall in the range 0.6 – 2.5w/m2. So that’s a strong link but it’s strength is meaningless when there’s a link in the chain with only a tenth of that strength. See what I’m saying?

May 8, 2011 12:06 pm

Martin Mason:
Ira, the Second Law of thermodynamics works at every level and you cannot transfer heat from a colder body to a warmer body. A black body cannot also reabsorb and re-emit lower intensity radiation that it has already emited.
You clearly do not know the basis for the Second Law, which is statistical physics. You seem to believe in some sort of magical version of the Second Law. The actual Second Law is based on the fact that although energy transfers occur in both directions, by simple statistics it becomes essentially astronomically improbable that the net flow of energy, what we call “heat” will be from the hotter body to the colder body for any macroscopic bodies.
In particular, applied to two radiating objects at different temperatures, the Second Law simply states that the amount of radiation from the hotter body that is absorbed by the colder body is always larger than the amount of radiation from the colder body that is absorbed by the hotter body. The laws of radiative transfer automatically satisfy the Second Law when applied correctly and all models of the greenhouse effect, be they toy models or advanced radiative-convective codes, satisfy the Second Law.

Greenhouse heating by back radiation is surely wrong because it says that if you put a frozen steak inside a vacuum flask with reflective interior that the steak would cook, that if you stand in front of a mirror you will heat up from the reflected rays

No…That doesn’t happen because of simple conservation of energy. The steak is not producing thermal energy or receiving it from a hotter object. So, all that the reflective interior can do is slow the cooling of the steak.
The earth is receiving lots of thermal energy from a hotter object, namely the radiation that it receives from the sun. Its steady-state temperature is set by the balance of what it receives and what it emits back out into space. The greenhouse effect, by affecting the rate at which the earth emits radiation back out into space for a given surface temperature, causes the earth’s temperature to warm in order to maintain radiative balance.

or that you can save on your heating bills by filling the loft with CO2?

Insulation in general does help you to save on heating bills. The reason that CO2 in particular does not is for a variety of reasons, one being that the heat loss from a house is due to many mechanisms besides radiation. (For the earth system as a whole, the only way it can lose heat is via radiation, although heat can be transferred between different parts of the system, e.g., surface to atmosphere by convection and evaporation/condensation.) Another is that, as discussed above, the full picture of how adding CO2 affects the greenhouse effect relies on the fact of the lapse rate in the troposphere, something that is not applicable in your example.

May 8, 2011 12:08 pm

Dave Springer says: May 8, 2011 at 10:23 am
The only possible explanation for why the average temperature of the ocean is 4C is because that is the average surface temperature of the earth taken over a period of time long enough for convection and conduction to equilibrate the entire volume.

Dave,
Wouldn’t the explanation be that the surface temperature where there are down-welling currents must be around 4C?
Currently (no pun intended) the major down-welling current is the gulf stream in the north Atlantic. As long as the north Atlantic is ~4 C, the currents along the bottom will be ~ 4 C. It is reasonable to speculate that down-welling currents would always be in cold regions, since that is where the surface water is most dense. (Add in the fact that evaporation of warm water as it heads pole-ward will increase salinity and increase density and we have a stronger reason for down-welling currents are in the polar regions.)

Charlie Foxtrot
May 8, 2011 12:11 pm

Several commenters maintained that a colder body cannot heat a warmer body. They may have confused heat with temperature. Heat can be transferred from a cold body to a hotter body, but a colder body cannot increase the temperature or heat content of a hotter body. However, a colder body can decrease the rate of cooling of the hotter body by radiation to the hotter body. Electrons jumping orbitals and producing radiation don’t know if there is a warmer or colder body out there somewhere, therefore the total instantaneous radiation from any body is independent of the temperature of the surroundings.
Something I did not see in the comments was mention of emissivity, which would have a direct impact on the sensible temperature of the air and surface of the earth. Good emitters (high emissivity) both absorb and emit energy better (grass, leaves, black bodies) than low emitters (most light colors, stainless steel). Emissivity will vary greatly from year to year depending on snow cover, clouds, crops, deforestation, etc.
Dr. Glickstein’s discussion is interesting, but I think is too simplified to convince any warmist that CO2 is not cooking the earth.
A couple of things that I am curious about that directly affect the emissions balance, and which have not appeared in any discussion I have read, are:
1. Energy is absorbed and converted to chemical energy through photosynthesis. I have no idea how much energy that would be (probably not much), but with higher CO2 levels, vegetation is growing faster so it is logical that more radiation is being converted to plant material than in the past, which is stored energy that will not be re-emitted until the material is burned or otherwise reverts to it’s former state.
2. The earth might correct temperature variations by moderating emissions, especially in the Arctic. When the northern oceans are warmer than normal, they freeze later in the season, and since they emit to an almost perfect black body (deep space with no sunlight) they release much more heat and take up very little, esp. in winter, helping to correct the global temperature. Ice and snow are poor emitters and also insulate the warmer ocean beneath whereas a dark ocean is a very good emitter. A change in ocean currents, for example diversion of a warm pacific current to the Arctic for a prolonged period, could cause global cooling by eliminating ice in the Arctic all year. I have no idea how significant this cooling would be, but suspect it is substantial. Perversely, an abnormally warm current to the Arctic would cause warming of the atmosphere while reducing heat content in the ocean, making it appear as if the earth is warming when in fact it is cooling. Do the GCM’s address this?

William
May 8, 2011 12:21 pm

As there are multiple periods in the geological record of tens of millions years in duration when CO2 levels were high and the planet was cold and when CO2 levels were low and the planet was warm, it appears there is a basic fundamental assumption in the model of atmospheric radiation that is incorrect or there is an omission of another mechanism from the standard models of atmosphere radiation.
One possibility is the greenhouse mechanism saturates such that the initial CO2 causes most of the warming.
Gas molecules transfer energy from molecule to molecule by collisions in addition to radiation.
As one moves higher in the atmosphere there are more ions due to galactic cosmic rays striking the atmosphere. An ion radiates continuously due to motion of the ion in addition to the band specific radiation that is emitted when an ion captures an electron. It is possible that radiation from ions provides the leak to cause the CO2 mechanism to saturate.
Atmospheric carbon dioxide levels for the last 500 million years
http://www.pnas.org/content/99/7/4167.full
Using a variety of sedimentological criteria, Frakes et al. (18) have concluded that Earth’s climate has cycled several times between warm and cool modes for roughly the last 600 My. Recent work by Veizer et al. (28), based on measurements of oxygen isotopes in calcite and aragonite shells, appears to confirm the existence of these long-period (_135 My) climatic fluctuations. Changes in CO2 levels are usually assumed to be among the dominant mechanisms driving such long-term climate change (29).
Superficially, this observation would seem to imply that pCO2 does not exert dominant control on Earth’s climate at time scales greater than about 10 My. A wealth of evidence, however, suggests that pCO2 exerts at least some control [see Crowley and Berner (30) for a recent review]. Fig. 4 cannot by itself refute this assumption. Instead, it simply shows that the ‘‘null hypothesis’’ that pCO2 and climate are unrelated cannot be rejected on the basis of this evidence alone.
http://www.nature.com/uidfinder/10.1038/nature01087
Despite these successes in linking variations in greenhouse gas concentrations to climate change in the geologic past, the oxygen isotope palaeotemperature record from 600 Myr ago to the present displays notable intervals for which inferred temperatures and pCO2 levels are not correlated1. One of these occurred during the early to middle Miocene (about 17 Myr ago), a time well established as a warm interval (relative to today), but with proxy evidence for low atmospheric pCO2 (ref. 2). Moreover, whereas climate models predict tropical warming in response to elevated pCO2, geologic data — in particularly the oxygen isotope record — indicate muted warming or even cooling at low latitudes while higher latitudes warm (the ‘cool tropicsparadox’10–11).

Martin Lewitt
May 8, 2011 12:24 pm

Bryan,
The 2.66 grams per square inch figure is considering the whole air column.
http://www.wolframalpha.com/input/?i=14.7+lbs+*+400%2F1000000
Alan D. McIntire,
The solar radiation hitting the earths surface is 184 W/m^2 per Trenberth 2009 or 198 W/m^2 per Trenberth 1997. The surface albedo must be applied to this to get the net gain. Back radiation adds another 333 or 324 W/m^2 respectively. Since Trenberth’s diagram don’t show any reflection at these wavelengths , it is unclear whether he has applied the albedo that corresponds to these wavelengths. The albedo of surfaces varies with wavelength, and for some surfaces like snow the variation is quite significant.
If the emission from a surface is blocked from escaping in some parts of its spectral emission range, then, everything else being equal, it must increase in temperature until enough of its spectrum is in the ranges that do allow the energy to escape. That would be its equilibrium temperature. Of course, everything else isn’t equal, it will never have to reach that temperature, because conduction, evaporation, convection, poleward heat transport, etc. will assist in transporting the heat away.

stumpy
May 8, 2011 12:42 pm

Energy in = energy out is not technically true though, the heat that reaches the surface is attenuated as the surface warms and cools, for example heating of the ocean surface and ground surface. When the sun first hits them in the morning they are not instantly at thermal equilibrium and radiating what they recieve straight back upwards, plus the deeper ocean can store and release energy – I know over time energy in = energy out, but it doesnt actually apply to any moment during a day, its why the coldest temps are night are the early morning, and why its warming in the afternoon, not when the sun is highest. I think the diurnal pattern of energy release is also something worth considering, as it does not match the incoming diurnal pattern.
I think to further peoples understanding the impact of surface attenuation and the affect on the ocean on the global energy flow need to be discussed, as the greenhouse effect doesnt quite hold if these affects are not considered when looking at a point in time on the earth

Super Turtle
May 8, 2011 12:50 pm

The issue at hand here is not that co2 can absorb energy, but what occurs and how does it contributes to AGW, and specific man’s co2 output.
If I suspend in the middle of my room a 4 foot by 6 ft by 2 inch thick lead plate in the middle of my room (not touching the walls), then while that lead plate has huge ability to absorb energy (1000’s of times more than co2), that introduction of the lead plate will NOT cause the room to warm up. So, any energy absorbed by this lead plate will be re-emitted back into the room.
If the above lead plate could heat up the room, the we would all purchasing big lead plates to suspend in the middle of the room for free heat.
On the other hand, increasing the insulation of the WALLS of the room is a different matter.
So, at the end of the day is not that co2 can absorb energy, but what occurs with that effect. As the lead plate example shows, just because physics says something can absorb energy does not necessary mean it affects the inflows and outflows of energy in a given system.
And in fact it not the co2 that holding this energy anway, but the other substances such as ground and ocean and other parts of atmosphere that represents the bulk of the “mass” of the atmosphere to hold that heat energy.
In fact, this explains why engine blocks and especially motor cycle heads are often painted black, not silver in color. The choice of black paint on the engine can cause the engine to cool better! In other words, placing black paint on an engine block INCREASES cooling effect yet black paint is known to absorb energy better!
In effect, since black is a good absorber of energy it’s also thus becomes a good emitter of energy. Black objects emit more radiant energy (cools faster) than a white or silver to its cooler surroundings.
The issue here never been that co2 can absorb energy, but the issue is not the slam dunk as to what effect this has in terms warming the overall system, or even as some claimed can in fact cause cooling of the system based on the same reasons why black paint on engines can help cooling!
Super Turtle

Nullius in Verba
May 8, 2011 12:53 pm

“If there were some mechanism that were taking volumes of air high in the atmosphere and transporting them lower in the atmosphere his thesis would have some merit but there is no such mechanism at work on Venus, Earth, or Mars. The pressure gradient is nearly static and if an actual volume of gas is neither expanding nor contracting there is no associated change in temperature.”
Hadley cells transport volumes of atmosphere from high up to lower down. Don’t confuse the motion of the fluid with its pressure distribution. The fluid can move across a pressure gradient, expanding and contracting as it goes, while the pressure at every point remains constant.
If there were no convection loops on Venus, it wouldn’t be so windy there.

HankHenry
May 8, 2011 1:20 pm

It seems that for incoming sunlight there is one surface (where sunlight strikes) but for the outgoing radiation there is another surface (at the top of the atmosphere for some wavelengths). One might say that the heat that consumes time migrating from the incoming surface to the outgoing surface represents the heat of greenhouse warming.
I always wonder about the 15 degree C number quoted as the earth’s surface temperature. It seems that this is the just the air temp nearest the surface struck by sunlight (assuming I don’t misunderstand the source of the number). Why wouldn’t it be more appropriate to attempt to average all temps (air, sea, and land) inside the regions somehow touched by solar radiation (but excluding deep earth temps influenced by heat of radioactivity).

Dishman
May 8, 2011 2:12 pm

Ron House wrote:

Dishman says:
In re Objection #2:
Gas molecules follow ballistic trajectories between collisions…
… unless all the work on Gravity since Galileo is wrong, and doesn’t apply at the molecular level, of course.
Ballistic trajectories are trajectories under the action of gravity. Note, he didn’t say they travel in straight lines.
However: the approximate speed of molecules in the atmosphere is about 1,700 km/h. The path isn’t very curved, in other words. And the mean free path of a molecule in air is about 65 nanometres. You’d be pretty amazing if you could measure the deviation from a straight line of a path segment with that length and curvature.

