## Hans Jelbring: The Stefan-Boltzmann Law and the Construction of a Perpetuum Mobile

Posted: February 25, 2012 by tallbloke in Astrophysics, atmosphere, Energy, methodology, solar system dynamics

Stefan-Boltzmann Law and the Construction of a Perpetuum Mobile

Hans Jelbring
BSc, meteorologist, Stockholm University, Civil engineer, electronics, Royal Institute of Technology, Stockholm, PhD, institution of Paleogeophysics & geodynamics, Stockholm University

Abstract

By analyzing two concentric spheres where one is placed inside the other some far reaching conclusions can be made concerning temperatures on two concentric spherical surfaces. The volume in between the inside surface of the outer sphere and the outside surface of the inner surface is of interest when there is vacuum in that volume.  These surfaces are assumed to emit electromagnetic radiation according to the Stefan-Bolztmann law. It is shown that the outer spherical surface will be heated by electromagnetic radiation to a temperature above the temperature of the inner surface.  The maximum difference of surface temperatures at power balance will depend on geometry, that is to say the ratio of the spheres.  The result implies that either it is possible to build a simple heat generator by small means for anybody or alternatively the Stefan-Boltzmann law is not applicable in a real world, at least not at close to ambient temperatures where the radiation isn’t directed towards empty space.

Background

Applications of the Stefan-Boltzmann law are important. The temperature of the surface of sun, temperatures in combustion chambers as well as the average temperature of earth as seen from space can be calculated (ref. 2).  The temperature measurements from satellites are adjusted to fit the power intensities predicted by the S-B law. The Planck derivation of that law can be found in reference 3. The history of its final derivation involves a number of steps:

“Josef Stefan in 1879 showed experimentally that the flux from a cavity in thermal equilibrium is proportional to the fourth power of the absolute temperature.” and “ Ludwig Boltzmann in 1884 derived this fourth power relation from thermodynamic theory. Until Planck’s work, there was no theoretical method of determining the constants of proportionality.”

Later Planck suggested that the power from a hypothetical “oscillator” was quantized and the modern S-B law was born and the “ultraviolet catastrophe” was removed.

There are a number of assumptions made to reach the final S-B law that fits the experimental data. In fact the S-B law had to be modified by Planck before fitting to experimental observations. It is hard to understand that any gas can radiate as a “cavity in thermal equilibrium”. It should also be hard for a perfect smooth black body surface to do so, too, especially at close to ambient temperatures in the atmosphere where convective processes always are at hand and mostly dominate the energy transfer between matter.

Professor Gerhard Gerlich (ref. 4.) stated two major arguments against the application of S-B law in the atmosphere:

• The constant σ appearing in the  T4law is not a universal constant of physics. It strongly depends on the particular geometry of the problem considered.
• The T4 law will no longer hold if one integrates only over a filtered spectrum, appropriate to real world situations.

In short, there are reasons to question the validity of the S-B law for physical situations which are not appropriate for its application, such as calculating the power of “Back Radiation” as done by Kevin Trenberth et al (ref. 5)

Despite these problems when applying the S-B law to all surfaces and gases in our environment the hope certainly is that Trenberth et al (ref. 5) are correct since then there should be technical possibilities to construct cheap perpetuum mobile machines for everybody.  It would be a service to mankind to provide energy almost as free as the air we breathe. It is admitted that such machines should be tested thoroughly before large scale production is to be started. Any hypothesis and theoretical model should be adequately tested before applied or declared true.

Methodology

A vacuum is assumed to exist between two concentric spheres. The two surfaces are designed of the same material and emits electromagnetic radiation close to that of a black body and their irradiance is equal to  P = e σT4 where   σ  is the S-B constant and e is emissivity.

The ratio between the temperatures of the sphere surfaces will be calculated when the spheres have reached an energetic equilibrium. Such is the case when each sphere is receiving/emitting equal total power according to the S-B law. Some postulates are needed:

1. Any point on both spheres emits photons in any direction within a half sphere with the same probability.
2. S-B law is valid in vacuum.
3. The radius of the inner sphere is R1. The radius of the outer is R2. R1 < R2.
4. The temperature of the inner and outer sphere is T1 and T2.
5. Any photon hitting a surface will get absorbed and not be reflected.
6. The two concentric spheres have equal masses m0 (for construction purposes only).
7. The surface area of a spherical cap is given according to figure 1.
8. The geometric configuration of the spheres is shown in figure 2.
9. The two spheres are perfectly insulated from both inner and outer surroundings.

Statement:

The spheres S1 and S2 cannot rest in a thermodynamic equilibrium, meaning that the temperatures of the surfaces of the spheres will be unequal.

Proof

Call the two spheres S1 and S2, their temperature as T1 and T2 and the power they emit P1 and P2. We are interested in what the temperatures will be when an energetic equilibrium has been reached between the spheres after an unspecified time.

Observe that for symmetry reasons the temperatures on both spheres have to be uniform, that is, constant on each sphere surface. The question at hand is therefore, whether T1 = T2 at equilibrium.

The power (P1) that radiates from S1 towards S2  is P1 = 4ᴨ R12 e σT14                                           (1)
The power (P2) that radiates from S2 is P2 = 4ᴨ R22 e σT24                                                                 (2)

All power emitted from S1 will reach S2 and get absorbed. The power emitted from S2 will only partially reach S1. Consider a photon being emitted from its surface as in figure 2. For symmetry reasons the sphere S1 will cover an equal space angle seen from any arbitrary point A on the surface of S2. Hence, a constant fraction of P2 will get absorbed by P1 for a specific α.  Denote that fraction by k, 0< k < 1. Observe that k = k(α).

In equilibrium P1 sends out as much power as it receives. We get:

4ᴨ R12e σT14 =  k 4ᴨ R22e σT24 which can be simplified.                                                                   (3)

(T1/T2)4  =  k (R2/R1)2  Observe  that R1/R2 = sinα according to figure 2.

(T1/T2)4  =  k /sin2α                                                                                                                                    (4)

The only way T1 can be equal to T2 is that k(α) = sin2(α)  which seems unlikely for any α.  Fortunately it is possible to calculate k(α)  which follows below.

The surface of a spherical cap can be calculated by the formula in figure 1. The spherical cap S in figure 2 can be calculated according to the same formula. The ratio of S to a half sphere with radius R2 cos α is equal to the proportion of photons from S2 that hits S1 and gets absorbed by it. That proportion is k(α).

S = 2ᴨ R2 cos α h

S = 2ᴨ R2 cos α (R2 cos α – R2 cos2 α)

S = 2ᴨ (R2 cos α )2(1cos α)   and hence                                                                                            (5)

k(α) = (1cos α)                                                                                                                                       (6)

Substitute equation (6) into equation (4) and we get:

(T1/T2)4  =    (1 – cos α)/sin2α                                                                                                                 (7)

This equation is valid for 0 <  α <  ᴨ/2 according to figure 2.  The right hand side of equation (7) is not equal to 1 for any α within the definition interval.  The monotonic function (1 – cos α)/sin2α has the end values 0.5 and 1.0. Hence T1 cannot be equal to T2 Q.E.D.  See also table 1.

Construction considerations

For constructing a perpetuum mobile some practical opinions might be of value. The inner sphere can be filled with water and the outer sphere can be surrounded by water. Water can be passed to the inner container from outside by a small pipe and then returned back. It will be chilled and the water surrounding the outer sphere will be heated. Equation (7) shows a temperature relation when an energetic equilibrium has been reached. It does not say anything about how long time that would take or how close to the equilibrium ratio it is possible to reach in a real construction.

Values in table 1 can be used for design purposes.

