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Two Basic Foundations

This article will be a placeholder article to filter out a select group of people. The many people who arrive and confidently explain that atmospheric physics is fatally flawed (without the benefit of having read a textbook). They don’t think they are confused, in their minds they are helpfully explaining why the standard theory is wrong. There have been a lot of such people.

Almost none of them ever provides an equation. If on rare occasions they do provide a random equation, they never explain what is wrong with the 65-year old equation of radiative transfer (explained by Nobel prize winner Subrahmanyan Chandrasekhar, see note 1) which is derived from fundamental physics. Or an explanation for why observation matches the standard theory. For example (and I have lots of others), here is a graph produced nearly 50 years ago (referenced almost 30 years ago) of the observed spectrum at the top of atmosphere vs the calculated spectrum from the standard theory.

Why is it so accurate?

If it was me, and I thought the theory was wrong, I would read a textbook and try and explain why the textbook was wrong. But I’m old school and generally expect physics textbooks to be correct, short of some major revolution. Conventionally, when you “prove” textbook theory wrong you are expected to explain why everyone got it wrong before.

There is a simple reason why our many confident visitors never do that. They don’t know anything about the basic theory. Entertaining as that is, and I’ll be the first to admit that it has been highly entertaining, it’s time to prune comments from overconfident and confused visitors.

So here are my two questions for the many visitors with huge confidence in their physics knowledge. Dodging isn’t an option. You can say “not correct” and explain your alternative formulation with evidence, but you can’t dodge.

1. Is the equation of radiative transfer correct or not?

Iλ(0) = Iλm)em + ∫ Bλ(T)e [16]

The intensity at the top of atmosphere equals.. The surface radiation attenuated by the transmittance of the atmosphere, plus.. The sum of all the contributions of atmospheric radiation – each contribution attenuated by the transmittance from that location to the top of atmosphere

Of course (and I’m sure I don’t even need to spell it out) we need to integrate across all wavelengths, λ, to get the flux value.

For the derivation see Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations. If you don’t agree it is correct then explain why.

[Note that other articles explain the basics. For example – The “Greenhouse” Effect Explained in Simple Terms, which has many links to other in depth articles].

If you don’t understand the equation you don’t understand the core of radiative atmospheric physics.

—-

2. Is this graphic with explanation from an undergraduate heat transfer textbook (Fundamentals of Heat and Mass Transfer, 6th edition, Incropera and DeWitt 2007) correct or not?

From “Fundamentals of Heat and Mass Transfer, 6th edition”, Incropera and DeWitt (2007)

You can see that radiation is emitted from a hot surface and absorbed by a cool surface. And that radiation is emitted from a cool surface and absorbed by a hot surface. More examples of this principle, including equations, in Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics – scanned pages from six undergraduate heat transfer textbooks (seven textbooks if we include the one added in comments after entertaining commenter Bryan suggested the first six were “cherry-picked” and offered his preferred textbook which had exactly the same equations).

—-

What I will be doing for the subset of new visitors with their amazing and confident insights is to send them to this article and ask for answers. In the past I have never been able to get a single member of this group to commit. The reason why is obvious.

Once again, this is not designed to stop regular visitors asking questions. Most people interested in climate don’t understand equations, calculus, radiative physics or thermodynamics – and that is totally fine.

Call it censorship if it makes you sleep better at night.

Notes

Note 1 – I believe the theory is older than Chandrasekhar but I don’t have older references. It derives from basic emission (Planck), absorption (Beer Lambert) and the first law of thermodynamics. Chandrasekhar published this in his 1952 book Radiative Transfer (the link is the 1960 reprint). This isn’t the “argument from authority”, I’m just pointing out that the theory has been long established. Punters are welcome to try and prove it wrong, just no one ever does.

211 Responses

1. You might also point them at the Green Plate Effect which has splattered a few and Izen’s new GIF

2. There is even video from Izen

• Hi Eli, you should run this blog. You have a good sense of free speech in your comment field. I had a lot of fun there with the blankets and such. It´s a shame you are a blanket-man though.
I saw you had a foul-mouth there recently. Just to be clear, it wasn´t me. I´m a bit more sophisticated than that, even though I like to compare the gh-effect with pissing your pants and getting your balls to catch on fire from re-absorbed heat from the urine.

3. Willis Eschenbach’s The Steel Greenhouse is yet another relatively simple example.

The comments are interesting. The last I checked, which was some time ago, only one person who initially thought that Willis must be wrong changed his mind.

• If you think a massive steel-shell is a simple description of a cold fluid cooling a hot solid, you are a very strange-minded man.

I posted my reply on my own blog since you are not so big fans of free speech here:

https://lifeisthermal.wordpress.com/2017/11/06/two-flawed-foundations/

• Lifeisthermal, If you think pissing your pants is a simple description of the greenhouse effect, then you are a strange-minded man. 🙂

The point here (since you seem to have missed it) is that these two questions (and the steel greenhouse) are EASIER than the real greenhouse effect. These are baby-steps along the way toward an understand of the greenhouse effect. If anyone can’t handle these simpler ideas, then there is no point trying to discuss the actual greenhouse effect.

• lifeisthermal,

In a comment on The “Greenhouse” Effect Explained in Simple Terms I asked you to answer these two questions.

Being pressed to answer yes or no to a couple of questions isn’t very pleasant.

Especially the second question. I found that every time I have asked it to a confident (and inappropriately-named) climate skeptic I have either got silence or a response like “don’t show me your climate science rubbish here”. Then I point it it is just an undergraduate heat transfer textbook.

Silence follows. Deafening silence.

I just can’t get a commitment to yes or no after that.

The heat transfer textbook is wrong?

Or you were wrong?

The right response would be:

a) the heat transfer textbook is wrong, along with another 6 from the university library posted in Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics (link above)

– or

b) oh, it turns out I have been talking rubbish all this time and never realized it

The first answer would demonstrate the inappropriately-named skeptic to be confused about basic heat transfer as taught in every undergraduate physics/engineering course in the world.

The second answer would require humility.

Cognitive dissonance kicks in. Silence.

• on November 7, 2017 at 8:00 amscienceofdoom

Yes, the radiative transfer equation accurately calculates the emission from the atmosphere in relation to emission from the surface. It is an accurate calculation of the reduction of heat emission in detail, exactly quantified for the wavelengths in the observed effective emission.

“You can see that radiation is emitted from a hot surface and absorbed by a cool surface. And that radiation is emitted from a cool surface and absorbed by a hot surface.”

But on this one, you make assumptions about absorption in relation to temperatures. What the s-b equation does, is calculating the emission from a body at a temperature, and, when in presence of other bodies at different temperatures, calculates the simultaneous transfer of energy according to differences in emissive power.
When you say “And that radiation is emitted from a cool surface and absorbed by a hot surface”, you need support for the part claiming that the hot surface absorbs the emission from the cold surface. There are no equations showing that there is an effect on the hot surface from the cold surface emission, there is only calculation of transfer and absorption from the hot surface in the cold surface. This is the “net”. “Net” is heat. The s-b equation only calculates heat, nothing else. You might think that there is a logic in a reasoning of transfer outside “net”, which is not heat. But your thoughts is irrelevant. Because the equation only shows “Net”. Nowhere is there a mention of any other energy being transfered, because calculations show only negative transfer from cold to hot.

Give a reference, experimental and observational is preferred, that supports the claim that there is a transfer of energy, which is not heat or work, that cause an increase in temperature.

If there is such principles of transfer of energy, which happens outside of the net-transfer, and doesn´t do work on the body, it must be universal and confirmed in observations of relations between bodies, independent of the gh-theory.

I have searched the litterature for a couple of years. I am certain that there is no support for a claim where that adding a cold body to a warm body, increases the temperature of the warm body. The only transfer that can increase temperature is heat, “net”. And a force doing work on the body.

Is co2 heat, or is it work?

If it is not one of those, it will not increase the temperature. Only heat and work can do that.

• lifeisthermal,

4. […] at the religious page scienceofdoom.com you can find this post where the priest is preaching about how he misunderstands science so badly […]

5. lifeisthermal,
I agree with you that there is definitely a misunderstanding of science taking place on the SoD site. However, I feel like I probably disagree with you on who it is that is misunderstanding science so badly.

6. I think this lifeisthermal fellow is not too earnest in his worship of free speech. It comes that I chose to comment on his article, but my comments vanished after short time, combined with side effects on the article itself.

I’d like to share my new comment here too, for I think that at least in one place it might survive then. And the complete story shows nicely, of what quality arguments and style of lifeisthermal are. Here it comes:

Since my first two comments do not appear anymore, and, even more miraculously, at least one certain passage of the article is modified compared to the original version ( as it happens on exactly the point I criticised ), I have the pleasure to comment on the new version again.

This is done best by quoting the errors:

“What we know is that the sb-law tells us that at equal temperature there is no transfer of heat between surroundings and solid.”

This is the very passage that has silently been modified after my first comments have vanished, but it has not significantly benefited from the rewriting. The SB-law was not, is not, and will never make statements about heat transfer. It’s about temperature and radiation of energy correlated to that temperature. Heat, temperature, and energy are, though connected, different concepts. The SB-law is about the connection of temperature an radiated energy, not heat.

It’s enough to point on this single error to show the quality of the argument, but one can even have an experimentum crucio for the physics established by lifeisthermal in a lab:

Put identical volumes of water in two containers of identical inner shape, one being a dewar, one being a standard beaker. Put identical immersion heaters that works with identical powers in each of these containers. Close the containers and bring the immersion heaters on.
Will the final temperature of each container that is established after time

a) be the same

b) differ

The physics of the author lifeisthermal leads by logical deduction to a). Real world experiments show b).

How can this discrepancy be explained?

• Maybe you should add that I your comments wasn´t approved, because I had not yet been able to read them. And, when approved, you are free to write whatever you want, and there has not been, or will ever be, any editing by me.

” Heat, temperature, and energy are, though connected, different concepts.”

Well, you are not completely wrong. But I think “aspects” describes them better.They are aspects of the same thing, constant flow of energy.

Temperature is heat flow through mass.The different parts in mass, molecules/atoms/volumes of bulk mass, is subject to heat flowing through it. Temperature is directly proportional to this internal heat flow, and to the heat flowing out from it.

Heat is the transfer of energy to surroundings, and it is determined by potentials in differences in temperature. A body emits at a rate proportional only to its temperature. If a body at a lower temperature is present, there is also an addition of transferred energy, called heat. In that case, emission and transfer is paralell. Emission at T^4+transfer at a rate T1^4-T2^4. Transfer and emission sums up to the total power needed to sustain the steady state.

Example: surface emission at 385W/m^2, T=287K, and transfer to the atmosphere which emits 244W/m^2 at 256K needs a power density from the heat source at 4pir^2*244+385=1361W/m^2. Perfectly balanced.

Energy is universal, it applies to everything. Mass, heat, temperature, work, force.

Heat is independently related to temperature, they have an exclusive relationship through T^4. The emission of heat/the radiation from a body, depends only on the internal state, measured as temperature of the emitter. Heat emission is proven to be dependent only on the internal state of the emitting body. This is a fact based on the observation that the intensity of the heat flow, the power density, is equal for all solids at the same temperature. This was a conclusion based on observations of heated solids, where they all start to glow at the same temperature, 798K. Practically all solids, emit the same power density in the same wavelengths, at the same temperature. This means that emission of heat/light, is independent of the mass of the emitter and the composition of the emitter, it depends only on the internal state of the emitting mass, which is measured temperature, average kinetic energy, which is a net of all energy and forces present inside the body.

• “Practically all solids, emit the same power density in the same wavelengths, at the same temperature.”

Not true. The wavelengths emitted will be similar, but the W/m^2, or power density as you say, will depend of the emissivity of the surface.

• LIT,

“Example: surface emission at 385W/m^2, T=287K, and transfer to the atmosphere which emits 244W/m^2 at 256K needs a power density from the heat source at 4pir^2*244+385=1361W/m^2. Perfectly balanced.”

What is the r that you used and why did you multiply the ToA emission by 4pir^2?

7. Eli Rabett,

“You might also point them at the Green Plate Effect which has splattered a few and Izen’s new GIF”

Perhaps I should take this over to Rabett Run, but your equations are incomplete/wrong.

Before you can rearrange your bunnies, you have to take inventory of your Watts. Since the green plate is radiating σ T2^4 to the blue plate, the blue plate is not radiating σ T1^4 in either direction but rather it’s radiating σ T1^4 + 1/2 σ T2^4 in either direction. So the green plate receives from the blue plate not 1/2 σ T1^4 but rather it receives σ T1^4 + 1/2 σ T2^4 from the blue plate… which means the blue plate receives from the green plate σ T1^4 + 1/2 σ T2^4 + 1/2 (1/2 σ T2^4). Every iteration of this process emits additional radiation out of the plate system. See below:

To simplify the nomenclature we’ll abbreviate σ T1^4 = H and σ T2^4 = R .
So, for example, σ T1^4 = 2σ T2^4 is written H = 2R

Step 1. 400 W/m^2 = H + R

Step 2. 400 W/m^2 = H + R + 1/2R

Step 3. 400 W/m^2 = H + R + 1/2R + ½(1/2R)

Step 4. 400 W/m^2 = H + R + 1/2R + ½(1/2R) + ½*½ (1/2R) = H + R + ½ R + ¼R + 1/8 R

Rearranging the bunnies ½H = R and substituting into Step 4 we get

Step 4. 400 W/m^2 = H + ½H + ¼H + 1/8 H + 1/16H

This is summing a geometric series which at its completion will give us 2H = 400 W/m^2. Replacing the nomenclature abbreviation we get 2σ T1^4 = 400 W/m^2, no change in temperature.

• LJ Ryan,

What is the formula for radiation from a surface?

Physics textbooks, going back probably a hundred years, but I have only seen ones going back 50 years, all say:

R = εσT4
where R = radiation from the surface, ε = emissivity (=1 if we assume it is a perfect emitter), σ = 5.67×10-8, T = temperature of surface.

Is this correct, or not?

If correct – emission of radiation is dependent on the temperature of the surface, and for a perfect emitter (a black body) nothing else. Consequences follow and we can examine them.

If not correct – please provide your physics textbook (this will take some time of course, something close to eternity but we are patiently waiting). In this case, while waiting for the Imaginary Textbook we are not on common ground (this blog is based on Textbook Physics).

Once you have clarified, we can review.

• LJ, I am not sure how you say we agree when we come to different answers!

In my answer, the addition of the green plate clearly causes the blue plate to get warmer then it was before. By ‘slowing the blue plate’s radiation loss’, the whole system warms. We can treat it as a series (and we could take smaller time steps to get a more accurate description of the temperature as a function of time), but as Green warms from 0K to 221K, Blue warms from 244K to 262K.

IF H & R are the actual, final answers once equilibrium is reached, then you can stop with Step 1: H+R = 400W/m^2. To this we can add the requirement that H-R = R since the power into the green plate (H-R) must equal the power out of the green plate (R). Two lines of algebra gives 3R = 400 or R = 133 and H = 267 (dropping the units for ease of typing).

IF H & R are the just initial estimates before equilibrium is reached, then Step 1 is already wrong. The power radiated to space is NOT necessarily 400 W/m^2. In this case we could iterate (or do some other more sophisticated approaches).

Suppose H0 = 200 and R0 = 0 (ie suppose the blue plate has stabilized, and then the green plate at 0K is introduced) . Then only 200 is radiated to space (200 from blue and 0 from green), while 400 is absorbed. The system must warm up. Well, since the green plate is absorbing 200 but radiating 0, we could make a next guess (jut like in the original post!) that R1=H1/2. IE that R1 = 100.
Then the blue plate receives 400 + 100. So it now receives 500, so it will arm up to radiate H2 = 250 from each side. We go on alternately warming one plate, then the other.

STEP 1: H0 = 200, R0 = 0
(H + R = 200)

STEP 2: H0 = 200, R1 = 100 (=1/2 H0)
(H+R=300)

STEP 3: H1 = 250, R1 = 100
(H+R=350)

STEP 4: H1 = 250, R1 = 125 (=1/2 H1)
(H+R=375)

STEP 5: H1 = 262.5, R1 = 125
(H+R=387.5)

STEP 6: H1 = 262.5, R1 = 131.25
(H+R=393.75)

STEP 7: H1 = 265.6, R1 = 131.25
(H+R =396.88)

STEP 8: H1 = 265.6, R1 = 132.8
(H+R=398.44)

Viola! The correct iterative approach converges to H=266.67 for the blue plate and R=133.33 for the green plate and 400 total. (of course, it was MUCH easier just to do the two lins of algebra).

(Is it cheating to give the answer before LJ agrees with the two conditions? 🙂 )

• tjfolkerts

I’m glad to see you agree with me! The green plate does not warm the blue, it simply slows the blue plates radiation loss. Said otherwise the green insulates the blue.

• LJRyan,

The plates will have to become warmer than they would be on their own in order for the rate of emission to match the rate of absorption. You can call it insulating, but that is really just a simple term that’s thermodynamic implications are that a cooler exterior is radiating energy back to a warmer interior, thus raising the temperature of the warmer interior.

8. SOD,

R = εσT^4 is correct.

• LJ Ryan

So the result should be pretty simple.. But to calculate it we need one more assumption – the confusing first law of thermodynamics (conservation of energy).

1. The energy out of the system must be the energy into the system once it is in steady state (that is, one the temperature of each component has stopped changing, that is, once internal energy is not changing).

– Correct or not?

2. The energy input is 400W/m2. This is defined.

– Correct or not?

3. The radiation from the green plate out to space is the only energy out.

– Correct or not?

4. So the radiation from the green plate to space must be 400W/m2

– Correct or not?

• SoD,

In regards to your questions 3 and 4, according to the graphic, isn’t the left side of the blue plate and right side of the green plate the radiative output?

• SOD,

1. Correct

2. Correct (input to the blue/green plate system)

3. Not. The blue plate also radiates

4. Not, the blue plate also radiates

• LJ Ryan,

The transmissivity of the green plate is zero.