Whether or not it’s directly measureable doesn’t matter. What matters is whether or not it’s happening.
If molecules follow ballistic trajectories, the minimum KE will occur at zenith and be higher at all other points.
A short MFP just means that equipartitioning will be rebalanced (on average) every MFP.
Assuming air (a diatomic gas with 5 degrees of partitioning), for a given delta-h, the thermal energy will be reduced by 1/5th of the energy required to lift molecules by that delta-h.
My point is that Objection #2 is not, in fact, correct.

May 8, 2011 2:24 pm

The solar radiation hitting the earths surface is 184 W/m^2 per Trenberth 2009 or 198 W/m^2 per Trenberth 1997. The surface albedo must be applied to this to get the net gain. Back radiation adds another 333 or 324 W/m^2 respectively.
There seems to be a whole lack of common sense going around these days.
So, according to T, the earth receives 184 w from the sun, radiates some of this back outwards, where some of it is absorbed by the atmosphere and re-radiated back to earth, where it provides 333 w.
In other words, 184 w ends up as 666 w, because the back radiation is omni-directional at at most 50% reaches the earth.
Of course, it is like two mirrors pointing in towards each other. Start out with 184w and after it reflects back and forth a bunch of times, you end up with 333w.
Only one problem. If you start out with 184w from the sun, you can NEVER absorb more that that. Where is the excess energy coming from? The sun is the energy source.
The atmosphere cannot manufacture 333 w of back radiation for a 184 w power source. The back radiation from the atmosphere must be less than the original 184 w from the sun.
Otherwise we could put an ingenious array of one-way mirrors around a 100 watt light bulb, and generate millions of watts of power. What climate science has invented is a perpetual motion device, where the energy out is more than the energy in.

May 8, 2011 2:31 pm

“If there were some mechanism that were taking volumes of air high in the atmosphere and transporting them lower in the atmosphere his thesis would have some merit but there is no such mechanism at work on Venus, Earth, or Mars.”
There is massive movement of air from high altitudes to low altitudes, as a result of an equally massive movement of air from low altitudes to high altitudes.
http://en.wikipedia.org/wiki/Hadley_cell

wayne
May 8, 2011 3:01 pm

Here is an experiment to disprove the existence of back radiation.
CAUTION: Only perform with parental supervision.
CAUTION: Treat any large Fresnel lens as if it were an open flame!
CAUTION: Since this experiment is indoors, pay close attention to ANY light or infrared (IR) source, which could be focused by the lens. That includes sunlight through windows, hot kitchen stove filaments, etc.

The only thing great expense in performing this experiment with be the Fresnel lens itself. One square meter Fresnel lenses are available through science supply sites. The lens need to be a plano-convex radial lens to focus parallel rays onto a one square centimeter spot (or even down to one square millimeter). I have heard an F-stop near one is best. This gives a one ten-thousand multiplication though that will to be reduced by up to ~50% due to refraction, scattering and optical imperfections in the facets.
You will need a very small black metal cap to hold one cm³ (1 gram) of water or ice. This will be placed at the focal point of the lens. You will need an infrared (IR) heat source, a space heater, large wattage hair drier, something to add additional warmth to a wall later in the experiment. You will later need an ice cube.
Similar to:
http://www.modulatedlight.org/optical_comms/fresnel_lens_comparison.html
http://www.scientificsonline.com/large-fresnel-lens.html?&cm_mmc=Mercent-_-Google-_-NULL-_-3052833&mr:trackingCode=CC6E1735-DB81-DE11-8C0A-000423C27502&mr:referralID=NA
Now, ready? If back-radiation proponents are correct, we should be able to use this huge lens to focus back-radiation from any warm wall within your house, at the ambient room temperature of say 20 ºC (68 ºF), in some room to boil water in seconds.
Use a small flashlight to illuminate the wall and so the exact focal point can be located. At this point place the black cap with the one gram of water. Wait. If they are correct the water should evaporate quickly, even boil, for that 70 ºF wall is giving off 5.67e-8*(273K+20K)^4 or 418 Watts per square meter that is going to focus down to 400 W x 5,000 multiplication factor or 2,000,000 Watts per square centimeter and we only need 2250 Joules to vaporize that gram of water. Wait, maybe there are some loses within your setup so be patient. Wait. Don’t know about your experiment results but mine is not working. Half an hour later I gave up, something mystical is blocking all of those back radiation photons from entering that lens. Watts up with that?
Now replace the water and cap with an ice cube. it should develop a hole slowly at the focus. Here the temperature at the focus is less than the temperature of the wall and radiation flows.
But I want to see the water boil, so cover the lens and warm up that wall in front of the Fresnel lens to a much higher temperature, let’s say 104 ºF (40 ºC). The watt flux should now be something near 5.67e-8*(313K-293K)^4 = 0.009 W/m^2 or 50% efficiency of 0.009 W/m^2 x 5,0000x = 45 W/cm^2. Uncover the lens. Well, it probably won’t boil but you should now evaporate that gram of water in about 2250 J/g ÷ 45 J/s or 50 seconds if you have a perfect setup with no loses. Let’s just say in a couple of minutes.
Careful, but now fill the cap and point that heat source, directly into the lens from five or six feet away, this is why Fresnel lenses can be dangerous, the water should now boil and quickly in seconds.
Thermal radiation heat going backwards does not exist in reality, at all, if you are a scientist type and not merely a AGWChurch member, run the experiment, prove it in reality.
Sorry back-radiation lover’s, you love is just in your minds (and wrong application of Stefan-Boltzmann equation).
That is also why Fresnel lenses are not dangerous in a very warm dark room at the focus. That room is at radiative equilibrium due to everything being at the same temperature and there is, in reality, no radiation at all, or you could focus it.

Martin Lewitt
May 8, 2011 3:11 pm

Ferd Berple,
You are just looking at the downward radiation, the upward radiation balances it, there is no energy problem. The mirrors on two sides of the bulb has some merits as an analogy, you wouldn’t doubt that the total flux of photons going back and forth between the mirrors sums to more than the 100 watts being input, but at equilibrium, only 100 watts escapes. A completely enclosed bulb is not a good analogy, because the energy is allowed to accumulate, the filament’s resistance will increase with temperature, reducing the amount of energy input, but still the bulb components will eventually melt.
In the climate case, the system is very complex, because the increase in temperature in response to the back radiation, increases a number of other responses to increased temperature before equilibrium is reached.
The idea that the CO2 effect is insignificant, and adiabatic explains all, fails for the common observation of another greenhouse gas, H2O. If it isn’t back radiation, how do you explain the difference in heat loss, on a clear calm humid night vs a clear calm dry night?

May 8, 2011 3:17 pm

DirkH;
David, this means that at -25 deg the upper atmosphere is no more colder than the ground; an equilibrium is reached. Of course, nothing happens from there; but your premise was to show that the people who say a cold object cannot warm a warm object are somehow wrong – your example shows that two equally cold objects are in equilibrium and does not contradict those people at all. In your example, the upper atmosphere is no more colder than the ground.>>>>
Read more carefully my friend. I said that at at 25,000 feet it is colder than that even at the equator. Two points:
1. According to AMSU-A http://discover.itsc.uah.edu/amsutemps/ satellite data the global average at that altitude is in the range of -38C. It would be warmer than that over the equator, and colder… much colder, over the arctic.
2. 25,000 feet isn’t “upper atmosphere”. In fact, it is only part way up the troposphere which is the lowest layer of the atmosphere. Troposphere ends at 11Km, then Stratosphere about 50Km, Mesosphere about 88Km and then the thermosphere about 500km and then the Exosphere.
Lot’s of other examples by the way. Build a house with no insulation and measure the temperature in the house and between the inner and outer walls with the furnace running at a constant rate. Then insulate the wall cavity. With the furnace running at the same rate the inside of the house will become warmer. The thermometer between the inner and outer walls will also read warmer, but still cooler than inside the house. The insulation absorbs outbound heat and reradiates it back into the house. Pile snow up around a house, and despite being colder than the inside of the house, the house gets warmer with the exact same heat sources inside, that’s how the early pioneers in NA insulated their sod shacks in the winter. That’s how the Inuit survived centuries in the far north by living the winter in igloos where the cold ice walls nonetheless kept them warmer than they would have been in the open…It’s even the same principle that a freezer works on as does an air conditioner. A kiln gets much hotter than an open fire, but put a thermometer inside the wall of the kiln and you will find that it is cooler than the temperature inside the kiln. Same with a forge. And no, these examples have nothing to do with convection, you can build a forge or a kiln with thin walls or thick walls and have the exact same air circulation, but the thick walled ones have a higher temperature inside, and the temperature of the walls is cooler.
I am a raging skeptic. An angry, they are misleading us with how they represent the results, producing results from tragically flawed and sometimes outright fraudulant “science” and are being used by lobbyists who are more interested in lining the pockets of their lobby than they are in what’s best for humanity, and I’m disgusted by the whole thing. But cold things radiate photons carrying energy and if they bump into something warmer, there are any number of factors that determine if they will be absorbed or not. But the relative temperature of the two bodies has zippo to do with it. The relative temperature of the two bodies only defines the magnitude and direction of the NET energy exchange, which can only be from warmer to cooler. But take the cooler body away, and the warmer body’s temperature must fall because it is no longer recieving any heat from the colder body.

wayne
May 8, 2011 3:38 pm

Martin Lewitt says:
May 8, 2011 at 3:11 pm
The idea that the CO2 effect is insignificant, and adiabatic explains all, fails for the common observation of another greenhouse gas, H2O. If it isn’t back radiation, how do you explain the difference in heat loss, on a clear calm humid night vs a clear calm dry night?
——
Simply the difference in the temperatures of one layer to the other (clouds in this case). The air above the cloud tops will be cooler than normal shielded by the water mass in the clouds and the air between the surface and the cloud base warmer than normal. Please, pick up a calculator and calculate these layer’s flux, the overall flux from surface to TOA when all added remains the same, but, the inter-atmosphere profile of the temperatures causing the flux at each layer will be different in the two cases.

Martin Lewitt
May 8, 2011 3:46 pm

Wayne,
For how much of the infrared black body frequency range is your Fresnel lens able to focus at that one spot? Is your lens material transparent to infrared?

May 8, 2011 3:57 pm

Joel Shore says:
…”Okay…So it is time to debunk the Postma nonsense again, even though we already did it in the last thread that Ira wrote.”……
Joel, in case any readers are unaware, is one of the authors of one of the most imaginative pieces of fiction ever to have made its way into a science publication.
Readers of Ira’s last thread will know that Joel postulated that;
1. Heat can move from a colder surface to a warmer surface spontaneously.
2. Invented statements and then proceeded to attack them.
3. Has the Stephan Boltzmann equation applying to gases.
Perhaps Joel is still a bit “rusty” about when and where the Stephan Boltzmann Equation applies.
It applies to black bodies issuing a continuous spectrum centred around a characteristic temperature.
It is the full integration of the Planck function.
It does not apply to the line spectra issued by H2O and CO2 making up less than 1% of the atmosphere.
Joels proof of the “greenhouse theory” involves getting rid of the Oceans and atmosphere.
Postma is absolutely correct in stating that the long wave radiation produced by the Earth system is released at an average height.
Some radiation from the ground leaves directly through the “window”.
Radiation also leaves from different heights above the surface.
The average height is about 5Km.
The lapse rate which is completely independent of the radiative effects then determines the surface temperature .
The greenhouse theory is like Joel’s paper, …… pure fiction.
Bare Rock Earth is Joel’s favorite topic and he rarely strays from it.

May 8, 2011 4:02 pm

Charlie Foxtrot: exactly. Well-said.
Dr. Glickstein: I appreciate your efforts to draw analogies to explain the CO2-induced warming of the Earth. However, I’m suspect the effort is doomed from the start because one cannot apply a steady-state model to a dynamic process. The Earth does indeed receive energy from the Sun, and radiates energy into space. But, the system is not at steady state due to Earth’s rotation around its axis, slight eccentricity of its orbit around the sun, greater or lesser cloudiness over time, absorption of energy into oceans or loss of energy from the oceans, and likely a host of other factors not mentioned in this list.
I’ve done a fair bit of outdoor cooking, over a campfire with a chunk of meat rotating on a spit above the fire. One would have a tough time creating a valid model of how the meat cooks using a steady-state model. Same thing applies to the Earth’s energy balance.

