"Miskolczi’s theory does not claim that increasing greenhouse gases won’t increase temperatures initially. In fact, his radiative relationship of temperature to atmospheric opacity is very similar to the one above with an additional exponential term: 1+τ+e-τ. "

But didn't you just show that there was no effect from increasing greenhouse gases (i.e., the addition of water made no difference)?

Thats one area where people get tripped up. Radiative eqns suggest temperatures must increase monotonically. However, the energetic constraints at the maximum forbid it.

Finally! Thank-you. Why are there so few tests proving/disproving the theory? This was obviously easy and inexpensive.

The whole kit and kaboodle costs less than the humidity/temperature transmitter I just bought I have yet to build the data logger and the supplier of an important part for that is out of stock.

However, the energetic constraints at the maximum forbid it.

The energy consuming bits are greedy the energy from the surface used to warm the atmosphere cannot be used to raise more water from the surface. It can however be used to retain it can also be (and is) used for the Orographic Lift that causes temperatures to fall and the water to precipitated out

This exercise proves nothing. IR on this scale has a very small role in heat transport, and even that is not modified by the increased opacity of a few cm of moister air (which you have not attempted to quantify).

On a scale of metres, heat transfer is dominated by conduction and convection. These modes respond to temperature gradients, which are several degrees per metre. In the bulk atmosphere, temperature gradients are degrees per kilometre, so much less heat can be transferred by those means. IR, however, does not depend on temperature gradient, and becomes the dominant mode on large scales. For your apparatus, the temperature is determined by internal convection, and conduction through the plastic, and the neighboring stationary air boundary layers.

On the opacity side, we've talked of an optical depth of about 2 for the whole atmosphere. That means that on a grey-body average, IR is attenuated by a factor of about 8 over an effective distance of about 10 km. It's logarithmic; over a meter, the attentuation is about 8^0.0001, or about 0.0002.

So the effect of IR was small anyway, and was attenuated by natural GHG by a factor of less than .0002. You might have doubled that attenuation. The temperature, determined predominantly by convection and conduction, is, as expected, unaffected.

David, I'm surprised how poorly your blog lives up to its professed faith in numeracy.

OK, before people jump on that, the attenuation factor is .9998. Afdter passing through the air, the IR is .9998 of its initial amplitude, or a reduction by a factor of .0002.

Nick:

"On a scale of metres, heat transfer is dominated by conduction and convection. These modes respond to temperature gradients, which are several degrees per metre. In the bulk atmosphere, temperature gradients are degrees per kilometre, so much less heat can be transferred by those means. IR, however, does not depend on temperature gradient, and becomes the dominant mode on large scales. "


Well, you got the first part right, but you gotta prove the second part. Got any evidence for that IR part?

Dave I get "Easy Tiger" when I click on CoRev's name I'm not sure that's intended.

Nick, If you have a reference to more accurate experiments demonstrating an increase in temperature due to increased attenuation of IR in a cavity with a fully developed greenhouse effect, please let us know.

Jan, it can't find the page. Thanks for alerting me to this.

#9 David
No, again because of scale, such experiments are unlikely. As I say, there are two issues - conduction/convection is almost certain to dominate radiation in heat transport, and trace levels of GHG will not significantly attentuate IR anyway over small distances.

Such experiments also don't reproduce other important features of the greenhouse effect - the lapse rate in temperature, which leads to IR emission from much colder parts of the atmosphere - and also the effect of back radiation from parts of the atmosphere at different temperature.

jae #7 Not sure what you want evidence of. Stefan's Law governs radiative transfer, and does not involve temperature gradient. As to IR being dominant - I'll just refer to your favorite K&T cartoon, and all the experimental evidence behind it.

Nick #11

I am sure a lot of chemists will be surprised to hear that the attenuation over small distances is too small for the cuvettes that they have been using for the past 100 or more years in spectrometric analysis.

CoRev, editor
http://www.globalwarmingclearinghouse.blogspot....

Jan #12
Again it's scaling. Chemists use powerful sources, much more than thermal emission levels. They would also nor
mally use higher concentrations of absorbing gases.

Nick #14

Again it’s scaling. Chemists use powerful sources, much more than thermal emission levels. They would also normally use higher concentrations of absorbing gases.

You seem to get your scales back to front quite a bit could it be because you've acutally never set foot in a chem lab?

Higher concentrations you say then look at the limitations near the end and see that .01 M is about the limit and CO2 is about .0165 - .017 M water at about .18 M jae might like to check the concentrations as it has been 40 years since I worked in a chem lab.

Transmittance is a ratio of incoming to outgoing power of the source is irrelevant. Sure the ratio of remission to absorbed is 1 but if you are going to invoke Kirtchoff's law here you perhaps should think twice before criticizing FM for doing the same thing.

Nick, the title of the post is purposely provocative. I think you are cool. There are two questions however that beg for answers:

1. Is there a maximum to the greenhouse effect?
2. If so, is it affected by IR attenuation?

No-one is denying radiative equilibrium on the atmospheric scale, least of all Miskolczi. In the atmosphere, maximum greenhouse effect is achieved only at the earth surface, and the atmosphere acts as a quasi-cavity, which makes it a different set-up. This questions are about the simplified system.

Jan/David,

Looking at the posts before this one, we are working with a chemical system?

Greenhouse effect is a physical problem, according to theory.

So how can we describe the problem in terms of chemistry when the object is a gas.

Or have I missed something.

Louis #17

Not chemistry per se but infrared (and other) spectrometry are tools much used b y chemists. Determining concentrations of materials is very much the chemist's domain.

#5 Nick Stokes
"David, I’m surprised how poorly your blog lives up to its professed faith in numeracy."

Go beat yourself with your negatives.
Even if you are smarter and more educated,
Cannot stand the stench in the beer parlor
Stagger and Barf outside !

Your job, on your high pedestal, is to teach
Pay back the high cost of your training.

Small systems can prove the principles of larger ones

For example, Cloud Chamber
Steam in the shower, low level fog above a swamp
Turn the system upside down, to eliminate convection

I would like to see SF6 remove something
Or be removed by CO2, H2O ?

Infrared camera, looking at diffraction grated spectrum, could be a very versatile input device

Nick:

"jae #7 Not sure what you want evidence of. Stefan’s Law governs radiative transfer, and does not involve temperature gradient. As to IR being dominant - I’ll just refer to your favorite K&T cartoon, and all the experimental evidence behind it."

Nick, I think the radiation cartoons attempt to show the radiation balance, which is a function of the temperature distribution. They say nothing about how that temperature distribution comes about. The radiation is an effect, not a cause. I think M is correct in his ideas about non-radiative inputs (K) causing the temperature distribution. To my thinking, this is where the climate scientists generally have it wrong, big time. They essentially igore convection KT gives all of 24 Watts/m-2 for convection, which they label as "thermals". That's nuts. The surface of the earth would be just like a closed greenhouse, were it not for convection.

Relative to a "maximum greenhouse effect," I'm still waiting for an answer as to why temperatures in very moist areas (tropics, e.g.) never exceed about 33 C. Shouldn't the maximum greenhouse effect exist in such humid areas? The maximum TEMPERATURE effects are found in the deserts, where there is only about 1/4 the greenhouse gases (usually somewhere around 5-8 g/m^3 absolute humidity). Maybe the maximum greenhouse effect is below 5-8 g/m^3?

Oh for heavens sake! What makes you think there isn't something fundamentally wrong with this experimental setup? (4)

Look at the red line (temperature inside the container) at about 16:36 - see the sudden "knee" where it stops rising and goes dead flat?

Temperature rises pretty steeply between around 16:33 and 16:36 and then goes dead flat?

C'mon what physical process does that? Unless there is some physical flaw in the experimental design. This is lent extra weight when you add the extra water at around 16:45 (the green line discontinuity).(1) (2) Despite the addition, nothing happens to any other measurement.(3)

And all this is supposed to happen over a period of what? 5-10 minutes?

What all this says to me is that the experiment is constrained somewhere so that the results are artificially held - possibly through measurement error - within bounds that don't reflect the actual process.

Can you say "broken"?


(1) And WTF is happening when the partial pressure of h2o is falling in the face of increasing temperature between 16:30 and 16:40? Did they repeal the gas law last week or something?

(2) "A small amount of warm water was added after time 16:45" and the temperature didn't increase? In a "closed" experiment?

(3) You don't provide a scale for humidity, but that looks like a helluva jump. In the face of rising temperature humidity falls and then suddenly returns to its starting value with the addition of a "small" amount of "warm" water? The word "small" communicates the idea of "an insignificant fraction of that originally present". How much exactly?

(4) Bit of a postscript. I checked the Atech product range and www.atech.hk and while I couldn't find an exact match, the picture seems to show a device in the WS-233-xxx range. This is specifed as recording "indoor temperature" between -10 and +50.

You maxed out at +50 right? Maybe your instrument just didn't have the range you wanted.

JM: You obviously have no idea how greenhouses work. Or a notion of relative humidity. Why don't you try to repeat the experiment?

Jae, it doesn't matter whether I understand greenhouses or not.

The measuring instrument is a consumer device specified to operate at no more than 50 degrees and this experiment was conducted at 50 degrees or above.

The results are worthless (even given the other objections I have to the experimental design).

a.) the device is not specified to operate reliably (or at all) above 50 degrees

b.) I doubt it even measures temperatures above 50 degrees on the off chance that it might give semi-reliable estimates out of its specified range.

Oh for heavens sake! What makes you think there isn’t something fundamentally wrong with this experimental setup? (4)

Look at the red line (temperature inside the container) at about 16:36 - see the sudden “knee” where it stops rising and goes dead flat?

Thats because there are only a few measurements.

(2) Despite the addition, nothing happens to any other measurement.(3)

No, the relative humidity (green line) increases.

(1) And WTF is happening when the partial pressure of h2o is falling in the face of increasing temperature between 16:30 and 16:40? Did they repeal the gas law last week or something?

Thats what you would expect the relative humidity to do.
http://en.wikipedia.org/wiki/Relative_humidity

(2) “A small amount of warm water was added after time 16:45″ and the temperature didn’t increase? In a “closed” experiment?


(3) You don’t provide a scale for humidity, but that looks like a helluva jump. In the face of rising temperature humidity falls and then suddenly returns to its starting value with the addition of a “small” amount of “warm” water? The word “small” communicates the idea of “an insignificant fraction of that originally present”. How much exactly?

Enough to just cover the bottom of the container.

(4) Bit of a postscript. I checked the Atech product range and http://www.atech.hk and while I couldn’t find an exact match, the picture seems to show a device in the WS-233-xxx range. This is specifed as recording “indoor temperature” between -10 and +50.

You maxed out at +50 right? Maybe your instrument just didn’t have the range you wanted.

Good question. I will try it again today with the transmitter part. Its sitting at 57.2C now as it bakes in my mini-oven.

"Enough to just cover the bottom of the container."

I presume you mean the original amount? Did you measure the quantity? And did you measure the quantity you added?

And also did you measure both the temperature of the original water and of the additional warm water?

"Its sitting at 57.2C now as it bakes in my mini-oven."

It's not specced to operate at that temperature so its readings cannot be reliable.

Good grief, GM, he isn't doing a master's thesis here. Suggestions, rather than rude snarks would be more appropriate. Anyway, it looks to me that the temperature hit 50 C for a very brief time, and I would imagine that the instrument doesn't abruptly stop reading at 50 C (the accuracy probably just goes down).

It would be interesting to put SF6 in the container, since it is something like 23,000 times more powerful as a GHG as OCO.

"Thats because there are only a few measurements."

David (it is David?), can I just ask you about this. I presumed that you were graphing a continuous series of real-time measurements so that there were vastly more than a "few".

Can I take from your comment here that you sampled the output on a schedule and the lines in the graph represent a fit to those samples rather than a continuous read-out?

JM: Read the post again. The data are provided. BTW, the temperature never actually hit 50 C.

Brilliant !

Try a movie making type steam gnerator
(Even a small unit can cloud up the corner of a room)

Blow in steam -- see if the steam removes itself
Or creates runaway, pizza oven heat, feedback effects
Will the results be IPCC dogma, runaway, rejected ?

JM, read the latest post. The specs on the transmitter go to "+60C" whatever that means. Sure I would like to do it with better gear. I am not asking you to believe it. I am asking you to try it.

<blockpost>For your apparatus, the temperature is determined by internal convection, and conduction through the plastic, and the neighboring stationary air boundary layers.</blockpost>

Agree in general, however I would say that achieving the theoretical maximum is limited by the factors above. Do YOU dispute a theoretical maximum of 1.5 times?

David, you lost me in the second experiment. How does ambient (30C) equate to OLR?

David #33 JM said

For your apparatus, the temperature is determined by internal convection, and conduction through the plastic, and the neighboring stationary air boundary layers.

as I did above. IR is not a significant influence on the temperature on this scale, and even if it were, a few cm of moist air will not attenuate it noticeably. The factor of 3/2 derives from IR theory, and has no relevance here.

I think the temperature measuring issue is a red herring. Your result is what is to be expected, and says nothing about AGW in the atmosphere.

Nick, you will have to do better than bald assertion. I think its exceptionally close. There is a central black body (the sensor), a convection zone as in the atmosphere (the inside of the jar), a conduction zone (the glass wall) just as in the upper atmosphere.

The purpose of a model is abstraction. Because a jar is not the atmosphere does not make it irrelevant.

David #36
Calculations were in the previous post. The absorption side is easier. With a gray-body optical depth of about 1.87, say, 10 km of atmosphere absorbs about 80% of IR. Then 1 m of air absorbs maybe 0.02%. You have much less than 1m. Adding water vapor increases this, but it is still negligible.

You can visualise this by thinking of a hazy day where hills a few km away are faint. IR absorption is on this scale. But you don't expect such a haze to affect your ability to see your hand.

On the importance of IR, in the atmosphere, IR dominates convection, but not by that much - say 24 W/m convection vs 240 W/m2 IR. But, as I said, convection responds to temperature gradient, while IR doesn't. And in your experiment, the temperature gradients are of the order of 1000 times greater.

These are big scale differences, and if you want to claim that your experiment does show something about AGW, I think the onus shifts to you to resolve them. Numeracy requires no less!

"On the importance of IR, in the atmosphere, IR dominates convection, but not by that much - say 24 W/m convection vs 240 W/m2 IR."

That is one place where the current theory is so screwed up and where M is so right. It's the other way around, as anyone who steps outside will notice. Again, if it weren't for convection, the earth's surface would be just like the inside of David's jar.

What ?? -- No runaway destructive feedback, imploding the container, creating a mini black hole, devouring Earth.

Shut this down David, immediately, before the God particle visits from another stringy dimension.

Protesters outside your convection oven can get very mean -- pull the power plud to destruction of humanity.

Nick, We were interrupted by down time. Your issue is that the path length is too short to produce detectable attenuation of IR in this experiment right? Interestingly, addition of water does produce a change, cooling not warming, so some effect is noticeable. This is as Noor says, the net effect of more vapor is cooling through increased convection. Presumably it is heated through attenuation of IR. It a way it illustrates that its too simplistic to talk about net positive or negative feedback. Attenuation increases temperature, which increases convection which decreases temperature.

But you miss the main points I think. There is a temperature gradient across the apparatus, analogous to a lapse rate. Heat is propagated. Whether it is by convection or conduction is not so important for this purpose. There is also an IR phase, but the ratios are different in the atmosphere.

The main point, which I would be grateful if you address, is what to you think the maximum greenhouse effect should be: 1.5, 2 or something else? This is clearly falsifiable. The RealClimate toy model says 2. FM says 1.5. I strongly suspect it is 1.5, and if so the RealClimate bots are demonstrable wrong in this aspect of their understanding.

<abbr>davidss last blog post..Greenhouse Heat Engine #3</abbr>

Nick#37

"On the importance of IR, in the atmosphere, IR dominates convection, but not by that much - say 24 W/m convection vs 240 W/m2 IR."

I had thought the Earth's true convective heat flux was of the order of 80 - 100 W/m2. Is this not true?

#41 Steve
I think that figure of 80-100 includes latent heat transport. LH transport requires both evaporation and condensation (rain). It's not clear that the condensation part of the cycle is happening here, so I quoted just the pure convective figure. But the figuring here is rough, and my argument stands if the convection is 80-100. The point is that in the whole atmosphere IR is not much more than convection, maybe a factor of up to 10, and on David's scale, temperature gradients, which drive convection, are maybe 1000 times higher. Then you can expect convection to dominate.

jae: The ambient is the lower boundary condition.

Nick #42

I think that figure of 80-100 includes latent heat transport. LH transport requires both evaporation and condensation (rain).

Yes

The point is that in the whole atmosphere IR is not much more than convection, maybe a factor of up to 10

aye? Is that convective transport is 10 times radiative? If so you had a different opinion earlier on.

and on David’s scale, temperature gradients, which drive convection, are maybe 1000 times higher.

In the container?

Then you can expect convection to dominate.

Then there is the interaction of the gases in the container with the container wall which conducts heat to the cooler atmosphere.

It's not so silly.

In my distant past we had a piece of electronic equipment imported from North America that was completely sealed and had some heat being generated inside by lasers & motors and such that used to keel over in over in aussie summer heat. All we did to fix it was screw a heat exchanger on the body and that fixed it.

Now the question is only does the extra water vapour inside improve the exchange of heat inside with the wall of the container as the water vapour does in the atmosphere since it has less vapour above to absorb it's radiation?

It’s not so silly.

Huh? Not nearly as silly as a discontinuity at the earth's surface.

It should be quite easy to fine tune this experiment to ensure that the ratio of the surface area of the water to the volume of air inside the container roughly matched that of the ocean to the troposphere (say).

The receiver may have to be suspended on a string above the water.

I also think the external temperature needs to strictly be the actual skin temperature of the container (use Bluetack or something) rather than the external air temperature as there will be convective cooling off the outside of the container. It would be wise to ensure external convective cooling was minimized by embedding container in ground up to water level.

The experiment should be run for a whole day (must get up/start at sunrise and put experiment in middle of large patio of flat field with no shade etc ;-).

IMHO results should not be considered 'valid/steady state' until/if/when there is some condensation visible on the inside of the container.

Steve, There was condensation on the sides, and a small amount of free water in the bottom at the end. Within the precision of this experiment, which isn't very high, you can only try to saturate the air. I think the ambient is the right boundary condition. I think the right setup is a clear, still, sunny day so as the ambient represents as near as possible blackbody temperature.

Agree with all that. But we also have a deep pool of largely un-irradiated seawater at the bottom of our optical cavity.

As Gorshkov and Makarieva (2002) point out, the only two stable states we can be really 'bank on' thermodynamically are a fully frozen earth or one with a fully evaporated ocean. They note (and I agree) there is a lot of evidence of another stable state of some sort somewhere in between these two conditions. They suggest a biotic effect. Again I agree.

The existence of this mysterious stable state is the fundamental basis of M Theory although I find M's 'reason' somewhat hard to swallow.

Whether that steady state actually exists (AGW orthodoxy 'computer says no' ;-) is the core issue.

Even if that state actually does exist, whether M has identified the correct dynamic of that state is also a moot point. This is where my own doubts creep in.

If you can get this little experiment to work OK with ordinary air and fresh tap water (which I guessed you used), then the logical next step would be to replace the water with seawater. This would impose the correct partial pressure of CO2 on the gas phase as the air cavity/surface skin of the water warmed up.

If M is right, the steady state situation (if it exists) should be pretty much the same regardless.

Next, the container could be flushed with CO2 and the experiment re-run.

If M is right, the steady state situation should still be pretty much the same regardless!

Then M either bows to thunderous applause or exits stage left.

You could also then really screw with everyone's head big time by inserting one drop of olive oil and running the experiment again.

What fun this would all be!

David #45

Not so silly as suggesting that convection in the jar unimportant. You answered my next question: "Was there condensation on the wall?"

There was condensation on the sides, and a small amount of free water in the bottom at the end.

We know where the heat went to drop the temperature. Water condensed on the side because the walls were cool being exposed to the atmosphere. (Was there as much on the glass jar if any?). It's the same with the atmosphere on a larger scale where we have evaporative cooling on the surface convection lifts it away (more likely cause a discontinuity than I(mu)=Io+Ii*mu ) to where it can export that heat (and entropy) to space more readily, assisted by Steve's biotic dimethyl sulphides.

I want some idea of the temperature gradient on the walls and what if the container walls were KI?

Next, the container could be flushed with CO2 and the experiment re-run. If M is right, the steady state situation should still be pretty much the same regardless! Then M either bows to thunderous applause or exits stage left.

Exactly. That would be an ideal definitive experiment. One could hope. Its amazes me it has taken so long for me to get this to the point where it can be put in a nutshell. Now I don't know why it took me so long. Part of it is that the paper goes off on so many intricate branches, which is OK, but the discussion has been frustratingly tortuous as a result.

Gorshkov and Makarieva: look I don't know and I haven't had time to study another intricate paper. I don't think snowball earth and dry oceans are consistent with the Su=3OLR/2 constraint, but unless I studied it more deeply and worked with the numbers and equations a bit more I don't have a view. Just looking at the way variance increased when I added water at the temperature maximum, I was struck by the appearence of long term pesistent behaviour in the temperature changes. Water is strange stuff, and the capacity to phase change in the 0-100C range gives it mysterious powers. So I am skeptical of the need to invoke biotic control - just skeptical, I don't know.

David #51

Water is strange stuff, and the capacity to phase change in the 0-100C range gives it mysterious powers.

would you believe the range is -100 - 100C

Kaye & Laby

Thanks Jan.

Regarding biotic control, David, I understand your 'skepticism'.

ALL cyanobacteria need iron as an essential nutrient. One of the reasons advanced for why cyanobacteria bloom more strongly in NH oceans (actually in bimodal seasonal peaks I have found - although no-one has bothered to publish this fact) and less well in the SH oceans (actaully unimodally I have noticed) is that there is more iron in NH oceans.

Maybe yes, maybe no, but it is also known that the terminations of ice ages start in the NH due to the obliquity issue.

But now this (I'm sure you can decipher the simple Italian):

http://omniclimate.files.wordpress.com/2008/11/...

Steve David

I rather suspect that once people stop letting their emotions drive their science that we'll find that there are several feedback paths all negative with the only "amplifier" high energy cosmic rays (it works more or less like a thermionic valve).

Positive feedback can't happen without an internal energy source. While earth has a geothermal energy source that's internal it's strong enough to drive a flea off a dogs back.

I don't see how this experiment can be meaningful. The absorption of IR by the walls of the container, and the convection of air around the walls of the container are going to be the largest influence on the temperature, even if the radiation source were held constant. In addition the radiation source, the sun, is not controlled. These factors will determine the temperature of the gas inside the container. The IR absorption of the small amount of green house gases in the container as, Nick points out will be miniscule.

There is no hope for this experiment to shed any light on the supposed saturation of the greenhouse effect. It is totally meaningless.

Eric:

The absorption of IR by the walls of the container, and the convection of air around the walls of the container are going to be the largest influence on the temperature, even if the radiation source were held constant.

I take it that your assertion is that there is no maximum temperature due to greenhouse effect. In that case your assertion is provably wrong.

<abbr>David Stockwells last blog post..Model of Global Warming</abbr>

Jan, When I was doing the experiment I was fascinated at how sensitive the temperature of the system was to clouds, puffs of wind, when the water was added. The 1.5 greenhouse factor is an amplification factor, but with the water the system was far more sensitive to variation than the dry container. Beats playing computer games.

<abbr>David Stockwells last blog post..Model of Global Warming</abbr>

David #57

Did you see as much condensation with the glass jar?

I can imagine it was fascinating to see but one supposes that with water molecules being slightly polarised when the collide with the container walls they like to stick there especially if there is a slight charge due to the breeze. If they stick there they give up all their thermal energy and condense leaving the heat to hitch a ride on the passing breeze. Any rotational or vibrational energy absorbed will go right along with it.

Back to tearing my hair out over imapd - after i take the terriers for a walk. This is no way to spend a vacation.

David #56
David, Eric isn't asserting what you say; he's just saying that this is a heat transfer problem mainly involving convection and effects at the plastic, including maybe some IR absorption there.

Ironically, there are AGW sceptics witht a bee in their bonnets about the greenhouse effect being misnamed - that greenhouses work just like your apparatus, by blocking convection and inserting an insulating boundary layer. They're right that the analogy with atmospheric IR is inexact, but wrong in thinking that the inexact name invalidates anything.

Even at an IR level, your reasoning doesn't work. FM's theory matches Su with OLR, which is the outward radiation at TOA. Crucially, TOA is the point where no IR is coming back. This is far from being true at the exterior of your apparatus. There is the additional difficulty that FM's theory is for planar radiation, which yours certainly isn't.

The other aspect of your response to Eric that I didn't understand was the reference to the existence of a maximum temperature as proving something. In this situation, there will always be a maximum, regardless of the heat transfer mechanism - otherwise you would have a wondrous furnace.

#59 Nick Stokes
"Even at an IR level, your reasoning doesn’t work. FM’s theory matches Su with OLR"

Please be precise, some might be misled, interpret "matches" with equals; see Zagoni;
(8) SU = 3 OLR / 2

One gas removes the effect of another
We have to break it down, make things simpler;

Drinking Bobbing Bird experiment
Several birds, different fluids
Buy different, easy to evaporate solvents, at your paint store
Alcohol from the drug store, even Vodka
But stay away from naptha

Use diamond drill to make a filling hole for the Bobber's fluid.

Adding a different working fluid Bobbing Bird, to the communal drinking bowl.
One removes bobs from another ?
Runaway, self-destructive , bobbing feedbacks, resonating, butterfryng, chaotically cooking one bird after another ?

Nick,
Thanks for saving me the trouble of pointing out that David hasn't answered my objectives.

Any scientific paper which proposes and experimental result has a section which deals with experimental error. This is supposed to prove the validity of the experiment by examining quantititively what other phenomena besides the relationship being explored by the experiment, could have contributed to the result, spoiling the experiment.

The phenomena we have listed would tend to overwhelm any radiation absorption properties of the gas inside the container that are being tested.
The burden of proof is on the experimental designer to calculate the contributions of all the other phenomena to the equilibrium temperature of the air inside the vessel, if he wants the results he is claiming to be accepted.

One cannot be allowed to claim that the experiment is scientific in any way, if this burden has not been met. In this case, it has not even been attempted.
To my mind, this is evidence that the writer is not either unaware of the requirement, or wants to pull a fast one and get away with something.

Oops I should have said objections, not objectives.

Older personal computers have 4 analog, joystick inputs.

In addition to the weather station, what extra sensors to add ?

Perhaps use a giant glass vine making bottle, well
insulated. But leaving radiation in and out the top.

Eric and Nick: I'm not smart enough to prove it, but I think the errors you are looking at are quite small, relative to the massive greenhouse effect. They are probably the reason that the actual temperature doesn't quite get to the theoretical temperature.