That is, radiation from inside cannot be transmitted through it.

The radiation from the blue plate towards the green plate is 100% absorbed by the green plate.

Now, given this constraint – 3. the radiation from the green plate out to space is the only energy out.

– Correct or not?

• SoD,

To be clear according to the graphic, the left side of blue is radiating out to space, correct? This is the equilibrium equation from eli’s site.

“Looking at the two plate system, the energy going in is 400 W/m2 and the energy going out is σT14 + σT24 Since these will be equal at equilibrium

400 W/m2 = σ T14 + σ T24”

Half of the blue plate emission goes towards green, the other half goes to space.

• Thanks Brad – for pointing out I haven’t read the graphic properly. (Thought it was a variant of my heating from the inside concept but now I have read the text as well and see how Eli is trying to stretch people’s conceptual thinking with an alternative approach..)

• LJ Ryan,

Thanks for spotting my mistake (I misread the graphic).

3. Energy out = σT14 + σT24

this is the radiation of the blue plate to space (left) plus green plate to space (right).

– Correct or not?

• SOD,

“You agreed with the formula at the start. Now you’ve come up with a completely different formula.

Which part of the reasoning above is wrong and why?”

The reasoning above is not necessarily wrong, but certainly incomplete.

The Blue/Green plate example starts with one plate defines variables and builds from there. When the green plate is added the variables are redefined but use the same names as was used prior to introducing the green plate.

The intended purpose is to “prove” cold radiation warms a hotter surface by adding steady-state variables together, but only once, Specifically, the green radiates to the blue while the unchanged blue radiation, conferred via the sun, is incident on the green. Either the variables are fixed in equilibrium or they are a step-variable, but cannot be both in the same instance.

In the plate assembly example, Eli steps the variables once, the green to blue radiation, then claims equilibrium.

If in fact, the blue plate absorbs radiation via the green plate, must it also radiate half of that absorbed green radiation back to the green plate?

• LJ Ryan,

If in fact, the blue plate absorbs radiation via the green plate, must it also radiate half of that absorbed green radiation back to the green plate?

Sure. And the green plate radiates half of that towards the sun, i.e. its temperature increases. At steady state, the governing equations are

σ(T₁⁴ + T₂&#2074) = 400W/m²

Tࠡ⁴ – T₂&#x2074 = T₂&#x2074

The first equation is conservation of energy. Energy in = energy out. The second equation is the energy flow between the plates (σ cancels out), which must equal the energy emitted from the back side of the green plate. You now have two equation in two unknowns, so you can calculate values for Tࠡ and T₂. From the second equation,

Tࠡ⁴ = 2T₂&#x2074.

T₂&#x2074 = Tࠡ⁴/2

Substituting,

3σT₂&#x2074= 400W/m²

T₂ = 220.21K

1.5Tࠡ⁴ = 400

Tࠡ = 261.88K

The blue plate radiates 266.67 W/m² from each side.

The green plate radiates 133.33W/m² from each side.

The blue plate radiates half the 400W/m² from the sun plus half the 133.33W/m² from the green plate from each side.

I hope I didn’t screw up the formatting, or this won’t make much sense.

• Rats. A semicolon must have been left out when I copied.

• And a σ here and there too.

• “In the plate assembly example, Eli steps the variables once, the green to blue radiation, then claims equilibrium.”

No, Eli steps the variables once and explicitly states this is NOT equilibrium, because the numbers don’t add up. Then he directly “steps to infinity” using knowledge of radiation and knowledge of conservation of energy. His results clearly work and are the one correct final answer.

• nitpick: There is no equilibrium, only steady state, in an open system like the blue and green plate example.

• DeWitt:

It should be a nitpick, but with these folks it’s an important point. Over at PSI and the “Sophistry” blog, the Slayer-in-Chief calls this steady-state condition “equilibrium” and concludes that there are no heat transfers (even “net”) in equilibrium.

My favorite example is his recent slideshow “disproving” the GHE, where he concludes that since the earth is in roughly steady-state conditions, it receives no “heat” (Q) from the sun. I didn’t know whether to laugh or cry…

9. ChrisA,

“Will the final temperature of each container that is established after time
a) be the same

b) differ

The physics of the author lifeisthermal leads by logical deduction to a). Real world experiments show b).

How can this discrepancy be explained?”

Ummm, if the delta T between ambient and the immersion (ta – ti) heater is zero the two samples will, in fact, be the same.

If delta T is not zero, the two samples may differ. A small delta T not much of a difference if any. A large delta T, a larger difference.

Cutting to the case of your thought experiment, the heated water in the dewar will not be warmer than the immersion heater. That is, the boundary to water temp after time is the heater, not the walls of the container. Likewise, the boundary to average terrestrial temperature is insolation, not the walls of the container, i.e. CO2.

• I’m not understanding why the water in the dewar wouldn’t be at a higher temp than the water in the beaker? After all, the radiative input jus the same. However the radiative emission would be drastically different, with the beaker having a higher emissivity. Would this not lead to the breaker system being cooler?

And unless the immersion heater turns itself off when it is over a desired temp, the water temp in the dewar would be much higher than the water temp in the beaker.

I took his thought experiment to be illustrating that the containers will be at different temps, not that they will be warmer than their internal heaters (although i still think that the water in the dewar will be warmer than the heater and water in the beaker) Lifeisthermal seems to think that a cooler outside surface doesn’t aid in heating a warmer interior. If physics works as lifeisthermal expects than the dewar and beaker temps should be identical, both internally and externally.

• LJ Ryan,

First let me make one restriction I have forgotten to mention in the original comment: There will be no security shutdown of the immersion heaters at any temperature. Thus, we can treat those heaters like an infinitely working heat source of constant power – much likely to the sun in Eli Rabbet’s GPE.

The difference is this: Since the sun will not be heated up further, the immersion heaters WILL be heated up differently in the different envoirenments, though fed with identical powers.
The reason simply is, that systems which have constant input of energy, but different “speeds” at which energies can leave those systems, must differ in their containment of total amount of energy. In the case of my water-in-containers example, this difference in internal energy manifests in a difference in temperature.

If there was no difference, the 1st law of thermodynamics would be violated.

Maybe it helps to remember that an immersion heater is no power source as such, but a heat device ( with a surface ) driven by a power source, which can be located outside the system easily.

• ChrisA,

Not sure this “restriction” changes my previous answer…. if the delta T between ambient and the immersion (ta – ti) heater is zero the two samples will, in fact, be the same.

Assuming the magnitude of delta T is large and negative, the thermos water will be closer in temperature to the temperature of the heater; that is, the temperatures will differ between the beaker and dewar.

Neither container, however, will get warmer than the heater; regardless of the infinite time on,

• LjRyan,

Assuming the power converted to heat by each heater is equivalent and remains constant, do you agree that the heating element and water in the dewar will be at a higher temp than the heating element and water in the beaker?

• LJ Ryan

The question was not if the water can be warmer than the surface of the heater ( it is never ), but if the final temperature of water in steady state would be the same in both containers.

Yes or No?

• ChrisA,

“The question was not if the water can be warmer than the surface of the heater ( it is never ), but if the final temperature of water in steady state would be the same in both containers.

Yes or No?”

It depends on the temperature of ambient surrounding the containers;

Yes if the heater is the same temp as ambient.

No, if the heater is warmer than ambient

No, if the heater is cooler than ambient

• LJ Ryan ( and Brad Schrag ),

Of course my question was made under the condition that the two containers will be in identical envoirenments, which are of course cooler than the heater.

Thought that to be clear from the beginning: If not identical envoirenments, there would be no comparability. If the envoirenment was warmer than the heater, the device is not even a heater anymore. But I defined it as one!

That is stating the obvious. The experiment is extremely simple under this conditions and leaves with a simple question ( to LJ Ryan ):

Will under this clarified conditions the water in the beaker at steady state be at a final temperature less than the final temperature at steady state in the dewar, YES/NO?

Can’t make it easier than this.

10. SOD,

“3. Energy out = σT14 + σT24

this is the radiation of the blue plate to space (left) plus green plate to space (right).

– Correct or not?”

Correct

• LJ Ryan,

σT14 + σT24 = 400

– Correct or not?

• SOD,

σT14 + σT24 = 400

– Correct or not?”

Correct

• SOD,

Sorry SOD but I’m not changing the equations. And my reply is no more mish-mash than the equations used in the plate assembly.

Your questions 1-5 pertain to equilibrium conditions.

Your questions 6-7 pertain to a step and/or variable conditions.

So I ask you again: If in fact, the blue plate absorbs radiation via the green plate, must it also radiate half of that absorbed green radiation back to the green plate?

• LJ Ryan,

Now we look at the green plate.

6. Energy input to green plate = σT14.

– Correct or not?

7. Energy output from the green plate = 2σT24

– Correct or not?

• SoD,

Minor point on formatting.

Lets see if the sub tag works too.

σT14

• Hmm. Neither sub nor sup works, so how did you get a superscript 4? I’ve only found 1 and 2.

• SOD,

6. Not. Energy input to the green plate is σT1^4 + 1/2 σT2^4 + 1/8σT2^4….

7. Not. Energy output from the green plate = 2σT2^4 + 1/4 σT2^4 + 1/16σT2^4….

• LJ Ryan,

Let’s start with #7 – emission of radiation from the green plate. You agreed earlier (November 8, 2017 at 4:12 am) that the formula for emission of radiation from a surface was:

R = εσT4 [ – edited this just after to make this just the temperature T rather than specifically T1]

We’ve defined the surface to be a perfect emitter so ε = 1 and the formula becomes R = σT4.

The green plate is at temperature T2 (which we want to work out).

Therefore, the green plate is emitting σT24 to the right and σT24 to the left.

You agreed with the formula at the start. Now you’ve come up with a completely different formula.

Which part of the reasoning above is wrong and why?

• DeWitt,

Here is how I get the formula to work when writing the comment:

I should have used the sub tag (imagine less than and greater than signs surrounding sub, just like for the sup tags in the graphic) to write T14 but was a little lazy.

• LJ Ryan,

Lets look at it another way. If T1 = T2, then no energy can flow between T1 to T2 because they are in thermal equilibrium. That can only happen if the back side of the green plate is perfectly insulated. The energy that is radiated by the green plate to space must come from the blue plate, which means there has to be a temperature difference between the plates.

• If I might interject, LJ seems to be treating T1 and T2 as fixed, initial temperatures. That is T1 =244K = the temperature initially with just the Blue Plate, radiating 200W/m^2 in each direction. And T2 = 205K = the temperature with the Green plate warming up enough to radiate out the initial 100 W/m^2 in each direction. These are NOT the steady-state temperatures!

From there, the radiation of the two plates provide feedback. Some of the green plate radiation hits the blue plate, causing it to warm. then the warm blue plate (at a temperature above T1) radiates back extra power, so that green gets warmer than T2.

Then the final temperatures of the plates (which I will call T(blue) and T(green) are indeed given by LJ’s infinite series:
σT(blue)^4 = (σT1^4 + 1/2 σT2^4 + 1/8σT2^4….)
σT(green)^4 = (2σT2^4 + 1/4 σT2^4 + 1/16σT2^4….)

OTOH, SoD’s discussion is assuming that T1 = T(blue) = the final temperature for blue and T2 = T(green) = final temperature for the green plate.

• LJ Ryan,

The reasoning above is not necessarily wrong, but certainly incomplete.

The Blue/Green plate example starts with one plate defines variables and builds from there. When the green plate is added the variables are redefined but use the same names as was used prior to introducing the green plate.

The intended purpose is to “prove” cold radiation warms a hotter surface by adding steady-state variables together, but only once, Specifically, the green radiates to the blue while the unchanged blue radiation, conferred via the sun, is incident on the green. Either the variables are fixed in equilibrium or they are a step-variable, but cannot be both in the same instance.

In the plate assembly example, Eli steps the variables once, the green to blue radiation, then claims equilibrium.

If in fact, the blue plate absorbs radiation via the green plate, must it also radiate half of that absorbed green radiation back to the green plate?

That’s just a mish mash of words.

You accepted a formula to be true. Now you have introduced a new formula which contradicts the old formula.

If the temperature of the green plate, T2 = 500K the radiation from each surface is 3,544 W/m2.
If T2 = 400K the radiation from each surface is 1,452 W/m2.
If T2 = 300K the radiation from each surface is 459 W/m2.

This is what we find in physics textbooks and what you accepted at the start.

Now at the end of your series of answers apparently it often isn’t this value.

So if T2 = 300K then – according to you – the radiation can be any kind of value depending on what is going on around it?

– My claim, if T2 = 300K then the radiation from each surface is always 459 W/m2.

You claim this is not true, contrary to your earlier claim that it is true. Hopefully onlookers can see how bizarre this is.

• Oops. The series for green should not have the first “2”:
σT(green)^4 = (σT2^4 + 1/4 σT2^4 + 1/16σT2^4….)

Or we can more simply state this all in terms of power. Then P1= 200 and P2 = 100 are the INITIAL powers radiated by the blue and green plates respectively:
P(blue) = P(1) + 1/2 P(2) + 1/8 P(2) + …
= 200 + 100/2 + 100/8
= 266

P(green) = P(2) + 1/4 P(2) + 1/16 P(2) + …
= 100 + 100/4 + 100/16
= 133

Or even simpler:
P(blue) = P(1) + 1/4 P(1) + 1/16 P(1) + …
= 200 + 200/4 + 200/16 + …
= 266

P(green) = 1/2 P(1) + 1/8 P(1) + 1/32 P(1) + …
= 200/2 + 200/8 + 200/32
= 133

• LJ Ryan,

Sorry SOD but I’m not changing the equations. And my reply is no more mish-mash than the equations used in the plate assembly.

Your questions 1-5 pertain to equilibrium conditions.

Your questions 6-7 pertain to a step and/or variable conditions.

So I ask you again: If in fact, the blue plate absorbs radiation via the green plate, must it also radiate half of that absorbed green radiation back to the green plate?

At the start I asked a question. You confirmed it.

On November 8, 2017 at 3:17 am I asked:

What is the formula for radiation from a surface?

Physics textbooks, going back probably a hundred years, but I have only seen ones going back 50 years, all say:

R = εσT4
where R = radiation from the surface, ε = emissivity (=1 if we assume it is a perfect emitter), σ = 5.67×10-8, T = temperature of surface.

Is this correct, or not?

Please confirm what equations hold under what conditions.

“Assuming the power converted to heat by each heater is equivalent and remains constant, do you agree that the heating element and water in the dewar will be at a higher temp than the heating element and water in the beaker?”

Yes, but only if the ambient surrounding both containers is cooler than the heater. If the ambient is temp greater than the heater, the beaker will be warmer; if the ambient is equal to the heater, the water temperatures will be equal.

• Fair enough. I guess for me the assumption was that if you were putting a heater into a container it was with the objective to heat the water than the ambient temp, else it wouldn’t be a heater.

• LJ Ryan,

that is correct now.

Of course the active heaters will always be warmer than the ambient of the containers. Never mentioned that, because I thought that to be clear from the beginning.

But how can you now state the correct answer? Above in a direct answer to me you stated: “No, if heater is warmer than ambient”

Yes to Brad Schrag, No to me, two opposing answers to the very same question.

Which one is correct now? Please confirm one of them.

• “Yes to Brad Schrag, No to me, two opposing answers to the very same question.”

Similar but not the same question:

Brad asked: “…do you agree that the heating element and water in the dewar will be at a higher temp than the heating element and water in the beaker?”

you asked: “…but if the final temperature of water in steady state would be the same in both containers.”

• Chris,
I think LJ is taking the approach that the final temps of each are dependant on the ambient temperature outside the vessels. That is, if the ambient temperature outside the vessels is warmer than the heating elements that the water will be warmer in the beaker than the dewar. While i haven’t tried to work out the math i think this is accurate. However, i think it is an incredibly extraordinary situation that we would have to clarify that the ambient temperature is cooler than the heater. After all, a heater by definition increases the temperature of a system so you’d expect the heater to have a higher temperature than ambient.

12. Maybe Unicode works

σT⁴₁

13. Many people are very confused on how to do heat transfer calculations. I explain some basics in Heat Transfer Basics – Part Zero.

It’s just basic stuff taught in the beginning of physics and engineering undergraduate courses.

14. SOD,

“At the start I asked a question. You confirmed it.

Hello SOD, confirm—confirm—confirm! The plate assembly, as conceived by Eli, neglects to include all radiation incident on green, via blue.

So once again SOD, If in fact, the blue plate absorbs radiation via the green plate, must it also radiate half of that absorbed green radiation back to the green plate?

• LJ Ryan,

The equation for radiation from a surface is dependent only on temperature.

You confirmed it. Then you claimed it was something else. Now you have “confirmed confirmed confirmed” it again.

This is why physics uses formulas. No ambiguity – at least to people who understand formulas.

The equation for radiation from a surface tells you everything you need to know for how radiation varies with temperature. Otherwise there would be other terms in it. This is how physics solves problems.

R = σT4

Not

R = σT4 x (dT/dt)
where dT/dt = change in temperature with time

or

R = σT4 + 1/2Tother
where Tother = another temperature somewhere else

or any other variation.

This is the whole point about the equation. It tells you ALL of the dependencies

You are trying to do heat transfer equations in some very strange way because you don’t understand how to solve for unknown values in the steady state condition.

We have two variables that we don’t know – T1 and T2.
We can form 2 equations for the steady state result that include T1 and T2 by using the conservation of energy.

I can tell you the answer. Eli told you the answer. You said he was wrong. I asked you to confirm each step. You confirmed the step. Denied the step. Confirmed the step.

I’ll do the solution for you.

a. The radiation into the system = the radiation emitted by the system (conservation of energy) – this was point 3.
b. The radiation from the green plate = the radiation absorbed by the green plate (conservation of energy).