Bill Illis
May 8, 2011 4:21 pm

1 Watt = 1 Joule/second
1 Watt/m2 = 1 Joule/second/m2
Now convert the whole concept into Joules and bring time into the equation – say a 24 hour period. It will look much different.
The problem with all the calculations is that the definition of “Watt” has a time dimension already built into it so we tend to forget that all this has to happen in time.
The Sun is beating down at 960 Joules/second/m2 at the height of the day but there is almost no energy coming at night. How come the average Earth temperature is only the equivalent of 418 Joules/second/m2 in the daytime and 364 Joules/second/m2 at night.

wayne
May 8, 2011 4:40 pm

Martin Lewitt, it’s an experiment for you to perform, look on the specifications when your lense arrives, or better, mid-IR lens are also available, specify your choice when ordering. Let us know your results. ☺

astonerii
May 8, 2011 4:46 pm

“However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
Maybe the effect of compressing the atmosphere might cause an increase in of 33 degrees C.

wayne
May 8, 2011 4:49 pm

I wish my fingers would learn that ‘lens’ is singular, ‘lenses’ are the plural. ☺

astonerii
May 8, 2011 4:49 pm

“Yes, ferd berple, when a gas is compressed its temperature increases. However, once compressed, that increase in temperature will radiate away.
For example, when you pump up your bicycle tire, the pump and the air in the tire will be warmer than the ambient (uncompressed) air in the room. However, the air in the tire, although it is compressed to several times the normal ambient pressure, will not remain warmer for very long. The heat energy will radiate and convect and conduct away until the tire reaches equilibrium with the ambient temperature of its surroundings.”
A tire is an enclosed space where the air pressure is the same in all locations. The atmosphere is not. The air circulates. As it warms, it rises and cools, and as it cools it falls and warms. Go ahead and bat away a fact of science and try and shove your less than fully thought out process down some other people’s throats.

astonerii
May 8, 2011 4:59 pm

And continuing.
I would like to see the back radiation proponents do the following.
Take a spherical black body object with a small heat source in its center and shine a light on it and read the temperature.
Take the same spherical object and enclose it with a black body enclosure that does not touch the surface except at some required points to center it around it. use the same lamp at the same distance, and measure the temperature at the surface of the original sphere. Now make another black body enclosure and repeat the procedure. Do this until you have 12 layers of black body. If your argument is correct, then the temperature should sky rocket to the point of starting a fire. According to back radiation proponents statement of how it works.

DirkH
May 8, 2011 5:04 pm

Ira Glickstein, PhD says:
May 8, 2011 at 4:26 pm
“The same is true of the Atmosphere. If we transported an Atmosphere’s worth of air to a planet that had no Atmosphere, the force of gravity of that planet would indeed compress the Atmosphere and warm it. However, before long, the Atmosphere would cool down to the temperature of the planet. If we then placed that planet in an Earth-like orbit around the Sun, short-wave Solar radiation would be absorbed by the surface of the planet”
You’re right so far.
” and the surface would re-emit long-wave radiation to its Atmosphere where some would pass through and out to Space, while some would be absorbed by the H2O and CO2 and other “greenhouse” gases in the Atmosphere.”
Here it starts going downhill. There are three modes of energy transport from the surface upward: LWIR Radiation, conduction and convection, LWIR radiation being the weakest. IOW, the oceans get warmed by visible and UV radiation and will give the energy to the atmosphere mostly by surface conduction; kinetically.
” The heated gas molecules would bump into other air molecules and warm them, and like any material above absolute zero, the Atmosphere would emit radiation at a variety of long-wave wavelengths in random directions, some of which would be absorbed by the surface of the planet, warming it further. Just like our Earth.”
The gas molecules who get excited by LWIR photons would reradiate some immediately; in some cases they would give the energy to neighbouring molecules, thermalizing the energy. But Kirchhoff’s Law states that for any number of thermalization events there must be an equal number of dethermalization events as long as the gas is in local thermal equilibrium, which it is in the lower atmosphere. So no heat is trapped there.
See
http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/
So, the consequence is that LWIR radiation is transporting energy to space. Some is re-emitted back to Earth from where it is re-emitted back upwards. Whether there is 0.03 % or 0.04 % of CO2 in the atmosphere only influences how often the photons get absorbed and re-radiated on their way to space – an increase in CO2 delays the process a little but does not change it fundamentally and *Does* *Not* *Trap* *Heat* any more than a sieve traps water.

May 8, 2011 6:16 pm

Ira,
Please detail exactly why you think Swedish climatologist Dr. Hans Jelbring and chemical engineer William Gilbert are “foolish” “non-scientific Disbelievers” and “irrational” in their explanation of the so-called “greenhouse effect” based simply upon the adiabatic lapse rate:
http://www.tech-know.eu/NISubmission/pdf/Politics_and_the_Greenhouse_Effect.pdf
as well as physicist Joe Postma who explains the same
http://www.globalwarmingskeptics.info/forums/attachment.php?aid=327

May 8, 2011 6:36 pm

Bryan says:

Readers of Ira’s last thread will know that Joel postulated that;
1. Heat can move from a colder surface to a warmer surface spontaneously.
2. Invented statements and then proceeded to attack them.
3. Has the Stephan Boltzmann equation applying to gases.

Actually, what readers of that thread will know is that Bryan is a purveyor of pseudo-science. He has absolutely no serious desire to discuss the science and, like all purveyors of pseudo-science, tries to distract people by nitpicking words…even after the wording has been corrected to everyone’s satisfaction. (See here for more discussion specifically on the tactics that Bryan, Gerlich and Tscheuschner and other purveyors of such pseudo-science employ: http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/#comment-649274 ) The existence of people like Bryan within the skeptic movement, willing to actively engage in such deception, certainly helps to keep you guys marginalized in the scientific realm.

3. Has the Stephan Boltzmann equation applying to gases.

Joels proof of the “greenhouse theory” involves getting rid of the Oceans and atmosphere.

More nonsense. Simple toy models of the greenhouse effect illustrate the concept and give one a qualitatively understanding of what is going on. Line-by-line radiations codes do the full calculations for the actual absorption lines as they exist in the atmosphere in order to nail down the effects quantitatively. Yes, the simple models don’t have all the gory details in them…That is the whole point.

Postma is absolutely correct in stating that the long wave radiation produced by the Earth system is released at an average height.
Some radiation from the ground leaves directly through the “window”.
Radiation also leaves from different heights above the surface.
The average height is about 5Km.
The lapse rate which is completely independent of the radiative effects then determines the surface temperature .

Fine…Everything you say is basically correct. The one thing that you and Postma are leaving out is what I said in my last post…that the average height of 5km is determined by the IR-absorptivity of the atmosphere, i.e., by the greenhouse gases in the atmosphere and clouds. (The technical statement is something like this: At any given wavelength, the height at which most of the radiation escapes to space is given by the height at which the “optical depth” [ http://en.wikipedia.org/wiki/Optical_depth ] of all of the atmosphere above that point to radiation of this wavelength is of order 1. More quantitative details can be found in books on atmospheric radiation, such as Ray Pierrehumbert’s book “Principles of Planetary Climate”.)

John Runberg
May 8, 2011 6:41 pm

I haven’t had time to read all of the comment but I haven’t seen any comment on land-use/botany. Even in the oceans there are plants. Much of the earth is a GREEN body that is converting solar energy into some form of biomass. How many watts does it take to produce a bushel of wheat or a 1,000 board feet of pine lumber or produce a nice turf (or Veg. Garden) at the White House? It isn’t being factored in!

May 8, 2011 6:55 pm

Ira has explained it here: http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655793
I have explained it here: http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655686
The only way that adiabatic compression could possibly explain the earth’s surface temperature would be if the atmosphere (and / or the earth itself) were undergoing continual gravitational collapse, which we know is of course not the case. Otherwise, such a hypothesis does not even satisfy the First Law of Thermodynamics (basically, conservation of energy): Without substances in the atmosphere that absorb terrestrial radiation, the earth’s surface at its present temperature would be emitting back out into space way more energy than it receives from the sun and hence would rapidly cool down.
The temperature structure as a function of altitude, i.e. the lapse rate, in the troposphere is set by the considerations discussed regarding adiabatic expansion and compression (basically because a lapse rate higher than the appropriate adiabatic lapse rate for the given water vapor content is unstable and leads to convection until marginal stability is restored).
However, this alone does not determine the temperature at the surface. That is determined by consideration of the absorption of the atmosphere of terrestrial radiation (and radiation emitted by the atmosphere), which essentially ends up determining at what altitude the temperature has to be determined via radiative balance between the Earth system (earth + atmosphere) and the sun and space [which for the earth system with its current albedo is ~255 K]. Then, by using the lapse rate, you can get the temperature at the surface.
However, as one adds greenhouse gases, one raises this altitude and hence the required surface temperature. (In the simplest case, this is under the assumption that the lapse rate doesn’t change, whereas the more complicated calculations assume that there is somewhat of a decrease in lapse rate, by the well-understood physics of the moist adiabatic lapse rate in the tropics. This decrease in lapse rate is a negative feedback that is included in all of the climate models.)

jae
May 8, 2011 6:59 pm

IRA: It is getting more and more conspicuous that you are completely ignoring the alternative explanations for the Earth’s temperature. You have completely ignored me and many others who have suggested some other theories to this point, unless I missed something. You need to respond. With some facts. Otherwise, my advice is to shut up.

wayne
May 8, 2011 7:06 pm

Thank you Hockey Schtick !!…
William C. Gilbert’s words are much better than the ones I have been using and now I have someone to attribute them to. He’s right, it is merely not so simple, but understandable, physics.

May 8, 2011 7:40 pm

wayne says:

William C. Gilbert’s words are much better than the ones I have been using and now I have someone to attribute them to. He’s right, it is merely not so simple, but understandable, physics.

It is clear that Gilbert has never actually read any elementary textbook on atmospheric physics or climate science or he would be aware that the argument about adiabatic lapse rate that he presents is not new to anyone in the field…It is discussed in all of these books. (Alternatively, maybe Gilbert is aware of this but is confident that the readers who he wants to convince have not.)
However, far from his claim that this temperature profile explains the 33 deg temperature difference between what is observed and what radiation balance requires, the fact is that it alone doesn’t explain any of it. You need to consider not only how the temperature varies with height but what then sets the constant that tells you what the absolute temperature is at some height in the troposphere. In the absence of absorption of terrestrial radiation by the atmosphere (and with the other caveats about still having the same albedo and such), that average temperature would have to be 255 K at the surface because of radiative balance and then the temperature would decrease with height at the lapse rate from there.

May 8, 2011 7:43 pm

jae says:

IRA: It is getting more and more conspicuous that you are completely ignoring the alternative explanations for the Earth’s temperature. You have completely ignored me and many others who have suggested some other theories to this point, unless I missed something. You need to respond. With some facts.

The errors in your explanations have now been explained to you multiple times in this thread. There is no excuse for you to continue to cling to them.

Stu N
May 8, 2011 8:03 pm

Joel:
“The only way that adiabatic compression could possibly explain the earth’s surface temperature would be if the atmosphere (and / or the earth itself) were undergoing continual gravitational collapse, which we know is of course not the case.”
Thank you! I’ve had big trouble trying to explain this to people before. I tried using the example of a bike tire, which heats up when you compress the air but then doesn’t stay hot (even though it stays compressed, but the elasticity of the innertube, analgous to gravity, is not reducing the volume of gas). Success rate with this explanation is surprisingly low.

jae
May 8, 2011 8:10 pm

Joel:
“The errors in your explanations have now been explained to you multiple times in this thread. There is no excuse for you to continue to cling to them”
?? Can you please point me to those explanations, sir? I must have missed them, or else, perhaps, probably, most likely, they were balogne? Come on, fella, provide some tangible thing that I can relate to, instead of your Orwellian insistance that there is, “out there,” something that proves your case/argument/status/being/etc..
WTF?

May 8, 2011 8:32 pm

“The only way that adiabatic compression could possibly explain the earth’s surface temperature would be if the atmosphere (and / or the earth itself) were undergoing continual gravitational collapse, which we know is of course not the case.”
Typical CAGW-practitioner gobbledygook
And the atmosphere does not behave as an enclosed bicycle tire. No wonder the “success rate” with that explanation is so low.
Here’s Dr. Jelbring’s more detailed explanation:
http://www.google.com/url?sa=t&source=web&cd=1&ved=0CBYQFjAA&url=http%3A%2F%2Fruby.fgcu.edu%2Fcourses%2Ftwimberley%2FEnviroPhilo%2FFunctionOfMass.pdf&ei=9V3HTbH4IIP4sAOLzuTqAQ&usg=AFQjCNHTux_zuvY-oHVCehfC6OOmfzUafQ&sig2=p-LjGfmce1A-Vqk5ybrjZg
Show me the exact error’s you claim, citing specific excerpts from his paper.

richard verney
May 8, 2011 8:39 pm

ferd berple says:
May 8, 2011 at 2:24 pm
The solar radiation hitting the earths surface is 184 W/m^2 per Trenberth 2009 or 198 W/m^2 per Trenberth 1997. The surface albedo must be applied to this to get the net gain. Back radiation adds another 333 or 324 W/m^2 respectively.
There seems to be a whole lack of common sense going around these days.
So, according to T, the earth receives 184 w from the sun, radiates some of this back outwards, where some of it is absorbed by the atmosphere and re-radiated back to earth, where it provides 333 w.
In other words, 184 w ends up as 666 w, because the back radiation is omni-directional at at most 50% reaches the earth……………………………………
///////////////////////////////////////////////////////
There does appeart to be a complete lack of commonsense in these figures.
If these figures were correct (ie., back radiation is approximately twice the power of solar), why are we wasting time with trying to capture solar energy rather than back radiation energy?
Why don’t we construct a large array of mirrors collecting and focusing this back radiation much like the Californian Solar power station: see
http://greenwombat.files.wordpress.com/2010/01/esolar-power-plant.jpg
This not only would produce more energy (viz about 330 cf about 190 W/m^2; back radiation cf solar), it would additionally solve the storage problems inherent with present day green energy projects since back radiation is available day and night etc.
The answer is simple, either this back radiation does not exist, or if it does exist it is incapable of doing any sensible work (ie., it lacks sensible energy).
The lack of any serious research into collecting and utilising the back radiation energy source strongly suggests that leading energy scientists regard the back radiation set out in Trenberth’s energy diagrams to be complete and utter nonsence!
Anyone who supports Trenberth’s diagram and considers that the back radiation theory is correct should be asked to explain why we are not utilising this fanastic energy source. After all, if it existed it should be the main focus of green energy projects.

