## The Thermodynamic Approach

From an ongoing discussion here, it becomes worthwhile to demonstrate the correct solution to the plane parallel plate hotbox problem. A similar problem is called the “steel greenhouse” and was solved in my book “In the Cold Light of Day“, where with spherical geometry the climate alarmist radiative greenhouse effect mathematically formally disproves itself by ending with a solution where 1 = 2.

Dispensing with graphics, we simply consider a box which has a lid which is transparent to shortwave solar radiation, allowing solar radiant heating to enter the box. At the bottom of the box is a surface with some albedo or reflectivity α. What doesn’t get absorbed is reflected and just goes straight back out through the transparent lid.

The lid however is not transparent to the longwave radiation which the bottom surface will emit as IR thermal radiant energy. The question is: What is the equilibrium temperature of this system, if we consider that the box is insulated from physical energy loss and can only equilibrate via radiation?

All that we need to solve this problem is the First Law of Thermodynamics, i.e. the law of conservation of energy:

dU = Q + W = m Cp dT

This first law states that in order for an object to experience an increase in thermal energy, dU, then you need to supply heat (Q) and/or work (W), and then the resulting change in temperature for the object will be given by the terms ‘m Cp dT’ which are the object’s mass, thermal capacity, and change in temperature.

The equilibrium condition is defined as the state where the temperature is no longer changing, which is an obvious definition, and thus the end-state of the system is where dT = 0, which via the First Law requires that dU = 0, which thus also means that Q = 0 since we do not have any work occurring upon this system.

So it looks like we’ll need the equation for heat Q, which for radiant heat transfer requires another equation called the Stefan-Boltzmann Law. So, that’s three equations we’ll need in total: 1) the First Law of Thermodynamics, 2) the equation for radiant heat transfer for a plane parallel geometry, 3) the Stefan-Boltzmann Law.

The Stefan-Boltzmann Law simply gives the what the radiant thermal emission from a body will be given its temperature, and takes the form of a flux or energy flux density:

F = εσT4

where emissivity ε is the ability of the surface to emit energy, which is unity (1) when it can emit perfectly but otherwise some value between 0 and 1 for non-ideal emission. Here we will use ε = 1 just because it makes the equations simpler to read; it models the ideal scenario but doesn’t fundamentally change the underlying mechanics.

We can use this equation for example for the emission of the lid, given that this lid is opaque to the longwave emitted from the bottom and thus will attain some temperature from this radiation. The question simply becomes: if the interior of the box absorbs some energy, then what does the exterior of the lid have to emit in order to balance this absorption?

We define the solar input as S, and what is absorbed of this input is (1 – α)S because some energy is not absorbed. Thus, this absorbed input needs to equal the emitted output from the lid and its temperature, in order to satisfy conservation of energy:

(1 – α)S = σTL4

So solve for the temperature of the lid TL :

TL = [(1 – α)S / σ]1/4

A nice simple solution for the equilibrium temperature of the lid, given that the lid has to emit outward what is absorbed in the interior of the box.

What about the temperature of the bottom surface of the box, where the shortwave input initially warms the box? The equilibrium condition is again obviously where its temperature is no longer changing, and so we once again refer to the First Law but now applied to this bottom surface: dTB must equal zero, which means that Q must equal zero for the bottom surface of the box. So what is Q here?

Let’s consider the heat flow between the bottom and the lid of the box (QBL), which for plane parallel geometry takes the form:

QBL = σ(TB4 – TL4)

Let us consider the sign of QBL : since it is the bottom of the box which is initially heated, and first attains a temperature, then TL4 will always be less or at most equal to TB4. That is, QBL ≥ 0 since the lid can never emit more than what it receives from the bottom surface; it is greater than zero when temperatures are changing, and becomes equal to zero when temperature stops changing as per the mathematical definition of the First Law. Thus, since QBL ≥ 0, then that means that the ceiling never sends heat to the bottom of the box since it always only receives heat from the bottom of the box. And since heat is what is required to increase temperature as per the First law of Thermodynamics, then we never need to be concerned about the lid of the box raising the temperature of the bottom surface of the box. And now since we know that Q = 0 for equilibrium, then QBL = 0 gives us for the bottom of the box:

TB4 = TL4

or

TB = TL = [(1 – α)S / σ]1/4

We can confirm this solution using another approach, that of setting the heat flow between the input and the bottom surface (QIB) equal to zero, given that there is no heat flow from the lid to the bottom surface which we need to consider:

QIB = (1 – α)S – σTB4 = 0

which results in the same previous equation for TB. Very nice and consistent, Q = 0 for all objects, and input equals the output. There is no other way to solve this given the definition of the First Law of Thermodynamics.

## The Climate Pseudoscience Radiative Greenhouse Approach

In climate science, which is a field of formal pseudoscience practiced at universities, the approach is taken to avoid reference to the equation and definition of heat flow, and all energy is instead treated as being able to increase temperature and cause heating, even if the energy comes from a cooler object. And so where we previously did not treat the emission from the lid as sending any heat to the bottom surface, now we treat all of the energy from the lid as being able to heat the bottom surface alongside the incoming shortwave radiation. While we note that the shortwave incoming radiation is high intensity and high frequency, and the emission from the lid will be longwave low intensity low frequency, in climate science this fact is put aside and all radiation is treated as having the same power to effect change and heat even if the frequencies are completely different.

And so because the climate science approach is to never use a heat flow equation, the approach here instead is to simply add the lid’s emission along with the shortwave input, and make this equal to the emission from the bottom surface:

σTB4 = (1 – α)S + σTL4

The climate pseudoscientists do at least still recognize that the lids output must equal the input to the box, and thus that (1 – α)S = σTL4 thereby giving

σTB4 = 2(1 – α)S

or

TB = [2(1 – α)S / σ]1/4

So they allow themselves to double the input in other words, and pretend that the lid is an additional source of heat and energy even though the lid only ever receives energy from the surface’s emission in the first place. There are of course other laws of thermodynamics, i.e. the Second Law, which explain that heat flow is not reversible, but nonetheless the climate pseudoscientists reverse the energy from the lid and put it back into the surface…

If climate pseudoscientists (i.e. climate scientists) would actually dare to refer to the First Law of Thermodynamics and the attendant equation for heat flow, as we will do here, then it becomes clear that this approach and its solution is inconsistent with the First Law of Thermodynamics. What did the First Law say about the condition of equilibrium and steady state temperatures?

dU = Q + W = m Cp dT

To have equilibrium and non-changing temperatures, then dT = 0, which requires Q = 0 (W is already 0) so that dU = 0. But if we dare apply the equation for heat flow between the bottom surface and the lid in this solution, we find:

QBL = σ(TB4 – TL4) = [2(1 – α)S – (1 – α)S] = (1 – α)S

Now since the input S is a constant greater than zero, then here QBL > 0, which means via the First Law that dT > 0, and hence, this is NOT an equilibrium solution even though it was intended to be solved as one! And so, the solution contradicts the logic of what it was supposed to solve for in the first place, thus exposing itself as a flawed and incorrect approach.

The approach is incorrect because it avoids the definition and equations for heat flow. And this is really the fundamental basis of the entire approach of climate science (pseudoscience) to studying the climate, given as they also avoid the fact that the Sun heats the Earth…notwithstanding that they also base all this on a flat Earth model!

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### 331 Responses to How to Solve the Dual Pane Parallel Plate Problem

1. Kenneth Pritchett says:

Excellent essay sir! Always amazing to me that thermodynamics and really basic math can be so easily discarded by the alarmists. Well, not discarded – ignored is a better word. I can see that it might be viewed by alarmists as counter intuitive when we state that the lid does not heat the bottom of the box – even though it is emitting radiation. My thought is that they confuse radiation with heat in all cases.

It would be interesting to see this worked into an actual physical experiment perhaps using some soda lime glass as a lid or better yet Pyroceram which has a nice fall off in transmission of LWIR. Placing thermocouples on each side of the lid and at the bottom of the box, perhaps covered with an inch or so of play sand.

Pitting the predictions of both treatments against one another in a real world test should put this to bed.

2. The horrible global Marxist U.N. IPCC money train is still gathering steam despite its whole rationale being a fake physics hoax. Now Pres. Biden is threatening a Holocaust for the U.S. with his all-embracing plans to sink the economy in leftist redistribution of wealth which is no longer even trying to hide its real intent of “climate justice”, which is what the IPCC was pushing all along instead of climate amelioration.

https://papundits.wordpress.com/2021/04/12/biden-environmental-justice-appointees-equate-energy-production-fossil-fuels-with-racism/

https://stopthesethings.com/2021/04/09/green-new-deal-pointless-posturing-joe-bidens-trillion-dollar-wind-solar-boondoggle/

As mad as that makes me, all I can do is teach real thermal physics in hopes of developing a new generation of “woke” physicists. Join the growing students of my Climate Science 101 course and arm your mind to fight IPCC lies:

http://historyscoper.com/climatescience101.html

3. Joseph E Postma says:

“It would be interesting to see this worked into an actual physical experiment”

R.W.(?) Wood did it a hundred years ago, and Joseph Fourier did it too 200 years ago. Recently Nasif Nahle with PSI did it too, finding the same result: the lid of a box does not heat the box!

Just get a Styrofoam cooler, paint the inside matte black, paste a lid over the top, and measure the temperature on the bottom of the cooler with it facing the sun with a thermocouple. It should reach much higher than the solar input, which can be either measured with a solarimeter or even calculated. The effect should be extreme.

All existing actual real greenhouses also perform the experiment daily. They don’t show temperatures rising above what the solar heat provides!

4. Kenneth Pritchett says:

Don’t get me wrong, I know the experiments were done and recorded, I’ve done two myself (though I had not heard of Nahle’s experiment, just blogs trashing the man’s paper like this one: https://daryanenergyblog.wordpress.com/2014/05/04/an-exercise-in-how-climate-denial-works/ ). But in today’s video driven culture it might reach more people resistant to the idea that their heros are lying to them. The BBC video here: https://youtu.be/SeYfl45X1wo fooled a whole lot of people. I know because it has been sent to me about 50 times as “proof” CO2 causes warming.

5. Joseph E Postma says:

Yes that video is so stupid. That a gas cab absorb radiation does not mean that the gas is making the source of the radiation hotter! We live in a world of sophistry.

6. esttom says:

Good discussion, but there are some issues here! For starters, this equation:

(1 – α)S = εL σ TL^4

Should be

(1 – α)S = εL σ TL^4 + εB (1-εL) σ TB^4

Namely, the total output (right hand side) is the sum of the radiative emission from the lid, plus whatever portion of the radiative emission from the base that is not absorbed by the lid.

I could go on, but once you fix this first thing you will find it affects your other equations as well!

Fun problem, hope you manage to solve it! Good luck!

7. “plus whatever portion of the radiative emission from the base that is not absorbed by the lid”

ALL emission from the base is absorbed by the lid. This simplifies the equations and the solution but of course either way doesn’t change the fact that the lid can send no heat to the base, etc.

Thank you, yes, the problem was solved! Good luck to you too! The climate science greenhouse effect disproves itself time and again!!

8. esttom says:

“ ALL emission from the base is absorbed by the lid.”

Absorptivity is equal to emissivity, so this cannot be the case unless εL=1… in general the amount absorbed is εL Tb^4 , and the amount transmitted is (1-εL) Tb^4 .

Hope this helps!

9. esttom says:

Should be the amount absorbed is εL εB σ Tb^4 , and the amount transmitted is (1-εL) εB σ Tb^4

10. CD Marshall says:

A video for this would be great, a simple experiment to conduct?

11. esttom says:

My previous comment was cut short. It should say:

“ ALL emission from the base is absorbed by the lid. ”

Emissivity equals absorptivity, so this is only possible if εL=1…

In general, the amount absorbed is εL εB σ Tb^4 , and the amount transmitted is (1-εL) εB σ Tb^4 .

12. Yes, I’ve changed the ε = 1 in all cases so that the equations are simpler to read, since what I want to demonstrate here is the basic solution which shows that there is no radiative greenhouse possible via the laws of thermodynamics.

13. CD Marshall says:

Joseph have you seen this? I stumbled across it in my Spenc searches it is older and was referenced at PSI.

for a troll demanded,
“You state: “Spencer believes cold can warm hot” So you didn’t study the experiment and you didn’t read it properly. Please quote where he said cold can warm hot? ”

So I referenced the original paper “Yes, Virginia, Cooler Objects Can Make Warmer Objects Even Warmer Still”

Any offhand citations you (or anyone) have to add to this would be great. I linked the paper, or a simple explanation on the physic he got wrong. I know you covered it thoroughly, I’m just trying to look for a concise short version.

Thansks

14. esttom says:

Joseph,

I appreciate the change, it’s a good idea to study the case of εL=εB=1 first. For even more simplicity let’s set α=0 because it just changes S anyway.

So now I have another question.

The box absorbs light from the sun, equal to S. The lid emits an equal amount, also S. There is zero heat transfer between box and lid.

So the box is absorbing all the incoming energy, but the lid is the one emitting all the outgoing energy. How does this make sense? If there were heat flow from the box to the lid I would get it, but there isn’t. What’s going on?

15. CD Marshall says:

When you are free to discard the original meanings of equations in physics it is easy to produce any number of outcomes desired, this is how they are able to maintain ghge myth.

16. CD Marshall says:

summation was a joke, maybe a bad one?

17. CD Marshall says:

…Robert Kernodle’s humor remains uncontested.

18. @esttom – I’ll help you work through this, so that you can develop an understanding of the solution as you solve it.

“The box absorbs light from the sun, equal to S. The lid emits an equal amount, also S. There is zero heat transfer between box and lid.

So the box is absorbing all the incoming energy, but the lid is the one emitting all the outgoing energy.”

Yes, all of that is correct!

“How does this make sense? If there were heat flow from the box to the lid I would get it, but there isn’t. What’s going on?”

OK, now this is very important and will help you comprehend the math. Here’s what I would like you to answer: What is the equation for the heat flow between the bottom and the lid? You correctly understand that the numerical value is zero, but what I need you to do is to write the equation out, OK?

Once you write the equation for the heat flow between the bottom and the lid (which equates to zero), then we can move to the next step. Just write the equation out, OK? 🙂

19. esttom says:

Joseph,

Of course the heat flow for arbitrary temperatures is Q=sigma (Tb^4-Tl^4). But since this heat flow is zero it cannot account for the energy which enters the box and leaves from the lid. It seems to me this solution has energy teleporting from one to another.

20. esttom – OK, that’s excellent! Good work so far. Now, look at the heat equation to answer the next question:

Q = σ(Tb^4 – Tl^4)

Question: Referring to the heat equation, what is the specific emission from the bottom directed toward the lid? (Hint: It’s the Stefan-Boltzmann Law.)

21. esttom says:

Feel free to speed things up, these questions are a bit basic. σTb^4

22. OK. So σTb^4 is the energy from the bottom to the lid. And the lid emits outward, as you said earlier, S, and Tb = Tl, which you also pointed out earlier.

The energy that the lid emits it gets from the bottom. The lid cannot lose energy to the bottom, only to space, and the energy the lid emits to space it gets from the bottom.

23. esttom says:

Very good. However, you have neglected the second term in the heat transfer equation. So let me repeat your questions back, except regarding the second term.

Referring to the heat equation, what is the specific emission from the lid directed toward the bottom? (Hint: It’s the Stefan-Boltzmann Law.)

24. “you have neglected the second term in the heat transfer equation”

This is a disingenuous comment given that I asked you to provide the heat transfer equation and thus I am fully in acknowledgement of all of its terms. Your comment also doesn’t provide any explanation as to how you think that I have neglected a term, or what effect you think that doing this would have, etc.

Do you think that you will be able to continue this discussion without seeding baseless innuendo? Do you think that you can continue this with just focusing on the math and the facts, and not interject things which have no apparent basis or quotations to demonstrate, etc? I would appreciate that if you could.

Moving on: Of course, the second term is the interior and also the exterior emission of the lid. The difference is that on the interior the energy cannot be lost because it is replaced immediately by emission from the bottom, and the emission from the bottom is sustained by S. Thus, there is only one place where the lid can lose energy, and that is its exterior: the energy which the lid loses on its exterior is provided by the emission from the bottom, which is sustained by S.

25. esttom says:

on the interior the energy cannot be lost because it is replaced immediately by emission from the bottom… the energy which the lid loses on its exterior is provided by the emission from the bottom”

My point is you are double counting the emission from the bottom here. The bottom emits only σTb^4 according to the Stefan-Boltzmann law. The lid emits σTl^4 from each surface according to the Stefan-Boltzmann law, for a total of 2 σTl^4=2 σTb^4 . Thus the radiation from the bottom does not replace the energy lost by the lid. The lid is losing energy.

Similarly, the top receives S from the sun. As you have pointed out the first term of the heat equation, Q= σ(Tb^4 – Tl^4), says the bottom loses σTb^4 to the lid, according to the Stefan-Boltzmann law. But the second term says the bottom receives σTl^4 from the lid, again according to the SB law. So the bottom receives too much energy.

So we are still left with the problem that the bottom is absorbing a net energy of S, while the lid is emitting a net energy of S. Which again is obvious: since you have said there is zero heat flow between bottom and lid, there is no way for the energy absorbed by the bottom to reach the lid.

26. You are running far too far along with your misunderstanding here. Just take a breath, slow down.

Can you write down the equation for the First Law of Thermodynamics? I will do it for you:

dU = Q + W.

dU is the change in energy of an object. We have a scenario with no work W here, only heat Q. So:

dU = Q

That’s the First Law here which says that for an object to change energy (dU), there must be Q. If Q is zero, then dU = 0 which means no energy change.

You already wrote down for us the equation for Q for the lid, and acknowledged that it is zero. Thus:

dU_lid = 0, when Tb = Tl.

Thus, the lid does not lose energy on the interior. The lid doesn’t lose energy on the interior because it receives energy from the bottom.

Whatever the lid emits on the interior, it doesn’t lose it, because it is perfectly equal to the energy it receives from the bottom.

“since you have said there is zero heat flow between bottom and lid, there is no way for the energy absorbed by the bottom to reach the lid”

There is an energy field with temperature Tb = Tl in between the lid and the bottom. This energy is indeed in contact with the lid, and with the bottom. The bottom is emitting σTb^4 toward the lid, and so the lid receives it, but it doesn’t change temperature at equilibrium because Q = 0, i.e., the energy cannot raise temperature any further.

The lid emits σTl^4 on both sides, but on one side the energy is immediately replaced, and doesn’t go anywhere, whereas on the other it loses the energy to space; thus it does not lose 2 σTl^4, but only σTl^4.

Question: Do you acknowledge the First Law of Thermodynamics for an object? Where:

dU = Q + W = m Cp dT

Do you acknowledge that equilibrium for any given object is when dT = 0, and thus that for the object Q must equal 0?

Do you acknowledge that if Q > 0, then dU > 0 & dT > 0 which is this NOT a steady state or state of equilibrium?

Do you prefer the second solution found in this OP, where Q > 0 between the lid and plate? How do you reconcile Q > 0, and thus dT > 0, and thus temperatures are changing, when you claim that this is a solution where dT = 0 and temperatures are not changing?

27. esttom says:

“The lid emits σTl^4 on both sides, but on one side the energy is immediately replaced, and doesn’t go anywhere, whereas on the other it loses the energy to space; thus it does not lose 2 σTl^4, but only σTl^4.”

I agree with this, the lid is losing σTl^4 overall. But that’s exactly why yours isn’t a steady state solution — it should be zero!

“Do you acknowledge that equilibrium for any given object is when dT = 0, and thus that for the object Q must equal 0?”

I agree, but you have to be specific what you mean by Q. There is a heat flow between any pair of objects. For example, in this problem the heat flow between bottom and lid is zero.

You can also add up all the heat flows into any one object, and it is this Qtotal which must equal zero. For the lid, Qtotal is the sum of the heat flow from bottom to lid, from sun to lid, and from space to lid. Your solution doesn’t have Qtotal=0 for the lid or the bottom.

28. “I agree with this, the lid is losing σTl^4 overall. But that’s exactly why yours isn’t a steady state solution — it should be zero!”

How can you possibly say that the lid should be losing zero energy? The exterior of the lid is exposed to space, and it emits to space σTl^4. As it emits that to space, it receives σTb^4 on its interior, which is equal to what it loses on its exterior.

There is no possible way here for you to claim that the emission from the lid should be zero. Emission from the lid being zero is not what makes a steady state solution! The steady state solution is when the temperature is constant, which is when dT = 0. So then we refer to the first law, as done above, to see that Q = dU = 0 as well.

“you have to be specific what you mean by Q”

Sure:

“Heat is defined as any spontaneous flow of energy from one object to another caused by a difference in temperature between the objects. We say that “heat” flows from a warm radiator into a cold room, from hot water into a cold ice cube, and from the hot Sun to the cool Earth. The mechanism may be different in each case, but in each of these processes the energy transferred is called “heat”.” – Thermal Physics, D. V. Schroeder, Addison Wesley Longman, 2000

“If a physical process increases the total entropy of the universe, that process cannot happen in reverse since this would violate the second law of thermodynamics. Processes that create new entropy are therefore said to be irreversible. […]
“Perhaps the most important type of thermodynamic process is the flow of heat from a hot object to a cold one. We saw […] that this process occurs because the total multiplicity of the combined system thereby increases; hence the total entropy increases also, and heat flow is always irreversible. […]
“Most of the process we observe in life involve large entropy increases are therefore highly irreversible: sunlight warming the Earth […].” – Thermal Physics, D. V. Schroeder, Addison Wesley Longman, 2000

“Heat is defined as the form of energy that is transferred across a boundary by virtue of a temperature difference or temperature gradient. Implied in this definition is the very important fact that a body never contains heat, but that heat is identified as heat only as it crosses the boundary. Thus, heat is a transient phenomenon. If we consider the hot block of copper as a system and the cold water in the beaker as another system, we recognize that originally neither system contains any heat (they do contain energy, of course.) When the copper is placed in the water and the two are in thermal communication, heat is transferred from the copper to the water, until equilibrium of temperature is established. At that point we no longer have heat transfer, since there is no temperature difference. Neither of the systems contains any heat at the conclusion of the process. It also follows that heat is identified at the boundaries of the system, for heat is defined as energy being transferred across the system boundary.” – Thermodynamics, G. J. V. Wylen, John Wiley & Sons, 1960

“The temperature of a body alone is what determines whether heat will be transferred from it to another body with which it is in contact or vice versa. A large block of ice at 00C has far more internal energy than a cup of hot water; yet when the water is poured on the ice some of the ice melts and the water becomes cooler, which signifies that energy has passed from the water to the ice.
“When the temperature of a body increases, it is customary to say that heat has been added to it; when the temperature decreases, it is customary to say that heat has been removed from it. When no work is done, ΔU = Q, which says that the internal energy change of the body is equal to the heat transferred to it from the surroundings. One definition of heat is:
Heat is energy transferred across the boundary of a system as a result of a temperature difference only.” – Classical and Statistical Thermodynamics, A. H. Carter, Prentice-Hall, 2001.

“How and why does heat energy flow? In other words, we need an expression for the dependence of the flow of heat energy on the temperature field. First we summarize certain qualitative properties of heat flow with which we are all familiar:
1. If the temperature is constant in a region, no heat energy flows.
2. If there are temperature differences, the heat energy flows from the hotter region to the colder region.
[…]” – Elementary Applied Partial Differential Equations, R. Haberman, Prentice-Hall, 1998

“You can also add up all the heat flows into any one object”

That is still just what the First Law is: dU = Q = m Cp dT => Q is the heat flow, all heat flow, into the object.

“and it is this Qtotal which must equal zero. For the lid, Qtotal is the sum of the heat flow from bottom to lid, from sun to lid, and from space to lid. Your solution doesn’t have Qtotal=0 for the lid or the bottom.”

There is no heat from the sun to the lid, nor from space to the lid. The lid is transparent to the shortwave solar input S, and space emits nothing. Perhaps this is what is confusing you so much here? Gosh it would be great if it was, if we could clear this up for you here finally. Let me repeat, as I thought that this was clear: the lid is transparent to shortwave solar energy.

In my solution: Q = 0 from sun to bottom; Q = 0 from bottom to lid; F = σTl^4 from lid to space and this equals the S from Sun to bottom.

You haven’t shown that your solution has Q = 0 for each object; you only have Q = 0 summed over all objects, but this doesn’t get around the fact of the First Law and that if for some object Q > 0, then its temperature cannot be constant.

Are you making progress with this here yet, with understanding the basic concepts and math involved?

29. CD Marshall says:

Joseph I copied the conversation we had with “Erik the climate scientist” I haven’t removed all the other comments not pertinent to the conservation, but the whole thread was over 23,000 words. That’s almost a 4th/5th of a typical fantasy novel (100k) where I started with him in the thread anyway (the entire thread hit 240 comments).