Now, using the equation you confirmed/denied/confirmed.

a. Radiation into the system = 400 = σT14 + σT24

(you confirmed this equation apparently on November 8, 2017 at 8:01 pm as point 5).

b. Radiation into the green plate = radiation emitted by the green plate, so:
σT14 = 2σT24

(you refuted these equations as points 6 & 7 on November 8, 2017 at 10:27 pm)

Somehow – according to you – point 5 was correct, but point 6 and 7 have different formulas.

The surface has no idea what is nearby. The molecules just radiate thermal energy according to the local temperature. (This is why the formula is only dependent on the local temperature, T).

Solving these equations to find T1 and T2.. we substitute b into a

400 = 2σT24 + σT24 = 3σT24

so T24 = 400/3σ, so T2 = 220K

Now using this value of T2 in equation b:

T14 = 2T24

T1 = T2 x 21/4 = T2 x 1.19 = 262K

This solution takes care of the fact that radiation is going back and forward. It uses the principle of conservation of energy and the equation for emission of thermal radiation.

In this solution energy is conserved. It is the final steady state condition.

• And in the worked example above, I said energy was conserved.

You can see this easily.

a. Energy into the system = 400.
Energy out of the system = radiation from left surface of blue plate + radiation from right surface of green plate = σT14 + σT24 = 5.67×10-8 x (2624 + 2204) = 400

Energy in = Energy out at these temperatures.

b. Energy into green plate = σT14 = 267
Energy out of green plate from left surface = σT24 = 132.8
Energy out of green plate from right surface = σT24 = 132.8
Total out of green plate = 267.

Energy in = Energy out at these temperatures.

Wow. Conservation of energy works. The Stefan Boltzmann law of thermal emission works.

• Just to be tedious seeing as there is a rounding error above (and 132.8×2 doesn’t quite equal 267), more accurate values from the equations:

T2 = 220.21K, T1 = 261.88K

Emission of radiation from system at these temperatures = 400.01 – check

Energy into green plate = 266.68
Energy out of green plate = 133.33 x 2 = 266.66 – check.

• SOD,

Uhmmm…there’s a problems with the reasoning leads to T1= 262K

Initially σT1^4 = 2σT2^4…okay fine.

Then the plates are declared in steady state so σT1^4 equal σT2^4…

Which is correct?

Does σT1^4 = 2σT2^4 or does σT1^4 = σT2^4?

• LJ Ryan,

Where do you see that σT₁⁴ = σT₂⁴ at steady state? I can’t find it.

• Ljryan,

The radiation exciting the system is t1 and t2. The amount of radiation in t1 equates to the entire amount that t2 emits, hence the 2t2. But remember hall’s of t2 goes back into the system. So at equilibrium, the left side of the first plate (t1) plus the right side of the second plate (t2) have to add up to 400. Total emission of the second plate quill still equal t1, out being split in half between each side means t2=t1/2

• DeWitt,

By definition, objects in steady state radiate to one another equally. If blue radiates to green more then green radiates to blue, green will continue to warm and blue will continue to cool until steady state is achieved.

Initially σT1^4 = 2σT2^4. By definition, when the plates achieve steady state, must equal σT1^4 = σT2^4. If you want to claim σT1^4 = 2σT2^4 still holds, the radiation incident on the blue plate via the green plate must equal 2σT2^4, not σT2^4 as is stated in the problem.

• LJ:

You are confusing a dynamic steady state condition with a static thermodynamic equilibrium. This is a common mistake, and one of the fundamental errors of “Slayers”.

Steady-state conditions simply mean that the temperatures and power flows do not change over time. It does NOT mean that these flows must be zero (net) or equal and opposite (gross).

In these problems, we are interested in the steady-state condition. With non-zero power flows into and out of the system (and subsystems), it is never in thermodynamic equilibrium.

• Lj,
they would radiate at each other equally except the left plate has radiative input, only half of that input is passed over to the right plate. Their inputs differ, can’t expect their outputs to be equal.

• LJ Ryan

You’ve invented a law of thermodynamics.

If you have only 2 bodies in an isolated system then yes they will be radiating to one another equally. This is not “by definition”. This follows as a consequence of conservation of energy in this particular setup.

If you have multiple bodies in a system this does not follow as a consequence.

I show the working out in my comment of November 9, 2017 at 4:12 am – the result there follows from conservation of energy.

You have an idea in your head of what the right answer is, and you are trying to work backwards to justify this “right answer in your head” by first inventing a new Stefan-Boltzmann equation and second inventing a new law of thermodynamics.

I still have very low expectation that you will come to realize this.

• LJ Ryan,

You’re assuming that steady state is the same thing as thermal equilibrium. It’s not. At steady state, temperatures don’t change with time, but you are allowed to have heat flow and thus you’re allowed to have objects with different temperatures in the system. At thermal equilibrium, as you say, by definition temperatures are equal because there cannot be heat flow. But this system has an effectively infinite capacity heat source at high temperature and an effectively infinite capacity heat sink at close to absolute zero. There can be no thermal equilibrium at normal time scales.

Let’s look at the consequences that entail from T₁ = T₂. In that case, there is no heat transfer from blue to green or from green to blue. That means that all of the incoming 400W/m² has to be radiated from the front side of the blue plate and thus T₁ = T₂ = 289.81K. Of course that means that either the green plate can’t be radiating any energy from its back side or there must be a heat source radiating 400W/m² to the back side of the green plate for energy to be conserved.

• Ed Bo, DeWitt and SOD,

You guys are all correct, I confused steady state with equilibrium. If however, you look at the Green Plate assembly as described by its author over at Rabett Run, the system in in equilibrium, not steady state. Also, numerous posts also use the steady state and equilibrium interchangeably.

DeWitt attempted to correctly standardize the system description as STEADY STATE on November 9, 2017 at 7:22 pm.

• For the diagram to depict equilibrium, wouldn’t both the green and blue plate need to have the same temperature, and thus emission? Trying to understand the terminology.

• Conventions depend on context. DeWitt pointed out to me a long time ago that I was using the term “equilibrium” to apply to “steady state” and I’ve tried to use “steady state” since. I’m sure that many examples and textbooks on heat transfer use equilibrium where they “should” be using steady state, but it’s not important enough to go digging through tomes to demonstrate it.. I’m happy to be wrong on that point. My memory is often faulty.

Thermodynamics equilibrium means that the system is closed, with no energy loss or gain.

The system we are considering has flows of energy and is not in thermodynamic equilibrium.

In fact, thermodynamic equilibrium is a very uninteresting place – useful only for demonstrating a few core principles of thermodynamics (as far as I can see). The climate, and simple examples to illustrate climate energy flows, are completely different from this condition.

• And to comment specifically on your question:

So once again SOD, If in fact, the blue plate absorbs radiation via the green plate, must it also radiate half of that absorbed green radiation back to the green plate?

If a body absorbs radiation it doesn’t have to radiate any given % of it back. Not as a fundamental principle. That is not how radiation from a surface works.

The formula in that case would be something like:

Rout = σT4 + Rin/2 (we don’t find this in textbooks)

Instead we form one or more equations from conservation of energy and find out the steady state result.

The requirement of conservation of energy generates the result -> what temperatures satisfy the conditions of conservation of energy and the Stefan-Boltzmann equation?

This is how we get the result.

You can see the end result in the equation (November 9, 2017 at 4:12 am):

σT14 = 2σT24, and therefore, cancelling σ:

T14 = 2T24 – this final result has the “factor of half” that you have some intuition about.

I’m guessing, but don’t know, that I can’t give you conceptual insight. This is just my expectation from a number of years of running this blog – providing simple examples with equations from textbooks, and their solutions, and watching confused people reply with confusion and imaginary physics.

—–

I can only demonstrate to readers that there is just one equation of thermal emission from a surface and it is always the correct equation and it always depends only on the temperature of the surface, not sometimes true and at another time a quite different equation.

If that was the case you would be able to find this alternative equation of thermal emission in textbooks.

Seeing as this alternative is not the case, and seeing as conservation of energy is true, we can easily prove mathematically the steady state solution for temperature for this problem. There is only one solution. It satisfies both conditions.

Some people don’t like the result.

15. Just to clarify:
* I think SoD is presenting a simple, correct analysis of the the Green Plate effect.
* I think LJ is presenting a different, more complicated, *almost* correct analysis. If LJ cleaned up a few things in his analysis & description, he would ALSO get the correct answer.

Both approaches should lead to the blue plate radiating 266 W/m^@ @ 262 K, and he green plate radiating 133 W/m^2 @ 220 K. The presence of the cooler, unpowered green plate is responsible for the blue plate warming from 244K to 262 K.

16. LJ Ryan

“So once again SOD, If in fact, the blue plate absorbs radiation via the green plate, must it also radiate half of that absorbed green radiation back to the green plate?”

I’m not SOD but: Yes. And You will find that this leads, if you do those steps iteratively ( half, half of half, half of half of half… ) you will end up at the values delivered by the GPE.

But you can have it even easier.

If you assume the system to be in steady-state and if you assume the 1st law to be correct, then the system will have to fulfil two requirements:

1) Input = Output for the entire system
2) Input = Output for every single plate

You will find that only ONE solution satisfies 1) and 2) together.

Or, to say differently: No rise in temperature at blue when green enters the system would violate the 1st law. It’s consequence of the 1st law that the presence of colder objects can cause further warming of warmer objects, if the system is fed by a heat source constantly.

17. ChrisA says

“It’s consequence of the 1st law that the presence of colder objects can cause further warming of warmer objects, if the system is fed by a heat source constantly.”

Not if the two objects are thermally isolated from their surroundings

Two metal blocks A and B sit separated inside a vacuum filled adiabatic enclosure.
Adiabatic enclosure consists of a perfect reflector face surrounded by a perfect insulator

Initially both at the same temperature. The zeroth law of thermodynamics applies.

Both emit and absorb equal amounts of radiation.
Neither one is said to heat the other.
Both objects remain at the same temperature

One block (A) has a power supply which is now switched on causing the temperature of the block to rise.
This in turn means that it will emit more radiation.
A will now heat B causing its temperature to rise.
B will in turn emit extra radiation but this ‘back’ radiation is caused by A.

Now comes the clincher

If B were not there at all the temperature A would be even higher.
So B cannot be said in any meaning of the word as a cause of heating A

• Sure. Yours is a closed system, unlike the plate thought experiment. It can reach thermal equilibrium. If you increase the heat capacity of the system by introducing another object, a fixed amount of energy added will result in a lower temperature increase. But that’s irrelevant to the plate result where the system is open and equilibrium is never reached, only steady state.

• Bryan

Please note that a system as designed by you will heat up one body faster without the presence of another body. True.
But if your energy source works eternally, equilibrium will never be reached and thus no final temperature can be defined.

Anyway, the systems used for simplifying are NOT adiabatically enclosed and thus showing a different behavior.

If enclosed like you suggested, a second body besides the heater will prevent some of the initial energy to be reflected from the walls. That slows down heating.

If not enclosed, a second body besides the heater will prevent some of the initial energy to vanish into space immediately. That slows down cooling.

We are talking about the latter case here.

And if you slow down cooling while a heat source gives a constant input, you will end up at some higher temperature than without the slow down.

If you state otherwise, please explain what happens to the energy difference of the warmer body between the cases source + one body / source + two bodies.

In the latter case the warmer body is unable to lose energy as fast as in the former case. Nonetheless it is fed constantly by the source. What is to happen then?

18. TJ’s comment about the dewar and beaker got me thinking yesterday. What happens to the temperature of the interior of a hollow vacuum sphere that is heated from the exterior? If I could suspend a second sphere in the middle of the first, would it be warmer than the outside ambient temperature that is heating the larger sphere?

Tried to make a graphic with applicable equations to help illustrate the question and results I ended up with better…

Click to enlarge

• Brad: The radiation emitted by the outer sphere is emitted inward in “all direction” – over a solid angle of 2Pi. Therefore some of the radiation emitted inward by the outer sphere misses the inner sphere and is absorbed by the outer sphere, not the inner sphere. So you need to correct for that. Depending on how you approach the calculation, you may also need to correct for the angle at which the radiation strike the absorbing surface using Lambert’s cosine law. When you get the “viewing angles” right, I presume you will find that second law of thermodynamics is correct and that both spheres will have the same temperature.

• It’s called Venus.

• This is NOT Venus. Frank is right (and for the right reasons) — both the shell and the inner sphere will be at the ambient temperature, Ta.
To be like Venus, the scenario would require:
1) a “sun”
2) a shell that allows incoming short wavelength sunlight and blocks outgoing long wavelength thermal IR.

• Brad Schrag asked: “If I could suspend a second sphere in the middle of the first, would it be warmer than the outside ambient temperature that is heating the larger sphere?”

Frank replied: ” When you get the “viewing angles” right, I presume you will find that second law of thermodynamics is correct and that both spheres will have the same temperature.”

I know for a fact that when the calculation is done correctly, it will give the same temperature for the two spheres. The laws of thermodynamics say so. Nice to see that Brad Schrag has confirmed that by correcting his calculation.

This points out a weakness with SoD’s challenge questions for the extreme skeptics. If a detailed calculation leads to a result that violates basic principles, there must either be an error in the calculation or an error in applying the basic principles. If I am certain I have not done the latter, then I may see no need to bother with finding the error in the calculation.

Many of the extreme skeptics base their arguments on what they think are basic principles that they either misunderstand or misapply. So challenging them to find a flaw in a detailed calculation will do nothing to convince them. One must show them the flaw in their approach.

I have tried this from time to time with extreme skeptics. Once it starts to become clear that they can no longer defend their position, they typically stick their fingers in the ears and start chanting something to block out the noise they don’t want to hear.

• Or they do like Vizzini in The Princess Bride, shout “Inconceivable,” or some other words to that effect.

• Mike, the actual aim of this isn’t to convince anyone of anything. It’s just to screen out the over-confident but confused who mostly resist answering either of these basic questions. If they don’t answer, they don’t get to write more comments.

• Thanks Frank, that does make sense. So in my correction I took the approach that if I were on the inside of s1 looking toward the center, s2 would take up a percentage of my field of view r22/r12. Correcting the irradiance received at the surface puts everything how it should be. Gives the right answer, not sure if it’s the proper method to solve this though.

A cavity with a small hole for observation is the best approximation we have for a blackbody. It would be perfect if there were no hole that lets radiation leak out or in, but then it wouldn’t be useful. If you do the calculations, it turns out that the absorptivity/emissivity of the wall doesn’t make much difference. The photon gas inside the box will still have a blackbody spectrum at the temperature of the wall. The AERI FT-IR spectrophotometer used to obtain atmospheric emission spectra uses two cavities at different temperatures for calibration.

• Let a=b and be a real, non zero number;
-Multiply each side by a

a^2=ab;
– next subtract b^2 from each side

a^2-b^2=ab-b^2;
– factor each side

(a+b)(a-b)=b(a-b);
– common (a-b) on each side, so divide both sides by that to clear it out

a+b=b;
Since a=b lets substitute b for a;

b+b=b;
2b=b;
2=1;

Funny math and funny physics can give very real dilemmas.

• Of course, this is not a “real dilemma” since it is predicated on a very simple error along the way. 🙂

• Maybe i should have said it’s a real personal dilemma, especially if you avoid learning from your mistakes.

• It does certainly illustrate how easy it is to be misled if you are not careful and/or not willing to look past the things that seem so obvious. (I assume most people here know the error … )

• For anyone who doesn’t know, it’s division by zero. (a-b) = 0 That will give you the #DIV/0! error in Excel. Before the division, the last equation gives you (a+b)*(0) = b*(0), which is correct, 0 = 0. 0/0 is still undefined.

19. SoD, while it may be satisfying to fire away at the sophist bloggery of amateur sceptics, you might find a little hardball scepticism a tad more challenging to critique.

The Sceptical Chymist

• Quondam,

If you have something to say, why don’t you say it? You serve no purpose to link to a wretchedly written anonymous document that rambles along with no clear starting point and no apparent destination.

• Quondam,

What’s the question? It looks like the author hasn’t read much atmospheric physics. If they had they would understand why using the adiabatic lapse rate as the environmental lapse rate is a good approximation – it what’s we find all the time in the tropical region in practice. Another way of saying it is the potential temperature is almost constant up through the tropical troposphere:

No one is ever going to produce a GCM that solves for turbulence – it’s just about the hardest problem in physics. So we use the constraints that we do know. Of course, everyone in climate modeling understands that convection and the resulting humidity, cloud formation and rain events is one of the big problems in atmospheric climate models. There isn’t some article of faith, some dogma that is accepted as a revealed truth.

For example, I recommend anyone interested in understanding the subject better to read Water Vapor Feedback and Global Warming, Held and Soden 2000. Reading one good review paper is worth doing, reading 100 papers is of course better.

Here is an extract:

There is a large gap between climate sensitivity experiments with compre- hensive climate models and computations with simple models like the radiative-convective model.

Because of the turbulent character of atmospheric flows, the complex manner in which the atmosphere is heated (through latent heat release and by radiative fluxes modified by intricate cloud distributions) as well as the rather complex boundary condition that the Earth’s surface provides, it has proven difficult to develop models of an intermediate complexity to fill this gap, and the continuing existence of the gap colors the sociology of the science of global warming. Building and analyzing climate models is an enterprise conducted by a small number of groups with substantial computational resources.

Many processes occur in the atmosphere and oceans on scales smaller than those resolved by these models.

These scales of motion cannot simply be ignored; rather, the effects of these small scales on larger scales must be approximated to generate a meaningful climate.

Some aspects of this closure problem have been reasonably successful, whereas others are ad hoc or are based on empirical relations that may not be adequate for understanding climate change.

And a graphic from their paper:

——————————-
Anyway, this post is aimed as a future filter. To avoid comments overwhelming threads from the “skeptics” who refuse to answer two basic questions while pretending they have great insight.

This post isn’t aimed at discussing the finer points of problems in climate modeling. There are many other articles which go into that subject in a lot of detail.