Jae said,
"Eric and Nick: I’m not smart enough to prove it, but I think the errors you are looking at are quite small, relative to the massive greenhouse effect. They are probably the reason that the actual temperature doesn’t quite get to the theoretical temperature."

No matter how smart you are, if the experiment is bogus, you can't find a way around that fact.

Ordinary glass absorbs an re-rediates in the IR. This is one of the heat loss mechanisms that takes place as result of windows in a house, which transmit heat through them. Here is a link which explains the heat loss mechanisms associated with glass. This is common knowledge among people who deal with household energy. This is why special E glass is recommended for energy efficiency.
http://www.socalgas.com/construction/builders/B...
"Some of the long-wave energy emitted inside the room finds
its way to the window surface. This long-wave radiation is
absorbed by ordinary window glass and then re-radiated or
emitted as heat, either to the outdoors or back inside, from

the glass surfaces. Window heat loss by this means can be
reduced if the glass is treated to absorb less and emit less
long-wave radiant energy.

The ability of a surface to reflect long-wave radiation is
Measured by its emissivity. Emissivity vanes from 1 (100%
of long-wave radiation emitted) to 0 (0% emitted). For glass,
the lower the emissivity, the lower the U-factor. Clear glass
has an emissivity of about .84 while bright aluminum foil has
an emissivity of .05."

An emissivity of 0.84 means that glass is absorbing most of the IR entering it an re-emitting it in all directions. The emissivity of the small amount of atmosphere in the jar is miniscule compared. The number of emitting and absorbing atoms is way too small. The IR will mainly bounce between the walls of the glass, where it is absorbed and remitted. The molecules of gas will get most of their energy from contact with the walls of the jar.

Eric:

"The molecules of gas will get most of their energy from contact with the walls of the jar."

I don't see how that's possible, since the glass remains rather cool--at a much lower temp. than the air inside--due to losses outside. Touch the glass in your closed car from inside. Then touch the dashboard.

The ability of a surface to reflect long-wave radiation is
Measured by its emissivity.

Since when?

jae #68

Eric:

“The molecules of gas will get most of their energy from contact with the walls of the jar.”

I don’t see how that’s possible, since the glass remains rather cool–at a much lower temp. than the air inside–due to losses outside. Touch the glass in your closed car from inside. Then touch the dashboard.

Don't you find it amusing to find laymen teaching science to scientists such as yourself?

Any way it's fairly obvious from the fact that the gas in the plastic jar cools over time after the introduction of warm water that the addition aids cooling. It's also obvious from the condensation on the wall that it's at the wall that the water is giving up it's heat at the rate $$\geq$$ 2500 J/gm.

jae #68

" Eric:

“The molecules of gas will get most of their energy from contact with the walls of the jar.”

I don’t see how that’s possible, since the glass remains rather cool–at a much lower temp. than the air inside–due to losses outside. Touch the glass in your closed car from inside. Then touch the dashboard."

The air inside your car is in contact with the dashboard also, which is hotter than the glass, because it absorbs practically all the solar radiation falling on it, while the glass is mainly transparent to the short wave radiation. The air in the car is heated by contact with the dashboard and convection within the car.

The experimenter has yet to provide an elementary analysis of the results of his experiment compared to the size of the effect he set out to measure.
I have yet to see an idealized version of the experiment analysed to tell what the temperature rise in an ideal bottle would be with and without the water vapor. That would be the first thing to do, to determine whether it is possible that one could learn anything from the experiment even with good equipment.

Let us try calculate the results of a simple 1 dimensional experiment, with perfectly transparent, non absorbing glass separating the earth's atmosphere from the controlled sample of atmosphere, and look at the radiation balance alone, assuming that heat conduction and convection were not factors.

Compare the calculated temperature achieved inside the atmospheric sample, assuming it was a uniform slab, at 0%, 50% and 100% relative humidity, assuming the slab is 1 meter thick.

Use a fixed solar radiation flux and a fixed downwelling IR flux from the atmosphere, to achieve thermal equilibrium between the earth's surface temperature and the atmosphere.

Recognize that the entire earths atmosphere will absorb a fraction, 0.6 of the IR radiation incident on it, and assume this is at 50% humidity, for illustrative purposes.
http://www.aps.org/units/fps/newsletters/200807...
(see equation 19)

We need an estimate of how what fraction of the IR radiation passing through the air sample will be absorbed and re emitted. This will depend on the number of absorbing molecules in our sample slab.

We know the number of molecules relative to the atmosphere as a whole.
The weight of 1 square meter of the atmospheric column above us is about 7000Kg. The weight of 1 cubic meter of atmosphere is 1.225KG (at sea level)
Taking Ln(0.6)*1.225/7000 we find that about 1/10000 of the IR incident on the sample air slab will be absorbed by it if it contained the average mixture of greenhouse gases in the atmosphere 50% humidity.

If the earth's surface underneath the slab is receiving and absorbing a solar flux of equal to S, and down-welling radiation D, with GHG's in the sample set same as the atmospheric concentration of GHG's, the upward radiation, assuming equilibrium U will obey the equation
U=S+D,
and the earth's surface temperature will be give by U=s*T^4, where s is the SB constant, assuming that the surface is a perfect emitter/absorber

If we remove or add GHG's from the sample to see what the new equilibrium temperature of the surface will be, the effective downwelling radiation has been reduced because of the absence of GHG's in 1 Meter of atmosphere.
Let us see what happens to the surface temperature under such conditions by using equation 16 of our reference:

Ts=(n+1)^0.25 * Ta
If we add change the value of n from 0.6 to 0.6001 we can calculate the change in new value of Ts, which we call Ts'.
Ts'-Ts=Ts*((1.60001)/(1.6)-1)=0.017C.

If the temperature of the surface underneath the bottle has changed by more than this, it was due to other factors in the experiment besides the content of the air sample. The detectable limit was actually 10 times the magnitude of the temperature change the experimenter was able to measure.

eric:

"Recognize that the entire earths atmosphere will absorb a fraction, 0.6 of the IR radiation incident on it, and assume this is at 50% humidity, for illustrative purposes."

Are you saying that the "window" radiation is then 40 percent of the total IR?

Jae 74 said
“Eric Said,
"Recognize that the entire earths atmosphere will absorb a fraction, 0.6 of the IR radiation incident on it, and assume this is at 50% humidity, for illustrative purposes.”

Are you saying that the “window” radiation is then 40 percent of the total IR?"

I don't believe this is what is meant in the reference. The authors are dealing with the problem of absorption and re-emission during the upward and downward propagation of the IR, so the spectrum of IR that doesn't interact with the atmosphere, which is the IR window, doesn't come into this calculation. To me more accurate, it is the fraction of a layer of atmosphere required to absorb nearly 100% of the radiation that it can absorb.
The authors say,
"The thickness of a layer is such that almost all the incident radiation is just absorbed in that layer...."

Eric: O.K. I guess I misunderstood.

Eric #70: If you are talking about the change due to addition of water, I would say the change was detectable, downward, and the condensation on the walls of the jar indicate an evaporative cycle enhancing cooling is far greater than any warming from increased absorption.

Addition of another layer/jar is a different experiment.

David,

The experiment could be improved by starting with a mixture of air at its existing composition, plus water and do the experiment you did.

It needs an increase of CO2. How to do? Do they make those Sparklet bulbs anymore?

I have an industrial grade temperature logger with wired probe which could do the trick in terms of measuring the temperature. All one needs then is one of those old soda water pourers into which one poured some water and screwed the CO2 cartridge into its valve system to carbonate the enclosed water.

David,

looking at the graphed measurements, the internal temperature plateaued. This seems a limit of detection factor for Dick Smith marketed instruments - weather equipment is only produced to measure things within a certain measurement range, for cost factors.

The abrupt change to the slowly rising data is the clue. Data truncation is an instrumentation issue, not measurement. John Brignell should be consulted on this ASAP.

Hi Louis, I think I know how to build a more accurate experiment after this first trial. Concurrent experiments to control for environmental variation, laboratory temperature sensors, and mount the whole think on EPS foam.

<abbr>davidss last blog post..Maxpower Datasheet Experiment 2</abbr>

jae writes:

The radiation is an effect, not a cause.

1. No matter how many times you say this, it still won't be true.

2. The statement is 100% at odds with Miskolczi, who believes, along with every physicist in the world, that radiation can affect matter.

Relative to a “maximum greenhouse effect,” I’m still waiting for an answer as to why temperatures in very moist areas (tropics, e.g.) never exceed about 33 C. Shouldn’t the maximum greenhouse effect exist in such humid areas?

There's this thing called "atmospheric circulation" which redistributes heat among latitudes.

The maximum TEMPERATURE effects are found in the deserts,

Deserts are actually distinguished by lack of moisture, not high temperature. The Gobi desert, for example, can get quite cold.

admin says, of his small-scale experiment:

The purpose of a model is abstraction. Because a jar is not the atmosphere does not make it irrelevant.

The attenuation coefficient of carbon dioxide in the 15 micron band averages 1.48 per reciprocal meter atmosphere (Essenhigh 2001). Let's assume your apparatus is one meter across and that the concentration of carbon dioxide starts at 0.000387, the present concentration. The optical thickness would then be 0.000573. Double the carbon dioxide concentration and it would become 0.00115. The attenuation, by the definition of optical thickness, would change from 0.9994 to 0.9989; i.e., not enough to measure with your equipment. The experiment is meaningless. It certainly DOES matter that your apparatus is not the size of the atmosphere.

Optical thickness = k x p x L where k is absorption coefficient in reciprocal meter atmospheres, p is partial pressure in atmospheres, and L is distance in meters. You can't get around the fact that your experiment uses a very short value for L, which makes the results nearly or completely unreadable.

Steve Short writes:

I had thought the Earth’s true convective heat flux was of the order of 80 - 100 W/m2. Is this not true?

The figures from K&T97 are 24 W/m^2 for pure convection/conduction ("sensible heat") and 78 W/m^2 for evaporation ("latent heat"), for a total of 102 W/m^2. Planetary astronomers confuse this a bit by conflating both as the "convective flux:" fc = fsens + flat.

BPL #78

The maximum TEMPERATURE effects are found in the deserts,

Deserts are actually distinguished by lack of moisture, not high temperature. The Gobi desert, for example, can get quite cold.

I think your comprehension is a little strange I read jae's remark a little differently "maximum temperature effects" as meaning extremes of temperature are to be found there. I believe you are telling only half the story here.

There is no upper limit to the greenhouse effect (not in practical terms, anyway; you might get an upper limit of 2000 K or so from the shift of peak wavelength due to Wien's Law). The lower layers may become saturated, but the uppermost layer is never saturated, and each layer affects every other layer. Here's a mathematical explanation as to how this works (just remove the hyphen):

http://www.geoci-ties.com/bpl1960/Saturation.html

BPL #82

2 things

1 your link isn't working
2 your not making sense.

#82 Barton Paul Levenson
"There is no upper limit to the greenhouse effect ... The lower layers may become saturated, but the uppermost layer is never saturated, and each layer affects every other layer. "

Tropopaused, uppermost, water is greater than saturation, lower layers can have desert low humidity. How can we reverse the runaway accelerating cooling ? Is there a limit before Zero Kelvin ?

Jan Pompe writes:

1 your link isn’t working

My bad. You have to remove the hyphen and cut-and-paste it into the address window in your browser.

2 your not making sense.

"You're." In what way am I not making sense?

BPL #85

There is no upper limit to the greenhouse effect

This is a start. There is a limit to how hot the sun can make the planet. You won't 2000k without some concentration generally if the sun is over head 75c is tops for a simple passive system. in Sydney it's something like 60C. Of course if you can get sufficient concetration of light the theoretical limit to that is 5788. None of this is without limit.

If you think you can get it hotter in your solar furnace without concentrating the light I suggest your own experiment and report the results.

BPL, 78:

"The maximum TEMPERATURE effects are found in the deserts,

Deserts are actually distinguished by lack of moisture, not high temperature. The Gobi desert, for example, can get quite cold."

See my comment to JamesG in the Greenhouse quiz thread.

Jan, I estimate that if there were 23 billion Pascals of CO2 partial pressure on the Earth's surface, the surface temperature of the Earth would be about 2013 K. This represents about 227,000 atmospheres of CO2.

Extra credit: Would there be a phase change in the CO2? Why or why not?

My point being, of course, that there's no close upper limit to the greenhouse effect, as admin (and Miskolczi) seem to think. Venus has 89 atmospheres of CO2 and its surface temperature is 735 K. We're a long way below Venus, which means the Earth's surface temperature has plenty of room to climb.

BPL #88

Extra credit:

Not for you old son :- you didn't do the prac :- that's a fail.

# 88Barton Paul Levenson

"Jan, I estimate that if there were 23 billion Pascals of CO2 partial pressure on the Earth’s surface, the surface temperature of the Earth would be about 2013 K. This represents about 227,000 atmospheres of CO2."

I am into Virial bouncing sub orbital profiles. The Virial Temperature is the one with no convection. Do you have a Virial derived, no convection, steady state temperature, for the Earth's gravitation induced pressure ?

Franko #91

Looks like my response to that got caught in the spam filter but essentially I as where to CO2 is going to come from to get the same amount of CO2 Venus has in its atmosphere?

WE can't even get back to Carboniferous era levels because land creatures and plants are thought to have converted to fossil fuel but ocean creatures and plants have converted it to calcite and limestone. Simple arithmetic suggests it's impossible by simply burning fossil fuels.

Looking at lot of revealing, spying Venus links

BPL is wrong about the effect of CO2
CO2, (at 10e^6/385=2,597 times Earth concentration, 50 km up, inside the 40-60 km H2SO4 aerosols)) , allows -100C, at twice the solar radiation.

Nick is correct; that the 40-60 km clouds do dominate,
Effectively forming, similar to Earth's solid crust,
a layer severely hampering the convective circulation.

The worry of CO2 is a worry of too little. Easily sequistered by plants. The negative feedback is indicated by the yearly slope. We have to keep pumping up the leaky CO2 tire. Life and plants became CO2 happier when the Himalayas rose, exposing Carbonate rocks.

On Earth, what extra Solar input to get a very thick H2O cloud aerosol cover ?

Jan Pompe writes:

Looks like my response to that got caught in the spam filter but essentially I as where to CO2 is going to come from to get the same amount of CO2 Venus has in its atmosphere?

WE can’t even get back to Carboniferous era levels because land creatures and plants are thought to have converted to fossil fuel but ocean creatures and plants have converted it to calcite and limestone. Simple arithmetic suggests it’s impossible by simply burning fossil fuels.

Well, there are about 60 atmospheres of CO2 in carbonate rocks, and a billion or two years from now, when the sun is noticeably brighter, the Earth will undergo a runaway greenhouse effect and end up much like Venus. But my point wasn't that Earth is going to wind up like Venus any time soon. My point is that more CO2 will mean more warming and there isn't any close upper limit. We're more than capable of doubling the CO2 in the atmosphere in a few decades, and if we do the Earth's mean global annual surface temperature will rise by about 3 K.

Franko,

I'd reply to your critique of my CO2 calculations, but I can't figure out what the hell you're calculating. In general I find most of your posts opaque. Could you try to be a little less poetic and a little more clear?

BPL #94

Well, there are about 60 atmospheres of CO2 in carbonate rocks, and a billion or two years from now, when the sun is noticeably brighter, the Earth will undergo a runaway greenhouse effect and end up much like Venus.

Ahah so a warming sun will drive a growth in CO2?

I think I can go along with that.

# 94 Barton Paul Levenson
We need your Math skills to calculate NonRadiative IsoVirial atmosphere distributions. Lot of little circulars and ellipticals would give a clue of a reference.

Jan Pompe writes:

"Well, there are about 60 atmospheres of CO2 in carbonate rocks, and a billion or two years from now, when the sun is noticeably brighter, the Earth will undergo a runaway greenhouse effect and end up much like Venus."

Ahah so a warming sun will drive a growth in CO2?

I think I can go along with that.

A little knowledge is a dangerous thing. Drink deep, or do not drink, of the Pierian spring. -Alexander Pope

The present increase in CO2 is caused primarily by fossil fuel burning. This is confirmed by its radioisotope signature. The radioisotope signature of fossil fuel CO2 was first detected in ambient air by Hans Suess in 1955.

The sun hasn't increased noticeably in brightness in the last 50 years, so it can't have caused the sharp upturn in global warming of the last 30.

CO2 is really good as a cooler gas. On Venus cools the Tropopause to -100C. For Earth, -56C.

On Earth, high altitude clouds insulate, equivalent to plugging the reast of the IR window.

If we use SuperGreen gasses SF6 NF3, CH4 -- that would eliminate high altitude clouds -- and make our ankles warmer compared to our forehead ?

A little knowledge is a dangerous thing. Drink deep, or do not drink, of the Pierian spring. -Alexander Pope

The present increase in CO2 is caused primarily by fossil fuel burning. This is confirmed by its radioisotope signature. The radioisotope signature of fossil fuel CO2 was first detected in ambient air by Hans Suess in 1955.

Pope is a wise man. You should pay heed. All the radio-isotope signature tells us is that fossil fuel burning changes the mix.

# 95 Barton Paul Levenson
"Franko,
I’d reply to your critique of my CO2 calculations, but I can’t figure out what the hell you’re calculating"

CO2 on Earth is 385 parts per million (e6)
Invert to get CO2 ratio to Earth, (385e6)^-1 = 2,597 times Earth concentration, 50 km up, inside the 40-60 km H2SO4 aerosols)) , allows -100C, at twice the solar radiation.

I was trying to make the point that, radiated from below and above, CO2 is doing a disservice to the temperature.

I am really wondering if CO2 is really a net negative, on Earth, effect. -- H2O steals some of the CO2 effect below. -- But, at the Tropopause, radiates extremely well.

Jam Pompe writes:

Pope is a wise man. You should pay heed. All the radio-isotope signature tells us is that fossil fuel burning changes the mix.

And you think they can't calculate quantitatively what the mix is???

You don't even need the radioisotope figures, that only confirms it. We burn so much fuel every year. So much gets taken up by the oceans, the biosphere, and the "mystery sink." The rest stays in the atmosphere. There really is no question at all where the increase in CO2 is coming from. There is nobody competent in the field who thinks it's coming from anywhere else but human technology.

"There is nobody competent in the field "

Perhaps not.

What the biological pump is doing is anybodies guess.

"What the biological pump is doing is anybodies guess."

Plants sequister all they can. CO2 is the prey -- plants are the predators -- In the prey predator ratio, include your automobile.

The reason that there is no O2 runaway, is that the plants cannot get enough CO2

Your conclusions are a bit overreaching, or at the least, insufficiently specific.

1. In a sealed old flour container, an increase in temperature due to the blocking of convection has a maximum.

2. At this maximum in this environment, adding additional water vapor does not significantly increase the temperature.

I would also disagree with your statement that the initial rise in temperature "shows the classic greenhouse effect". This is not showing the effect, warming by borrowing outgoing LW IR, it is showing the actual behavior of a greenhouse, warming by blocking convection.

That said, the chart is interesting. There is a slight rise in temperature that goes along with the increase in humidity at 16:46, then the temperature falls a bit and the humidity declines more than it did at the time around 16:30 but less than at around 16:35. The temperature continues to decline slightly past 16:46, but the humidity starts going back up again at 16:52.

My conclusion would be that this shows that in this closed environment, water vapor moderates itself (so to speak) to regulate energy levels and keep a fairly constant temperature.

One interesting question would revolve around the level of total energy as work is done to evaporate the liquid water, as well as the heat contained within the water vapor. Temperature is not the only issue here. Latent and sensible heat matters are very important.

One of the main assumptions of AGW is that the temperature will continue to increase as the opacity of the atmosphere increases due to carbon dioxide gas (CO2). We can test the theory that increasing opacity increases temperature very easily.

I think you are off the mark here. How does an experiment increasing water vapor in a sealed container test anything about increasing carbon dioxide in an atmosphere? Carbon dioxide opacity is not water vapor opacity, even ignoring the other behaviors of water vapor that are not available to the other GHG, namely the ability to phase change to and from albedo affecting materials in various concentrations and being dependent in behavior according to temperature and altitude.

I contend this neither proves nor disproves anything about "global warming theory". There is something here, but that's not it.

Here's a basic question that you've probably already answered, but I'm looking for an answer in plain old "Poly Sci Major" terms. This topic obviously has large political ramifications.

The common (as I have found) explanation for the excessive warming effects of man-made greenhouse gas is that: 1) Radiation (visible and infrared) from the sun comes through the atmosphere during the day and warms the earth. 2) The earth emits infrared heat back through the atmosphere to space at night. 3) The recent increase in man-made greenhouse gasses reduces the atmosphere's ability to transmit the infrared heat back to space, thus trapping the heat on/in the earth.

Q: The sun emits radiation across the electromagnetic spectrum, to include a great deal of infrared heat. If greenhouse gasses are warming the earth by blocking the retransmission of infrared heat from the earth, wouldn't the greenhouse gasses in the atmosphere also block an equivalent amount of incoming infrared heat from the sun?

I am interested in any modeling that has been done on the warming properties of atmospheres with excessive CO2, CH4 and CFCs.

My hypotheses are:

1. The last 200+ years of warming are a continuation of an ongoing trend that is controlled by the sun. This trend has .2 degree perturbations atop an overall trend up since the late 1700's.

2. Man-made greenhouse gasses are affecting the Specific Heat AND the Radiative properties of the atmosphere, but only to a degree that is measurable in a lab experiment.

3. The earth's temperature will eventually make a short down-turn, much as it did from 1940 to 1980, and the wisdom of going-green from a pollution and sustainability standpoint will fall victim.

BC: some observations:

(1) The sun doesn't emit the long-wave IR that is absorbed by the CO2, so you got that part of the AGW theory wrong.
(2) The primary greenhouse gas is water (which does absorb some of the shorter-wavelength IR in sunlight, as well as in the longer wavelengths). The INCREASE in CO2 caused by humans represents about 1/1000 of the GHGs in the atmosphere. IMHO (which isn't worth much :)), the quantity of CO2 increase is far too small to make any difference, despite all the mental gymnastics that are going on.
(3) The "blanket effect" has not been measured by radiosondes or satellites, casting doubt on the current theory. The only thing that supports the current theory is Global Circulation Models, and they are not tracking temperatures over the last 10 years. Something's wrong with the physical assumptions in the models. (And anyone who puts their faith in these complex "parameterized" models should be an easy mark for any con man).
(4) Indeed, it is wise to "go green" and conserve resources, but only at a rate that is ECONOMICALLY SUSTAINABLE (people seem to keep forgetting that nothing good will happen if we are not, first, well-off economically. Compare the advanced nations to the third world in terms of pollution, forest degradation, etc., if you don't believe that). The free-market has pushed conservation and "greening," and that is how it should be done.

Thanks JAE,

How exactly do the global warmers relate increases in CO2, CH4 and CFCs to increased temps? All of the "low end" websites and texts claim that the aforementioned gasses are trapping the earth's IR emissions. Is this scientifically feasible? Shouldn't they conduct/convect and radiate equally well in both directions?

BC: You are asking the wrong guy. The hypothesis has never looked right to me, despite studying it a lot. I have some fundamental trouble with treatises that try to explain everything by radiation, including Miskolczi's hypothesis (I'm basicly a complete outcast, since I have a gut-feeling that the whole "greenhouse effect" boils down to storage of energy by the atmosphere and oceans--and that the radiation merely "reflects" this fact. But I'm trying to recover from my heresy here at Skeptics Anonymous. :) ). You might check RealClimate.org and search for their "learning from a simple model" post. I think that provides a good basic analysis of the hypothesis. Arthur Smith has a good paper on it, also; and perhaps he can provide the link again.

Jae, BC, sure, here it is: http://arxiv.org/abs/0802.4324

Jae, I have a question for you, an honest one, I'm interested in your perspective on it, based on your convictions regarding heat storage and warming. What do you think is the average temperature of the oceans (surface and deep ocean), and why do you think it is so high, low, or whatever it is, relative to the average surface temperature?

jae & BC

Here is the other

Arthur, as you know, I am a complete amateur in this area. But I keep trying to learn more, simply because I'm interested in the world around me. But, since you asked, what the hell, I'll speculate and give you some of my thoughts, FWIW:

Of course, I can't say anything about how "high" or "low" the water temperatures in the oceans are; all I can say is that they are "just right" to keep the planet roughly in balance all the time. Since sunlight penetrates water to a considerable depth and water absorbs all that energy, the oceans are vast reservoirs of heat (heat storage), and I think they dominate the planet's temperature. For example a Westerly wind on the Left Coast near where I live makes the shoreline temperatures about 50-55 F always, winter and summer. The oceans have had millions of years to develop certain properties (heat content, current flows, etc.). We see clearly how the oceans can affect the temperatures of the whole world, when we have an El Nino or La Nina. And now it looks like we will have cooling due to the negative phase of ENSO. And then there's the chaotic behaviors, which make it probably impossible to predict climate with much precision.

The heat stored in the oceans is constantly released as sensible and latent heat and that heat is critical to our survival. Fortunately, most of that heat is stored in the upper part of the ocean, or we would be frozen toast. Multi-decadal warming or cooling trends may well be related to how much of that heat goes to deeper water.

I think Lindzen and Douglass have it right in that the oceans keep the planet's temperature in balance, through cloud formation and negative feedback (and this probably also occurs over "soggy" land areas, as well). The water on the planet serves as a great big buffer system. This is especially evident in the tropics, where the temperature never climbs much above 30 C.

Jae (#112) - but of course average ocean temperature is measurable, here's one claim on that:
http://www.windows.ucar.edu/tour/link=/earth/Wa...

in particular

The average temperature of the ocean surface waters is about 17 degrees Celsius

but

the majority of our ocean water has a temperature between 0-3 degrees Celsius

because

90 % of the total volume of ocean is found below the thermocline in the deep ocean

.

If the ocean is such a store of heat, why is it so cold? Isn't the fundamental form of the second law of thermodynamics the proposition that a cold body cannot warm (transfer net heat to) a hotter body? So how could that cold, cold, deep ocean be warming the surface? Or, if it's just the surface waters that are warming us up, that's not really very much ocean, is it? Isn't that huge deep ocean acting in the other direction to cool us down?

And how did that deep ocean get so cold in the first place? Why hasn't it warmed up in the 8000 years since the end of the last ice age?

Arthur, 113. Gee, you ask hard questions. :)

I guess the overall answer is water is a marvelous substance. It's density decreases both above and below 4 C, unlike any other material. Therefore both warm water and ice "float."

"The average temperature of the ocean surface waters is about 17 degrees Celsius"

Kinda interesting that that is the average temperature of the planet, eh?

"the majority of our ocean water has a temperature between 0-3 degrees Celsius "

Remember that it would be -18 C if it were not for the "greenhouse" effect. So it is being kept warm.

"If the ocean is such a store of heat, why is it so cold?"

It isn't ALL so cold, as you've noted.

"So how could that cold, cold, deep ocean be warming the surface?"

It obviously isn't. The Sun is warming the surface.