Table 1   (α  in degrees and T in Kelvin according to equation 7)

α                                                   T1/T2

1                                        1         0.841

2                                        5         0.841

3                                        10      0.843

4                                        20      0.847

5                                        30      0.856

6                                        40      0.867

7                                        50      0.883

8                                        60      0.904

9                                        70      0.929

10                                     80      0.961

11                                     90      1.000

The size of S1 need to be quite big to include a water reservoir (by preference) and hence, α should be rather big. For reaching the highest possible temperature difference between S1 and S2 α should be small. There has to be a compromise. Let´s just investigate an example where R2 is 2 meter and α is 50 degrees. Then R1 is 1.29 meter.  Assume that water of a temperature 10C (283K) is circulating through the smaller sphere in one circuit and water at the same temperature surrounding S2 in a second circuit. Interestingly it is hard to tell which if the outer sphere will heat or/and the inner sphere will cool (or both will happen) since the energy source is unknown.  Assume water within S1 will not cool and the water surrounding S2 will heat.  The theoretical maximum temperature difference is then 283/0.883K or 320.5K or 47.5C. Let´s assume that heat exchangers produce one type of temperature drop plus other practical obstacles produce others.  Still, it might be possible to achieve and maintain a 30K temperature difference between input water and output water in this trial unit of a perpetuum machine.  Several units can then be used in series to achieve higher end temperatures if needed. I would be very glad if anybody could get a perpetuum mobile to work if constructed along the suggested design.

It is acknowledged that the author is insecure of exactly why the contradictions between the results above and the first law of thermodynamics emerge. However, the results in this article should be seen in a wider context than what has been done in the postulates. Below are some possible explanations for the derived result and alternative consequences:

1. My calculations are wrong
2. The S-B law is not applicable at ambient temperature in an enclosed vacuum chamber
3. The S-B law is not applicable at (close to) ambient temperatures  in our atmosphere
4. The first law of thermodynamics is not valid
5. A perpetuum mobile can be constructed and work

As can easily be understood the consequences for science as such is enormous if my calculations happens to be correct.  In that case it is of utmost importance to discuss point 2-5 above very carefully by responsible scientists and apply this new knowledge to a multitude of inaccurate models now in use in the scientific community.

References

1.  From Encyclopedia Britannica

Stefan–Boltzmann law, statement that the total radiant heat energy emitted from a surface is proportional to the fourth power of its absolute temperature. Formulated in 1879 by Austrian physicist Josef Stefan as a result of his experimental studies, the same law was derived in 1884 by Austrian physicist Ludwig Boltzmann from thermodynamic considerations: if E is the radiant heat energy emitted from a unit area in one second and T is the absolute temperature (in degrees Kelvin), then E = σT4, the Greek letter sigma (σ) representing the constant of proportionality, called the Stefan–Boltzmann constant. This constant has the value 5.6704 × 10−8 watt per metre2∙K4. The law applies only to blackbodies, theoretical surfaces that absorb all incident heat radiation.

3.   J.S. Penn, Sonoma State University,  http://physastro.sonoma.edu/people/faculty/tenn/P314/BlackbodyRadiation.pdf

4. Gerhard Gerlich and Tscheuschner, Ralf D., Falsification of the Atmospheric COGreenhouse  Effect within the frame of Physics, version 2.0, July 24, 2007.

5. Kiehl, J. T. and trenberth, k.E., Earth´s Annual Global Mean energy budget, 1997, Bull. Amer. Meteor. Soc., 78, 197-208.

1. Nick Stokes says:

Firstly, this can’t be right. A cavity of whetever complex shape is in radiative equilibrium at uniform temperature.

The fallacy here is
“The ratio of S to a half sphere with radius R2 cos a is equal to the proportion of photons from S2 that hits S1 and gets absorbed by it.”

It assumes that a surface element at A emits uniformly in all directions and so the proportion of radiation captured is the area of S divided by a hemisphere.

But that is not so. The element at A is a disk, and does not emit uniformly in all directions. Seen from a point on S, the radiation from A varies according to the apparent area of the disk (at A) as seen from that angle. You have to integrate that over S.

It’s late here, and I would probably get it wrong – maybe in the morning.

2. wayne says:

Hans: by saying in the condition “5. Any photon hitting a surface will get absorbed and not be reflected.”, am I right assuming you mean the emissivities of both sphere surfaces are both exactly one?

For if the emissivity of a gas, at a given spatial concentration, is not strictly one the other portion not absorbed is either reflected or transmitted. Right?

3. I offer a much simpler, and well-verified, understanding, here.

For my latest proof, in the context of the true energy balance of the Earth+Atmosphere system, see:

“The True Energy Balance of the Earth_Atmosphere

4. Richard LH says:

1.

Looking across the inside of the outer sphere it would be impossible to tell how far away (if it even exists) of the inner sphere is. The inside has to be in thermographic equilibrium. i.e. the same temperature

5. Brian H says:

Simplified, some of the photons from S2 miss S1 and are reabsorbed by S2, whereas all photons from S1 hit and are absorbed by S2.

Since S2’s area is larger, this would seem to be necessary if temperatures are equivalent and held so.

6. B_Happy says:

Fundamental mistake : it is assumed that the spheres are in equilibrium when the temperatures are equal. No, the spheres are in radiative equilibrium when the POWERs are equal.

7. Nick Stokes says:

The field of view is as shown in Fig 2. I will take to to show the fate of radiation emitted by an element dA located at A. Because of spherical symmetry, integrating over dA is easy.

The total emission from dA is, by S-B,
dA σ T2^4.

Consider another element dS on S. The radiation received from dA is got by integrating over all such elements dS. Now Hans has assumed that all such elements receive an equal amount, so you get it by dividing the area of a hemisphere by the sub-area that receives the radiation.

There’s axial symmetry, so let dS be a ring, characterised by the angle θ made with the vertical. θ varies from 0 to α. Then
dS=2 p r^2 sin θ dθ ( r=R2 cos α)
The proportion is got by dividing the integral over S by the integral over the hemisphere. If you do that, you get a factor (1 – cos α) (as in Eq 6) which does not match the sin^2 α from k.

But in fact the radiation received by dS has to be scaled by a factor cos θ. This is the view factor; at dS the element dA is seen obliquely, and goes to zero when θ reaches p/2.

When you integrate
dS=2 p r^2 cos θ sin θ dθ
from 0 to α you get the factor (1 – cos^2 α) instead of (1 – cos α) as in (6). Then Hans’ expression (7) becomes
(T1/T2)^4 = (1 – cos^2 α)/sin^2 α = 1.

8. B_Happy says:

Sorry, but the temperatures cannot be the same. Without using any maths, just common sense, you can show this just by imagining the distance between the inner and outer surface being increased. As the outer surface is pulled away, the radiation from the inner surface is more and more spread out. So the power per unit area at the outer surface is less, so the temperature is less. If you have doubts just think about what happens when the outer surface is so far away that the inner sphere looks like a point. So only time the temperatures can be the same is when the surfaces are so close together that they are effectively in contact.

9. Nick Stokes says:

B_Happy says:
“Sorry, but the temperatures cannot be the same. Without using any maths, just common sense…”

No, common sense says they must be the same, and so does the second law of thermodynamics. As Hans suggests, if not you could run a perpetuum mobile.

But it is common experience. Furnaces, ovens – you don’t get cool bits because of the way objects are organised. An insulated furnace will be at uniform temperature, regardless of geometry. The fact that the view factor maths which Hans essayed gives that result is something of a miracle, but it must.

10. B_Happy says:

No, you cannot run a perpetual motion machine, because the POWER is the same. It’s the POWER that has to be balanced, NOT the temperature.

And, it’s not the uniform temperature of your furnace that is the thing to think about, but the temperature as experienced at a distance from it.

11. tchannon says:

I suspect the difficulty is using simplified SB

Opened out to parallel plates the paradox melts.

A geometry effect from curvature.

Lets switch to a square or cube. For a lot of that there are parallel plates and the corners become curious.