• Quondam

The author of your linked article doesn’t know much about physics. Fancy verbiage that is definitively quite hollow. Let me just comment on the ansatz and one obvious error :

1/ Only linear near equilibrium thermodynamics is addressed by Onsager in his article and work. That includes for instance ordinary thermal conduction (Fourier’s law) or diffusion (Fick’s law).
None of the relevant processes in atmospheric heat transfer namely radiation, convection and latent heat transfers belong even remotely to his theory. Instead they are much more difficult and complex processes and belong to far from equilibrium thermodynamics (google for instance Ilya Prigogine). Equation 1 about local entropy production ( 2 law of Th in fact) and the article definitively fails to demonstrate where GHE theory might be at odds with it. Finally there is no relationship between heat flux and local temperature gradient in such far from equilibrium processes.

2/Quote from article: Photons are bosons, molecules aka electrons are fermions.

This is false.
Photons are indeed bosons, yet molecules and atoms may be either bosons or fermions. He_4 is a boson, He_3 is a fermion. H2 (hydrogen molecule) is a boson.

By the way it is also wrong to claim that :

Photon fluxes are intrinsically bi-directional in contrast to those of molecular origin.

In ordinary thermal conduction in a gas the molecular energy fluxes are of course as bi-directional as the photon fluxes. See the theory of thermal conductivity in a textbook.

• As GC has taken time to read this essay, a brief reply seems not inappropriate.

Eq. 1 is the 2nd law as phrased by Onsager. It is in no way restricted to equilibria, steady states, or linear dissipative processes. Dimensionally it’s ds=dQ/T.

Eq. 2 is the Carnot Equation it is limited to two-terminal steady-state problems but not linear processes.

The Navier-Stokes Equation proper contains, in my textbook, no thermal gradient terms but a linear viscous dissipation term (L&L Eq.15.7). It can be modified to include linear thermal and visco-thermal dissipations (Eq. 56.4) of limited relevance when free nonlinear convection is present.

The radiative transfer models of Manabe et al. do seek self-consistent thermal profile calculations except for that portion of the troposphere where GHGs are important (e.g. Fig 1, Held & Soden 2000). If one is willing to assume the troposphere is not in thermodynamic equilibrium, adding CO2 can only push it further away from equilibrium and require an increased rate for entropy production. By the 2nd law, this means an increase either in an energy flux or an increase in the thermal gradient. The former is commonly assigned a solar-fixed value.

The core message of this essay is in the title, Thermal Dissipation. In the crudest description of atmospherics, 1 joule of energy enters from a 5000K source and departs as 1 joule from a 200K source. First law, no problem. Second law, 96% of the incoming energy has been dissipated. A very rough picture might be that for every 1ev photon coming in, twenty-five 0.04ev photons depart. The ‘ubiquitous’ thermodynamics of climate modeling is 1st law observant and 2nd law oblivious.

It may be worth noting that no assumptions of linearity have been made in the less trivial cases discussed. Fluxes are defined as gradients of unspecified functions of temperature. Nonlinear solutions follow from global variational dissipation calculations. That these so easily converge to a common value is characteristic of variational solutions. When seeking forcings to increase surface temperatures, one should not be too surprised to find Mother is looking for ways to neutralize your efforts.

• Quondam,

If one is willing to assume the troposphere is not in thermodynamic equilibrium….

Is the atmosphere isothermal No, it’s not. Therefore, it’s not in global thermal equilibrium. That’s a fact, not an assumption.

• Quondam

Let’s be quite clear.
The atmosphere and whole climate system is a dissipative structure and a system far from thermodynamic equilibrium. Energy is dissipated and entropy produced. Nobody argues about this !

Yet,
1/ there is unfortunately no known, established extremum or variational principle that rules systems far from equilibrium. Prigogine could not discover one and none could ever be discovered since. Entropy production isn’t.
2/ In convection in general and in atmosphere in particular a defined average temperature gradient doesn’t imply a defined convective and latent heat flux. Same gradient may result in different heat fluxes. This is in sharp contrast with a relationship such as the Fourier law where the heat flux is uniquely determined by the gradient and the thermal conductivity.
3/ The thermal gradient close to the wet adiabatic in troposphere is an emergent property, essentially independent of GHG’s provided there are enough of them to make the air columns in troposphere unstable with respect to convection. This is an observational fact ! In spite of various GHG concentrations the vertical temperature gradient in troposphere is measured to be close to the adiabatic be it on Earth, Jupiter or Mars.

So when you say:

But, more importantly, this critical thermal gradient has always been assumed to be invariant to greenhouse gas perturbation and thus not a variational parameter. No justification has been offered, except the tangential assertion of convective equilibrium. That a gradient may be calculated with equilibrium parameters is not proof this gradient characterizes an equilibrium system − a fundamental thermodynamic paradox. Should we dope a copper wire with a trace of antimony we expect an increased resistance due to reduced electron mean free paths and, under constant current conditions, an increase in internal potential gradients. By analogy, there should be a thermal gradient increase on tropospheric doping by trace amounts of greenhouse gases due to shortened photon mean free paths.

think about it. The reasoning with the doped copper is inappropriate in case of convection and latent heat. It’s near equilibrium linear thermodynamics.

• Quondam

Quote:

The venerable equations of Navier- Stokes and Schwarzschild appeared to be the only tools in the arsenal of this science – the former dealing with isothermal viscous phenomena and the latter with radiative fluxes in defined thermal profiles. The dormant chemist within asked, “Isn’t this basically a problem of energy transport in thermal gradients? Where’s the Thermodynamics?”

Wrong in every respect.

Navier Stokes equations do not of course assume isothermal fluids. See a textbook on fluid mechanics.

Radiative transfer does not assume defined thermal profiles. The latter is calculated self-consistently.

And obviously thermodynamics is already everywhere in “this” (sic, GHE) science . In particular it is merely an essential part of the equations of fluid mechanics.

And by the way convection is not just a problem of thermodynamics. The heat flux in convection is not a simple function of local thermal gradients as in ordinary conduction. It’s a global phenomenon that depends for instance on Rayleigh number in Bénard convection..

Obviously, the chemist asleep didn’t even notice.

20. SOD wrote: “I can only demonstrate to readers that there is just one equation of thermal emission from a surface and it is always the correct equation and it always depends only on the temperature of the surface, not sometimes true and at another time a quite different equation.”

To be more accurate, SOD might have said: “I can only demonstrate to readers that there is just one equation of thermal emission from a surface IN LOCAL THERMODYNAMIC EQUILIBRIUM and it is always the correct equation and it always depends only on the temperature of the surface, not sometimes true and at another time a quite different equation”.

Some great scientists have invented a number of devices (LEDs, lasers, fluorescent lights) that produce far more visible light than normally expected given their temperature. They all involve tricks that place more molecules in an excited state expected for a Boltzmann distribution of energy over all states. I suspect that SOD used the phrase “thermal emission” to account for this contingency.

There are several other caveats that might be mentioned. One is emissivity. Another is that the derivation of Planck’s Law starts with the assumption that radiation is in equilibrium with the material emitting that radiation – which is true for all surfaces relevant to climate emitting radiation. When we observe the thermal radiation emitted by our planet from space, however, we don’t observe a blackbody spectrum because different wavelengths come into equilibrium (by repeated emission and absorption) at different altitudes/temperatures.

I’ve made many mistakes by forgetting that the fundamental physics of absorption and emission of photons is determined by Einstein coefficients. Unfortunately, the standard physics education doesn’t lead us from Einstein coefficients to the Schwarzschild eqn (which requires matter in LTE) and then to Planck’s Law (which requires thermodynamic equilibrium between radiation and matter) and finally to the S-B equation (which adds a wavelength-dependent emissivity term, which I assume involves symmetrical reflection at the surface so that absorptivity equals reflectivity). Beer’s Law applies when emission is negligible, ie when powerful (hot) sources of radiation is absorbed by much cooler matter.

None of this changes the correct answer to the simple issues SOD and others are trying to straighten out above.

“which I assume involves symmetrical reflection at the surface so that absorptivity equals reflectivity)”

I’ve always thought sb law doesn’t include absorptivity because it is focused on the power of the radiation emitted by an object. When it’s reversed to calculate the temperature based on a known incoming flux, the absorptivity of the surface is used and emissivity is dropped since it doesn’t apply. A=E for a particular wavelength I’d a surface, right?

21. Having a discussing on a different forum in regards to GHE and the divergence theorem came up, yippy can perhaps guess who. Anyway, this was my response as i understand the divergence theorem and it’s implications. Am i accurate in my thoughts and reasoning?

It’s been 20 years since I used the divergence theorem so I had to brush up on it to make sure I understood it’s implications. There is no breaking of the divergence theorem by the GHE. Divergence Theorem states that the amount of flux crossing the border of an enclosed surface is equal to the sum of all the sources and sinks on the inside of the volume. There are no implications that the incoming has to equal the outgoing. For a given object, there can be a positive divergence, negative divergence, or zero. Let’s use the satellite measurements of incoming and outgoing radiation to see what this means for the earth and our atmosphere. Measure incoming from the sun is 1365W/m^2, or in terms of total power hitting the disc of the earth, 1.74E17 Watts. Measured outgoing LW (which by the way, measured reflected solar flux is 101, which gives the albedo value of .3, but were not concerned with that, just the LW since it is what the surface and atmosphere emit), annual mean, is 238W/m^2, or when adjust for the entirety of the sphere of the earth is 1.21E17 Watts. If I were to relate this to the Divergence Theorem, I would note that there is a negative divergence on the earth in terms of radiative flux. The amount coming in minus the amount going out equates to 5.26E16. What that means is that there have to be sinks absorbing that amount of Wattage within the surface and/or atmosphere, or to put it differently, there is 4.54E21 Joules of work being done by the incoming radiation. Converting that back to W/m^2 equates to 103. If there isn’t something absorbing those watts, creating that negative divergence within the volume of the earth/atmosphere then you would be correct. However, our measurements are very clear that there is more watts coming in then going out, which leaves watts to do work on the atmosphere and surface in some way.

• “measured reflected solar flux is 101, which gives the albedo value of .3, but were not concerned with that, just the LW since it is what the surface and atmosphere emit

The “flux crossing the border” would also include the outward, reflected shortwave sunlight. This light travels outward through the theoretical border around the whole earth, so must be included.
* There is an inward flux of ~(341 W/m^2)*A of SW radiation (sunlight).
* There is an outward flux of ~(101 W/m^2 + 238 W/m^2)*A = ~ (339 W/m^2)*A of SW radiation (reflected sunlight) and LW radiation (IR earth-glow).

The ‘negative divergence’ is only the ~ 2 W/m^2 of net inward flux. It is this small imbalance that drives the warming.

• tjfolkerts,

You are nearly right, the best numbers (Stephens et al. , 2012) I know of are: flux from the sun is 340.2 +/- 0.1 W/m^2 (averaged over the sphere),
reflected short wave is 100.0 +/- 2.0 W/m^2
outgoing long wave is 239.7 +/- 3.3 W/m^2
total out is 339.7 W/m^2.

The imbalance is 0.5 W/m^2. As you can see that is nearly an order of magnitude too small to get from the satellite data. It comes from measurements of ocean heat content. I think that has been updated to 0.6 W/m^2.

• Ah right. I should have reduced the incoming by the measured sw that is leaving.

22. Have you seen this?

23. on November 23, 2017 at 7:53 pm | Reply Steve Titcombe

Hi, this is Lampacres accepting the challenge (from Skeptical Science).

I have never seen this site before: excellent site name. Conjures up the image of Tim the Magician guarding the bridge at the end of the Monty Python film “the Holy Grail”. As Tim found out, you must be sure to know the REAL answer to your own question before asking it.. as the response to his question “What is the air-speed velocity of a swallow” proved to be his down fall.

If you’d be so kind, can I submit my full response by email: there’s lot’s of font settings (superscript on stuff) that will get messed up.

I won’t abuse the email address (but you can create a tempory one if you doubt me). I’ll trust you to publish the document here by correcting only formatting issues.

Isn’t this exciting….

Best Wishes
Steve Titcombe

• Steve,

African or European?

Sure, you can email to scienceofdoom – you know what goes here – gmail.com.

• Steve’s essay graphically reproduced (5 pages):

Click to enlarge

Click to enlarge

Click to enlarge

Click to enlarge

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• Steve,

You agree with the equation of radiative transfer. Awesome.

This equation predicts that when we reduce the transmittance of the atmosphere – for example, with an increase in GHGs – the outgoing longwave radiation reduces (for an atmosphere with a decreasing temperature with height).

Correct?

• “2. If the object (now at thermodynamic equilibrium} is a perfect blackbody, then it will be radiating photons ”

Steve,
Am I understanding this statement by you to mean that an object only radiates once it reaches thermodynamic equilibrium?

• This is really much text, but the basic error included begins in the very design of the thought experiments. It is easiest to see with the case of the two infinite plates.

They are designed to go to thermodynamic equilibrium, which means they neither cool to their envoirenment, nor are they heated by some energy source. Thus, they have to be in an adiabatic enclosure.
For this, the claim of thermodynamic equilibrium ( uniform temperature after certain time ), is perfectly true.

However, regarding the real earth we have to look for a strictly non-adiabatic case, for it is heated by the sun on one hand, and constantly emits energy to space on the other hand. The system surface-atmosphere is therefore NEVER to reach thermodynamic equilibrium in the sense of the infinite plates. It is a steady state system, equilibratet TO some input. The analogy fails totally.

For a correct analogy the two plates need an energy source ( representing the sun ) emitting onto one plate ( representing surface of earth ), and the ability to lose energy to their envoirenment ( representing emission from the system to space ).

And you will find, that for very simple reasons, namely the 1st law of thermodynamics combined with mere book keeping of energy quantities, the introduction of a second plate ( representing a long-wave active atmosphere ), will lead to temperature rise of the first plate ( the one the “sun” shines on ).

The system will always go to a state where output into space is equilibrated to input from the source, thus:

a) INPUT = OUTPUT for the entire system.

Furthermore, every single plate will go to a state where also

b) INPUT = OUTPUT for every single plate.

If you find the values for a case where a) AND b) are true and that also does not violate the 1st law, you will find only ONE possible state, and in in this the temperature of the first plate has risen compared to the case where the second plate is not present.

It is very simple.

The error done in the text is conceptual, because the dependencies of the parts of a system are to match if an analogy is constructed. Here, they do not so. It would be necessary to show, that something heated by an energy source will not get warmer, even if insulation of that something is increased.

You can enclose the CPU of your computer with room temperature cold wool, its temperature will nonetheless increase compared to no wool. ( please try it maybe AFTER you replied ).
The reason is very simple: If you put energy into something at a constant rate, and then diminish the rate at which this energy can leave to the environment, you will produce an accumulation of internal energy and thus higher temperatures.

For real earth in very raw principle it’s like this: Sun radiates to surface, but surface cannot radiate to space. The radiation to space happens at altitudes above the surface from the atmosphere. While energy comes to the surface at light speed, it can leave there only by much slower processes, mainly convection, towards those altitudes. To maintain the balance INPUT = OUTPUT, the slower transport needs to be denser of energy compared to the faster one. That makes it warmer than without that slow down.

The same goes with two plates, if the system is open and a energy source is there.

• I find Steve Titcombe’s text to be nearly illegible, but he seems to suffer from a number of misconceptions. (1) He seems to think it is possible to have a temperature gradient at equilibrium. (2) He does not seem to understand the difference between a steady state and thermodynamic equilibrium, with the Earth being in more-or-less a steady state, but nothing like thermodynamic equilibrium. (3) He seems to know that heat, energy, and temperature are different things, but does not seem to know what they are. (4) He seems to think that the equations of radiative transfer are somehow irrelevant to radiative transfer through the atmosphere. There is probably more, but it is too much a mess to untangle (like the irrelevant blathering about bosons).

• Steve,
Yes, I should have read further.

Let’s look at your sphere thought experiment. For a system to be in steady state (ss) means that it is in radiative balance when power in=power out or that Absorption=Emission. I think you’ll agree here. Let’s look at the whole system then.

We know that the surface is in ss when A(surf)=E(surf) so we can say the system is in balance when:

E(surf)=E(shell)

I’m going to start by looking at E(shell). Ideally we want to represent the E(out) as a function of the temperature of the orginating surface, will make solving for that temperature easier later on.

E(shell) is split in two:

E(shell) = E(out) + E(inward)

As you stated, the shell radiates both inward and outward evenly.

So the shell is in radiative balance when it’s absorption, A(shell), = emission, E(shell) so:

E(shell) = A(shell)

How much is it absorbing? That is based on the surface temperature, T(surf) and the surface areas of the sphere and shell, SA(sphere) and SA(shell)

E(surf) = sb*T(surf)^4 * SA(sphere)
A(shell) = E(surf)/SA(shell) or A(shell)=sb*T(surf)^4 * SA(sphere)/SA(shell)

so E(shell) is:

E(shell)=sb*T(surf)^4 * SA(sphere)/SA(shell) or sb*T(surf)^4/4

Now we can represent E(out) as a function of the surface temperature to be:

E(out) = E(shell)/2 or E(out) = sb*T(surf)^4)/8

Ok, E(out) is solved. We need to look back at our original balance equation and analyze the other side:

E(surf)=E(out)

So what’s the A(surf):

E(surf) = A(power source) + some amount of E(inward)

Remember, the outer shell is radiating toward the surface, as well as the internal power source. Thus the surface temperature is no longer based on a single source of radiation. So how of E(inward) does it receive? I can break this down a bit more or you can find my thought experiment from a couple weeks ago on this site, but in short the amount of power received at the surface is E(inward). It will receive the same amount of W/m^2 as E(inward) represents. As you said, most of the radiation emitted will miss the sphere but what is received still has the same energy density as what is emitted on the inside of the shell. So:

E(surf) = A(power source) + E(inward)

We know E(inward)=E(out) so:

E(surf) = E(power source) + sb*T(surf)^4/8

Now we have the left and right sides solved as functions of T(surf) so let’s put them together and solve, also E(power source) is 459W/m^2, so:

E(surf) = E(shell)

459 + sbT^4/8 = sbT^4/4 :moving the t^4 to the same side we get:

459 = sbT(surf)^4/8 :multiply by 8:
3674 = sbT^4 :divide by sb:
6.48E10 = T^4 :quarter root:
T(surf) = 504K

Without a shell the surface temp steady state was 300, now it’s 504. Why did it warm up? I had to go through and correct some of my variable names, so apologies if I missed some corrections, but I feel the end math is correct.