"Or, if it’s just the surface waters that are warming us up, that’s not really very much ocean, is it?"

Must be enough to do the trick!

"Isn’t that huge deep ocean acting in the other direction to cool us down?"

Certainly.

"And how did that deep ocean get so cold in the first place? Why hasn’t it warmed up in the 8000 years since the end of the last ice age?"

Reaffirms your belief in God, eh?

Post for "If the ocean is such a store of heat, why is it so cold?" created here http://www.landshape.org/phpBB3/viewforum.php?f=5

Here's how the greenhouse effect works.

1. Of the sunlight striking the Earth, about 31% is reflected away, 19% or so is absorbed in the atmosphere, and 50% hits the Earth's surface.

2. The Earth's surface radiates infrared light.

3. Greenhouse gases in the atmosphere absorb the infrared light.

4. The greenhouse gases heat up.

5. The hot greenhouse gases radiate infrared.

6. Some of the infrared radiated by the greenhouse gases goes back down to the Earth's surface and heats it above what it would be from sunlight alone. You've got both sunshine and "atmosphere shine" heating the ground.

Some quantitative details:

The flux density (power per unit area) absorbed by the Earth's climate system is:

F = (S / 4) (1 - A)

Here F is in watts per square meter. S is the "Solar constant," the flux density from the Sun at Earth's orbital distance. A is the Earth's "bolometric Bond albedo," the fraction of sunlight reflected away at all wavelengths.

For Earth, S = about 1,366 watts per square meter (Lean 2000) and A = 0.306 (NASA 2007). This gives F = 237 watts per square meter.

Inverting the Stefan-Boltzmann law, the Earth's radiative equilibrium temperature is then:

Te = (F / σ)^0.25

where σ is the Stefan-Boltzmann constant (5.6704 x 10^-8 W/m^2/K^4 in the SI). This predicts a temperature of 254 K for the Earth.

But water freezes at 273 K. If sunlight were the whole story, Earth would be frozen over. It's the greenhouse effect that raises Earth's mean global annual surface temperature to about 288 K.

BPL: Once again. I think your explanation (as with that of most of the "community") completely ignores the kinetic and potential energy contained (stored) in the atmosphere. This energy has to be manifested as a rise in temperature above Te at the surface, via the ideal gas law.

6. Some of the infrared radiated by the greenhouse gases goes back down to the Earth’s surface and heats it above what it would be from sunlight alone. You’ve got both sunshine and “atmosphere shine” heating the ground.

impossible.

You effectively have 4 bodies in series

[hot sun] ---> [cool surface] ----> [cooler atmosphere] ---> [cold space]

or 3 when atmosphere is in equilibrium

[hot sun] -----> [cool surface + atmosphere] ----> [cold space]

Do try a little harder not to be fooled by the fact that the sun and space are in the same general direction from the surface. This way you can be quite clear about the direction (sign of) of the fluxes.

No, Jan, it's not impossible at all. That's the standard explanation of how the greenhouse effect works. Greenhouse gases radiate, and some of the radiation goes down. K&T97 estimate the surface gets 168 W/m^2 from sunlight and 324 W/m^2 from atmospheric back-radiation.

Back-radiation from the sky can be measured (and is) with an instrument called a pyrradiometer, and is calibrated by something called the Idso-Jackson formula. Measurements generally confirm what I say above.

In fact, the back-radiation from the atmosphere is the "E-sub-D" value in Miskolczi's model. Are you going on record as saying that both the consensus of climatologists and FM are wrong about how the greenhouse effect operates?

Oh, I shouldn't neglected the pyrgeometer, which also measures infrared back-radiation. They have a nice wiki page about it:

http://en.wikipedia.org/wiki/Pyrgeometer

jae, temperature storage has nothing to do with the overall balance. Over geological time, or even historical time, all the energy coming in goes out again. That's why you can get a very close first approximation of the surface temperature just considering energy balance.

The GreenHouse effect simplified;
Block off half of your car radiator
The other half has to radiate all the heat.
Hence the higher temperature.

BPL:

"No, Jan, it’s not impossible at all. "

Maybe the standard hypothesis is not impossible, but the way you stated it seems impossible. The zero-degree air does not HEAT the warmer surface. It seems to me that that violates the second law of thermodynamics. I think your "theory" merely says that the 324 watts.m-2 radiation (consistent with a BB at about 0 C) keeps the surface from cooling as fast, that's all.

But I also think the standard hypothesis is flawed, because it completely ignores gravity, potential energy, and kinetic energy. I am simply shocked that we have real physicists calculating Te as some 255 K, ignoring the atmosphere and water, as if the planet were just like the moon. The presence of an atmosphere and an enormous heat sink (water) are what cause the surface to be about 33 C higher than it would be otherwise. The role of the GHGs is to simply help thermalize the rest of the atmosphere. IMHO.

BPL

No, Jan, it’s not impossible at all. That’s the standard explanation of how the greenhouse effect works.

My mistake make that imbloodypossible. Is that better.

Greenhouse as jae says

C) keeps the surface from cooling as fast, that’s all.

lets just take the surface to air flux

Warm Surface ---> cool atmosphere.

The heat transfer equation $$T_h$$ is warm surface temperature and $$T_c$$ the surface air temperature will give us a heat transfer equation assuming BB all round:

$$Q_{lost} = \sigma \left( T_h^4 - T_c^4 \right)$$

It is the sort of equation engineers like the ones who build pyrgeometers and other non contact heat measuring devices have succsefully have been successfully using for years. Some physicist by the name of Dr Ferenc Miskolczi (whoever he might be) did some work on calibrating the power spectrum of the pyrgeometers in the early 90s does anyone here think he might not know how to actually use one? BPL obviously does.

Just to show I did not pull the equation out of the air here is a link.

Now I have asked both Nick and Arthur what sign the net flux absorbed by the warm body will be neither has been able to so far perhaps Barton can help.

Of course all the IR from the surface is not absorbed some goes through the IR window but even for an opaque atmosphere the best BPL can hope for is that the atmosphere is warmed by the surface until that atmosphere temperature is equal to the surface temperature. Now it is completely normal that when opposition to flow whether fluid, charge or heat energy is obstructed for the potentials i.e. pressure, voltage or temperature respectively to rise but this rise in potential is always note always in passive systems accompanied by a decrease in flow, current or flux respectively but in terms of energy dissipated (entropy maximised) the rate can be optimised as is seen in maximumpower theorem for passive electric circuits or equation 20 in Miskolczi 2007 for radiative transport.

jae writes:

The zero-degree air does not HEAT the warmer surface.

Yes it does, unless the zero you mean is absolute zero.

It seems to me that that violates the second law of thermodynamics.

No. You're thinking no heat can go from a colder source to a warmer one. What the second law says is that in equilibrium, not net heat can go from a colder source to a warmer one.

But I also think the standard hypothesis is flawed, because it completely ignores gravity, potential energy, and kinetic energy. I am simply shocked that we have real physicists calculating Te as some 255 K, ignoring the atmosphere and water, as if the planet were just like the moon.

That's how you calculate Te. That's how Te is defined.

The presence of an atmosphere and an enormous heat sink (water) are what cause the surface to be about 33 C higher than it would be otherwise. The role of the GHGs is to simply help thermalize the rest of the atmosphere. IMHO.

YHO is wrong. The presence of the atmosphere and oceans as a heat sink have nothing to do with it. They simply would not retain the energy over long periods of time unless something else was heating them. You're right that the presence of the atmosphere warms the surface, but it warms it because it contains greenhouse gases. If the atmosphere were pure nitrogen and oxygen, the surface would be at 255 K (assuming albedo didn't change; it reality it would, of course, and temperature would be even colder).

Jan writes:

My mistake make that imbloodypossible. Is that better.

No matter how vehemently you say it, it's still wrong.

The heat transfer equation is warm surface temperature and the surface air temperature will give us a heat transfer equation assuming BB all round:

Jan, you're confusing heat transfer with net heat transfer. Nobody doubts that the net movement of heat is from the surface to the atmosphere. But that does not mean that no heat goes the other way. The surface is as warm as it is partially because it's receiving a lot of energy from the atmosphere.

Again, FM shows this in his model as E-sub-D. You're saying E-sub-D has to be zero. It doesn't, and it isn't.

BPL #126

Jan, you’re confusing heat transfer with net heat transfer

I'm confusing nothing net heat transfer is the only one worth considering a reminder of a most basic form of the second law of thermodynamics is that heat only flows from hot area to cold ones unless work is done it.
The atmosphere cannot heat the surface to a higher temperature that it is itself.

So can you identify the heat pump in your scenario?

BPL #127

You're saying E-sub-D has to be zero.

I'm not saying that at all. Where on earth do you get such a silly idea?

This equation assuming black body absorber and emitter specifically does NOT say that:
$$Q_{lost} = \sigma \left( T_h^4 - T_c^4 \right)$$

$$\sigma T_c^4$$ is the back radiation is it not?

Now what is the sign of $$Q-{lost}$$ which is the radiation heat lost from the surface?

Let me try a simple demonstration of what I'm talking about. This model is called the slab atmosphere or glass slab model.

It consists of space, the atmosphere, and the ground.

Space contributes sunlight, which moves down. The atmosphere is transparent to sunlight, but completely absorbs thermal infrared light. The ground absorbs both.

Balance for space: Output is F downwards, input is Fa from the atmosphere.

Balance for the atmosphere: Input is Fs from the ground, output is Fa up and Fa down.

Balance for the ground: Input is F from space and Fa from the atmosphere.

Here's a diagram:


[SPACE__ F_|____^____________]
___________v____|____________
[ATMOSPHERE__Fa_|_Fa_|_____^_]
___________|_________|_____|_
[GROUND____v_________v__Fs_|_]


Sorry for the underscores, but I couldn't get spaces to line up even by manipulating fonts and the nbsp character.

Let's make it quantitative. Give F the same value as the flux density absorbed by the Earth's climate system, 237 W/m^2.

The equations for the space, atmosphere and ground flux density balances are:

F = Fa {1}
Fs = 2 Fa {2}
F + Fa = Fs {3}

Equations {1} and {2} lead to:

2 F = Fs {4}

So we know Fs is 474 W/m^2.

For the atmosphere, equation {1} implies that Fa is 237 W/m^2.

So we have F = 237, Fa = 237, Fs = 474.

Using the Stefan Boltzmann law, these lead to temperatures for the atmosphere and surface of 254 K and 302 K.

Notes:

1. The surface is warmer than the atmosphere.
2. Nonetheless the surface is receiving energy, and thus heating, from the atmosphere.

Now, let's say the atmosphere does not absorb infrared energy any more than it absorbs sunlight (i.e., it has no greenhouse gases in it). In that case the input to the atmosphere is zero, its output is also zero, and Fs = F = 237 W/m^2 and the surface temperature is 254 K.

Thus the presence of greenhouse gases warms the surface through back-radiation. Q.E.D.

Note that I used only conservation of energy and the Stefan-Boltzmann law. Nothing fancy, nothing about molecular absorption or heat storage. Just the simplest possible physics.

Jan writes:

"You’re saying E-sub-D has to be zero."

I’m not saying that at all. Where on earth do you get such a silly idea?

From the fact that you said back-radiation from the atmosphere heating the surface was "imbloodypossible." Remember?

Look, let's put this in terms even a moron can understand.

1. The atmosphere is warm.
2. The atmosphere gives off infrared light in all directions.
3. Half of those directions are "down."
4. The ground absorbs infrared light.
5. If an object absorbs infrared light, it heats up.

Do you see a mistake there?

The Earth gets both sunshine and "atmosphere shine" and BOTH heat it up, because BOTH are absorbed by it. That's how the greenhouse effect works.

BPL #130

From the fact that you said back-radiation from the atmosphere heating the surface was “imbloodypossible.” Remember?

Of course I remember I wrote it, it is still there and what I wrote it about is also still there and just so that you will be quite clear what is impossible:

6. Some of the infrared radiated by the greenhouse gases goes back down to the Earth’s surface and heats it above what it would be from sunlight alone. You’ve got both sunshine and “atmosphere shine” heating the ground.

It's still impossible what is cooler cannot heat what the sun has already made warmer. Heat flows one way.

Now about that heat pump that you were going to identify :- where is it?

1. The atmosphere is warm.
2. The atmosphere gives off infrared light in all directions.
3. Half of those directions are “down.”
4. The ground absorbs infrared light.
5. If an object absorbs infrared light, it heats up.

Do you see a mistake there?

No I don't because your error is one of omission you leave out that the ground emits more than it absorbs from the atmosphere so the atmosphere is not heating it up. Do try to get your algebra right.

So, BPL, according to your view, my hands are now HEATING my cup of coffee. Which means, by definition, that the coffee should be getting warmer. But I grarantee it is not.

I think the reason so many folks get it in their head that the back-radiation is "heating" the surface is because of the way the radiation cartoons, such as the Kiehl-Trenberth one, are drawn. They indicate that the only way the surface can emit at 15 C (about 390 Watts.m-2) is for the "backradiation" to add to the solar insolation (324 + 66) to yield the 390. This is crazy, misleading, and I think it violates the second law. 324 w.m-2 is consistent with zero C, and zero C air is not HEATING the surface. This is one primary reason Miskolczi makes much more sense in my mind. We KNOW FROM THERMOMETER MEASUREMENTS that the average AIR temperature of Earth at about 2 m elevation is close to 15 C. THAT is the source of the 390 w.m-2, not back-radiation! And the AIR at 2 m is in equilibrium with the surface (Su = Ed; backradiation = surface radiation). And the reason that the average temperature at 2 m elevation is 15 C is because the atmosphere and the water are storing solar energy, the amount of which is explained by the Ideal Gas Law. That energy = the potential and kinetic energy that keeps the atmosphere from falling on the surface.

Hi Jae - "324 w.m-2 is consistent with zero C" - only for a black body, and air is not black, remember? The windows of transparency of air to thermal radiation guarantee the emission will be somewhat less than the black-body number.

Arthur: Agreed on that.

jae #134

I think where the error lies is in the notion that the world would only only be 255K without greenhouse gases. This starts with the simplification that Richard Lindzen describes

here

It is commonly claimed that the natural component of this blanket keeps the earth about 33 degrees Centigrade warmer than it would be in the absence of this blanket. The claim is a little inappropriate insofar as it requires getting rid of the greenhouse impact of clouds while retaining them to reflect sunlight. Getting rid of clouds as reflectors would reduce this difference substantially. This, however, is a relatively minor point.

The real "equilibrium" temperature if the clouds were treated equally for incoming and out going, irrespective of the different microphysical processes involved, is ~279 K this number if you multiply it by the (1-albedo)^.25, that is treating the clouds equally for outgoing as incoming, gives the ~255 apparent temperature as obtained by measuring the OLR.

This is just the beginning of compounding errors caused by approximations and simplifications Richard Lindzen lists a few more and Ferenc Miskolczi some more that Richard Lindzen doesn't mention. I'm sure if we dig deep enough even more can be found.

Jan writes:

"You’ve got both sunshine and “atmosphere shine” heating the ground."

It’s still impossible what is cooler cannot heat what the sun has already made warmer. Heat flows one way.

No, Jan, that is wrong. NET heat flows one way.

You've agreed that infrared light from the atmosphere strikes the ground.
Presumably you agree that it is absorbed by the ground.
Where does the energy go? Infrared light carries energy, you know. Back-radiation from the atmosphere constitutes 324 Watts per square meter according to K&T97. What happens to the ground when it receives that much energy?

Where does the energy go, Jan?

jae writes:

So, BPL, according to your view, my hands are now HEATING my cup of coffee. Which means, by definition, that the coffee should be getting warmer. But I grarantee it is not.

Yes, jae, your hands are heating your coffee. The coffee is cooling overall because it's losing more heat than it's getting.

jae posts:

I think the reason so many folks get it in their head that the back-radiation is “heating” the surface is because of the way the radiation cartoons, such as the Kiehl-Trenberth one, are drawn. They indicate that the only way the surface can emit at 15 C (about 390 Watts.m-2) is for the “backradiation” to add to the solar insolation (324 + 66) to yield the 390. This is crazy, misleading, and I think it violates the second law.

You're wrong on all three counts. I'll ask you the same thing I asked Jan: if the ground is absorbing 324 watts per square meter of infrared light from the atmosphere, what happens to that light when the ground absorbs it? Where does the energy go?

324 w.m-2 is consistent with zero C, and zero C air is not HEATING the surface.

By the Stefan-Boltzmann law, 324 W/m^2 corresponds to 275 K, which is 2 C, but that's a quibble. Zero C air WOULD be HEATING the surface. Air at any temperature other than absolute zero would.

As for the second law reference, which you're still making after being corrected about it, I'll repeat the refutation -- the second law says that NET heat cannot flow from a colder to a warmer object. It does not say that NO heat can.

Scientists get the three laws of thermodynamics in freshman physics. Do you really think that, all over the world, they've missed a simple violation of it?

Jan writes:

I think where the error lies is in the notion that the world would only only be 255K without greenhouse gases. This starts with the simplification that Richard Lindzen describes

254 K is what it would be if the albedo stayed the same, Jan. Lindzen correctly points out that if the albedo changed, so would the temperature. This has nothing to do with the fact that the temperature is higher with an atmosphere containing greenhouse gases.

Barton Paul Levenson #139

Yes, jae, your hands are heating your coffee. The coffee is cooling overall because it’s losing more heat than it’s getting.

That is either most unphysical or you a redefining portions of the english language.

BPL #141

254 K is what it would be if the albedo stayed the same, Jan. Lindzen correctly points out that if the albedo changed, so would the temperature. This has nothing to do with the fact that the temperature is higher with an atmosphere containing greenhouse gases.

I think I see your problem :- you speak a different language to the rest of us. Lindzen says:

The claim is a little inappropriate insofar as it requires
getting rid of the greenhouse impact of clouds while retaining them to reflect sunlight.

Does not in the least resemble what you are saying he says.

BTW BPL

Merry Christmas

I'll think of you when playing Santa Claus in the hospital tomorrow.

BPL: I am certain in my mind that K&T 97 is completely wrong or is being misunderstood by you. We KNOW, from actual thermometer measurements all over the planet, that the average temperature of the AIR (2 meters up) is about 15 C. Therefore, the average radiation in the air at 2 m has to be about 390 w/m2. It cannot be 324 W/m2 as K&T seem to have it. It's 390 (plus the window radiation) up and 390 down. Su = Ed + St, just as Miskolczi says.

You are completely wrong in believing that zero-degree air is in any way HEATING 15 C air. Unless, you have a different definition of HEATING, as Jan suggests may be true. The 324 watts of radiation can keep the surface from cooling below 0 C, and it can slow the loss of heat but it cannot heat it.

BPL: Maybe we can picture it this way. The 324 watts in the K&T figure "heat" the surface to 0 C, and the other 66 watts from the Sun "heat" it on up to the 15 C ?

But I still say the diagram is nuts. The 390 watts is up and down; it is just a measure of the air temperature of 15 C. Period. No magic addition of radiation from the air and from the surface is needed. The 390 watts is just a characteristic of 15 C worth of heat storage, that's all.

jae writes:

You are completely wrong in believing that zero-degree air is in any way HEATING 15 C air. Unless, you have a different definition of HEATING, as Jan suggests may be true. The 324 watts of radiation can keep the surface from cooling below 0 C, and it can slow the loss of heat but it cannot heat it.

The surface receives 324 W/m^2 of infrared light from the atmosphere. It absorbs that light. What happens to the ground when it absorbs light? What happens to any material body when it absorbs light? Where does the energy go?

Barton Paul Levenson #147

The surface receives 324 W/m^2 of infrared light from the atmosphere. It absorbs that light. What happens to the ground when it absorbs light? What happens to any material body when it absorbs light? Where does the energy go?

The energy gets reradiated at a faster rate because the surface is already hotter further more hat atmosphere cannot back radiate more than the surface at the same temperature because the atmosphere doesn't absorb what goes through the IR window.

Net back radiation unless the atmosphere is very much hotter than the surface will always be less than the upward radiation therefore it cannot heat the surface. The net radiative surface/air effect will always be towards cooling.

BPL:

"What happens to the ground when it absorbs light? "

The surface would get to zero C, with that amount of "light." You need to add another 66 watts to get to the 390 watts that is supposedly equivalent to 15 C. Kiehl and Trenberth 97 say that that other 66 W comes from the "ground." The 66 watts is what's left over from the 168 watts of sunlight that strikes the surface, after you subtract the 102 watts for convection and latent heat. It all adds up very nicely, and seems to make perfect sense, but I think it is the wrong picture of what actually happens, as I said earlier.

Per Wien's displacement;
Your microwave oven at 2.4 GHz = 0.015K

From such a low temperature, how can a microwave oven heat your leftover pizza slice ?

Franko #150

calculate the effective temperature at 600 watt monochromatic radiation. Wien law does not apply to monochromatic radiation.

Ok, I got it figured,
The pizza slice would get so hot, emissivity derated, to radiate 600 watts back into the microwave oven

Franko

The pizza slice would get so hot, emissivity derated, to radiate 600 watts back into the microwave oven

I doubt it will water will unfreeze and cheese will melt - all irreversible entropy increases.

Jan, completely missing the point, writes:

Net back radiation unless the atmosphere is very much hotter than the surface will always be less than the upward radiation therefore it cannot heat the surface. The net radiative surface/air effect will always be towards cooling.

Yes, Jan, that is true for net radiation. It is not true for radiation. You don't appear to understand the difference. Your "therefore" in the first sentence above is a non sequitur. There is no "net back radiation" in the first place, there is just back radiation, and it does heat the surface.

You seem to think that all the heating of the surface comes from the sun. Sorry, but that can't be true. As I have demonstrated numerous times on this blog, sunlight alone will only get the surface up to 254 K. To get it up to the actual 288 K, you need the back-radiation from the atmosphere as well.

Here's the energy budget for the Earth's surface, going by K&T97:

INPUT
Solar energy: 168 Watts per square meter
IR from atmosphere: 324 W/m^2
Total: 492 Watts per square meter.

OUTPUT
Radiated to the atmosphere: 350 W/m^2
Radiated to space: 40 W/m^2
Sensible heat loss: 24 W/m^2
Latent heat loss: 78 W/m^2
Total: 492 W/m^2

All the input goes to heating the surface.
All the output goes to cooling the surface.
(Over the long run) input and output are equal, therefore
(over the long run) the surface temperature stays at the same level.

OK, BPL, you could say that the 324 watts is heating a 255 K planet up to 0 C, and that the other 66 watts brings it up to 15 C, as per K&T. The problem with all that is that you have to ASSUME the 255 K number (you cannot measure it). I maintain that this number is completely bogus, because it fails to account for the presence of water or an atmophere, both of which are quite transparent to SW. If you go way out in space and "take a radiation picture" of Earth, it would be 255 K, because that is the temperature of the radiation out and equals the Solar radiation in. But that is not a measure of the surface temperature (and radiation), which is purely a function of the kinetic and potential energy of the atmosphere (which is, in turn, affected by the heat capacity of the oceans). The GHGs serve to rapidly thermalize the bulk of the atmosphere. The radiation is the effect of the heat storage, not the cause.

And I still say that the K&T cartoon is nuts, because it says the air averages zero C (aka 324 w), when we KNOW it averages about 15 C. The way the cartoon is drawn shows the surface discontinuity that M shows is wrong.

BPL #154

Yes, Jan, that is true for net radiation. It is not true for radiation. You don’t appear to understand the difference. Your “therefore” in the first sentence above is a non sequitur. There is no “net back radiation” in the first place, there is just back radiation, and it does heat the surface.

No point is missed net back radiation is negative i.e. more is going forward than back hence the cooling.

Jan writes:

No point is missed net back radiation is negative i.e. more is going forward than back hence the cooling.

Can you possibly unscrew your head from "net back radiation?" We're not discussing net radiation. Or rather, you are, but I'm not. I'm simply talking about the fact that the infrared the ground receives from the atmosphere contributes to the temperature of the ground. This is a point no scientist, anywhere, disputes. You've simply glommed onto jae's crackpot view of the subject out of a knee-jerk desire to support him and oppose me. Plus, apparently, a completely inability to admit you're wrong about anything.

Back-radiation from the atmosphere helps heat the ground. Deal with it.

Insulate or heat ?

On a local scale; warm Chinook Wind, lightning, rain on a snow covered field -- all warm the surface up.

Temperature in a perpetually shady location can be warm -- courtesy of atmospheric storage and transport

Does the sleeping blanket warm you up ?

BPL is showing his fangs again:

"This is a point no scientist, anywhere, disputes. You’ve simply glommed onto jae’s crackpot view of the subject out of a knee-jerk desire to support him and oppose me. Plus, apparently, a completely inability to admit you’re wrong about anything.

Back-radiation from the atmosphere helps heat the ground. Deal with it."

It's true that no scientist disputes that the radiation can warm, but it must be in an amount consistent with the amount of radiation. But I am a scientist, and I dispute the K&T scemes. The K&T-type cartoons are a CROCK, because they defy the second law of thermodynamics. They insist that an AVERAGE radiation of 235 w.m-1 from sunlight can somehow WARM the surface above -18 C. That cannot happen, no matter how you "bounce" radiation back and forth between the surface and the air. The MAXIMUM HEAT produced by 235 w.m-1 is -18 C. Period. See my little essay on the Forum for more details.

BPL #157

Can you possibly unscrew your head from “net back radiation?” We’re not discussing net radiation. Or rather, you are, but I’m not.

Sorry BPL this is the second time that you've shown yourself to be algebraically challenged.

Anyone else see a developing pattern here?

"Anyone else see a developing pattern here?"

When a moth flies near the candle flame
The temperature of the flame increases, just a little
The temperature of the moth -- Kirchoff enclosed

Franko

Please do not perpetuate fairy tales in a science forum.

The blanket on your bed does not keep you warmer until you have yourself warmed the blanket.

Jan:

"The blanket on your bed does not keep you warmer until you have yourself warmed the blanket."

Not so, according to AGW theory. The blanket "back-radiates" all the energy you spent in heating it, thereby cooking your butt.

Speaking of backradiation, if the sun ever shines here again, I'm gonna take some mirrors outdoors and try to cook myself with reflected backradiation.

Space blanket not only stops convection, but also reflects back your body's radiation. Reduces your cooling, commonly referred to as warming.

The greenhouse effect actually reduces the heat energy into the atmospheric system. By increasing the surface temperature; surface radiates more directly to space.

All radiation, no matter how hot, does have back radiation, even if it is cold as the Cosmic microwave background.

Franko

All radiation, no matter how hot, does have back radiation, even if it is cold as the Cosmic microwave background.

This is true but the second law must still apply and that is that heat flows from hot to cold and that is why we must speak of net flow.

that net is
$$Q = \sigma \left( \epsilon_h t_h^4 - \epsilon _c t_c^4 \right)$$

since atmospheric transmissivity $$T_A is $$1-\alpha$$ where α is absorptivity ($$\epsilon _c$$) then we can rewrite the equation for surface cooling

$$Q = \sigma \left( \epsilon_h t_h^4 - \left(1-T_A) t_c^4 \right)$$
and $$1 \geq T_A \geq 0$$ so net cooling of the surface is always $$Q \geq 0$$ and there can be no "heating" of the surface by the atmosphere.