What happens at the apex of a corner?

12. B_Happy says:

“Opened out to parallel plates the paradox melts”

Of course it does – because then the areas are equal. It’s the difference in areas that causes the power per unit area to be different! There is no paradox when you allow for the different areas.

13. Nick Stokes says:

tchannon says:

“I suspect the difficulty is using simplified SB”

No, the form of temperature dependence plays no role here. It’s pure geometry and ray theory.

14. wayne says:

This seems but the classic blackbody camber example with a second surface, S2, within this larger first surface S1. No radiation can get within S1 and no radiation can escape outside S2. The temperatures are going to be identical even if the emissivities of S1 and S2 differ. I see no perpetuum mobile here in this example.

I have always viewed the SB law in radiation transferred as:

$\large \begin{matrix} flux_{actual}=\left [ \varepsilon _{warm}\,|\,\varepsilon _{cool} \right ]_{min}\; \sigma \; \left ( T_{warm}^4 - T_{cool}^4 \right ) \\ or,\;depending\;on\;the\;directed\;orientation \\ flux_{actual}=\left [ \varepsilon _{warm}\,|\,\varepsilon _{cool} \right ]_{min}\; \sigma \; \left ( T_{cool}^4 - T_{warm}^4 \right ) \end{matrix}$

and that leads you to believe that:

$\LARGE \begin{matrix}flux_{net}=|\varepsilon _a,\varepsilon _b|_{min}~\sigma\left(T_a^4-T_b^4\right)\\ _{WHEREAS}\\ power_{possible,\,unidirectional}=\varepsilon\sigma\left(T^4\right)\end{matrix}$

(Hope latex understands me here!) 8)

If the areas in consideration are also different then an area term must also be added.

It seems maybe a more inquisive example would be to let the emissivities be different. Let’s say the ε of S1 is 0.95 and the emissivity of S2 is 0.05, or the opposite. Are they then also identical temperatures? Even better, let exactly one-half of the radiation from S2 to be let out of the system as with our atmosphere. Then what would the example show.

To me the emissivities of gases are not like emissivities of solid surfaces, they also depend on the concentration or density of the gas if you are going to view them as a surface and transmission is added to the possibles. The emissivity of our atmosphere decreases radially as you move up toward space as the density is decreasing with altitude. That is how the temperature at the surface in a dense atmosphere gets to be higher than the more-simple black body calculation assumptions.

Or Hans, maybe I am missing your entire point in this article. If you see that, explain that a bit more.

15. B_Happy says:

Wayne,

What are the units of the Stefan-Boltzmann constant sigma?

16. wayne says:

Well, [W m−2 K−4] or [J/s/m2/K4]. Your point B? Stefan-Boltzmann relation in most real calculations are of two opposining powers, like two monter trucks nose-to-nose and the rate of transfer, (distance/second) in the trucks case, is which one has the greater power when both emissivities are one.

When the emissivities do not match the net flux rate is always limited by the weaker of the two. Just take two opposing walls in a square room, one black emissivity 0.95 and the other white of emissivity 0.05, they still maintain identical temperature when at equilibrium. So what is the flux rate between them and why. To me ‘THAT’ is where simple climate science discussions including SB jump the tracks. Might as well bring in Kirchholff’s law here for it is bound to come into this discussion.

Hans: I really like your post here, great topic! It’s going to make everyone think down to the very essence of radiation. Maybe myself, for I still find me questioning whether my view is exactly correct in some strange cases. Any and all input is welcome here.

17. greg elliott says:

Nick Stokes says:
February 25, 2012 at 9:22 pm
But in fact the radiation received by dS has to be scaled by a factor cos θ. This is the view factor; at dS the element dA is seen obliquely, and goes to zero when θ reaches p/2.

This looks like the correct answer. The sun at the poles does not have the same effect as at the equator. Thus the energy received by S1 is not simply a function of the area given by S, but also the angle of intersection.

18. B_Happy says:

Wayne,

My point is that you (and others) have omitted to notice the m^-2 in there. i.e it is the flux per area. Now go back and put the areas into your formula and rethink the conclusion.

Look it is easy to disprove this whole thing using a proof by contradiction. It is claimed effectively that the temperature of two separated bodies is independent of the distance between them. OK so
take two concentric spheres radius R1 and R2 as in the example. They are each emitting a creation amount of energy/second. Now make the outer sphere have radius twice as big. The area has increased by a factor of 4. So if the temperature is the same, then energy/second has increased by a factor of 4. This is obviously impossible. So the premise is incorrect. So the temperature cannot be the same.

19. wayne says:

B, you didn’t read my comment close enough, the comment was within about the areas (the latex was getting wide enough without them).

Of course, if you are now speaking of both surfaces not being ‘per square meter’ you would have to also include these areas. And if the time period was not ‘per second’ that would also have to be included to then get the amount of energy transferred, no longer power, but one second of transfer is usually just assumed. Wouldn’t you agree?

20. B_Happy says:

Wayne,

I did read your comment – my point is that it is wrong if the areas are not equal. In Jelbring’s example the areas are not equal. So the assertion that the temperatures are equal is false. And that I’m afraid is that.

21. Hans says:

Wayne says:
“February 25, 2012 at 3:07 pm
Hans: by saying in the condition “5. Any photon hitting a surface will get absorbed and not be reflected.”, am I right assuming you mean the emissivities of both sphere surfaces are both exactly one? For if the emissivity of a gas, at a given spatial concentration, is not strictly one the other portion not absorbed is either reflected or transmitted. Right?”

I guess you caught me being inconsistent about the emissivity. To make the reasoning easier let emissivity be 1.000 (a true black body). The problem will remain the same. The simplest answer to your gas question is that there is no gas at hand in the PM construction and I will get away easy.
To tell the truth, I put in emissivity when realizing that it would disappear directly when calculating the ratios between the absorbed powers at hand without thinking much more. The role of emissivity is of course important, One idea I did get is that geometry also cause a part of measured emissivity. Consider a smooth surface and one filled by drilled micro half spheres. These two surfaces don´t have the same areas so you directly have to introduce an “average” surface to get the S-B law to work if they have equal temperatures(?). Are surface roughness affecting the emissivity used in real life even if the micro surface is made of exactly the same material? This question is of interest but it was not what I wanted to discuss in this thread and I should have put emissivity as 1.0000 as I intended from the beginning.

As B points out, this is a pure geometrical problem. When you have accepted that the power absorbed from each sphere has to be equal in an balanced situation and that electromagnetic radiation is sent out with equal probability in all directions from any arbitrarily point (in a half sphere) the paradox seems unavoidable. I am very interested in getting more opinions from all of you since this paradox is a challenge to resolve and I have no direct answer myself. I would like to remind you of the reference to professor Gerhard Gerlich´s statement about the S-B law and geometry. I don´t know about the background of his statement but it seems very proper. He might have the solution to the problem at hand.

22. Hans says:

I have found a calculation error which will change the construction of the perpetuum mobile. It will not change the paradox that the spheres have to get different temperatures in a energy balance situation, though. I will be back on this topic later.

23. wayne says:

I’m sorry B but I don’t see your conclusion. The area missed by S2’s radiation is exactly the proportion of the S2/S1 area ratio of radiation actually passed between the two. Are you thinking this is not so? To me Hans’ “k” has to be equal to 0.6427876 in his example of R2=2 and a=50° and his k/sin^2(α) seems where he veered. I’m still trying to find what must be Hans’ flaw in the geometry logic but don’t have the time right now. Look to the 1+ex-secant, for in celestial navigation’s dip of the horizon; that is the same ratio if I recall, the rotation into three-d should not matter in this case.
http://en.wikipedia.org/wiki/File:Circle-trig6.svg

24. B_Happy says:

After using a long bus trip to think about this,and doing algebra in my head,I have decided I was incorrect! Nick Stokes back at 9:22 has the correct algebra, and the areas do match. But Jelbring is incorrect, and there is no paradox.