24. on November 24, 2017 at 9:33 am | Reply Steve Titcombe

No, you are not understanding this statement correctly. The paragraph to which you are referring to, stated;

2. If the object (now at thermodynamic equilibrium} is a perfect blackbody, then it will be radiating photons across the entire spectra corresponding to the Wien’s Displacement curve. If it’s not a perfect blackbody the wavelength distribution of the photons will not correspond to the Wien’s Displacement curve – but the object will still meet it’s Stefan-Boltzmann requirements for Power Density emission from it’s surface.

The purpose of this note was intended to explain the difference in photon frequency spectra between a perfect blackbody and a body which is not. However, it is apparrent that my reference to thermodynamic equilibrium (going back to an earlier point) may have thrown you. If you had proceeded further into my essay you would have seen several mentions of emissions occuring at all times (if not, how else could thermodynamic equilibrium be achieved between the two objects). Hopefully, the following paragraph, mid-way through, should re-assure you;

The heat flow equation states that object T1 emits ENERGY fully, at: sigma * T14, all of the time; it never has it’s energy “stopped up” inside of itself. And, the ENERGY from the 2nd object (if it’s cooler than the 1st object) can’t act as HEAT. The 2nd object does not stop the 1st object from emitting. As the 2nd object radiates it’s own ENERGY commensurate with it’s own TEMPERATURE, HEAT will flow across the boundary to the cooler object. This HEAT (Q) leaving the first object is the same HEAT which is passed to the 2nd object (there is only one boundary between the two objects).

• You have to note that the people describing the greenhouse effect do not demand “heating” of the surface of earth by the atmosphere, if your distinction of energy and heat is taken as given.

They are claiming that the sun heats the surface to higher temperature when long wave absorbing gases are in the atmosphere, compared to the case with no such gases.

And the mechanism of this is as I described above: The outgoing energy can’t leave the surface at the same high speed ( or better: rate ) than it’s coming in – however: if there were no long wave active gases or even no atmosphere at all, then it could do so. And this difference comes simply from the presence of certain gases. Their mere presence leads to a higher temperature due to sunshine. No direct heat transfer from atmosphere to surface is necessary for this ( again your distinction between heat and energy taken as given ).

• Heat and work, they are the only things that can increase the temperature.

Which is the atmosphere, heat or work? Which is co2? Heat or work?

“The outgoing energy can’t leave the surface at the same high speed ( or better: rate )”

You need a reference to support this claim. It must include experimental data showing how the rate of transfer and temperature is reduced(transfer) and increased(temperature). As an effect of adding a cold fluid, or co2, to a heat source.

Your claim is contradictive in itself, since an increasing temperature of the source(surface), inevitably leads to a higher rate of transfer to the surroundings.

The real relationship goes like this;

adding a low temperature fluid to a heat source, is an addition of mass to the constant heat flow. Without the fluid layer, all heat is emitted by the solid. With the fluid, the solid body must supply the energy for emission from both solid and fluid, at any point in time.

The emission of the atmosphere and surface, must be supplied with heat from the source(surface) at the same time. The surface emits according to its temperature, and transfers heat to the atmosphere at the same time.

Without atmosphere, no transfer. With atmosphere, added transfer.
It is easy to see that an atmosphere, which cause an additional amount of heat to be transferred, causes more energy to be drained from the source.

Which is perfectly logical. Everybody knows that adding cold fluids to hot bodies causes the to cool down. It is crazy to think that cold fluids and dry ice would have opposite effect on bodies in space, to what we observe in every day life.

Give one single example of a situation where adding a heat absorbing low temperature fluid to a heat source, causes increasing output of power from the heat source when it heats the fluid.

• Lifeisthermal,

A surface has a temperature of 300K.

Can you tell us what the radiation emitted by the surface is if it’s is radiating into a:

1) a 0 K vacuum

And
2) a shell that is 250K

Better yet, if you don’t want to run the numbers, just grace us with your opinion of which one will have a higher radiative transfer from the surface.

• lifeisthermal said:

You need a reference to support this claim. It must include experimental data showing how the rate of transfer and temperature is reduced(transfer) and increased(temperature). As an effect of adding a cold fluid, or co2, to a heat source.

Something like this?

Give one single example of a situation where adding a heat absorbing low temperature fluid to a heat source, causes increasing output of power from the heat source when it heats the fluid.

Since the sun is the heat source in real GHE and it’s power output is not affected by GHE, the sense of your question needs further explanation.

Maybe it is as simple as DeWitt Payne states: You do “forget” the sun when it suits your purposes.

In any way I would find it interesting to see you explain what my linked videos show.

• lifeisthermal,

Without atmosphere, no transfer. With atmosphere, added transfer.
It is easy to see that an atmosphere, which cause an additional amount of heat to be transferred, causes more energy to be drained from the source.

Only if the atmosphere also emits to space. If it didn’t, the transfer would eventually stop and the surface temperature would go back up the the same level as without an atmosphere.

It’s actually pretty easy to demonstrate the single layer, non-reflecting (more or less) atmosphere greenhouse effect. Build two well-insulated boxes with one end open, the sides coated with aluminum foil or aluminized mylar film and the bottoms painted black. Put thermocouples in the bottoms of each box, cover the open end of one box with cling film and the other with glass. Point the boxes at the sun. The temperature of the bottom of the glass covered box will be higher than the cling film covered box because the cling film is nearly transparent to thermal IR and glass is nearly opaque to thermal IR. Or see here: https://scienceofdoom.com/2014/06/26/the-greenhouse-effect-explained-in-simple-terms/#comment-120211

Horace-Bénédict de Saussure showed that this worked in 1767 with his hot box, an early example of a solar oven.

• Steve,
This quote from you seems like the basis of your misconceptions

“The heat flow equation states that object T1 emits ENERGY fully, at: sigma * T14, all of the time; it never has it’s energy “stopped up” inside of itself. And, the ENERGY from the 2nd object (if it’s cooler than the 1st object) can’t act as HEAT.”

Radiation is radiation. Whether or not radiation is absorbed by a surface is dependant only on the absorptivity of the surface in regards to the incoming radiation wavelength. As long as the absorptivity is not 0, the surface will absorb some of it, or all of it if it has an absorptivity of 1.

That is why the equation for net transfer involves the surface temperature and the temperature of the surroundings P=σ(To 4 Ts 4)

The temperature of the surroundings will have an impact on the emitting bodies rate of heat loss. The reason the surroundings have an impact is because of the radiation the are emitting. The hotter surface isn’t going to not absorb incoming radiation because it is from a cooler object. You are correct that it isn’t considered heat because it is coming from a cooler surface. However, it is still energy that is absorbed. This incoming and absorbed energy reduces the overall rate of energy loss by the hotter object.

Because the hotter objects rate of energy loss has been reduced the result is it that it will warm up. As it warms this causes the surface to increase how much radiation it emits. Eventually it will warm enough that it’s emission matches the entirety of the incoming to the surface, and then it will be in steady state.

25. on November 24, 2017 at 9:54 am | Reply Steve Titcombe

To Science of Doom (4:06 AM)

How the EMR (as LWIR radiation) from the Earth’s surface propogates through the Earth’s atmosphere is not an issue for me. The radiative energy flow equation is not a concern to me (providing that it’s not misappropriated like the Heat Transfer equation has been). So, to eliminate any further discussion about question 1: I most certainly agree that the Earth is not a perfect blackbody so the spetra of wavelengths emitted by the Earth’s surface does not correspond to the Wien’s Displacement curve (even so, the Earth’s full budget of energy, as mandated by S-B, is emitted) and, most certainly, I agree that specific wavelengths within that EMR spectra which is emitted from the Earth’s surface do get attenuated more than other wavelengths due to the LWIR-active molecules in our atmosphere (even so, the full budget of energy is emitted out into the 0K envirnment). I hope that we can now agree that we both respect the 1st LoT.

• Steve,

You accepted the equation is true. If a) the transmittance of the atmosphere reduces (more GHGs) the result b) is a reduced OLR.

This is a consequence of the equation.

You can “blah blah blah” but if the equation is correct then b) follows a) as a mathematical certainty.

Maybe you don’t think the equation is true?

Or maybe you think i) the equation is true, ii) if we reduce transmittance in an atmosphere where temperature reduces with height then the consequence is NOT that OLR reduces.

That would be a fascinating mathematical piece of work.

PS This is why I ask people to confirm or deny this equation.

26. on November 24, 2017 at 7:59 pm | Reply Steve Titcombe

To Science of Doom (11:01AM)

I suppose that I’m taking the third option that you’ve offered me.

Yes, the OLR is affected by the increased quantity of LWIR-active molecules in the atmosphere, but only insofar as the spectra of wavelengths of photon emissions towards the 0K environment is further distorted from what would otherwise be seen if a perfect blackbody were radiating out directly into a 0K environment i.e. only the profile of the OLR emission spectra is modified. The actual outgoing longwave radiation budget, as an energy value, is NOT affected. The entire Earth system (comprising the Earth’s surface and all parts of it’s atmosphere) MUST continue to radiate ENERGY to the 0K environment at the power density commensurate with it’s ‘system’ TEMPERATURE- it has no ‘choice’, the S-B law mandates that it must.

I have previously referred to the cool molecules near the top of the troposphere, so let’s say that they’re at a TEMPERATURE of 215 K. Let’s also say that a photon with a LWIR wavelength is emitted from the Earth’s surface (at TEMPERATURE 288K) and makes it nearly all the way to the top of the atmosphere but then strikes a LWIR-active molecule (at 215K). The ENERGY transferred to the molecule will, as kinetic energy, cause it to vibrate more and thus we say it’s TEMPERATURE has increased. Before this molecule can emit a photon with the same ENERGY value, it is highly likely that it will collide with another molecule (not necessarily an LWIR-active molecule). This other molecule may then radiate photons at a wavelength representing it’s own (cooler than Earth) TEMPERATURE – in this way the OLR budget is guaranteed but the Wien’s displacement curve is now ‘low-end shifted’ i.e. more ENERGY radiated at the even longer wavelengths.

The TEMPERATURE of the molecules in the atmosphere are indeed represented by their own (even longer wavelength) LWIR emissions in all directions – and that’s both outward (to the 0K environment) and downward (towards the Earth’s surface). However, please see my treatment of question 2 to see why this downward emission of ENERGY does not manifest as HEAT upon any warmer molecule that it eventually collides with.

Finally, no HEAT is “trapped” by LWIR-active molecules and no HEAT is “amplified” by LWIR-active molecules and HEAT is not a conserved quantity (it reduces the zero at thermodynamic equilibrium). Again, please refer to my treatment of Question 2 to understand why this is so.

• I don’t think you understand the role of an equation in this process. It isn’t garnish to the main meal.

Now you need to demonstrate that the equation produces the opposite of what it does produce. This will be interesting. I suspect you have no idea of what this equation means or how it was derived.

27. SoD,
Is there a guide or link for reference for typing special characters in the comments?

28. on November 24, 2017 at 8:45 pm | Reply Steve Titcombe

Thanks for persevering with my essay (I can’t deny that it’s readability is poor when compared with all the other excellently written contributions to this site).

Let’s take your revised numbers of Tsphere = 322 K and Tshell = 227 K as agreed. This means that the sphere is now compelled to be radiating at 610 W/m^-2. To do so, over it’s entire surface of 12.56 m^2, would mean that it has 7655 W at it’s disposal. But we previously calculated that the sphere had an internal power source of only 5768 W. Somehow, our new system is now generating an additional 1887 W – that’s an extra 1887 J every second. It is this additional energy which is the issue for me– and it can not be claimed that the shell causes the sphere to have a “back log” of emissions (causing itself to heat up as a result of this backlog) – that was the reason for my ramblings about photons being bosons early in the essay. The sphere has no difficulty in meeting it’s S-B commitments commensurate with it’s temperature – it has no ‘choice’ but to do so, always.

And the other thing that I need help understanding in this “restricted emissions” argument: If the sphere now has 7655 W at it’s disposal, what is preventing this ‘additional energy’ going around again (and again), ever growing at each iteration?

And finally, when the radius of the shell ≈ the radius of the sphere (where the shell is not quite touching the sphere) your explanation provides 6000 W for the sphere to dispose of and yet when the radius of the shell = radius of the sphere such that the sphere and shell are in conductive mode, the model returns to normality i.e. the new external surface of the sphere (formerly the external surface of the shell) returns to radiate at a temperature of just 300K again. And if I reverse the process so that the radius of the shell ≈ the radius of the sphere again, I can ‘gain’ energy. This is what really troubles me: this point where Radius sphere = Radius of shell.

• Steve,
In a very sloppy manner I have arrived at the stable state for the environment in your thought experiment. The surface has a temperature of 300K. A shell is introduced, what happens.

The shell, let’s assume is not at 0K, but let’s assume it’s much colder than the surface. So it emits some radiation, both in and out. But because it’s rate of emission is lower than what is coming from the surface, it’s temperature increases. What is the result of an increased surface temperature? Emission increases. This process happens until the emission from the shell=absorption of the shell, or in other words the shell is in steady state (not equilibrium. As a side note, the strict usage of the word HEAT is a pet peeve of mine but if you are going to prescribe to it I’d suggest brushing up on usage of the word equilibrium).

At the point that the shell is in equilibrium we can then work backwards and calculate how much radiation that amounts to at the surface. Remember, SB law isn’t about heat transfer. It relates surface temperature to radiation emission. That means that the surface will be in steady state when the absorbed radiation equates the emitted radiation. We can therefore say that the surface temperature that is receiving 7600W, or 602W/m^2 is going to be 322K.

What I gave you isn’t a step by step process, it is the final steady state result of your experiment. So when you say what is the what keeps it from growing at each iteration, this is the end result of it growing until power in = power out.

Emission inside a shell is different than emission outward from a surface. Inverse square law discusses emission outward, shell theorem discusses emission inward. It states that the distribution throughout the inside of a shell will be a constant amount. So if the inside of the shell is emitting at 100W/m^2 that what the power/area will be at any point throughout the shell.

Going to recheck my math…..

• So I still didn’t get the math right.

balance of shell

power in = power out

emision from shell = sbT2^4 so

sbT1^4/4 = sbT2^4

emission from one side of shell is

sbT2^4/2

balance of inner sphere

power in = power out

internal + back radiation = sbT1^4

459 + sbT2^4/2 = sbT1^4

459 + sbT1^4/8 = sbT1^4

3672 = 7sbT1^4

t1 = 310K

sbT1^4/4 = sbT2^4
T1^4/4 = T2^4
310^4/4 =
t2 = 219K

29. Steve,

Many people can’t understand simple heat transfer.

I asked a question of one of our regular commenters with a longstanding inability to understand this simple subject – A Challenge for Bryan and more interesting for you is the solution: A Challenge for Bryan – The Solution.

It should be simple for you to find the mathematical flaw, given that the equations are derived from conservation of energy. If the system is “creating energy” then you will be able to find it.

No one can find it, because the system conserves energy. But that doesn’t solve people’s conceptual problems.

Your brain is telling you that this solution creates energy, but the equations tell you that it doesn’t create energy. I can only imagine the frustration.

Anyway, this is a typical simple undergraduate heat transfer problem that you get when you do your first class in heat transfer.

30. Buried in a thread long after the question so I put my answer here.

lifeisthermal replied (November 26, 2017 at 12:21 am):

Yes, the radiative transfer equation accurately calculates the emission from the atmosphere in relation to emission from the surface. It is an accurate calculation of the reduction of heat emission in detail, exactly quantified for the wavelengths in the observed effective emission.

So as the transmittance reduces, for an atmosphere with a temperature that declines with height, the consequence is a reduction in OLR.

This is a necessary consequence of the equation.

Correct?

You can see that radiation is emitted from a hot surface and absorbed by a cool surface. And that radiation is emitted from a cool surface and absorbed by a hot surface..

But on this one, you make assumptions about absorption in relation to temperatures. What the s-b equation does, is calculating the emission from a body at a temperature, and, when in presence of other bodies at different temperatures, calculates the simultaneous transfer of energy according to differences in emissive power.

When you say “And that radiation is emitted from a cool surface and absorbed by a hot surface”, you need support for the part claiming that the hot surface absorbs the emission from the cold surface..

Exactly why I extracted a textbook showing this specific point and why I asked and you “answered” by not answering.

So – as far as I can tell from your waffle – you disagree with the extract from Incropera and DeWitt, and with all the other heat transfer textbooks shown in Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics.

My “support” is that this is exactly what all the heat transfer textbooks say.

It should be very easy for you to confirm it or to claim it is wrong. The equations are given in the reference multiple times.

• lifeisthermal,

Is co2 heat, or is it work?

If it is not one of those, it will not increase the temperature. Only heat and work can do that.

You conveniently forget about the sun when it suits your purposes. It’s the sun that transfers energy to the surface and atmosphere of the Earth. That’s why the temperature of the system is greater than 2.725K.

CO2 can and does affect how that energy is transferred again to space from the surface and atmosphere. The Earth is not a closed system at thermal equilibrium. It is an open system at approximately steady state. There is a flow of energy in and a flow of energy out. If you reduce the flow of energy out with the same energy in, then energy will accumulate and increase the temperature, reducing the imbalance. That’s the greenhouse effect in a nutshell.