The exception being on the odd occasion at the odd location where the atmosphere is actually significantly warmer than the surface.

The space blanket may find an analogy with clouds scattering IR but not with clear sky.

jae

Speaking of backradiation, if the sun ever shines here again, I’m gonna take some mirrors outdoors and try to cook myself with reflected backradiation.

According to AGW theories it should work equally well indoors;)

Backscatter concept is another way of looking at resistance to energy flow. Kirchoff is amazing, different absorptions and emissions, -- an extreme example of equipartition.

Trying planck's equation to find out temperature of pizza slice to emit 600W between 2.4 and 2.5 GHz

Franko

I don't think backscatter and Kirchhoff's have much common ground. The former is reflection and the latter absorptivity and emissivity.

Let us know how you go with the pizza temperature.

Jan writes:

But I am a scientist,

I find that very hard to believe.

and I dispute the K&T scemes. The K&T-type cartoons are a CROCK, because they defy the second law of thermodynamics.

No, they do not. And they are not unique to K&T; every such diagram has been similar since Aldrich published the first one in 1917. Miskolczi has one in his article, too.

They insist that an AVERAGE radiation of 235 w.m-1 from sunlight can somehow WARM the surface above -18 C.

No, they do not. They suggest that 168 W/m^-2 from sunlight PLUS BACK-RADIATION FROM THE ATMOSPHERE can warm it above -18 C.

That cannot happen, no matter how you “bounce” radiation back and forth between the surface and the air. The MAXIMUM HEAT produced by 235 w.m-1 is -18 C. Period. See my little essay on the Forum for more details.

You're ignoring thermal IR.

Jan, I'll ask this again. I've asked it several times without yet getting a straight answer from you, or any answer.

IR photons to the tune of 324 W/m^2 are absorbed by the ground from the atmosphere. What happens to the energy once it enters the ground? Where does it go?

Jan writes:

Franko

Please do not perpetuate fairy tales in a science forum.

The blanket on your bed does not keep you warmer until you have yourself warmed the blanket.

Once again, Jan, you have net radiation confused with radiation. You're also missing the fact that blankets work primarily by suppressing convection, not by radiative means. But even a blanket colder than you would keep you warmer than no blanket. You are just flat-out, dumbass wrong on this point no matter how strongly you believe it or how many times you repeat it. Heat can flow from colder objects to warmer ones, and does, all the time.

Jan writes:

Sorry BPL this is the second time that you’ve shown yourself to be algebraically challenged.

Oh my gosh, Jan, please don't tell my old professors that I'm algebraically challenged! They might take back my degree in physics!

BPL: You got me confused with Jan above.

The Keihl-Trenberth-type cartoon is truly a crock. There is no possible way to turn 235 watts of power into 390 watts of power, without ADDING energy somewhere. Back-radiation is not ADDITIONAL energy, and that is the Achilles Heel of these cartoons. There is only so much power being put into the system by the Sun in these diagrams--235 watts; therefore, there cannot be more power in the system than 235 watts. You cannot keep adding the same power together to produce more power! It's so plain! It is truly staggering and amazing how many otherwise bright people have been bamboozled by this perpetuum mobile of the second kind. It is clearly dead wrong, no matter how famous and often-quoted it is.

I guess every generation has to have its favorite perpetuum mobile. It was the 200 mpg carburetor for mine. It appears to be magic radiation cartoons for the save-the-earth generation.


And BPL, I am a scientist, a PhD polymer chemist. However, I'm getting old and rusty, so I'm fair game.

BPL:

"IR photons to the tune of 324 W/m^2 are absorbed by the ground from the atmosphere. What happens to the energy once it enters the ground? Where does it go"

Well, first of all, the 324 watt figure is a figment of faulty reasoning. But IF there were actually 324 watts, they would be absorbed by the surface, tending to warm it to about 2 C. K&T actually only have 168 watts hitting the surface directly and 235 total watts to play with. So the maximum temperature they can get under their "average conditions" is -18 C.

As I tried to explain on the Forum, it is not physical to analyze the earth's heat budget by using such "averages."

BTW, Miskolczi does NOT use numerical average radiation budgets and is therefore a different can of worms.

BPL #171

you have net radiation confused with radiation.

I have nothing confused I know the difference and I don't ignore the fact that the net radiative heat loss from the surface is positive and therefore cannot lead to or be considered a gain in heat which should not be confused with either energy or temperature for the the surface even if the temperature goes up because of it.

It's not a question of blind faith that cool things do not heat things that are warmer it's a matter of training (a BSc in physics) and many years experience in process control engineering. Not to mention the experience of getting colder initially of getting into a cold sleeping bag while camping in sub zero temperatures.

please don’t tell my old professors that I’m algebraically challenged! They might take back my degree in physics!

Whether you have ever deserved that degree might be questionable but it's fairly obvious that you have never had to apply what you learnt in a practical sense, and you have now shown yourself unable to get your head around simple algebra for the second time.

Reflected, backscattered -- call it insulation
Other radiation, Sun or CMB -- call it heating

Well, BPL, I have to eat crow. I read K&T again, and I've completely mischaracterized it. Apologies. They do not infer that 235 watts produces 390 watts, like I thought I remembered. No problems with the second law. They provide some measured values (235, e.g.) and some modeled values (390 watts, e.g.), and try to put them all together in a "budget."

I still think it's a crock, though, for the following reasons:

1. The downward radiation, which is given as 324 watts should match the 390 watt surface radiation. It makes no sense that the surface should not be in equilibrium with the air near the surface. I think this is what M says.

2. As pointed out by someone heree (Jan?), K&T "adjusted" the humidity of the 76 atmosphere by 12 % to make their modeled values match measured values for global mean clear sky flux.

3. I think the radiation is simply a measure of the temperature, and the radiation does not cause anything (my own bias).

Importantly for this discussion, I can see no place that they say that the 324 watts of back-radiation is doing any "heating" of the surface.

jae writes:

The Keihl-Trenberth-type cartoon is truly a crock. There is no possible way to turn 235 watts of power into 390 watts of power, without ADDING energy somewhere. Back-radiation is not ADDITIONAL energy, and that is the Achilles Heel of these cartoons. There is only so much power being put into the system by the Sun in these diagrams–235 watts; therefore, there cannot be more power in the system than 235 watts. You cannot keep adding the same power together to produce more power! It’s so plain! It is truly staggering and amazing how many otherwise bright people have been bamboozled by this perpetuum mobile of the second kind. It is clearly dead wrong, no matter how famous and often-quoted it is.

I replied to a similar argument by John Dodds here:

http://www.geocities.com/bpl1960/Dodds.html

I'll ask you what I asked Jan, jae. Infrared light from the atmosphere, to the tune of 324 watts per square meter, is absorbed by the ground. Where does that energy go? What effect does it have?

jae writes:

Well, first of all, the 324 watt figure is a figment of faulty reasoning.

It has been measured. I don't know how accurate the figure 324 is, but it's not far off. Google "pyrgeometer" to find out how climatologists measure back-radiation from the atmosphere.

But IF there were actually 324 watts, they would be absorbed by the surface, tending to warm it

Exactly. Thanks for finally admitting that I'm right and you're wrong.

The 168 watts from the sun plus the 324 watts from the atmosphere would heat the Earth to 305 K, except that the Earth cools by 102 watts per square meter from sensible and latent heat losses.

to about 2 C. K&T actually only have 168 watts hitting the surface directly and 235 total watts to play with. So the maximum temperature they can get under their “average conditions” is -18 C.

Yes, jae, 235 watts per square meter can only heat the Earth to 254 K. Very true. Now why do you think the total flux absorbed by the climate system is the same as the flux absorbed by the surface?

As I tried to explain on the Forum, it is not physical to analyze the earth’s heat budget by using such “averages.”

Not sure what you mean here.

BTW, Miskolczi does NOT use numerical average radiation budgets and is therefore a different can of worms.

He damn well does. Go look at his diagram again. Note, also, that his E-sub-D is back-radiation from the atmosphere.

Jan, whose mindset appears to be stuck in a vice, writes:

you have net radiation confused with radiation.

I have nothing confused I know the difference and I don’t ignore the fact that the net radiative heat loss from the surface is positive and therefore cannot lead to or be considered a gain in heat which should not be confused with either energy or temperature for the the surface even if the temperature goes up because of it.

Yes, Jan, net radiation is from hotter object to cooler object. But back-radiation is not net radiation. Why you are unable to grasp this point eludes me. I can only speculate that you realize that acknowledging it blows away your whole stupid argument.

please don’t tell my old professors that I’m algebraically challenged! They might take back my degree in physics!

Whether you have ever deserved that degree might be questionable but it’s fairly obvious that you have never had to apply what you learnt in a practical sense, and you have now shown yourself unable to get your head around simple algebra for the second time.

What degrees in physics do you have, Jan?

My algebra is just fine. I'm just pointing out that you're fixated on the net radiation equation for a problem which doesn't involve net radiation. That's not "being unable to get around simple algebra." That's "pointing out that some idiot is using the wrong equation for the problem." Try to understand the difference.

I'll ask again, Jan -- the ground absorbs 324 watts per square meter of thermal IR from the atmosphere. What happens to the energy carried by that radiation when it is absorbed by the ground? Where does the energy go?

BPL #180

Yes, Jan, net radiation is from hotter object to cooler object.

Thank you

But back-radiation is not net radiation.

Please show where I have said it was.
You seem terribly confused about this.

the ground absorbs 324 watts per square meter of thermal IR from the atmosphere. What happens to the energy carried by that radiation when it is absorbed by the ground? Where does the energy go? Where does the energy go?

It all goes to space eventually but the ground radiates 390 watts /m^2 and the rest including what has come from the sun is carried aloft to where it is radiated by latent and sensible heat convection. The net total passing through the entire system is ~235 W/m^2 and this is what constrains temperature at the surface which is really just part of a continuum from the effective orbital temperature of ~ 400K to ~2.7 K and all the GHGs can really do is change where on that continuum we might find the surface currrently it's at around 286-7K average.

Now it should be patently obvious with a little arithmetic that the net Long wave IR radiative effect with 324 W/m^2 incoming and 390 W/m^2 outgoing the net effect is cooling there is really no sense it separating the back radiation from the forward radiation as it's all included in the radiative cooling equation:

$$Q = \sigma \left( \epsilon_{surface} t_{surface}^4 - \epsilon_{air} t_{air}^4 \right)$$

the algebra is straight forward why are you having such difficulty with it if:

My algebra is just fine

aye??

Why can't you see that M7 Eq (1)
$$F+P+ K+A_A - E_D - E_U = 0$$
and empirically verified Eq 4
$$ A_A = S_U A = S_U \left( 1 - T_A \right) = E_D$$

combines with Eq (1) to give

$$F+P+K = E_U$$

if as you say

My algebra is just fine

?

BPL #179

Google “pyrgeometer” to find out how climatologists measure back-radiation from the atmosphere.

If you want to find a bit more detail out bout the accuracy of pyrgeometers you would probably be better off googling "pyrgeometer Miskolczi" and instead of getting a wikipedia entry you will get a peer reviewed paper.

Yes, jae, 235 watts per square meter can only heat the Earth to 254 K.

But 235 watts does not reach the surface only 168 W/m^2 does and this can only heat the surface to 233K really goes to show what a lot of nonsense is being purveyed by the paper and the source of that nonsense is the way the averaging is done. Fact of the matter is that with a solar constant of ~2366@/m^2 is capable of heating the surface to about 390K at the tropics and not at all at the poles (the incoming radiation sails past without being absorbed) and the diurnal temperature variation at the tropics quite wild the atmosphere with it's GHG and the oceans actually smooth this diurnal and spacial variation and Dr Steve Short has been searching papers and provided quite a few links that actually looks at the processes by which this occurs.

You really aught to take a look at some of them on I particularly like is Ozawa & Ohmura Journal of Climate 1996 "Thermodynamics of a Global-Mean State of the Atmosphere—A State of Maximum Entropy Increase"

BPL:

"It has been measured. I don’t know how accurate the figure 324 is, but it’s not far off. Google “pyrgeometer” to find out how climatologists measure back-radiation from the atmosphere."

LOL. And just where on Earth would I place that pyrgeometer to obtain the AVERAGE downward radiation? I suggest you re-read Kiehl and Trenberth, too, since you don't seem to understand your own "hypothesis."

Jan writes:

the ground absorbs 324 watts per square meter of thermal IR from the atmosphere. What happens to the energy carried by that radiation when it is absorbed by the ground? Where does the energy go

It all goes to space eventually but the ground radiates 390 watts /m^2 and the rest including what has come from the sun is carried aloft to where it is radiated by latent and sensible heat convection. The net total passing through the entire system is ~235 W/m^2 and this is what constrains temperature at the surface which is really just part of a continuum from the effective orbital temperature of ~ 400K to ~2.7 K and all the GHGs can really do is change where on that continuum we might find the surface currrently it’s at around 286-7K average.

Hand-waving. I'll ignore the fact that 235 W/m^2 can't be what constrains the temperature of the surface since it could only heat the surface to 255 K. I'll ask again -- air gives off IR, the ground absorbs the IR. Where does the energy of the absorbed IR go?

jae writes:

“It has been measured. I don’t know how accurate the figure 324 is, but it’s not far off. Google “pyrgeometer” to find out how climatologists measure back-radiation from the atmosphere.”

LOL. And just where on Earth would I place that pyrgeometer to obtain the AVERAGE downward radiation?

Jesus, do I really have to explain this? You measure it in a lot of places, equally weighted by area, and take the average. You measure it all through the day, equally weighted by time interval, and take the average. Do you really not understand what an average is?

bARTON #184

I’ll ignore the fact that 235 W/m^2 can’t be what constrains the temperature of the surface since it could only heat the surface to 255 K. ... Where does the energy of the absorbed IR go?

Ignoring constraints is good. Gets you far. With upward radiation of 390 W/m^2 from the surface along with the latent and sensible heat so that exactly what comes in from space is radiated back to space and you have to ask? Come now. It would be much more sensible to ignore the resonant like effects at the surface than the fact that the temperature is constrained by available energy.

BPL:

"Jesus, do I really have to explain this? You measure it in a lot of places, equally weighted by area, and take the average. You measure it all through the day, equally weighted by time interval, and take the average. Do you really not understand what an average is?"

Can you point me to that data? I've been looking for it for a long time. Only Miskolczi seems to use much actual data for these analyses. K&T state on page 200:

"We must rely on model calculations to determine
the surface radiative fluxes. As described above,
longwave radiation emitted from the surface is absorbed
and reemitted by greenhouse gases and clouds
throughout the earth’s atmosphere. The transfer of
longwave radiation depends on both the local temperature
of the gaseous absorber and the efficiency of
the gases to absorb radiation at a given wavelength.
This absorption efficiency varies with wavelength."


There''s a lot in that paragraph, Barton. The last two sentences are particularly intriguing and seem to be ignored by those who simply want to add radiation from different sources to end up with some magic (288K) number.

BTW, I don't think you can get to 288 K by adding radiation from colder surfaces (say, 324 watts from zero degree air + 66 watts from 20 C surface), because of the Wein Displacement Law and the dependency of temperature on wavelength.

Jan, IR from the air hits the ground. The ground absorbs the IR. Where does the energy of the IR go when it gets absorbed by the ground? What happens to it?

BPL #188

IR from the air hits the ground. The ground absorbs the IR. Where does the energy of the IR go when it gets absorbed by the ground? What happens to it?

Please explain how many times you need to be told? you have been answered in

118, 124 (even an equation which would explain it all if you weren't so algebraically cHallenged), 127, 128, 148, 156, 160 (most important), 166 (some of got messed up can produce the detail again if needed), 175, 181 and 186.

BPL:

"IR from the air hits the ground. The ground absorbs the IR. Where does the energy of the IR go when it gets absorbed by the ground? What happens to it?"

Well, Barton, I don't think the radiation does anything, if the surface is above the temperature of the air. It certainly cannot warm a warmer surface. If the surface happens to be colder than the air for some reason, then the radiation warms the surface.

And I wish someone would explain to me how you can add 324 watts from the air (denoting a radiation source around 0 C) to 66 watts from a 15 C surface to produce 390 watts that supposedly indicates a surface temperature of 15 C. I think the dependency of emissivity on temperature and wavelengths makes this impossible.

jae writes:

“IR from the air hits the ground. The ground absorbs the IR. Where does the energy of the IR go when it gets absorbed by the ground? What happens to it?”

Well, Barton, I don’t think the radiation does anything, if the surface is above the temperature of the air.

No, jae, "I don't think the radiation does anything" cannot be the right answer. The most basic principle of modern physics is conservation of energy. Energy can neither be created nor destroyed. It can't just vanish. Try again.

Barton Paul #191

Energy can neither be created nor destroyed. It can’t just vanish. Try again.

Good you know the law of conservation of energy I think jae does too now can you quote the law of conservation of radiation flux densities?

Barton, it's my understanding that you can surround a 1000 C surface with 1,000,000 500 C surfaces, and you still cannot increase the temperature of the former. Why don't YOU explain it to ME? I think the Wein Displacement Law and the dependence of emissivity on wavelength and temperature are involved.

BPL, here's perhaps a better example of my quandry: You can melt steel with an oxy-acetylene torch, but not with a propane torch . The oxy-acetylene torch gets much hotter. Now, I can direct 20 propane torches at once to the piece of steel, and it still won't melt, despite the fact that the total wattage from the 20 propane torches (the sum of the watts from all torches) is much greater than the wattage from the oxy-acetylene torch. So, you cannot just add wattages from lower temperature objects to produce higher temperatures. Why?

# 194 jae “cannot just add wattages from lower temperature objects to produce higher temperatures. Why?”

Equipartation Principle, magically adjusts, via any radiation coupling, (Kirchoff enclosed radiation) the energy of motion (temperature) of otherwise isolated particles. Even if coupled only by lower frequencies, the harmonics produce the higher ones. The exact quantum jump to quantum jump, --somehow the one way mirrors cannot decrease entropy.

Thanks, Franco, but I think more than equipartition is involved. Perhaps we are talking about which energy levels can be filled at various temperatures.

jae and Franko

First equipartition and entropy are related concepts also long wave is higher entropy than short-wave. Entropy can however be reduced in a sense by concentration. For example the microwave concentrates all it's energy in a single frequency band so even if it is low temperature high entropy the entropy is not as high as it otherwise be if it was just a peak (Wien Law) of a wide band emission. Another concentration that occurs in the microwave is the fact that the 600 watts is emitted from (typically) 9cm^2 aperture so the flux density is 6.67e5 W/m^2 (and effective temperature of ~1800K). More than enough to melt the cheese of the pizza.

Yet another reduction is that the emitted energy van only be dissipated by molecules (thermalised) with a dipole moment (eg H20) or free electrons like metals and maybe some other bits and pieces.

I hope this helps.

I hope this helps.

#196 jae
”I think more than equipartition is involved”

You could think of it as quantum states, discrete steps, raching each other up. If the two objects are coupled, at equal temperature, backward and foreward radiation energies are equal. The ratching energies between quantum states balance.

If coupled at only one frequency, (one quantum energy) -- then the colder emits less, until temperatures are equal.

I have always wondered if there is some way around this, bypass the equipartation principle; a molecular one way valve, trap door to let in, but not out. Is it possible to have a molecular self-powered robot to shovel a fluid in one direction only ? Molecular generator and propeller, to violate entropy ? Some kind of magical filtering, diffraction grating ?

jae writes:

Barton, it’s my understanding that you can surround a 1000 C surface with 1,000,000 500 C surfaces, and you still cannot increase the temperature of the former.

Your understanding is wrong.

Barton arm-waves:

"Your understanding is wrong."

You gotta do more than arm-wave to convince me. Try the oxy-acetylene and propane torch experiment. Bet you can't melt steel with any number of propane torches and any number of watts.m-2 (forced air and pure oxygen not allowed).

The answer is in the shining in the firefly
In the ocean, all the way to violet.
All colour, but no fire

Franko:

"In the ocean, all the way to violet."

YES, I think that is on target, somehow.

Franko

Is it possible to have a molecular self-powered robot to shovel a fluid in one direction only ?

yes you could be thinking of Maxwell's Demon

Interesting there's no background radiation in the radiation budget cartoon on more in NASAs teaching site.
picture:
http://eosweb.larc.nasa.gov/EDDOCS/images/Erb/c...
alas the full version includes for an unquantified back-radiation:
http://eosweb.larc.nasa.gov/EDDOCS/radiation_fa...

So my conclusion is that when you are teaching kids you have to avoid counting the same energy twice because they'll notice such trickery, but you can easily fool "smart" scientists because they help you out by seeing just exactly what they want to see and ignoring logic, arithmetic, rationality and physical laws.

In BPL's defense though I read that Boltzmann argued that the 2nd law was only statistically correct and wasn't inviolate, and Planck eventually came to accept that idea. But then again, probably they both expected that the numbers in needed to match the numbers out.

James, back-radiation from the atmosphere does not entail any breaking of any law of thermodynamics. To think it does shows that you don't understand those laws.

If back-radiation violates the first law of thermodynamics, please compute by how many Joules it does so.

If back-radiation violates the second law of thermodynamics, please compute by how many Joules per Kelvin it does so.

If you don't know what I mean by "computing" a violation of a law of thermodynamics, then you probably got your understanding of that law from a popular science or pseudoscience source which put it entirely in verbal, qualitative terms and got it wrong.

Oh, and before you accuse scientists of " ignoring logic, arithmetic, rationality and physical laws," you really ought to learn something about logic, arithmetic, rationality and physical laws.

BPL
Firstly, you need to take stock of what you've said here and elsewhere about some scientists before getting on your high horse about me. Well I happen to agree with you that a lot of scientists talk a lot of crap some of the time and certain scientists all the time. And realclimate.org, quite frankly, is where you'll find they regularly violate these 4 tenets I mentioned. I've seen two of them getting the Stefan-Boltzmann equation wrong for starters. Add in the regularly repeated nonsense about if there was really an MWP then it would mean a higher climate sensitivity (which only partly makes any sense if you think the atmosphere is really a simplistic electric circuit) or the absolute nonsense about the ice-cores telling us anything at all about positive feedback. Oh but there are so many more gems...just see the hot-spot, feedback and aerosol hand-waving arguments to see many examples of illogicalities, irrationalities and mutually-exclusive, contradictory arguments. That site is a virtual crap mine.

Secondly, try reading properly. I supported you in writing that Boltzmann and Planck finally agreed that the second law is only statistically correct. Which would mean that back-radiation doesn't break the second law. However if a giant like Planck was so reluctant to accept that, then I don't know why you'd feel so confident about it: It's clearly not obvious. While there are several professors of Physics who would - and do - argue that back-radiation breaks the second law - I wasn't in fact arguing the same: I was being more general. However I'm certain that if I did a radiative model (as I do from time to time) it would show that no radiative heating of the ground could occur from the atmosphere. That wouldn't of course mean it wasn't possible, just that the standard theoretical treatment (ie net heat flow) doesn't allow for it in models - as perhaps you already know - or maybe you didn't.

I was of course referring to a proper radiative model, correctly constrained and with minimal independent variables - not the joke models that the IPCC uses with about 30 adjustable parameters which allow you to produce virtually anything you like - even massive cooling if you go overboard on that aerosol parameter.

# 207JamesG

People genuinely look for climate information
Misled; End up in a church with a motivation
Promising Salvation, Churches do not have to pay taxes
Crimatists consume Tax dollars, promising a reward

Believe the Charismatic used climate salesman
Relax, and let the priest do your thinking
Or get berated, excommunicated
While the crimatists pig out on your tax dollars

JamesG #207

It's not the back radiation that breaks the second law it's the decoupling from forward radiation that BPL has been attempting that does. Back radiation does nothing more than reduce the net forward radiation from the surface.

Jan
Thanks. Eli started arguing a similar thing to BPL in answer to G+T's paper but then he relented to something that at least we can all agree on - that the greenhouse effect is really just a prevention of cooling.

But I just realized that if there were a tropical tropo hotspot, ie hotter than the surface, then back-radiation would heat the surface there. Maybe that was the original idea but it got lost with all the planet-wide averaging.

Another thing about most of those cartoons that is annoying is that they always show a radiation arrow from a GHG layer to the ground, yet just as much radiation should go upwards too. Hence another reason to prefer the NASA one.

Use the basis of Planck Wien, and Boltzman
Equipartition via Maxwell's Demon
Packet of radiation comes into the surface
Packet goes to Aa (atmosphere absorbed)
By Demon of Equipartition -- up or down
If down the process repeats
There is no loss, because the one that went up
Got replaced by radiation incoming

James G:

"Thanks. Eli started arguing a similar thing to BPL in answer to G+T’s paper but then he relented to something that at least we can all agree on - that the greenhouse effect is really just a prevention of cooling."

Which prevention of cooling can more easily be explained simply by the heat storage of the atmosphere and water, IMHO.

"Another thing about most of those cartoons that is annoying is that they always show a radiation arrow from a GHG layer to the ground, yet just as much radiation should go upwards too. Hence another reason to prefer the NASA one"

I agree. If you are going to have 390 watts going up, you must have 390 watts going down, or there is a surface discontinuity problem.

JamesG posts:

While there are several professors of Physics who would - and do - argue that back-radiation breaks the second law -

Name one. Name one person with a degree in physics who thinks atmospheric back-radiation violates the second law of thermodynamics. And if by some miracle you do locate such a person, please provide contact information so I can ask him what the hell he's thinking.

I wasn’t in fact arguing the same: I was being more general. However I’m certain that if I did a radiative model (as I do from time to time) it would show that no radiative heating of the ground could occur from the atmosphere.

I do radiative-convective models of planetary atmospheres all the time and back-radiation does warm the ground. That's how the greenhouse effect works.

Jan writes:

Back radiation does nothing more than reduce the net forward radiation from the surface.

Jan, the ground absorbs IR given off by the atmosphere. The IR carries energy (E = h ν and all that). Where does the energy go when the ground absorbs it? What happens to it?

jae writes:

I agree. If you are going to have 390 watts going up, you must have 390 watts going down, or there is a surface discontinuity problem.

You're forgetting non-radiative heat transfer and sunlight. Yes, the net power has to be zero for the surface to have a steady temperature. But no, it does not have to be in radiative balance. Here's a flux density budget for the surface (K&T97):

Input: 168 W/m2 from sunlight, 324 W/m2 from atmospheric back-radiation. Total: 492 W/m2.

Output: 390 W/m2 thermal radiation, 24 W/m2 sensible heat exchange, 78 W/m2 latent heat exchange. Total: 492 W/m2.

For steady temperature, input = output.

BPL:

"Jan, the ground absorbs IR given off by the atmosphere. The IR carries energy (E = h ν and all that). Where does the energy go when the ground absorbs it? What happens to it?"