25. Hans says:

Wayne says:
February 26, 2012 at 8:24 am
“I’m sorry B but I don’t see your conclusion. The area missed by S2′s radiation is exactly the proportion of the S2/S1 area ratio of radiation actually passed between the two. Are you thinking this is not so?”

Yes, I am thinking it is not so if I understand you correctly. The half sphere of interest around any point on the outer spherical surface A has the area 2pi (R2cos(alfa))^2 and the radius R2cos(alfa) as in figure 2.

The point is that S is a part of the half sphere above and the area of
S is 2pi R2cos(alfa)(R2cos(alfa)-(R2cos^2(alfa))). The parenthesis is h in figure 2. (see also figure 1).

The latter is simplified as 2pi R2^2 cos^2(alfa) (1-cos(alfa))
The ratio of these two surfaces is 1-cos(alfa). In the figure alfa is approximately 45.3 degrees
making (1-cos(alfa)) = k; k = 0.297. 29.7% of the emitted photons from any arbitrary point A at S2 will get absorbed by S1. If I am wrong here I am certainly wrong altogether. The mistake I discovered in my calculations is another one. I will be back tomorrow due to time limitation. Hopefully I got it right above.

26. wayne says:

“One idea I did get is that geometry also cause a part of measured emissivity. Consider a smooth surface and one filled by drilled micro half spheres.”

Heh.. yeah, a golf ball surface! 🙄 That might make a great future post on the effects of emissivity in the real world surfaces and how that plays into the SB when you properly use it. If you look at a sandy surface with a good magnifying glass you also see many dark near ‘black body’ deep cavities and also that most radiation from the grains right on the surface radiate against the neighboring grains side surfaces, not upward and out. To me that plays directly into why there is no such thing as “back radiation” somehow “slowing” radiation. Didn’t mean to drag it OT Hans, it’s just there are so many questions right here on this subject that never get asked, or if asked, never really get answered.

But on this subject, I do think you have a mistake somewhere, like, what geometry brings in you sin^2(a). I couldn’t follow what equation you hinged that on. Seems more of a exsecant, 1/cos(a)-1, relation to me. Do look up a few sites on “dip of the horizon” for that I think is really the factor, ratio wise, we are speaking of.
As: http://mintaka.sdsu.edu/GF/explain/atmos_refr/dip.html
It is this ratio that ‘misses” S1 from S2. Well, just see the diagrams.

27. AusieDan says:

Are there no practical people here who have actually tried to make something work?
What about the energy to heat one lot of water and to cool the other?

At very best, the energy required would exactly equal the energy generated, if as well as having perfect spheres, perfect vacuums in between and perfect black bodies, and I nearly forgot, the ability to keep the two spheres perfectly aligned and perfectly insulated from all outside influences.

No, on second thoughts, I will float an new mega power (no liability) company based on this good theory.
I will take subsctiptions from each of you, for shares in said company in billion dollar lots or in such smaller magnitudes as may equal but not exceed your net worth on the very practical planet.

Someone once said that a good theory should contain tno more than the very least number of variables necessary to accurately describe the project. But no less either.
Tennis anyone?

Now back to dreaming (I meant climate sciencecy sort of thinggy).

28. Genghis says:

This is all too funny. Of course the center sphere will be hotter. It is exactly the reverse of a parabolic dish, or exactly like a magnifying lens focusing the suns rays.

It also explains the temperature of the earths core and the ‘greenhouse’ effect.

And of course it is a perpetual motion machine, exactly like the Earth is perpetually orbiting the sun and it will stay a perpetual motion machine as long as no work is done.

29. wayne says:

Hans, after reading your last comment I now see what your diagram was saying, from point A inward. Funny, I was concentrating on what was missing S1 from point A. Different strokes I guess. Though after reading Nick’s breakdown of the geometry, I can now see his logic… a much more elegant approach.

30. 40 Shades Of Green says:

It appears you won European blog of the year See @bloggies tweet feed.

A bit of a sceptical sweep. Jo Nova has won for Oz and NZ and Steve McIntyre for Canada

The responses above to Hans Jelbring’s piece are very depressing, diving as they do into needless mathematical complexities. Hans is not proposing that a perpetual motion machine is feasible using his technique. He is using the obvious violation of the laws of thermodynamics that it would entail to challenge people to find a flaw in his geometrical/mathematical reasoning.

Here is my response to his challenge. It is very general in its claim and relies on physics rather than on mathematics.

ASSERTION
Two bodies of any kind (black, grey, blue, green, any size, any shape, and made of any material) which are enclosed (including one within the other) anywhere in an evacuated container ‘universe’ (whose walls are by definition 100% reflective) and which start out at different temperatures will in time equilibriate at an intermediate temperature.

PROOF
1. Any body above absolute zero temperature continually emits streams of photons from its surface.

2. For every photon that is emitted from the body, its ‘energy budget’ is fractionally reduced.

3. For every photon that is absorbed by the body, its ‘energy budget’ is fractionally increased.

4. For every photon that is emitted and then absorbed by the same body, there is no change in its ‘energy budget’.

5. On average (Wien’s Law) a hotter body emits photons that are of a higher frequency than those emitted on average by a cooler body.

6. Higher frequency photons contain a larger amount of energy than lower frequency ones.

Over time, as the bodies exchange packets of energy, it is obvious that the two bodies must equilibriate. The hotter body will continually receive (averagely) lower energy packets from the cooler body. The cooler body will continually receive (averagely) higher energy packets from the hotter body. So over time the hotter body will contain less and less energy, thus cooling down, and the cooler body will contain more and more energy, thus warming up. This will go on until the two bodies are at the same temperature, whereupon they will happily carry on exchanging photons of (averagely) the same energy for ever.

End of debate.

32. tchannon says:

33. DocMartyn says:

I suspect that the best way to know if a scientific theorem is correct and valid is to examine the technology that results from it.
My paternal grandparents had a coal fire, a legacy of the retirement package of coal miners.
As children we would also hear the playful voices of the adults:-
“leave the damned poker alone”
We would of course place the iron poker in the hottest part of the fire and observe the tip become red hot. The odd thing is one could touch the handle, it would be hot, but not burn the skin.
Now it is quite obvious that there was a temperature gradient along the length of the poker, the tip in the hot coals would be glowing red and the far end would be uncomfortable to hold.
So what?
Well, what we were observing was a steady state, which some people often mislabel as an equilibrium. One end of the bar is heated, heat is transferred through the iron atoms. Those at the surface can radiate and cool the poker. There is a continuous gradient of heat and a continuous gradient of S-B emission spectra. However, if we were to take a cross section of the poker at any point we would find that there was a second temperature gradient. The center of the iron mass would be hotter than the outer surface. It is far easier for energy to leave the surface than it is for the center.
The S-B equation works up to the moment that the system is measured. Take a pure silver sphere with 10 thermocouples at 1/10th radius and warm it to 20 degrees at the international space station. Throw it out of the air lock and monitor its cooling. The outer surface is going to cool far more quickly than the surface. The thermal conductivity is not high enough to transfer heat from the interior to the outer surface, which is radiating IR.
Now, during the cooling process, calculate the ‘average’ temperature and the surface temperature, from the S-B emitted spectra. Notice anything? Indeed, the ‘average’ temperature is far higher than the S-B temperature.
Same with our poker. Our poker’s emission spectra is the sum of the emission spectra of the temperature gradient along the surface, it ignores the temperature along its bulk inner phase. S-B is instantaneous with respect to the surface, but not to the bulk phase.

34. Hans says:

“End of debate.”