31. DeWitt has proved yet again that the sun did it all. 🙂 Well, plus El Niño while ignoring La Niña.

32. on November 29, 2017 at 3:06 pm | Reply Steve Titcombe

Firstly, an admission regarding the SoD article: “Two Basic Foundations”. As you (SoD) have suggested: I don’t understand the significance of the Energy Transfer Equation: Iλ(0) = Iλ(τm)e-τm + ∫ Bλ(T)e-τ dτ with respect to the Radiative Greenhouse Effect, which we are discussing. As I first stated, the energy transfer equation appears sound in every respect but can you provide a link to where it’s significance to the RGHE is further explained?

Having now read the SoD Article: “A Challenge for Bryan – The solution”, I admit that it does seem plausible. If I were an undergraduate student being shown it for the first time, I could imagine that I would have no hesitation in accepting this method as “just the way this problem is solved”. I could easily perform the maths to determine the appropriate answers – not least because my pass grade would depend upon doing so. However, the situation for me isn’t so simple as imparting new knowledge on a blank canvas. I first have to let go of what I already have accepted. And that won’t happen unless I am confident that it is indeed better than that which I already have.

It is evident that “the solution” acknowledges that the surface of the shell comprises both an inside surface (at 4πRshell^2) and an outside surface (also at 4πRshell^2 i.e. 8 πRshell^2 ) rather than considering (like the RGHE deniers) only the outside surface of the shell (just 4 πRshell^2 ) as it’s effective surface area from which it emits all of it’s heat energy. Whilst at first glance “the solution” appears plausible, it is evident that it is flawed: Specifically, when Rshell is diminished to equal Rsphere , then Tshell should increase to equal Tsphere but “the solution” does not provide this outcome. Surely you can acknowledge that this hints at a problem somewhere within “the solution”? That’s not to say that “the solution” is finished but, at the very least, it does need a ‘tweak’. To satisfy my own curiosity about it’s provenance, do you know who is credited with “the solution”?

The SoD article titled: “Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics” was very useful as it showed many references to the Heat Flow equation.

It re-assured me that my two text books: “Momentum, Heat and Mass Transfer”, by Bennet and Myers, from McGraw-Hill 1982 and “Transport Phenomena”, by R. Byron Bird, Warren E. Stewart and Edwin N. Lightfoot, from John Wiley 2005 weren’t both written by scoundrels, as they also say exactly the same as your many examples; that heat flow is always from the warmer object to the cooler object. Being more assured that I’d find it in my two books, I thought I’d look for “the solution” equations. Now that I knew what I should be looking for, I was confident that I would find it – right under my nose- having been there all the time. Odd thing was: “the solution” equations weren’t there – in either book.

SoD stated: “I’m relying on the various proponents of the imaginary law because I can’t find it in any textbooks. Feel free to correct me if you understand this law in detail”. To clarify, to SoD, the position of the RGHE deniers, in respect to statement 2e, it is not true that it is said: “In a radiative exchange, the hotter body does not absorb the energy from the colder body as this would be a violation of the second law of thermodynamics”. Instead, for 2e, the RGHE denier community believes: “In a radiative exchange, the hotter body absorbs the energy from the colder body but the greater energy emitted by the hotter body always results in a net loss of energy by the hotter body”. Whilst, for 2f, it is said by the RGHE denier community: “In a radiative exchange, the hotter body absorbs the energy from the colder body but none of this acts as heat upon the hotter body as this would be a violation of the second law of thermodynamics”. This (2f) might be the fundamental difference between RGHE proponents and RGHE deniers.

I have now given more thoughtful consideration to the very particular phrase “net heat flow”: Why did these authors simply not choose to use the shorter phrase “heat flow”. Perhaps the deliberate inclusion of the additional word “net” is, after all, significant. I have previously dismissed this rather odd word “net” because I had assumed that it simply meant the difference between the energy density emitted by object 1 (being σ * T1^4) and the energy density emitted by object 2 (being σ * T2^4), but again, this is just “Heat Flow” – it does not explain the inclusion of the word “net”. So, perhaps “net heat flow” does actually mean something different to just “heat flow”. If that is true, then perhaps “net heat flow” really does allow for “some heat does occur on the surface of the hotter object but the final, overall (net) effect is that heat always flows from the hotter object T1 to the cooler object T2”. If this is indeed the case, then many authors on this subject have done a decent job in wrong-footing the RGHE denier community. However, before submitting this response, I have scrutinised again the use of the word “net” in all of those copied pages from your referenced text books and I still see no inclination by the authors to further qualify their use of this term “net”. There’s no note, footnote, suggestion, hint or inference that might suggest that the hotter body will also get hotter. In fact, your preferred example of “Fundamentals of Heat and Mass Transfer” is even less convincing to me that you might indeed have a case, when it says: “Since there are only two surfaces, the net rate of radiation transfer from [italic font] surface 1, q1, must equal the net rate of radiation transfer to [italic font] surface 2, -q2, and both quantities must equal the net rate at which radiation is exchanged between 1 and 2”. Incropera and his colleagues appear to be using the term “net” solely as meaning the difference (by subtraction) between the two ‘opposing energy flows’ and nothing more.

If you can provide the evidence that one or more of your referenced text books do include “the solution” equations then I can, with some justification, accuse the author(s) of those that didn’t include “the solution” equations of a dereliction in their duty by omitting vital details from their text books: and everyone at this web site will be happy and I’ll be wiser (once that tweak to “the solution” has been applied). However, if none of your referenced text books include “the solution” equations then I am certainly going to stay as a RGHE denier until otherwise convinced.

• Whilst at first glance “the solution” appears plausible, it is evident that it is flawed: Specifically, when Rshell is diminished to equal Rsphere , then Tshell should increase to equal Tsphere but “the solution” does not provide this outcome.

If Rshell = Rsphere then we are no longer talking about radiative transfer, and as such equations for radiative transfer no longer apply.

Being more assured that I’d find it in my two books, I thought I’d look for “the solution” equations. Now that I knew what I should be looking for, I was confident that I would find it – right under my nose- having been there all the time. Odd thing was: “the solution” equations weren’t there – in either book.

Did you textbooks have any problems with their solutions in them? Often times the textbooks are to teach the fundamentals so that people will have the understanding of how and why to apply the right equations. They aren’t going to contain a solution to every conceivable problem.

I have now given more thoughtful consideration to the very particular phrase “net heat flow”: Why did these authors simply not choose to use the shorter phrase “heat flow”.

So many people that fall into the denier group for GHE tend to latch onto the definition of heat and hold to it too tightly. Heat, as it’s defined, is the net flow of energy in the positive direction between objects. That is why heat can only travel from the warmer to the colder object. That doesn’t mean though that there isn’t energy transfer happening in each direction, which there most certainly is. That’s the very reason that heat transfer equations include the temperature of each surface. The cold surface does make an energy contribution to the warmer surface, it’s just not near as much as the warmer surface makes to the colder. That is why the net transfer, and thus heat, is less as the two objects temperature get closer. The colder surface is almost contributing the same amount as the warmer surface is losing.

When we stick to this strict use of the word heat it becomes more difficult to have open discussions about energy transfer, in my opinion. I think it’s a very important fact to understand though that the SB Law doesn’t state what the temperature of a surface is in relation to its heat transfer. It states what the temperature of a surface is in relation to its radiation, or energy transfer. This is important to understand because it helps illustrate why I warm surface can stay warmer in the presence of a cold surface, than it would without the cold surface nearby. After all, the cold surface is probably much hotter than the 2.6K avg temp of the universe that we live in.

• Steve,

I’ll pick one small part of your comment to expand on:

Specifically, when Rshell is diminished to equal Rsphere , then Tshell should increase to equal Tsphere but “the solution” does not provide this outcome. Surely you can acknowledge that this hints at a problem somewhere within “the solution”? That’s not to say that “the solution” is finished but, at the very least, it does need a ‘tweak’. To satisfy my own curiosity about it’s provenance, do you know who is credited with “the solution”?

Once Rshell = Rsphere then they will be physically in contact and will conduct heat, rather than radiate heat. Then the equations will no longer be describing the physical process. Instead we will need to write a different equation for heat transfer.

So, no, this doesn’t hint at a problem in the solution described.

I have no idea who is credited with the solution.

Anyone who understands:

a) the first law of thermodynamics
b) the Stefan-Boltzmann equation for radiation

– will be able to work it out for themselves.

Item a) energy in-energy out = energy gained. So in steady-state, energy in = energy out.
– Do you accept item a)?

Item b) radiation from a surface = εσT4 where ε = emissivity (a material property we can measure), σ=5.67×10-8, T = temperature of the surface in Kelvin.
– Do you accept item b)?

The point is, crazy as it might seem, there is no special understanding of anything needed, no revealed truth, no amazing insight.

If the example I gave is incorrect then you will be able to find the mathematical error or that the solution doesn’t satisfy the first law of thermodynamics or that the solution doesn’t satisfy the Stefan-Boltzmann equation.

It’s no more amazing than the temperature of a room going up when you add insulation to the room (e.g. close windows and doors) with the same heat source.

Confused people who learnt their “physics” at fantasy climate blogs keep claiming “you are creating energy“.

Take a look back at the comments for that example, and I can give you a few more simple thought experiments with worked solutions if you like (please just ask) – and read the comments there also – each case people say “you are inventing energy”.

No. Energy is conserved. So the point is, the confused arrive and imagine that energy is being created because of a conceptual flaw in their thinking. Look back at the example. Is energy being created?

The equation used to find the solution is “energy in = energy out”.

• Another article where you can read confused replies is Do Trenberth and Kiehl understand the First Law of Thermodynamics?.

Another simple problem worked out from basic heat transfer equations – radiation and conduction, and from energy in = energy out. I invented that one myself.

You will see the replies suggest everything under the sun.

All because confused people are unable to see that no energy is created.

Their brains tell them energy is created so off they go to explain the mistake in very random and contradictory directions.

I have to say it is both entertaining and instructive. Not everyone is cut out to do basic heat transfer.

If you are prepared to accept the astounding revelation that insulation can increase the temperature of an internally heated body you might begin to understand that all the contradictory replies are just demonstrating how little all these confused commenters understand about heat transfer. Or about the relationship between energy stored and temperature.

33. Steve T wrote: “I don’t understand the significance of the Energy Transfer Equation: Iλ(0) = Iλ(τm)e-τm + ∫ Bλ(T)e-τ dτ with respect to the Radiative Greenhouse Effect, which we are discussing.”

I was always mystified by the dilemma: Why doesn’t doubling the concentration of a GHE in the atmosphere double both emission and absorption and therefore have no effect on outgoing OLR? (390 W/m2 at the surface, 240 W/m2 at space). (Answer, it does, but only to a first approximation.)

I didn’t really understand the GHE until I learned about this equation, but I found the differential form (before integration) much easier to understand:

dI = n*o*Bλ(T)*dz – n*o*I*dz = n*o*[Bλ(T) – I]*dz

where I is the incoming radiation intensity at wavelength λ, dz is the increment of distance the radiation passes through causing a change in I of dI, n is the density of absorbing molecules and o is their absorption cross-section at wavelength λ. The first term is emission from the increment dz and the second is absorption passing through dz. T the local temperature for dz. (n*o*dz is dτ the incremental change in optical depth.) This version is sometimes called the Schwarzschild eqn.

If the emission term is negligible (say in a spectrophotometer), this reduces to Beer’s Law for absorption.

If dI = 0, then I = Bλ(T). In other words, when absorption and emission have come into equilibrium, the result is radiation of blackbody intensity. When both absorption and emission are important, but radiation has not reached an equilibrium with the medium it is traveling though, then we need to numerically integrate the Schwarzschild eqn. For example, from the surface to space and then over all relevant wavelengths. Or from space to the surface for DLR.

If the incoming upward radiation I (from the surface or GHGs below) is has intensity greater than Bλ(T), then dI will be negative. Upward intensity drops as radiation passes through the atmosphere from hotter to colder for this reason. In the stratosphere, the upward radiation can arise mostly from altitudes below that are colder. Outgoing OLR on its way through the stratosphere increases in intensity slightly (and rising GHGs there cause cooling).

The GHE only exists because our atmosphere mostly gets colder on the way to space at the altitudes where absorption and emission are most important. There would be no GHE in an isothermal atmosphere.

This software can automate the process of integrating the Schwarzschild eqn:

http://climatemodels.uchicago.edu/modtran/

34. The claim there is a radiative greenhouse effect is the claim that a cold nitrogen bath is a heater,

and that the core of the giant magical heater is the

light blocking insulating phase change refrigerant,

accelerating conduction-chilling of not just the surface and it’s associated features, but the cold nitrogen bath, too.

When someone tells you the greenhouse gases warm the planet

go to any chart of sunight top of atmosphere vs mean sea level, and point out to them the green house gases,

overwhelmingly create the

reduction in surface energy density,

noted in the charts showing lobes of energy refracted away, by the green house gases.

When you have ascertained he can read well enough to understand that,

tell him, to tell you, how many percent more light come out of the light-warmed earth,

for every otherwise available percent of warming sunlight never reaching the Earth due to refraction by green house gases.

Tell the green house gas radiative effect believer that you want to see clear, definitive work showing precisely how much more light leaked out of the planet,
when green house gases stopped one percent, from leaking into it.

Tell him to show you definitively his mathematics.

Next ask him why his Church’s teachings don’t match the temperature for the planet etched into scientific regulatory and physical standards stone, as the

International Standard Atmosphere.

Ask the fake why his Church’s pseudo-calculations come up 33 degrees short and ask him to show you where his church uses properly processed gas law to account for the part of the temperature calculation that only gas law can take care of.

Have him name the law of thermodynamics governing the temperatures of gases and have him point out to you which of the factors of the equation of the Law,
he claims,

makes more energy leave a light warmed rock, less light reaches.

When he can’t tell you the name of that law,
ask him to show you his chart of gas physics law, that assigns CO2 higher energy than Standard atmospheric air.

Tell him the real chart, part of the gas law governing gas temperatures, assigns CO2 a lower energy constant in equations, than Standard atmospheric air and therefore by definition,
must cool
any volume of Standard atmospheric air it’s mixed into.

Have the purveyor of the fakery explain to you precisely why his Church’s faked pseudo-mathematics, comes up precisely 33 degrees short, for calculation of Earth temperature.

Tell him to show you the precise place in the mathematical progression the 33 degree difference appears and when he starts mumbling about how he ain’t no Kl’eye muh tahllughJisT

tell him to name the element of the mathematics that Stefan Boltzman can’t take care of, which Gas Law is written specifically to account.

When he stalls and fidgets and talks sh** tell everyone there, that the reason the Church comes up with a fraudulent global atmospheric temperature is because their leader Hansen’s programs

faked proper calculation of global atmospheric temperature,
by not accounting for,

an element named the hydrostatic condition: intrinsic to compressible phase fluids, also known of to the layman, as

PRESSURE.

When the modern magic gas quack claims to have calculated global atmospheric temperature he has actually not invoked proper processing of the Gas Law written to

account for the intinsically variable density,
of compressible phase matter: compressible phase fluids, in instance of gases.

When the proper accounting is done,

*snap* the calculation comes out precisely equal to the International Standard Atmosphere,

and the quack telling you cold light blocking refrigerant refracting 20% of otherwise available warming spectra to space, is a magical heater,

can then explain to you why his Church doesn’t get the same temperature as the REAL global atmospheric temperature,

etched into our calibration and warranty and liability regulations worldwide, for everything from table lamps to space-ship parts.

Anything producing heat, ultimately has part of it’s warranty and specifications certification, based in the International Standard Atmosphere.

We know that temperature is right, and we know the Law invoked along the way in calculating the Standard is right.

We also know the Magic Gassers telling people light blocking refrigerant is a magic heater for an even larger heater – which is actually a cold nitrogen bath –

can’t hit that temperature when they claim to ‘calculate’ global atmospheric temperature.

There’s only one place in all thermodynamics, processing of temperature of gases can turn to,

in order to calculate their temperatures right.

We know for a fact the Magic Gas Quack James Hansen’s followers, come up with a temperature exactly 33 degrees short.

WHAT happens if you refuse to use Gas Law to account for the

3 3 d.e.g.r.e.e.s. compression warming intrinsic to compressible phase fluids, gases?

James Hansen’s and all these other frauds’ erroneous 33 degrees of fraud, happen.

What happens when you use the proper gas Law to solve for the 33 degrees of compression warming intrinsic to compressible fluids at Earth volumes, and pressures, etc?

The Magical Gaissiness, is all cleared up,
and the fraud has been set straight on why there is a GAS LAW
to TAKE CARE of that ERRONEOUS 33 DEGREE SHORTFALL

his hick @\$\$ came up with,

told him thim pewdurs wuz so faist, thay could calculate the temperatures of compressible phase matter,

without using compressible phase matter law,

to account for that 33 degrees of c.o.m.p.r.e.s.s.i.o.n. warming.

All magic gas barking hicks, are science darkening frauds who can’t even calculate the temperature of the global atmosphere properly.

If one should tell you he thought he could,

simply have him tell you how much more energy leaks out of a light warmed rock,

each time insulation makes another percent less, go into it.

He’ll talk about how he can’t understand what you’re saying. Remind him it’s because he’s the one who doesn’t know whyStefan-Boltzmann fraud comes up 33 degrees short.

All magic gas barking frauds can be ridiculed into sullen silence without ever even needing to crack a book.

Just tell him to show you that ”other” time in allllll thermodynamics,
a cold nitrogen bath left a light warmed rock warmer,
than if there was no cold nitrogen bath.

• You didn’t answer the questions. Your post makes no sense. It’s all hand-waving, and not even very good hand-waving. The technobabble on Star Trek is better than that. However, it doesn’t make any sense either.

• So you think that pressure has the cause of a 33K GHE? Please keep the answer short, like a yes or no will suffice.

• “can be ridiculed into sullen silence without ever even needing to crack a book.”

Pretty much tells us where Mr. Eltor is coming from. And going.