May 8, 2011 8:45 pm

“The only way that adiabatic compression could possibly explain the earth’s surface temperature would be if the atmosphere (and / or the earth itself) were undergoing continual gravitational collapse, which we know is of course not the case.”
Typical CAGW-practitioner gobbledygook
And the atmosphere does not behave as an enclosed bicycle tire. No wonder the “success rate” with that one is so low.
Here is a more detailed explanation from Dr. Jelbring:
http://www.google.com/url?sa=t&source=web&cd=1&ved=0CBYQFjAA&url=http%3A%2F%2Fruby.fgcu.edu%2Fcourses%2Ftwimberley%2FEnviroPhilo%2FFunctionOfMass.pdf&ei=9V3HTbH4IIP4sAOLzuTqAQ&usg=AFQjCNHTux_zuvY-oHVCehfC6OOmfzUafQ&sig2=p-LjGfmce1A-Vqk5ybrjZg
Show me exactly where he is in error, citing exact quotes from the paper, without any hand waving.

jae
May 8, 2011 8:45 pm

Luuuucy?????? Ira?????? we are missing some comments here. WTF? Are you THAT desperate????

May 8, 2011 9:05 pm

While you’re at it, Dr. Noor van Andel’s paper also explains the “GHE” based upon the adiabatic lapse rate. Again, show me exactly where he is in error, citing exact quotes from the paper, without any hand waving.
http://climategate.nl/wp-content/uploads/2011/02/CO2_and_climate_v7.pdf

UncertaintyRunAmok
May 8, 2011 9:13 pm

Wow. Once again, misconceptions abound on both sides.
Ira, just because it is the only seemingly rational suggestion that has been posited publicly does not mean that it actually describes the physical process accurately. If it did, the term “backradiation” would not have had to have been coined in the first place.
It is a simple matter to increase the absorption in a gas above the level of the incident energy, without ever having to increase the relative concentration of the gas, and is done on a daily basis in my field, and has been for many decades, but we do not call it “backradiation”.
Some (no, I don’t know exactly how much) of the reflected solar energy has already been absorbed by the atmosphere before being reflected. The statement that it contributes nothing to the atmospheric temperature profile is thus at least partly incorrect.
So Ira, if you really are sincere about wanting to know where the 33K difference comes from, you could start by calculating the Mie backscatter cross section of a 10um dielectric sphere, and I would suggest calculating it with a 9.25um or 10.59um em wave incident on it. I suggest those 2 wavelengths because measurements have already been done with them, and you will have something to compare the calculated values to, although the measured values were normally given in steradians, which can also be calculated.
You might also consider how it is possible to treat two entities that are physically coupled (think about it) as if they were two separate “blackbodies” radiating “at” each other.
Cheers.
BTW, N2 DOES have resonance lines in the solar UV region.

David
May 8, 2011 9:21 pm

A warm atmosphere will emit radiation. Some of it will be backradiation.
Even without IR absorption the atmosphere will warm.
http://earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
Convection.
Evaporation.
Incoming energy absorption (visible light).
With this in mind, and accepting the backradiation mechanism, how can the “greenhouse IR” component be ascribed the full 33 C difference?
Surely it can’t be?

martin mason
May 8, 2011 10:01 pm

From what I see the proponents of the Greenhouse theory simply keep stating it as fact without explaining the observed phenomena that seemingly falsify it such as.
– You cannot heat a hot body from a colder body which you have to be able to do.
– The temperature at the surface and at height can be explained without recourse to radiative theory which to me means that the temperature is determined by the fact that we have an atmosphere not necessarily because we have an atmosphere with GHGs. The atmospheric column is nothing like a bicycle tyre in physical terms, that is a red herring.
– In the past when we have had multiples of current CO2 levels we have not had runaway warming.
– There is patently no runaway warming now while CO2 levels ramp up inexorably. There is no runaway warming on venus.
– If back radiation is so powerful and it is a heat source, why don’t we use it?
It’s the answers to the simple things that are important not the circular academic arguments. The consensus view is that AGW is happening because of a trace GHG, the only problem is that there is no unequivocally correct theoretical or observational data that supports it.

May 8, 2011 10:39 pm

John of Kent says:
May 8, 2011 at 3:09 am
Ira Glickstein does not seem to understand how matter interacts with radiation.

On the contrary he seems to understand it fairly well, you however appear not to!

Martin Lewitt
May 8, 2011 10:58 pm

Wayne,
“The air above the cloud tops will be cooler than normal shielded by the water mass in the clouds and the air between the surface and the cloud base warmer than normal.”
The problem is for those doubting back radiation to explain the greater night time head loss in dry clear skies vs humid clear skies. You don’t seem to be able to do it without clouds, are you doubting the humid vs dry clear sky difference exists?
BTW, no huge lens covering a wide enough range of the infrared spectrum exists to conduct your hypothetical experiment.

Darren Parker
May 8, 2011 11:27 pm

How does this fit in with the Holographic Universe theory?

May 9, 2011 12:07 am

According to the more reasonable reconstructions of past temperature (Moberg, Loehle) the surface temperature has been rising since 1700. This is 230 years longer then co2 has been rising to any great degree.
What is Ira’s explanation for this earlier warming?
The Earth is a big heat engine. Most of the effective solar heating of the ocean occurs in the tropics, whereas most of the effective radiation of heat to space occurs near the poles. Most of the energy between the equator and temperate latitudes is shifted by the ocean. From the temperate latitudes to the poles, most of the energy is shifted by the atmosphere. This movement of energy is achieved mainly by convection of one sort or another, and since the temperature of the ocean is little affected by back radiation, changes in co2 concentration can’t have much effect.
Why does Ira refer simply to the Earth’s surface as an amorphous entity instead of taking into consideration the fact that 70% of it doesn’t absorb much energy from back radiation?
over 90% of downwelling radiation striking the ocean surface has been emitted a kilometre or less above it, because of re-absorption and re-emission. Since this re-emitted radiation is travelling at any random angle wrt to the ocean surface, most of it is travelling at angles which will either miss the surface or be reflected by it. The back radiation which is entrained into the ocean bulk by wave action is minimal, because wave action subducts water which is well below the level back radiation can penetrate to (little more than its own wavelength). Most of the radiation measured as coming from the ocean surface is actually either been reflected or is being emitted by molecules evaporated from the ocean surface.
Since the ocean surface temperature changes precede surface air temperature changes by several months, and since the top two metres of ocean contain as much heat capacity as the entire atmosphere above it, it is clear that surface temperature and atmospheric temperature is strongly influenced by the ocean, which is heated by the sun, not by back radiation.
Solar variation affects albedo, and that affects how much solar energy the ocean absorbs. This is why the number of sunshine hours correlates with surface temperature over the last 130 years much more closely than co2 level.
http://tallbloke.files.wordpress.com/2010/06/soon2009.jpg
In my opinion, surface temperature variation is due to the variation of the Sun. Co2 is along for the ride.

Robert Clemenzi
May 9, 2011 12:26 am

Why is “the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet”?
As dr.bill said, this is because the Planck equation is actually an integral. When you change the x-axis, the peak actually does shift. I have written a full explanation at
http://mc-computing.com/Science_Facts/Blackbody/Blackbody_Equations.html
There is also a plot showing the difference in the shape of the two 300K curves and their peaks.
biological evolution has “tuned” our visual system to be most sensitive where the light energy is maximized
Not quite, that is where the intensity per wavelength interval is brightest, not where the energy per frequency interval is largest.
This demonstrates why people who believe in Global Warming always plot spectra versus wavelength. When you want to analyze the energy in the system, all plots should be versus frequency (normally expressed as a wavenumber) to show an honest plot. In that case, most of the energy from the Sun is in the IR. When plotted versus wavelength, the plots incorrectly imply that most of the energy is in the visible and UV spectra.

May 9, 2011 12:52 am

The one big attractive feature of the “greenhouse theory” is its amazing elasticity.
The climate getting warmer is a sure proof of CAGW
But then so is the climate getting colder, say some IPCC advocates.
Joel Shore has refined this even further to say Postema is a believer, only Postemo hasn’t quite realised it.
Perhaps with the same elasticity Joel will prove that Gerlich and Tscheuschner also are really IPCC advocates.
Joel goes on to say of me;
…..”He has absolutely no serious desire to discuss the science”…..
Quite the opposite Joel;
On page 9 of your co-written Halpern et al comment paper you state that the Stephan Boltzman Law can be used to work out the thermal energy between atmospheric shells.
I brought your attention to it in my post above;
….. ” 3. Has the Stephan Boltzmann equation applying to gases.”……
Its no good trying to weasle your way out of it.
Your comment was printed in a specialist Journal with readership almost entirely of Physicists.
They would have no problem understanding your point.
Do you still maintain this position or will this be your third major repudiation of this notorious document.
In fact you should perhaps repudiate the document in its entirely!

May 9, 2011 12:52 am

Ira, I’d really like some of the “back radiation” now. Just stepped outside my house in Kamloops BC and pointed an IR thermometer straight up into the sky and got a temperature reading of -10 F. When I pointed the thermometer horizontally down the street, got 34 F. Pointing it at the outside of my house gave a temperature of 44 F as did pointing it at my driveway. The airport temperature now is given as 48 F. The altitude of the airport is about 200′ lower than where I live. I should note that the sky is perfectly clear at this time with no clouds in sight (just wish it would do this in the daytime).
The surface of the earth heats up considerably during the day and this heat is released during the night. Of course having heat buffers of variable capacity greatly complicates models. The -10 F temperature that I’m seeing with my IR thermometer is the average temperature of the atmosphere from 1600′ above sea level to space. What portion of this would be related to “greenhouse gasses” and what portion is the result of heat flux from sun-warmed surface and air to space?

wayne
May 9, 2011 1:16 am

Martin Lewitt, you are right, I do not “believe” in “back radiation”, just plain radiation as taught in physics.
Clouds make warmer nights because the clouds are usually warmer than the normal air temperature at that altitude and therefore the surface’s rate of loss by radiation upward will be less leaving you with a warmer than normal night. It is also that same radiation keeps those clouds warmer than te air at that altitude would normally be. Just using Stefan-Boltzmann will get you very close if you know the temperatures and emissivities involved. There are even rare cases when the clouds are actually warmer than the surface and then the clouds are radiating downward to the surface and also to space but as you should see, this condition is short lived since it has cooler on both sides unless continually replenished with warmth to maintain the temperature differential.
BTW, that experiment does not require all frequencies to be valid. I’ll just leave it for each here to decide it’s merits, sounds like most here can handle high-school science correctly.

Vince Causey
May 9, 2011 1:29 am

From the comments here, a lot of people seem to believe that the greenhouse effect cannot exist because it violates the 2nd law of thermodynamics. To my shock, I am finding myself agreeing with Joel Shore on the elementary physics. I have just finished reading the Postma essay, and it contains the same strawman argument that crops up every time this matter is discussed.
Postma described a black body warmed by a heat source and claimed that you cannot make it warmer by reflecting that emitted radiation back. Whilst acknowledging back radiation exists, he asserts that this energy cannot cause the black body to increase in temperature. He states that a body cannot raise its own temperature. Well of course not. The reason this is a strawman fallacy is because nobody is making that claim. The claim is that the back radiation reduces the rate of cooling of the earths surface because the net outflow in watts is slightly less than it would have been without the back radiation. It is the sun that causes the warming because energy is now coming in faster than it is leaving, so the temperature will rise until a new equilibriumis reached.
An analogy is a tub with a ‘V’ cut near the top. The higher the water level is above the ‘V’, the faster it flows out. This is an analogy of the SB law of blackbody radiation. Suppose the sun is represented by a running tap. The water will rise above the ‘V’ just enough so that the rate of input is equal to the rate of overflow. Let us now represent the back radiation by blocking some of the ‘V’ which represents the reduced outflow of radiation from the earths surface. The water level (temperature) will now rise such that the outflow increases to balance the inflow. If you turned the tap off (remove the sun from the climate system), the water level (temperature) would start to decline just as you would expect from thermodynamics.