Interesting you bring up E = hv. That is exactly what intrigues me about this addition of fluxes from sources at two different temperatures, in order to calculate a total flux—and then use that total to back-calculate another temperature. I think this relationship is key to understanding why there is no HEATING of the surface by the radiation from the "cooler" air, like you keep insisting on.

We know from the SB Distribution that the peak frequency for the photons (and therefore their energy) increases with increasing temperature. Thus, the average frequency of the photons from the cooler air is lower than the average frequency of photons from the warmer surface. So isn't the average energy of the photons from the cold air less than the average energy of the photons from the warm surface? There appears to be something inherently "different" between, say, 10 watts from a 0 C surface and 10 watts from a 15 C surface. A different number of photons (and the average energy level of those photons) comprise the 10 joules/sec. in the two cases. The flux is the same, but I don’t think you can calculate a temperature, based on the addition of fluxes from sources at two different temperatures, because you cannot get the proper peak photon frequency level for that calculated temperature by doing this. Maybe it’s the effect of temperature on emissivity?

Back to my post 194 re: the oxy-acetylene vs. propane torch issue (that you seem to be avoiding?). Why can’t you just add watts.m-2 for a whole bunch of propane torches to get the temperature you need to melt steel? That question is no different than the one you posed; i.e., what happens to all those watts?

Anyway, I think that’s basically why K&T still doesn't make sense to me.

Maybe I’m all confused here (again), but something just doesn’t compute for me.

BPL

Jan, the ground absorbs IR given off by the atmosphere. The IR carries energy (E = h ν and all that). Where does the energy go when the ground absorbs it? What happens to it?

Still trying to decouple back radiation form forward sorry you can't do that because without the greater forward radiation there will be no back radiation.

The equation is a simple one I've showed it to you a number of times prehaps you need a simple analogy of a voltage divider $$R_{in}$$ and $$R_{out}$$ $$V_{out}$$ at the node. Then
$$V_{out} = \frac {R_{out}} {R_{in}} V_{in} - \frac {R_{out}} {R_{in} }V_{out}$$

This equation too looks like the "back current" in the second term in RHS can be decoupled from the forward one represented by the first term in RHS. It also looks like there is more current into the node than out but Kirchhoff's law tells us that the current into the node must sum to zero so the current all flows from $$V_{in}$$ source to ground.

jae writes:

I think this relationship is key to understanding why there is no HEATING of the surface by the radiation from the “cooler” air, like you keep insisting on.

It's not just me that insists on that, jae. It's every professional climatologist in the world, and even Ferencz Miskolczi.

Well, Jan continues to twist and turn to avoid answering my question. I don't want this to go on forever. I'm beginning to feel like I'm whipping a chimpanzee to get it to learn how to type. The poor thing just isn't capable of understanding what's desired.

The air radiates thermal infrared radiation (hereinafter "IR"), and the IR is absorbed by the ground. The IR carries energy, to the tune of 324 watts per square meter averaged over time. The ground absorbs the energy. What happens to the energy? I'll give the answer Jan seems unable to figure out.

The answer is, the energy from the photons is absorbed by the molecules of the ground. Carrying more energy, those molecules move and vibrate faster -- i.e., the ground heats up. It heats up a predictable amount:

dT = dH / (cp m)

where dT is the change in temperature, dH the change in energy content, cp the ground (or any object's) specific heat at constant pressure, and m the object's mass.

Let's put some quantitative flesh on those bones. The specific heat (at constant pressure) of granite is about 800 Joules per Kelvin per kilogram. Let's say we add 1,000 Joules of energy to a granite block massing one kilogram. The temperature of that bit of rock will increase by (1000 / [800 x 1]) or 1.25 kelvins.

Let's examine our block in a simplified universe in which only radiation is exchanged. Let's say our block has a bolometric Bond albedo of A = 0, which means it has an absorptivity/emissivity of 1. One side of the block receives 168 watts per square meter of sunlight from a hotter source, like the sun. Our block is one square meter in area, let's say -- it's a flat plate, like a flagstone.

It absorbs all that radiation. To stay at an even temperature, it must radiate as much as it receives -- conservation of energy. Its radiative equilibrium temperature would then be

Te = (F / σ) ^ 0.25

where Te is the temperature and F the absorbed flux density. The Stefan Boltzmann constant, σ, has a value of 5.6704e-8 watts per square meter per kelvin-to-the-fourth in the SI, so the temperature of our block is 233 K. Pretty chilly.

Now let's say our block also receives 324 watts per square meter from a cooler source. The total input is now 168 + 324 or 492 watts per square meter. At equilibrium, our block of granite radiates away the same amount, and its radiative equilibrium temperature is 305 K.

It's hotter than it would be in the absence of the radiation from the cooler source.

Note why this is so -- the block of granite does not know that the source of one stream of energy is from a hotter object and one is from a cooler object. To the granite, it doesn't matter. Radiation is radiation. The more radiation pouring onto the block, the hotter the block gets, no matter what the source of the radiation is or what temperature that source is at.

The Earth is warmer than it would be in the absence of atmospheric back-radiation. The back-radiation constitutes the misnamed "greenhouse effect." That's how the greenhouse effect works.

Nobody with a smattering of elementary physics and/or climatology disputes this. Only a crackpot like jae would take issue with it, and only a slavish supporter of anyone on his side like Jan would side with a crackpot like Jae, and continue to do so after he is proved wrong again and again and again. But I'm not a psychologist. I'm a guy with a degree in physics. All I can do is demonstrate why jae & Jan are wrong. Getting them to agree that they're wrong is beyond my capabilities, or possibly beyond anyone's capabilities.

BPL: If you are correct, then why can't I melt steel with the propane torches?

The problem with BPL's equations is that he first has to assume that a cold source can heat a hot one before applying any equations. I presume that's how his home-made models all work. However, if this were possible then it's not yet been observed in anything we've yet seen. Calling people a crackpot for using basic logic is just dogmatic. Faraday, Pasteur and many other giants of science were also considered crackpots by the dogmatic consensus but they were right. The guy who proved that a real greenhouse wasn't heated by radiation but was prevented from cooling by lack of convection was likely also an anti-consensus crackpot, but he was right. The guy who said malaria wasn't from bad air was surely a crackpot. Etc, etc. Anyone who closes his mind and dogmatically accepts illogical, unphysical explanations can never advance science. You have to seek an explanation that makes sense.

But it's unquestionable that the atmosphere heats the surface. The problem is that most folk seem to reduce the problem to a 1D, static flat plate, when it's a 3d, continuously rotating magnetic orb in a vacuum, heated on one side only. The dark side is obviously only ever heated by the atmosphere. In the morning the surface is also colder than the atmosphere. At the tropics, the moist adiabat ensures that there the atmosphere is likely hotter than the ground. So there is no real need to invent a physically unrealistic mechanism whereby a cold, low-energy device can heat up a hotter, high energy device. Anyone who does so is being illogical and is most certainly misapplying the Stefan-Boltzmann law.

Nor does it surprise me in the least that the majority of climate scientists cannot think in 3d, just as it didn't surprise me that the majority of the worlds economists thought that all this compounded debt was no problem. People, the intelligent as well as the stupid, seem to persuade themselves to believe exactly what they want to believe and they will make up the most ludicrous explanations to cover their dogma.

Most scientists in fact are happy to settle for the explanation that the greenhouse effect is a restraint on cooling and go no further than that. Few people actually bother to argue like BPL, that a cold substance can heat a hot substance. That is a perpetual motion argument and anyone should feel silly arguing it. Would that it were really true.

Barton Paul Levinson #220

Continues to build his perpetual motion machine of a second kind. Trouble is that it is never going to work because the upward flux is greater than the downward flux.

Barton: There are actually quite a few of us crackpots around. Here's a take from another one:

"If, however, the earth is heated by the sun and also by the radiant transfer performed by trace gases, then its surface temperature must necessarily be higher than the temperature the sun could make it, due to the extra radiant energy impinging on it at any time of the day. Yet the simple fact is that the earth’s sunlit surface temperature is perfectly consistent with solar irradiance alone. Ergo, greenhouse theory is demonstrably false. Radiant energy from trace gases cannot heat the air, nor does it heat the earth. "

http://www.ilovemycarbondioxide.com/FAQ.html

jae #224

Bart's perpetual motion machine works fine in thought experiments and possibly in digital computers as long as the program runs but they just don't work in the real world or in analogue computers which are subject to times arrow (entropy).

Another crackpot says:

"Fig. 1 reminds us that the temperature in the troposphere decreases with increasing height, and may also call to mind the Second Law of Thermodynamics (in Clausius’ formulation): “Warmth can never pass from a colder to a warmer body unless another related change occurs at the same time” (cf. Baehr, [11]), or more simply “Warmth can never spontaneously pass from a body of low temperature to a body of higher temperature” (see Schmidt, [12]). The Law may raise doubts about how “atmospheric backradiation” can work. In fact, properly considered, it already suggests that “atmospheric backradiation” is a mirage.

Even disregarding the Second Law, close inspection of the mathematical boundary conditions of the radiative fluxes represented in Figure 1 also suggests that “atmospheric backradiation” is nonsense. Both terrestrial radiation and claimed “atmospheric backradiation” are radiant fluxes. As such, they are definable as to direction and magnitude; i.e. as vectors. Now the combination (in this case, addition) of vectors produces a new vector (see Bronstein-Semendjajew [13]), with the proviso that two exactly equal vectors of contrary direction sum to zero.

In the light of the laws of vector analysis, it is useful to split the flux of terrestrial long-wave radiation shown in Fig. 1 (393 W/m²) into its components, namely heat dissipation by radiation into space (45 W/m²), and re-radiation of “atmospheric backradiation" (348 W/m²). Now it is clear that only the 45 W/m² is thermal radiation of the earth's surface, see Fig. 2. The remaining 348 W/m² and the “atmospheric backradiation”, which is equal in size but opposite in direction, sum to zero. So this breakdown into outgoing radiation from the surface and “atmospheric backradiation” is sheer physical nonsense. "

http://www.geocities.com/atmosco2/backrad.htm

jae #225

but jae you are really being unfair crackpots like Rudolph Clausius who have the mistaken idea that “Warmth can never pass from a colder to a warmer body unless another related change occurs at the same time” or that <PE>/<KE> = -2 did not have Bart's superior intellect to guide him.

JamesG writes:

Few people actually bother to argue like BPL, that a cold substance can heat a hot substance. That is a perpetual motion argument and anyone should feel silly arguing it. Would that it were really true.

I note that you again fail to provide a quantitative analysis.

I am not misusing the Stefan-Boltzmann equation, I am using it. Is it your contention that if a cooler object is near a warmer object, the cooler object doesn't radiate?

The Stefan-Boltzmann equation, as modified for gray or band radiation, applies to EVERY MATERIAL OBJECT. Not just warmer ones. The atmosphere is a material object, and it sure radiates in a lot of bands. Some of the radiation goes down and hits the ground. The ground warms as a result. If you deny that, you are the crackpot, not me.

Jan writes:

Continues to build his perpetual motion machine of a second kind. Trouble is that it is never going to work because the upward flux is greater than the downward flux.

Jan, if I am violating the second law of thermodynamics in my description, please say BY HOW MUCH. If you can't do that, you're talking out your euphemism. The second law of thermodynamics is an equation (or rather, a series of equivalent equations). If you want to show a violation, show your work. If you can't do that, shut up.

jae quotes some other crackpot:

“If, however, the earth is heated by the sun and also by the radiant transfer performed by trace gases, then its surface temperature must necessarily be higher than the temperature the sun could make it, due to the extra radiant energy impinging on it at any time of the day. Yet the simple fact is that the earth’s sunlit surface temperature is perfectly consistent with solar irradiance alone. Ergo, greenhouse theory is demonstrably false. Radiant energy from trace gases cannot heat the air, nor does it heat the earth. ”

This is simply wrong. The Earth's sunlit surface temperature is NOT consistent with solar irradiance alone. Solar irradiance alone would take it to 254 K, not 288 K. I have demonstrated this over and over and over again. But it no longer surprises me that it doesn't sink in where you're concerned.

jae writes:

The Law may raise doubts about how “atmospheric backradiation” can work. In fact, properly considered, it already suggests that “atmospheric backradiation” is a mirage.

It's a mirage which is routinely measured by pyrgeometers. jae, it's easy to check this. Buy one of the damn things and set it up in your back yard. Then read what it says. If it reads zero, you're right and there's no back-radiation from the atmosphere. If it reads above zero, you're wrong.

I missed this gem of JamesG's before:

Nor does it surprise me in the least that the majority of climate scientists cannot think in 3d

You've never opened a textbook on atmospheric radiation, have you? Opening Goody and Yung, I find a discussion of the Earth's shape and how it bears on radiative equilibrium on page 2. The discussion of the equation of radiative transfer begins on page 16 and immediately discusses the fact that intensity includes a component of solid angle, which is of course an extension of angle to two dimensions, and of course path length adds the third. JamesG, it doesn't surprise you that climate scientists can't think in 3D because their not thinking in 3D is a delusion springing from your massive ignorance.

Bart #229

Jan, if I am violating the second law of thermodynamics in my description, please say BY HOW MUCH.

Haven't you read Miskolczi & Mylnzcak 04 and Miscilkczi 07 yet I think it's amplu explained there remember Aa=Ed? and Su = Aa + St so using TIGR profiles with a surface temp of 286 and Ed = 311.4 W/m^2 then

absorption of LW by the surface is

$$q = -\sigma \epsilon t_s^4 + 311.4 = 5.67e-8 \times .96 \times 286^4 + 311.4 = -56W/m^2$$

So you see there is a long wave absorption deficit of -56W/m^2 so that is how much the surface cools due to radiation. This number is of course balanced by insolation = OLR and K in Miskolczi's paper.

If that is not enough for you even your favourite K&T97 has a LW radaitive deficit of 66W/m^2. Now if you are claiming that the surface is absorbing 324 W/m^2 then you are out by 390 Wm^2 according to K&T 97.

Are you beginning to understand it now? I have not been avoiding answering I have been answering I have been giving you equations and all now you have numbers too. Are you happy?

Barton: I think you misunderstood the quote, when you say:

This is simply wrong. The Earth’s sunlit surface temperature is NOT consistent with solar irradiance alone. Solar irradiance alone would take it to 254 K, not 288 K. I have demonstrated this over and over and over again. But it no longer surprises me that it doesn’t sink in where you’re concerned.

Actually the Earth's SUNLIT surface is far COOLER than it "should" be, thanks to convection. At high noon in July in, say Denver, the total measured solar insolation at the surface is about 900 w m-2. Using the S-B equation, that is consistent with about 82 C (180 F)!! (About the temperature in a well-insulated greenhouse). Thanks to convection, it doesn't get that hot, or nobody could live in Denver. Now, you neo-physicists are trying to tell me that there's an additional amount of "heat" coming from backradiation, on top of this solar radiation, which would dirve the temperature even higher? Hogwash.

Bart #231

It’s a mirage which is routinely measured by pyrgeometers. jae, it’s easy to check this.

It's why it's called a mirage and not an illusion :- it's a trick of the light rather than the mind. To resolve the problem one needs two pyrgeometers looking simultaneously upward and downward.

Typo in 233 the equation should read:

$$q = -\sigma \epsilon t_s^4 + 311.4 = -5.67e-8 \times .96 \times 286^4 + 311.4 = -56W/m^2$$

Jan writes:

Now if you are claiming that the surface is absorbing 324 W/m^2 then you are out by 390 Wm^2 according to K&T 97.

Yes, Jan, for the 9,000th time, the ground radiates more to the atmosphere than the atmosphere radiates to the ground. The net transfer is from the ground to the atmosphere.

How does this prove that the ground doesn't absorb the 324 W/m^2 it gets from the atmosphere?

Where does the energy go, Jan?

jae writes:

Actually the Earth’s SUNLIT surface is far COOLER than it “should” be, thanks to convection. At high noon in July in, say Denver, the total measured solar insolation at the surface is about 900 w m-2. Using the S-B equation, that is consistent with about 82 C (180 F)!! (About the temperature in a well-insulated greenhouse). Thanks to convection, it doesn’t get that hot, or nobody could live in Denver. Now, you neo-physicists are trying to tell me that there’s an additional amount of “heat” coming from backradiation, on top of this solar radiation, which would dirve the temperature even higher? Hogwash.

jae, the figures I'm using are averaged over the day and over the Earth's surface.

Bart, 237:

"Yes, Jan, for the 9,000th time, the ground radiates more to the atmosphere than the atmosphere radiates to the ground. The net transfer is from the ground to the atmosphere."

Hmm, have you trapped yourself, here? By definition, something that radiates more is warmer. So the ground is warmer, and the air is cooler. So, do you still say that the air is HEATING the ground (i.e., there is a flow of heat from a cold source to a hot one)??

Bart #237

How does this prove that the ground doesn't absorb the 324 W/m^2 it gets from the atmosphere?

It's simple arithmetic how much a body warms or cools depends on how much heat it gains or looses that means net heat absorbed so if it emits more than it absorbs it cools.

Bart, 238:

Yeah, and my figures are NOT averaged. So what does the backradiation (which has to be a hell of a lot more than the average 324 watts) do in the situation I outlined? Can't I add it to the solar insolation to produce more heating, just like K&T do in their radiation cartoon and like you do?

Bart #238

jae, the figures I'm using are averaged over the day and over the Earth's surface.

So the the net of all fluxes over the day and surfaces for a steady state model (and model it is) will be zero.

Here's an interesting quote from Warwick Hughes' site:

"Arrhenius took over this formulation in his celebrated paper of 1896 that remains the cornerstone of the anthropogenic global warming (or climate change) movement, by asserting that while atmospheric carbon dioxide “increases in geometric progression, augmentation of the temperature will increase in nearly arithmetic progression”. Arrhenius won a real Nobel for proceeding to calculate that if carbon dioxide increased by 50 per cent from the level in 1896, global average temperature would increase by between 2.9 and 3.7 degrees, depending on season, latitude and hemisphere, with a global annual mean of 3.42 degrees. The level of carbon dioxide has nearly increased by 50 per cent since 1896—faster it is true than Arrhenius expected—but global temperature according to the Goddard Institute has increased by just 0.73 degrees."

http://www.warwickhughes.com/blog/?p=194#more-194

jae writes:

“Yes, Jan, for the 9,000th time, the ground radiates more to the atmosphere than the atmosphere radiates to the ground. The net transfer is from the ground to the atmosphere.”

Hmm, have you trapped yourself, here? By definition, something that radiates more is warmer. So the ground is warmer, and the air is cooler. So, do you still say that the air is HEATING the ground (i.e., there is a flow of heat from a cold source to a hot one)??

Yes. The ground is hotter than it would be if it weren't receiving the heat from the atmosphere. The NET transfer is from ground to atmosphere, but there is transfer from atmosphere to ground.

Jan writes:

How does this prove that the ground doesn’t absorb the 324 W/m^2 it gets from the atmosphere?

It’s simple arithmetic how much a body warms or cools depends on how much heat it gains or looses that means net heat absorbed so if it emits more than it absorbs it cools.

Yes. Very true. And still completely irrelevant. The ground is still absorbing energy from the atmosphere, and that energy heats the ground. Your comment does nothing to reply to what I said. It's a complete non sequitur.

Bart

The NET transfer is from ground to atmosphere, but there is transfer from atmosphere to ground.

Getting better now all you need to do is realise that the radiation for the atmosphere surface layer is isotropic sooner or later you are going to realise that the the net fluxes from the surface sum to zero.

Bart #245

Your comment does nothing to reply to what I said.

It does it says you are wrong.

Barton #121 #179

Oh, I shouldn’t neglected the pyrgeometer, which also measures infrared back-radiation. They have a nice wiki page about it:

wiki's are good manufacturers instructions are better:

In order to calculate the incoming LW irradiance at the detector, the temperature of the pyrgeometer body must be known. This is measured using a thermistor (type YSI44031), which is located beside the cold junctions of the thermopile. The downward (or upward) longwave radiation is then calculated using the following formula :-

$$LW = \frac {U_{emf}} S + \left( 5.67E1-8 \times T_b^4 \right)$$

where $$U_{emf}$$ is the output voltage from the thermopile, S is the calibration constant of the instrument, and $$T_b$$ is the pyrgeometer body temperature, measured by the thermistor, in degrees kelvin. Note that for an upward facing pyrgeometer, the thermopile output voltage will in most instances be negative. This is because the upwelling irradiance from the pyrgeometer is likely to be greater then the incoming irradiance from the sky.

I've added a little emphasis but it should make it quite clear that the net radiation usually cools.

Bart, 244:

"Yes. The ground is hotter than it would be if it weren’t receiving the heat from the atmosphere. The NET transfer is from ground to atmosphere, but there is transfer from atmosphere to ground."

Ahhh, you are finally getting it! I guess you just have a different view of what "heating" means than I do.

BPL
If you can step back from the name-calling I'll tell you where you are going wrong and why. Clearly you do planetary calculations using 1D equation and ignoring convection. Whereas I do 3D finite element model calcs for reactor physics using real shapes and including convection. The difference being that I have to correctly simulate the real temperature of a real situation whereas you don't. Hence like many tabletop physicists you make the classic mistakes as follows:
1. 1d just doesn't reflect the real world. It's a simplifying assumption to enable simplistic calcs - no more, no less. But real radiating bodies interact and feedback on each other and the incidence angle of the radiation source and sinks are extremely important - to the extent that we use ray-tracing algorithms to find them.
2. You use the wrong form of Stefan-Boltzmann (S-B) which involves only one temperature, ie T^4. But the only useful form of S-B in models involves a difference (Ta^4 - Tb^4), where Ta is the source temp and Tb represents the surroundings temp. This determines where the radiative heat flow goes. Hence when used properly, S-B doesn't allow heat transfer from a cold to a hot medium. You can quibble as much as you like about that - but this is the form of the equation that reflects real life and it is the only acceptable form for finite elements - where we do real benchmark tests to prove it's correctness.
3. You ignore the fact that radiation only becomes important at very high temperatures. For low temperatures is acceptable to ignore radiative heat transfer completely and still get the right answer. And that is indeed what we normally do.
4. You ignore convection. This is an easy yet classic mistake. Put simply, in real life if you ignore convection you won't get the right answer. This is because there is a convective feedback loop which affects the radiative sources and hence the directions and magnitudes of heat flow. Secondary convection in reactors usually turns out to be more important than the radiative source that spawned it.

Now taking that on board, you'll see why I don't reproduce simplified equations - it's because they are too simplified to be relied upon. For most radiation problems a 3d model is essential to get an answer approaching correctness. The important difference between climate models and nuclear models is accountability: We have to be right but climate modelers don't.

Now in fact I gave you a good out from your illogical position (ie that cold sources can make hot objects hotter) and you refused to take it. I said that it was obvious that for one side of the planet (the dark side) the atmosphere must be hotter than the surface, therefore must heat it. this means that the Trenberth et al. and other cartoons aren't actually wrong - they just oversimplify a time-dependent and geometry-dependent problem and they therefore confuse even individuals with degrees in physics. It's merely your own confusion that leads you to argue for perpetual motion (cold flowing to hot) rather than accept a more rational and correct argument. If the atmosphere to ground heat flow was shown as coming from underneath then it would be a slightly better representation.

Also if you now note that heat transfer via convection in the atmosphere is probably more important than has been assumed up to now, then you're coming into line with current thinking in the climate modeling field. and if you don't believe that then you can talk to Judith Curry about it. This of course was something Lindzen had been saying for many years. And this is the probable reason why the modeled hotspot isn't found in the measurements.

JamesG:

"Now in fact I gave you a good out from your illogical position (ie that cold sources can make hot objects hotter) and you refused to take it. I said that it was obvious that for one side of the planet (the dark side) the atmosphere must be hotter than the surface, therefore must heat it. this means that the Trenberth et al. and other cartoons aren’t actually wrong - they just oversimplify a time-dependent and geometry-dependent problem and they therefore confuse even individuals with degrees in physics. "

Hmm, I'm not sure I agree. It seems to me that the surface and the atmosphere directly above it are pretty much at equilibrium during the day. That is over the wet parts of the Earth. Doesn't the Clausius-Clapeyron Equation demand this? In dry areas the surface is definitely hotter, when the sun is shining.

Also, I think M and some others are considering "surface" to be the atmosphere at 2 m height??

JamesG:

I think what you say about essentially ignoring radiation at low temperatures (room temperature, e.g.) is true. I have read articles by several physicists saying the same thing. Radiation from gases doesn't get important until you get near combustion temperatures. Convection rules at lower temperatures, and

On reflection I've been quite unfair to climate modelers. I don't really assume that they think that a cold source can heat a hot one - it was BPL who suggested that. I'd prefer to think that it is so obvious to them that the atmospheric to surface heating happens on the dark side of the planet that they don't bother to even mention it. It is obvious after all! I've only seen tabletop physicists on the web come up with that perpetual motion argument; though enough of them to make me unfairly generalize.

Nor do I think accountability is the only difference between climate modeling and general heat transfer modeling: The massive scale of the problem makes modeling the climate an almost impossible task and one which necessitates simplifying assumptions. The problem only comes when they start to believe that these assumptions have been justified because the model tells them what they want to see. I have certainly seen that attitude a lot but it may be just the loudest voices I'm hearing. Many loudmouth modelers (you know who I mean) seem to be well aware of model limitations and their underlying, biased assumptions but feel their overriding messianic mission makes it ok to say that the models are good enough to base policy on. I'd say the models aren't even adequate. And the idea that 20 models each producing a very wrong answer can be combined, called a "robust" result and used to predict local droughts or floods is probably the biggest misapplication of models in the entire climate circus.

OOps, JamesG, you said DARK side. I agree, of course.

Jae
"It seems to me that the surface and the atmosphere directly above it are pretty much at equilibrium during the day. "

I don't believe I suggested otherwise. However I suspect that it depends on whether you are at the Arctic or at the tropics.

In fact, it seems to me that the atmosphere and the surface are normally close to equilibrium at all times, EXCEPT during the early morning hours, when the surface radiation finally cools it below the air temperature.

I just have difficulty perceiving any kind of equilibrium in a chaotic, spinning, vortexing, shearing gassy system :)

JamesG #250

I think you put that rather well however this in your next post I don't quit agree with:

I’d prefer to think that it is so obvious to them that the atmospheric to surface heating happens on the dark side of the planet that they don’t bother to even mention it. It is obvious after all!

Even at night the back radiation is less than the upward radiation the difference is because of the radiation transmitted through the atmospheric window. The air has to be quite a deal hotter than the surface to compensate that.

Jan
You'll notice I didn't mention back-radiation did the surface-heating. This ties in with my previous comments about radiation not being too important and convection usually dominating at low temperatures. If the atmosphere is warmer than the earth then it's convection that's warming the surface, not radiation. There's just too much focus on radiation by people who've never had to compare their calculations to real life. I think they get carried away with the T^4 term in Stefan-Boltzmann and think "ooh a power of 4; it must dominate". But it just plain doesn't dominate until you get to appreciably large temperatures.