30. Edited my comment above because I see that you were referring to Q for the lid from all sources…and the error is in thinking that the lid gets Q from the Sun. You weren’t summing Q over disparate objects there…although, I suspect you might still intend to do that, and so in that case I put the clipped text here:

The 1st Law applies to single objects: the change in energy of a single object is given by:

dU = Q + W = m Cp dT

Thus, it is Q for single objects, and for a single object to be steady state it individually requires its own Q = 0, so that its dU = dT = 0. The definitions above also show that Q applies and has meaning only to single objects, particularly when we consider that Q occurs at a surface boundary: it is the surface of a given single object, not over multiple disparate objects. Summing Q over multiple objects doesn’t have any meaning. And in any case, at equilibrium all you would be doing is summing zeros anymore, given that Q = 0 for all objects at equilibrium.

31. “Qtotal is the sum of the heat flow from bottom to lid, from sun to lid, and from space to lid”

Again: there is no energy from the sun to the lid, nor any from space to the lid.

The lid outputs to space, of course. And what it outputs to space it gets from the bottom, and the bottom receives its energy from S.

32. Wow CD, good work. Worth reading? Maybe edited?

33. esttom says:

“The exterior of the lid is exposed to space, and it emits to space σTl^4. As it emits that to space, it receives σTb^4 on its interior, which is equal to what it loses on its exterior.”

Please just add up all of the emission and absorption fluxes for the lid. It’s just the Stefan-Boltzmann law. Since Tb=Tl, I’ll write them both as T.

Lid absorbs from interior: +σT^4
Absorbs from exterior: 0
Emits to interior: -σT^4
Emits to exterior: -σT^4
Total: -σT^4

To be clear, the lid should be emitting. But this must be balanced by equal absorption so the lid doesn’t lose energy overall. That is the condition for a steady state, and is equivalent to Qtotal=0.

“There is no heat from the sun to the lid. The lid is transparent to the shortwave solar input S.”

Agreed, heat flow from sun to lid is zero.

“In my solution: Q = 0 from sun to bottom; Q = 0 from bottom to lid”

No, the bottom is absorbing heat form the sun! Q=S from sun to bottom, Q=0 from bottom to lid, and Q=0 from bottom to space. Qtotal=S for the bottom, which is again not zero!

You should have Qtotal=0 for each object. Instead you have Qtotal=-S for the lid and Qtotal=+S for the bottom.

34. “Emits to interior: -σT^4”

But does not lose to interior, because it also gets σT^4 from the interior. Hence: no energy is lost from emission to the interior.

And yes, let us look at your sum:

“Lid absorbs from interior: +σT^4
Absorbs from exterior: 0
Emits to interior: -σT^4
Emits to exterior: -σT^4
Total: -σT^4”

So, the total energy loss from the lid is -σT^4. That is exactly what should be found, because the lid has to emit the input energy from S. And at equilibrium S = σT^4. Thus your Total: -σT^4 plus S equals 0. No energy loss, steady state. You are demonstrating that my solution is correct, because emission from the lid balances the input from S. Thank you!

“To be clear, the lid should be emitting. But this must be balanced by equal absorption so the lid doesn’t lose energy overall. That is the condition for a steady state, and is equivalent to Qtotal=0.”

It is balanced, by the emission from the bottom to the lid. The bottom is constantly supplying the energy to the lid which is lost by the lid to space.

“No, the bottom is absorbing heat form the sun! Q=S from sun to bottom”

This is incorrect…a major mistake here! For the bottom, the First Law applies: dU = Q = m Cp dT; thus if Q = S on the bottom, then dU > 0 and dT > 0 for the bottom and hence it is not at steady state, but this is what we’re supposed to be determining. So again you contradict the first law.

Q for the bottom is Qb = S – σTb^4, and this is ZERO to get dTb = 0, so that S = σTb^4 = σTl^4. Q goes to zero for thermal equilibrium…it cannot remain constant and have constant temperature unless Work was involved, which it is not here.

“You should have Qtotal=0 for each object”

I do have that 🙂

“Instead you have Qtotal=-S for the lid and Qtotal=+S for the bottom.”

This statement simply seems to be made up! The lid, of course, has to emit the energy S for conservation of energy, hence “Qtotal=-S for the lid” is exactly what should be found: the lid emits S to space.

And for the bottom, Q = 0, since this is what defines its equilibrium temperature of Tb = (S/σ)^(1/4).

Perhaps you are confused about what Q is. Q is heat, and is the energy which crosses the surface boundary causing an increase in temperature. Q is something which is performed upon an object, just as work W is something performed upon an object:

dU = Q + W

That is: an object’s change in energy is given by the heat put in to it, and the work done upon it, and of course heat and work are the same things just occurring on different scales.

Thus, the emission from the lid to space is not heat!

The emission from the lid to space is simply energy, and this energy, as you have shown, is equal and opposite to the energy input thus satisfying conservation of energy. You seem to be getting confused about summing your Q’s, and what a Q is. Q is never an object’s emission…this is not what Q is at all! Q is the energy crossing an object’s boundary, going into the object, just as work is performed upon an object. I think I can see why this was so confusing for you. Apologies for not paying close enough attention to what you were saying, to identify your error earlier: the emission from the lid to space is NOT a Q. With that resolution, you should be able to correct your misunderstanding here.

35. Further clarification as to why the emission from a body is not heat:

The emission from a body is F = σT^4. However the heat flow into a body in a plane parallel is given by Q = σ(T1^4 – T2^4). Thus we can see that heat is the difference between emissions, and hence emission itself is not heat.

In the case where the emission is into space, space is not material and does not attain a temperature, hence Q is undefined for emission into space. Q only has meaning where temperatures may exist since Q is a function of difference of temperatures, and temperature may only exist in material bodies.

In the case where the material body has T2 = 0, then initially all of the emission is heat but as T2 then begins to rise then only a portion of σT1^4 is heat.

For a spherical body emitting σT1^4 into space, then the purely spatial vectors of the emission have undefined heat flow since space cannot attain a temperature; if some of the emission σT1^4 reaches a material body, then just that portion of energy intercepted by the body may act as heat and thus again it is clear that emission from a body into space is not heat.

Thus, this resolves your accounting of the emission from the lid to space as heat. The emission from the lid to space is not heat.

36. The temperature of space is undefined, hence:

Q = σ(Tl^4 – undefined) = undefined

37. But we CAN define for the lid that Fl = σTl^4 into space, which is what conserves energy. So there you go.

38. esttom says:

“So, the total energy loss from the lid is -σT^4. That is exactly what should be found, because the lid has to emit the input energy from S. And at equilibrium S = σT^4. Thus your Total: -σT^4 plus S equals 0. No energy loss, steady state.”

No, my total is already the sum of all of the energy fluxes into and out of the lid. You cannot arbitrarily add another flux of S. The total energy flux for the lid is -σT^4, period. And this total energy flux is the same as the total heat flux for the lid, Qtotal=-σT^4.

“Thus, the emission from the lid to space is not heat! The emission from the lid to space is simply energy”

This is important: all energy fluxes are either work or heat. This is clearly expressed by dU=W+Q. The emission from the lid to space is not work. So it is heat.

“Q is never an object’s emission…this is not what Q is at all! Q is the energy crossing an object’s boundary, going into the object”

Positive Q is heat going into an object. Negative Q is heat emitted from an object. Both are heat.

39. boomie789 says:

40. “Positive Q is heat going into an object. Negative Q is heat emitted from an object. Both are heat.”

Not all of an object’s emission may be intercepted by another object, hence not all of an object’s emission can be heat. And an object’s emission into space is not heat, since temperature is undefined for space. The temperature of space is not zero, but undefined, hence an object’s heat flow into space is undefined. An object can however have emission into space, but this cannot be equated to heat.

“This is important: all energy fluxes are either work or heat. This is clearly expressed by dU=W+Q. The emission from the lid to space is not work. So it is heat.”

Wow…this is really going off the rails!

“all energy fluxes are either work or heat”

Wrong. Flux is σT^4, whereas heat is (for plane parallel) Q = σ(T1^4 – T2^4). Thus, obviously, heat is a difference of fluxes…not a flux emission itself.

“This is clearly expressed by dU=W+Q”

Where Q = σ(T1^4 – T2^4), and hence is not a flux emission.

Just work on understanding that there is no heat flow into space, because Q is a function of difference of temperatures, and space has undefined, not zero, but undefined temperature. This resolves your difficulties here! 🙂 I mean it is not something discussed or encountered very often in physics, but it is really quite fascinating: temperature is undefined for space; it is not that the temperature of space is zero, but rather, it is undefined. Only material, matter, can be ascribed temperature. I hope you can see how this resolves your questions you’ve brought here – they were actually quite good…you were sticking to the definitions really quite well in fact! The only problem was that emission to space is not heat. Once you fix that, all of your concerns vanish. Thanks! It was really neat thinking about the undefined nature of the temperature of space.

41. esttom says:

“The temperature of space is not zero, but undefined, hence an object’s heat flow into space is undefined.”

The temperature of space is 2.7 K, which can be approximated as 0 for this problem. There is no problem having heat transfer to space.

“Thus, obviously, heat is a difference of fluxes…not a flux emission itself.”

You can always group the Stefan-Boltzmann fluxes in terms of heat transfer. For the lid, all of the emissions and absorptions are (as before)

Lid absorbs from interior: +σT^4
Absorbs from exterior: 0
Emits to interior: -σT^4
Emits to exterior: -σT^4

dU=+σT^4-σT^4-σT^4

Now group the terms appropriately

dU=(σT^4-σT^4) – (σT^4)

The first parentheses are the heat transfer from the bottom, the second parentheses are the heat transfer to space.

So it’s equally valid to sum up the heat transfer from the lid to each other object, or just to add up all the Stefan-Boltzmann emission and absorption.

“your questions you’ve brought here – they were actually quite good”

Thanks!

42. Joseph E Postma says:

“This is clearly expressed by dU=W+Q. The emission from the lid to space is not work. So it is heat.”

However, dU has to equal zero in order for the lid to have constant temperature, since dU also equals m Cp dT, and we need dT = 0 for temperature to be constant.

Hence, this is why we must appreciate that Q is not just 0 to space, but is undefined – no value at all. And thus, the emission from the lid to space is just that: emission, of σT^4.

This was a really fun conservation! It was so great resolving the difficulty here with the appreciation of the curious nature of space!

43. esttom says:

If you’re still worried about heat transfer into space, then we can focus on the bottom, which doesn’t emit to space.

To demonstrate my previous point, I’ll calculate the total heat flux in two ways: 1) by adding the heat flux from each other object, 2) by summing up all the S-B fluxes.

1) For the bottom, heat flux from the sun is +S, and heat flux to the lid is 0. Qtotal=+S
2) There is an incoming flux of +S from the sun, an incoming flux of +σT^4 from the lid, and an S-B emission of -σT^4. Qtotal=S

44. “The temperature of space is 2.7 K, which can be approximated as 0 for this problem. There is no problem having heat transfer to space.”

2.7K is the equivalent blackbody temperature of the cosmic microwave background radiation. This is NOT the temperature of space! You are missing a crucial and fundamental point here. Radiant energy can be equated to a temperature which a material blackbody would have in equilibrium with it via the Stefan-Boltzmann Law. This is NOT the temperature of space! This is really important for you to understand. You’re making a fundamental error here, thinking that space may have a temperature. You have to discard that idea.

“the second parentheses are the heat transfer to space”

There is no heat transfer to space. I know you would prefer to ignore what was just explained to you and continue on with your existing points as if they weren’t affected…but, there is no heat transfer to space. This resolves your difficulty. You must upgrade your understanding of temperature and space, and whether space can have a temperature (it cannot). Your sums and interpretation of them are incorrect.

Until you resolve this error you’re making here, ascribing a temperature to space, there is little more progress we can make. But it definitely has caught you out. Space does not have a temperature. Heat flow into space is undefined since heat flow requires temperature differences, and there is no defined temperature difference between a body and space. Just get there!! That’s the last piece of the puzzle you need!

45. “1) For the bottom, heat flux from the sun is +S, and heat flux to the lid is 0. Qtotal=+S”

Incorrect, because the heat flux from the Sun is 0 (zero) at equilibrium. But yes, it then follows that at equilibrium the heat flux is zero to the lid: Qtotal = 0 + 0 = 0, then! 🙂 Yay!

2) There is an incoming flux of +S from the sun, an incoming flux of +σT^4 from the lid, and an S-B emission of -σT^4. Qtotal=S

The flux from the Sun gives Q = 0 from the Sun; the flux from the lid gives Q = 0 from the lid as well: Qtotal = 0 + 0 = 0. And yes, there is an emission of σT^4 = S from the lid to space.

We’re getting exactly the solution you demand via my analysis and solution! Excellent progress!

46. esttom says:

“Incorrect, because the heat flux from the Sun is 0 (zero) at equilibrium.”

You’re assuming the conclusion here. We are discussing whether your solution is a valid steady state solution. You cannot assume that it is a valid steady state, then use that to say all heat fluxes are zero.

It’s a ridiculous conclusion at face anyway. Obviously there is an incoming energy flux from the sun. It’s not zero.

If you’re going to set some fluxes incorrectly to zero, and insist that it’s impossible to other fluxes into space (it’s not)… then you win, I guess.

47. I appreciate your attempt to side-step the fact of your fundamental error of ascribing a temperature to space, however:

“You’re assuming the conclusion here. We are discussing whether your solution is a valid steady state solution. You cannot assume that it is a valid steady state, then use that to say all heat fluxes are zero.”

We assume only that a state of thermal equilibrium will be reached for this system, where temperatures are not changing (dT = 0), and then we apply the First Law of Thermodynamics along with the equation for heat flow to determine the conditions of that end-state. So it’s not quite so asinine as you pretend here.

“It’s a ridiculous conclusion at face anyway. Obviously there is an incoming energy flux from the sun. It’s not zero.”

And of course no one ever said that the incoming flux from the sun is zero – it is S, and constant. The flux from the Sun, S, is constant, and the heat flow Q between S and the bottom goes to zero, appreciating that Q is not the flux S from the Sun.

“If you’re going to set some fluxes incorrectly to zero”

They never were. Heat flow was set to zero since this is what the First Law requires for equilibrium. And heat flow is not flux emission, but the difference of those.

“and insist that it’s impossible to other fluxes into space (it’s not)”

I quite clearly repeated that there is flux emission into space from the lid, but that this is not a heat flow for all of the reasons stated. Again you confuse flux with heat flow…notwithstanding the error about space having a temperature.

“then you win, I guess.”

I do win, yes.

48. Malcolm Hornsby says:

Hi Joseph. Do you know of Dr Robert Holmes? He makes the following claim.

“This formula collapses the ‘Greenhouse effect’ and proves it does not exist;
Te = ∜0.523 x Tv
The surface temperature on Earth is easily calculated – from Venus.
The fourth-root of the TSI difference times the temperature in the Venus atmosphere at 1atm = Earth’s temperature!”

He also backs this claim with a paper.

https://www.researchgate.net/publication/324599511_Thermal_Enhancement_on_Planetary_Bodies_and_the_Relevance_of_the_Molar_Mass_Version_of_the_Ideal_Gas_Law_to_the_Null_Hypothesis_of_Climate_Change

49. Malcolm, we’re all fans of Dr Holmes here 😀

50. CD Marshall says:

I haven’t even had time to look it over thoroughly. That guy could really spread the cow patties couldn’t he? Yet if you cut all of that down I doubt he had very much actual science he talked about.

51. Kev-In-ZA says:

Thanks Joe for a very fundemental and instructive topic. Reading through the comments, particularly @esttom’s difficulties, had me also thinking very deeply about the apparent conundrum. Perhaps I can throw in some of my own observations which might help clarify matters a bit ito where @esttom might be getting stuck, and also add the higher Law which has been inplicitly used here, but not defined.

Firstly, 3 laws are involved:
– Law of Conservation of Energy (LCE) (∑E=0=Ein – Eout – Eaccumutation)
– 1st LOT (dU = Q + W = m Cp dT)
– SB Law (σT^4)

The LCE is a more general law and must first apply. The 1-LOT is a special form of the LCE, and deals LCE within thermodynamics of processes, hence focus on Heat (Q) and Work and Internal Energy (U). And obviously all Heat is Energy, not all Energy is not Heat (ignoring Work).

As Joe has been at pains to point out, Heat (Q) is by definition that Energy exchange which causes an object’s temperature to change. So when in Steady State, T = Constant, and hence dU = 0, and therefore Q=0 for all objects (bottom and lid).

Now looking at the LCE for each object:
-bottom: Ein = S = Eout = σTb^4
-lid: Ein = σTb^4 = Eout = σTl^4 (with Tb = Tl = T)
From Radiative Heat-Transfer equation Qb->l = σ (Tb^4 – Tl^4) = o [because Tb = Tl)

But now the conundrum. Seemingly, the lid is following SB up and down. But if that SB-down flux is applied as pure Energy as part of the LCE for the lid, then Ein = σTb^4 while Eout = σTl_up^4 + σTl_down^4 = 2.σT^4

1 = 2…..? Not..!!!

So what to make of this conundrum…deep thoughts.
Using an electrical analog, perhap SB-law defines not a Flux (current) but a Potential. Using an incompressible flow analog perhaps SB-law sets up a Potential but that Energy only “flows” when “object quantum states” allow that flow. So perhaps Energy does flows from bottom to lid via the SB-law to fill the state “voids” of the lid created by up-emission. While the lid down-emission creates addition “voids” in the lid but cannot find “voids” in the bottom (which are simply in=out transitions from sun to emission). So the photon of down-emission spontaneously return from the bottom and refill the “voids” in the lid producing no net effect ito Energy for the lid. So perhaps what is seemingly Energy emission per SB is not actually.

But, hey, I’m just an Engineer trying to understand what I know intuatively to be true. So perhaps a quantum physics Neanderthal explanation, but the down emission just cannot act as Heat or Energy, or the Laws go out the window.

52. That was excellent Kev!!!!!!!

53. CD Marshall says:

@Erik Davis “GHGs increase average temperature by slowing down radiation loss to space. It is that simple. If less energy is lost per day, more energy remains in the system when the sun comes back around to add more energy.”

54. But the sun still doesn’t create the climate and nevermind the flat earth……

55. CD Marshall says:

Interesting he actually played on the E and avoided the Q the opposite of what they normally do and avoided the entire controversy of energy verses heat.

56. CD Marshall says:

Which can only “change” the surface temperature if “backradiaiton” was real. So they still bank on something they can’t prove exists and then claim that’s not what they are doing as they are actually doing it.

57. esttom says:

Kev, it seems we came to the same conclusion,

“But now the conundrum. Seemingly, the lid is following SB up and down. But if that SB-down flux is applied as pure Energy as part of the LCE for the lid, then Ein = σTb^4 while Eout = σTl_up^4 + σTl_down^4 = 2.σT^4”

Namely that Joseph Postma’s solution does not conserve energy!

You tried to think of alternative explanations, but the simple truth is you’re right: this solution does not conserve conservation of energy. As you nicely showed, all it takes to prove this is summing up the Stefan-Boltzmann fluxes for the lid.

58. CD Marshall says:

Troll:
“Spectroscopic analysis clearly shows heat passing through C02 in the atmosphere and radiating back to earth.” which was a quote

No matter how hard I tried to get him to see the grammatical error in the statement he couldn’t see it.
Me:
“IR can’t go through a gas and back. You guys are not geniuses.”

59. Lol 🤣 esttom – Kev gave a detailed explanation for why he agrees with my solution whilst pointing out what was your conundrum in understanding it 🤣

60. Kev-In-ZA says:

@esttom, as Joe has pointed out, I agree with his solution. The only thing you and I agree on is that a conundrum of understanding exists about something somewhere. And I have suggested that it is in the automatic assumption that SB-law implies emission of Energy per temperature, and that that Energy so emitted must “Act” as a true Energy (or Heat) flow. Further, I am very inclined to trust the conventional understandings of LCE and 1LOT (and for that matter 2LOT as Heat hot to cold only) over what the SB law is really implying.

An alternative interpretation of this is that SB emission up and down is correct. Then the lid is always at a higher temperature than the bottom such that your “SB interpretation” of down-emission as Energy is preserved. The solution for a single lid becomes EL_up = (σTb^4)/2 = (σTL^4) = EL-down = (σTb^4)/2 = (σTL^4)

Now E is conserved for the lid, while the LCE for the bottom then becomes:
Eb_in = S + (σTb^4)/2 = Eb_out = (σTb^4).
Or: S = (σTb^4)/2 ==> 2 . S = σTb^4 ==> S = (σTb^4) x (1/2)
So effectively the bottom temperature has to be (2)^0.25 times higher than the lid, or 1.189x higher.

Now for a little thought experiment. What if we add a second lid directly above the first. Then this second lid (SL) will also need to conserve Energy. ESL_in = (σTb^4)/2 = ESL_out. So up and down emission of the second lid is (σTb^4)/4.
Hence the first lid must be 1.189x higher temperature than the second lid, and the bottom in turn must be 1.89x higher than the first lid… etc… etc… etc…
4-lids and you double the bottom temperature compared to the top lid. But why stop there. Tell me what temperature do you want, and I’ll tell you how many lids you need. This experiment has been conducted. And multiple lids of IR radiatively resistant material failed to produce even a hint of this radical thought experiment alternative interpretation result.

Clearly this alternative interpretation is wrong, and easily disproved by experiment.

61. Kev-In-ZA says:

“…..Then the bottom [scratch – lid] is always at a higher temperature than the lid [scratch – bottom] such that your “SB interpretation” of down-emission as Energy is preserved….”

62. tom0mason says:

“The temperature of space is 2.7 K, which can be approximated as 0 for this problem. There is no problem having heat transfer to space.”

My own reading of this very mistaken idea (from my school-boy physics lessons) is —
Yes there maybe energy left over from the ‘big bang’ but in the vacuum of space there is NO temperature (just energy). There is no temperature because it requires the energy to interact with matter to cause thermal changes. Without matter there is no temperature, just an equivalent energy level.
As you say “Space does not have a temperature.” and it is an argument I have had too many times with so many ‘educated’ people.

63. tom0mason says:

I’ve had to look it up (after 50 years or so my memory ain’t that good) …
Heat capacity formula (a misnomer if ever there was one) –
Q = m•C•ΔT
where Q is the quantity of heat transferred to or from the object,
m is the mass of the object,
C is the specific heat capacity of the material the object is composed of,
and
ΔT is the resulting temperature change of the object.

It is obvious to me that as m tends to zero so does the result Q.
So for a vacuum (i.e. no mass, m=0), Q equals 0. A vacuum has no capacity to transfer heat.
(To think at one time I used this fairly regularly — heatsinks in electronics)

64. esttom says:

@Kev-in-ZA , the math you’ve presented is exactly correct. The steady state temperature of the bottom should be 2^(1/4) higher than that of the lid.

Of course this is not the answer Joseph Postma got, so something has to give. It’s either 1) the Stefan-Boltzmann law is wrong, or 2) Joseph Postma is wrong. I know which horse I’m betting on.

As for the experiment you’re describing, a very similar thing is done all the time for spacecraft insulation https://en.wikipedia.org/wiki/Multi-layer_insulation . The more layers of radiative insulation used, the warmer the spacecraft can be for a given amount of onboard heating. Often up to 40 layers are used.

It’s hard to do similar experiments on earth because convection is much larger than the radiative effect you’re trying to see. But a body in space (including the earth as a whole) can ultimately only lose energy through radiation.

65. Great comments guys kev & tom0! This fellow seems dead set on purposefully missing the point! lol 😆

66. esttom says:

@tom0mason , it’s true that defining the temperature of space can be confusing. On the one hand it doesn’t much matter in this situation: the lid in this problem definitely radiates outward into space according to the Stefan-Boltzmann law, whatever you may think about the temperature of space.

[JP: Yes, it radiates out to space. Thank you for acknowledging that since you seemed to really struggle with this fact earlier. And of course for all of the reasons stated to you, this is not a heat transfer since heat may only transfer into material bodies.]

Still, it’s an interesting question. Your argument is that since Q=m C dT, and m=0 for space, there can be no heat transfer. I wonder if that is the right equation to use here. It brings to mind the momentum of light; momentum is p=mv, so one might conclude that light has no momentum since its mass is zero. But in fact this is just the wrong equation to use: with more general equations from relativity, you can see the momentum of a photon is p=E/c=h/λ .

[JP: In this case there are no other analyses or equations for heat: temperature is something only a material body attains to. There is nothing in all of thermodynamics, unlike in physics with momentum for massless particles, where something non-material can have a temperature. The closest thing is radiation converted to temperature via the Stefan-Boltzmann Law, but this still applies to the temperature which a body would have if it emitted this energy, etc.]

Finally, here is another way to define temperature: if a cup of coffee sitting in a room settles to a temperature of 20 C, then the room’s temperature is 20 C. Meanwhile, an object in space settles to a temperature of 2.7 K (as long as it’s far from stars). So the temperature of space is 2.7 K.

[JP: Your analogy is flawed because it goes from two material bodies (cup of coffee and surrounding air in room), to one material body and the non-material body of space. It is not space that has that temperature of 2.7K: this is the blackbody temperature of the radiation filling space, which a blackbody would attain.