May 9, 2011 1:34 am

Ira,
Here is a schematic from a search, nothing special, merely from the first site that had a lapse curve, month of June, 45 degrees North latitude.
http://i260.photobucket.com/albums/ii14/sherro_2008/Lapse.png?t=1304929624
Source was Bigg 2005, blog was http://scienceofdoom.com/2010/04/24/tropospheric-basics/
The coldest temperature measured on earth is about -89 deg c at Vostok. This is colder than any part of the lapse curve below about 75 km above Earth.
How can it get so cold? What stops heat getting to Vostok?

RJ
May 9, 2011 1:41 am

“Anyone who supports Trenberth’s diagram and considers that the back radiation theory is correct should be asked to explain why we are not utilising this fanastic energy source. After all, if it existed it should be the main focus of green energy projects.”
“- If back radiation is so powerful and it is a heat source, why don’t we use it?”
Good questions. Although what about this free energy oven. Does it qualify?
http://www.slayingtheskydragon.com/en/blog/111-a-pictures-worth-a-1000-words#comments

Allan M
May 9, 2011 2:04 am

Joel Shore says:
You clearly do not know the basis for the Second Law, which is statistical physics.
The basis for the second law (and the first) is careful and accurate measurement, mostly paid for by wicked 19th century capitalists, who wanted to know how to get more from their steam engines, and also how to avoid being sold snake oil.
There is no need for a statistical ‘fudge’ unless you wish to introduce special pleading for “IPCC Approved” photons.
If I were to dump a billion microscopic ice particles into a tank of hot water (di-hydrogen monoxide), your statistical approach would insist that a few of them may have warmed the water (hydrogen hydroxide), especially using a computer model.

Robert Stevenson
May 9, 2011 2:16 am

I use spreadsheets to calculate Earth IR emissions and absorption by CO2. Using mean beam lengths and CO2 partial pressures, I use PcL charts by Hoyt C hottel to obtain the emissivity of CO2 in the atmosphere. For current CO2 partial pressure or concentration I calculate a distance of 3600 m of traverse through the earth’s atmosphere by ‘IR radiation’ before the absorbable IR is ‘filtered out’ and reduced to zero. I then calculate that by doubling the atmospheric CO2 to 700ppm or so the ‘absorption distance’ is reduced to 2000 m but with no additional IR absorption by CO2. Therefore I conclude no global warming from increasing atmospheric CO2.

Bomber_the_Cat
May 9, 2011 2:53 am

Ira,
“But, that still does not explain why the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet and other graphics on the Internet”
Maybe I could help, or perhaps just confuse you further.
If you plot the radiant energy against wavelength for blackbody around 287K, then you get a peak intensity at about 10 micron. The graphs reproduced from Petty show the peak intensity at about 18 micron. This apparent anomaly arises because the vertical axes of the 2 graphs use different units.
The Plank distributions plotted against wavelength (the way I like to see them) have a vertical scale of ‘Watts per Square metre PER MICRON’. The graphs shown by Petty are plotted against wavenumber (not wavelength) and the vertical axes have units of ‘Watts per Square Metre per Steradian PER WAVENUMBER’.Note: graphs plotted in these different ways are NOT simply mirror images of each other. [Forget the Steradian bit, this is just a multiplying factor which doesn’t effect the peak or shape]
Say that the peak radiation on the wavelength plot occurs at 10.5 micron. Because the energy is plotted per micron it represents the total energy between 10 micron and 11 micron. In terms of wavenumber, 10 to 11 micron is equivalent to a range of 909 to 1000 per cm. So the energy for the 10.5 micron peak has to be dispersed over 91 different wavenumbers on the wavenumber plot. If we consider another point on the wavelength plot, say between 18 and 19 microns, this is represented by a wavenumber range from 526 to 555. So the energy around 18 microns has only to be shared between 29 wavenumbers whilst the energy around 10 microns has to be shared over 91 wavenumbers. As a consequence, the peak intensity no longer occurs at 10 micron, it now occurs at 18 micron. That is why the two methods of plotting give different peak intensities.
Did that help, or just confuse everyone further?

Martin Lewitt
May 9, 2011 3:02 am

Boris Gimbarzevsky,
Thanks for sharing your thermometer readings It would be interesting to know if on clear nights, sky readings taken at the same ambient temperature but different absolute humidities varied due to increased water vapor greenhouse effect.

May 9, 2011 3:25 am

Dishman says:

If molecules follow ballistic trajectories, the minimum KE will occur at zenith and be higher at all other points.
A short MFP just means that equipartitioning will be rebalanced (on average) every MFP.
Assuming air (a diatomic gas with 5 degrees of partitioning), for a given delta-h, the thermal energy will be reduced by 1/5th of the energy required to lift molecules by that delta-h.
My point is that Objection #2 is not, in fact, correct.

That may be your point, but nothing you say backs it up. You say “the thermal energy will be reduced by 1/5th of the energy required to lift molecules by that delta-h.”
A quick calculation shows that the average molecular speed, assuming constant g, is sufficient to lift a molecule about 11 km. The mean free path of a molecule in air is about 65 nanometres. Dividing these, we see the energy needed to lift the molecule by your delta-h is about 1.7*10^-11 of its total energy. In other words, diddly squat. Gravitational effects make no difference to the result that could possibly be measured.

Smoking Frog
May 9, 2011 4:52 am

Charles Nelson said: Gotcha Henry! He thinks CO2 is like a toxin in the atmosphere! An amazing toxin which is NOT TOXIC at 380ppm but suddenly BECOMES TOXIC at 400ppm. There in a single post is the fuzzy thinking of the Warmists! Fantastic. Textbook example of ‘Carbonaphobia’!
I don’t read him that way. I think he’s making the point that the tiny size of the fraction, by itself, doesn’t tell us anything. If he’s making the point that you think he’s making, he’s doing a poor job of it.

May 9, 2011 5:10 am

Ira,
you do not consider conduction and convection, nor non-GHG absorbers/reflectors of radiation nor spatial issues. Simply saying that the difference in temperature between a perfect blackbody and the incoming radiation is due to “greenhouse” effects is disingenious. You can convince yourself by a thought experiment: would the atmosphere really be at zero degree Kelvin were it not for GHGs? I’d say obviously not and hence your simple difference doesn’t apply.
best regards,

Smoking Frog
May 9, 2011 5:21 am

Martin Mason said – If back radiation is so powerful and it is a heat source, why don’t we use it?
Because the earth’s surface is already at the temperature to which the back-radiation has raised it.

Dave Springer
May 9, 2011 5:35 am

Allan M says:
May 9, 2011 at 2:04 am
“If I were to dump a billion microscopic ice particles into a tank of hot water (di-hydrogen monoxide), your statistical approach would insist that a few of them may have warmed the water (hydrogen hydroxide), especially using a computer model.”
Some actually will warm the water. Things get more than passing strange when you leave to domain of bulk matter and look at what’s going on with individual atoms and molecules. Some of the water molecules in your tank of hot water are colder than ice and some of the water molecules in the ice are more than boiling hot. Those molecules are few and far between so for most practical considerations they are ignored. Temperature is an artifact of motion. In bulk matter where the molecules are all bumping into each other chaotically some small fraction will happen to be moving very fast and some very slow. A good analogy would be what happens to a set of billiard balls after the break. Initially all the motion is in the cue ball but when it hits the rack the energy of its motion is distributed amongst all the balls but not equally. The cue ball could stop dead in its track only to get whacked back into motion by one of the other balls a moment later or if it’s a glancing blow the cue ball might remain the fastest mover for a few moments. But because energy can be neither created nor destroyed the total kinetic energy in all the balls on the table will always equal the initial energy in the cue ball. Temperature is analogous to the total kinetic energy of all the billiard balls.

Dave Springer
May 9, 2011 5:49 am

Robert Stevenson says:
May 9, 2011 at 2:16 am
“Therefore I conclude no global warming from increasing atmospheric CO2.”
This was disproven by expermental physicist John Tyndall over 150 years ago. Back radiation from gases that absorb IR is quite real. Suggest you read the original work here (it’s free):
http://www.archive.org/details/heatconsideredas00tyndrich

Dave Springer
May 9, 2011 5:57 am

RJ says:
May 9, 2011 at 1:41 am
“Anyone who supports Trenberth’s diagram and considers that the back radiation theory is correct should be asked to explain why we are not utilising this fanastic energy source. After all, if it existed it should be the main focus of green energy projects.”
“- If back radiation is so powerful and it is a heat source, why don’t we use it?”
Good questions. Although what about this free energy oven. Does it qualify?

We do use it all the time. One example is the smudge pots used to limit freeze damage in orange and avocado groves. The heat of the smoke from the smudge pots is inconsequential. The effect comes from the black carbon particles in the smoke which absorb upwelling IR from the ground and send a portion of it downwards and sideways. It works as an insulator. This is the same mechanism by which CO2 molecules insulate. Anyone who contests this is simply wrong and needs to take an introductory class in physics.

May 9, 2011 6:12 am

“The water will rise above the ‘V’ just enough so that the rate of input is equal to the rate of overflow. Let us now represent the back radiation by blocking some of the ‘V’ which represents the reduced outflow of radiation from the earths surface. ”
The problem is that Mr “T” says the energy from the sun is 184 watts and the amount being back radiated is 333 w. In other words, you are blocking more water than the maximum flow of the outlet, which is physically impossible in you analogy.

Dave Springer
May 9, 2011 6:31 am

Boris Gimbarzevsky says:
May 9, 2011 at 12:52 am
Ira, I’d really like some of the “back radiation” now. Just stepped outside my house in Kamloops BC and pointed an IR thermometer straight up into the sky and got a temperature reading of -10 F.

You’d like more back radiation. You’re getting quite a bit of it now. The temperature of the cosmos is 3K and if it weren’t for back radation when you point your IR thermometer at the sky on a clear night you’d get a reading of -454F.
Back radiation doesn’t come from just CO2. When a CO2 molecule absorbs upwelling IR it thermalizes surrounding non-GHG molecules via kinetic transfer. A common fallacy is that CO2 molecules emit a photon of the same frequency as that absorbed. That’s true in a very thin gas but in a cold dense gas the excited CO2 molecule almost instantly bumps into a neighboring molecule (likely a nitrogen molecule in the case of the earth’s atmosphere) and transfers some of its newly acquired energy to the neighbor.

May 9, 2011 7:05 am

Ira Glickstein, PhD says:
May 8, 2011 at 4:26 pm
“If we then placed that planet in an Earth-like orbit around the Sun”
Thanks Ira, what I read you to be saying is that if the atmosphere was pure N2O2, without any H2O, that it would have no effect upon the surface temperature of the earth. That since N2O2 do not absorb longwave radiation, then they cannot keep the earth warm.
I’m saying that this is simply wrong. I’m saying that an atmosphere of pure N2O2 would change the surface temperature of the earth significantly as compared to an earth without an atmosphere.
This is the calculation that must be undertaken first. How much of a difference would we see in the surface temperature of the earth due to N2O2.
The next calculation is the effect of phase change of H2O in the atmosphere. Not the GHG effect of H2O, but the effects of evaporation and condensation in transporting heat vertically.
Then, when you have completed those two calculations, what you have left is a candidate for the GHG effect.
From what I’ve seen, when you apply these calculations to real models, such as venus, earth and mars, then the first calculation, the effects of the atmosphere without regard to the composition provides nearly the same amount of heating on venus, earth and mars.
These are real observations, not theoretical models. And, observation trumps models, no matter how correct we think our models are. I don’t see this calculation in your work. I don’t see where you calculate the effects of a N2O2 atmosphere that does have any H2O or CO2.
As to why the atmosphere of venus, earth and mars show the same amount of heating regardless of composition, that is a matter of theory. Put 10 experts in a room and you will have 20 opinions. None of these opinions will change the facts.
The atmosphere of venus, earth and mars show a significant heating effect that is independent of the composition of GHG and this is not accounted for in your model.
As to why this is, my theory is that it is due to compression. That the molecules i the atmosphere all have roughly the same amount of energy due to convection, but since there are many more of them per cubic meter at lower altitudes, there is more energy per cubic meter, which we register as increased temperature.
For example, we could have 1 air molecule per cubic meter. When we went to measure the temperature of this meter of air, we would average out the energy of this one molecule over the volume and end up with a low temperature.
Now, if we were to have a whole lot more molecules per cubic meter, all will the same energy as our first example, we would see this as an increase in temperature. However, in both examples the molecules are at the same energy level, yet the samples taken from lower down in the atmosphere will appear to have a higher temperature.
However, this may not be the reason for what we observe. And it doesn’t really matter. What does matter is that FROM OBSERVATION the atmospheres of venus, earth and mars show similar warming at similar pressure, independent of their levels of CO2.

May 9, 2011 7:11 am

Ira,
Here’s why your bicycle tire analogy fails: you instead need a leaky tire that has to be inflated continuously. Would the air in that situation remain hot? Yes
The open atmosphere essentially acts as a continuous ‘air conditioner’ or pump. Cold air continuously descends to the surface and is heated by the surface as well as from compression. That air then ascends and expands and cools until the process repeats. Thus, adiabatic compression is able to maintain a higher surface temperature on a continuous basis, the so-called GHE.