It is far from an end of the debate in this case. Look at my next comment of you prefer qualitative statements.
I notice that Socrates….. won´t approach the heart of the problem which is that the S-B times an area tells exactly what the power emitted from that area is. I have nothing against qualitative arguments as long as they can be followed easily. Calculations as the one I have performed have the benefit to be possible to check easily by those who knows how to use to the mathematical language. I propose that Socrates…. do tell what is wrong with the calculations that are done before introducing new red herrings. The calculations are done for a reason and that reason is to facilitate checking of the statement made. You are welcome to tell where I am wrong and then we can discuss your proof.

35. Hans says:

Qualititive arguments to prove that the outcome of the S-B law depends on geometry.
P = e σT^4 has to be valid of the S-B law has any meaning. All surfaces have emissivity = 1.0
The surface are considered to have no radiative losses at the edges of the surfaces. The surfaces are thermally insulated from the surroundings.

Case I
Consider two large square plane surfaces with areas AxA at a distance d from each other. A >> d.
We don´t bother about small “energy” leakage around the edges of the surfaces. An energy balance can be found for the same T at the two surfaces when P1 = P2.
In this case it is not hard to defend that P1 = e σT1^4 AxA and P2 = e σT2^4 AxA for P1 = P2 leads to T1 = T2.

Case II
Consider two long pipes with radii R1 and R2 where R2 > R1 and with the same length L. The smaller is concentrically placed inside the bigger one so they face each other everywhere. We don´t bother about small “energy” leakage at the ends of the pipes.
Approximately P1 = e σT1^4 2pi R1 L and P2 = e σT2^4 2pi R2 L. The basic condition is that P1 has to be equal to P2. When this is the case it follows that T1^4 R1 = T2^4 R2 or
(T1/T2)^4 = R2/R1.
Since 0 < R1 < R2 it follows that T1 cannot be equal to T2.

Case III
This is the case presented at the head and it has the advantage that I don´t need to treat any "energy leakage" at the ends fo the surfaces as has to be done in case I and II. I stand by what I have stated in the head of this thread and that there exist a paradox that has to be solved. If the calculations are wrong it should be easy for a professional mathematician to find the error(s).

Game over and the start of a serious debate of the applications of S-B law in science and especially in climatology is warranted.

36. Hans says:

wayne says:
February 26, 2012 at 10:58 am

“…..what geometry brings in your sin^2(a).”

Excuse me for limiting my answer. See my other two comments and come back if you are not satisfied.
sin(a) is by definition R1/R2 according to figure 2 in the head.

37. Berényi Péter says:

“The only way T1 can be equal to T2 is that k(α) = sin^2(α) which seems unlikely for any α.”

However unlikely, it is the correct answer. You simply have to take into account Lambert’s cosine law.

38. tchannon says:

DocMartyn,

The critical point, a generalism, is SB is for a surface which is unknown unless a very definite connecting law with a body is known.

Worse, this varies with wavelength.

I don’t have a publishable image to hand (would break ethics) of the effect of a very thin layer.
You will know store printed supermarket stick on barcode labels. Some of these use thermal printing. The paper is black with a very thin white overlayer.

Unfortunately an IR barcode reader sees right through the white layer, red isn’t too good either so these are best for the blue/green laser scanners. This why scanning is often poor with this kind of label.

I wanted to point at layered IR filter construction but it’s hard to find anything, all the manufacturers hide the details. Spluttering of various materials under vacuum to form a sandwich of many layers, might be dozens.

39. Nick Stokes says:

Berényi Péter says: February 27, 2012 at 4:19 pm

“However unlikely, it is the correct answer. You simply have to take into account Lambert’s cosine law.”

Thanks, B – that is the factor I was using here. I didn’t know it had a name.

Fig 1 of that Wiki reference shows how the radiation received on S is variable. And the integration I did is the same as this, except that my outer limit for integration was α instead of π/2.

Hans’s Case II with the long pipes can be done similarly. There he has even neglected the fact that P2 is not equal to P1, because some of the radiation from the outer pipe does not impinge on the inner pipe. This at least was correctly treated in the sphere case. If you take account of that, and Lambert’s Law, you will get the correct result that temperatures must be equal.

40. I don’t see why application of the SB equation or indeed any maths is necessary. At equilibrium, the outer sphere is radiating according to its temperature. It’s receiving radiation from two sources, itself, and from the inner sphere, and the inner sphere is simply radiating what it receives from the outer sphere, so there is no net flow between the spheres, hence equilibrium. As the inner sphere is receiving radiation from the larger outer sphere, but has a smaller surface area, its temperature will be higher than the outer. Equal temperature is not necessary for thermal equilibrium, just equal energy flows, and they are equal.

It’s akin to a spherical mirror “focussing” reflected energy onto a smaller area. I haven’t yet spotted the fallacy in the argument that the outer sphere must be hotter than the inner.

Uniform temperature wrt to flat parallel surfaces is only obtained if the emissivities are identical; if they are not then equilibrium will be obtained with differing temperatures, the surface with the lower emissivity becoming hotter in order to emit what it receives.

41. Hans says:

Berényi Péter says:
February 27, 2012 at 4:19 pm
“The only way T1 can be equal to T2 is that k(α) = sin^2(α) which seems unlikely for any α.”
However unlikely, it is the correct answer. You simply have to take into account Lambert’s cosine law.

Stating that my calculation is wrong doesn´t help very much especially when it is possible to show that your statemnet is wrong which already is done in the proof text. I also stated “Fortunately it is possible to calculate k(a) which follows below”, and equation (7) is the result. It shows that (T1/T2)^2 = (1-cos(a))/sin^2(a). It means that T1 cannot be equal to T2 since the right hand side cannot be 1 for ANY a.

Please, show why you state that my calculations are wrong and wait with such a comment until you have shown that it is a fact. If you can prove what you claim, do so and I will be greatful. Follow the equations (4) to (7) and point out where and why I am wrong. Also explain why Lambert´s cosine law is applicable in this case.

42. Hans says:

MostlyHarmless says:
February 27, 2012 at 7:26 pm

“I don’t see why application of the SB equation or indeed any maths is necessary.”

“It’s akin to a spherical mirror “focussing” reflected energy onto a smaller area.”

“Uniform temperature wrt to flat parallel surfaces is only obtained if the emissivities are identical”.

Yes the emissivities are equal and for simplicity you can consider it 1.0
The discussion on this thread makes sure that math is needed. The derivation of the S-B law itself is based on a lot of math and postulates. Math is needed to scrutinize its validity as far as I can see.
I don´t see what your second comment has to do with the situation at hand. All emitted photons are absorbed and none is reflected. Any photon that is emitted is directed homogenously in a half sphere.
The basic relation that has to be paid attention to is what relates to power emitted and received from the two spheres.

Equation (3) is based on the fact that the absorb power on S1 has to be equal to emitted power from S1 at a power balance. If there is no power balance one of the spheres would be heated without limits. S-B law deals with power/area.

The calculations shows that that at a power balance T1 cannot be equal to T2.

43. Nick Stokes says:

Hans, I have been explaining that over and over, quite precisely, but you seem to be deliberately ignoring it.

The ratio of S to a half sphere with radius R2 cos a is equal to the proportion of photons from S2 that hits S1 and gets absorbed by it.
is wrong. It is at this point that you must apply Lambert’s Law. Different points on the surface of S see the surface element at A at different levels of obliquity, and you must allow for this in the integral with a factor of cos θ. So your Eq 5 is wrong; it should have cos^2 α, not cos α.

44. tchannon says:

Hans,
What Nick says is part of what I have in mind. SB often silently includes assumptions about shape, ignorable sometimes.

45. DocMartyn says:

Take a solid Pu/Pb sphere in vacuum with a surface area of 1 m2, which thanks to radioactive decay has a temperature of 269.7 K and so is radiating at 300 W/m2.

Take a pair of hemispheres, 1,000 atoms thick, with a combined surface area of 2 m2 and encase the central sphere.