In any case, let’s just set the temperature of the background radiation to zero, and your solution still ends with Q > 0 thus dT > 0 thus not at equilibrium, while you claim it is a solution for equilibrium. How you can sit here and claim that dT > 0 equals dT = 0 is ridiculous…]

67. “The steady state temperature of the bottom should be 2^(1/4) higher than that of the lid”

But then Q > 0, which via the first Law requires that dT > 0, which thus means that the system is not in steady state equilibrium, which is what you present the solution as. Thus, your solution contradicts its intention, is not in equilibrium, and thus is not the correct solution.

dU = Q = σ(Tb^4 – σTl^4) = m Cp dT > 0 since your solutions ends with “the bottom should be 2^(1/4) higher than that of the lid”, hence this is not steady state. How you can just sit here and ignore that dT = 0 means no temperature change, while claiming that dU > 0, Q > 0, dT > 0 means no temperature change! lol!!! dT > 0 means temperature change; Q > 0 means temperature change; dU > 0 means temperature change; “the bottom should be 2^(1/4) higher than that of the lid” means temperature change.

Of Course, Kev has repeatedly explained his position, which you amazingly just ignore and put words in his mouth. Amazing that you think you can get away with this.

“It’s either 1) the Stefan-Boltzmann law is wrong, or 2) Joseph Postma is wrong”

Of course, it is you that is wrong since you claim that dT > 0 equals dT = 0, which is so utterly ridiculous.

“spacecraft insulation”

Of course insulation stops something from cooling as fast. On the dark side of the orbit the temperature drops precipitously. Insulation slows down that drop. That’s not the greenhouse effect.

R.W. Wood conducted the appropriate experiment, as did Joseph Fourier, as did Nasif Nahle, as do all greenhouses in existence: NONE of the experiments demonstrate the bottom of the greenhouse becoming hotter than the solar input!

68. esttom says:

[JP: snip

esttom: I have saved your comment and will post it if you publicize who you are, that we may confirm your credentials if you have any. If you’re unable to put your face to your claims then it is clear that you are merely trolling here.]

69. CD Marshall says:

So this was a wild ride on the comment section. Every time you disprove the ghge with hard physics these guys come out of the wood work. Weird.

So I wonder what they’ll say next about an actual greenhouse disproving the ghge theory? I’ll help them, they can say because it stops convection it prevents backradiation from occurring. LOL yeah I just helped them out.

70. Barry says:

Got awfully quiet here for some reason

71. Troll has been caught out six ways from Sunday…but they don’t relent just continue with more sophistry.

72. justgivemeall says:

Never seen anyone quit just being asked for credentials. These people really are cowards

73. esttom says:

I have to decline your request to reveal my identity, other than to say I am a postdoctoral student in a major physics department. This is a consequence of the unhinged harassment demonstrated in the Reddit thread you linked at the start of the article ( https://www.reddit.com/r/RealClimateSkeptics/comments/mf6abw/toy_models_free_body_diagrams_and_the_debunk_of/gu9x48b?utm_source=share&utm_medium=web2x&context=3 ). I do not want to open myself up to that.

74. Kev-In-ZA says:

@esttom. Well your response had me thinking for a while, and also had me retracing steps as to what might be causing the blockage and how to dislodge the blockage. It seems that you might be right and wrong at the same time.

In your example case of MLI insulation of a satellite in the vacuum of space, the result is correct with each layer providing a theoretical maximum (1 / (2)^(0.25)) drop in temperature at equilibrium. AKA multiple layers of radiative insulation in series. If the Satellite during “night” has a heat source at equilibrium, then sudden addition of more MLI will cause the satellite temperature to rise up to a new equilibrium temperature. However with no Heat source there is no equilibrium temperature and sudden addition of more MLI to a Satellite without heating will just slowdown the rate of cooling of the satellite. Therefore, some additional factors and logic at play. And interestingly as Joe has pointed out, SB is not a Heat flow by definition.

What seems wrong with your reasoning, is the application of it to a simple dual pane-plate situation on earth under normal Insolation. I did have to go back to Joe’s OP to see if he explicitly referred to the situation of a vacuum between the plates or not, and it appears unclear, but does refer to “only equilibrate via radiation”. Now at this point I am going to specifically change the assumption, and assume the gap is filled with dry N2 (thus IR transparent). I’m also going to add some values to S and T to contextualise.
Assuming: Albedo of 30%; SWIR transmission through lid of 85% and zenith Insolation Cs=1370W/m2
S = (1370W/m2 * (1-0.30) * 0.85 = 815W/m2 and rest the same.

Now the lid must emit S = σTL^4 for total system LCE neglecting convection/conduction. Hence TL = 346K or 73C. But per @esttom understanding the lid is also supposedly radiating σTL^4 towards the bottom.
Hence, as a first pass, the EB_in = S + σTL^4 = 2.S So theoretically, the bottom from reverse of SB is given by EB_out = 2 x 815 = σTB^4 ==> Tb = 412K = 139C
(Sidebar: This don’t seem logical as it implies that you should be able to boil water in a simple well insulated pot with a glass lid on it by just placing it under bright sun at zenith.)

But in our thought experiment, the N2 between the bottom and the lid will largely equilibrate the bottom temperature to the lid temperature via convection and conduction with such a high temperature differential. Since the bottom is perfectly insulated per assumption, and the lid only radiates what the total system receives. Achieving thermal equilibrium is after all just what objects strive towards, and heat transfer from the bottom to the lid will only stop when they are at equal temperature. And when they are at equilibrium, no further Heat transfers from the bottom to the lid, nor from the lid to the bottom. So they are in a form of “equipartition energy equilibrium” in the final state.
So we are back to … TB = TL for this specific scenario:
Where LCE per our understanding is:
Eb_in = S = Eb_out = σTb^4
EL_in = σTb^4 = EL_out = σTL^4

And LCE per @esttom understanding is:
Eb_in = S + σTL^4 = 2.( σTL^4 ) and Eb_out = σTb^4 (With TB = TL then 2 = 1)
EL_in = σTb^4 and EL_out = 2 x σTL^4 (With TB = TL then 1 = 2)
So we are back to our original little conundrum…..

75. Oh come on be a big boy now. You’re not a physics post-doc…anyone on the internet can say that… In fact if you actually had a face you would get more respect…anonymous trolls trolling never get nor deserve respect on the internet…are you new here?

Yes, I called you a retard after you repeatedly stated there, like you keep doing here, that dT > 0 equals dT = 0. That’s a retarded thing to say. It’s retarded to keep saying it. And there is absolutely nothing wrong with insulting your enemy; it’s rather quite a disgusting position of sophistry to continue to troll and demand that no one exhibit righteous indignation.

Go read those comments with me telling this troll off…that thread actually just shows several comments, whereas this discussion went on for dozens and dozens – but look at its last statement:

“Thus there is a net energy out flux from the ceiling in your mistaken solution.”

It is here stating that a net output of energy from the ceiling to space is incorrect and a mistaken solution! Of course there has to be emission from the ceiling to space!

After that It came here and started into the “all heat must be zero, but emission to space is heat” sophistry, which of course has been explained above with excellent input from others.

But hey…if you want to pretend fear of insults on the internet (lol) as an excuse to not put your face to your position…but that’s not how anyone gets respect, particularly when they keep saying that Q > 0 & dT > 0 equals dT = 0.

I’m totally public, and can put my position to my face and my identity, etc; so why should I have respect for anonymous trolls who won’t do the same.

76. @Kev, yes, I thought it would be clear that it is all a vacuum scenario, etc.

“(Sidebar: This don’t seem logical as it implies that you should be able to boil water in a simple well insulated pot with a glass lid on it by just placing it under bright sun at zenith.)”

Exactly. There is no empirical experimental demonstration of the supposed effect of downstream energy coming back to increase temperature, etc.

Here’s a relevant quote:

“If a physical process increases the total entropy of the universe, that process cannot happen in reverse since this would violate the second law of thermodynamics. Processes that create new entropy are therefore said to be irreversible. […]
“Perhaps the most important type of thermodynamic process is the flow of heat from a hot object to a cold one. We saw […] that this process occurs because the total multiplicity of the combined system thereby increases; hence the total entropy increases also, and heat flow is always irreversible. […]
“Most of the process we observe in life involve large entropy increases are therefore highly irreversible: sunlight warming the Earth […].” – Thermal Physics (pg. 82) (Schroeder 2000)

The temperature which the ceiling attains cannot send heat back to the bottom.

And equilibrium is Q = 0 = dT, and if the planes have different temperature then Q > 0 hence dT > 0, which is NOT the solution for dT = 0, obviously.

And yes that’s good to point out the 1 = 2 problem here again, consistent with how that was demonstrated for the spherical greenhouse in my Cold Light of Day book.

77. Joseph E Postma says:

You guys want me to do a video discussion/presentation on this?

78. esttom says:

@Kev-In-ZA , looks like you know your stuff. I think I’m not able to comment here anymore, but feel free to shoot me an email at [] if you want to discuss further.

79. You can comment esttom – just demonstrate that you can actually can put skin in the game, or even just do this: show us how your solution has Q = 0 between the bottom and ceiling when they’re not at the same temperature, using the heat flow equation.

80. CD Marshall says:

If you do don’t hold back on the physics. If you have to do a simpler version later fine but don’t hold back on the original presentation.

81. tom0mason says:

esttom said,
“Still, it’s an interesting question. Your argument is that since Q=m C dT, and m=0 for space, there can be no heat transfer. I wonder if that is the right equation to use here. It brings to mind the momentum of light; momentum is p=mv, so one might conclude that light has no momentum since its mass is zero. But in fact this is just the wrong equation to use: with more general equations from relativity, you can see the momentum of a photon is p=E/c=h/λ .

It is very obvious from this comment that you esttom, can not differentiate temperature and thermal effects from energy flow in the vacuum of space.
The formulation you cite, p=E/c=h/λ, is all about energy and not heat flow — heat flow requires material mass — where in your cited formula is the mass?

82. esttom says:

@Kev-in-ZA , your analysis of the MLI example is spot on. In particular this sentence is important: ‘If the Satellite during “night” has a heat source at equilibrium, then sudden addition of more MLI will cause the satellite temperature to rise up to a new equilibrium temperature.’ To put it simply, heat source + extra insulation from cold = higher equilibrium temperature. That’s the core idea of both MLI and the greenhouse effect.

[JP: MLI on satellites is actually mainly used to protect them from the Sun, but also to slow cooling at night. It’s modulates the temperature swing. Extra insulation from cold is not the climate greenhouse effect.]

You analysis of the two-plate problem with convection is also correct. In this case convection makes Tl=Tb, and the temperature Tb ends up exactly what it would be without the lid. That is to say, convection totally kills the greenhouse effect here. On earth the situation is more subtle — this is a whole complicated story which I could go into, but the end result is that the greenhouse effect survives. The basic idea is that the atmosphere is tall enough that convection can no longer perfectly equilibrate the temperature of the atmosphere (lid) with the temperature of the surface (bottom). But there are other complications, it’s very interesting.

(This also addresses your sidebar: a pot with a lid on it doesn’t have a significant greenhouse gas because convection destroys it).

[JP: All modes of heat transfer follow the same rules and laws…of thermodynamics. Both physical transfer and radiant transfer of heat are electromagnetic processes…the exact same electromagnetic process.

If we could simply evacuate an enclosure above a pot of water and boil the water underneath with a sufficiently insulated thermos system, we would. There is no experimental or laboratory evidence that exists for such a thing, and it wouldn’t be difficult to get it. Thermodynamics has been understood for a long, long time now. Imagine the application for camping, for hikers, for survival situations where you need to boil water, etc: it would have been engineered long ago, starting when Fourier and deSaussure were exploring this. You just need an insulated thermos, filled with water, with a plane-parallel enclosure on the top used to “trap heat” and make the bottom of the enclosure in contact with the water much hotter than the solar input. And you could engineer this enclosure with multiple layers, so that the bottom doesn’t have to come to a temperature to emit twice the solar input, but three times, four times, five times, etc.

The climate greenhouse scheme has never been empirically demonstrated – it has only been interpreted from a starting point of the Sun not being able to create the climate via flat Earth theory with solar input only -18C globally, and ignoring the fact of the lapse rate and where the average must then be placed, etc. The climate greenhouse effect only exists outside of empiricism, and existing facts.]

83. esttom says:

@Joe : my solution does not have Qlb = 0 (heat flux to lid from bottom), because that is not the right requirement for a steady state! The correct requirement for a steady state can be stated in two equivalent ways: 1) Each object O has energy influx = energy outflux. Or 2) Each object O has total heat influx = total heat outflux. To check 1), you just sum the Stefan-Boltzmann fluxes into and out of O. To check 2), you take every object besides O, calculate its heat flux to or from O, and add these up. Both methods give the same answer.

So Qlb is not zero, but Qlb + Qls = 0 . That is to say: the heat flow to the lid from the bottom is exactly balanced by the heat flow from the lid to space, so the lid gains 0 energy overall. A similar statement holds for the bottom.

Final note: your method is similar to 2), but you are trying to find a solution where the heat flow between every possible pair of objects is zero, or Qij=0 for each i,j . This would be the correct condition for thermal equilibrium. However, it is impossible to find a thermal equilibrium solution here, because thermal equilibrium means all objects are the same temperature. As long as the system is caught between the sun at 6000 K and space at 2.7 K, there cannot be a thermal equilibrium solution.

[JP: Yes we know your position. All you have to do is provide empirical evidence…not interpretation from climate science…but actual laboratory evidence where the bottom surface becomes much hotter than the shortwave input alone can explain.

In any case, that’s quite a convenient way to sidestep thermodynamics, just denying that thermal equilibrium can exist. Of course thermal equilibrium can exist, and it doesn’t require anything being equal to 6000K because the geometry doesn’t require it – the inverse square law applies and thermal equilibrium requires equilibrium with the local flux of sunlight, not the temperature of the Sun itself. View factors apply! This is such a basic mistake and lack in your comprehension of the physics here.

And finally of course, my correct solution has Qbl = 0, and lid emits S to space to balance the input. There is no heat flow into something which cannot receive heat – space literally receives NO HEAT, because to receive heat requires mass. QED.

Qlb + Qls = 0 + 0 = 0]

84. Consider this:

On the left we simply affix a SW-transparent, LW-absorptive layer to a surface which is SW-absorptive (and insulated so no heat transfer through). The affixed LW-absorptive layer has “infinite” (very high) conductivity. What temperature will it attain?

Obviously the layer will attain a uniform temperature of T = (S/σ)^(1/4)

Now just split that layer in half and create a vacuum gap, as on the right. Does the simple presence of a vacuum gap, allowing the gap to exchange radiation, cause the piece remaining on the left to increase in temperature? There is nothing fundamentally different happening: on the left side the molecules were transferring and sharing energy via electromagnetism, whereas on the right they’re actually still doing just that too.

Here’s the scenario for 5 layers:

Wow, imagine how useful that would be!? High temperatures are the basis of the modern world in electricity production. Look at how easy it would be to increase temperatures to values useful for industry and power production, etc!

Here however is the correct solution:

Each layer is 91C degrees. In a plane parallel geometry, there is no possibility other than equal temperatures everywhere…that’s the requirement of the geometry. Thermal equilibrium with the Sun isn’t found in being equal temperature to the Sun, because the view factor i.e. the inverse square law reduces the heat flux with distance; you have to be in equilibrium with the heat supplied. Only when objects are touching do the temperatures becomes equal. There is no view-factor for plane parallel geometry as the layers are in full view of each other, thus no reduction in heat with distance, thus the temperatures will be equal here.

85. If space could have a temperature it would be infinite by now…warming up and up and up for billions of years from stars, etc.

The space around the Sun should be hot as heck by now…the space of the solar system should be getting as hot as the Sun! Does the Sun leave a heat path through space? If we had heat vision goggles would we see the path that the Sun left through space with its heat and temperature imprint on space? Uh oh…that’s absolute space we’re getting into!!! What detector can we use to see where hot object used to be in space? If space can be heated and attain a temperature via radiation, why not through conduction? The Sun is touching space! There should be a scorching red path through space where the Sun used to be. But what is the Sun’s path through space? What’s the reference frame!?

Instead…only view factors for the radiation applies, and you attain the temperature from the radiation present…not from the space present.

In any case, the summing Q’s approach where we include that heat is something sent into space fails in and of itself:

Heat is something that flows between objects, from one object to another, occurring at an object surface boundary. How do you assess what heat is flowing? From the heat equation of course. So consider three objects, one hottest, one intermediate, and one coolest:

Consider the intermediate object: is the intermediate object being heated by, or is it heating, the coolest object? How do you assess that? We can easily understand that the hottest object is heating both, and we should also understand that the coolest object can be heating neither; the intermediate object thus must only be being heated by the hotter, while it is also heating the cooler. The way you assess this is of course through the heat equation as a difference of emission between two objects (and here in this example we would need to apply view factors – inverse square law). But this makes it clear: heat is assessed between pairs of objects, and YES, I do agree that you can sum heat in this fashion.

Of course the climate alarmist solution and esttom’s is to just directly add in the flux from the ceiling along with the solar input…but this is incorrect, as we have just demonstrated.

Qb = (S – σTb^4) + (σTl^4 – σTb^4) = S + σTl^4 – 2σTb^4 = 2S – 2σTb^4

But if as per esttom we have σTb^4 = 2σTl^4 = 2S, then:

Qb = 2S – 4S = -2S

which obviously simply makes no sense at all anymore.

Given that heat flow is assess between pairs of objects, and heat flow induces temperature change for any object, then the only solution where Q’s will sum to zero is if they are all in fact equal to zero in each instance.

Qbl can equal zero because Qls (lid to space) equals zero. The σT4 is simply the flux emission from the lid to space, the Stefan-Boltzmann Law. This is not heat. Heat is a function of difference of temperatures, thus if there is no temperature to have a difference to, then there is no heat. Space has no temperature.

86. Kev-In-ZA says:

@esttom, Part A) while your response to the MLI is fine in most cases, that is to say “….heat source + extra insulation from cold = higher equilibrium temperature…”. Not much rock science there. But a rider to that statement is that there must be a “viable” Heat-Source driving that higher equilibrium temperature state. If the Heat-Source is not “viable” as a valid source of Heat, or as the equilibrium temperature approaches the temperature of a bulk Heat source, then the statement may not be true.

In your response to my analysis of the two-plate problem with high convection, you fail to answer to the LCE conundrum that follows when Tb = TL, and you continue to consider the SB emission as valid Energy from the lid to the bottom, that the LCE of the lid is in deficit (1.S in coming from bottom vs 2.S out up and down) while the bottom is in excess (2.S in vs 1.S out), yet there is no other way for the energy to get from the bottom to the lid especially via Heat transfer.

JP has responded to the sidebar (with almost the exact practical idea I had for boiling water in a thermos with a 2-plate lid partly evacuated to really get the back radiation “cooking”) and the conundrum of theoretically being able to generate just about any temperature desired with multiple LWIR resistant plates which he illustrates nicely (so much for “Dispensing with graphics…..” but a picture does paint a thousand words…). Now that would be truely useful for us engineers to apply to some practical uses. So I’ll leave that angle as already well responded to a needing a rebuttal.

So I’ll leave the general comments and linkage to GHE for part B)

87. CD Marshall says:

You know as I kid I left a tape recorder on the back window of our car in the 80s during a summer day, it melted. Yes it actually melted on one side. Now imagine that kind of thermal heat well insulated.

88. Kev-In-ZA says:

@esttom, Part B) as to the statement of “…. heat source + extra insulation from cold = higher equilibrium temperature. That’s the core idea of both MLI and the greenhouse effect” I partially agree, but again with riders and complications. Specifically, the atmosphere or CO2 “back-radiation” never Heats the surface (except in special circumstances like during Inversions and Katabatic Winds). The Sun HAS to do that Energy-In part.

And I abhor the title GHE, and would much prefer something like “Atmospheric Temperature Profiling Effect” as it then covers the multiple effects involved in the atmospheric column and how these affect and control the local surface temperatures (such as energy distribution, clouds, latent heat, lapse rate, convection, radiative insulation, and ultimately local temperature surface/cloud/atmospheric emission to space). Overall, a pretty complex beast.

As to CO2, I accept that it has a radiative insulation effect in the atmosphere just like water-vapour and clouds. And I believe it is band specific with most effect around the 15micron band, and future that it is largely saturated at about 50-100ppm. I understand that doubling CO2 at current level, all else equal, would produce about 0.6-1.0K global surface temperature increase, but that negative feedback will likely lower that quite a bit. (I might differ slightly from JP on this.) But I am much persuaded by the summary of Howard Hayden informed by the work of William Happer and William van Wijngaarden as presented in this treatise. http://www.energyadvocate.com/co2_clim.pdf

As to the foundations of climate science, there is some real bad pseudoscience there. Flat earth starting point with no temporal and heat capacity dynamics. Bad Toy Models that can lead people astray, and a starting base that gets the temperature of the Moon wrong even with no GHE/ATPE. And the crap that get peddled as science and evidence such as trends in GMST, SLR, extreme weather, lots of gibberish and fudged data. And I have zero faith in the phantom fantasies of GCM’s for accuracy and predictions. Overall view, there is not Climate Emergency, and more CO2 in the atmosphere is good for the Earth.

89. CD Marshall says:

0.06 can easily be attributed to nothing more than molar mass contribution. The water phases decreases temperatures or maintains a decreased temperature longer it does not influence higher temperatures.The water cycle in all of its phases with the Coriolis effect manages our planet fueled by real time solar density of the Sun and a multifarious secondary influences such as gravity, cosmic rays, terrestrial/solar brightening/dimming, planetary harmonics, galactic harmonics, geothermal and magnetic field strength/decay both solar and terrestrial to name a few.

You cannot look at our planet simply at radiation alone. You can for emissions out to space but the “energy” budget used at the surface and troposphere and the energy in and out of a system should be demonstrated as two distinct models to limit confusion.

90. boomie789 says:

James Lindsey is the guy who exposed peer review journals a while back.

the “Dog humping in parks perpetuates rape culture” study, lol. They actually published it too.

91. boomie789 says:

Skip to around 28mins and watch for a bit. When he talks about the communist OSS agent.

92. esttom says:

There are a lot of points here now, and I got busy, but I’ll try to touch on everything. In the next comment I’ll address why we don’t see technology using a greenhouse type effect (with multiple glass plates) to create elevated temperatures.

@JP : re the vacuum gap scenario. There is indeed a fundamental difference between two conductive plates touching and separated by a vacuum gap. Vacuum is a very good insulator. In fact there is a fundamental bound on the amount of radiative heat transfer between objects: the same sigma*(Th^4-Tc^4) that we’ve been discussing. I don’t know of an upper bound on heat conduction.

Re the 5 plate scenario: it’s nice to see your solution drawn out explicitly. It is very clear here that the leftmost plate has 2000 W/m^2 energy flux in and only 1000 out, so it cannot conserve energy. Similarly the rightmost plate has 1000 W/m^2 in and 2000 out.

[JP: The temperature of the pane to the right of the bottom-most pane does not represent an additional input. Treating it as an input violates the 2nd Law because this reduces the entropy of the system, and it also violates the 1st Law because it can’t act as heat, and finally it wouldn’t even be consistent with your demand to sum all Q’s since you’re not calculating the Q from the pane on the right and adding it this way but are only fully adding in its energy with the solar input. Thanks for helping make these facts clear!]

Re thermal equilibrium with view factor: thermal equilibrium with the sun does require being at the same temperature of the sun. View factor does not affect this. Radiative heat flow between two objects with emissivity 1, including view factor, is Q1->2 = sigma A1 F1->2 (T1^4-T2^4) . It is clear heat flow only goes to zero if T1=T2 or if the view factor is zero (no line of sight).

[JP: The view factor with point or spherical sources becomes the inverse square law, and heat flow goes to zero at different temperatures because flux decreases with distance. With plane parallel geometry then heat flow goes to zero at equal temperatures. Thermal equilibrium is defined as Q = 0.]

Re temperature of space: you argue space should warm up because of all the stars. However, you might know that space is very large, large enough to make the heat input from stars negligible.

[JP: Well of course you just want to say whatever you can here…to try to pretend legitimacy after claiming the ridiculous position that space can have a temperature…lol. Space doesn’t have a temperatures, and it cannot receive heat.]

Re heat flow into space: of course objects emit heat into space. You yourself said a spacecraft cools down in space. This can only be explained by it emitting heat.

[JP: Emission is not heat. Emission is simply the Planck emission law or the S-B Law, etc. It emits energy. But this is not heat for space. It can be heat for other objects potentially…but it is not heat for space. It is simply emission of electromagnetic energy.]

@Kev-In-ZA : re the two plate problem with convection: convection can transfer a large amount of heat with very small temperature differences. So Tb will be very slightly larger than Tl, almost equal, and convection supplies the missing energy flux from the radiation analysis. As you pointed out the bottom appears to have excess energy influx of S while the lid appears to have a deficit of S; this is because convection transfers S from bottom to lid.