Bomber_the_Cat
May 9, 2011 7:14 am

We seem to have an unusually large number of comments from the anti-science brigade today who seem to have invented a whole new branch of previously unknown physics.
Imagine an ice cube, at zero degrees Celsius, alone in the vacuum of space (which is at, say, absolute zero). What happens to that ice cube? Well, it radiates away heat (as everything above absolute zero does) and eventually cools to absolute zero itself, asymptotically. Now imagine an another ice cube, this time at -10 degrees Celsius, being placed near to the original ice cube. This colder ice cube also radiates heat; some of it will impact the first ice cube and ‘warm’ it. The original ice cube still continues to cool, but not as fast as it was doing before because it now has a heat input which was not there in the first case. The colder ice cube therefore helps to keep the original ice cube warmer. Of course, it never gets above its starting value of zero degrees Celsius, but that is how the presence of cold objects can keep hot objects warmer i.e. they are still warmer than what would be in their place if they were not there (in this case the cold vacuum of space at absolute zero).
The 2nd Law of Thermodynamics? – Yes – it fully supports this – no contradictions. .

May 9, 2011 7:31 am

martin mason says:

From what I see the proponents of the Greenhouse theory simply keep stating it as fact without explaining the observed phenomena that seemingly falsify it such as.
– You cannot heat a hot body from a colder body which you have to be able to do.

In any model of the greenhouse effect, the net heat flows are from the warmer earth to the colder atmosphere. The atmosphere is not heating the earth itself…what it is doing is slowing down the cooling of the earth (for a given surface temperature). Since the steady-state temperature of the earth is determined by the balance between what it receives from the sun and what it emits back out into space, an IR-absorbing atmosphere will in fact cause the earth’s steady-state temperature to be higher than it would be if the atmosphere did not absorb IR. In a colloquial sense, one might call this “heating” the earth, although it is best to avoid such language since it leads to confusion and to instead state clearly what the effect of the atmosphere is, as I have explained above.

– The temperature at the surface and at height can be explained without recourse to radiative theory which to me means that the temperature is determined by the fact that we have an atmosphere not necessarily because we have an atmosphere with GHGs. The atmospheric column is nothing like a bicycle tyre in physical terms, that is a red herring.

No…Such a hypothesis does not even satisfy energy conservation! What can be explained without recourse to GHGs is the lapse rate in the troposphere. What can’t be explained is what the temperature at any altitude is. It is as if you told me the slope m of a line of the form y = mx + b and claimed that I could now compute y for any x. In fact, I would need the value of b to do so. The value of b is in essence what is determined by the absorption of IR radiation by the atmosphere.

– In the past when we have had multiples of current CO2 levels we have not had runaway warming.
– There is patently no runaway warming now while CO2 levels ramp up inexorably.

This is a strawman argument. (Almost) noone is claiming that there will be “runaway warming”. [Hansen is talking about the possibility under certain circumstances and has specific arguments as to why the current case may be different from what the earth has experienced in the past, so let’s leave him out of this.]
What we are talking about is a climate that is quite sensitive, but not unstable, to perturbations. See, for example, here for a discussion of what the paleoclimate record tells us: http://www.sciencemag.org/content/306/5697/821.summary

There is no runaway warming on venus.

Really? You think the temperature on Venus is very pleasant? Did the runaway go on forever? No…An object that is linearly unstable will tend to find a point where stability is restored. However, a runaway did occur to produce Venus’s current very hot climate. Fortunately for earth, we are not as close to the sun and thus not subject to the runaway scenario that occurred on Venus.

– If back radiation is so powerful and it is a heat source, why don’t we use it?

How are you proposing that we harness it? As has already been noted by Smoking Frog, the back-radiation is already heating the earth. (Because most objects tend to be almost perfect blackbodies in the mid- and far-IR, almost all of the back-radiation that is received by the earth is absorbed.) Also, the back-radiation that we receive is diffuse (coming from all different angles) and so it cannot be focused in the same way that solar radiation (which is approximately from a point source) can. Furthermore, since the energy of an individual photon is proportional to its frequency, the back-radiation…unlike solar radiation…cannot be harnessed by photovoltaic devices, which rely on having photons of sufficient energy to cause electronic transitions (in particular, energies much larger than thermal energy of ~kT where K is Boltzmann’s constant). Finally, although the back-radiation we receive from the atmosphere is, in the global average (e.g., day and night at all latitudes), larger than the radiation we receive directly from the sun, the amount of radiation that we receive from the sun when it is actually shining and close to overhead is a fair bit larger.

Dave in Delaware
May 9, 2011 7:37 am

Ira – regarding your summary comment 4) at May 8, 2011 at 7:51 pm
my comment – NO, the atmosphere does NOT emit LWIR across a distribution of wavelengths like a blackbody , see my earlier comment at Dave in Delaware says: May 8, 2011 at 7:00 am
Ira Glickstein, PhD says:
“4) As I understand it, the ~15μm radiation from the Surface to the Atmosphere is absorbed by H2O and CO2 molecules which, when excited, bump into nitrogen and oxygen and other air molecules, and heat the air. The result is emission of long-wave radiation at a range of wavelengths (not just in the ~15μm region, but including the ~7μm and ~10μm regions) in random directions, some of which finds its way back down to the Surface and is re-absorbed. That conversion of a portion of the ~15μm radiation into other wavelengths may explain why there is a “bite” out of the ~15μm radiation that seems to be missing from the radiation out to Space. ”
——————————————
This is the part of your comment I disagree with –
“The result is emission of long-wave radiation at a range of wavelengths (not just in the ~15μm region, but including the ~7μm and ~10μm regions) in random directions, …”
My reply – If CO2 absorbs at 15μm, it emits at 15μm. The Nitrogen and Oxygen in the atmosphere neither absorb nor emit in the LWIR wavelengths of interest. That is why there is an Atmospheric Window – the atmosphere neither absorbs nor emits in that 10μm region.
Think about the satellite-looking-down readings – the 10μm atmospheric window corresponds to the surface temperature. Over the Sahara, the 10μm readings show a surface temperature around 320 K. Over the Mediterranean, the 10μm readings show a surface temperature around 280 K. Over the Antarctic, the 10μm readings show a surface temp just below 220 K. If the atmosphere (mostly N2 and O2) emitted LWIR like a blackbody, the atmospheric window readings would look like some middle atmosphere temperature – not like the surface. There is an Ozone band in the middle of the atmospheric window, I expect from ozone in the upper atmosphere; if N2 was emitting blackbody LWIR there should be other N2 bands in the middle of the window, and there are not.
———————————-
Ira comment – “That conversion of a portion of the ~15μm radiation into other wavelengths may explain why there is a “bite” out of the ~15μm radiation that seems to be missing from the radiation out to Space. ”
My reply – Recall that the satellite-looking-down readings are Emissions; their temperature tells us something about where those emissions originated. Looking at the CO2 band satellite 15μm readings, the temperatures are around 220 K and nearly the same for the Sahara, the Mediterranean and the Antarctic. That looks to me like the Tropopause. It does not necessarily tell me how the energy got to the Tropopause, so I am not sure I could estimate a ‘bite’ value. That emitted energy could have arrived from lower atmosphere CO2 emissions, or from atmospheric lapse type convection, or possibly from the tops of thunderstorms.
My comments are referenced to these satellite IR spectra at
http://mensch.org/5223/IRspectra.pdf
Slide (c) Western Tropical Pacific is particularly interesting because it compares ‘clear’ with ‘Thunderstorm Anvil’. Note that the CO2 band temperature is identical in the clear and thunderstorm curves. Also note that the atmospheric window 10μm region temperatures are MUCH DIFFERENT for clear vs thunderstorm anvil. This suggests to me that while the atmosphere does not emit like a black body, the thunderstorm (liquid water) does.
Best regards to Ira – I hope my comments are constructive and instructive, and will help your ‘mental models’ and descriptions.

May 9, 2011 7:56 am

Allan M says:

The basis for the second law (and the first) is careful and accurate measurement, mostly paid for by wicked 19th century capitalists, who wanted to know how to get more from their steam engines, and also how to avoid being sold snake oil.
There is no need for a statistical ‘fudge’ unless you wish to introduce special pleading for “IPCC Approved” photons.

Experiment did provide the evidence for the Second Law. However, our modern (20th century) understanding of where the law comes from is based on statistical physics. To dismiss an entire field of physics that has successfully explained a huge wealth of empirical observations as “a statistical ‘fudge'” because one implication of it does not line up with your ideology is bizarre.
This is essentially the project that Claes Johnson, one of the “Slaying the Sky Dragon” authors, has embarked upon. He wants to base things on the bizarre proposition that an artifact of numerical calculations of the differential equations actually governs the universe instead. The irony is that although he claims to have gotten rid of “back radiation” by coming up with a different interpretation of the terms in the equation governing radiative transfer between two bodies, he has in fact done nothing to change any numerical result using those equations…including all of the numerical results that support the existence of the greenhouse effect! What he has done is snookered a few people who want to believe the nonsense that he is peddling.

If I were to dump a billion microscopic ice particles into a tank of hot water (di-hydrogen monoxide), your statistical approach would insist that a few of them may have warmed the water (hydrogen hydroxide), especially using a computer model.

It depends what you mean by “microscopic”. It doesn’t take a very large particle size before the statistics make it vastly improbable for such a thing to occur. But, yes, if you particles were truly only a few molecules large, then you could detect behavior contrary to the 2nd Law. (Of course, the distinction between “ice” and “water” begins to lose meaning at such a small length scale since phases of matter are themselves a macroscopic concept.)

Dave Springer
May 9, 2011 7:57 am

Circa 1850 Tyndall experiments with greenhouse gases in a nutshell.
Radiation flow:
heat_lamp -> -> adjustable shade -> salt crystal window -> mirrored tube filled with test gas -> salt crystal window -> adjustable shade -> galvanometer
Tyndal ran literally thousands of experiments with different gases and mixtures at different pressures, different infrared light sources, and different lengths of tubes. His setup took up a lot of space but I marvel at the cleverness he used to increase accuracy and precision using the technology of the day. His analog gear was so sensitive he read the reading from the galvinometer through a telescope because getting a warm body anywhere near the experiment mucked it up.
His best most stable lamp was a black painted face of a vessel filled with boiling water. His galvanometer response is not linear so the primary purpose of the adjustable shades was keep the radiative energy input to the face of the galvanometer in a range where you got maximum needle deflection from minimal change in radiation.
In lectures he used to take his galvanometer to the podium in the hall. He’d point it at a blank wall on the other side of the hall and ask someone from the audience to walk over to that wall. The galanometer needle would move dramatically when a person entered the scene.
Anyhow Tyndall found a great many gasses, most famously water vapor, that would dramatically lower the galvanmeter reading when present in the tube versus the same mixture without it. He also used some rather ingenious ways of processing his gasses prior to filling the tube with them including means of completely drying a sample of the atmosphere.
If one wants to dispute the concept of back radiation one must first explain why Tyndall’s galvanometer would read progressively lower as the absolute humidity of the atmospheric sample in the tube rose. No one will. Every attempt is fatally flawed in some way. Tyndall didn’t discover anything that wasn’t already predicted by the theoretical physics of the day. He confirmed the theory via experiment or in other words he was doing science the way it’s supposed to be done.
Tyndall did discover some things that no one had worked through on paper at the time. One notable thing he discovered was the non-linear absorptive response as the partial pressure of an IR-absorbing gas went up. He found it to be linear at the lowest partial pressures then progressing to exponential or in other words when it comes to IR absorbing gasses increasing the amount of it is a case of diminishing returns after a certain amount.
This is why the climate boffins talk about CO2 doublings and temperature rise. At the current atmospheric CO2 partial pressure its absorptive response is in the exponential range. The linear response range is all below the first 100ppm. It’s sort of like people picking low hanging fruit in an apple orchard. They can all pick as fast as they can for a while but as the number of people picking increases at some point they’ll be competing with each other for the same low hanging fruit. So putting more pickers in the orchard will always increase the rate at which fruit is picked but at some point it becomes a case of diminishing returns.

Robert Stevenson
May 9, 2011 7:59 am

Dave Springer
You did not read what I had written – no additional heat absorbed- no temperature increase- no global warming. 150 years ago Tyndall new nothing about CO2 absorption bands

May 9, 2011 8:02 am

Charles Higley says:
May 7, 2011 at 9:33 pm
“It should also be pointed out that, as CO2 partitions 50 to 1 into the oceans, we would have to emit 50 times more CO2 that required to simply double atmospheric CO2. There is not enough carbon available to do this. If we really tried, we might be able to do 20%.”
It could also be pointed out that CO2 partitions 100,000 to 1 between the rocks and the air, so we would have to add 100,000 times more CO2 to just double atmospherics CO2…
This is wrong, as the ocean comparison is wrong because of time-scales. It takes the oceans centuries if not millenia to absorb additional CO2 so that partitioning will take effect eventually, for rcok/air the timescales are more like millions of years…

Philip Peake (aka PJP)
May 9, 2011 8:02 am

One little point about assumptions this is all based upon.
Incoming energy is spread over HALF the earth’s surface.
Outgoing energy is spread over the FULL surface of the earth, with an uneven distribution, probably (?) higher rate of outgoing radiation on the night side of the planet.
So its entirely incorrect to say that incoming = 240W/m^2 and outgoing = 240W/m^2.
Get the energy distributions right, then try again.