JamesG starts out wrong and continues wrong all through:

Clearly you do planetary calculations using 1D equation and ignoring convection.

Clearly you don't know what you're talking about. How the HELL did you get the idea that I'm "ignoring convection?" I told you I wrote radiative-convective models of planetary atmospheres. Do you know what "convective" means? Hint -- it has something to do with convection.

JamesG, continuing to live in a fantasy world, babbles:

I don’t really assume that they think that a cold source can heat a hot one - it was BPL who suggested that.

And every physicist since the problem began to be seriously studied in the 19th century. Let's review the definition of how the greenhouse effect works from some real scientists. First, Henderson-Sellers and Robinson (1986, p. 60):

"The term greenhouse effect had passed into the literature to denote the process in the atmosphere whereby solar energy passes almost unimpeded to the surface, creates a warm surface which emits longwave radiation which in turn is absorbed in the atmosphere only to be reradiated back to the surface. The net effect is to maintain the surface at a higher temperature than would occur if the atmosphere were as transparent to longwave radiation as it is to shortwave energy."

Sellers (1965, p. 43):

"In the absence of atmospheric counter radiation the earth's surface would be 30 to 40 {deg}C cooler than it is."

Petty (2006, p. 142):

"...we expect the surface temperature of the earth to be substantially warmed by the absorption and re-emission of longwave radiation by the atmosphere. This warming effect due to the presence of an atmosphere is commonly known as the Greenhouse effect."

Hmm, four scientists all of whom agree that back-radiation from the atmosphere heats the surface, and I didn't even have to get up from my desk to pick up the reference books involved.

I’d prefer to think that it is so obvious to them that the atmospheric to surface heating happens on the dark side of the planet that they don’t bother to even mention it.

Lots of crackpots would prefer to think that the authorities they envy secretly agree with them.

It is obvious after all!

Wrong, though.

I’ve only seen tabletop physicists on the web come up with that perpetual motion argument; though enough of them to make me unfairly generalize.

So you haven't actually read any of the relevant literature, have you? BTW, there's nothing perpetual-motion about it. I'll ask you what I ask Jan, and which he never answered -- if heating of the surface violates the second law of thermodynamics, please compute by how much and give me the quantity of the excess in Joules per Kelvin.

JamesG writes (for the second time):

I said that it was obvious that for one side of the planet (the dark side) the atmosphere must be hotter than the surface, therefore must heat it.

No, James, the entire dark side of the Earth does not have an atmospheric temperature inversion. This effect exists only in your mind. Fourier in 1824 had a clearer picture of how the night sky and the ground interact than you do. You might want to pick up his book some time. You have to start somewhere.

BPL
Why don't you slink back to the self-help group at realclimate where you can collectively fling insults, feel self-righteous for no good reason and get Stefan-Boltzmann wrong together. At least Arthur took it on the chin when I pointed out his basic errors but you just keep shaking your rattle like an angry child.

Here's another idea. Go to one of the 1000's of real scientists who you say agree with your ideas and tell them that you think back-radiation means that heat flow can go from cold to hot but some other crackpot thinks that the heat flow from the atmosphere to the surface happens at night-time and in the morning when the atmosphere is actually hotter than the surface. If by some miracle anyone thinks your idea is more plausible than mine then bring him here and we'll argue with him instead of you. You're just too unpleasant.

JamesG, continuing his reign of error, writes:

You’ll notice I didn’t mention back-radiation did the surface-heating. This ties in with my previous comments about radiation not being too important and convection usually dominating at low temperatures.

Let's see, the Earth warms by 168 watts per square meter from direct sunlight and 324 W/m^2 from atmospheric back-radiation. It COOLS by 24 W/m^2 sensible heat (pure conduction and convection), 78 W/m^2 latent heat (evaporation), and 390 W/m^2 radiation. So in both heating and cooling, radiation is the dominant mechanism.

A recent revision by Trenberth et al. (2008) gives slightly different figures, but the same story -- Illum 161 W/m^2, Ed 333, Lsens 17, Llat 80, Fs 397.

jamesG fumes:

Why don’t you slink back to the self-help group at realclimate where you can collectively fling insults, feel self-righteous for no good reason and get Stefan-Boltzmann wrong together.

Stefan-Boltzmann is too simple to get wrong, James. It's just a simple product. Flux equals emissivity times S-B constant times absolute temperature to the fourth power. Even you should be able to use it.

At least Arthur took it on the chin when I pointed out his basic errors but you just keep shaking your rattle like an angry child.

Militant ignorance does tend to make me angry.

Here’s another idea. Go to one of the 1000’s of real scientists who you say agree with your ideas and tell them that you think back-radiation means that heat flow can go from cold to hot but some other crackpot thinks that the heat flow from the atmosphere to the surface happens at night-time and in the morning when the atmosphere is actually hotter than the surface. If by some miracle anyone thinks your idea is more plausible than mine then bring him here and we’ll argue with him instead of you. You’re just too unpleasant.

If you can't stand the heat, whether radiative or convective, stay out of the kitchen. Too bad about you if you feel insulted. You were rude first, pal. You came in here insulting scientists left and right, accusing me of not understanding simply physical laws (when I have a degree in physics and you don't), and generally acting snide and superior in your each and every post. I'm polite to people who are polite to me, and to the scientists who bust their ass every day trying to figure out how some aspect of nature works, instead of bloviating about it on blogs. You act like a jerk, you'll get treated like a jerk.

Just in case anyone thinks I'm making it up when I talk about JamesG's conduct, let's examine a few quotes from this thread. #204:

you can easily fool “smart” scientists because they help you out by seeing just exactly what they want to see and ignoring logic, arithmetic, rationality and physical laws.

#207:

a lot of scientists talk a lot of crap some of the time and certain scientists all the time. And realclimate.org... just see the hot-spot, feedback and aerosol hand-waving arguments to see many examples of illogicalities, irrationalities and mutually-exclusive, contradictory arguments. That site is a virtual crap mine.

#222:

Nor does it surprise me in the least that the majority of climate scientists cannot think in 3d...

Few people actually bother to argue like BPL, that a cold substance can heat a hot substance. That is a perpetual motion argument and anyone should feel silly arguing it.

#250:

Now in fact I gave you a good out from your illogical position...
It’s merely your own confusion that leads you to argue for perpetual motion (cold flowing to hot) rather than accept a more rational and correct argument.

#253:

Many loudmouth modelers (you know who I mean) seem to be well aware of model limitations and their underlying, biased assumptions but feel their overriding messianic mission makes it ok to say that the models are good enough to base policy on.

#263:

Why don’t you slink back to the self-help group at realclimate where you can collectively fling insults, feel self-righteous for no good reason and get Stefan-Boltzmann wrong together. At least Arthur took it on the chin when I pointed out his basic errors but you just keep shaking your rattle like an angry child... You’re just too unpleasant.

Rational scientific speaker or hostile, self-important crackpot? You be the judge!

Bart the terrible:

"Stefan-Boltzmann is too simple to get wrong, James. It’s just a simple product. Flux equals emissivity times S-B constant times absolute temperature to the fourth power. Even you should be able to use it."

You just got it wrong again. You forgot the emissivity, and that is VERY critical. It may even explain why you are so wrong on how the surface gets heated.

Barton
I'm only telling you what anyone who does heat transfer calculations for a living - and has to get it right - would also tell you. How do I know your capabilities? Because you've advertised them here. The fact that you didn't know the correct difference form of Stefan-Boltzmann tells me everything I need to know about knowledge of heat transfer. The four points I mentioned above are graduate level errors which are ironed out by experience of working with real heat transfer models. Go talk to someone who does real radiative 3d heat transfer modeling and tell them your ideas about cold body radiation causing hotter items to get even hotter. They'll laugh at you.

I'm not pushing my convection versus radiation theory on anyone - you can believe what you like. Number-wise it doesn't actually matter where the heat comes from. It's only you who has a bee in your bonnet about it. It doesn't even matter. I don't expect that any climate scientist will realize that convection dominates at low temperatures until they talk to some people who know about these things. Certainly none of the climate models seem to include it yet. But yes Judith Curry did say on climateaudit that there was a growing realization that they should. When they do bother with convection they'll get a few revelations - probably in about 5 years.

Bart: If you haven't done so, I suggest you read the paper by Gerlich and Tscheuschner: http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1...

G&T are very well-respected and famous physicists.

Some exerpts:

"Rather, the atmospheric Greenhouse mechanism is a conjecture, which may be proved or
disproved already in concrete engineering thermodynamics [95{97]. Exactly this was done
well many years ago by an expert in this eld, namely Alfred Schack, who wrote a classical
textbook on this subject [95]. 1972 he showed that the radiative component of heat transfer
of CO2, though relevant at the temperatures in combustion chambers, can be neglected at atmospheric
temperatures. The in
uence of carbonic acid on the Earth's climates is denitively
unmeasurable [98]."

"It cannot be overemphasized that a microscopic theory providing the base for a derivation
of macroscopic quantities like thermal or electrical transport coecients must be a highly
involved many-body theory. Of course, heat transfer is due to interatomic electromagnetic
interactions mediated by the electromagnetic eld. But it is misleading to visualize a photon
as a simple particle or wave packet travelling from one atom to another for example. Things
are pretty much more complex and cannot be understood even in a (one-)particle-wave duality
or Feynman graph picture."

" Second law of thermodynamics:
{ Heat cannot move itself from a cooler body into a warmer one.
{ A heat transfer from a cooler body into a warmer one cannot happen without compensation.
A ctitious heat engine which works in this way is called a perpetuum mobile of the
second kind.
Clausius examines thoroughly, that the second law is relevant for radiation as well, even if
image formations with mirrors and lenses are taken into account [178, 179]."

And while we're talking about books, here's one you can find on Google books:
An Introduction to Continuum Mechanics By J. N. Reddy

6.4.1
...Radiant Energy exchange between surfaces or between a region and it's surroundings is described by the Stefan-Boltzmann law, which states that the radiant energy transmitted is proportional to the difference of the fourth power of the temperatures of the surfaces. The proportionality parameter is known as the Stefan-Boltzmann constant...

And it gives a nice equation showing the difference too. Of course any textbook on heat transfer will tell you the same thing. Why yes it is indeed simple, which makes it even more unbelievable that someone with a degree in Physics can get it wrong. Ah so Physics isn't turned upside down after all - using the correct equation tells us Barton is wrong. Rational behavior would involve being embarrassed at such an obvious error. Not Barton however - he of course knows better. Incidentally the other two members of realclimate (Gavin and Rasmus) got this simple equation wrong in 2 other different ways. You can plunder the realclimate archives for those. So, simple or not, there are at least 3 ways for apparently smart people to be wrong on even such a simple equation.

So the points I made above about with my remarks about "some scientists" is once again proven by BPL - even apparently clever people will persuade themselves that something utterly illogical is true if it gives them the answer they want. It's not an insult - it's a mere observation.

JamesG #259

I think they get carried away with the T^4 term in Stefan-Boltzmann and think “ooh a power of 4; it must dominate”.

I think they just never make the comparison, and forget just how small the S-B constant makes the resulting resulting radiation.

The real difficulty is in making it clear to someone who does not want to know. A little while back Barton was talking about being able to measure back radiation with pyrgeometers.

The point I've been trying to make is that they can't do that all they can measure is the difference between the upward radiation (from the body of the instrument) and backward radiation to the thermopile. One can go out and buy a cheap non-contact thermometer and see the effect for themselves. The chief differences between a cheap thermometer and a pyrgeometer are the filter and the calibration factor.

One thing I have a great deal of difficulty in understanding, even though I work in mental health, is why some people get angry and angrier with the increasing evidence coming in that tipping points and runaway temperatures are unlikely instead of being relieved.

Lastly, no I didn't search for Fourier's ideas about temperature inversion. I simply reasoned that Earth's dark side without an atmosphere would be about as cold as the dark side of the moon (-110C). So if no atmosphere makes for a cold surface and an atmosphere makes it warmer then the atmosphere must be warming the surface on the dark side. Too simple?

JamesG #271

There is this too scroll down to Net Radiation Loss Rate.

Bart: Here's a story about a famous Princeton physics professor, who is an expert on radiation, who disagrees with your "concensus." http://icecap.us/index.php/go/political-climate

JamesG #272

Too simple?

It is a bit. While on the dark side (midninght) I actually did some measurements.

Surface temperature 25C
air temperature (~1.5m) 22.5C
With unfiltered uncalibrated radiometer estimated upward radiation ~444W/m^2
Same tool back radiation (overcast) 401 W/m^2
Then with big hole in clouds 359 W/m^2
Body of the radiometer was 26C

Expected BB radiation from a 22.5 body is 433W/m^2

I would make too much of the the absolute values but the relative values are interesting i.e.
t4surface > t4cloudy > t4clear

it's just preliminary stuff I've being doing prior to doing some surface/near surface measurements over 24 hour periods.

jan, 275:

Dry surface, correcto?

jae #276

Mostly dry we had rain a couple of days ago so not exactly desert conditions. There wasn't any dew about. Surface had only thin patchy cover. Next time I do a night shift I might take the stuff with me and measure closer to dawn.

I should mention (remind) the moon's surface has longer to cool than the earth's so an airless earth might not get as cool (or as hot) as the moon's. There was also a slight breeze about and the chill factor was such that I would have guessed a lower temperature for the air than 22.5 (about 19-19C).

Jan:

"Next time I do a night shift I might take the stuff with me and measure closer to dawn."

Good. I would bet on a reversal at dawn. That's why frost forms.

Oh, frost AND dew.

jae #279

Good. I would bet on a reversal at dawn. That’s why frost forms.

I have done some prelims here too.

I am particularly interested in temperature inversions.

Yes I agree the temperature wouldn't be the same as the moon's and that's likely not nearly a correct temp for the moon either. I'll stick with "it would be a lot colder".

I guess Barton isn't arguing too differently to us but for me that difference calculation in S-B controls the direction of heat flow while he seems to separate it out as 2 directional heating, one greater than another. For me that's like saying some of the river flows uphill but most of it flows downhill.

I got some support for my night time hot spot here at the Uni of Utah:
http://ocw.usu.edu/Forest__Range__and_Wildlife_...
where it says:
What causes these differences in temperature lapse rates in the atmosphere? Well, there are several factors that contribute to varying temperature distributions, but the most important is heating and cooling at the earth's surface. (See graphic above) As the earth loses its heat at night through radiation, the air in contact with the ground also cools. Conduction becomes an important process here, as air at lower levels cools faster than the air above. This creates a condition of cool, heavier air below warmer air. The layer of cool air above the surface deepens as the night progresses. In mountainous terrain, this condition is even more pronounced in the valleys, as cool air drainage from the slopes above helps to deepen the layer of cold air. The deepening of the cool air layer is also affected by the amount of cloud cover at night. Clear nights cool faster than cloudy nights. Warmer air just above a cool air layer creates a very stable air condition. Smoke or any other parcel of air that is forced to rise will stop when the warmer air is reached. This inversion layer discourages vertical air movement.

"Conditions usually begin to reverse after sunrise. As the earth's surface is heated by solar radiation, it warms the air in contact with it and above it by conduction and convection. The stable air at lower levels warms until it is no longer colder than the air above, and the temperature lapse rate approaches the dry adiabatic rate. Inversions usually disappear sometime before noon as unstable conditions continue to develop. Air near the surface is not much warmer than the air above, thus making it buoyant and able to rise. Any lifted parcel of air will continue to rise until it reaches a level of equal temperature. A smoke column could rise many thousands of feet."

Also on the Wikipedia discussion on this, where there was a very similar argument to this one, the final word seems to have gone to this comment:
"It is actually simpler than that. In the morning the ground is colder than the atmosphere a few hundred feet above. (See Lapse Rate - Plot and Definitions, drag the zoom controls in the lower panel to see the morning temperature inversion.) Thus, every night heat flows from the atmosphere to the ground. This keeps the ground from getting as cold as it would without the atmosphere and results in an increase in the average surface temperature. The confusion results from applying static equations to a dynamic problem."

The lapse rate comment refers to this page here containing a lapse rate applet that I, alas, can't get to work on my browser:
http://mc-computing.com/Science_Facts/Lapse_Rat...

Hope that helps. With all due respect to the genius Fourier (who may have been misinterpreted by BPL), we have actually learnt a few things since his time.

jae, who apparently can't read, posts:

“Stefan-Boltzmann is too simple to get wrong, James. It’s just a simple product. Flux equals emissivity times S-B constant times absolute temperature to the fourth power. Even you should be able to use it.”

You just got it wrong again. You forgot the emissivity, and that is VERY critical.

jae, "emissivity" is the 17th word in the paragraph you quote.

jae quotes that famous vaudeville duo, Gerlich and Tscheuschner. jae, check here:

http://rabett.blogspot.com/2008/03/formal-reply...

If they were ever "respected" physicists, they are respected no longer, any more than William Shockley, William Turner, or Thomas C. van Flandern is, and for the same reason. They are not, by the way, climatologists.

JamesG, trembling on the verge of a great discovery, writes:

Lastly, no I didn’t search for Fourier’s ideas about temperature inversion. I simply reasoned that Earth’s dark side without an atmosphere would be about as cold as the dark side of the moon (-110C). So if no atmosphere makes for a cold surface and an atmosphere makes it warmer then the atmosphere must be warming the surface on the dark side. Too simple?

And the atmosphere warms the surface how, James? You're almost there, if you think about what you just said a bit.

James writes:

With all due respect to the genius Fourier (who may have been misinterpreted by BPL), we have actually learnt a few things since his time.

You haven't. You still don't understand what Fourier wrote about the subject in 1824, since you're still insisting that back-radiation can't warm the ground, or perhaps as Jan and jae argue, that it can't exist or is some kind of mirage.

Hmm, looking more carefully at what he wrote, James apparently still believes that the atmosphere is always warmer than the surface at night.

For those who care about what thermometers say: It isn't. Unless there's an inversion. I don't think more than 10% of the Earth's surface, if that, is ever in ground-atmosphere inversion at any one time. Probably more like 1%.

But that old atmosphere warms the ground at night anyway. Amazing.

During the day, too.

Bart #285

Except that it’s not the correct equation, since, as I’ve said approximately a billion times, I’m considering back-radiation in isolation, not net radiation.

HOw many times and different was do you need to be told and shownthat it's a mistake to do so before it finally sinks in?

JamesG #282

For me that’s like saying some of the river flows uphill but most of it flows downhill.

I drew a parallel between heat flow and electric current even gave an equation of the same form for a voltage divider all to no avail.

however I think that with the exception of the odd temperature inversion (I don't see all that many where I live) that whether or not the atmosphere warms the surface at night will depend on latitude and season as much as the time of day.

Then we have to mess about with averaging to get the "static" atmopsheric results.

BPL, 283, apologies for the oversight.

BPL, 284, the rabbit is rabid, and like you, not open to anything that might go agaqinst the phony (and totally unexplained) "physics of crisis." The guy is basicly an obstructionist, IMO.

BPL, 287: "During the day, too." LOL. So the temperatures during the day in Scottsdale during a summer day would be over 100 C, at least, according to your nonsense (add solar insolation of 1000 wm-2 to whatever you think the "backradiation" is during the day). :)

Bart: I forgot a point:

You say "They are not, by the way, climatologists." Thank God!. But, snarks aside, can you give me a definition of that profession? What are the requirements to be a "climatologist" in your mind? Does a "climatologist" have to understand at least as much about the atmosphere as a meterologist or a physicist? Is a "climatologist" basically a political scientist?

jae #291

“They are not, by the way, climatologists.” Thank God!.

He forgot to mention that neither Arthur Smith nor the bunny are either.

BPL
Well Barton I doubt we'll ever agree but I'll try to achieve some rapprochement and I'll try to be more respectful. The main thing I contest is that heating, by definition, implies that a hotter item heats a cooler item. The negative part of the Stefan-Boltzmann difference equation refers to a surrounding temperature, not the surface of the planet - hence even highly locally the heat flow must be always hot to cold. If you were to describe the whole effect as a prevention of cooling then we might even agree on that point. I don't think I'm being contentious there because everyone else seems to prefer that definition as being more elegant. There is in fact no contention in saying that the atmosphere traps the heat and that prevents the surface from cooling. I only say that the simplified cartoon of Trenberth is misleading in showing a heating (ie hot to cold) process that doesn't actually occur, though in terms of net Joules it may still be correct and maybe that cartoon helps some people understand the numbers involved in this cooling constraint better. For something closer to reality I prefer the NASA cartoon as it makes more physical sense. However all cartoons are merely a simplification of a time-dependent, geometry-dependent problem. While I prefer to think of things in terms of the real situation - that of a rotating orb heated on one side. You however seem to argue from the standpoint of a spatially and temporally averaged situation which to me is illogical.

On night-time heating, if you'd read properly this part above, which was a quote from elsewhere; "This creates a condition of cool, heavier air below warmer air. The layer of cool air above the surface deepens as the night progresses" then you'd realize that I'm not actually out on a limb with saying the atmosphere is hotter than the surface at night: It seems to be accepted fact. Whenever something is hotter than something else it must therefore heat the colder part. And apparently that is the case for the entire night. Of course the hot part of the atmosphere will be losing heat too to the cooler part above it, but it will still be heating (ie hot to cold) the surface. I'd contend that such night-time heating would be mainly by convection rather than radiation - which I'd have thought was obvious.

Back-radiation as a term (meaning cold items heating hot ones) does seems to me to be utter nonsense. I'd rename it "cold re-radiation". I don't rule it out because nature is a funny beast but as we've never observed such a phenomenon on Earth, it's likely only a simplified abstraction of reality which helps some people and confuses others. On that we'll continue to disagree.

That said i suspect we'll continue to disagree.

Let's examine two objects in isolation, each with an internal heat source. They are granite blocks, each one meter square, with 10 meters distance between them. They radiate only to each other; their backs are insulated and their sides are too thin to matter. Each is a perfect blackbody, their absorptivity and emissivity is a = ε = 1.

[ A ] => <= [ B ]

Start off with A at a temperature of absolute zero. Each of these objects being perfect blackbodies, they radiate power (infrared light) by the Stefan-Boltzmann law:

F = ε σ T^4 (1)

where F is the flux density in watts per square meter, ε the emissivity, σ the Stefan-Boltzmann constant (5.6704e-8 W/m^2/K^4) and T temperature (K). A radiates zero watts per square meter, B radiates 1 x sigma x 250^4 or about 221.5 watts per square meter. Let's say B is supplied continuous power in this amount by a resistance coil inside it, which in turn is plugged into a wall outlet. Energy is conserved, so it doesn't cool down or heat up.

The inverse-square law is:

I = F / r^2 (2)

where illumination I is in watts per square meter, luminosity (source flux density) F is in W/m^2, and distance r is in meters. The blocks are separated by ten meters. A therefore receives 2.215 watts per square meter from B, which heats it up slightly. Its temperature rises to (2.215 / sigma) ^ 0.25 or 79.1 K. A perfect blackbody, it radiates 2.215 W/m^2 right back at B.

B is a perfect blackbody. The distance between A and B is ten meters. B receives 0.002215 watts per square meter from A as well as 221.5 from its internal power source. Its incoming energy is now greater than its output, so it must heat up. Its new temperature is (221.502215 / sigma) ^ 0.25 or 250.000625 K. Not a great difference, but A has succeeded in raising B's temperature just a bit.

Let's give A an internal heat source, like B's, so that its initial temperature is 200 K. This requires a power source of 90.7 watts. A, being a blackbody, has to radiate this much.

New radiation budgets for each object:

A:
Input 90.7 watts internally, 2.215 watts from B, total 92.915
Output 90.7 watts.
Temperature 200 K to start.

B:
Input 221.5 watts internally, 0.907 watts from A, total 222.407 watts.
Output 221.5 watts.
Temperature 250 K to start.

Each is taking in more than they are radiating, so they will have to heat up and radiate more. Recalculate with temperatures based on total input, output rises due to increased thermal radiation, etc. An iterative model (code on request) stabilizes with the following budgets:

A:
Input 90.7 internal + 2.224 B = 92.924
Output 92.924
T 201.2 K

B:
Input 212.5 internal + 0.929 A = 222.429
Output 222.429
T 250.3 K

So A managed to raise B's temperature even though A is cooler than B. How is this possible?

It's possible because B doesn't "know" that A is cooler than it is. Radiation is radiation. B absorbs the infrared photons from A, they increase the internal energy of B's molecules, those molecules increase their root-mean-square velocity through the laws of statistical mechanics, and B increases its temperature.

Now, J, J and J maintain that A can't radiate to B. Back-radiation, they say, violates the second law of thermodynamics, which states that "heat cannot flow from a cooler body to a warmer one."

Except that it doesn't state anything of the kind. What is states is that "NET heat cannot flow from a cooler body to a warmer one."

What is the net heat flow in this situation? It's 2.224 watts from B to A, minus 0.929 watts from A to B, for a net heat flow of 1.295 watts from B, the warmer body, to A, the cooler body. Nothing is being violated, but cooler A is still radiating to warmer B and raising its temperature. In the absence of the cool block A, the warm block B would be at a temperature of 250.0 K. In the presence of the cool block A, B's temperature rises to 250.3 K. Radiation from a cooler source is heating a warmer source.

And that's how the greenhouse effect works. The warmer Earth is receiving radiation from the warmer sun, and a lot more radiation from the cooler atmosphere. In the absence of an atmosphere, if the same albedo were retained, Earth would be at a temperature of 255 K. With an atmosphere providing back-radiation, its temperature is 288 K. That's the greenhouse effect, and that's how it works.

The warmer Earth is receiving radiation from the warmer sun, and a lot more radiation from the cooler atmosphere.

There you go again heat flowing up a temperature gradient, like water flowing uphill.

James, I appreciate your effort to make peace. I really have been letting posts on this blog get me angry, which is stupid and counter-productive. I still think you're badly mistaken, and you clearly think the same of me, but I'll try to be more polite about it.

On whether any temperature inversions are net positive - don't forget to also include for the tropical hotspot from the moist adiabat. While I dispute the magnitude of that hotspot only because the temperature measurements say that it isn't as significant as the models predict, there is no denying the moist adiabat should exist across a large swathe of the widest part of the planet, completely independently of GHG's. Furthermore there is a delicious irony in Barton invoking the satellite temperature measurements when it suits him yet he like many others in the realclimate tribe dispute those same measurements when they don't reflect what he expects. "For those who care about these things" - yes indeed! As usual in this debate we have a total lack of consistency from the pro-AGW camp.