Re greenhouse effect: I agree the name is unfortunate. I’m glad you believe the greenhouse effect exists and CO2 doubling would have some effect. The exact numbers are a more difficult conversation.

[JP: It’s psychotic to imply beliefs in people who have clearly stated opposing opinions.]

Re moon temperature: I’m curious what you mean about the temperature of the Moon being wrong. My guess is that the actual average temperature is lower than the predicted blackbody temperature? If so I have a very satisfying mathematical explanation.

[JP: Trivial stuff. But just as there is a sufficient and satisfying mathematical explanation for why the bottom of the atmosphere is warmer than the average temperature. :)]

93. Kev-In-ZA says:

@CD Marshall, excellent additional points. Absolutely agree with all those other points that maketh the and influence the climate. And one other biggy causes of variability through ocean circulations,

As to a cassette tape melting, not likely in the true sense as a tape is likely made of polystyrene which melts at 270C, but has a glass temperature of 100C which would allow it to start flowing a bit. So just possible under native sun with perhaps a slight magnification effect. thru the glass.

94. CD Marshall says:

Wasn’t the tape it was the player and yes it warped to the point the player was worthless. This was the 80s things weren’t as durable as they are now especially rubber and plastic and cars were giant hunks of heat conducting steel. If you dropped a player back then it broke, buttons popped off and everything. A record could just as easily melt to a lump back then.

They are all connected, Sun to the water/wind cycles move thermal energy around globally from the climate is complex with many moving parts.

The more you learn about climate the less you end up knowing for one subject opens up to a library of other data you simply didn’t know before.

95. Kev-In-ZA says:

CD, yes, the mind does indeed start warping at the complexity of climate dynamics. I think Happer recently described it, apart from the brain, as one of the most complex systems he could think of.

96. CD Marshall says:

Solar Vacuum Tubes with Heat Pipes – How They Work
https://www.123zeroenergy.com/how-solar-vacuum-tubes-work.html

97. boomie789 says:

[video src="https://files.catbox.moe/itm4ke.mp4" /]

98. And here we are.

99. Snitch for public safety it might save lives you’ll be a hero.

100. boomie789 says:

Westerners and east Asians evolved to police their commons. This actually useful trait is being hijacked as a weapon against us.

101. justgivemeall says:

I see a lot of police in Ontario not going along with the new rules for stopping and questioning for no reason. Maybe a little hope although I don’t have much faith in our current society. Our idiot PM is guilty of destroying Canada and should be hung for treason.

102. CD Marshall says:

We have a puppet in the US who can’t even remember his own name.The globalists are ready to move in. The great reset is pretty much the great global takeover.

A “vaccine passport” really what’s next? You have have be chipped and sterilized like cattle?

103. CD Marshall says:

Does any proof on the net exist about Maurice Strong helping to form the IPCC or has it all been erased?

104. CD Marshall says:

Joseph,
I found another physicist who is agaisnt gw and is not afraid to reveal himself,
” … PhD from Cornell, a postdoc and over 25 years as a professor in universities and colleges, I am a physicist, yes. If you don’t understand that when it comes to climatology is not an exact science and you think that their predictions are solid, you have proven to have never looked at the research in that field.”

He’s smacking down a troll with that comment. Hoping he’s not another Roy.

You can find him under his real name, Patrick LaBelle, in the Friends of Science posts…

105. CD Marshall says:

Joseph,
“As a physicist, I am appalled by the brainwashing behind the climate hysteria.
…PhD from Cornell, a postdoc and over 25 years as a professor in universities and colleges, I am a physicist, yes.”

Found him on Friends of Science posts a Patrick LaBelle, who is not afraid to show his real identity. He hasn’t made any videos in a few years though, I see.

106. boomie789 says:

107. justgivemeall says:

It’s a start anyway, now if we could get the military to take over parliament and then set up an actual govt run by people and not political parties lining their own pockets

https://wattsupwiththat.com/2021/04/17/atmospheric-energy-recycling/

I did not wade through the article there either, but it looks like the author is endorsing the “simple model” serves to represent the more complex reality, and I’m tempted to sum it up as BS.

109. justgivemeall says:

Wow that is one for Joeseph again with enough atmospheric layers the planet could be warmer than the sun! Seriously I have no education and I can figure that one out and this guy claims to have a phd. I’m now going to light a candle and put many layers of glass over it I should be able to overheat the house. I’ll let you know how I make out with that.

110. Kev-In-ZA says:

@JP, I see @esttom is taking a bit of a sabbatical in responding, but I have continued to mull on certain details on the 2-pane with a vacuum setup, some of which are very clear, but some parts that I don’t understand the true “nature” of. Perhaps you can help my thinking/understanding. My Heat-Transfer and Thermo knowledge should be adequate based on Mech Eng degree, and while I tend to think about these issues in a theoretical as well as practical sense, always trying to honour fundamental laws and truths, I will acknowledge hardcore Radiation and Quantum Mechanics theory is a bit Greek to me.

I am well aware of the Wood and Nahle experiments which covered single LWIR resistant pane evidence, but with conduction/convection active. This was still fairly compelling. But less familiar with the Fourier and deSaussure work which I think covers multiple LWIR resistant pane evidence.

Now considering the 2-pane setup, but with a vacuum gap and full vacuum environment, all the Insolation energy reaching the bottom (say S=1000W/m2) must be lost at equilibrium via the lid. Hence lid up surface emission must be S = σTL^4 = 1000W/m2 Hence TL = 91C. Now there are two ensuing alternative assumptions that follow ito how to look at the bottom vs the lid:
i) the bottom and the lid are 2 independent objects thermodynamically; or
ii) the bottom and the lid are essentially the same control-volume thermodynamically and share the same internal energy state (obviously ignoring the thickness of the lid ito practical heat-transfer dynamics)

If i), then the bottom is receiving S energy flux, and must loss S energy flux which is does via radiative heat-transfer to the lid. Since R-HT is given by Q1-2 = σ(T1^4-σT2^4) and given that Heat-Transfer is also Energy Transfer, all is good with the LCE and 1-LOT and 2-LOT when Eb_in = S = Eb_out = σ(TB^4-σTL^4) => Hence σ(Tb^4) = 2.S and so Tb = 160C. Now this would seem to be a pretty easy experiment to prove or disprove.

If ii) and the bottom and the lid are essentially the same control volume, then all is good from an overall system view, and starting the system from STP condition will involve transient HT from the bottom to the lid until everything is at 91C condition and up-emission puts the total system into control-volume equilibrium. The lack of understanding I have then is how is energy mobility within the control volume (most of which is just a vacuum between the plates anyway) thought of..??

Option i) is compelling from a simplistic point of view, but allows plain silica-glass (a remarkable good LWIR resistant material) to become a impressive temperature magnifier just by splitting the same glass sheet multiple times within a vacuum.

Option ii) is a little quirky for some minds, but then so much of physics has similar problems when you peer too far into the gearbox.
Would very much appreciate further insights..??

Would also appreciate your insight on a further issue with radiation from the Sun. Under normal HT rules, energy from a colder object can never increase the Internal Energy (U) of an initially hotter object. So doesn’t matter whether radiation or conduction HT involved. That radiation from a cold object can limit the energy loss from a hotter object according to the 4th power of the temperature difference and corrected for viewing factors and areas etc of course. So “Cold” radiation can do radiative insulation, but Heat-Transfer….Neyt!!! So how should Solar radiation flux be thought of in this context…!

111. Two posts by Bob Wentworth at WUWT that you all may want to comment on.

Atmospheric Energy Recycling
https://wattsupwiththat.com/2021/04/17/atmospheric-energy-recycling/

Deconstructing Wilde and Mulholland’s Analysis of Earth’s Energy Budget
https://wattsupwiththat.com/2021/04/18/deconstructing-wilde-and-mulhollands-analysis-of-earths-energy-budget/

112. justgivemeall says:

the Second Law doesn’t say no energy can flow from cooler to warmer. It simply requires that the heat flow (i.e., the net energy flow), must be from warmer to cooler. As illustrated in the heat flow illustrations (Figure 4 and Figure 6), even with energy recirculation, heat always flows from warmer to cooler.

This seems to be a contradiction in terms,first you have back radiation heating the earth but the earth is warmer than the atmosphere so it can’t. Am I just not getting what Bob is saying? This would be like boiling water in a pot and saying that the hot water rising recirculates to the bottom and makes the pot overall energy higher but if you shut off the burner this doesn’t continue to get hotter. In Bobs explanation the hot water layer at the top would keep heating the cooler water on the bottom that just makes no sense.
Cheers Barry

113. esttom says:

Now, both @JP and @Kev-In-ZA have asked why we don’t see technology using a series of glass plates to create a greenhouse effect in a thermos, achieving very high temperatures. This is a great question and I enjoyed thinking about it. I think the are several very satisfying reasons:

1) There is already a superior solution which works better and is much easier: solar cookers. Existing solar cookers can apparently reach hundreds of degrees Celsius (!) and cook a steak in minutes, using just a parabolic reflector dish

[JP: parabolic reflector dish…magnifying the input! lol…thanks for the example and the physics which debunks the GHE!]

and a glass top. Even a glorified box lined with foil can reach well over 100 degrees Celsius, enough to boil water. Improving this with a greenhouse effect would require a high vacuum (a very difficult requirement: I wouldn’t want to bring a scroll pump on a backpacking trip),

[JP: Don’t be ridiculous and invent some insanity to make the practical demonstration needlessly impossible. High vacuum in sealed enclosures is very easy to create and exist in any thermos bottle. It doesn’t need to be “perfect” vacuum, but simply enough to reduce convection/conduction, and this is achieved with very little vacuum in fact.]

multiple glass plates with very little conduction between them, glass that could attain high temperatures without shattering… and a good solution already exists. I should note the glass plate on solar cookers prevents some radiative heat loss,

[JP: The glass plate is of course to stop convection. Too bad they don’t demonstrate the GHE…all you need is backradiation from them, and you cannot possibly argue the convenience that convection just so happens to take away ALL of the GHE enhancement that should be there and keeps the system to appear and respond the same way as if no backradiation GHE existed…lol!!!!]

but this is probably not too imporant because…

2) Solar cookers are almost certainly conduction limited. All the greenhouse effect does is isolate a system radiatively from the outside (at IR wavelengths). With a larger number of glass panes it would be theoretically possible to achieve closer and closer to zero heat loss through radiation. But if there is any conductive heat loss through the walls of the thermos, that will become the limiting factor, and bringing the radiative heat loss to 0.000001 W/m^2 doesn’t help. Think of resistors in parallel — if you have 1 ohm in parallel with 1000000 ohms, increasing the latter does nothing. To actually reach temperatures approaching those of the sun, the system would have to accept heat from the sun, while perfectly removing ALL sources of heat loss.

[JP: Again: You cannot possibly argue the convenience that convection just so happens to take away ALL of the GHE enhancement that should be there and keeps the system to appear and respond the same way as if no backradiation GHE existed…lol!!!!]

3) Finally, there is a better solution than insulating against radiative heat loss. Consider the hot region at the center of a solar cooker. This hot region has some field of view to the sky, where it can lose heat via radiation. Now, this heat loss can be reduced by introducing glass planes across the whole field of view. But it’s even better to introduce mirrors at these locations which redirect the sun’s light toward the hot region. The logical conclusion of this process is a parabolic reflector — an object at the focal point sees the sun in every direction it looks. Now, instead of insulating the field of view to prevent heat loss, we have replaced the whole field of view with a heat source!

[JP: A parabolic mirror does not demonstrate the climate science radiative greenhouse effect…lol.

https://climateofsophistry.com/2016/08/11/how-a-magnifying-glass-debunked-climate-alarm/

https://climateofsophistry.com/2019/07/12/how-magnifying-glass-physics-debunks-climate-alarm/%5D

114. J Cuttance says:

Might I suggest that anything in Wentworth’s graph that contains β⋅S is, in fact, B.S.

115. CD Marshall says:

Joseph you should do another post covering this,

Deconstructing Wilde and Mulholland’s Analysis of Earth’s Energy Budget
https://wattsupwiththat.com/2021/04/18/deconstructing-wilde-and-mulhollands-analysis-of-earths-energy-budget/

It’s a good direction to keep going.

116. CD Marshall says:

I was told that photons can be considered to have some mass under gravity is that true? I would say no for gravity may effect the path of a photon it does not effect the impact transference of energy. A faster photon does not transfer more energy. So besides effecting escape velocity and path I see gravity not actually changing the photon structure.

117. I have, by no means, read it all, but, I have chosen just one paragraph from Wentworth’s atmospheric-energy-recycling article to look at, because it immediately caused me problems. Here’s the paragraph I have chosen:

Because the surface of the Earth is assumed to be neither gaining or losing net energy (when averaged over a day or a year), the amount of power absorbed by the surface must lead to an equal amount of energy leaving the surface. The power leaves the surface via a combination of thermal radiation and convective transport of latent heat (water vapor) and sensible heat (hot air).

Notice how he says that the amount of POWER absorbed by the surface must lead to an equal amount of ENERGY leaving the surface. POWER has units “joules per second”. ENERGY has units of “joules”. The problem for me is that he is trying to equate a power quantity to an energy quantity or to equate “joules per second” to “joules”. These are different units of measure.

I could NOT equate “miles per hour” to “miles” could I ?

So, this alone indicates a serious underlying confusion permeating his entire analysis.

118. justgivemeall says:

I always have a problem with energy having to equal energy out, it should be energy in minus work done equals energy out. You cannot measure the exhaust in a car and say that you know how much energy was put in to the engine that did work.

119. CD Marshall says:

It is and always has been Joe’s original root of the error in climate science, heat in equals energy out, using “power” is eluding to heat. They try and justify entropy as heat and entropy is energy, the Universe doesn’t care how that energy is used, as long as it still exists. Yes in the Earth’s energy balance heat is used in work. The second is assuming the troposphere is in thermal equilibrium and using energy once again to justify ‘heat’ forcing.

All they do is put new lipstick on the same GHGE pig and say, ‘SEE ITS NOT THE SAME THING!”

120. CD Marshall says:

121. CD Marshall says:

For clarification, “heat in equals energy out” is correct and not the climate science version which is heat in equals heat out which is why they justify “forcing”.

122. justgivemeall says:

Thanks CD still trying to get my poor old brain around all this terminology
Cheers Barry

123. CD Marshall says:

It can be frustrating Barry but when it “clicks” it’ll get easier, trust me. I had no physics education 2.4 years ago. These guys are great educators.

The fundamentals are the hardest to learn, because the bricks for the rest of the foundation don’t fall in place until yo get the foundation, then the structures made from the foundation seems so much clearer.

Like anything, you look at the final product and you think wow, but learn the blueprint and start at the beginning and wow becomes, “oh that’ why.”

124. From an email:

Hope you are well.
Am just catching up on your “Thermodynamic Approach” article.

You may have already seen this, but just in case,
The Influence of IR Absorption and Backscatter Radiation from CO2 on Air Temperature during Heating in a Simulated Earth/Atmosphere Experiment

The experiment shows that:
1. There is measurable back-radiation / backscatter from the Greenhouse Gases in the atmosphere
2. There is no measurable increase in the temperature created by that back-radiation / backscatter radiation.
The simplest explanation is that radiation fluxes cannot be added – as you have said all along.

The experiment has now been reviewed (?) here: Watts: Review of Seim and Olsen paper and I have a comment under the name Zagzigger (near the beginning).

Already, Watt’s Lukewarmists and others are thinking up fantastical explanations why it can’t possibly be true.
However, all the explanations tend to be more complex than “you can’t add fluxes” – but we’ll say where it goes from here.

Like the GHE effect, there is no experimental proof you can add fluxes together – although you would think that would be easy to show.
Maybe, we should design an experiment for others to perform.

However, I’d be interested in your views.

Best wishes,

125. @Kev & All,

So yes, regarding this graphic…it really does come down to how we solve this problem:

or

In the latter approach we get a temperature gradient…WOW that’s just like the atmosphere! Success! Err….but the temperature gradient in the atmosphere is caused by the adiabatic effect, NOT from radiation. You see the conflict here? Talk about being able to claim evidence or proof of your derivation by analog to the atmosphere…when in fact the thing in the atmosphere is caused by something else. The end result looks the same – temperature gradient with bottom being warmest – but the mechanism is in fact entirely different! So it’s ripe for the installation of simulacrum science/physics here. Already of course they say that the temperature gradient is due to the GHE, and they ignore the fact of the adiabatic effects, etc.

What about thinking in terms of the other analysis: what if all the layers are touching? Well then of course they must all be the same temperature, 91C for 1000 W/m^2.

And so then we split them up into layers – now the bottom is multiple times the temperature of the top?

Like we’ve said, this could be easily demonstrated. It would be easily discovered. And it does not require a “perfectly” engineered scenario – it just needs to work well enough, and you would see it. Fourier and de Saussure experimented on it extensively…it never affected Fourier’s work that Fourier would show that temperature depends on backradiation – this isn’t part of the thermodynamics Fourier developed, and this not modern thermodynamics outside of what climate science has done with it.

Here’s another approach: model the system with an actual partial differential equation (PDE) in time and space, and evolve it computationally. You do run Matlab? I’ve coded this in Matlab.

Because the PDE reduces to a Fourier Transform computation, you need to apply boundary conditions in order to get the computation started. Literally, boundary conditions at the layer surface boundaries. For example, Dirichlet and Neumann boundary conditions, etc.

We actually have names for certain types of boundary conditions: for example, INSULATION is a Neumann boundary condition where the first derivative at the boundary of the domain with respect to space (the surface layer) is set equal to ZERO. In our case the domain is the temperature of the layers as a function of space, where each layer is divided into N “infinitesimals”.

So…isn’t this what multi-glazed glass panes is called? Insulation? If it’s insulative, then we know what boundary condition we need to apply when we compute the solution vs. time. When you do that, you get the solution of equal temperatures. And when you compute this you get a smooth asymptotic rise to equilibrium.

On the other hand the climate alarmists say that backradiation from each layer must me added back in to the layer before it. In this case you must use the Dirichlet boundary condition where you must instead compute the temperature at the surface layer boundaries directly, where the temperature is given as resulting from a sum including backradiation. In this case you get the solution with the gradient, which attains much higher temperature at the bottom then the solar input. In this computation, the layers do not smoothly asymptote to an equilibrium, but their function of temperature vs. time gets “pumped” repeatedly as the higher layers attain temperature. At least I wouldn’t expect a natural phenomenon to exhibit behaviour like this.

I mean this is a classic physics problem, and although people think that physics and science always just has the right answer, it doesn’t. This is what happens. You have some scenario thought up, you have several different ways that you can mathematically solve it. But which is the correct math? The correct math eventually becomes sacrosanct and no-one is told the story of how there were so many different competing solutions. Well…we go to empiricism to find out what math to use. Of course sometimes we have the empirical result first and then the math needs to be correctly solved for the first time, as in the example of Planck and the blackbody, etc.

So what does empiricism say of this solution? What does empiricism tell us of which math, which boundary conditions, to use? If the multi-layers of glass is insulation, then we know the math to use for that. We CALL it insulation, and it literally has its own math for that – a Neumann boundary condition where the slope of the temperature vs. space at the boundary is zero – this is what insulation mathematically MEANS.

On the other hand, we don’t call multi-glazed windows “temperature amplifiers”. Do we? No we call it insulation. We WOULD have called multi-glazed windows “temperature amplifiers”, not “insulation”, if the climate backradiation GHE existed and operated.

The solution to the conundrum, if the existing state of things doesn’t provide a sufficient answer, is then to provide empirical laboratory evidence. All existing empirical data does not demonstrate the existence of the backradiation GHE: NONE!

If you say that the temperature gradient of the atmosphere is empirical evidence…that conflicts with the adiabatic physics…so it’s not valid empirical demonstration here. Just present a laboratory example with controlled conditions, etc.

Here is the PDE computed as a function of time for either boundary condition (unlisted on my channel):

Insulated:

In any case…either of those are still just proposed math models. Which math is correct? In science and physics and engineering we use the math which reality shows us to use. As it stands, reality shows us to use insulation math, not RGHE math. We CALL THIS scenario insulation, we don’t call it “temperature amplification”. No experimental evidence exists suggesting us to use the RGHE math! None, that is, outside of the INCORRECT interpretation that the temperature gradient in the atmosphere is due to the RGHE…as this is actually due to the adiabatic effect, etc.

126. Joseph E Postma says:

I discussed all this about the PDE solution in my first book.

127. justgivemeall says:

Doesn’t this pretty much prove what you have always said that there is radiation but it doesn’t cause warming. I realize the experiment isn’t perfect because of lack of convection but interesting

128. Joseph E Postma says:

You just need radiation and an enclosed space…you just need to see the effect A LITTE BIT, given imperfect conditions.

But they don’t see the effect EVEN A LITTLE BIT!

It would have ALWAYS been working A LITTLE BIT throughout all history. It would be part and parcel of all equations of heat flow.

The heat flow equations would have never been able to look like

Q12 = s*(T1^4 – T2^4)

It could have never happened. You would have always had to include a backradiation term.

129. Look at the plot of the derivative of the bottom layer’s surface temperature as a function of time when the RGHE conditions are computed for the PDE:

This derivative plot shows the rate of change of temperature vs time. Typically, intuitively, we would expect a smooth asymptotic function as heat diffuses through the system. There would be no bumps, just a smooth decrease to zero temperature change. But with the RGHE conditions the rate of increase of temperature gets bumped back up when the higher layers attain a temperature.

So here are two empirical measures that can be performed:

1) the temperature should far exceed the solar input temperature if the RGHE exists (has NEVER been demonstrated!)

2) if the RGHE exists then the rate of change of temperature increase on the bottom layer should not be asymptotic, but demonstrate bumps or oscillations when the higher layer(s) attain a temperature provided by the heat from the lowest layer. You can check this just by taking the element-wise differential of the thermocouple’s readings as a function of time. Say you have the data logger recording temperature every 2 seconds…just take the element-wise differential of that data and see if the resulting plot is smooth and asymptotic, or shows humps superimposed on the asymptotic curve.

130. Could you imagine if scientists and climate scientists (LOL…clowns) actually did real experimentation like this? I have a whole book on experimental methods in physics…Melissinos’ “Experiments in Modern Physics”.

We all note how climate science has no laboratory evidence for its GHE…aside from misinterpretations of the existence of the lapse rate, and claiming that IR absorptivity of a gas “is the GHE” which it of course is not.

131. My thought is in relation to a comment by Willis at WUWT: I have never stood in front of a campfire before dawn, directly in line with the rising sun, but I’m guessing if I did this, then I would sense the radiation of the fire and never feel the radiation of the sun directly in line with that of the fire. Now radiation from the sun might hit AREAS of my body that the radiation from the fire misses, but this is just a GREATER AREA receiving more direct radiation now, NOT an additive effect of fluxes. … A diagram of the geometry might clarify, but I’m not taking the time, at the moment, to illustrate what I mean.

132. CD Marshall says:

So WUUT effectively denies the denial of the ghge. Like every other source they used the tree Ds, yet endorsed the plate experiment. No doubt WUUT is here to sow dissension.

133. CD Marshall says:

three Ds, not trees.

134. Controlled opposition, as I always said.

Also: briarpatching – if you accept the fundamental premises of your enemy’s position, you’ve already lost the war. Stuck in the briar patch. They lose on purpose at wuwt.

135. CD Marshall says:

“The cause of climate change is a subject of theoretical physics, not of ecology. Ecology makes three invalid assumptions: “natural” is good, the climate is fragile, and human influences are bad. Physics makes no such assumptions.

IPCC’s claim “With a very high level of confidence…” reverses the scientific method. No one can prove a theory is true. Science can only prove a theory is wrong.

IPCC and climate alarmists have abandoned the time-tested scientific method. They exclude evidence that proves their theory is wrong. This is “confirmation bias” and it contradicts the scientific method.

In true science, proof that a theory is wrong prevails over all opinions, consensus, and so-called evidence used to support the theory.” Dr. Ed Berry, Climate Physics

WUUT definitely has a “confirmation bias” they already support the ghge as if it were already confirmed.

136. Joseph E Postma says:

Found this in an old file on my computer…a comment I saved a long time ago from someone:

“One must remember that thermal radiation cannot raise the temperature of anything to a level above that of the radiating body. If sunlight alone can raise the temperature of a black surface to 100°C by concentrating the energy with a magnifying glass we can easily raise this temperature to 1000°C, but no matter how much we concentrate this energy we are limited to heating the black surface only to the temperature of the sun’s radiating surface of about 5700 K.
On the other hand if an ice surface is radiating upwards from a surface temperature of 273.15 K (0°C) no matter how large a magnifying glass we used even if we focussed this radiation to a million times its concentration we could still not heat any overhead body to a temperature greater than 273.15 K.
Since back-radiation first emanates from the surface below it cannot drive temperatures to a level above that from which it was derived, and since the surface below must be at least as warm as the body generating the back radiation there can be no heating of the surface below from back radiation from above.
Furthermore there are two possible cases both of which negate any net back radiation.
The first case is that the temperature of the radiating body is exactly the same as the Earth’s surface below. Since the radiation flux is identical for identical radiating temperatures (assuming that the surface of both bodies have the identical radiating capacity) the net result would be back radiation identical to upward radiation from the surface with no net radiation transfer. This case is not physically possible because in order for the body to absorb radiation from below it must be at a lower temperature than the Earth’s surface below otherwise there would be no radiation absorption and therefore no back radiation.
The second case which is what really happens is that the upper radiating surface is at a lower temperature than the ground below and the back radiation is at a lower rate than the upward radiation from the Earth’s surface resulting in net upward radiation or to put it another way; zero net back radiation and zero net back radiation from the GHE is incapable of: “”according to GHE advocates, ALL the liquid water on the entire planet, ALL the oceans, ALL the rain, etc, is created from back-radiation and the GHE. Sunlight of -18C on average just can’t cut it”

Norm K.”