May 9, 2011 8:11 am

Here is a reference showing that Jupiter has 60K higher temperature than expected.
33K higher than expected on earth. 60K higher than expected on Jupiter. Co-incidence?
http://burro.astr.cwru.edu/stu/advanced/jupiter.html
•As seen in the table below, the average temperature of Jupiter is approximately 160 K. However, due to the equation for thermal equilibrium (below), it should only be about 100 K.

Dave Springer
May 9, 2011 8:15 am

Also one disputing back radiation must explain how many instruments that measure CO2 concentration manage to work. The basic design of these:
15um infrared light source -> beam splitter
one side of beam splitter -> sealed sample of air with known CO2 concentration -> IR photo transister
other side of beam splitter -> sample of ambient atmosphere -> IR photo transister
CO2 concentration is obtained by comparing the output of the two photo transisters.
Pretty much the same as Tyndall’s experimental setup only tweaked for one particular gas in a narrow range of partial pressures and put into a space the size of a thimble using modern technology.

wayne
May 9, 2011 8:24 am

Dave Springer says:
May 9, 2011 at 5:57 am
RJ says:
May 9, 2011 at 1:41 am
“Anyone who supports Trenberth’s diagram and considers that the back radiation theory is correct should be asked to explain why we are not utilising this fanastic energy source. After all, if it existed it should be the main focus of green energy projects.”
“- If back radiation is so powerful and it is a heat source, why don’t we use it?”
Good questions. Although what about this free energy oven. Does it qualify?
We do use it all the time. One example is the smudge pots used to limit freeze damage in orange and avocado groves. The heat of the smoke from the smudge pots is inconsequential. The effect comes from the black carbon particles in the smoke which absorb upwelling IR from the ground and send a portion of it downwards and sideways. It works as an insulator. This is the same mechanism by which CO2 molecules insulate. Anyone who contests this is simply wrong and needs to take an introductory class in physics.
———–
Dave, you and I structurally disagree, but end at the same conclusion. To me, and I am sure it is correct by physics, is the effect comes from the black carbon particles in the smoke which absorb upwelling IR from the ground increases the temperature at that altitiude (10 m) and by SB using delta T (LBL if you want) between the surface and these particles decreases the upward flux making it warmer than it would have been without the carbon particles in this thin layer where the smoke is.
You do not need a special kind of “back radiation” that flows from cold to hot to explain it, it only scrambles minds away from what is actually happening. If you care about people’s minds at all, stop wording it like you just did. There is no special radiation called “back radiation” or “backward radiation”.

May 9, 2011 8:28 am

“It takes the oceans centuries if not millenia to absorb additional CO2 ”
This does not appear correct. CO2 absorbtion at the surface of the ocean is limited by the surface area. However, the surface area of H2O in the clouds is fantasitically greater than the surface of the ocean. In addition, the water in the clouds is cold, which increases its capacity to hold CO2.
This is where CO2 is absorbed and returned to the oceans. High in the couds, not at the ocean surface. Near the equator, where the oceans are warm this CO2 will be released back into the atmosphere as the rainwater warms. In higher lattitudes where the oceans are cold this CO2 will not be released. The rain water will mix with the cold seawater and sink to the bottom of the ocean, carrying the CO2 with it. The polar oceans should be less caustic, more Ph neutral as a result, as compared to the tropical oceans. Which has been reported.
Eventually this cold water will upwell in the tropics and release its CO2. Thus the CO2 being released during La Nina years was deposited at higher latitudes many years before. This accounts for the cause and effect lag observed between warming and CO2 release, which Al forgot to mention in his movie. This is strong evidence that temperature drives CO2. There is no similar observation that temperature lags CO2 in the paleo record, which is evidence that CO2 is not a driver of temperature.

Dave Springer
May 9, 2011 8:30 am

ferd berple says:
May 9, 2011 at 8:11 am

Here is a reference showing that Jupiter has 60K higher temperature than expected.
33K higher than expected on earth. 60K higher than expected on Jupiter. Co-incidence?
http://burro.astr.cwru.edu/stu/advanced/jupiter.html
•As seen in the table below, the average temperature of Jupiter is approximately 160 K. However, due to the equation for thermal equilibrium (below), it should only be about 100 K.

“This extra heat is generated due to gravitational contraction – the planet is slowly shrinking in diameter. This way, by compressing by only a few millimeters every year, it can generate heat by increasing the pressure of its constituent gas.”
So Ferd, is it your assertion that the earth and venus are shrinking? I already explained to you that temperature only increases during the process of compression and the earth, Venus, and Mars (unlike Jupiter) aren’t shrinking.
What part of that don’t you understand?
Let me fix that for you by adding the explanation for the extra warmth on Jupiter from the article you quote:

wayne
May 9, 2011 8:38 am

Robert Stevenson says:
May 9, 2011 at 2:16 am
For current CO2 partial pressure or concentration I calculate a distance of 3600 m of traverse through the earth’s atmosphere by ‘IR radiation’ before the absorbable IR is ‘filtered out’ and reduced to zero.
—-
Robert, I come up with a close conclusion, not exact, but close. However, the 3600 m is more than 36 times that is given in other conversations I have read speaking of it’s absorption very near the surface. Do you have an average absorption coefficient you are using, or a mean optical thickness for CO2 from which that is calculated? Maybe Hottel specifies emissivity direct then what equation are you using to get the 3600 m extinction distance?

Dave Springer
May 9, 2011 8:40 am

Wayne
If you want a bit deeper understanding of the radiative energy flow one needs to understand that all matter above absolute zero radiates and where there are two bodies at different temperatures there’s a net transfer of energy between the two from warmer to colder. The net transfer rate is all the matters for most practical concerns but at the deeper level the radiative transfer goes two ways. The warm object doesn’t stop the cold object from radiating so energy from the cold object is indeed reaching the warm object just as energy from the warm object reaches the colder one. It’s simply a matter of the colder object receiving more energy from the warmer than the warm object receives from the colder.
That doesn’t seem like proverbial rocket science to me but you may perhaps be right that it might as well be rocket science for many people.

May 9, 2011 8:46 am

Explain why the surface of venus, which only receives some 2.5% of the sun’s energy due to the albedo effect of the clouds can have a temperature of 500C.
Why do we use a figure of .7 when calculating the solar energy reaching the earth’s surface? If we did this on venus, then the surface temperature should be very cold indeed.
The missing 33K comes from the assumption that only 70% of the solar radiation is available at the surface due to albedo. This is an incorrect assumption, as is clearly demonstrated by the surface temperature of venus.
97.5% of the solar radiation never reaches the surface of venus, yet it has a high temperature. Thus, the .7 figure used when calculating solar radiation at the surface is refuted by observation.
Consider 240k / 288K. This is .8. Thus, the missing 33K may not be missing at all. It may simply result from our asusmption that albedo is .7. If albedo is not what we assume it to be, for reasons that are apparent on venus, then the problem is not where is the missing 33K, but rather why temperatures are not hotter than they are.
fully available in our assumption

May 9, 2011 8:48 am

So Ferd, is it your assertion that the earth and venus are shrinking?
Earth’s Shrinking Atmosphere Baffles Scientists
An increase in CO2 could be one reason why a layer of Earth’s upper atmosphere went through its biggest contraction in 43 years.
http://news.discovery.com/earth/earth-atmosphere-shrinking.html

May 9, 2011 9:14 am

“This extra heat is generated due to gravitational contraction – the planet is slowly shrinking in diameter.”
What is the mechanism by which this is possible? Why has the planet not reached equilibrium? Why does the temperature of all the planets and the sun increase towards the center? If the temperature increase is a result of continued shrinkage, then this should long ago have ended on some, if not all planets.
If this efect is unique to Jupiter, why is it so similar to what we see on other planets. Is is reasonable to suggest that a similar observation has unique and different causes on each different planet? Or is it more reasonable to assume that these unique explantions are a result of a lack of understanding. That they are modern day epicycles.
We see a surface temperature of venus that is higher than expected if we used the same methodology for albedo as used on earth. We see a surface temperature for Earth that is higher than expected if we use .7 for albedo, but NOT higher than expected if we use for example .8 or .9. We do not have to increase this above 1, so it is possible the missing 33k is simply due to an error in our understanding about how albedo actually works.
We see a surface temperature for Jupiter that is higher than we expect. This follow a similar pattern as with venus and earth. On each of the three planets science explains this by using 3 different methods, and proposes that these are unique to each planet.
That may well be, but occam tells us that this is not the likely explanation. Since a simpler explanation exists, namely that we are not properly accounding for the effects of clouds in our calculations. The observations of venus suggest that clouds do not block solar radiation the way we think, or the surface of venus would be much cooler. When we apply this logic to earth, the”missing” 33 K is no longer missing. It is simply a result of us using .7 when a more accurate number would be .8 or .9. Jupiter is also covered with clouds. It could well be that the missing 60K on jupiter is due to the poorly understood clouds, not due to continued shrinkage.

Dave Springer
May 9, 2011 9:16 am

ferd berple says:
May 9, 2011 at 8:28 am

“It takes the oceans centuries if not millenia to absorb additional CO2 ”
This does not appear correct. CO2 absorbtion at the surface of the ocean is limited by the surface area.

The effective surface area is greatly increased or diminished by wind driven waves. It’s also limited by how far out of equilibrium it is. It’s also limited by how fast the surface water mixes with the deeper water which is governed by both winds and convective currents. The amount of CO2 that can be held in water increases with depth as well. In deep ocean trenches CO2 has been observed in liquid droplets. So the rate at which surface water is sequestered at depth is also very important.
It’s been a bit of a surprise for the climate boffins that no matter how much faster humans introduce CO2 into the atmosphere only about half of it remains in the atmosphere. The mechanisms that serve as carbon sinks for the atmosphere (we didn’t even start talking about biological and chemical sequestration) are diverse and not well enough understood to predict whether this will remain true going forward.

wayne
May 9, 2011 9:21 am

Dave Springer,
You see, I know radiation in many cases can and does radiate in any and all directions, but that depends on the temperature and radiation from the surrounding matter. To radiate all directions the matter all around must be colder than the source that is radiating. The main difference you and I have is that radiation never actually flows from cold to hot matter, it doesn’t even flow from equal temperature pieces of matter or you could burn yourself with a Fresnel lens in a dark warm room as my little experiment should show you, if you will get curious and perform it yourself. Planets and stars radiate against basically nothing so you can let the other temperature in the Stefan-Boltzmann equation be zero, but always subtracting 2.7K would be more proper. You are one tiny step from moving your rather unrealistic view into what actually happens.
Here is what happens because of the Stefan-Boltzmann equation that screws so many people up (sigma=5.6704r-8, epsilon=1):
ε σ ( 288K^4 – 280K^4 ) = 41.572 W/m^2
or
ε σ ( 288K^4 ) – ε σ ( 280K^4 ) = 41.572 W/m^2
The top one will always show you the correct and actual flow of radiation between two pieces of matter. The latter one implies that matter always radiates at all times in all directions even if the temperatures are identical and the last term gets this quirky term applied to it, “back radiation”. You will end up with the same numeric answer but it leaves you with the wrong impression that two things have occurred when in reality only one thing has occurred, radiation from the warmer to the cooler or not at all if the two temperatures are the same.
This same logic applies to your split beams and it is why such instruments must be constructed to tightly control the temperatures inside. By the second set of equations temperatures of different pieces of the instrument do not matter at all, by the top equation surrounding temperatures of the pieces of the measuring instrument matters greatly. Go lookup an operation manual and see if I am not correct.
Please, drop the “back radiation”, it really does not exist, use the top equation.

May 9, 2011 9:23 am

Dave Springer says “I already explained to you that temperature only increases during the process of compression ”
What you and Ira and Joel fail to consider is that there is continuous compression of cold descending air which heats that air until it becomes hot enough to then ascend, expand, and cool until the process repeats. The atmosphere thus acts as a continuous compressor (and expander).
In addition to the 4 links I provided above, here’s 2 more peer-reviewed papers that explain this as well:
http://www.climatephysics.com/PDFs/Chilingar%20-%20Cooling%20due%20to%20CO2.pdf
http://docs.google.com/fileview?id=0B74u5vgGLaWoYTAyYTc4NmUtZDFkMC00ODg3LTgwYzAtOTg3OTJhZjI5MDVk&authkey=CKyD1r8H&hl=en

Dave Springer
May 9, 2011 9:26 am

ferd berple says:
May 9, 2011 at 8:48 am

So Ferd, is it your assertion that the earth and venus are shrinking?
Earth’s Shrinking Atmosphere Baffles Scientists
An increase in CO2 could be one reason why a layer of Earth’s upper atmosphere went through its biggest contraction in 43 years.
http://news.discovery.com/earth/earth-atmosphere-shrinking.html

Does this change air pressure at sea level significantly?
That’s what’s happening on Jupiter. Pressure at depth is constantly rising. That isn’t happening on the earth. Atmospheric pressure at sea level rises and falls to some degree as horizontal pressure ridges sweep along but the average pressure remains the same at 1 bar.