Barton
Jan said that he didn't believe the 2nd law was being broken. I said it too. As far as G+T are concerned I'm sure they just reject the notion of back-radiation, not the net effect. I'm not even sure the second law applies to chaotic systems and there are many oddities in nature that cause flow to the opposite direction than they should. Hence in the sense of "net" - we all agree. But imagine instead of your 2 bodies radiating, you have a string of molecules from the surface to the atmosphere. Each one radiating more than the one above it and each also being subject to convective flow from all directions. Now that is a bit different than the simplified system you describe. And the question is - does that lower energy molecule radiate towards the higher energy molecule below it? If it does then what does quantum mechanics have to say about it? I'm aware that Stefan-Boltzmann is an approximation to the real quantum packet movement and that remains a mystery to most of us: we cannot even separate it out as particles or waves. However I really don't see why it so important to you whether it is heating or prevention of cooling. As I said, heating, by definition, is from hot to cold; only the net effect counts and the net effect is heating upwards. The separation in the Trenberth cartoon is likely only there to show the numerical importance of the GHG effect, not necessarily to show a heating effect from cold to hot.

I remember a similar argument about aerofoils when one person said it's not the bottom air pushing, it's the top air sucking and my then supervisor said "who the hell cares - it's the difference that counts". Incidentally, in that research project we managed to prove that the Bernoulli effect on wings could be broken by inducing vortices, so I'm not averse to the breaking of the odd law if there's a good explanation. Back-radiation just isn't good enough for me.

James G:

"Incidentally, in that research project we managed to prove that the Bernoulli effect on wings could be broken by inducing vortices, so I’m not averse to the breaking of the odd law if there’s a good explanation. Back-radiation just isn’t good enough for me."

Interesting. I read somewhere recently that the Bernoulli effect is now not considered the main cause for lifting, but only a relatively minor secondary effect. Is that true?

Bart: besides the obvious straight physics that states that colder bodies cannot heat warmer ones, I don't know how to get around Thieme's observation that the "radiation flows" in the K&T-type cartoons are vector quantities (direction and magnitude), and they therefore cancel each other.

It's about angle of attack too, otherwise you couldn't fly upside down. I've not heard of another effect. I had to move onto another project so I never got that wing design through the wind tunnel. I might return to it when I retire.

Jae
This is where I went wrong - looking for some sense in the cartoon. So I went looking for extra Joules from atmosphere to surface. I'm still sure there are a lot there that aren't counted but it's clearly not what Trenberth meant. However I see now that all Trenberth did was use the classic Stefan-Boltzmann, like Barton above, to get bi-directional heat flow. This does seem to be the standard way that Physicists explain the greenhouse effect. Yet radiation in all directions regardless of temperature would need an upward arrow too. So it remains just an artificial split-up of the heat flow that heat transfer specialists just wouldn't do. At least I got one realclimate devotee to admit that instrument measurements must be the final arbiter of a theory so maybe there was a point to all this.

#301: I think its common knowledge that the proportion of lift attributed to upper surface curvature vs attack angle varies according to design. Jets are mostly lifted by attack angle, but slow moving wings by the curvature.

Jan Pompe writes:

There you go again heat flowing up a temperature gradient, like water flowing uphill.

Jan, temperature gradients are only relevant to heat transfer through conduction or convection. Radiation can go anywhere it damn well pleases as long as the medium is transparent or translucent.

JamesG writes:

the question is - does that lower energy molecule radiate towards the higher energy molecule below it?

Yes. Actually, a single molecule at one time can radiate in pretty much any direction, and will, at random. 50% of the time that will be "down."

If it does then what does quantum mechanics have to say about it?

It tells you what wavelength(s) the molecule will radiate at. Absorption/emission bands are a quantum phenomenon.

jae writes:

Interesting. I read somewhere recently that the Bernoulli effect is now not considered the main cause for lifting, but only a relatively minor secondary effect. Is that true?

Yes. Airplanes work mainly by shoving air down.

Bart: besides the obvious straight physics that states that colder bodies cannot heat warmer ones,

There is no such physics. You've got the second law of thermodynamics wrong.

I don’t know how to get around Thieme’s observation that the “radiation flows” in the K&T-type cartoons are vector quantities (direction and magnitude), and they therefore cancel each other.

They cancel each other because the system is in long-term thermal equilibrium, not because they're vector quantities. And the system would not be in equilibrium if no back-radiation were coming down from the atmosphere; the surface would cool rapidly until equilibrium was reached again.

Bart: LOL, you definitely belong in the AGW believer's camp.

Bart #304

Jan, temperature gradients are only relevant to heat transfer through conduction or convection. Radiation can go anywhere it damn well pleases as long as the medium is transparent or translucent.

Which is precisely why one should work with the net radiation (that is use the difference equation) to determine which direction the heat is flowing. Try not confusing heat with energy.

#306

They cancel each other because the system is in long-term thermal equilibrium, not because they’re vector quantities.

Would like to try convincing Nick Stokes of that? I tried some time back without success.

Well, Jan, technically "heat" only exists in material bodies, so we aren't receiving any more heat from the sun than from the atmosphere. Are you sure you want to go there?

Bart #309

Well, Jan, technically “heat” only exists in material bodies, so we aren’t receiving any more heat from the sun than from the atmosphere. Are you sure you want to go there?

It doesn't really worry me Bart the state of NSW pays me to deal with people who suffer serious delusion, hallucination and confusion or have problems with anger management and impulse control.

You are still confusing heat with energy.

No, Jan, I'm really not. Thermal energy is a subset of energy in general, consisting of randomized molecular motion in material substances. I may use "heat" as a shorthand for "thermal energy" or even "radiation generating thermal energy," but it doesn't affect any of the arguments I have made here or elsewhere. In fact, the whole issue is a red herring.

What does your being an MH/MR professional have to do with it? Are you implying that I'm mentally ill because I disagree with you about global warming? Great argument.

Barton #311

I may use “heat” as a shorthand for “thermal energy” or even “radiation generating thermal energy,”

It is not helpful if you have and use your own private definition of things. Heat is energy in motion or energy transferring it is not what you use it as shorthand for. If we are talking about energy being transferred (i.e. heat) then we must talk in terms of net flux which only goes from warm to cool.

What does your being an MH/MR professional have to do with it?

What it means is that I'm perfectly happy to deal with private definitions such as your "shorthand" for thermal energy.

You did ask if I want to go there after all and I'm saying why it doesn't worry me.

Now that we are in that territory I have to ask: do you think the anger that you have displayed, (and admitted to at leats twice) when someone shows that CO2 is not the bogeyman you think it is, is entirely rational or objective?

Bart: you still haven't explained to me why I cannot melt steel with any number of propane/air torches, even though the wattage could easily "add up" by your procedure to a sufficiently high equivalent temperature.

Bart #314

Like I said I'm perfectly happy to deal with you private 'short hand' for thermal energy but you need to realise that your private definitions might not mean the same thing or anything to others, and like I said I'm quite used to that sort of thing. I am also quite patient I don't get angry over the silly things some people say and do.

Try sticking with the program. Hence when we say that heat flows from hot to cold we mean precisely that the net energy transferred is from hot to cold and if you are going to talk only about the energy in one direction only instead of the net energy transferred you are going to get it wrong.

Now the atmosphere when it is in equilibrium it will radiation exactly the same rate it absorbs it cannot radiate more what ever the surface radiates and absorbed wavelengths is replaced by the back radiation the net effect is zero energy transferred from the surface at those wavelengths.

The idea of back radiation warming the surface is a mirage because $$\delta T = \frac {\delta Q} {cm}$$ and δQ=0. Any warming or cooling that might occur at the surface with variations in optical depth will be entirely due to reduced transmission through the atmospheric IR window. Of course nay change in temperature will increase the radiation from the surface by $$dq = 4 \sigma t_s^3$$ for 288K that will be 5.42 W/m^2K.

Jan writes:

when we say that heat flows from hot to cold we mean precisely that the net energy transferred is from hot to cold

Right. But back-radiation is not NET energy. Why you can't grasp that simple point eludes me. I suspect it's an MH/MR thing.

The idea of back radiation warming the surface is a mirage because and δQ=0. Any warming or cooling that might occur at the surface with variations in optical depth will be entirely due to reduced transmission through the atmospheric IR window.

Nonsense. Only 40 W/m^2 gets through the atmosphere now. The greenhouse effect amounts to 155 W/m^2. Do the math.

Bart #315

Nonsense. Only 40 W/m^2 gets through the atmosphere now. The greenhouse effect amounts to 155 W/m^2. Do the math.

Even from this statement we can see the the total flux is 195W/m^2 and the 155W/m^2 is warming the atmosphere not the surface.

oops "is warming" -> "has warmed"

Sunlight alone: Surface is 254 K
Sunlight plus atmosphere: Surface is 288 K
Conclusion: The atmosphere helps to warm the surface.

Q.E.D.

Sorry, Jan, the climatologists and the physicists and the meteorologists are right and you're wrong.

Bart:

"Sunlight alone: Surface is 254 K
Sunlight plus atmosphere: Surface is 288 K
Conclusion: The atmosphere helps to warm the surface."

No, not Q.E.D., but P.P.L (Piss-Poor Logic).

If A is true (254K), and B is true (an atmosphere), and C is true (288K), that does not mean A and B CAUSE C. It's possible, but not Q.E.D.

Bart #318

your first premiss is incorrect

Jan --I know the Earth's surface doesn't really have the albedo of the whole Earth system.

Let's assume that the Earth has no atmosphere, no ice caps, and no oceans. It would probably have an albedo around 0.15 rather than the present 0.306. At least that's the surface albedo figure used in most models.

The Earth's temperature from sunlight alone would then be 267 K, still below freezing. {[(1366.1/4)(1-0.15) / σ]^0.25} It would not be 288 K.

The argument stands. The Earth's surface is hotter than it would be without an atmosphere, and the reason the atmosphere makes the surface hotter is back-radiation.

There are only three methods of transferring heat: Conduction, convection, and radiation.

Conduction follows the thermal gradient, which is from warmer Earth to cooler atmosphere.

Convection moves warmth up and cold down. Convection can only warm the atmosphere and cool the surface.

That leaves radiation. Q.E.D.

Bart #321

The Earth’s temperature from sunlight alone would then be 267 K, still below freezing. {[(1366.1/4)(1-0.15) / σ]^0.25} It would not be 288 K.

Sorry but you have forgotten the emissivity which due to Kirchhoff's law will be equal to 1-0.15 so we have $$ \left(\frac {1366\left(1-.15\right)}{4\left(1-.15\right) \sigma}\right)^{.25} = 279K$$ which is the black body equilibrium temperature. You keep leaving bits out in what appears to me to be a selective inattention to detail.

The reason the world gets hotter is not the back radiation that because that implies that there is more energy in the system than came in which is a violation of the first law and that does not happen. You can work this out with simple arithmetic.

The reason for the rise in temperature is secondary to absorption /remission of upward radiation the ratio of that is always 1 but the ratio of that amount that is absorbed to that transmitted through the IR window may vary. If you decrease the amount of radiation through the window the temperature rises to compensate. Like most second order effects it is weak and there is a cost, and that cost is that is radiation output from the planet decreases so you get a higher temperature at the surface and a lower effective temperature of the planet and the equilibrium temperature remains unchanged.

There are two analogies you can compare this with one is the garden hose where if you stick your finger on the end the pressure in the hose rises but the volumetric output rate decreases. Or the electric potential where if you increase the load resistance the voltage across the load increases but the current through the circuit decreases.

I'll leave it to you to do the actual sums. Let me know if you run into trouble.

Jan, the temperature with an albedo of 0.15 can't be the same as the blackbody temperature. By definition, a blackbody has an albedo of 0.00.

Oh, and the absorptivity/emissivity for an opaque body will be the complement of the albedo, not the albedo itself.

Oh, and did I mention that the emissivity for infrared light will be different from the absorptivity for visual light?

Bart #323 #324 #325

Kirchhoff's law and absorptivity = 1 - albedo and it's not wave length dependent even for fluorescent bodies when the body in equilibrium energy absorbed = energy emitted so integrate ove the entire spectrum and Kirchhoff's law will hold. Try it sometime.

Sorry, Jan, but Kirchhoff's Law only holds for a particular wavelength. The visual absorptivity of a planet and its infrared absorptivity will almost never be the same. "Planets aren't gray." For the Earth's surface, the albedo is about 0.15, for an absorptivity of 0.85, but the IR absorptivity/emissivity is 0.95-1.00, depending on whose model you use (K&T still favor 1.0, I usually use 0.95 in my semigray models).

Bart #327

Sorry, Jan, but Kirchhoff’s Law only holds for a particular wavelength.

Who told you that nonsense ? Even for fluorescent media where it absorbs at one frequency and emits at another it must hold. It is essentially a restatement of the first las of thermodynamics.

If a body is in thermal equilibrium it's temperature isn't changing so it's thermal energy content isn't changing. Mathematically $$ \frac {dT}{dt} = \frac 1 {mc} \frac {dE}{dt}=QA=0$$ where mc is the thermal capacity, E is energy content T the temperature and t time thus Q is rate of heat transfer in watts/m^2 and A area in metres.

Remember the radiation heat transfer equation

$$Q= \sigma \left (\alpha_cT_h^4 - \epsilon_c T_c^4\right) = 0$$ on equilibrium. Subscript h indicates heat source and c indicates the sink.

So at equilibrium $$T_h = T_c$$ and $$\frac {T_h^4} {T_c^4}=\frac {\epsilon_c} {\alpha_c} = 1 $$ and there is no dependence on wavelength in any of these equations

My son and I repeated the experiment as mentioned and we the same results. We then used two glass jars, one as a control and one with water vapor and got the same results. We tried the two jar experiment again, but his time we stayed indoors and used a heat lamp and got the same results. In our fourth experiment we use one jar as a control and added vinegar and baking soda to the second jar to produce CO2. After and hour into the experiment we added even more vinegar and backing soda to create even more CO2 and yet again the temperature did not increase. The mean control jar temperature was 34.87 while the experimental jar was 35.43. The mean humidity for the control was <20% (we could not measure blow 20%) and the mean humidity of the experimental jar was 42.73%.

Hey, JQ, please present the actual data. Somehow, I think this kind of data are important to the overall "mystery."

jae, I don’t know much about climate, my area is physiology, but I thought it would be fun to try this with my son. We happened to find this site while searching for ideas for my son’s science fair project. We did 5 experiments and I would be more than happy to share the results (we just finished the last one about an hour ago). We have a lot of data and I’m not sure where to post them. I could e-mail them to you if you like or tell me how to post them here. It would be great to have some criticism by people that know more than I do.

Hi JQ Public. Its great to hear about your efforts to do science with your son. I have some other results yet I need to write up, done with the tile and reaching over 120C in temperature. What would be good to would be to write an equation that predicts these temperatures. I hope in the near future to do this in a post. Then there would be a nice set of experiments for such situations. Cheers

David #332

What would be good to would be to write an equation that predicts these temperatures.

you might find this comparison of insolation models a good start.

Ugh. a Dozen different models. I was think more along the lines of making rigorous an informal relationship like:

Max temp = Black body temp + Ambient - Losses(temp)

David #334

That's a different POV entirely.

However one can see from that link just how varied opinion (i.e. messy) can be regarding how to calculate how much solar heating can be expected in place for which there is no data.

Jan writes:

Sorry, Jan, but Kirchhoff’s Law only holds for a particular wavelength.

Who told you that nonsense ?

Every book on radiation physics I've ever read (want me to list some quotes?). I've posted the quotes before. You're wrong again, Jan; not every body is gray. Kirchhoff's Law only holds for a given wavelength or for a gray body. There is no a priori reason to think absorptivity is the same at one wavelength as it is as another. You are essentially denying that spectra exist -- absorption spectra, emission spectra, any kind of spectra at all.

Bart #336

Every book on radiation physics I’ve ever read

In that case you should be able to give a rigourous proof that the first law of thermodynamics is wrong and energy is not conserved.

Congratulations.

You’re wrong again, Jan; not every body is gray. Kirchhoff’s Law only holds for a given wavelength or for a gray body.

Are you saying that coloured or fluorescent bodies are capable of creating or destroying energy?

You'll have to prove that too.

Bart #336

You’re wrong again, Jan; not every body is gray. Kirchhoff’s Law only holds for a given wavelength or for a gray body.

I forgot phosphorescent bodies must also be capable of creating energy?

You'll have to prove that too. How's your quantum mechanics?

Jan, even you can't be this stupid. You really think the portion of electromagnetic energy absorbed by every material object is the same at every wavelength??? You can't be this stupid. It's some kind of internet game, isn't it, poking at people you dislike?

Bart #339

You really think the portion of electromagnetic energy absorbed by every material object is the same at every wavelength???

You think that is what I said or are you setting up a strawman?

If Jan's idea were true, absorption spectra would not exist. There would be no wavelengths at which a gas or a liquid (or a solid) absorbed electromagnetic energy more strongly than at other wavelengths. All those spectra you see in astronomy books wouldn't have any of those dark lines. They would all be nice rainbows where one color gradually blended into another. Astronomers would never have been able to detect the cosmological red shift, or any other red or blue shift, because no lines would exist to measure the shift of. We wouldn't know the universe was expanding. Radial velocity techniques wouldn't have been able to discover exosolar planets. We wouldn't know about spectroscopic binary stars. The Doppler effect would only apply to sounds.

Jan is either playing some kind of head game or he is really devoted to militant ignorance.

Oh, and there wouldn't be any such thing as a band model, or a correlated-k distribution. The Lorentz and Doppler and Voigt line profiles wouldn't exist, because there would be no lines to apply them to.

I could go on like this all day. Jan wants to eliminate much of what we've learned about radiation physics since the early 19th century.

Bart #341

If Jan’s idea were true, absorption spectra would not exist.

Oh really integrating over the entire spectrum causes it to cease to exist?

How does that work?

Jan writes:

You really think the portion of electromagnetic energy absorbed by every material object is the same at every wavelength???

You think that is what I said or are you setting up a strawman?

Jan, that is EXACTLY what you said when you said the emissivity was the same at all wavelengths. Emissivity is absorptivity. You insisted that if Earth's surface albedo was 0.15, implying visual absorptivity of 0.85, its infrared absorptivity had to be 0.85 as well. That's exactly the value you used in your temperature equation run-through.

The emissivity is the same at the absorptivity at a given wavelength -- that's Kirchhoff's Law. Or at all wavelengths for a gray body. It is NOT the same as every other emissivity for the same body at every other wavelength for a nongray body. That's why some things can be described as "nongray bodies," and why gray body approximations are APPROXIMATIONS. Planets are not gray. Neither are any real objects, with the exception of black holes.

Bart #344

Jan, that is EXACTLY what you said when you said the emissivity was the same at all wavelengths.

Are you trying to reinterpret what I said to create a strawman? I think you are it's either that or you have a severe comprehension problem.

In #326

Kirchhoff’s law and absorptivity = 1 - albedo and it’s not wave length dependent even for fluorescent bodies when the body in equilibrium energy absorbed = energy emitted so integrate over the entire spectrum and Kirchhoff’s law will hold. Try it sometime.

With some added emphasis.

Do you understand what that means?

Nothing like what you are saying it means I can assure you.

"In quantitative terms, the relationship between absorptivity a and emissivity ε is embodied succinctly in Kirchhoff's Law,, which states that

ελ(θ,φ) = aλ(θ,φ). (6.12)

Note that this equivalence is strictly valid only for monochromatic radiation at a given wavelength λ and when the viewing directions θ and φ are specified..."

Petty, Grant W. 2006. A First Course in Atmospheric Radiation (2nd Ed.). Madison, WI: Sundog Publishing, p. 126. Emphasis added.

"...there is a very simple relationship beween emissivity (ελ) and absorptance (αλ)

ελ = αλ [2.5]

This is known as Kirchhoff's Law and indicates that, at the same wavelength, good emitters are equally good absorbers."

Henderson-Sellers, A. and Robinson, P.J. 1986. Contemporary Climatology. NY: John Wiley & Sons, p. 35. Emphasis added.

Bart #346

“In quantitative terms, the relationship between absorptivity a and emissivity ε is embodied succinctly in Kirchhoff’s Law,, which states that

ελ(θ,φ) = aλ(θ,φ). (6.12)

I am quite familiar with that characterisation but it doesn't allow for fluorescence which is where EM energy is absorbed at one wavelength and emitted at another. This breaks the generality and Kirchhoff himself was no doubt aware of that when he said that the law cannot be dependent on wave length.

Total energy absorbed $$Q_a = \int_0^{\infty}q_a\left(\lambda\right)d\lambda$$

Where

$$q_a\left(\lambda\right) = q_a \alpha \left(\lambda\right)$$


and emitted $$Q_e = \int_0^{\infty}q_e\left(\lambda\right)d\lambda$$

Where
$$q_a\left(\lambda\right) = q_a \epsilon \left(\lambda\right)$$

and

qa and Qe determined by Planck's law for source and sink respectively and α(λ) and ε (λ) is determined using Einstein coefficients.

Then Kirchhoff's law is better expressed by

$$\int_{0^+}^{\infty} \alpha \left(\lambda\right) d\lambda = \int_{0^+}^{\infty}\epsilon\left(\lambda\right) d\lambda$$

Oops I just looked at my watch and was reminded of phosphorescent materials that emit light more slowly than they absorb (fun stuff) so better integrate over time as well but I leave that as an exercise for those interested enough to do it.

another correction

qa and Qe determined by Planck’s law for source and sink respectively and α(λ) and ε (λ) is determined using Einstein coefficients.

$$q_a$$ and $$q_e$$ determined by .....

Bart: Wicki is your friend:

"At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity.
Here, the absorptivity (or absorbance) is the fraction of incident light (power) that is absorbed by the body/surface. In the most general form of the theorem, this power must be integrated over all wavelengths and angles. In some cases, however, emissivity and absorption may be defined to depend on wavelength and angle, as described below"

http://en.wikipedia.org/wiki/Kirchhoff's_law_of...

Hmmm....Here are some interesting observations concerning K's law:

"Kirchhoff's Law of Thermal Radiation was formulated in 1859. It is at the same time the simplest and least understood law in physics today. Kirchhoff's law was describes how an object radiates when placed in thermal equilibrium within the confines of an enclosure."

"Unfortunately, many believe that a temperature can be obtained from a thermal spectrum without knowledge of whether or not the object of interest was in fact in thermal equilibrium with an enclosure. This can lead to serious errors in that the temperatures obtained can be completely invalid."


"In actuality, Kirchhoff's Law is valid only for the perfect absorber. The consequences are serious. As a result, blackbody radiation becomes critically dependent on the nature of the object in question. "

http://www.thermalphysics.org/kirchoff/kirchhof...

jae #351

Yes there are a few tricks to using Kirchhoff's Laws however

“In actuality, Kirchhoff’s Law is valid only for the perfect absorber. The consequences are serious. As a result, blackbody radiation becomes critically dependent on the nature of the object in question. ”

This is not quite correct and engineers have for the past couple of decades quite successfully using devices like this for non contact thermometry for several decades now and they at the very least depend on Kirchhoff's holding for the thermocouple array as well as the object whose temperature is being measured. It's not at all unusual for some academic to pop up and say we can't do what more practical people have been doing for a while.

For instance I know when my non contact thermometer measures 37C on my stainless urn because the reflectivity is about .52 (emissivity therefore .48) the temperature on the surface is about 99C I've set the thermostat just below boiling point. I leave it as an exercise to check the numbers.

Point being that while nothing is ever ideal there are ways to calibrate for and work around the less than ideal circumstances and the emissivity can be determined fairly readily from reflectivity which is readily directly measured using kirchoff's law.

Jan / jae,

Jan writes:

It’s not at all unusual for some academic to pop up and say we can’t do what more practical people have been doing for a while.

It seems to me that BPL scores a lot of points by defending his view with excerpts from introductory textbooks on atmospheric radiation and climatology. Yet, I noted that other engineers (e.g. Gary Gulrud at Anthony Watts' blog) have argued in a similar vein, i.e. that we have here a case that the emerging hands-off science of climatology has just inherited one of its laws in a simplistic manner, and has failed to understand what is obvious to the average hands-on engineer. If so, why can't someone respond to this Kirchhoff confusion with excerpts from engineering textbooks?

Or failing that, why can't someone address the actual proof that Ferenc gave on M7, p. 24, instead, since it seems to me that trying to win this by attributing the result trivially to Kirchhoff just isn't convincing people.

On M7, p. 5, Ferenc has specifically stated that the result is not trivial, "The validity of the Kirchhoff law – concerning the surface and the inhomogeneous atmosphere above – is not trivial. Later, using the energy minimum principle, we shall give a simple theoretical proof of the Kirchhoff law for atmospheres in radiative equilibrium." He refers to p. 24, where, after mathematically deriving the relationship, Aa = Ed, from the classical equations of radiative transfer, he states, "This is the proof of the Kirchhoff law for the surface-atmosphere system." I'm no expert, but it looks to me that the only controversial assumption along the way of this proof is the procedure for setting dB0/dtauA=0 according to the energy minimum principle. And it seems to me that critics have shut up about the EMP principle after a large body of literature has been found that the IPCC has mysteriously forgotten to discuss.

Frankly, it is the addition of this proof on p. 24, along with the empirical evidence that the law holds generally on the Earth and Mars, that has managed to convince me that Aa = Ed really is a newly-discovered natural law, and apparently unaffected by the addition or subtraction of CO2.

Alex #353

Yet, I noted that other engineers (e.g. Gary Gulrud at Anthony Watts’ blog) have argued in a similar vein, i.e. that we have here a case that the emerging hands-off science of climatology has just inherited one of its laws in a simplistic manner, and has failed to understand what is obvious to the average hands-on engineer.

That is the crux of the problem but it's not just Kirchhoff's law it's feedback too not to mention first order differential equations with two boundary conditions at a single boundary.

The complication Ferenc mentions really has little to do with where BPL is going wrong BPLs $$\epsilon_{\lambda}=\alpha_{\lambda}$$ is trivially true for most absorbers/emitters including atmospheric absorbing species (hence $$A_A=E_D$$ at equilibrium) but it isn't the whole story and jae (#351) has already addressed it by his reference to Wikipedia which contains with in it:

Here, the absorptivity (or absorbance) is the fraction of incident light (power) that is absorbed by the body/surface. In the most general form of the theorem, this power must be integrated over all wavelengths and angles. In some cases, however, emissivity and absorption may be defined to depend on wavelength and angle, as described below.

I put some mathematical meat on it in #347 this isn't clear enough for you? I kept it simple just by referring to all wavelengths and not worrying about angles (the parallel planar layer approximation) it's just another integral though.

EMP is a different issue again it refers to non-equilibrium models so will have importance in GCMs but isn't really relevant in FMs static equilibrium model.

I distinctly remember our friend Arthur Smith on Andy Revkin's blog, ridiculing a paper which had a novel interpretation (with equations) of atmospheric climate, by saying (paraphrasing) 'he completely ignores the Kirchhoff law, so his analysis is invalid'. I believe I replied that Arthur's was perhaps an unjust criticism because Kirchhoff's law could only be used for the atmosphere if you made several very gross assumptions. The question was - are these assumptions justified, ie what does the data say? So it seems AGW proponents are not all in agreement on this issue. Yet I could have sworn I read that Kirchhoffs law is used in climate models for the lower atmosphere at least. One supposes that in the final analysis if the data roughly agrees with the law then it can be presumed to hold can it not?