I did a blog post about this once. With a magnifying glass, the best that you can do is perfection and so if you can perfectly focus the sun’s light and return it to the density it has at the surface of the Sun, then you can get 5700K. That’s perfection, and you can’t do better than perfection.

With the stacked panes backradiation scheme, though, there is no limit in theory at all: you just keep stacking panes and you can get any temperature that you want…you can do any number of times better than the input. And so we see here that we could then use a stacked-panes device like this to get as hot as we want it, hotter than the input, then simply throw it back into the sun and make the sun hotter! We could set up these devices outside of a furnace, then once they’re hotter than the furnace we throw them into the furnace to make the furnace hotter still.

Anyway…still awaiting empirical demonstration of the climate RGHE. Amazing, we live in a state of science today where world-wide global political prognostications are made…without a single laboratory proof of the basis of the claims.

137. “With a magnifying glass, the best that you can do is perfection and so if you can perfectly focus the sun’s light and return it to the density it has at the surface of the Sun, then you can get 5700K”
Joseph,
That is really interesting. Does this mean that in theory we can use a telescope focused onto a thermocouple device to directly measure the temperature of the Sun and thereby confirm the S-B equation?

138. Joseph E Postma says:

Well, confirming the S-B Equation and the Planck curve has already been done in controlled laboratory settings. Something that the RGHE has never been able to be done with…

139. CD Marshall says:

Just for the fun of it you should sent you paper rejected by AMS to Dr. Berry and see what he thinks about it? He has worked with them before (I think).

140. CD Marshall says:

This is a great video…

141. CD Marshall says:

It’s short like all of his videos.

142. CD Marshall says:

You know I actually read a paper that said CO2 reverses the ghge at Antarctica. This was published in “a credible scientific journal”.

It made the upper layers warmer than the surface I believe was the point. Which is just inversion and has been well known in meteorology for years. As I discovered many variants of inversions exist.

143. ashemann says:

i read about 20 posts in to the esttom and joe debate, then scrolled down to make this reply/comment and saw there was alot more.
He doesn’t see that the incoming energy isn’t heat when it passes through the lid to the bottom of the box, its only heat when it gets to work on the bottom of the the box. until then its just energy in a vacuum.
And the underside of the iid is absorbing heat from the bottom of the box as that emission from the bottom of the box is creating work in the underside of the lid.
The lid is emitting energy to space from the top matching the energy the bottom is receiving from the S,………………..so the lid is emitting the same energy as it let in…….the energy coming in is only used as heat/work internally and does not exist as heat when outside of the box in the mass-less vacuum no matter which direction it is travelling.

144. CD Marshall says:

So sea levels “rise” is not accurate at all, due to the planet and it’s physics. I can’t seem to find the word I’m looking for though. Anyone? Antarctic water rising will not effect the North Pole. Just not how the physics of the planet operates.Oceans rise and fall all the time with tides, The Coriolis effect shapes ocean levels, the center of mass and the shape of the planet forms the rest. Throw on top of that land mass, drift, tilt, erosion, precipitation, evaporation and what not. Temperature expands and decreases ocean volume.

A hurricane will dramatically change local sea levels.

Where would the main bulk of ocean water increasing go? To the equator?

145. Greg E says:

Just read through the discussion with esttom. It’s very clear he doesn’t understand JP’s point: radiation to space is neither heat nor work, but an unknown form of energy which is not subject to the 1st law of thermo.

146. justgivemeall says:

CD on the BC coast I have seen that our oceans apperantly have risen 6 mm in the last twenty years but oddly enough tide guage in Alaska a few hundred miles away. So I’m not sure I have much faith in ocean measurements and what about natural cycles,once again more questions than hard truth.

147. justgivemeall says:

That should be tide guage in Alaska was 6mm lower

148. CD Marshall says:

Does anyone have a legit explanation for atmopsheric e-folding and its relationship to the hydro-static balance of the atmopsheric layers? I’m getting mixed information in climate science.

149. justgivemeall says:
150. MP says:

@ CD Marshall -:2021/04/21 at 6:19 PM

A study compared 33 year old sand beach coastline sattelite images worldwide with current day.

Red dots is shrinking beaches
Green dots is growing beaches
No dots is almost no growth or shrinking

Most red dots are because of sea sand mining

https://www.nature.com/articles/s41598-018-24630-6

151. CD Marshall says:

Thanks MP.

152. CD Marshall says:

JP this artist left a link on one of your old videos.

153. Steve Titcombe says:

Joe,
It’s been a while since I last commented on your site and, at the time that I did, I was a committed disbeliever in the concept of back radiation. However, I now see where we have been getting it wrong. For far too long I’ve held back and not shared what I now know to be true: Willis Eschenbach’s conclusions to his “steel greenhouse” thought-experiment are actually correct.

For many months I, too, struggled with steel greenhouse thought experiment: How could something that contributed no additional power (the shell), increase the temperature of the sphere? Surely, that would require additional energy – and yet the shell provides no additional energy.
But what is being overlooked, by those that continue to struggle with that thought experiment, is the effect of Time: Specifically, during that period when the sphere continues to radiate it’s own source power (towards the shell) whilst the surrounding shell is not radiating this same source power out into free space. During this time, the sphere and shell have not reached their respective steady state temperatures. During this time, additional energy is being built-up within the system.
To help, I’m going to offer an analogy. I know that many dislike analogies, but this one is good, so please bear with me and let the (necessarily long) story unfold….

A person has been raised to believe that they should give away 10% of their wealth (which is represented by the funds held in their checking account) to an external charity, each month.
On the first day of the first month, when this person starts their first paid job (say at a monthly rate of \$1000), they have zero dollars in their checking account, so they give away nothing to charity. At the end of “month one”, they receive their \$1000 salary paid into their checking account.

At the commencement of “month two”, they now have \$1000 in their checking account and so give away a \$100 check to charity. At the end of “month two”, they again receive their \$1000 salary paid into their checking account.

At the commencement of “month three” they now have \$1900 in their checking account and so give away a \$190 check to charity. At the end of “month three”, they again receive their \$1000 salary paid into their checking account.

At the commencement of “month four” they now have \$2710 in their checking account and so give away a \$271 check to charity. At the end of “month four”, they again receive their \$1000 salary paid into their checking account.

You will see that eventually their checking account will reach \$10,000 and they will give away \$1,000 each month to charity, and they will earn \$1,000 for the next month – the state of financial equilibrium has been reached.

Many years pass and this hardworking and generous person rears a child and this child is also raised with a very charitable nature. However, unlike their hardworking parent, this child (now entering adulthood) does not become an independent worker in their own right but instead, remains wholly dependent upon their generous parent for their entire income. The parent, believing that “charity begins at home”, continues to give 10% of their wealth away but now, all of their charitable giving goes to their dependent child, rather than the external charity.

On the first day of the first month (donation day), when the dependent child leaves home, the dependent child has no wealth in their own checking account and so gives no charitable donation to any external recipient and similarly gives no charitable donation to their hard-working parent either. However, the parent has \$10,000 in their checking account, so the parent now gives away 10% of their wealth (as a \$1,000 check) posted to their dependent child – and, on arrival, this check is paid into the dependant child’s own checking account, so the dependant child’s checking account jumps to \$1,000 dollars, whilst the parent’s checking account drops to \$9,000. However, by the end of the first month, the parent is paid their regular \$1,000 salary so the parent’s own wealth is again restored to \$10,000.

At the commencement of “month two” (donation day), the parent posts another \$1,000 check to the dependent child. For the first time, the dependent child similarly posts a check for 10% of it’s own wealth (the \$1,000 gratefully received last month) to an external charity and also posts a check for 10% of it’s own wealth back to their generous parent (the child, like the shell in the thought experiment, has two directions for giving). So, the child posts a check for \$100 to an external charity and posts a check for \$100 check back to their parent (leaving the child \$800). The two checks sent between the parent and the dependent child always cross in the post. On their arrival, the dependent child will now have \$1,800 whilst the Parent will now have \$9,100 but, by the end of “month two” the parent also receives their regular \$1,000 paycheck, so the parent’s own wealth is restored, but this time to \$10,100.

At the commencement of “month three” (donation day), the parent posts a \$1,010 check to the dependent child. Similarly, the dependent child posts a check for 10% of it’s own wealth (now \$1,800) to an external charity and also posts a check for 10% of it’s own wealth back to their generous parent. So, the child posts a check for \$180 to an external charity and posts a check for \$180 check back to their parent (leaving the child \$1,440). The two checks sent between the parent and the dependent child again cross in the post. On their arrival, the dependent child will now have \$2,450 whilst the Parent will now have \$9,270 but, by the end of “month three” the parent also receives their regular \$1,000 paycheck, so the parent’s own wealth is restored, but this time to \$10,270.

At the commencement of “month four” (donation day), the parent posts a \$1,027 check to the dependent child. Similarly, the dependent child posts a check for 10% of it’s own wealth (now \$2,450) to an external charity and also posts a check for 10% of it’s own wealth back to their generous parent. So, the child posts a check for \$245 to an external charity and posts a check for \$245 check back to their parent (leaving the child \$1,960). The two checks sent between the parent and the dependent child again cross in the post. On their arrival, the dependent child will now have \$2,987 whilst the Parent will now have \$9,488 but, by the end of “month four” the parent also receives their regular \$1,000 paycheck, so the parent’s own wealth is restored, but this time to \$10,488.

You will find that, eventually, the parent’s checking account will grow to reach \$20,000 and that the parent (whilst still earning a salary of only \$1,000 per month) will be required give away \$2,000 each month to the dependent child. The dependant child’s checking account will grow to reach \$10,000 and the dependent child will give a \$1,000 check to their ‘external’ charity and will give a \$1,000 check to their parent – the new state of financial equilibrium has been reached. All the ‘additional’ money in the system is accounted for (it only ever came from the gainful employment of the working parent and yet, by the introduction of a wholly dependent child, the parent has, over time, become wealthier – twice as wealthy in fact – whilst still only earning the same \$1,000 salary each month. Neither the parent nor their dependent child has fraudulently created any fake money. The wealth held in the two checking accounts is entirely genuine – but the (wholly dependent) yet generous child’s “back-giving” has allowed the parent’s own wealth to increase.

A dollar, when given by a poorer person to a richer person, must inevitably make the richer person one dollar richer. However, the richer person will always ‘outgive’ to the poorer person – the net flow of dollars is always from the wealthier parent to the poorer child.

The same is true with the energy conveyed by the photons from a colder object to a warmer object – the radiant energy from these photons will be thermalized by the warmer object (like the dollar from a poor person, each Joule has to be accounted for). However, the warmer object will always ‘outgive’ the amount of energy it gives to colder object. The net flow of energy is always from the hotter surface to the colder surface.

The colder object does not prevent the hotter object from emitting all of it’s radiant exitance at the level prescribed by the S-B law. Similarly, the hotter object does not prevent the colder object from emitting all of it’s radiant exitance (on both it’s surfaces) at the level prescribed by the S-B law.

I’m not sure how well this submission will be received, but I really do hope that it helps those that endeavour to read it through.

Cheers,

154. Joseph E Postma says:

Steve, that argument is well known and is really the standard scheme…so nothing we haven’t seen before. It is based on fermion statistics rather than boson statistics…and of course photons are bosonic not fermionic. Also keep in mind that the steel greenhouse example debunks itself with its solution where 1 = 2…thus we cannot possibly think that simply coming up with an analogous argument gets around the fact of the mathematical analysis where 1 = 2. And finally: NO empirical demonstration exists. To repeat: no empirical demonstration exists.

155. Wow look at that power of sunshine CD!

156. ashemann says:

Steve Titcombe.

This is horse shit, and it is the one sentence in your whole 1000 word post relies on to make your assumptions correct instead of bs like willis’s steel greenhouse bollocks.

prove me wrong, demonstrate the mechanics, show me a real world example of low frequency photons thermalising in a higher frequency environment.

”The same is true with the energy conveyed by the photons from a colder object to a warmer object – the radiant energy from these photons will be thermalized by the warmer object”

157. ashemann says:

Not all photons are equal, only in the fantasy RGHE And Willis’s SG are they equal.

1 Joule of energy from shortwave radiation absorbed and re-emitted as long wave radiation involves tens of millions less photons absorbed as short wave than photons emitted out. as longwave.

158. ashemann says:

The best you can hope for with your back radiation bollocks is this.

Every pulse and flux of photons from either warm or cold object is full spectrum for their relative temperatures.

The warm objects flux contains the higher frequency photons that the cold object does not contain and cannot emit and they are the only photons thermalised in the exchange between both objects.
All other photons being exchanged are simply replacing photons emitted by both objects prior pulse of photons.

Only the photons representing the frequency of the temperature difference are being thermalized and exchanging energy between both hot and cold objects.

159. CD Marshall says:

JP,
Been dealing with a complete idiot over here if you want to pop in and Joe Incognito.

On Simon’s channel, a flat earther calling me a flat earther lol. What a nutjob all he does is spin the same crap.
It’s under my name:

CD Marshall
2 months ago
You meant to say “crucial moment for climate activists” for climate action doesn’t change global temperatures and neither does CO2. I would explain the physics behind that but I see from your comments you wouldn’t understand anyway.”

160. CD Marshall says:

You’d think that the core of Venus would be far warmer than half of Earth’s, unless you factor in that it is not as well insulated and that energy is escaping more to the surface. Over a million surface volcanoes would confirm that.

161. Philip Mulholland says:

CDM
“A car driving at 25 mph does not equal 100 miles an hour after four hours.”
Brilliant, I am so stealing that.

162. That is good!

163. CD Marshall says:

This guy I’m dealing with sound exactly like the PhD Joseph deal with, it’s like things glaze right over his head. Math is more real to him than direct observation. In his mind, a car going at 25mph for 4 hours does equal a car going a 100mph for one hour, 25×4=100, 1×100=100. The math says it’s the same!

Listen to how separated from reality he is…
“Note that the tropics only receive up to about 420 watts/m², enough to support 20°C on average assuming zero absorption in the atmosphere, a pitch-black surface and zero heat export to higher latitudes!
The greenhouse effect is the only thing that can explain why the average temperature in the tropics is about 27-28°C since the second largest energy source on Earth (geothermal heat) only supplies about 0.1 watt/m².”

How removed from reality do you need to be to think that is true?

164. CD Marshall says:

I welcome any critique
Phillip I’ve re-red some of the archived comments I have from you and I am still blown away by water “ice” super heated. Physics is beyond fascinating in how pressure and compression can form water phases in ways we simply can’t predict.

https://physics.stackexchange.com/questions/60170/how-does-the-freezing-temperature-of-water-vary-with-respect-to-pressure

165. CD Marshall says:

Joseph,
This is Simon’s thesis he got a PhD for…
YT summary:

Thesis:

166. CD Marshall says:

Joseph someone on another site asked these questions and I doubt he’ll get the answers on that site but they are great questions:

1) How does the greenhouse effect work in general and why is the additional CO2 enhancing it to exactly the degree the science say (how is this calculated)?
2) How is the average surface temperature of earth calculated from the average heat radiation of that surface (averaging over T is not the same as averaging over power that goes with T^4, so how is the value of the GHE of +33K calculated)?
3) Why is water vapor for sure a feedback to initial rises in CO2-levels and not the other way around?
4) How is it made sure that the additional CO2 since the beginning of the industrialisation is anthropogenic, not biogenic.
5) How good is the quality of the most widely used climate models in reproducing the existent proxy data of the past climate?
6) What proxy data ist there for past climate, how are they evaluated and how precise do they show past temperatures?

167. Philip Mulholland says:

CDM
That Phase Diagram by Martin Chaplin is packed with information.
https://archive.is/eQwox#selection-3237.53-3241.1

168. CD Marshall says:

Eureka! I found ANOTHER flipping physicist! They are drawn to me like stink on a politcal climate scientist. Another one who claims Dr Holmes is incorrect and Heller is nothing more than a con man but Potholer is right.

How dare I claim the ghge does not follow the COE laws! I’m such an arrogant bastard. LOL. I love these guys, free entertainment.
So his next reply (I can almost guarantee you) is going to be the classic, “you just don’t understand the physics.” Always my favorite.

169. CD Marshall says:

Okay this physicist grilled me pretty good. I felt like I was taking an exam. He’s “different” from most of these guys. I think you two could have a conversation and he wouldn’t be condescending (maybe?). He did not like Holmes because Dr. Homes (according to him) didn’t answer his question. We all know how that goes and it usually goes, “he didn’t answer the question how I wanted to hear it.”

Physicist:
“That‘s all there is and that was first discovered 1824 by Joseph Fourier and never been falsified ever since. And now comes Holmes and does all that without any energy at all just by gravitational pull and the ideal gas law (which he desperately ignores to be an identity, not a causality, which is mathematical nonsense), claiming the temperature difference could show up without the atmosphere radiating energy.

HOW exactly then he secures the energy flows to be in place that are necessary to hold coe and 2nd law like I described above: He was asked multiple times, he never answered. He just referred to his own calculation (which is wrong just by simple logic, because he set a causality where there mathematically is none) and says: Look what I have done! Why don‘t you see it? Well, yeah, I DO see it and I‘m telling you the obvious mistakes… I mean, come on, how can one follow such baloney if not because one wants to BELIEVE it must be something? Believe is free to choose, but it is never scientific.”

170. CD Marshall says:

Of course he hasn’t replied back to me on this yet,
Physicist:
“This extremely fundamental fact forces the GHE to be in place for any planetary atmosphere that contains gases which can absorb/emit thermal radiation at frequencies matching the planetary temperatures. They separate the import point of solar energy (at the surface) spatially from parts of the planetary energy export to space (from the atmosphere). The second law of thermodynamics claims any planet to go to the steady state power(in) = power(out), the law of energy conservation tells us the quantity of those flows. If those flows are accessing/leaving from different locations, the space between those locations must show a gradient in temperature with one warmer end and one cooler end.”
Me:
This entire argument can be presented without the ghge and still be true.

Physicist:
“That is GHE: Being the surface near layer of atmosphere the warmer end.“

Me:
Disagree. Warmer at the surface is due to gravity. Closer to the center of mass the warmer it gets (compression). Jupiter, Saturn, Neptune and Uranus all have hot cores no greenhouse gases needed. It gets warmer closer to the surface due to the adiabatic effect which the Kinetic Theory of gases does obey the COE laws. The surface is part of the troposphere, connected at the bottom and as an average is typically found at the middle for a temperature gradient the hottest part is typically the lowest. The Sun is heating the surface, thus the origin of the thermal heat energy conversion.

171. Guys check out the total psychosis and inversion of reality going on here:

https://andthentheresphysics.wordpress.com/2021/04/25/mind-your-units/

And I don’t know who “Dr Roys Emergency Moderation Team” is but they’re facing the same psycho here it looks like:

https://www.drroyspencer.com/2021/04/uah-global-temperature-update-for-march-2021-0-01-deg-c/#comment-664941

172. Joseph Postma says:

@ 1) How does the greenhouse effect work in general and why is the additional CO2 enhancing it to exactly the degree the science say (how is this calculated)?

Well we have the flat Earth diagrams which demonstrate the idea. Although to pin them on how it works is tricky: they start with backradiation, but then when you point out that heat cannot recycle, they switch to “slowed cooling/reduced emission” and what have you, etc. There is no greenhouse effect at all, other than in real greenhouses which stop convection – the open atmosphere can’t do that.

@ 2) How is the average surface temperature of earth calculated from the average heat radiation of that surface (averaging over T is not the same as averaging over power that goes with T^4, so how is the value of the GHE of +33K calculated)?

33K is the difference between the effective radiant temperature of the entire planet, calculated via conservation of energy between emission and input, and the physical near-surface air temperature. The effective radiant temperature is found at the average of the atmosphere, ~5-6km altitude. They claim that the difference between the average temperature and the warmest near-surface air temperature is due to a radiative greenhouse effect, but it is in fact due to the adiabatic lapse rate.

@ 3) Why is water vapor for sure a feedback to initial rises in CO2-levels and not the other way around?

It’s not for sure…it’s just made up. High concentration of water vapour actually keeps temperatures lower, not higher.

@ 4) How is it made sure that the additional CO2 since the beginning of the industrialisation is anthropogenic, not biogenic.

It is only assumed.

@ 5) How good is the quality of the most widely used climate models in reproducing the existent proxy data of the past climate?

There are so many tuneable parameters that the models can be made to fit anything. The climate data is now being manipulated as well.

@ 6) What proxy data is there for past climate, how are they evaluated and how precise do they show past temperatures?

There are many across several disciplines – geology, biology, paleontology, etc. The problem is that you cannot measure to higher precision than the internal variation over a time period…as it is, past variation is equivalent to modern variation, and in fact modern variation is mild compared to past variation.

173. CD Marshall says:

Chris (the physicist) showed me how to cross communicate posts, so I gave you an invite to our conservation. NO more struggling to find comments anymore.

All you have to do is comment with @username and it sends a ping that should notify the user. YT might be blocking some of your replies though so hope it works?

174. CD Marshall says:

I’m not entirely convinced he’s no a bot or uses bots 🙂 or has a bot give automated replies to some things. Actually you could use a few bots couldn’t you?

175. MP says:

@ Joseph

I posted this rhetorical question on andthentheresphysics.wordpress.com

It was deleted within a minute lol. Cognitive dissonance of climate change brainwashed from childhood people is strong.

176. Willard says:

Good morning, MP.

Just woke up. The first comment on a WP blog usually goes in the pending bin. I released it.

If you keep arguing by rhetorical questions, your stay may not last long.

Welcome to Climateball!

177. But you can’t answer the question, can you Willard… Because the answer debunks the basis of climate science as a field: you can’t cook something at 1/4 the power for 4 times as long.

178. Willard says:

You’re not a BBQ fan, Joe, are you:

What’s 365.5 divided by 2, again?

179. Slow cooked BBQ is still quite hot, Willard.

Don’t avoid the question Squirt…you’re a big boy now, you can directly answer questions here with us, can’t you Squirt?

Can -18C melt ice? Can -18C over a flat Earth do the same thing as +121C on a hemisphere? Are these the same thing, Squirt?

180. Willard says:

What’s 365.5 divided by 2, Joe?

181. CD Marshall says:

Joseph I tried ‘pinging’ you on YT did it work?

182. @CD: Yes, to a single short comment of yours.

183. Squirt, you can use your computer calculator to get the answer: 365.5 / 2 = 182.75.

I see that you struggle with really basic levels of communication, and math, Squirt.

184. CD Marshall says:

This physicist is well, read these comment…
“Energy absorbed is energy absorbed and if absorbed, it will effect the internal energy and thus the temperature of the absorber. Completely regardless of the wavelength of the absorbed photon. There is no theory in physics that states something else, so I wonder where this mistake comes from.”

Just wow.

185. Willard says:

Why use a calculator when one can use a graph of yours, Joe?

186. CD Marshall says:

“Physics demand a rise in temperature due to the additionally absorbed radiation.”
I’m having doubts of his validity as a physicist…well at least a functioning one anyway.

187. CD Marshall says:

I always have such high hopes then they all let me down. Are you the only sane physicist left on planet Earth?

188. Squirt, I see that you are continuing to have a very difficult time with basic communication. Are you able to make any points, or are you going to continue with rhetorical questions here after you admonished MP for the same, Squirt?

189. Yes, I am CD…haha 🙂

But really many seem to be insane. Did I tell you that I had a PhD in Condensed Matter Physics tell me that they were unqualified to vet the difference between flat Earth cold sunshine vs. round Earth hot sunshine? lol!

190. If you keep arguing by rhetorical questions, your stay may not last long, Squirt.

Welcome to ClimateofSophistry!

191. ” There is no theory in physics that states something else, so I wonder where this mistake comes from.”

hahahahha…they’ve never heard of the photoelectric effect!!!!!!???????hahahha

They’ve never heard of the heat flow equation Q = s*(Th^4 – Tc^4)? LOL!

192. ““Energy absorbed is energy absorbed and if absorbed,”

IF

Oh and guess what, there’s an equation that answers “IF”:

Q = s*(Th^4 – Tc^4)

193. CD Marshall says:

If you asked a Solar physicist, he’d ay he’s only qualified in discussing the Sun not the effects it has on Earth. I sense a pattern here of, “I’m not going against the climate religion, we know the punishment is career death.”