Under equilibrium conditions the outer sphere surface must be at 226.8 K and so radiate at 150 W/m2 outward. It must also radiate 150 W/m2 inward.

What is the equilibrium temperature of the inner sphere?

46. Hans says:

Nick Stokes says:
February 27, 2012 at 8:17 pm

No Nick, I am not ignoring your statement but I have not understood it. you seem to accept that the power P2 exist as I have written it.

Will you please, calculate the ratio of the power P2 emitted from the sum of any A (all the S2 surface) that is absorbed by S1 according to how you see the physics in this case. I believe that we agree that there has to be a solution to the problem and that it is possible to express quantitatively. This means that you also admit that S1 is absorbing a part k of P2. Now it is your turn to calculate
k(a) and show how you do it.

47. Hans says:

DocMartyn says:

February 27, 2012 at 9:02 pm

“Take a solid Pu/Pb sphere in vacuum with a surface area of 1 m2, which thanks to radioactive decay has a temperature of 269.7 K and so is radiating at 300 W/m2.
Take a pair of hemispheres, 1,000 atoms thick, with a combined surface area of 2 m2 and encase the central sphere.
Under equilibrium conditions the outer sphere surface must be at 226.8 K and so radiate at 150 W/m2 outward. It must also radiate 150 W/m2 inward.
What is the equilibrium temperature of the inner sphere?”

This example is great! The reason is that proposed solutions might help to illustrate how and why we understand the physics at work differently when applying the S-B law to power exchange between solid bodies at ambient temperatures.
A benefit of DocMartyn´s experiment is that it can be constructed and tested in real life although it will be cheaper to replace the radioactive material with a light bulb inside the inner surface or power from resistors in the inner surface.

Personally I would like anybody (many of you) to solve the problem designed by Docmartyn and let all of us have a look at the solutions. It should be noticed that DocMartyn assumes all(?) surfaces involved being black body emitters. The outer shell is so thin that it has a single temperature on each side. Meanwhile I will work on the solution myself.

48. Nick Stokes says:

Hans says: February 28, 2012 at 6:57 am
“Now it is your turn to calculate
k(a) and show how you do it.”

OK, I’ll write it out differently. I’ll show that for uniform temperature T, P1=P2.

Taking your Fig 2 to represent the radiation from an element dA placed at A, the amount actually absorbed by S (and hence S1) is

∫α0 I(θ) dS = ∫α0 2π r2 I(θ) sinθ dθ

where I(θ) is the radiant intensity in direction θ. You assumed this was independent of θ, but Lambert’s Law says it is proportional to cos θ.

I=Imax cos θ

and so the power from dA

dP2 = ∫α0 2π r2 Imax cos θ sin θ dθ = π r2 Imax (1 – cos2 α) = π r2 Imax sin2 α

We can work out Imax by extending S to a hemisphere, so α goes to π/2.

Then total rad from dA = dA σ T4 = π r2 Imax

Eliminating Imax:
dP2 = sin2 α dA σ T4

Integrating over dA is easy because of spherical symmetry. Just replace by area 4π R22:

P2 = 4π R22 sin2 α σ T4
= 4π R12 σ T4 = P1

So for uniform T, the power P1 emitted by S1 is equal to the power absorbed from S2, and there is equilibrium.

49. Nick Stokes says:

OK, it seems sub/sup don’t work for comments here. I’ll use Latex style:

Taking your Fig 2 to represent the radiation from an element dA placed at A, the amount actually absorbed by S (and hence S1) is

∫_α^0 I(θ) dS = ∫_α^0 2π r&sup2; I(θ) sinθ dθ

where I(θ) is the radiant intensity in direction θ. You assumed this was independent of θ, but Lambert’s Law says it is proportional to cos θ.

I=I_max cos θ

and so the power from dA

dP2 = ∫_α^0 2π r&sup2; I_max cos θ sin θ dθ = π r&sup2; I_max (1 – cos&sup2; α) = π r&sup2; I_max sin&sup2; α

We can work out I_max by extending S to a hemisphere, so α goes to π/2.

Then total rad from dA = dA σ T^4 = π r&sup2; I_max

Eliminating I_max:
dP2 = sin&sup2; α dA σ T^4

Integrating over dA is easy because of spherical symmetry. Just replace by area 4π R2&sup2;:

P2 = 4π R2&sup2; sin&sup2; α σ T^4
= 4π R1&sup2; σ T^4 = P1

So for uniform T, the power P1 emitted by S1 is equal to the power absorbed from S2, and there is equilibrium.

50. Nick Stokes says:

And it doesn’t recognise &sup2; – hope readers will.

51. Nick Stokes says:

No, it’s too hard to read. Here once more with ^2. Hope someone will rub out the earlier efforts:

OK, I’ll write it out differently. I’ll show that for uniform temperature T, P1=P2.

Taking your Fig 2 to represent the radiation from an element dA placed at A, the amount actually absorbed by S (and hence S1) is

∫_α^0 I(θ) dS = ∫_α^0 2π r^2 I(θ) sinθ dθ

where I(θ) is the radiant intensity in direction θ. You assumed this was independent of θ, but Lambert’s Law says it is proportional to cos θ.

I=I_max cos θ

and so the power from dA

dP2 = ∫_α^0 2π r^2 I_max cos θ sin θ dθ = π r^2 I_max (1 – cos^2 α) = π r^2 I_max sin^2 α

We can work out I_max by extending S to a hemisphere, so α goes to π/2.

Then total rad from dA = dA σ T^4 = π r^2 I_max

Eliminating I_max:
dP2 = sin^2 α dA σ T^4

Integrating over dA is easy because of spherical symmetry. Just replace by area 4π R2^2:

P2 = 4π R2^2 sin^2 α σ T^4
= 4π R1^2 σ T^4 = P1

So for uniform T, the power P1 emitted by S1 is equal to the power absorbed from S2, and there is equilibrium.

52. DocMartyn says:

Nick, I wonder if you have a solution to my Pu/Pb sphere problem?

53. Nick Stokes says:

DocM
Yes. There’s an easy way. The geometry has fixed effect, so we can say, in HJ’s notation, P1=P2+300
and P1=4πσR1^2T1^4; P2=4πσR1^2T2^4

Since you’ve obligingly worked out the in space temp for S1, it’s just a pythagoras-like formula
T1^4 = T2^4 + 269.7^4 = 226.8^4 + 269.7^4
T1= 298.48

54. DocMartyn says:

I am a bit slow Nick, so can we do this slowly.

My inner sphere initially radiates 300 W as it is at 269.7 K,

Starting at 4K the outer shell starts to warm, radiating inwardly and outwardly at the same power.

As we have a simple system I know that when the system comes to steady state, the outer shell will radiate away 300W, and as the external surface area is 2 m2, thus the temperature of the outer shell has to be 226.8 K. At this temperature the outer shell will radiate 150 W/m2 outward, so we have the conservation of energy, and 150 W/m2 inward.

You state that the inner Pu/Pb core will arrive at a steady state temperature 298.48K, thus it radiates a total of 450 W, (300 W from radioactive decay and 150 W from heating from ‘black radiation’)

The inner surface (2 m2) therefore gets 225 W/m/2 and and so attains a steady state temperature of 251 K, yet the outer surface has to radiate 150 W/m/2, not 225 W/m/2, outward

So Nick, I don’t see how the inner surface gains 225 W/m2 and emits 225 W/ms and the outer surface emits 225 W/m2 and gains essentially nothing from space. We know that the outer surface MUST radiate 150 W/m2. If we have a black body, then the inside MUST also radiate 150 W/m2.

So can you tell me the heat fluxes in your steady state solution?

55. B_Happy says:

DocMartyn,

Your problem is not specified completely, Even though the outer sphere is only 1000 atoms thick, that is enough to prevent heat transfer by radiation. So heat must pass from inner to outer surface by conduction. You need to know the thermal conductivity, and given that you can then calculate that the inner surface of the outer sphere is hotter by some amount than the outermost surface. Then you take it from there….