A similar thing happened with the discussion on G+T and the second law on the Atmoz blog when the host completely escaped the argument by saying that it only applied to closed systems. Neat escape I thought.

Now Jan talks about the first law. Hey can we introduce the zeroth law here too? My Physics lecturer taught me that the zeroth law signified that heat flows from hot to cold, which is why it's considered fundamental. Of course it's more subtly framed than that.

James - I assume you're referring to my commentary on Nichol's paper here:

http://dotearth.blogs.nytimes.com/2008/04/17/so...

Let me quote it here:

On the paper by Jon Nicol posted by H. Nelson here (#24) - I find it interesting that our “skeptic” readers accept with such enthusiasm (#32, #33) a mysterious new paper that happens to track their biases so well. Note that (A) this paper has apparently not been peer-reviewed, unlike the paper Andy was blogging on here, (B) it cites no references, and in fact seems ignorant of some fundamental basics of the field (like how the basic greenhouse effect even works!), and (C) there are many claims highlighted here, some or all of which are clearly wrong.

On (C), for example look at the box on page 16 (what I’m looking at was missing the equations and figures, and had 28 pages, so your page numbers may be different) - “Not all of the CO2 molecules in the atmosphere are being used to collect energy,[...] the actual number of internally excited molecules will depend only on the total power being radiated by the earth’s surface which in turn will only depend on the solar insolaton.”

The total power radiated by Earth’s surface does *not* depend only on solar insolation - rather the solar component is greatly increased by the thermal radiation back from the atmosphere. That’s what the greenhouse warming is all about - radiative increases in energy transfer at all layers of the atmosphere, leading to warming of the surface.

Nichol seems to be ignorant or deliberately avoiding the fact that in thermal equilibrium, absorption and emission must match (Kirkhoff’s law). He actually tries to claim that this doesn’t apply: “This redistribution of the energy into LTE considerably reduces the average momentum of individual atoms, removing to a very large extent, the possibility of reciprocal collisions returning the energy to internal excitation of molecules.” and further arguments along those lines seem to be completely denying this very fundamental aspect of thermal radiation.

So, in short, the paper is wrong, and it’s conclusions have no foundation in reality. Now why were our “skeptics” not at least a little skeptical here? Hmmm.

James, that was almost a year ago. What exactly has happened to Nichol's wonderful new theory you guys were all touting back then? I don't think even Tom Vonk believes in it any more - Nichol got some obvious mathematical things wrong (units on some of his equations were completely off), in addition to the basic theoretical problems I pointed out in this comment. Nichol's treatment, for instance, is completely contrary to Miskolczi's, which is at least much more subtly wrong (M is basically right in his treatment of radiative fluxes and the Kirchhoff implications, but wrong in his unfounded generalizations from specific observations).

And James, you actually would tout the G&T paper still too? Skepticism requires understanding and real thinking, you know, not blind following of whatever the latest absurdity is that matches the biases in your worldview.

Arthur says:

"Skepticism requires understanding and real thinking, you know, not blind following of whatever the latest absurdity is that matches the biases in your worldview."

after saying:

"(M is basically right in his treatment of radiative fluxes and the Kirchhoff implications, but wrong in his unfounded generalizations from specific observations)."

Now, I wouldn't have a problem with this, if you had added IMHO at the end of the second statement. But how do you state, unequivocally that M is WRONG and that the generalizations are UNFOUNDED? Did I miss a proof of all this somewhere?

LOL. Nicol has some mistakes, Thieme has some mistakes, Miskolczi has some mistakes, G&T have some mistakes, and maybe even Arthur Smith has some mistakes.

Yes indeed, I have some mistakes - I spelled Kirchhoff's name wrong. Can't find any others though :)

Arthur #359

Can’t find any others though

Yes this can be very difficult if you are wearing the glasses of your own conviction.

Arthur #357,

I recall you agreeing with Nick Stokes in CA many months ago that M's eqn.4 made 'absolutely no sense at all'. More recently, in NM, you seem to have accepted eqn.4 but only in an approximative sense, contrary to Ferenc's own view. Here, you say that his implications from Kirchhoff's law are basically correct. Is the spelling of Kirchhoff's name, which just about no one seems to get right, truly the only point you've changed your mind about? If there's more, it would be nice to assist in convincing BPL, so that this discussion about Kirchhoff's law might finally be resolved, once and for all.

James G #355

My Physics lecturer taught me that the zeroth law signified that heat flows from hot to cold

That's the second law. Zeroth law:

The "zeroth law" states that if two systems are at the same time in thermal equilibrium with a third system, they are in thermal equilibrium with each other.

is necessary for thermometers to work.

Kirchhoff's law talks only about a body in thermal equilibrium which broadly means it's temperature and it's internal energy is not changing and since that means the first law:

The change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

must hold.

It's just the principle of conservation of energy but I like to keep it in the family.

Jan #354,

Actually, the maths in #347 is so simple that I can almost follow it. But the point I'm making is that for people who can't follow it, the references to standard textbooks are very impressive, and they trump anything that can be brought up from memory, or even the Wikipedia. It's a point about the art of persuasion that I'm making, not about maths.

However, I am more interested in this point:

EMP is a different issue again it refers to non-equilibrium models so will have importance in GCMs but isn’t really relevant in FMs static equilibrium model.

This has completely thrown me. Firstly, didn't Ferenc just clarify this and say that Aa = Ed holds regardless of the equilibrium condition? He did, and said that Aa = Ed should hold for all 1761 of the TIGR profiles, that means for the TIGR2000. Secondly, how on earth can you say that the EMP (energy minimum principle) isn't relevant to Ferenc's model? Eqn.28 assumes that the EMP principle holds. I just quoted Ferenc's own words, that the 'proof' of the atmospheric Kirchhoff law follows from assumption of the EMP. Are you saying his proof on p. 24 is wrong?

Correction, some missing words above, "...that means for the TIGR2000, and any other atmospheric profiles we might find."

Jan reminded me of this post: http://landshape.org/enm/greenhouse-heat-engine... which actually I hadn't forgotten about. To which I responded, this shouldn't alter the deductive form of the proof on p. 24. To which Jan asked, "Sometimes I wonder what's going on inside your head." :) The way I see it, regardless of the name of the law we apply when setting dB0/dtauA=0, the result of Aa = Ed seems to provide strong empirical support for its existence/application here. OR, the application of this Hamilton principle, or EMP, or whatever we finally agree is the right name, finally proves the result Aa = Ed, deductively, from the classical equations of radiative transfer. The beauty of this double proof seems to greatly increase the probability that the theory as a whole, or those parts of it held together here (eqn.4, eqn.21, TD problem, eqn.28) are true. I can't see why this hasn't been discussed, pointed out, more often.

Alex #365

Just like to assure Alex that I quite enjoy our discussions.

finally proves the result Aa = Ed, deductively, from the classical equations of radiative transfer.

all fine as long we recall that those classical equations also describe empirically determined (statistically correct) laws so only have a high probability of being correct in the first place;-)

Jan, continuing to miss the point, posts:

The complication Ferenc mentions really has little to do with where BPL is going wrong BPLs is trivially true for most absorbers/emitters including atmospheric absorbing species (hence at equilibrium) but it isn’t the whole story and jae (#351) has already addressed it by his reference to Wikipedia which contains with in it:

Here, the absorptivity (or absorbance) is the fraction of incident light (power) that is absorbed by the body/surface. In the most general form of the theorem, this power must be integrated over all wavelengths and angles. In some cases, however, emissivity and absorption may be defined to depend on wavelength and angle, as described below.

Let's get back to the original point, shall we? You chastised me for not taking emissivity into account for my estimate of the Earth's surface temperature without an atmosphere. I pointed out that the IR emissivity of the Earth's surface was higher than its visual absorptivity. You then went on a kick about how they had to be the same. To defend this incredible piece of idiocy, you introduced two red herrings:

1) Kirchhoff's Law applies when integrated over the whole spectrum. I.e., for a gray body.

We were discussing how the absorptivity/emissivity varied between visual and infrared ranges. Why the HELL would we want to integrate over the whole spectrum???

2) Kirchhoff's Law doesn't apply to fluorescence.

Are you implying that the Earth's atmosphere is fluorescent???

My point stands, the Earth's absorptivity/emissivity is different for the thermal IR than it is for the visual. Same thing for greenhouse gases, which is why greenhouse gases exist.

I'll run through my estimate again, taking into account the Earth's surface emissivity as well as its albedo.

Assumptions:

1. Earth's albedo in the absence of an atmosphere would be about 0.15, the usual figure used in GCMs. This is comparable to the Moon (0.11) and the desert areas of Mars (0.15-0.22).

2. Earth's infrared emissivity is about 0.95.

3. The Solar constant at Earth's distance from the sun averages about 1366 watts per square meter.

The Earth's radiative equilibrium temperature is then

Te = {[S (1 - A)] / (4 ε σ)}^0.25

With σ = 5.6704 x 10^-8 W/m^2/K^4, this gives Te = 271 K. My estimate without taking emissivity into account, or assuming ε = 1, was 267 K. I was off by 4 K.

But 271 K is still, of course, below freezing (Tf = 273.15 K). So in the absence of an atmosphere, the Earth would be much colder, which is the point.

Heat can be transferred by conduction, convection, or radiation. Conduction follows the thermal gradient from surface to atmosphere. So does convection. That leaves radiation.

It is radiation from the atmosphere -- "back-radiation" -- that warms the Earth's surface above freezing. That's how the greenhouse effect works.

BPL

I don't believe you are properly reading what Jan is saying?

1) Kirchhoff’s Law applies when integrated over the whole spectrum. I.e., for a gray body.

2) Kirchhoff’s Law doesn’t apply to fluorescence.

He definitely didn't state 2. But said instead you would need to integrate over time as well.

As for 1), your language is unusual: Jan isn't stating WHEN Kirchhoff's law can be applied, which is what you are implying. It is quite clear what Jan is saying, and you have repeatedly avoided it, or mis-represented it.

Jan may be wrong, but if he it is more likely to be in his application of the law rather than getting the law itself wrong. It may be helpful therefore to agree on what the law is first; after which I'd guess your discussions will be more productive.

Arthur
Just to be clear, I've never personally touted Nichol's, G+T's or M's theory.

Jan
That's a great site. See here:
http://hyperphysics.phy-astr.gsu.edu/Hbase/ther...

But then there's the apparent oddity of the atmosphere being in thermal equilibrium by maintaining differences in temperatures - in possible violation of the zeroth law one might say. Or is it even in equilibrium? Striving for equilibrium more like. It's these initial assumptions that fascinate me. You can often invalidate a whole paper by pointing out an assumption made on the first line that's rather iffy.

Bart says yet again:

"It is radiation from the atmosphere — “back-radiation” — that warms the Earth’s surface above freezing. That’s how the greenhouse effect works."

Again, NO!

But, before you violate the second law of thermo any more with your back-radiation (which originates from a colder spot than the surface), tell me whether or not you believe in thermalization. If so, you must agree that the oceans and the atmosphere store heat from the sun. And if they do, wouldn't that, alone, increase the temperature beyond the 0 C you calculated using K's law? Even the moon calc's are probably off a little, since the green cheese stores some of the sun's heat during the day. But the cheese cannot store anywhere near as much as water and air, since it is opaque, while the air and water are quite transparent to visible but not so transparent to IR. Do you deny that the oceans largely control the climate on Earth? If you disagree, you disagree with about everyone. If they do have a large influence on Earth's climate, it MUST be that they do it with stored energy, no? And that stored energy doesn't exert its influence through radiation alone. It does it mostly through convection (winds).

Many in the AGW/CO2 crowd appear to be so focused on radiation, that they are forgetting all other forms of heat transfer.

Bart #368

1) Kirchhoff’s Law applies when integrated over the whole spectrum. I.e., for a gray body.

No Bart for coloured bodies that's what LBL's like HARTCODE and MODTRAN are for.

Why the HELL would we want to integrate over the whole spectrum???

To get the sums right Bart you do understand that don't you?

Are you implying that the Earth’s atmosphere is fluorescent???

Do you deliberately miss the point Bart? So now you are suggesting that the earths surface does not absorb SW and emit LW? Kirchhoff's law is not wavelength dependant it refers to sum of absorbed energy = sum of emitted energy that's why the LBLs and the convolution integration over the spectrum.

You need to get this right before you can get the rest right.

Gary Moran #369

I think in this case it's just as important to understand why the law says what it does and that is conservation of energy.

James G #370

Yes I saw that when I posted the link. The thermometers that measure the surface temperature are 1.5 m (in some cases 2 m from the surface) so the thermometer is in equilibrium with the air whose temperature it's reading and the assumption is that the air is in equilibrium with the surface and therefore the thermometer is in equilibrium also with the surface.

Then

in #371

But then there’s the apparent oddity of the atmosphere being in thermal equilibrium by maintaining differences in temperatures - in possible violation of the zeroth law one might say.

Isn't climate science fun?

The discontinuity arises from the assumption (or is it an approximation?) that earths optical depth is infinite at the surface leading to an infinite temperature gradient at the surface for which the 'work around' is to set up two boundary conditions at the surface boundary for a first order differential equation this is quite odd to say the least.

Gary writes:

I don’t believe you are properly reading what Jan is saying?

I believe I properly read what Jan was saying when he said that back-radiation from the atmosphere to the surface was, quote, "impossible," unquote. I believe I properly read what Jan was saying when he said that the emissivity of the surface in the thermal infrared had to be the same as the absorptivity in the visual due to Kirchhoff's Law, which clearly shows he doesn't believe that Kirchhoff's Law is specific to wavelengths. Let me quote the sequence of posts, in case you missed them:

Me: "Sorry, Jan, but Kirchhoff’s Law only holds for a particular wavelength."

Jan: "Who told you that nonsense ?"

If Jan says that Kirchhoff's Law doesn't only apply to a particular wavelength, then he's just plain dumbass simpleton wrong, my friend. No question of misinterpretation. One party (me) is right, and one party (Jan) is wrong. It really is that painfully simple.

jae writes:

Bart says yet again:

“It is radiation from the atmosphere — “back-radiation” — that warms the Earth’s surface above freezing. That’s how the greenhouse effect works.”

Again, NO!

But, before you violate the second law of thermo any more with your back-radiation (which originates from a colder spot than the surface),

If I'm violating the second law of thermodynamics, kindly state by how many Joules per Kelvin I am doing so. I have asked you this before.

tell me whether or not you believe in thermalization. If so, you must agree that the oceans and the atmosphere store heat from the sun. And if they do, wouldn’t that, alone, increase the temperature beyond the 0 C you calculated using K’s law?

Sure they store heat. And they release it through radiation. And the time to equilibrium is much less than geological time, and for the Earth's surface it's less than a day, which is why heat storage is IRRELEVANT to the question. (See, I can type in all caps too!). Heat storage will not sustain temperatures above radiative equilibrium. Hot things radiate. They also, if material, conduct and convect.

And Jan says it again -- pay attention, Gary:

Kirchhoff’s law is not wavelength dependant

You got that, right, Gary? Is there any other way to read it than "Kirchhoff's Law is not wavelength deped[e]nt?" I can't think of one. And since Kirchhoff's Law is subscripted with a lambda in every radiation physics textbook in the Universe, and since lambda stands for wavelength, I'd say Kirchhoff's Law bleeding well WAS wavelength-dependent, wouldn't you, Gary? And that would mean Jan was wrong, don't you think, Gary?

Jan goes on:

The discontinuity arises from the assumption (or is it an approximation?) that earths optical depth is infinite at the surface leading to an infinite temperature gradient at the surface for which the ‘work around’ is to set up two boundary conditions at the surface boundary for a first order differential equation this is quite odd to say the least.

There is no "work around." In a radiative equilibrium model, there is a discontinuity in temperature at the surface. There isn't one on Earth because the Earth system is not in radiative equilibrium, it is in radiative-convective equilibrium. Conduction and convection remove the thermal discontinuity when you have an atmosphere, aside from very rare situation such as those in deserts where you can sometimes have a thermal gradient of 30 K per millimeter (no, I didn't get the figure wrong).

And the optical thickness of the atmosphere is never infinite at the surface in a radiative-equilibrium model (see Houghton, chapter 2 for a simple example with diagrams). The optical thickness of the Earth itself is indeed infinite, because Earth is what we physicists like to call "opaque."

Want the math? The demonstration of how a radiative equilibrium model works is really simple if you know algebra.

Aw, heck, make it simple: Here's a link to such a demonstration:

http://envsci.rutgers.edu/~weaver/Pubs/weaver_r...

Bard:

Now, didn't you see #350, or did you see it and forget it, or...?? Wicki says:

"In the most general form of the theorem (Kirchoff's), this power must be integrated over all wavelengths and angles. In some cases, however, emissivity and absorption may be defined to depend on wavelength and angle, as described below”

I know Wicki isn't infallible, but I think I'll accept it over your opinion.

You say:

"Heat storage will not sustain temperatures above radiative equilibrium. Hot things radiate. "

That seems to be an untenable position. If you heat up some rocks in a fire, then take them out of the fire and put the hot rocks in a pit with a pig, they store the heat and cook the pig. Their temperature will stay above the "radiative equilibrium" of the surroundings for quite awhile. If you were to keep heating them again, before they cool enough to get back to "radiative equilibrium," they will stay above "radiative equilibrium" as long as you keep heating them. I submit that the Sun does this to the atmosphere and oceans, especially the tropical oceans and tropical atmosphere on a continuing basis, thereby storing heat. And this can easily explain why the global average temperature is above your calculated 0 C for a non-existing, dead, waterless, atmosphere-less planet Earth (incidently, I thought the proper calculation was -18 C and that GHGs supply a full 33 C on average).

You don't need "backradiation" to explain a greenhouse effect, although there is nothing wrong with showing the radiation, since it is a "signature" of the temperature of the atmosphere at any given time. IOW, quantitatively showing radiation levels is completely equivalent to showing temperature levels, ala SB.

Now since I'm in a mood to argue and shoot my mouth off, I want to ask you the same question that I have posed at Lucia's:

It is easy to prove that a true greenhouse works by greatly limiting convection, not by the glass in any way "trapping" or "blocking" or "absorbing" IR. And the climate scientists are usually very quick to point out that the atmospheric "greenhouse effect" is not to be confused with the real greenhouse effect and that, in fact, the term is a misnomer and should not be used. But then those same people turn right around and insist that the greenhouse heating from the atmosphere results from the atmosphere "blocking" or "traping" IR, just like they agreed the glass in the real greenhouse could not. IOW, they are actually using the very same mechanism that they agree is incorrect for the real greenhouse. Many even use the term "blanket," which goes directly back to the real greenhouse effect, since blankets work mainly by resisting convection. My question is this: If the glass in a real greenhouse can't "block" IR to cause heating, just how does the atmopsphere "block" IR to cause heating? How does the glass in the real greenhouse differ from the thick "glass" of the atmosphere?

You will note that David's experiments indicate that you cannot "trap" enough radiation, even with 5 layers of glass, to raise the temperature inside a greenhouse.

There's probably some error in my logic here. I know you will be happy to help me find it, Bart.

oops, 376 should be addressed to Bart, not Bard. Sorry for the typo.

Bart #374

I believe I properly read what Jan was saying when he said that back-radiation from the atmosphere to the surface was, quote, “impossible,” unquote.

Not quite Bart you said back radiation was warming the surface :- again the half truth. This is not possible because the net flux is from the warmer surface.

If I’m violating the second law of thermodynamics, kindly state by how many Joules per Kelvin I am doing so. I have asked you this before.

and it has been explained in some detail before see #233.

Bart #375

Right there in equations 6 a & b you have the temperature discontinuity due to the rather odd two boundary conditions for a first order DE. They do a little better than most though and seem to be able to reduce the temperature discontinuity by modifying the Schwartzschild grey model to include an IR window in the atmosphere. Would do better if they didn't assume an semi-infinite atmosphere to begin with. The interesting conclusion they reach is that we cannot expect runaway in an atmosphere with an IR window a definite step in the right direction.

BPL #375, Jan #378

Jan writes:

The interesting conclusion they reach is that we cannot expect runaway in an atmosphere with an IR window a definite step in the right direction.

He seems to be citing this section (p. 11,587):

Once the optical depth of the grey portion of the atmosphere is sufficiently large, the surface temperature depends only on the size of the window. For the surface temperature to increase, the size of the window must decrease. In other words, the modified SGM [Schwarzschild Grey Model] predicts that a runaway greenhouse effect is not possible in an atmosphere with a spectral window, no matter how small that window is. Another way to summarise this result, which is of particular relevance to the Earth's atmosphere, is as follows: in an optically thick atmosphere, the surface temperature is limited by the size of the spectral window.

The other point that Jan raises is that W&R assume a temperature discontinuity at the surface.

The don't seem to mention a caveat found in Collins [2003], p. 256 here http://bifrost.cwru.edu/personal/collins/astrob...

One further complication must be dealt with before we can use this description of a stellar atmosphere. In general, stellar atmospheres can be regarded as being infinitely thick. Since the influence of the lower boundary diminishes as exp(tau-tau0), and since this optical depth will exceed several hundred within a few thousand kilometers of the surface for main sequence stars, we can take it to be infinity.

So they can disregard this term, only because the stellar atmosphere extends is more than 'a few thousand kilometers' thick.

Yet this assumption is how WR's eqn.6a & b were derived.

It's looking very hard to deny Miskolczi is right about the surface temperature discontinuity.

jae #376

I know Wicki isn’t infallible, but I think I’ll accept it over your opinion.

Here is another one with more detail:

Kirchhoff's Law has a corollary: the emissivity cannot exceed one (because the absorptivity cannot, by conservation of energy), so it is not possible to thermally radiate more energy than a blackbody, at equilibrium. This has two caveats, however. First, if the surface is diffractive, so that incident energy at one angle is partially reflected to another angle, then the emissivity at one angle can exceed unity, but not the emissivity of power integrated over all angles. Second, if the object is nonlinear (e.g. fluorescent), so that incident power at one wavelength is re-emitted at another wavelength, then the emissivity at some wavelengths can exceed unity, but not the emissivity of power integrated over all wavelengths.[emphasis mine]

Pretty much cover the case of the surface (in fact the surface/atmosphere combination) I think.

Forget it. I'm drowning in the seas of stupidity here. Have a nice day.

BPL #381,

How about answering my question, then, about the temperature discontinuity.

You've directed us to the Weaver & Ramanathan [1995] paper for 'a demonstration of how a radiative equilibrium model works'. However, they assume without any demonstration that the 'solution' to their equation 1 is 6a and 6b.

Miskolczi has shown in M7, Appendix B, that they have left out a term, Bg*exp(-tau_A), in eqn.B3. Collins [2003] explains why the term was omitted, and the reason given is that after several thousand kilometers of stellar atmosphere, tau exceeds several hundred.

I have never seen you comment on this, other than to assert that he solved a non-existent problem.

I posted the data from our experiments at http://tnn93w.tripod.com/ give it a look if you like. Refer to post 329 for a little more details. Remember we are not scientist, so please don't be harsh :)

Thanks for that. Very impressive. Don't worry, I delete any abusive posts. One issue is that of the hypothesis formulation. Its very hard to say anything from the observation of no difference. That is, the effect of water and CO2 added could be too small to detect, so you can't really say that you have detected a maximum effect. OTOH, if you observe a significant difference then, you can say you have seen something. You could say, you were unable to detect any significant increase in temperature, due to CO2 or water vapor.

JQ: Yes, very interesting data. I have the feeling that we can still learn a lot from greenhouse experiments, but I don't know for sure what. :)

One thing that intrigues me is that there appears to be no place for "back-radiation" here. As David's multi-layer experiments seem to indicate, the amount of heat stored (or absorbed, trapped, etc.--all words used to describe the GHE) by the glass seems to have absolutely nothing to do with the temperature inside. So the question becomes, does the presence of hundreds of "layers" of GHGs in the real atmosphere have any effect on the temperature at the surface, because of "back-radiation?" I still hypothesize "NO."

While one difference between the experiment and the atmosphere might be, that the temperature of the surface of the earth is elevated over black body temperature (BBT) expectations, the temperature of the tile is consistent with black body expectations. So the experiment does not 'show the greenhouse effect' in that sense. However, what you say about the cause of the greenhouse effect (surface temps elevated over BBT) might be true to: that its not really accurate to say 'backradiation' does the surface temperature elevating, but rather the atmospheric profile conforming to the constraints of LTE say (equivalence of flux density and temperature) over a wide range of atmospheric pressure.

David: Whew, I'm more confused than usual. Could you elaborate a little?

Are we not inventing a lot of complicated equations and theory which say, ultimately, "yeah, the water and the air store some of the solar energy for a rainy day."

Another way of saying it is, the temperature of the ends of the experiment, the tile and the glass, are constrained only by the incoming solar flux, the material they are made of (absorption, reflectance, opacity), and heat losses (convection conduction for the glass, conduction for the tile). What is between them, its IR opacity, is irrelevant, and can't affect the temperature at the end points, because they are constrained.

David, this subject has just assumed another angle with the PM offering free insulation batts in homes. I have not done my homework fully yet, but as I understand it, it's heat in, heat out, circulation versus radiation again.

The main deficiency (of insulation) appears to be in Queensland and I think that is the free market at work. The batts do not work well. They cause the ceiling space to heat up more quicky by preventing heat flow to the rooms below. After a while, the insulation itself reaches the same temperature as inside the ceiling and the insulation effect starts to fail as similarly, the room below starts to reach equilibrium also. Like in your work, one more layer only delays the inevitable.

They are more popular down south here because of the dynamics. They prevent some heat buildup in rooms until the cool of the evening takes over in an average scenario. The last hot period of 4 days over 40C and hot nights would possibly yield, on analysis, the conclusion that they were no use after a day or so.

On the negative side of the economic ledger, the eforts and funds of people in manufacture, transport (many trucks for few batts), installation and fireproofing precautions add to the GHG burden if people want to go down that path.

I'm sure there are more useful ways to spend taxpayer money, like dropping stamp duty or payroll tax or reducing GST and/or income and company tax. It's because these have been too high that the PM can make these silly promises about "green homes".

Maybe he needs a good science advisor.

Maybe he needs a good science advisor.

Large sectors of the economy insolvent and oil cheap as chips, so taxpayer funded green initiatives are the answer?

Jae

“yeah, the water and the air store some of the solar energy for a rainy day.”

That seems wrong. Isn't it the air that stores the water for a rainy day?

Nick

That seems wrong. Isn’t it the air that stores the water for a rainy day?

Yes that too!!

i like to eat pen ink

My son repeated the experiment mentioned and also repeated the experiment with CO2. He also did the experiment with a control jar using water vapor and CO2. He won first place in South Plains Regional Science Fair. He received a certificate of outstanding achievement from the American Meteorological Society for the experiments. Thanks to the administrator of this web site for the great idea. The data for the experiments can be found at http://tnn93w.tripod.com/

That's great! Congratulations to both of you. Science is fun!