194. CD Marshall says:

The sad part is he mentioned Fourier

195. You would never look at this and think that there was a problem:

What do equal temperatures do to each other? Not a damned thing, of course!

But the presence of arrows really seems to mess with the mind!:

That’s the exact same thing as above, but now people want to change it to:

Don’t think I have a graphic that changes those fluxes to temperatures…but anyone can solve them.

196. Willard says:

Joe,

I asked rhetorical questions to make a point.

Speaking of communication, you still fail to own your miscalculation. You are still holding my comments in the pending bin. You are not calling me by my name. And you are still hiding behind your flying monkey.

Funny that he thought of a chicken:

> But the same surface

Have I ever told you the Kentucky-Kiev Paradox, Clint?

If you give me one chicken, I will bring you two.

https://www.drroyspencer.com/2021/04/uah-global-temperature-update-for-march-2021-0-01-deg-c/#comment-664189

Do you happen to know Clint, by any chance?

To answer your and his silly question, a planet can receive light and still be really cold.

You know where to find me.

197. “I asked rhetorical questions to make a point.”

So did MP you absolute psycho…

“You are not calling me by my name.”

As you do to others, Squirt. Are you so psychotic that you demand to be able to do to others what they cannot do in return? You are psychotic with your double standards.

I think we have a mental-health issue here, guys…lol.

Thanks for letting us know that you’re also esttom…that explains so much!! You are a psychotic troll.

I gave you the invite to tell us whatever points you wanted to make, I’m letting your comments through…but then you run because you of course have nothing to say. Come on Squirt, where are you going with your divide by 2 question? Tell us what you mean with it, and the point.

Gosh it is SO invigorating and rewarding to know that the only people who defend climate change science are psychotic trolls and incompetent/fake scientists who can’t vet basic questions about empirical reality.

Gosh I’ve really and truly WON this one for humanity! 🙂

198. Squirt: “To answer your and his silly question, a planet can receive light and still be really cold.”

No shit Sherlock. OMFG psycho.

199. Willard says:

So did MP you absolute psycho…

As predicted:

I should have clarified that I asked rhetorical questions to make a point about rhetorical questions, and predict that Joe won’t get it.

If you ever cook a chicken, Joe, please make sure not to let it rest on the counter before putting it in the oven. Also, if your chicken is twice the size of the one in your recipe, chances are that you’ll need to cook it for a longer time, not half the time as your current calculations might lead you to infer. Always better to check with a thermometer.

200. Willard says:

As you do to others

That tells me you’re a sock puppet, Joe.

Thank you.

See you at Roy’s.

201. You asked rhetorical questions after admonishing people about rhetorical questions to make the points that you’re a psycho – got it, Squirt!

MP’s rhetorical question makes a point…yours aren’t making any point and don’t seem to be leading anywhere. There’s a difference, Squirt.

“If you ever cook a chicken, Joe, please make sure not to let it rest on the counter before putting it in the oven.”

But resting it on the counter should fully cook it, given enough time, according to climate science flat Earth physics.

“Also, if your chicken is twice the size of the one in your recipe, chances are that you’ll need to cook it for a longer time, not half the time as your current calculations might lead you to infer.”

You are a complete psychotic. You’ve had a psychotic break at some point in your life and now you spend time on the internet. Great.

Why make the chicken twice the size? The Earth is fixed in size. Climate science makes the surface area that sunlight falls upon twice the size with quarter the power.

I don’t to anything for “half the time”, I demonstrate it as it actually empirically physically exists…unlike the flat Earth theory of climate science.

Your attempts to INVERT what I demonstrate, and you similar INVERSIONS of your very own behaviour and interaction with people, indicate complete and total psychosis, Squirt.

202. Willard says:

It’s so nice to see you, Clint.

203. YOU LINKED that comment thread, you absolute psycho. I don’t know who “Roy’s Moderation Team” is, but they do a damned good job. Roy can enjoy his site with his psychotic followers for all I care.

204. Life must be scary for you, Squirt. I’m not “Clint” either.

Rent free.

205. Willard says:

I’m not “Clint” either.

It’s possible. Just as it’s possible you read a forest of comments where I used “kiddo” for a guy you do not know. It’s also POSSIBLE that you’re using caps lock just like Clint for the fun of it. That your prosody matches his could be pure coincidence.

Everything is possible, including doubling the flux by doubling the area it covers.

Stay safe.

206. “Everything is possible, including doubling the flux by doubling the area it covers.”

That is the most psychotic inversion of what I demonstrate which I have ever come across, Squirt.

Climate science dilutes the power of sunshine by spreading it over four times the surface area which intercepts it.

No one, ever anywhere, has doubled the flux by doubling the surface area.

207. “It’s possible. Just as it’s possible you read a forest of comments where I used “kiddo” for a guy you do not know. It’s also POSSIBLE that you’re using caps lock just like Clint for the fun of it. That your prosody matches his could be pure coincidence.”

You’re really afraid that there are others out there like me, who know what I know, and who can demonstrate it independently as well I can, aren’t you Squirt?

You are in a psychotic paranoid state of fear, Squirt. You must not feel safe. Sad.

208. Nice to have you here “Dr Roys Emergency Moderation Team”! You do excellent work!

209. Thanks! There are lots of moderation emergencies at Dr Spencer’s which require attention but this Willard character has been really making a nuisance of himself lately. He seems to think that you are doubling the flux by doubling the area it covers and can’t seem to understand that in fact you are halving the surface area (spreading the sunlight over a hemisphere as opposed to over the whole sphere). Did you notice ATTP’s calculation? Willard linked to it with the word “Enjoy” at 10:10 am.

And yes…this Willard seems like they might actually have a mental health issue going on…perhaps should go easier on them. Nahhh.

Yes I mentioned that flux thing in a response to Willard above – that’s the craziest reinterpretation and inversion of what I/we present which I have ever witnessed.

Yes I did see the calculation…not sure what the point of it is as it just shows that the intercepted energy is that from a disk cross-section of the sphere, and it also shows the area of hemisphere. Everything in it demonstrates my model and the point of my model…so not sure what they’re trying to imply in that comment. Often times, at the very end I’ve noticed, they’ll present something truthful and they’ll even present what you yourself present, but pretend that it is theirs and that it doesn’t show what it factually actually directly shows – it is an insane level of sophistry when they do that! They’ll present your argument as if it is theirs, but then ignore what the result of the argument is! It’s just amazing, really. It’s amazing to see it go that far.

211. None of that OP at ATTP makes the slightest bit of sense or reason. Is that site is this Willard character’s?

A flat plane is zero dimensions? Who knew!? hahaha 🙂 Idiotic. It must be this Willard’s…

212. Dr Roys Emergency Moderation Team says:

No, it’s ATTP’s site, but Willard is a regular commenter there who makes occasional “guest posts” I think. That was one of Willard’s posts, you are right. Although apparently ATTP has endorsed at least part of it as it says that Willard contacted him through personal email to check some details before it was posted. I’m not sure whether to believe any of it is genuine confusion or deliberate distortion. Who can tell these days? I have spent a considerable amount of time trying to explain the “divide by 2” principle to Willard but he just always comes back to “you are doubling the flux by doubling the area it covers, which is impossible”. I point out that the area is in fact being halved, and it just seems to be ignored.

213. Oh I see it was Willard’s OP…ok.

Yes Willard’s comments nor his OP make any sense at all.

214. Steven says:

Here from ATTP. The energy balance argument isn’t arbitrarily dividing the solar flux by 4. It starts from the fact that power in = power out for the earth, then proceeds by simple math.

Power in is of course Pin=S(1-a)pi*R^2, where S is the solar constant, a is albedo, R the radius of earth.

Power out is easy to calculate if there are no greenhouse gases. Since there is nothing to absorb outgoing IR, power out is just the integral of Stefan Boltzmann flux. Pout=sigma epsilon integral T^4 dA (using constant emissivity epsilon for simplicity). This is equal to Pout=sigmaepsilonA * (the average value of T^4). A is the earth’s surface area, 4 pi R^2 . So Pout=sigmaepsilon4piR^2*(the average value of T^4).

Equating Pin=Pout gives (the average value of T^4)=S*(1-a)/4 . This is where the factor of 4 comes from.

Of course in reality (the average value of T^4) is much greater than predicted here. The reason for the discrepancy is of course in the calculation of Pout. In reality the absorption and emission by greenhouse gases changes this calculation.

215. Kev-In-ZA says:

@Joe, Thanks for some of the detailed answers to my questions (now way back up in the comments). Will chew on for key understanding.

And I see things have moved in a few strange directions. The Willard (Enjoy) link on ATTP is just bizarre. He seems to think that people with advanced degrees in physics and engineering have failed to understand issues of basic geometry, when it is their own stupidity that is plainly on display that if something is a function of a 4th power (and also includes temporal dynamic), you cannot work with simple averages and expect to reflect reality. What a dumb-ass.

216. Well said Kev.

217. @Steven – Yes, we understand that the ratio of area between a disk and a sphere with the same radius is 1/4. We understand that power in equals power out. And in fact, “power in equals power out” is true regardless of the composition of the body in question, and so your remark that the composition changes Pout is incorrect.

And yes, we agree with T^4 = S(1-a)/4, and for the Earth this works out to -18C, which is in fact the measured and known empirical result for the effective temperature of the Earth. There is NO discrepancy between T^4 = S(1-a)/4 and the measured effective temperature of the Earth…they are in perfect agreement.

Of course, given that T^4 = S*(1-a)/4 is an effective average temperature, then we would never expect this temperature to be physically located at the slice of atmosphere closest to the surface, since this location does not represent the physical average location of the atmosphere. The physical temperature of -18C is found at the average of the atmosphere, which is found physically at 5-6km altitude. Then, when you apply the adiabatic lapse rate, you determine the average temperature of the slice of air nearest the surface, which is +15C. The near-surface air would be warmer than the expected average regardless of the composition of the atmosphere.

What we’re interested in here though is physics. We wish to understand what creates and drives the climate. Climate science believes that the Sun has nothing to do with creating the climate because it uses the T^4 = S(1-a)/4 as an input, where of course S has been divided by 4. This is not the input but is only the output, and while the input and output total energies are equal their fluxes are not, and it is flux that drives the system, not merely total energy. We must characterize the flux correctly in order to understand what energy is capable of doing. S(1-a)/4 does not correctly characterize the input flux – real time, average, or otherwise – it is only the output, and is not equal in flux to the input.

218. Steven says:

My argument applies at the surface because that is where all outgoing radiation would originate from in the absence of greenhouse gases. The average of T^4 at the surface would be S(1-a)/4 . Not 5 or 6 km above.

219. MP says:

Posted this post on ATTP. It was posted directly without moderation waiting, but deleted in 10 seconds. Lol

Could get the catch back for a screenshot by returning browsser page

220. @Steven – The lapse rate applies only against the thermal capacity of the air and the strength of gravity, independent of the air’s radiative properties. It is thus impossible for the the average or effective temperature to be physically found near the surface.

In addition, if it is GHG’s which cause or allow the atmosphere to emit, then GHG’s provide a vector by which the atmosphere to directly cool; they would increase atmospheric emissivity and thus cool the atmosphere. If the atmosphere cannot emit without the presence of GHG’s, an atmosphere with no emissive power would become very, very hot indeed; but of course, anything will generally emit, and with an atmosphere the emission is not restricted to the surface only. The atmosphere is part and parcel of the thermal system of Earth and its presence naturally moves the radiative surface from the ground surface and to altitude in the atmosphere.

If GHG’s increase the emission of the atmospheric body, then they cool that body.

Again – we’re interested in physics here, and what drives the climate. T^4 = S*(1-a)/4 cannot be used as an input…because it is NOT the input.

221. Joseph E Postma says:

Nice capture MP!

222. MP says:

Now it is posted again on ATTP, they decided to argument it.

223. Joseph E Postma says:

I wouldn’t even bother with this insane, psychotic Willard kid.

Look at the comment you’re responding to: Squirt (Willard) is claiming that spreading the solar input only over a hemisphere is equivalent to doubling the surface area which solar radiation impinges upon? In fact…one can barely even read Squirt’s comment:

Squirt: “So to divide by 2 the way you do is in effect doubling the surface of the Earth’s surface over which are spread incoming solar variation.”

Huh? WTF…

And Squirt says this in the midst of dividing by 4 and spreading solar radiation over area it never spreads upon! lol

224. Willard says:

You are more than welcome to come tell that at AT’s, Joe.

Sooner or later you’ll get short of flying monkeys.

225. It is quite astonishing to read nonsense like this, It is almost as if there is no understanding at all of what division actually is. I suspect that part of the problem is that the divide by four trick is hidden in plain sight. They never seem to notice that divide by four is applied before the energy has even entered the climate system.

226. Squirt you haven’t provided a single argument, explanation, or position on anything here yet clearly. We’re waiting.

227. Dr Roys Emergency Moderation Team says:

Yeah, Willard is either dishonest or delusional. Either way there is no point continuing to discuss it with him. I could barely get comments posted at ATTP, so I called it a day there. Over at Dr Spencer’s he seems to believe ATTP’s “Enjoy” calculations refute your point, even though it kind of seems like he still doesn’t understand that point (if we go with him being “delusional”).

228. Yes Philip and if you look at the inversion now, it is disturbingly insane: spreading sunshine over a hemisphere is now doubling the flux of sunshine and doubling the area it spreads upon 🤣

229. Steven says:

My argument shows the average value of T^4 at the surface. The lapse rate or the temperature at 5 km up (which you call the “effective temperature”) aren’t relevant if all outgoing radiation comes from the surface, as it would without GHGs. The incoming power is equal to the outgoing power, and the outgoing power is entirely a function of the temperature at the surface. So the result is (the average value of T^4 over surface)=S*(1-a)/4 .

Again, this mathematical result would be exactly true without greenhouse gases. The fact that the actual average surface T^4 is higher can only be explained by the presence of GHGs.

230. “My argument shows the average value of T^4 at the surface.”

No…it gives the effective radiative temperature of the Earth…which is what is confirmed by measurement. You have no idea what it even means.

“The lapse rate or the temperature at 5 km up (which you call the “effective temperature”) aren’t relevant if all outgoing radiation comes from the surface, as it would without GHGs.”

The atmosphere would still emit without GHG’s because anything emits…there is no such thing as emissivity = 0. If GHG emit, if they allow the atmosphere to directly emit MORE energy, then they cool the atmosphere. You people have no idea what you’re doing.

“The incoming power is equal to the outgoing power, and the outgoing power is entirely a function of the temperature at the surface.”

No…the outgoing power is a function of the entire system as a whole, because it is a single thermal system. The atmosphere must be included as part of the emission.

“So the result is (the average value of T^4 over surface)=S*(1-a)/4 .”

This is called the effective temperature, which is the temperature of a blackbody emitting the necessary energy. You said “which you call the “effective temperature””, indicating that you have no familiarity with the term whatsoever. It is called that because that’s what it is, because this is a defined term in basic undergrad physics. You people have no idea what you’re doing, or talking about. You literally have no clue about the most basic concepts in physics…climate science is truly a selective simulacra of physics.

“Again, this mathematical result would be exactly true without greenhouse gases.”

This result is precisely what is measured from space, which is how you would confirm this result. The mathematical calculation for the effective temperature of Earth is precisely what is empirically measured.

“The fact that the actual average surface T^4 is higher can only be explained by the presence of GHGs.”

Given that you have no idea what you’re doing or talking about, aren’t even familiar with the term “effective temperature”, I am pretty sure that we can safely assume that your conclusions are bonkers…lol.

The lapse rate necessitates a distribution in temperature in the atmosphere. Given that an average can not be the extremities of a distribution, but will be found around the middle of the distribution, then it is impossible that the average of the distribution would be found at the hottest end of the distribution. The lapse rate absolutely establishes that the near surface air must be warmer than the average air temperature.

And finally, solar heating is not T^4 = S(1-a)/4 or -18C, but is in fact much, much hotter than this. T^4 = S(1-a)/4 is not the temperature that the planet is heat to…it is heated to much higher temperatures than this. T^4 = S*(1-a)/4 explains NOTHING of the climate or physics that occurs…it is simply an output, or more precisely, it is the effective temperature of the Earth.

231. MP says:

ATTP banned my last Dr Robert Holmes derived insight post lol

It is changed in [Enough peddling, MP. – W]

232. Squirt, I JUST explained to the fine and respectable folks here how the absolute state of psychosis of climate science greenhouse effect believers is to now recite what I myself present, while not acknowledging what the explanations explain…lol. We all here are all about the effective temperature of the Earth and what it means.

And we can add on top of that another level of psychosis:

Squirt: “I’m not talking about a virtual blackbody in space.”

Squirt, next sentence: “…what we’re determining is the effective radiative temperature of the Earth. This is the temperature of a blackbody that would radiate…”

TLDR: “I’m not talking about a virtual blackbody. I am talking about a virtual blackbody.”

You’re psychotic. No one speaks the way that you do. No one says things like this, and contradicts themselves from one sentence to the next. No one says that half the area is twice the area; no one says that flux should be doubled. No one does that. Except for you, Squirt…and the other climate greenhouse freaks. You people are wrecked. You’re wrecked. Something wrecked you. I would not like to know the damage you have wreaked upon people throughout your life with the absolute state of psychosis you continue to demonstrate here.

Squirt: “I’m not talking about a virtual blackbody. What I am talking about is a virtual blackbody.”

233. They’re coming out in full force it looks like with the psychotic ownership inversion deflection (OID):

“Entropic Man says:

There’s an unspoken difference in approach here.

DREMT and his fellows at Roy Spencer are talking about measuring the instantaneous surface energy input of the dayside. Hence his division by 2. This is fine if you want to know the flux at 56N 7W at 4pm local time. It is useful for weather forecasting, which plugs the local flux into short term planetary model grid squares hour by hour.

The rest of us are thinking about climate and are more interested in the annual average flux and the planetary energy budget.. Hence our division by 4 to cover the whole planetary surface; and averaging of daily, seasonal and longer term variations.”

And that’s what I just said above…being interested in the physics of the climate and what creates it and sustains it. The “annual average flux” doesn’t explain how the climate is created the Sun, if your interest was actually “thinking about climate”. lol You see what they did there?

If you were actually thinking about the climate, you would want to know how it is created and sustained, and you can’t do that with the average solar flux of 1/4 power what it actually is.

Do you guys see the level of psychosis that they’ve devolved to, just to try to defeat me?? hahahaha

234. “The rest of us are thinking about climate and are more interested in the annual average flux and the planetary energy budget. Hence our division by 4 to cover the whole planetary surface; and averaging of daily, seasonal and longer term variations”

Yes because averaging out the climate to a uniform state really tells us about the climate…lol!

Psycho. Whatever this “ATTP” and this “Willard” and the other commenters there are all about…it clearly reduces to one thing: total and absolute psychosis.

Yes…oh you’re so interested in the climate…hence we should average out the climate to a completely uniform state because this tells us the most about the climate! lol!!!

You guys enjoying the comedy or what!?

235. Steven says:

Greenhouse gases are those with any emissivity in the longwave. Here I am considering an ideal atmosphere with no GHGs, not even weak ones, i.e. emissivity zero. Thus all emission occurs from the surface.

In this case, the average of T^4 taken across the surface is S*(1-a)/4=(255 K)^4. This is mathematical fact, as proven in my first post.

When the atmosphere emits and absorbs radiation (with GHGs), this is no longer true, and the surface average T^4 can be different. In your way of looking at things, the height at which average T^4=S*(1-a)/4 moves up. Then the surface can be hotter.

It is only the inclusion of GHGs that allows the surface average T^4 to differ from (255 K)^4. Since the actual surface average is higher, it is clear GHGs warm the surface, not cool it.

236. “DREMT and his fellows at Roy Spencer are talking about measuring the instantaneous surface energy input of the dayside. Hence his division by 2.”

Is Roy actually doing that? Wow…he must have learned something from me finally!

“This is fine if you want to know the flux at 56N 7W at 4pm local time”

Cute…but it’s actually useful for understanding how a climate is created at all, and in what manner it will be created, and what features it will contain, etc. Which is of course NOT what the annually-averaged climate could ever tell you!

237. “Here I am considering an ideal atmosphere with no GHGs, not even weak ones, i.e. emissivity zero. Thus all emission occurs from the surface.”

In this case the atmosphere would get hot as heck, because it can’t emit thermal energy. This might actually make the surface much hotter too. And still, the adiabatic lapse rate would still necessitate that the near-surface air be the hottest air, and the average would be at altitude.

If you add emission to the atmosphere, it will cool it.

“In this case, the average of T^4 taken across the surface is S*(1-a)/4=(255 K)^4”

In reality there is no such thing as zero emissivity. Take a gas giant like Jupiter or Saturn with no GHG’s: they get hotter the further down you go into their atmospheres too, and yet their effective temperatures are still precisely what is calculated. So no, there’s no necessity that -18C would be found at the surface when an atmosphere is present, irrespective of presence of GHG’s as demonstrated by Jupiter, Saturn, etc.

“the height at which average T^4=S*(1-a)/4 moves up”

That’s not what the GHE is…but I see you’re employing the old “use different definitions of the mechanism” trick.

If GHG’s allow the atmosphere to emit, then they cool the atmosphere. I mean at this point you freaks just state opposite world clown things…so whatever…lol.

238. MP says:

@ Steven says 2021/04/26 at 7:09 PM

Banning righteous post on your hubs and ignoring arguments here, while repeating same old circle jerking of debunked arguments?

239. These people are the epitome of psychotic clowns…lol!

240. Steven: “Greenhouse gases are those with any emissivity in the longwave.”

F = e*sT^4

T^4 = F/(es)

If e = 0, an atmosphere without so-called “greenhouse gases” (there is no such thing as this phrase, “greenhouse gases”…it is a meaningless expression), then T goes to infinity due to the divide by e = 0.

Clearly, low emissivity means higher temperature.

But now make e > 0, or, make e2 > e1, and the temperature is lower, i.e. gas 2 with higher emissivity e2 than gas 1 with emissivity e1 results in T2 < T1.

A gas with higher emissivity has a lower temperature, due to the inverse relationship between emissivity and temperature.

241. Steven says:

“In reality there is no such thing as zero emissivity.”

The point is: we know from first principles that the zero emissivity case would have surface average T^4 colder than what is observed in reality.

[JP: No, this is not first principles. It is only first principles for a fiction you’ve created out of the non-reality of zero emissivity. Of course you also consider it to be “first principles” when you average the solar input over the entire surface of the Earth as an input to the climate…but this only a fiction as well. There are approximations and idealizations, and then there are outright fictions. It takes intelligence to know the difference I guess.]

In fact it would be exactly (255 K)^4, when in reality it is higher.

[JP: In fact it couldn’t be exactly that physical temperature because the system is a sphere, not a uniform plane, and the averages over the nonlinear quantities do not work out the same way.

However, it IS exactly that effective radiative temperature.]

Thus increasing emissivity from zero (which is, by definition, adding greenhouse gases) increases temperatures.

[JP: See: https://climateofsophistry.com/2021/04/12/how-to-solve-the-dual-pane-parallel-plate-problem/#comment-71823

Increasing emissivity causes cooling. Basic laws of physics here.]

“Take a gas giant like Jupiter or Saturn with no GHG’s”

Jupiter and Saturn have GHGs.

[JP: https://en.wikipedia.org/wiki/Jupiter#Composition

TRACE amount of methane. That’s not what creates the temperature increase with depth on Jupiter…again it is the adiabatic gradient which does it.]

242. Little Squirt here seems to really want us, or me, to click the links to get that other site OP, and also to comment there. Given their psychotic behaviour here, I would highly recommend NOT visiting that site!

243. CD Marshall says:

“Could you please lead a short further explanation to Q = s*(Th^4 – Tc^4)? What are the parameters and what is the context to the question discussed?”

Joseph could you please get your “A beautiful mind” over here and talk to this guy. LOL. I can’t read him, he can relate better to your level not mine.

Undergrad level of physics isn’t going to convince him of anything other than I need more education. (Which is actually true so yeah…)

The ping was for him btw. With over 18k comments I doubt you’ll find me w/o the ping.

244. I replied with this:

“Could you please lead a short further explanation to Q = s*(Th^4 – Tc^4)? What are the parameters and what is the context to the question discussed?”

This is the basic heat flow equation for a plane-parallel system. It demonstrates that heat flows only from from a body with a hot temperature (Th) to a cool one (Tc). It incorporates the Stefan-Boltzmann Law to provide the total thermal radiant emission from each body, and it is the difference in this emission between two bodies which determines which body’s energy can act as heat, and thus raise temperature, for the other body. It is typically used in conjunction with the First Law of Thermodynamics dU = Q + W, which says that a change in a body’s internal thermal energy (dU) is equal to the heat flow and work performed relevant to the body.