56. Nick Stokes says:

Fluxwise, you can work it out this way.
S1 emits 300+P W/m2. S2 emits 150 W/m2, of which P W is incident on S1. What is P?

You know that because if S1 had no power source, and S2 was held at 226.8K, then P, the amount incident on S1, would be the same as the amount emitted. It’s determined by source and geometry. And we know (or should) that S1 would also be at 226.8K. So that gives P as the emission from S1 at that temp.

So that’s why the pythagoras-type formula works. It’s good for general shapes, as long as they conduct well enough to be at uniform temperature. You can just balance for S1:
If T0 is the temp for S1 radiating to space (300 W/m2)
T1 the temp for S1 to shell (300+P)
T2 the temp for S1 radiating just P (226.8)
then each flux is refated by the same S-B coefficient
flux = αT^4 (doesn’t matter what α is, as long as it’s constant)
Then
αT1^4 = 300 + P
αT0^4 = 300
αT2^4 = P
So T1^4 = T0^4 + T2^4
You can say that without knowing P. But in your case, it’s just 1 m^2 *σ 226.8^4 = 150 W.

57. Nick Stokes says:

DocM,
I thought of a simpler, pure flux argument. Suppose the Pu is in the shell rather than the center. The shell still radiates 300W out and in (150 W/m2), and is at 226.8K

Then S1 is also at 226.8K (equilibrium) and so emits and receives 150 W.

What if the Pu is in S1. The shell is at 226.8K and so S1 still gets 150W. So it emits 450W, and that determines its temp.

58. DocMartyn says:

Sorry Mick, B_wise is closest to the truth. In steady states there will not be a uniform distribution of heat in either system. The sphere will have a hot core and a thermal gradient as you move to the surface. Again with the second shell, the inner side will be hotter than outer face.
We actually observe these discontinuous gradients in places like steel mills where the outer surface of a bar cools a glows in the IR, but the inner bulk is red/orange. The interior of the planet is warmer than the surface, yet maintains a constant steady state gradient.
You cannot get a functional solution to the problem if you assume that the outer shell has a uniform temperature.

59. Nick Stokes says:

DocM
” the inner side will be hotter than outer face”
This is moving the goalposts a lot. You said yourself that both faces emit 150 W/,2.

In fact the temperature gradient inside S1 doesn’t matter to the solution.

60. B_Happy says:

DocMartyn

“In steady states there will not be a uniform distribution of heat in either system”

Which system are you referring to? Jelbring’s system is fully specified, but not physically realistic, and the temperatures of both spheres are equal.
Your system is not fully specified, unless you assume infinite conductivity, in which case Nick Stokes answer is correct. If it does not have infinite conductivity, then the spheres are even hotter inside, but by an amount that cannot be stated unless you assign a value to the conductivity.

61. DocMartyn says:

The solution to S_B ignores heat transfer rates from the bulk mass to the radiating surface. S_B is only applicable to an equilibrium state, and not to a steady state.
At steady state the outer sphere surface must emit 150 W/m2, however, the inner surface cannot radiate at 150 W/m2 and yet have influx/efflux rates that are equal

62. D J Cotton says:

Incorrect assumptions about how to apply the S-B Law lie at the heart of incorrect physics which abounds in the climate debate.

The greenhouse conjecture will not be debunked for a long time by actual climate data. But it can be debunked right now by new physics which is extending the work of Einstein and Planck.

Firstly we must recognise that radiation from a cooler atmosphere to a warmer surface is comprised only of standing (or stationary) waves which may be thought of as opposing waves along the same path between two particular points, one on the surface and one in the atmosphere. These opposing waves interfere iff they have the same frequency and amplitude.

In Wikipedia we read … It can arise in a stationary medium as a result of interference between two waves traveling in opposite directions. In the second case, for waves of equal amplitude traveling in opposing directions, there is on average no net propagation of energy.

In addition to the standing (or stationary) wave, there is also one way radiation from hot to cold and its frequencies are represented by the area between the Planck curves, which is the same as SBL effectively calculates by subtracting the area under the smaller (cooler) curve from the area under the larger (warmer) curve.

Standing waves cause resonant “vibration” between energy levels and the energy required to excite = energy emitted on relaxation for such standing waves.

So how could any extra energy appear from nowhere and get converted to thermal energy? A whole new and different process is required for that conversion. Climatologists seem to keep imagining physical vibration causing friction or something. It’s not like that. Energy cannot be created in the process of resonance associated with standing waves.

All radiation from the cooler atmosphere to the warmer surface comprises standing waves transferring no net energy either way. Only the additional “top portion” of the radiation from the warmer surface is separate radiation which does cause heat transfer from warm to cool.

I warned you at the outset that Claes Johnson’s Computational Blackbody Radiation is ground breaking physics extending the work of Einstein and Planck. You are not going to find it in textbooks, but that doesn’t mean it’s wrong. There is far more to it than just imaging a lot of identical mass-less photon particles crashing into surfaces and transferring thermal energy.

Any textbook which tells you that radiation between two plates transfers the full SBL amount in each direction is wrong, because there simply cannot be any transfer of thermal energy along a different path from cold to hot as it violates the Second Law. “Net” radiation has no corresponding physical entity and is thus meaningless.

Only standing waves have an identical path and can thus interfere with each other if they have equal frequency and amplitude, as explained in Wikipedia.

The Second Law applies to every individual path between two particular points. Standing waves may be considered as two opposing waves, but they do of course have the same path, and that makes all the difference. It’s up to you whether you want to take an interest in these new developments in physics or stick to your old beliefs so you can feel good trying to prove the IPCC wrong using climate data and yet still agreeing with them that heat transfers from the atmosphere to a warmer surface. It doesn’t..

63. Nick Stokes says:

DocMartyn,
“S_B is only applicable to an equilibrium state, and not to a steady state.”
Nonsense. S-B simply relates radiation to temperature, at a point in time. There is no restriction to equilibrium.

The shell conductivity is a red herring. You specified 1000 atoms – in the micron range. 300 W/m2 going through iron, say, conductivity 55 W/m2/K would create a temp diff of about 0.00003 K.

Do you have any ideas about a solution? To propose a problem, press for a solution, and then say, tricked you, I didn’t specify everything, seems pretty juvenile.

64. DocMartyn says:

I asked you for the solution Nick. You stated that the inner sphere radiated 450 w/m2, and so that the inner surface of the shell received 225 w/m2 and that the outer surface radiated at 150 w/ms.
A decaying radioactive source is not at equilibrium, it continually converts mass to heat.
The idea of a thought experiment is to make one think. S_B does not describe this system very well does it Nick?

65. Nick Stokes says:

Doc
“You stated that the inner sphere radiated 450 w/m2, and so that the inner surface of the shell received 225 w/m2 and that the outer surface radiated at 150 w/ms.”

No, I said nothing about 225 W/m2. In fact the inner surface receives 300 W/m2, as it must. 225 W/m2 from the central sphere, and 75 W/m2 from itself. Of the 150 W/m2 it radiates, 75 W/m2 hits the centre body S1, and 75 W/m2 hits the shell.

That actually says nothing about S-B. And the half-life of Pu is irrelevant. You posited a power of 300 W. It’s just juvenile quibbling.

66. DocMartyn says:

‘In fact the inner surface receives 300 W/m2, as it must”

Sorry Nick, I must have framed my question very poorly.

The inner surface of the shell is twice that surface area of the sphere; as the inner surface of the shell initially receives 300 W in total; influx begins at 150 w/m2.

We know that at steady state the inner surface influx must be equal to the outer surface efflux as the shell is thin and has the same surface area on both sides and is at a constant temperature; 226.8 K.