35. Magic Gas frauds pretend to be among the world’s mathematicians, and their original Fraud Daddy, James Hansen, confidently told the world that ‘Climate Mayuth’ could calculate the temperature of the global atmosphere and come up thirty three degrees short, and that it wouldn’t be found to be error if checked,

but – just ONE pass through the calculations that lead to the REAL atmospheric Temperature – the international regulatory and physical standard named the

International Standard Atmosphere,

and the

and what’s the first thing you notice?

There’s a 33 degree shorfall to the fakes’ pseudo-calculation, and where does this

mandatory 33 degree error arise?

It arises when the Magic Gassers don’t invoke Gas Law at the proper point along the way to

account for our Atmosphere’s

33 degrees of compression warming.

Ask your local Magic Gasser to tell you why there even IS a gas law to calculate gas temps, if another law can do it correctly without the processing gas Law provides.

He’ll stutter and talk about his malice toward people who catch him trying to pass scam error off as real mathematics,

but he won’t calculate the temperature of the Global Atmosphere

until he invokes Gas Law to account for the 33 degrees of compression warming,

Fraud Daddy Hansen swears isn’t error.

It is error, we know the correct atmospheric temperature, and we know the step his

mis calculation

omits.

The hydrostatic equation,
where the
hydrostatic condition – pressure in compressible phase fluids – gases –

is accounted.

Everything they say is fakery based on what’s called an ‘inversion scam’ – which is why every step along the way you find the eggregious violations of Conservation of Energy, a child can point out.

Like – why can’t you even calculate the temperature of the Global Atmosphere right? In math, you either arrive at the correct answer for the temperature of something or you don’t,

and Magic Gas Brigade Quacks, come up 33 degrees short.

• I fell like you aren’t going to answer a single question posed to you.

So let’s clear up the pressure issue. The only way to get thermal energy from a gas is to compress it. When compression stops there is no more production of thermal energy.

Likewise, when a gas losses pressure exerted on it it loses thermal energy.

So as a gas parcel descends from high altitude toward the surface you would be correct that it is undergoing an increase in pressure and subsequently gaining thermal energy.

However, simultaneously there is a packet ascending from the surface toward the upper atmosphere that is losing thermal energy. How do we know this?

Because if it didn’t happen that way the atmosphere would either be increasing or decreasing in volume. And specifically, since you claim the atmosphere is a source of heat, it actually means the volume would have to be decreasing. The side effect of this would be a constantly increasing pressure.

Because the total volume that the atmosphere takes up is basically constant, we know that it isn’t producing thermal energy due to compression.

36. Why don’t you all tell me the name of the law of thermodynamics governing gas and other compressible phase matters’ temperatures?

Write the equation of the law for us all here, and tell us which one of the factors of the Law,

persuade you there is light blocking insulation that can be mixed into a bath,
that makes more and more light leak out

every time it makes less and less leak in.

Write the equation of the Law,

name the factors of the Law,

tell us what each factor stands for.

When you can’t,

it’s because you’re a bunch of posturing frauds,

who can’t competently name the Law of Thermodynamics governing the temperature of the Atmosphere

your Church calculates wrongly: erroneously: arriving at the mandatory 33 degree shortfall that occurs,

when Gas Law isn’t properly processed along the way.

When you can name the Law,

and write the equation,

you need to be prepared to be asked to show us where the Law assigns CO2
a higher energy constant than Standard atmospheric air:

because the REAL Law – is comprised of the Chart of Energy Constants of Gases, in part –

and it specifically names CO2,
and it specifically names Standard air mix,
and it specifically assigns CO2 less energy than Air.

Meaning all addition of CO2 into any Standard class air mix,
must make that volume of compressible phase fluid’s temperature,

go down.

You all need to show me a chart of Law, assigning CO2 a higher energy constant,

than Standard Atmospheric Air.

When you can’t it’s because you’re a bunch of frauds.

I expect to see your Chart of Law – assigning CO2 a higher energy constant than air,

and you need to all tell me if even one of you can,

the name of the Gas Law that governs the temperatures of gases hence the Atmosphere.

And you also need to show me that ”other” time, in allll thermodynamics,

when a cold light blocking nitrogen bath,

heated the light-warmed rock it was conduction chilling.

• Let’s assume the 33K GHE is in fact driven by the change in pressure due to gas law. We need to identify how much energy is required to raise atmosphere temps at the surface by 33K.

Using the specific heat of air at approx 285K, for a constant pressure is just about 1.0040 kJ/kg*K. We also need to know how much mass we are raising the temperature of.

Mass of atmosphere: 5.148E18

So if we multiply Cp*K*mass that will yield the number of kJ needed to raise temps by 33K
1.70E20 kJ

That seems a bit high. I’m realizing that maybe we don’t need to try to raise the temp of the entire volume, just the part closer to the surface. So what if we take a different approach and use that

w=p*Δv

We can use the stefan boltzmann constant (incorrectly I might add) to determine how much radiation would be needed to sustain a difference of 33K.

So (288^4-255^4)*5.67E-8 = 150W/m^2
Multiply that by the surface area of the earth:
150W/m^2 * 3.14*6.37E6^2 = 1.912E16 j/s

Still a big number, but maybe this is more manageable and realistic. How much of a change in volume is needed to support 1.912E16 j?

1.912E16j = 101325 j/m^3 * Δv
Δv = 1.887E11m^3

Seems like a lot of change in volume. But the atmosphere is HUGE. So that shouldn’t be a problem. Given a height of 100km and internal radius of 6.37E6 the volume of the atmosphere is approx 5.18E19m^3

Since we need a change in volume of 1.887E11 cubic meters every second to sustain a 1.912E16 production of energy we approx 2.74E8 seconds of atmosphere left. Also note, that to get a 1K change in air actually takes approx 1kJ, so we just lost 3 decimal places off that seconds count. Instead of 9 years we have 3 days until the earth’s atmosphere is collapsed to sustain a 1.192E16j/s

If I messed up on my math somewhere or used any gas laws incorrectly please let me know. Just trying to be an open minded learner.

37. It shouldn’t be hard for anyone familiar with gas energy mechanics and atmospheric chemistry, – for all of you,

to engage in conversation related to the name of the Gas Law that governs the temperature of what you claim to understand about.

I want to see how many of you can name the Law of Thermodynamics that governs atmospheric and gas temperatures,

I want you to write the equation,

I want you to tell me, us, people reading, what the factors stand for.

Is that, that hard for you all to do?

I want you to point to the Factor you claim is responsible for your Church’s claim

that mixture of CO2 into a volume of Atmospheric Air can make it’s temperature rise.

I want to see YOUR chart
with the Energy Constant for the two gases, the Chart of Law just like the REAL Gas Law has,

listed as having CO2 possess a higher energy constant than Air such that additional admixture of CO2 will make that volume of gases’ temperature,
rise.

Because the REAL law governing gas temperatures is comprised of the Equation and the Chart of gas Energy Constants,

and the REAL Gas Law – the one that governs what you all claim to be experts on –

assigns CO2 the lower Energy Constant.

The NAME of the LAW
Governing GAS TEMPERATURES: WHAT IS it?

What is the equation,

and which is the factor,

that convinces you all,

that a cold nitrogen bath is a magical heater,

and that the frigid evaporation/condensation phase change refrigerant,
blocking 20% of otherwise available warming firelight to the

light warmed planet,

is the magic core
of the magical cold nitrogen bath that’s a heater.

This shouldn’t be hard, girls.

I want to see how many of you can even name the Law of Thermodynamics,

governing gas and atmospheric temperatures,

doesn’t arrive at the identical temperature of the International Standard Atmosphere.

Because there’s only one place a 33 degree error can occur in that,
and it’s to

refuse to use properly processed Gas Law,

to calculate the temperatures of compressible phase material: gases.

• Allen,

It doesn’t look like you are going to answer the questions posed.

1. The answer to question 2 is always revealing.. Just kidding. The lack of answer to question 2 is in fact revealing.

2. There are a set of equations for the atmosphere. But they aren’t solvable in any form because they are for turbulent flow, which doesn’t have a solution.

If we combine the ideal gas law, the equation for pressure with height (due to the weight of the atmosphere) and the first law of thermodynamics under adiabatic expansion we can derive an equation for how temperature reduces with height when the atmosphere is subject to convection. It turns out that the actual tropical atmosphere temperature profile (we call this the lapse rate) pretty much follows this theoretical lapse rate. In the extra-tropics less so.

Potential Temperature and Temperature Profile in the Atmosphere – The Lapse Rate you can ask questions about that.

If you are very interested in the derivation I can scan in a few pages of equations from an atmospheric physics textbook.

3. I attempt to give a simple explanation of turbulence and parameterizations in Turbulence, Closure and Parameterization. It’s a big subject, not easy to condense into one article. Of course, if you have a solution to the Navier-Stokes equations for the atmosphere I won’t be the only one interested.

4. You didn’t answer question 1 either. This covers radiative transfer through the atmosphere. This is equally important for explaining the energy balance of the climate. In the article I link to the derivation of the equation. You are invited to explain what is wrong with the derivation.

Of course, running with the odds based on your comments, you most likely haven’t the slightest clue what all those symbols mean.

I’m just running with the odds and would be delighted to have you, the first person with rambling diatribes to visit this blog to actually say – “oh yes here is my explanation of equation 1 and here is why it is wrong/right“.

—-
The reason I wrote this article was to filter out people who don’t understand equations or physics from cluttering up other articles with lengthy incoherence and so far it is working beautifully.

Remember, you do have to answer the questions.

Here is your chance to shine.

• Allen,

One of your comments got delivered by WordPress into the trash due to insults and invectives. Another one perhaps due to the ratio of capitals to correct case being very high. One capital letter to start a sentence is usually sufficient. Perhaps one word in caps in an occasional sentence, for emphasis, might be ok. When you hit 50% caps to lower case it’s a sign that, well, you know what I’m saying, I don’t need to spell it out.

You might have confused us with a ranting blog.

If the answer is no, please show us the courtesy of respecting this blog’s strange rules and instead visit a ranting blog – there are so many excellent ones to choose from.

Call it censorship if it makes you sleep better at night.

38. That’s IGNORANCE ITSELF:

PART of the ATMOSPHERE’S TEMPERATURE in FACT IS in FACT

that 33 DEGREES.
=============

Because the total volume that the atmosphere takes up is basically constant, we know that it isn’t producing thermal energy due to compression.”
==============

It’s WHY your Church * * *doesn’t get the right answer when you calculate the temperature of the Global Atmosphere.* * *

39. The ludicrous claim that 33 degrees of Earth’s temperature isn’t due to compression of the atmosphere is MANIFEST IGNORANCE of the elements making up Global Atmospheric temperature.

Name the Gas Law Brad Shrag
that governs gas, hence Atmospheric temperature.

You already failed MASSIVELY, not knowing where your Church’s erroneous 33-degree shortfall comes from.

You tried to claim OUTRIGHT: in plain black and white text, that there is no warming of the atmosphere accounting 33 degrees of it’s temperature.

That is PUREST falsehood, 33 degrees of Earth’s global atmospheric temperature is DIRECTLY created through compression.

Furthermore, you ALSO ignorantly claimed gases that are compressed hold more energy.

Wrong answer, COMPLETE inversion of the ACTUAL physics gas energy mechanics operate under.

Compression of gases makes them hold LESS energy.

That’s why when pressurized gas is released from compression it cools.

When compressed, it transmits energy easily, and conduction removes energy more readily than when there is room for the gas molecule to expand.

These are both MASSIVE fails about even the BASICS of how gases operate,

and you still can’t even name the law of thermodynamics governing gas temperatures.

Name the law of thermodynamics, governing gas, hence atmospheric temperatures.

Or you’re even more obviously the fake I told everyone here you’ll turn out to be.

40. Dewitt Paine, Mike M, Brad Shrag:

Tell these readers the name of the law of thermodynamics governing the temperature of the Atmosphere.

Tell us why there has to be a Gas Law governing it’s temperature, and why someone can’t simply use the Law of Physics governing the temperature of some sand or wood or ice on a beach,

the Atmospheric Air none of you can name the law governing the temperature of

sits over.

Tell these readers between you all, you know why there has to be a law of thermodynamics governing the temperatures of gases.

Tell us all why your Church’s KooK-0-Dynamics come up that

erroneous 33 degrees short,

so that when you’re through telling us all about how ”we don’t understand about the cold nitrogen bath scrubbing the planet of energy actually being a big giant heater,”

and tell us all,

how much more light leaks out of the planet,
every time the green house gases, make another percent less, leak into it.

Explain to us all how that happens. There is a cold nitrogen bath, conduction chilling a light warmed rock,

and then, light refractive insulation mixed into the bath, that stops 20% of otherwise available warming firelight from the sun, from ever joining Earth’s physical or mathematical systems,

is the ”Magic Heating Core,” of the ”magic cold nitrogen bath that is a heater.”

What seems to be your peoples’ problem, answering a few fundamental questions, about the Atmosphere you insist is a giant magical heater?

You can’t tell me the name of the law governing your Kook claims,

You can’t show me another instance of insulation mixed into a bath chilling a rock,

causing sensors to detect and depict more light leaving an object
the insulation makes less light reach,

You’re all in here, wishing you could explain your Quack Claims,

and – all of you TOGETHER – can’t tell us the name of the law governing the Atmosphere’s temperature.

You call that intellectual competence?

To tell anyone who’ll listen a cold nitrogen bath is a heater,

and the light blocking refrigerant,

stopping a fifth of the potential energy from ever joining the system,
conduction scrubbing the planet right alongside the nitrogen
evaporation/condensation phase change refrigerating

not just the surface but the overall nitrogen bath,

is the ”magical hot heater core of the magic heater that is really a cold nitrogen bath.”

And you expect others to show some respect or regard for the intellectual processes that brought you all, as a group, to these conclusions?

A cold nitrogen bath,
conduction chilling a light warmed rock,

actually – heats it.

And the light blocking refrigerant in the bath,
reducing surface energy density by 20% before the light even arrives,
conduction chilling the planet alongside the other single phase gases,

is also phase-change evaporation/condensation refrigerating
but
it’s rhillie
a big ol’ magical heater core,

in the big cold nitrogen bath that’s secretly a heater,

to believers.

You people are at the point where if someone asked you all ”What happens to the temperatures of light-warmed rocks, 20% less light warms?

Your answer, is that “if the magical gaissiness stops light from getting to the sensors on the rock, they will detect and depict more light leaving the rock,

each time the magical gassiness makes a percent less go in it.”

Each time the light blocking insulation
stops more light from going in,
the magical gassiness,
makes more come out.

That’s your story, and you’re sticking to it.

and the refrigerant chilling the bath,
reducing surface energy density on the rock by 20%

is the magic core

And you think a child of ten can’t see the Conservation of Energy violations in that.

• Allen Eltor says: “Tell these readers the name of the law of thermodynamics governing the temperature of the Atmosphere.”

It is possible Allen Eltor is referring to the Ideal Gas Law. Unfortunately, the volume of the Earth’s atmosphere isn’t fixed, so the Ideal Gas Law places no constraints on the Earth’s temperature. Pressure and Temperature are free to vary, and any change is easily accommodated by a change in Volume. Think in terms of a gas in a cylinder with a movable piston; not a sealed container.

Therefore the Earth’s temperature isn’t determined by pressure.

Pressure is simple: The pressure at any altitude is produced by the weight of the atmosphere above that altitude. At a given altitude, pressure usually varies only slightly (a few %?) with weather and geography.

Temperature is controlled by the rate at which energy enters and leaves. When more enters than leaves, it becomes internal energy – rising temperature. This is commonly known as the First Law of Thermodynamics or Conservation of Energy.

For the planet as a whole*, radiation is the only mechanism by which energy enters or leaves the planet. 90% of the energy that leaves our planet (as long wavelength infrared radiation or LWR) is emitted by GHGs in the atmosphere. GHG’s also absorb LWR on its way to space.

Twice as much of a GHG CAN cause twice as much absorption of LWR and twice is much emission of LWR. To a first approximation, these phenomena offset each other. However, the surface of the Earth is emitting an average of 390 W/m2 of blackbody radiation upward, but only 240 W/m2 is coming out of the top of the atmosphere. This is the REAL greenhouse effect: a 150 W/m2 (39%) reduction in radiative cooling on the way from the surface to space. If you were truly interested, this phenomena would be relatively easy to understand.

Translating the simple 150 W/m2 GHE into a 33 degC GHE is complicated, because it involves making some assumptions.

* Within the atmosphere, energy is also transferred by latent hear (evaporation) and work done expansion of gases (w = PdV) when the pressure changes. However, for the planet as a whole (Global Warming), we don’t need to be concerned about these complicated internal transfers of energy.

• Frank, regarding your assertion pressure and temperature are not related in the atmosphere,

you might want to rephrase it till it becomes

“pressure and temperature in gases are directly related.”

Nathaniel

• Nathaniel,
It is completely correct to say that the earth’s temperature isn’t determined by pressure. The only way to get dependant relationship between pressure and temperature involves a gas in a sealed container. Do you believe the atmosphere to be a sealed container such that when it heats up it presses against the walls of the container? If you do believe this can you please enlighten me, and probably the rest of us, as to the height of the top of the container. That would be necessary for calculating temperature change based on a static volume and changing temperature/pressure.

41. Allen Eltor,

You can see an answer on physics and a comment on Etiquette from me already.

I see your most recent comment above and so I’ll make a change in settings so your further comments go into moderation.

You can try and respond to my question while meeting blog guidelines and if you are successful your comment will make it out of moderation.

If not you can declare victory and leave. Luckily, your comments so far will be kept as a testament to your insights / inability to answer the two foundation questions.

42. Allen’s most recent comment wasn’t actually posted.

He posted one where he answered your question whether the laws of radiant transfer work, saying of course they do,

and that the math of emissions inhibition and absorption are functionally identical

due to the principles that cause cause Conservation of Energy in all electromagnetic energy transactions.