The equation for heat flow is important because it demonstrates certain impossibilities with regard to the approach of climate science to understanding the climate. For example, climate science has this idea of a “greenhouse effect” where radiation from the cooler atmosphere will cause the warmer surface to warm up some more, thus implying heat flow from the cold atmosphere to the warm surface; this is actually impossible given the heat flow equation and thus given the First Law of Thermodynamics, and it is also impossible via the 2nd Law of Thermodynamics too, if you get into that.

Energy density, or flux, or energy flux density, in the context of radiation and heat, is given by the Stefan-Boltzmann Law, and the units are Joules per second per square meter, or Watts per square meter, W/m^2. It is a cross-sectional density, a rate-density of flow through a surface area. So it is not density in the material sense of cubic space; it is the rate of energy per unit time per unit area. It is proportional to temperature via the S-B Law: F = sigma*T^4 (W/m^2).

245. CD Marshall says:

Thank you he does know his physics (right or wrong is the question?) a lot more than I do to influence him. I find him hard to read, I’m sure you don’t, seeing how you have a higher level of understanding these guys.

246. angech says:

Will try.
Are we talking about a recognised amount of solar input?
Everyone seems to agree on the output/blackbody /effective temperature.
And the fact that it has to equal the incoming solar input.
[sort of].

Both sides bright scientific backgrounds, surely?

An error of two claimed by both sides seems on the surface impossible.

Semantics? Misinformation?

Or poor presentation?

Looking at Postma’s side the maths seem to indicate a doubling of the energy per unit area along with a halving in size. ie dividing by 4.

Looking at ATTP it is 4 times the area for the incoming energy dividing by 4.

ATTP works out an average temperature, whole planet.
Postma works out a different average for half the planet.

Both claim justification for different concepts of heating the surface.

I think that both sides should acknowledge that the heat in equals the heat out.
That the difference in their models equates simply to the difference between local and general maths.
Else I would ask for a more scientific examination and formal rebuttal by either side of the others basic maths concepts.An error of 2 by either side in maths alone seems extremely unlikely.

247. CD Marshall says:

His concept of all energy equals heat has no foundation in physics, just in climate science and at my undergrad level of physics, he is using thermal equilibrium to justify entropy increasing the temperature.

So he does not understand thermal dynamics or bought into the “revised incorrect climate science version.”

On the watts per meter per second, what was my error again? He quite confused me. I swear a physicist could write a book while sleeping if you could just harness stream of consciousness while you slept.

248. CD Marshall says:

I should be letting you sleep or get back to family time. 🙂 Cheers!

249. Joseph E Postma says:

Climate of Sophistry
5 seconds ago
@Chris A.

Refer to the First Law of Thermodynamics, with no work: dU = Q = m Cp dT

To get an increase in temperature, a body requires heat, Q. It doesn’t merely require energy, because not all energy is heat. Heat is a specific form of energy, and this specific form of energy we call heat is what can increase temperature. Where work is not involved, heat is the only form of energy which can increase temperature. It’s the First Law of Thermodynamics: dU = Q = m Cp dT.

So there’s solar energy, which certainly acts as heat because solar radiation at Earth is +121C or 1370 W/m^2.

And then there’s radiation from the atmosphere: can it act as heat? If radiation from anything is raising a temperature somewhere else, then that radiant energy has to be heat. Of course whether or not energy can act as heat is given by the heat equation. The heat equation for the atmosphere & surface, where the surface is generally warmer, thus indicates that the cooler atmosphere cannot send heat and thus cannot raise the temperature of the warmer surface.

The standard diagrams of climate science do indeed depict radiation from the atmosphere as heating and increasing the temperature of the surface – this is precisely what they depict. It is even given a label, called “Backradiation”, and the function of backradiation, i.e., radiation from the cooler atmosphere to the warmer surface, is depicted as increasing the temperature of the surface and thus as heating the surface.

I am referring to precisely what is there, in the diagrams, of course.

Climate of Sophistry
1 second ago
@Chris A. To understand the reality of the First Law of Thermodynamics, and that heat is required to increase temperature, simply imagine this scenario:

A temperature gauge in Earth orbit receiving 1370 W/m^2 from the Sun, reading +121C. Now, place a transparent sheet of ice at -18C in front of the thermometer’s view toward the Sun: do you really think that a VIS-transparent sheet of ice at -18C (and thermally emitting IR at that temperature) in front of the thermometer’s view toward the Sun is going to increase the thermometer’s temperature reading above +121C? Seems intuitive to me that it couldn’t.

So apply the heat flow equation: certainly the 1370 W/m^2 from the sun is maintaining the thermometer at +121C. If the thermometer is at +121C, then will the introduced sheet of ice send heat to the thermometer, or will the thermometer itself at +121C actually send heat to the sheet of ice? Heat flow is one-way only, so either the sheet of ice at -18C sends heat to the thermometer, or the thermometer at +121C sends heat to the sheet of ice. Heat doesn’t flow both ways, but only one way from stronger to weaker as a difference between them.

Thus it is entirely clear: the cold object cannot increase the temperature of the warmer object via its radiation.

250. CD Marshall says:

LOL I just replied to him as well, I probably got him confused he seems to only speak physics and not sure what variation he is speaking?

251. CD Marshall says:

…Or how honest he is at the answers.

252. Philip Mulholland says:

I’m not talking about a virtual blackbody. I am talking about a virtual blackbody.”
Where does this come from? Is a linguistic problem?
I’m is the same as I am so the sentence resolves into:
I am not talking about a virtual blackbody. I am talking about a virtual blackbody.”

So we have here a statement where Negative equals Positive

253. angech says:

“” To get an increase in temperature, a body requires heat, Q”
One can either increase the energy in or reduce the energy out.
Blanket principle
“It doesn’t merely require energy, because not all energy is heat.”
A lot of different energies are capable of creating heat in a body apart from radiation.

254. CD Marshall says:

This statement is redundant.
” I do not claim any photon equals any other. I claim that energy fed to matter due to absorption of radiation will rise the inner energy according to the amount of energy fed, regardless if the very same amount of energy comes from two longwave photons or one shortwave photon. This is true as long as a photon has no such high energy that it destroys the matter. And identical rises of inner energy lead to identical rises in temperature.

And yes, absorption always lead to rise in temperature, because it will not be transformed into some kind of work like dV without transit through a dT. That is possible, however first the temperature is altered by absorption of energy.

So, it‘s not about if photons are different if they are of different frequency, but about what the power they carry does. Or to put it simple: You cannot have photons be absorbed without a temperature effect. Its energetically impossible.”

He just claimed all photons are the same it’s like climate science saying we aren’t saying GHGs raise the surface temperature but we are actually saying that ghg raises the surface temperature.

I archived Geran’s fine example of the futility of this concept (I can remember all of this stuff like you guys do off the top of my head) but once I reference the text (or are remined by you guys) it “clicks” back in.

Geran:
Of course infrared impacting the surface does NOT imply absorption. You CAN add energy to a system without increasing temperatures. Bring a bowl of ice cream into your den. The ice cream is emitting infrared. You have “added energy” to the system (den), but do you expect the room temperature to increase? Or, add a liter of water at 40 degrees to a large bowl already containing water at 45 degrees. You have added energy to the large bowl (system), but the average temperature DROPS. Thermodynamic “heat” has two components: 1) energy transfer, and 2) from “hot” to “cold” (ΔT). Without both components, there is no “heat” just a transfer of energy. No heat=no raise in temperature.

255. @angech 2021/04/26 at 10:13 pm

“Are we talking about a recognised amount of solar input?”

Of course. What else?

“Semantics? Misinformation?”

Yes. Following:

“Looking at Postma’s side the maths seem to indicate a doubling of the energy per unit area along with a halving in size.”

That is not what my “side” shows at all. I show sunlight falling on the Earth, as it actually falls on Earth. I show what exists.

“Looking at ATTP it is 4 times the area for the incoming energy dividing by 4.”

That is accurate, yes.

“ATTP works out an average temperature, whole planet.
Postma works out a different average for half the planet.”

I work out the average for the whole planet, with sunlight input to half the planet since this is what actually exists. It makes a difference in the physics because in my depiction based on reality then the Sun creates the climate; in the ATTP /4 depiction, the sun does not create the climate. We are interested in the climate, and its physics, hence we are interested in what creates the climate, which you cannot learn from the average climate.

“Both claim justification for different concepts of heating the surface.”

Does sunlight fall on the entire surface uniformly as an input? Or is there such a thing as day and night, and latitude, etc?

“That the difference in their models equates simply to the difference between local and general maths.”

It is a difference between modeling what exists and hence being able to explain what exists (my approach), vs. modelling what does not exist and hence not being able to explain what exists without fudge-factor corrections (the greenhouse effect).

“Else I would ask for a more scientific examination and formal rebuttal by either side of the others basic maths concepts.”

If you’re not psychotic, you can do this yourself: A) does sunlight fall uniformly on the entire surface of Earth as an input, at a feeble -18C heating potential unable to create the climate, or B) does sunlight fall over a hemisphere with high intensity directly the sun diminishing to zero at the terminator, with local intensities which create the climate? YOU can perform the examination!

“An error of 2 by either side in maths alone seems extremely unlikely.”

It is. I have no error of two. The other side doesn’t either, but it DOES spread the incoming sunshine over the entire sphere as an input, diluting the solar flux to a point where the sun cannot be thought to create the climate.

256. CD Marshall says:

Physicist Chris is not giving up his fantasy physics.

257. @angech 2021/04/27 at 7:04 am

“One can either increase the energy in or reduce the energy out.
Blanket principle”

A blanket works by stopping convection. It doesn’t heat you…you heat it! Then it feels warmer on your skin because you have a warm blanket against you rather than cooler convecting air.

258. Philip Mulholland says:

The energy to heat the blanket comes from the body and not from the blanket.
The energy to heat the air comes from the sun and not from the air.

259. CD Marshall says:

They are nuts everywhere.
“Physics Chris”…
““Photons returning at the surface from the atmosphere at these levels of radiation does not increase the surface temperature and never could from the source. You aren’t adding new energy to the system, you are recycling energy already in the system.“ Yes, you are indeed recycling energy in the system which is AT THE SAME TIME fed with additional energy from the sun. Answer one question: Will a system that can recycle energy and is irradiated by the sun have the same temperature like a system that cannot recycle energy and is irradiated by the sun in the very same way? Answer this!”

260. CD Marshall says:

I said from water vapor sure, not ghgs.

​ @Chris A. Don’t be cute…intuition was not the extent of my previous comments as I also demonstrated to you the First Law of Thermodynamics, the definition and equation for heat flow, and how they work, giving you an example to help it make sense to you.

The sheet of ice at -18C cannot heat the thermometer at +121C to higher temperature, but in fact the thermometer at +121C will heat the sheet of ice. Heat flow is one way, and heat is what is required to increase temperature.

Your youtube link is simply a thought experiment postulating that heat can flow from cold to hot. Irrelevant.

A reflector reduces the emissivity of the system, and a reduction in emissivity requires an increase in temperature. This is not the greenhouse effect, so your scenario is irrelevant. GHG’s increase the emissivity of the atmosphere, they’re not about reflection. It makes a difference.

Here are the facts and the actual math of thermodynamics, which I will leave you with, not that you can understand it:

First Law: dU = Q = m Cp dT

Thus, to get an increase in temperature, i.e. a positive dT, we require Q, which is heat.

What is heat?

Q = s*(Th^4 – Tc^4)

Heat is the form of energy which flows one way from a hot object to a cool object; not all energy is heat, and the energy from the cool object is not heat hence cannot increase temperature, since heat is what is required to increase temperature. Only the difference in energy between a hot object and cool object is heat, and this energy flows one way only from hot to cold. Really basic stuff, but climate alarmists are unable to understand such simple facts given over as they are to flat Earth theory – flat Earth theorists, such as yourself, cannot really be expected to understand logic, math, physics, etc.

Thus in the scenario with the thermometer heated to +121C, and then you place a sheet of ice at -18C transparent to shortwave in the view of the sun from the thermometer, can the sheet of ice send heat to the thermometer? Of course it can’t, it’s colder than the thermometer. However, the sheet of ice can be heated by the longwave IR emission from the thermometer. The ice can receive heat from the hotter thermometer. So the thermometer heats the ice. If the thermometer is heating the ice, and given that heat flow is one way, then the ice cannot be heating the thermometer.

But look Chris, I know that you cannot accept these things and will merely continue to come up with side-arguments and deflections to pretend to ignore them. The reason you can’t understand such basic facts as heat flow being one way, as ice not being able to boil water (do you think that ice can boil water? lol…you do, to be consistent…you think that adding ice to a system can tip water over 99C and make it start boiling…lol!), is because your mind has been wrecked with flat Earth theory. You are a flat Earther and you’ve been brain-damaged with flat Earth theory. Climate science is the science of flat Earth theory, where solar input is diluted by a factor of four over a surface area four-times as large as is intercepted by the Earth, where the surface is a flat plane. You believe this model implicitly and this flat-Earth accounting of the sun-Earth interaction in embedded in all of climate science; thus to have been exposed to climate science and believe in its prognostications is to make one a flat Earther whether they are aware of it or not. Climate science is pregnant, is imbued with flat Earth theory throughout all of its parts, and thus the insanity of flat Earth theory becomes inculcated into the mind of anyone who takes climate science and alarmism seriously and believes in it. There will have to come a time when we will spend year re-educating people as to the falsehood of flat Earth theory.

262. CD Marshall says:

The ice sheet was brilliant by the way, because you could easily do an experiment with that to prove it or something very similar.

263. CD Marshall says:

I think it’s pretty obvious now his issue with Dr. Holmes was he didn’t like his answer.

264. “Yes, you are indeed recycling energy in the system which is AT THE SAME TIME fed with additional energy from the sun. Answer one question: Will a system that can recycle energy and is irradiated by the sun have the same temperature like a system that cannot recycle energy and is irradiated by the sun in the very same way? Answer this!””

ALL systems “recycle” energy, as in having the energy which they have accumulated flowing all about. The difference is that the “recycled” energy is not NEW energy, is not energy input, and cannot re-input itself – it is “downstream” energy, and it is at the same temperature of the system. Heat cannot be recycled, cannot be used for another round of temperature increase after the first time it has been used. They’re saying that the energy input to an ice-cube placed into a cup of coffee will recycle out of the ice cube back into the coffee and make the coffee hotter.

The Sun is a source of new raw energy. Passive objects only transfer energy, they certainly do not “recycle” energy in the sense of bumping themselves up with their own energy, etc.

265. Consider a pot of water at 99C. Just below boiling. Now, introduce ice to this system…in whatever manner.

Introduce the ice physically into the water. Will this tip the water over 99C? Of course not.

OK then. Put the ice beside the pot. Will this tip the water over 99C? Nope.

Try again. Put the ice above the pot, in view of the water. Will this now make the water boil? Nope!

266. CD Marshall says:

He won’t stop, he’s still trying. Even after explaining photons are everywhere bouncing around the system everywhere all the time night and day they aren’t ALL being absorbed. In fact very few of them are in comparison.

He wants an answer to his questions but he won’t accept the answers anyway.

They have mirror systems, very expensive ones, that can reach the temperatures of the Sun or close to it and of course they do that by manipulating the density flux. They aren’t “adding” new energy or recycling.

Just like that guy who does Sun art.

267. CD Marshall says:

If this keeps up we’ll be forced to drive cars with solar panel masts over them and a cycling system incase it dies like the Flintstones.

268. CD Marshall says:

More from Chris:
“Sure with the water cycle. Deserts prove no to ghgs.“ That‘s not an answer to my very clearly stated question, as you will be well aware of. I will gladly answer your points if you answer this YES/NO question. Until then, i chose to believe you have some reason to be evasive.

269. Joseph E Postma says:

Well it’s like I said above, and also put this in consideration of the absolute total psychosis we witnessed from Willard yesterday…the absolute state of psychosis:

We really must accept, as a matter of fact from hereon, that we’re dealing with brain-damaged people. The absolute anti-empirical and anti-rational insanity of flat Earth theory is imbued in climate science, throughout climate science. The whole field is infected with a virus borne out of flat Earth theory, and this virus infects everything and anyone who engages with the field without a critical virus-destroying filter first put in place. Those who engage with the field and believe in it cannot help but accept the mental irrationality virus into their mind, and it proceeds therefrom to distort all manner of their thought processes and comprehension abilities. The virus inverts the thought process, it inverts the ability to comprehend or acknowledge anything opposed to it. It takes control of the mind much like these examples of fungus spores which infect insects and take control of them like zombies. Perhaps these people have been zombified by this flat-Earth virus. They certainly seem to lack all self-awareness when they say things like “It’s not what we believe, it is just what we teach.”, or when they say that ‘having sunlight fall over a hemisphere doubles the flux from the Sun’, etc. These people are zombies!? They’re not self-aware. They cannot acknowledge that spreading sunshine diluted to freezing over a plane does not represent the Earth…who cannot acknowledge that? Of course, THE VIRUS stops them from acknowledging that, because this is the very origin of the virus; acknowledging this would destroy the virus, so the virus does everything it can to deflect away it.

This might be the first example of a mental-meme virus which attacks its host and causes its host to kill itself, and its world! This virus, as we have seen, is now getting people to hate the life molecule of CO2 and seek to eradicate it from the atmosphere! We’ve been infected with a mental virus which is going to destroy the planet.

270. CD Marshall says:

Yeah so I hit him with a sock full of quarters. After he “comes to” he’ll just rehash the same thing a hundred different ways and still get the same answer and still deny the answer like a mental patient.

271. Willard says:

Get help, Joe.

272. Steven says:

@Joe , after that rant about zombies I wonder how you accuse others of psychosis with a straight face.

273. Get help, Squirt, understanding that the Earth is round, that sunshine falls on a hemisphere, and that sunshine is hot enough to create the climate. Get help with acknowledging basic facts about reality Willard.

274. Willard says:

I know of at least one Sky Dragon sock puppet who keeps appealing to reality, Joe. Wanna know who?

Once more, Bob says hi.

275. @Steven – well you see, I’m into the physics of characterizing the reasons for things, and why they exist the way that they do. This is why I accept that the Earth is round and that sunshine is hot, for example.

Thus, when I am confronted with people who reject, in their analyses if not in their beliefs, that the Earth is round and that sunshine is hot, then I seek to also characterize them such as to explain them.

And so after years of encountering people like you and Squirt, who cannot acknowledge that the Earth is round and sunshine is hot, and who say things like “It’s not what we believe it is just what we teach.”, then the explanations of the physics which describes and explains you has to equal the extraordinary state of your position: extraordinary evidence requires extraordinary explanations. It is entirely extraordinary to encounter people who cannot acknowledge basic facts about reality such as the Earth being round and not flat, and thus, it requires extraordinary explanations to characterize entities like you and the lil’ squirt Willard.

276. Are you still in a paranoid psychotic state believing that every commentator who knows that the Earth is round and sunshine is hot is me, Squirt? You are obsessed with me…wow.

There are many other people out there, lil’ squirt Willard, who independently understand these basic facts about reality and their implications.

You’re still here psychotically not presenting anything…just trolling along in full psychosis mode.

Willard…I’m letting your comments through…make a fn point already would you? Or is all that you have is paranoid psychosis? lol

277. Willard says:

Considering your past traumas, Joe, I think you should seek help. As long as you’re only abusing people like me who have no qualms dealing with bullies, you might have found a less destructive way to deal with your monsters. Still, don’t wait until it’s too late. Violence may destroy your life and those around you.

As for the point, Bob repeated AT’s:

The geometry says that the sunlight passing through 1 m^2 measured perpendicular to the sun’s rays gets spread over a larger horizontal area on the earth’s surface of 1/cos(z) m^2, where z is the zenith angle (the angle between where the sun is in the sky and the vertical line perpendicular to the earth’s surface). So, the flux is 1368*cos(z) W/m^2 on 1m^2 of the earth’s surface. (Well, at the top of the atmosphere.)

I have no idea where you got your “>90% zenith flux” but I’m quite sure you can’t support it.

You’re not fighting me, Joe. You’re not even fighting reality. You’re fighting geometry. You’ll lose.

[JP: You psycho…lol…sunlight falls on a hemisphere. There’s math for that. Nice little attempt at OID again though…as if you own the geometric argument…you DONT! lol]

278. CD Marshall says:

Joseph after having a physicist fight over internal energy increase equals heat nothing amazes me anymore. They do teach this for I saw on a climate science doc they called internal energy a form of heat. They have to for the ghge has no purpose w/o the auto forcing mechanism climate science created.

279. Guys, just take a look here for example at this example of OID (ownership inversion deflection):

We start with DREMT stating the basic facts of the geometry and the factors involved:

Dr Roys Emergency Moderation Team says:
April 26, 2021 at 7:55 pm
“…[W]hen you divide by 4 you are spreading the incoming solar radiation over the entire Earth’s surface. When you divide by 2 you are spreading the incoming solar radiation over only the lit hemisphere. The lit hemisphere has only half the surface area of the entire sphere (obviously). That’s why I’m saying the surface area is being halved. The flux is higher when spread over only a hemisphere than over the entire sphere because the surface area of the hemisphere is half that of the entire sphere.”

Now let’s look at how lil’ squirt Willard pretends to take ownership of these facts, but then inverts them such as to deflect away from the meaning and implications of them:

Willard says:
April 26, 2021 at 8:38 pm

“No, the entire Earth’s surface is equal to 1. If you divide the Earth’s surface by 2, you get a hemisphere. If you divide by 4, you get a disc. That’s the geometric point.

So to divide by 2 the way you do is in effect doubling the surface of the Earth’s surface over which is spread incoming solar radiation. If that gets you twice the flux, that means something is wrong with your calculation.”

Look at the first sentence: did it actually say anything different than what DREMT said? He says the exact same thing: that if the area of a sphere is 1, then a hemisphere is found by dividing by 2, and a disk is by 4. This is exactly what DREMT just said, but the psycho prefaces his re-statement with “No, …”, thus implying that DREMT said something incorrect and then by re-stating it and thus inserting themselves as the owner of the identical statement. You see how that worked?

And so after pretending to refute DREMT’s statement with his preface of “No, …”, but then re-stating the very statement to pretend ownership of the same facts we continue to Willard’s second sentence:

to divide by 2 double’s the surface area over which is spread incoming solar radiation” – this is the inversion and deflection, now. This takes DREMT’s statement which was then restated by Willard, but now they invert the entire meaning of it, a literal inversion mathematically, where a division by two equates to a doubling, not a halving but a doubling, of the surface area over which sunlight falls. It then goes to a second layer of inversion where Willard states that a doubling of surface area “gets you twice the flux”, when in fact a doubling of surface area would equate to half the flux, not double the flux.

Willard then ends with “that means something is wrong with your calculation” which is of course the final deflection which is merely borne out of the inversion Willard itself created.

So here’s the algorithm:

1) Take ownership of a factual statement from your opponent by a) implying that they didn’t state the fact correctly, then b) re-stating the very same factual statement yourself

2) Invert the statement you just re-stated to pretend that it says the inverse of the very statement itself

3) Use the inversion to deflect interest away from the original and re-stated statement

Guys, WITH are we dealing with here? I mean – what in the good f have we encountered here?!

Do you guys want to know what Willard’s email (sans domain) is? This is what it is apparently, as he enters it to comment: “languageisasocialart@………”

“Language is a social art”, hey lil’ Squirt Willis? Is that what you’re doing here with this insane OID algorithm?

280. Kev-In-ZA says:

@Joe, I am at a loss for words as to how bizarre the logic of @Willard has played out. Is he still trying to instruct well qualified people on elementary geometry, or does he really believe that simple averaging of Insolation over the entire earth surface all at once is good enough for anything useful (other than basic first-order earth energy balance)…?

281. Philip Mulholland says:

“if the area of a sphere is 1, then a hemisphere is found by dividing by 2, and a disk is by 4.”
Well that would seem to be the end of geometry and calculus..
Euclid and Newton felled with a single blow.

282. @Kev – they can’t even give us that sunshine is hot and falls on a hemisphere…they can’t even give us that…!

283. CD Marshall says:

So does the surface absorb photons that are a longer wavelength than the surface temperature is emitting ? We know the Sun doesn’t, but what about the Earth? I’ve always wondered about that in the E budgets?
Sure some IR is being reflected downwards and they measure the flux, but is it really absorbed or just scattered?

284. Joseph E Postma says:

They treat all energy as being absorbed – the heat flow equation shows that it isn’t.

285. CD Marshall says:

IR photons are typically not absorbed at ambient room temperatures. Which make prefect sense or we couldn’t see anything. Room temps is around 68F/20C. Clearly photons are being scattered outside as well off of the ground, clouds, buildings…

In order to appease QM the incoming solar radiation would have to be lowered in climate science to justify absorption. This was well thought out thus the “geometry” excuse was born. Does make you wonder doesn’t it?

286. CD Marshall says:

You know I have heard that argument before, once the energy is absorbed into a solid matter it no longer follows Bosonic energy laws it follows fermions. I believe that was on Potholer’s site. So that justified all energy increasing temperature. They might very well teach that these days.