Dave Springer
May 9, 2011 9:36 am

Jupiter is still undergoing gravitational contraction that began when the solar system was born. This process stopped for the earth billions of years ago. The sun stopped its gravitational contraction when temperature increase from ongoing compression lit off a sustained fusion reaction. The outward pressure from fusion counteracts the inward pressure from gravity and an equilibrium point was reached where they are equal and opposite in force and the sun has a relatively stable diameter that will persist until it starts running out of hydrogen fuel to sustain the outward pressure.

May 9, 2011 9:43 am

Hockey Schtick says:

What you and Ira and Joel fail to consider is that there is continuous compression of cold descending air which heats that air until it becomes hot enough to then ascend, expand, and cool until the process repeats. The atmosphere thus acts as a continuous compressor (and expander).

We consider that…but it can’t change the result. You are talking about things that can only redistribute energy within the earth-atmosphere system. You still must satisfy conservation of energy when you look at the interaction of the Earth System with the sun and space. And, the combination of a surface temperature above the 255 K blackbody temperature and an atmosphere transparent to terrestrial IR radiation doesn’t do this: There is much more energy going out of the system than coming into it. The result of this would be rapid cooling.

richard verney
May 9, 2011 9:53 am

Ira
I always enjoy reading your articles and the comments that they generate.
I am sceptical as to how useful it is to consider just one element of how the atmosphere may work ignoring other factors which are known to exist and which in some circumstances are definitely more important factors (I have in mind convection, evaporation etc). Leaving that aside, I have problems in envisaging that the back radiation plays any significant role in the workings of the atmosphere.
I have no problem with the argument on net energy flow and that if the atmosphere is warmer it slows down the heat loss from the warmer earth (or ocean). The problem I have is whether the atmosphere is warm simply because of the absorption of solar energy (a combination of absorption of incoming solar radiation and absorption of reflected and radiated solar energy from the bottom up) and adiabatic lapse, or whether in addition back radiation from GHGs plays a significant role.
You will see the point I made at (richard verney says) at May 8, 2011 at 8:39 pm
Smoking Frog sought to answer the point. He comments at May 9, 2011 at 5:21 am:
“Martin Mason said – If back radiation is so powerful and it is a heat source, why don’t we use it?
Because the earth’s surface is already at the temperature to which the back-radiation has raised it.”
However, with respect that explanation cannot be correct. We know as fact that solar (possibly with the help of back radiation) heats the earth’s surface to the temperature to which solar energy has raised it, yet not withstanding this we can still extract work from the incoming solar energy. We can collect and focus the 184 w/m^2 of incoming solar energy and use it to melt salts etc in solar power plants. That being the case, it begs the question why can’t we collect and focus the 333 w/m^2 of back radiation and extract useful work from it.
The fact that we cannot do this suggests that the ground is not being bomb barded with 333 w/m^2 of (back) radiation.
I have a black and white marble path. On sunny days, this path is warm to walk on. The black marble slabs are considerably warmer than the white marble slabs. Not surprising since black is a better absorber than white and therefore the black marble stones absorb more of the incoming 184 w/m^2 of solar energy. IF back radiation existed and if there was some 333 w/m^2 of back radiation beating down on the ground, the black marble slaps should be significantly warmer than the white marble slaps at night or on cloudy days. However, they do not appear to be. In particular when left over night to bask in the supposedly 333 w/m^2 of back radiation, the black marble slaps do not heat up whereas they do heat up with only 184 w/m^2 of incoming solar energy!!
Ira, I would like to see someone conduct a very simple experiment along the following lines;
1. Get two identical metal slabs (same dimensions for area and thickness) save that one is painted brilliant white and the other matt black,
2. Pre heat the slabs to the forecasted daily temperature (say picking a day when the estimated temperature is going to be 20degC).
3. On a clear sunny day (ie the day estimated to be 20 degC) place the slaps in the sun and measure there respective temperatures every half hour.
4. On a mild summer night (one estimated to be 20 degC), pre heat the slaps to the forecasted night temp of 20C and then leave them out after the sun has fully set and measure there respective temperatures every half hour.
5. Compare the results.
6. In particular does the matt black slab gain heat at night from the down welling back radiation. If it does not, why not?
Ira, I would appreciate your comments

May 9, 2011 9:54 am

Bryan says:

Joel Shore has refined this even further to say Postema is a believer, only Postemo hasn’t quite realised it.
Perhaps with the same elasticity Joel will prove that Gerlich and Tscheuschner also are really IPCC advocates.

No…What I am saying is that Postma and G&T mix in a lot of correct science that is already known with some incorrect science…or just glossing over things…to reach completely incorrect and absurb conclusions. That tends to be how peddlers of pseudo-science operate.

On page 9 of your co-written Halpern et al comment paper you state that the Stephan Boltzman Law can be used to work out the thermal energy between atmospheric shells.
I brought your attention to it in my post above;
….. ” 3. Has the Stephan Boltzmann equation applying to gases.”……

It is sad to see that the pull of pseudo-scientific arguments are just too strong for you to resist making them again and again. The shells that we consider are blackbody shells, so yes, the S-B equation applies by definition. This is a model. Is it a good model? Well, yes, it is quite a fine model to illustrate the basic features of the greenhouse effect. When you want to do detailed quantitative calculations, you adopt a better model. Is that concept so difficult to understand?
It is not science to attack a model for being to simple when it is meant to be a simple representation and where more refined models are available and can be used to show that the qualitative predictions of the simpler models for which they are used are correct. That is instead the realm of pseudo-science, a realm that is apparently just too irresistible to you. That you cannot even distinguish the difference between scientific arguments and the pseudo-scientific nonsense that you peddle is very very sad for you, the people reading this thread, anybody who respects the scientific enterprise, and even AGW skeptics who don’t want to be considered complete crackpots. I actually pity people like Ira and David M. Hoffer for having to put up with people like you on their side of the AGW debate.

JAE
May 9, 2011 9:54 am

Heh, and LOL. Unless I missed it, nobody has yet tackled the “venus thing” presented above by a commenter (which suggests that the GHE theory isn’t valid):
http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html
BTW, it holds for other planetoids with an atmosphere (I can provide links if anyone cares).
Despite the Orwellian chants about “multiple lines of evidence,” the case for a GHE threat is really hurting these days, due to the total lack of any such lines of evidence. In fact the multiple lines of evidence appear to go the other way:
1. No warming for 10-15 years, despite all the increases in the terrible GHGs (and not just OCO).
2. Ice core records show no correlation between GHGs and temperature.
3. The temperature swings during the past couple of millenia indicate the climate changes naturally without changes in GHGs. (We don’t need Mann’s proxies to know what happened during the MWP and LIA; it’s recorded in the history books, for crying out loud!)
4. No “hot spot” in the mid-troposphere in the tropics, as predicted by ALL the climate models. (So much for treating their output as “data.”)
5. No laboratory demonstrations or other real-world empirical demonstrations of any GHE effects.
6. Areas with the most greenhouse gases (tropics) have maximum temperatures that are no where near as high as some desert areas with the lowest amount of GHEs (yeah, yeah, it’s the water evaporation and clouds; but that is precisely the point on Earth).
7. Articles such as the one cited above, which show clearly that temperature in a gaseous atmosphere has nothing to do with the “greenhouse effect.”

JAE
May 9, 2011 10:02 am

Dave Springer:
“We do use it all the time. One example is the smudge pots used to limit freeze damage in orange and avocado groves. The heat of the smoke from the smudge pots is inconsequential. The effect comes from the black carbon particles in the smoke which absorb upwelling IR from the ground and send a portion of it downwards and sideways. It works as an insulator. This is the same mechanism by which CO2 molecules insulate. Anyone who contests this is simply wrong and needs to take an introductory class in physics.”
Well, I question this. Since the backradiation is coming from cooler air above, it cannot be warming the orange grove.
The Clean Air Act killed the smudge pots, so they now just blow air through the orchards or add heat by spraying water. Maybe they didn’t ever really need that backradiation?

Dave Springer
May 9, 2011 10:05 am

Hockey Schtick says:
May 9, 2011 at 9:23 am
“What you and Ira and Joel fail to consider is that there is continuous compression of cold descending air which heats that air until it becomes hot enough to then ascend, expand, and cool until the process repeats. The atmosphere thus acts as a continuous compressor (and expander).”
Not in my case. I’ve discussed it in depth many times at WUWT. The following search reveals some of it.
http://www.google.com/search?hl=&q=tunderstorm+heat+engine+springer+site%3Awattsupwiththat.com&sourceid=navclient-ff&rlz=1B3GGGL_enUS290US290&ie=UTF-8
The big picture:
The sun heats the ocean, the ocean heats the atmosphere, the cosmic void cools the atmosphere.
This thread is about radiative transfer in and out which is a detail in the big picture. Convective transfer is a different detail but also very important. The water cycle in general is exceedingly important in basic understanding of atmospheric physics. Vast amounts of energy are lifted from the surface by evaporation in what’s called “latent heat of vaporization”. It’s called “latent” because it won’t register on a thermometer. When the vapor condenses the latent heat is released. As the earth gets warmer (for whatever reason) the speed of the water cycle increases. This is a negative feedback mechanism and it’s the most glaring flaw in the whole warmist narrative. There has never been a runaway greenhouse on the earth. Never. The hypothetical amplification of anthropogenic CO2 forcing by increased water vapor is a wholesale fabrication without a shred of evidence to support it and with mountains of contrary evidence. The water cycle is a negative feedback not a positive feedback. Most of the more informed global warmists and skeptics know that this amplification is the heart of the controversy. None dispute that when everything else is equal more CO2 will result in a warmer surface and that this is would be about 1.0C for each CO2 doubling starting from a base point of 280ppm. They also acknowledge that a warming surface will speed up the water cycle. The controversy is over whether a faster water cycle is a positive or negative feedback. The evidence says it’s negative. The alarmist must have it as a positive feedback because otherwise the only warming that’s gonna happen is a modest welcome amount that is of great net benefit to the biosphere. Living things don’t tolerate ice very well so the less of it there is the more life can flourish on this third rock from the sun.

Dave Springer
May 9, 2011 10:25 am

Climate boffins have pretty much conceded the debate over water vapor amplification. As this concession ocurred it manifested to the unwashed masses as “global warming” was first replaced by “climate change” and now is morphing yet again into “climate disruption”.
The climate does nothing if not change and it has been doing so since long before humans entered the equation so “climate change” failed to elicit the needed level of fear response in the masses. Climate disruption has more fear factor in it.
An inconvenient amount of climate disruption is a distinct possible result of anthropogenic activity but once again the evidence is all stacked against it. By the same token radically modifying anthropogenic activity to lessen any imagined disruption has an immediate negative consequence of slowing down the global economy which is driven in large part by fossil fuel consumption.
Thus we have a risk/reward decision to make and in my view trying to reduce CO2 emissions is essentially all risk and no reward. Real pollutants that immediately and directly effect our health and well being are a much greater concern. Heavy metals in our water, particulate pollutants in our air, ozone, and lots of other things are legitimate red flag events but CO2 is plant food and the more of it we have the better off we are.

May 9, 2011 10:27 am

“In deep ocean trenches CO2 has been observed in liquid droplets. So the rate at which surface water is sequestered at depth is also very important.”
That is intersting. The ocean pressure goes up 1 atmosphere per 30 feet of depth. At 300 K CO2 becomes a liquid at 100 atmospheres. The deep oceans are all below 300K and 100 atmospheres is only 3000 feet – much less the the depth of the average ocean basins.
So, what this means is that CO2 can exist as a liquid across most of the ocean bottoms. As liquid CO2 should weigh more than H2O, the question really should be why the bottom of the oceans is not all liquid CO2? Except that the oceans are not saturated, and thus liquid CO2 would be absorbed.
http://en.wikipedia.org/wiki/File:Carbon_dioxide_pressure-temperature_phase_diagram.svg
http://oceanservice.noaa.gov/facts/oceandepth.html

May 9, 2011 10:37 am

http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html
well worth the read, as well as the follow on explanation. what I find fascinating is that albedo has no effect on the temperature comparison between earth and venus, which accounts fully for the missing 33K in Ira’s analysis.
quote from the article:
For example, in the analysis, not only does the amount of CO2 not enter in (Earth has 0.04%, Venus a whopping 96.5%), but the albedo (from either cloud tops or the planetary surface) does not either (Venus has dense clouds that reflect much of the incident visible radiation, while Earth does not, and Earth’s surface is 70% deep ocean, while Venus is solid crust). The real atmospheres don’t care at all about these great differences in the two atmospheres and planetary surfaces, they only care, and quite precisely, about their distances from the Sun.

May 9, 2011 10:45 am

http://2.bp.blogspot.com/-4R16sH96XBk/TX5hIW8MADI/AAAAAAAAAD4/lhiu503gY58/s1600/venus%2Bearth%2Bcomparison.JPG
This cannot simply be coincidence. How can two planets with completely different atmospheres, albedo’s and surfaces show almost identical temperatures as a function of pressure if the standard climate model is correct? Quite simply they can’t. These observations are either false or climate science has it fundamentally wrong.

MartinGAtkins