• So you are saying that radiation flowing from a lower intensity source inhibits the emission of a higher intensity source?

If not, maybe provide a link for a source with a better explanation.

• Nathaniel,

There is no such thing as radiation inhibition, except in the minds of people like a person who’s initials are C and J. I’m afraid I’ll trigger moderation if I used his name.

• DeWitt,

the mathematics of radiant transfer mandate that emissions are either being inhibited,

or electromagnetic energy is flowing into regions already charged to an equal/higher concentration energy gradient.

This is strictly forbidden in electromagnetic energy mechanics.

The emissions suppressed are identical to the energy required to suppress them,

thus leaving conservation of energy principles properly in place

governing electromagnetic energy transactions where gradient forces meet.

This principle governs complementary-symmetry circuits in electronics as well.

When electromagnetic energy gradient is distributed equally on both sides of a specific point in the circuit, flow ceases,

so these circuits can remain charged,

yet have zero flow

until another event

Electromagnetic energy doesn’t flow toward a more concentrated energy distribution gradient. It’s part of the definition of what electromagnetic energy acts.

Nathaniel

• Nathaniel,

I see a comment from him in moderation. Another rant.

I asked for an answer to question 1 and question 2. Question 1 – apparently it’s ok. Question 2 – no answer. A rant instead, and not a David Mitchell style rant, an incoherent rant.

This is how it goes with people confused about heat transfer. Does the hot body absorb radiation from the cold body? No answer. Is the heat transfer textbook wrong? No answer.
And so on.

There’s a nice example with one of our earlier commenters on this article who said “I’ve already answered” but still couldn’t commit to question 2.

No point demonstrating proof to people who don’t accept what is in heat transfer textbooks. And won’t even say they don’t accept what is in heat transfer textbooks (for obvious reasons).

Also, I don’t accept ranting and insulting comments on this blog. I don’t need to repeat any of it as an example.

There are much better blogs for ranting people.

43. on December 6, 2017 at 12:29 am | Reply Steve Vertelli

You stopped publishing Allen’s replies to you, so he published it elsewhere with a scathing rebuke to your violations of thermodynamics :

I agree with him regarding conservation of energy governing electromagnetic field interactions.

All agree, conservation of energy governs electromagnetic energy interactions.

It’s well documented and very logical, that
the energy of suppressed emissions,
and the energy suppressing those emissions, are identical.

If they weren’t, the principles driving Conservation of Energy in all electromagnetic interactions

wouldn’t govern the ones from your example 2.

This is what makes symmetrical-design,
and comparator circuits work.

When electromagnetic potential equalizes within circuits, flow stops.

This zero-flow, charged condition,
makes it possible for them to remain on, yet conduct nothing,

until an external event,
occurs.

The same simple principles,
govern the fact that energy required to suppress emissions

matches identically,

the energy supressed.

• Steve,

Would you like to comment on question 2?

Is it correct? Does the hot body absorb radiation from the cold body – or is the heat transfer textbook wrong?

• scienceofdoom,

I’m curious what part of this is still unclear to you?

You know the underlying, defining characteristic of electromagnetic energy is that it won’t migrate into an equal or higher concentration energy field.

Faced with that inviolable mandate, that won’t ever be negotiated away – by a man or a whole planet full of them,

you are forced to look round for reasons

authors of books for students,

would refer to the energy transaction as absorption.

Students are being trained not to forget where energy is being held, and for practical purposes, reference to absorption is acceptable language for students.

However those same students

are also told specifically that definitionally: this can not in fact be true absorption:

the defining parameters which make an entity be able to be

‘electromagnetic energy’

specifically and formally forbid,

actual physical absorption

by any entity already charged to equal

or higher charge concentration.

So you’re left with the apparent, actual answer, – that for students,

the mathematics and intuitive grasp of the concept of absorption,

combined with the fact that their customers
are all going to refer to these energy flows as
absorption,

are a reasonable framework, to teach radiative transfer, and expect the students to be able to deal with references to it, in the work place or classroom, by others.

For instance in mechanical engineering this type energy transfer is actually – I think – taught as absorption.

The last time I heard about it from someone was a few years ago, and a mechanical engineer was discussing energy handling by matter.

At some point the conversation worked its way around to the fact that mechanical engineering people use this model constantly – and for their field(s) of endeavor, the mathematics and referencing absorption, work fine.

• on December 6, 2017 at 3:05 am Steve Vertelli

Scienceofdoom, I’m of those who say the term absorption, fits the needs of many less radiation-transfer-centric fields of endeavor, to a T,

as far as they need to process the mathematics of energy transfer.

I can’t really see that there’s any room for a question.

We know there are many fields that don’t process energy handling past the buildup and removal of heat, and the definitions surrounding absorption fit these peoples’ needs.

I think there’s not really anything mysterious about it.

• Nathaniel or Steve

Can whether of you please provide a reference for this:

“underlying, defining characteristic of electromagnetic energy is that it won’t migrate into an equal or higher concentration energy field.”

I’ve searched to no avail. This would seem to be the basis of misunderstanding on one party or the other in these conversations. A source for this would be helpful.

• Nathaniel,

The part that is unclear to me is whether you agree with the heat transfer textbook or not?

The part that is clear is you don’t know anything about thermal radiation – as taught in a field called physics. But I won’t waste everyone’s time trying to pick apart your ramblings that are unknown in physics textbooks.

By the way, we do have a policy that this blog accepts physics as taught in physics textbooks. From The Etiquette:

Basic Science is Accepted – This blog accepts the standard field of physics as proven. Arguments which depend on overturning standard physics, e.g. disproving quantum mechanics, are not interesting until such time as a significant part of the physics world has accepted that there is some merit to them.

The moderator reserves the right to just capriciously delete comments which use as their premise that standard textbook physics is plain wrong.

This is aimed to reduce the continual stream of unscientific rubbish that gets placed here as comments.

As explained in this article, commenters are asked to accept or reject two foundations, dodging is not acceptable.

44. definition of *how E.M. acts, sorry

Nate

• The laws governing electrons and electric currents can’t be blindly applied to electromagnetic waves. They simply aren’t the same thing. If I’m incorrect, please provide a source.

The laws governing electrons and their currents are what create and define

their waveform propagation characteristics. That’s what wireless signaling is.

Using the laws governing electrons and their currents, to generate

All electron-current mediated signaling is based on this.

45. DeWitt,

one of the definiting traits of electromagnetic energy

you’re made to list on a test question regarding electromagnetic energy,

is that it will not

spontaneously flow
into a higher gradient energy distribution.

Another definition of electromagnetic energy’s behavior

is the question:

” As energy gradients converge, what important flow condition is established

when equalization of unequal charge distribution
is complete?”

The answer you are to give, is ” Zero-flow conditions occur: energy flow ceases since there is no differential to create movement of energy.”

This is why the mathematics of transfer only admit to any difference flow.

Electromagnetic energy doesn’t flow into higher charge gradients.

When it won’t flow into higher charge gradients

you’re left with the answer classical physics describes.

Energy emission is curtailed, emissions are inhibited on a one-to-one basis,

and difference flow proceeds until energy charge inequalities are neutralized: in other words equalized.

• The flow of heat can’t be spontaneous, not the flow of radiation.

And when two objects come to the same temperature, it isn’t that they stop transferring radiation back and forth. Energy is still transferred, it’s just transferred at an equal rate so there is no flow of heat from one object to the other.

that’s actually incorrect about spontaneous emission.

Spontaneous emission is defined as an electron dropping to a lower energy level as it emits a photon.

Also

when two objects come to the same temperature they do in fact stop transferring radiation back and forth.

This is the characteristic of electromagnetic energy that makes complementary-symmetry
and comparator type electromagnetic circuits work.

When energy gradients equalize on both sides of a given point, all flow stops.

This allows such electromagnetic setups to remain equally charged,

and active,

but with zero flow,

until an outside event establishes an energy gradient imbalance.

Cheers

Nathaniel

• Nathaniel,

Do all surfaces over 0K spontaneously emit radiation from their surface as a function of the surface temperature? Or is there some other process that must happen before radiation is emitted?

• when two objects come to the same temperature they do in fact stop transferring radiation back and forth.

Really? So, if I try to measure a radiative energy flow between two plates of identical temperature in vacuum with a tool also at this very temperature, I will find a value of zero?

What’s with LEDs? They do emit visible light without getting glowing hot.
Let’s have one LED with a surface temperature equal to the surface temperature of my eyeball: Will I lose the ability to see the light of the LED for this special case?

These are very easy questions, which I think an expert like you can answer very easily with his impressive skills on radiative energy transfer…

• Hogwash. You do not cite any legitimate references for your contention because there aren’t any. You only wave your hands and misapply physical principles that don’t apply to photons. I say again, there is no such thing as inhibition of thermal radiation by a hotter surface. Only temperature and emissivity of the emitting surface determines the intensity of thermal EM radiation. A photon impinging on a surface can only be absorbed, reflected or transmitted. It’s original emission is not dependent on whether or not it impinges on another surface. You can find this in any physics textbook.

• Dewitt Payne:

You’re obviously trying hard with your ”general physics lite” electromagnetic transfer background.

• Nathaniel,

one of the definiting traits of electromagnetic energy you’re made to list on a test question regarding electromagnetic energy, is that it will not spontaneously flow into a higher gradient energy distribution.

Cite please. I want to see your source for that statement. Your assertion has no merit without a reference.

Photons are not electrons. They don’t have a charge. Hence there is no way for a photon to ‘know’ the energy level of a surface. Inhibition of emission is essentially action at a distance. It would require some mechanism other than photon transfer for information transfer between the surfaces. That doesn’t exist either.

Also, have you heard of the cosmic microwave background? That’s thought to be the residue of the big bang.

The CMB has a thermal black body spectrum at a temperature of 2.72548±0.00057 K.

If EM radiation from a cold source cannot be absorbed by a higher temperature surface by whatever mechanism, how do we know that this microwave radiation exists? The answer, of course, is that photons do not have a temperature.

• Nathaniel: If you knew anything about the field of statistical mechanics, you would realize that the Second Law doesn’t apply to the behavior of individual molecules and photons. The 2LoT is the result of the NET behavior of millions of molecules and photons following the laws of quantum mechanics.

In your electrical circuits, electricity (the bulk movement of electrons) doesn’t flow in the absence of a potential. Nevertheless, individual electrons continue to move. In antenna and other electrical circuits, the movement of elections produces electromagnetic noise (which are often referred to as having a temperature).

When two molecules collide, the laws of quantum mechanics allow a slower-moving molecule to collide and occasionally transfer kinetic energy to a faster-moving molecule. That is why we have a Boltzmann distribution of molecular speeds in a gas, not a uniform speed. And why we don’t define the concept of thermodynamic temperature in the absence of a large group of colliding gas molecules. The kinetic energy of a single gas molecule changes a billion times a second. However, when two large collections of molecules with different temperatures interact, the NET FLUX of kinetic energy is from the hotter group to the colder group.

The laws of quantum mechanics don’t prevent a photon from being emitted by a GHG in the colder atmosphere and absorbed by the warmer surface. However, the laws of quantum mechanics do ensure that the NET FLUX of radiative energy – heat – between two groups of molecules will always be from the hotter surface to colder atmosphere.

To avoid confusion, we say that energy can (and does) flow in two directions, but we define “heat” the NET FLUX of energy between to groups of colliding molecules large enough mean kinetic energy that doesn’t change every nanosecond – to have a temperature. That NET FLUX, heat, always goes from hot to cold. And we avoid talking about the temperature of single molecules (the ones that emit and absorb photons) and never apply the 2LoT to the one-way-flux of radiant energy. Quantum mechanics controls the behavior of individual molecules and photons; thermodynamic controls the flow of heat between two groups of molecules having a temperature.

46. SoD, are we sure Nathanial is not a sock puppet of the recently departed Alan? I have deja vu reading his posts. I am awaiting the appearance of smart photons and pseudo scattering to round out the entertainment

• David,

Good call.

Nathaniel’s IP address jumps around from comment to comment and in two cases is the exact same IP address as recently-arrived commenter Steve Vertelli. And interesting style parallels to the recently blocked AE. Sort of a toned-down non-insulting version with more attention to grammar. Perhaps a coincidence. 3 new commenters all with similar fantasy physics ideas.

Anyway, we await to see if the textbook question is answered. I don’t have high expectations.

As long time readers know I don’t expect the illuminati to change their minds. The benefit is for other readers.

• Steve Carson,

I’ve met you one time, and I’ve already caught you lying twice in one day.

The first time was when I caught you claiming Allen Eltor wasn’t answering you regarding your questions.

posted his answers up at Climatofsophistry,

and told everyone you were already dodging debate,

refusing to let him answer you,

and he said that shortly

you’d be claiming he wouldn’t answer you.

Sure enough a couple of hours later, there you were, lying:

Here we are just a few hours later, you’re lying, trying to claim

You were answered, you’re simply dodging any debate.

How many places does one go on the internet, and get lied to twice in one day, by the same character?

Well, that depends on how many magic gas frauds’ websites one visits, I suppose.

When you grow a pair, and can debate your questions without resorting to

lying every time the truth gets inconvenient to you,

you won’t be another lying, dodging, internet poser who can’t defend your

beliefs.

You’ve got a herd of nerds who can’t between all of you,

even name the law of physics governing gas and atmospheric temperatures.

You, as their Magic Gas barking Nerd Herd leader,

can’t even manage to keep from

lying,

when you know people are watching you.

Knowing Allen Eltor’s style of dealing with magic gassers, I knew you weren’t going to debate him.

Having him predict to everyone at ClimateofSophistry that you were immediately going to start dodging,

trying to keep him from posting,

I knew the chance of your actually having the guts to defend your “Energy runs backwards!” claims were less than zero.

I did everything a person could be expected to do,

speaking objectively, giving you every chance

and your teaching that Electromagnetic Energy flows backward,

into higher concentration gradients from lower ones.

I met you once, and caught you lying twice in the same day.

I guess once you’re a dyed in the wool fraud, it becomes so addictive it’s an

actual thrill to be caught lying.

• Nathaniel/Steve Vertelli/AE,

Thanks for your thoughtful contributions. Your insights will be kept here for posterity and anyone interested can see where to find out more.

As this blog has some rules, already explained, it will be clear that your contributions are no longer welcome.

Basically everything you asserted is wrong.

Temperature and pressure of gases especially in the troposphere are directly related.

Whether the gas is in a jar or open to the vacuum of space above,

the temperature is related to pressure.

Examples are endless. Clouds forming after striking rising cliffs are a direct result
of pressure reduction
creating temperature drop

in a mix open to the vacuum of space above,
and bounded by conditions as variant as planetary parameters can create.

When you drive your car upward into the hills then the mountains, the specific reason the temperature falls,
is because the pressure falls.

If you’ve ever been to a hot-air balloon festival, in the morning and evening, when people open the center vent to descend, plumes of water vapor forming,

make the balloons look like smoke is coming from their top.

That isn’t smoke, it’s water vapor chilled to the point it condenses to liquid
when released into the lower-pressure open atmosphere.

Nathaniel

—————–

It is completely correct to say that the earth’s temperature isn’t determined by pressure.

The only way to get dependant relationship between pressure and temperature involves a gas in a sealed container.

Do you believe the atmosphere to be a sealed container such that when it heats up it presses against the walls of the container?

If you do believe this can you please enlighten me, and probably the rest of us, as to the height of the top of the container.

That would be necessary for calculating temperature change based on a static volume and changing temperature/pressure.

• Yes, temperature and pressure are linked but in an open atmosphere they agree not dependent on each other as they are in a closed container.

Furthermore, a closed container of gas only changes temperature when work is done. If a gas were compressed in a closed container and then left in that compressed state it would radiate away is heat until it matched the external environment temp. It doesn’t maintain temperature because it is compressed.

You also keep speaking about electromagnetic circuits and equating it to electromagnetic radiation. Electromagnetic fields aren’t the same as electromagnetic radiation. I feel as though you’ve been told that all squares are rectangles so you have taken the stance that all rectangles have 4 equal length sides. You can’t apply any principal you want from circuitry of electromagnetic circuits just because it has EM in the name.

This will be the at least the 3rd time I’ve asked, and I do so because I truly want to learn, please provide a link for reference. It can help clear up lots of misunderstandings.

• Temperature and pressure are most certainly not directly related. There are times and places where the pressure is one atmosphere and the temperature 40 C. Other times and places where the pressure is one atmosphere and the temperature is – 40 C. At the bottom of the ocean, pressures can be hundreds of atmospheres, but the temperature is barely above freezing. The lapse rate in the troposphere can be close to 10 C/km, but except in the boundary layer it is rarely so large. There can even be temperature inversions; where I live, the morning temperature often increases as one goes up the side of the mountain from the valley floor. The stratosphere has a permanent temperature inversion. It takes a special sort of willful blindness to see a universal rule in something that rarely occurs.

48. on December 6, 2017 at 2:28 pm | Reply nobodysknowledge

3 questions to 3 sock puppets:
Do you know the difference between energy and net energy?
Do you know the difference between energy and net energy?
Do you know the difference between energy and net energy?

49. SoD,
“As this blog has some rules, already explained, it will be clear that your contributions are no longer welcome.”
.
Good decision, since someone like that only wastes people’s time. The guy gives even trolls a bad name.

• Steve F.,

Good decision, since someone like that only wastes people’s time. The guy gives even trolls a bad name.

Yes and no. I think there are people who lurk here. Some rebuttal may be useful for them, but there’s no point in dragging it out. I hadn’t thought of the CMB detection example to counter the warmer object doesn’t absorb radiation from a colder object trope.

I found his formatting extremely annoying even when just skimming, so +1 for the bad name comment.

50. For SOD acknowledgement see the URL for article by myself and David Archer on improvements to the accuracy of the free website “Modtran Infrared Light in the Atmosphere”. Published as part of education section of Annual Meeting American Meteorological Society Austin Texas, 2018.