287. Philip Mulholland says:

We live at the bottom of an ocean of air. The average pressure of which is 101,325 Pascal. Now air pressure is always with us but we cannot sense it, we can experience the effects of change in pressure, the force of the wind for example, but not pressure itself.
If we have a container with air inside at high pressure, we can release this air from the container and use the pressure difference to do work. However, once the difference in pressure is lost then no more work can be done, but there will still be air inside the container, except that this air is now at the same pressure as the air outside. The air inside the container still has a pressure but the difference in pressure has gone

So, we have here a perfect analogy for heat. Heat is difference in temperature and that difference can be used to do work. Once the temperatures are the same then there is no more heat and no further possibility of doing work. The air inside the container still has a temperature but the difference in temperature has gone.

288. That’s excellent Philip!

289. God these idiots. No one denies surface temperature, no one denies geometry. They DO deny that sunlight falls on a hemisphere as an input though! They DO deny the adiabatic gradient is caused by gravity! The temperature distribution is not due to radiation but to gravity. The radiation emitted by a passive gas is a RESULT of its temperature, not the cause of its temperature. Nice inversion too…all of this is a result of THEIR denying the role of sunlight on a hemisphere.

291. CD Marshall says:

FromNorway:
That’s the spectrum of outgoing IR radiation from Venus as measured by the Soviet Venera 15 probe in 1983-84.
Can you explain why the red curve labelled “Venera” mostly stays between the Planck curves for 210 K and 260 K?

If the 735 K surface temperature was caused by the high pressure working as a true heat source and there was no absorption of outgoing IR radiation in the atmosphere, the spectrum measured by the Venera probe would look something like the black curve here:
64.media.tumblr.com/0edb6738accf52de741a0b72d478557f/*tumblr_oarp7fDbqD1t4esr2o1_500.jpg

If you accept both the Venera measurements and the high surface temperature, you can’t deny that the atmosphere must be blocking about 99 % of the heat loss to space. How can that happen without having a massive impact on the surface temperature? That would be a gross violation the first law of thermodynamics!

And if your next response is to deny the high surface temperature altogether, then you have just killed your own “Venus has a 92 bar atmosphere….” argument!

All this boils down to you denying the very basic result of the spherical Earth having 4 times larger area than a circle with the same radius!”

How is an IR detector going to pierce Venus’ clouds and defect surface IR. Isn’t that only detecting IR bouncing back off the clouds? These guys with this endless “gotcha”.

292. angech says:

Joseph E Postma said
angech
“An error of 2 by either side in maths alone seems extremely unlikely.”
“Joe

It is. I have no error of two. The other side doesn’t either, but it DOES spread the incoming sunshine over the entire sphere as an input, diluting the solar flux to a point where the sun cannot be thought to create the climate.”

Thank you for the opportunity to post.

My comment was up at ATTP as well for nearly 2 days but has since been moderated [removed].

Willard seems to be happy to keep stating your maths is out by 2 without explaining why.

The removal of my comment elsewhere fails what Australians call the pub test.

I hope Willard will allow it back up

The arguments on both sides re GHG are not solved for me.

293. CD Marshall says:

K math magicians, is this verified?
The Latent Heat of evaporation is a function of water temperature, ranging from 540 cal per gram of water at 100C to 600 cal per gram at 0C.

294. angech says:

The problem is, real or not, that GHG’s exist.
That is there are molecules that absorb and then emit infra red.
A truism.

I trust.

The question then is do such molecules behave in a different way for heat transfer than molecules or atoms that do not have this property.
All the physics in the world will show that pressure is strongly correlated with temperature.
As you demonstrate.
Heat flow or energy flow is governed by the speed at which it can proceed.
Hot to cold.

But how fast is the question and do GHG make a difference in the speed of transfer and hence the ultimate temperature at that pressure and gas concentration with GHG in.

My contention is that it is a serious and worthy question.

Not one to be swept aside by saying pressure or only pressure governs it all.

Both concepts have validity.
Assigning the balance between them cannot be waved off by merely stating hot flows to cold but by how fast it does so.

295. CD Marshall says:

I guess the plate experiment is catching on? From Chris…

“Let the heat source of the plate be irradiation at the frequency of a 6000K star, the steady state temperature of the plate be 333K, everything be placed in vacuum and now we have by „thought experiment magic“ 3/4 of the thermal radiation of the plate recycled to the plate, thus radiating onto the plate where it came from. Let the plate be of copper oxide with a thickness of 1mm. We have to cases now:
a) plate irradiated on side A, no thermal radiation coming back onto it
b) plate irradiated on side A, 3/4 of its thermal radiation „recycled“ back onto side A by our thought experiment
Question: Will the surface of side A of this plate have the same steady state temperature in both cases a) and b)? If not, which one is warmer? I like to have this clear. Reasons for the answer are interesting after the answer. Lets do this like an argument usually is build up: „It‘s so an so, because…“

Ok! Is the thermal radiation, that is recycled back, absorbed?”

296. @CD – thermal capacity (Cp – specific heat capacity) is actually a function of temperature too. So when we have dU = m Cp dT, Cp looks like a constant but this is only true for small dT. In general, though, or more accurately, dU = m Cp(T) dT.

297. @angech

The problem is, real or not, that GHG’s exist.

Since there is no radiative greenhouse effect, then GHG’s do not exist.

Briarpatching: if you accept the fundamental premises of your enemy, then you’ve already lost. This is why the traditional debate goes nowhere and only gives the alarmists all of the power and the win. This is why the gatekeepers/fake skeptics will never reject this, despite the obvious evidence that they should, etc.

Briarpatching: a rabbit is safe from predators in a briar patch. On its way back home to the briar patch, a fox lies waiting for the rabbit outside of the briar patch. When the rabbit approaches, the fox jumps out and taunts the rabbit that he’s about to eat it. But the rabbit retorts: “you can’t catch me out in the open, I’m way faster than you! If you had’ve let me get into the briar patch there’s no way that I could have outrun you, you simpleton!”. Taken aback, the fox let’s the rabbit get into the briar patch, believing that he’ll have an easier meal. However, of course, the fox can’t navigate the briar patch at all, and the rabbit is safe and gets away.

There is no such thing as greenhouse gases, and the phrase means literally nothing. Inside a greenhouse there are gases of course, the air/atmosphere…but why call the air inside a greenhouse “greenhouse gases”? It’s just air…held in place inside the greenhouse prevented from convection.

In climate science, of course, the terms means gases which interact with radiation. But why call these greenhouse gases either? In astronomy and physics there are all manner of molecules which can create gases which can interact with radiation…we’ve never called this phenomena “greenhouse”.

So, “greenhouse gases” are named as such by climate science due to their function, the mechanism which they supposedly facilitate. Not merely absorption/scattering because we already just call that absorption/scattering. What is the function or mechanism implied then in the climate science “greenhouse effect”?

The mechanism is reverse heat flow, it is backradiation from a cold source acting upon a warmer body to make the warmer body even warmer, which is actually heat flowing in reverse. Climate science invents this scheme because it dilutes the solar input over the entire surface of Earth, resulting in a theoretical solar power of only -18C heating potential, thus unable to create the climate. In their approach to the climate, the Sun does not create the climate because it cannot heat anything to above -18C. Thus, they invent something which they call a “greenhouse effect”, even though it is not how a greenhouse works, to get the climate to be created via reverse heat flow. In climate science the Sun does not create the climate, but reverse heat flow from the cold atmosphere to the warm surface is what creates the climate.

However, obviously, the sun create the climate.

So no, the greenhouse gas/greenhouse effect concept does not have validity, because the entire theory is borne only out of the model of flat Earth theory where the Sun does not create the climate.

The mechanism of the climate science “greenhouse effect” can be empirically tested in controlled condition in a lab, and all such tests empirically refute, do not demonstrate, the mechanism. It is empirically refuted, as well theoretically refuted because it violates really basic thermodynamics laws, etc.

298. CD Marshall says:

what EXACTLY is he trying to prove with this?

“Let the heat source of the plate be irradiation at the frequency of a 6000K star, the steady state temperature of the plate be 333K, everything be placed in vacuum and now we have by „thought experiment magic“ 3/4 of the thermal radiation of the plate recycled to the plate, thus radiating onto the plate where it came from. Let the plate be of copper oxide with a thickness of 1mm. We have to cases now:
a) plate irradiated on side A, no thermal radiation coming back onto it
b) plate irradiated on side A, 3/4 of its thermal radiation „recycled“ back onto side A by our thought experiment
Question: Will the surface of side A of this plate have the same steady state temperature in both cases a) and b)? If not, which one is warmer? I like to have this clear. Reasons for the answer are interesting after the answer. Lets do this like an argument usually is build up: „It‘s so an so, because…

Ok! Is the thermal radiation, that is recycled back, absorbed?
It it is the very same experimental setup with two states: state a) without any „backradiation“ of the thermal radiation of the plate, state b) with „backradiation“ of 3/4 of the thermal radiation onto the plate. In both cases the side irradiated by the energy source is the same. In case b) this very same side of the plate is also irradiated by the „backradiation“. So we are just ignoring the backside of the plate.”

299. @CD: It’s exactly the discussion had with esttom, regarding parallel plates. Refer to my diagrams earlier of the parallel plates, etc.

One would never see anything wrong with equal temperatures defining equilibrium for plane parallel geometry:

The above diagram is the exact same thing as the following one:

However, now the mind seems to reel at the presence of these arrows. Uh oh…don’t you have to add the arrows together? Well, do we add temperature together? If you take one pane at 91C and introduce another pane at 91C, do we now get two-times 91C? Of course not!

No one would EVER add temperature together. No one would ever do that. But you change temperature to arrows, and all of the sudden people think that you should:

If you add arrows together, wow, look at that gradient…that explains the temperature distribution of the atmosphere where it is warmest at the bottom! Amazing!

Oh wait…the temperature distribution is caused by the adiabatic effect.

300. esttom says:

“ However, now the mind seems to reel at the presence of these arrows. Uh oh…don’t you have to add the arrows together? Well, do we add temperature together? If you take one pane at 91C and introduce another pane at 91C, do we now get two-times 91C? Of course not!”

The arrows represent energy fluxes, not temperatures. So it is correct to sum them.

301. The arrows represent temperatures, so it is not correct to add them…lol 🙂 The arrows are not additional inputs…the only input is from the Sun, and all other arrows are downstream passive energy, representing temperatures, which we do not add together.

302. “The arrows represent energy fluxes, not temperatures. So it is correct to sum them.”

Energy fluxes between pairs of objects are never added together, only subtracted, to give heat flow.

303. Anyway…who cares about a silly argument going “we should not add temperature together”, “we should add temperatures together”, “no we shouldn’t”, “yes we should”

lol!

Just provide some empirical evidence that the backradiation greenhouse effect exists, under controlled conditions.

As it stands, the empirical evidence sides with me: the radiative greenhouse effect has never been demonstrated over many, many instances to measure it and several hundred years of attempts to demonstrate it.

No evidence where evidence should exist = does not exist 🙂

304. “The arrows represent energy fluxes, not temperatures. So it is correct to sum them.”

This divorces the arrows from their origin, which is a temperature. The arrows represent the temperature. The arrows are a result of the temperature. To sum the arrows is to sum the temperatures which they represent.

The arrows do not represent temperature? Where do they come from then? How is their magnitude determined? What makes them come out of the surface? Why are they directed away from the surface? Why do they have the values that they do? Why have them at all? What is their connection to the diagram at all and why were they put in the diagram in the first place? This position is entirely untenable and absurd.

Let us again imagine the thermometer at +121C and the sheet of ice coming in view of the Sun:

So while the sheet of ice isn’t touching the thermometer but only has “its arrows” directed toward the thermometer, then it is making the thermometer get even hotter than the +121C from the Sun.

But then the sheet of ice touches the thermometer, and now makes the thermometer cool down.

Imagine a pot of water held at 99C. Now “shine the arrow” from an ice cube at it…wow that tips it over 99C to start boiling! Bring the ice cube really close to the pot so that it gets the full brunt of the ice-cube’s arrow – wow really boiling now!

Oooops! We dropped the ice cube into the pot and now the boiling stopped. I guess being dropped into the pot killed the ice cube’s arrows!? How are arrows killed, exactly? I for one would like to know how to kill arrows! I guess being immersed in water kills arrows?!

You see…the entire argumentative structure of climate science greenhouse effect breaks down with simple anlaysis. Get out of the briarpatch – there is no radiative greenhouse effect.

305. esttom – don’t be cute…the arrows represent the temperature which creates them.

Again…”no they don’t”, “yes they do”, etc…lol.

You have no empirical evidence for your claims. I do.

Empirical evidence refutes you.

Yes, it sure seems “not physical” that if you add two ice cubes together, you don’t get twice the temperature. Wow, that sure seems unphysical…don’t quantities always add!? How can you add two things together but not get twice the quantity?

Please familiarize yourself with the concept of temperature 🙂

306. CD Marshall says:

Fluxes don’t add so the highest flux sets the temperature.
Too simple?

307. Correct CD.

308. CD Marshall says:

You don’t add fluxes it’s a measure of joules in space and time, particularly per second per meter squared.

309. esttom says:

“ The arrows represent temperatures, so it is not correct to add them…lol”

No, you are wrong. The arrows are clearly labeled with units of W/m^2, which is not a unit of temperature. So they cannot represent temperature. They are energy fluxes.

“ Energy fluxes between pairs of objects are never added together”

Energy is a locally conserved quantity, so it is correct to sum all energy fluxes into an object. The sum should be zero. Adding up the arrows into the leftmost plate in your diagram clearly shows a violation of conservation of energy.. 1000+1000-1000=1000 W/m^2. That is not physical.

You are trying to obfuscate this extremely simple fact with all kinds of wild claims — that energy fluxes are actually temperature, or that energy cannot be summed. These are wrong. The simple truth is that the energy fluxes into each object, represented by the arrows, must sum to zero.

[JP:

esttom – don’t be cute…the arrows represent the temperature which creates them.

Again…”no they don’t”, “yes they do”, etc…lol.

You have no empirical evidence for your claims. I do.

Empirical evidence refutes you.

Yes, it sure seems “not physical” that if you add two ice cubes together, you don’t get twice the temperature. Wow, that sure seems unphysical…don’t quantities always add!? How can you add two things together but not get twice the quantity?

Please familiarize yourself with the concept of temperature 🙂]

310. Joseph,

311. esttom says:

Joseph, local conservation of energy is always true. The sum of energy flux into a given volume equals the rate of energy increase inside (here zero). This is always true, regardless of any details of the problem. So I could answer all of your questions of where the fluxes came from, why they exist, etc., but it will not change the answer: the sum of energy fluxes (arrows) into the leftmost plate in your diagram should be zero. But it is not, so your diagram is wrong.

This is the beauty of a conservation law: it disproves your solution without having to calculate through all of the details. All one needs to do is sum the fluxes.

Anyway, now you are supporting the idea that “fluxes don’t add.” Well, the precise statement of local conservation of energy is: the rate of change of energy inside a volume is equal to the integral of energy flux through the boundary. So, conservation of energy requires integrating (summing) energy fluxes. By denying that energy fluxes add, you are denying conservation of energy.

Aside from you inventing new schemes of “the statement of local conservation of energy” (lol), you give yourself away with this: “the rate of change of energy inside a volume is equal to the integral of energy flux through the boundary”

However:

“Heat is defined as the form of energy that is transferred across a boundary by virtue of a temperature difference or temperature gradient. Implied in this definition is the very important fact that a body never contains heat, but that heat is identified as heat only as it crosses the boundary. Thus, heat is a transient phenomenon. If we consider the hot block of copper as a system and the cold water in the beaker as another system, we recognize that originally neither system contains any heat (they do contain energy, of course.) When the copper is placed in the water and the two are in thermal communication, heat is transferred from the copper to the water, until equilibrium of temperature is established. At that point we no longer have heat transfer, since there is no temperature difference. Neither of the systems contains any heat at the conclusion of the process. It also follows that heat is identified at the boundaries of the system, for heat is defined as energy being transferred across the system boundary.” – Thermodynamics

For energy to cross a boundary, it must be heat. Not merely any old flux, but heat.

And thus to correct your silly little invention of “the statement of local conservation of energy”, I will simply refer you back to the actual statement of conservation of energy, the First Law of Thermodynamics:

dU = Q = m Cp dT

The change of energy inside a volume is equal to the heat flow through its boundary.

312. ashemann says:

FFS whats so hard.

Divide by 4 the out going flux is quite correct.
The incoming solar flux is divide by 2 as a square meter of lit surface has to be absorbing twice the joules as are emitted over the whole surface.

For every joule emitted by the whole surface over 12 hours 2 joules have to be absorbed on the lit half of the surface one emitted away and one stored until emitted on the dark side.
every 12 hours is the same, which makes every 24 hours the same.

313. AS part of my ongoing physics education I have been looking at the issue of Heat Capacity and have found that it is measured in units of Joules per Kelvin (as also is Entropy Q).

We have an equation for measuring potential energy (PE = mgh) and a critical point in the use of this equation is that we must define a datum level (for example the surface of the planet).

When we use the equation for Entropy it is always PER Kelvin and NOT Kelvin, in other words; is there is no datum point for the calculation of Heat Capacity or is zero Kelvin the datum point? If so why is there not an equation that measures the internal energy store of a body in the same was as the potential energy equation does for mass in a gravity field?

314. CD Marshall says:

@Philip Mulholland

Multiplicity is a tricky beast to tame. Let me know when/if you do so you can save me the headache.

315. CD Marshall says:

“Teaching thermal physics
is as easy as a song:
You think you make it simpler
when you make it slightly wrong.”
_ Mark Zemansky

316. Kev-In-ZA says:

@Phillip, not sure your level of knowledge, but perhaps a few points to your query.

First, Heat Capacity is J/(kg.K) not just Joules per Kelvin. Heat Capacity is a parameter of matter, and hence a datum is not applicable. However, Internal Energy (U) of matter has a hypothetical datum of Zero Kelvin, at which point matter at Zero Kelvin has no Internal Energy. But because Heat Capacity is not linear with Temperature, you should think of Internal Energy (U) a little like PE with only a local datum. It is just a property of matter that defines the Heat transfer per kg required to raise or lower that matter temperature by 1 Kelvin.

You also mention Entropy and seem to link to symbol Q. If you are a novice stay away from Entropy. As to Q, that symbol is normally reserved for Heat in Thermodynamics. W is typically used for Work, while U is typically used for Internal Energy. Symbols apply to JP 1LoT equation as in: dU = W + Q , and when ignoring Work (W), then: dU = Q = m.Cp.dT

Hope this help, and forgive if you know most of this already

317. Kev-In-ZA says:

@Joe, I have been scouting for empirical evidence or parallels around the Dual Pane setup. Solar panel collectors represented a simple analogy in that they virtually never exceed about 70-80C, but most have too much conduction or convection to confirm if a theoretically perfect Dual Pane system with vacuum can achieve the hypothetical ca 160C rather than 91C from 1000W/m2.

I found some references to Evacuated Flat Plate Solar Collectors, and this one certainly has me pondering. The important aspect of the Evacuated Flat-Plate collectors is that there can be no concentration of sunlight.
They seem to be achieving temperatures well above 150C quoting achieved efficiencies with the system up to dT of about 180K above Ambient of 15C. So they are going well beyond the 91C mark with the evacuated Dual Pane setup.

However, as noted in the paper, they are applying special measures to achieve this, namely:
-special coatings on the absorber which preserve SWIR Absorption coefficients around 0.93-0.96; while having an extremely low emissivity quoted as 0.06.
-special back plate high reflectivity foil to cut down emissivity from back plate
-evacuating the collector volume to low enough pressure to eliminate convection
-filling the collector volume with special gases still at fairly low pressure to reduce conduction
-they also increase the pressure in the water tubes so as to ensure good heat-transfer from the absorber plates to the working fluid (preventing steam which is a poor fluid for heat-transfer in the tubes)

I am still looking at this angle, but would be interested in your thoughts on this technology as empirical evidence that standard Sun potential of 1000W/m2 is not limited to 91C (nor full ToA Insulation of 1370W/m2 having a maximum potential of 121C).

318. nabrid51 says:

@Kev-In-ZA
Thanks for your comment. I have slowly been build up a spread sheet of physical properties and their dimensionality. To date my list contains 36 line items, physics is a complex subject. In the course of my research I have at #23 Specific Heat Capacity Cp with an equation of state of
SH = Q/M*K
Units of Joules/kg.Kelvin
Dimensions L^2T^-2K^-1
So yes you are correct.
However at #31 I also have Heat Capacity with an equation of state of
lim dT -> 0 dQ/dT Either this equation is wrong or Specific Heat Capacity and Heat Capacity are subtly different concepts.

The equation i have for #26 Entropy is dS = dQ/T with Units of Joules/Kelvin
Elsewhere I have #32 Heat absorption with an equation of state of
Q = mcdT and units of Joules.

From this I have assumed that Q is Heat measured in Joules.

I am still learning, so please continue to correct me where I am wrong.

319. Kev-In-ZA says:

@nabrid51, your #23 and #31 are the same, just different. The first is giving you a direct practical equation which you can use (within certain limits of applicability), while the second is the technical derivation of Specific Heat Capacity Cp at any given point of Temperature. #31 is a calculus description of the instantaneous rate of change of Heat addition with respect to Temperature increase. In plan-speak, it is the slope of the Heat Added (Joules) plotted against Temperature Change (K) at a given temperature. Which if you think of a simple x:y graph at any given point is y = A.x + C (A being the slope, and C being the Intercept, while y is the Heat change Q and x is the Temperature change T in the graph). Hence, Q = A.T + C with the slope A = m.Cp and the intercept C is arbitrary.

The Entropy equation is correct for a reversible process, but stay well clear if you don’t understand. I studies Thermodynamics at honours level as part of Engineering Degree, but I doubt I truly understand it. I doubt you will ever need it.

320. Kev-In-ZA says:

@nabrid51, I should have included that #32 is the same as 21, just swapped around and with different symbols. SH is Cp, M is m, K is dT; then SH = Q/M*K is Cp = Q/m.dT or (Q = m.Cp.dT)

321. Philip Mulholland says:

@Kev-In-ZA
Thanks for the useful clarifications.
At this point in my studies I am working in what I call librarian mode. The aim is to collect and organise information. In my analysis I am trying to drill down to the basics of the dimensional structure of the equations, understanding comes later.

322. @Kev-In-ZA 2021/05/02 at 2:03 pm

Yes, the important factor is this:

“-special coatings on the absorber which preserve SWIR Absorption coefficients around 0.93-0.96; while having an extremely low emissivity quoted as 0.06.”

So it’s via manipulation of emissivity. The coatings reduce emissivity of the surface itself, thus requiring higher temperature. This is all good and it isn’t the RGHE – CO2 does not coat the ground in such a way as to reduce the emissivity of either the ground surface or water surface, etc. GHG’s do not reduce the emissivity of the IR emitter.

In fact, the physics of these systems, if understood and as applied, demonstrate that GHG theory is inverted to reality: GHG’s are supposed to emit where the atmosphere cannot, thus they increase the emissivity of the atmosphere and via the physics this would then have to cool the atmosphere, not warm it or cause it to become warmer, etc.

323. Kev-In-ZA says:

@Joe, thanks. That was my understanding as well. They are exploiting the fact that absorption and emission of a surface, while typically the same for a given wave-length/frequency, can be very different between SWIR and LWIR. And then they spend a lot of effort minimizing convection and conduction within the practicalities of managing a vacuum without having to run a vacuum-pump continuously.
That insight, together with your previous comments, also helps my understanding of the maximum temperature potential of Insolation. The penny is dropping.
So my understanding is then that, 121C is the maximum potential of full Insolation (1370W/m2) on a surface with similar absorption and emission coefficient together with 100% conduction/convection insulation (or vacuum).
However, the Sun
-can heat to a higher temperatures if the LWIR emission can be manipulated to limit the emission at a given temperature,
-or can heat to a higher temperature if the flux can be concentrated but then only to a maximum of the temperature of the sun (5700K)

324. @Kev – yes this is something that the Slayers and I all discussed very early on, innocently assuming that they didn’t understand the physics of emissivity rather than what we now that they’re just purposefully lying and obfuscating everything.

For an absorbed flux F which needs to be reemitted, then F = esT^4, or T^4 = F / (es); and so if e is engineered to be below unity or below the absorptance then you get higher T. Nothing wrong with that physics of course, but it does show that the RGHE physics is doing the opposite, etc.

And yes of course if you concentrate the light, as with a mirror or magnifier glass, you can generate high T.

Thanks for doing the due diligence and finding the comments about the emissivity coating! A crucial factor which the liars would love to leave out.

325. Kev-In-ZA says:

@Joe, thanks, andl I learned quite a lot in this little engagement. So, it was well worth the head scratching that always ensures at the limits of one knowledge comfort zone. Self-doubt and curiosity are great teaching partners.

Hopefully I am also feeding back some useful insights to the many others that engage here.

Till later.

326. Kev-In-ZA says:

Seems @esttom has dropped off the radar, or no counter to empirical evidence of evacuated plat solar collectors (2021/05/02..2:03pm.) disproving multi pane simple flux addition theory.