So, at t<0, the inner sphere radiates 300 W/m2, and gains nothing from the 4K background temperature of space.

Then we add the shell and wait for the system to come to steady state.

At steady state the outer surface of the shell, which has a total surface area of 2 m2, must be radiating at 150 W/m2, i.e. a total of 300 W into space, and gains nothing from the 4K background temperature of space.

At steady state the inner surface of the shell, with a total surface area of 2 m2, must be radiating at 150 W/m2; i.e. a total of 300 W inwards.

The ratio of cross sectional areas is 2:1, so the inner Pu/Pb sphere receives 200W (2*(2/3*150 W)) back radiation from the inner shell and the inner shell receives 100W (2*(1/3*150W)) back radiation from the inner shell.

This suggests that the 1 m2 sphere surface, which initially radiated 300 W, must now be radiating 500W.

So what is the steady state temperature of the inner core?

67. David Socrates says:

D J Cotton says: February 29, 2012 at 11:14 pm

If you want to contribute something to the debate, the best thing to do is to explain yourself properly. I read through you piece three times and decided that your piece was incomprehensible. I do suspect that Claes Johnson is on to something but he makes no effort either in his writings to explain properly his theory.

If you really know what you are talking about, how about trying again. I am sureTallbloke would like a (coherent) article from you. If there is really something in what you (and Claes J) say this could be highly material not just to the debate associated with Hans Jelbring’s ‘paradox’ being discussed here on this thread but also much more widely in relation to the new science being undertaken by Nikolov and Zeller and others.

I think you and/or Claes owe the skeptic community a proper exposition of your ideas.

All the best
David S

68. Nick Stokes says:

DocM,
Your problem as originally stated was fine. It was a clear and soluble problem, and you went a long way toward solving it yourself. The issue is the extra things you brought in afterward. Resistance of the shell, decay rate of plutonium, and now I see 4K mentioned. If you want those taken into account, they should be specified in the problem statement.

In your latest, I think the reasoning involving 2/3 is wrong. It’s really very simple, and can be stated for fairly general shapes and arrangements, as long as each body is at (its own) uniform temperature.

Suppose power source P watts, inner body surface area A1, shell A2 (all SI units). The shell emittance must be P/A2 out, and also in (same temp). If A1 did not have a power source, but A2 was still emitting P/A2, then A1 would have to be at the same temp as A2, and hence emitting and receiving P/A2 W/m2. The fact that the power comes from A1 does not affect the amount received, so the power emitted is P + A1*P/A2. The emittance is P*(1/A1+1/A2), or in your case, 1.5*P W/m2.

Then it’s just a S-B calc to get T.

69. DocMartyn says:

So, the temperature of the inner sphere rises to a steady state level of 298.5K, pumping out 450 W; hence 450 W/m2.

This means that 225 W from my inner sphere falls on each m2 of the inner shell.

Which is more than the 150W the inner sphere must be radiating if it is at steady state.

70. Nick Stokes says:

If A1 (inner) had no power source, and the outer shell was at 226.8K, then it would be at 226.8K too, and radiating and receiving 150W (from S-B and power balance).

With a 300W power source, it still receives 150 W – same environment. But it emits 450W.

71. Frank says:

Hans wrote: “The constant σ appearing in the T4 law is not a universal constant of physics. It strongly depends on the particular geometry of the problem considered.”

Where in Reference 4 do you think Professor Gerlich makes such an absurd statement?

σ is a constant. It’s equal to 2*PI^5*k^4/15*c^2*h^3, where k is the Boltzmann constant, h is Planck’s constant, and c is the speed of light in a vacuum. These are all constants.

The only thing that changes is emissivity (e). W = σ*T^4 for an (ideal) blackbody. W = e*σ*T^4 for all objects including blackbodies where e=1. Emissivity is always less than or equal to 1. Emissivity is often close enough to 1 (for example, for much of the surface of our planet) that assuming e = 1 is a reasonable approximation. e is not a constant for a given material, it varies with the angle radiation is emitted and with temperature. This happens because there is a different “emissivity” for every wavelength (determined by the absorption cross-section for that wavelength and the amount of power emitted at any wavelength varies with temperature (Planck function). However, emissivity doesn’t change rapidly with temperature, so all the emissivity of materials here on earth is effectively constant whether they are in the topics or poles.

72. Reed Coray says:

I haven’t read all of the above comments in detail, but basically I agree with Nick Stokes. In particular, the first assumption:

“1 .Any point on both spheres emits photons in any direction within a half sphere with the
same probability.”

Energy Rate from Outer Sphere …to… Inner Sphere = 4 * pi * sigma * RI^2 * TO^4

where “sigma” is the Stefan-Boltzmann constant and has units of Watts per meter^2 per Kelvin^4. Note: This formula contains the radius, RI, of the outer surface of the inner concentric sphere, not the inner radius of the outer concentric spherical sphere.

The total rate energy is radiated from the inner surface of the outer sphere is

Total Energy Rate From Inner Surface Of Outer Sphere = 4 * pi * sigma * RO^2 * TO^4

The difference between these two rates, 4 * pi * sigma * TO^4 * (RO^2 – RI^2), corresponds to the rate that energy radiated from the inner surface of the outer sphere is radiated in the direction of the inner surface of the outer shell.

Unlike radiation from the inner surface of the outer sphere, which will in part be directed towards both the outer surface of the inner sphere and the inner surface of the outer sphere, all radiation from the outer surface of the inner sphere will be directed towards the inner surface of the outer sphere. That rate is given by

Energy Rate from Inner Sphere …to… Outer Sphere = 4 * pi * sigma * RI^2 * TI^4

Thus, the difference in energy rates (Outer Sphere to Inner Sphere …minus… Inner Sphere to Outer sphere) is

Energy Rate from Outer Sphere to Inner Sphere …minus… Energy Rate from Inner
Sphere to Outer Sphere = 4 * pi * sigma * RI^2 * (TO^4 – TI^4)

The above equation implies (a) that the net energy flow is from the object at the higher temperature to the object at the lower temperature, and (b) that if the two spheres have the same temperature (i.e., if TO = TI), then the net rate of energy flow is zero.

I believe that when applied to two identical but separate differential planar surfaces (dA1 and dA2), the second law of thermodynamics requires a COSINE(theta) radiation dependence. This is because the solid angle subtended by a differential planar surface area, dA, as seen from a point is the ratio of (a) the projection of dA onto the spherical surface that is normal to the line joining the point and dA …to… (b) the square of the distance from the point to dA. The projection of dA onto the spherical surface that is normal to the line joining the point and dA is the product of dA and the COSINE(alpha) where alpha is the angle between the normal to dA and the line joining the point and dA. If you place dA1 at the origin of a coordinate system such that the nomal to dA1 is in the z-axis direction and you place dA2 on the positive z axis at a distance d from the origin such that the normal to dA2 points in the x-axis direction, then from the point dA1, the solid angle subtended by dA2 is zero. Since dA2 has no solild angle as viewed from dA1, no energy from dA1 will fall on dA2. However, from the point dA2, the solid angle subtended by dA1 is dA1 / d^2. This means that if energy is radiated isotropically from each planar surface, some energy from dA2 will be directed towards dA1. Thus, we will have created a situation where independent of the temperatures of the differential surfaces (assuming both are non-zero), energy will flow from dA2 to dA1, but no energy will flow from dA1 to dA2. This would violate the second law of thermo dynamics which states that in the absence of work being performed on a system heat cannnot flow from an object at a lower temperature to an object at a higher temperature This violation of the second law is avoided by including the COSINE(alpha) term in the radiation law.

73. Frank says:

Hans wrote: “There are a number of assumptions made to reach the final S-B law that fits the experimental data. In fact the S-B law had to be modified by Planck before fitting to experimental observations. It is hard to understand that any gas can radiate as a “cavity in thermal equilibrium”. It should also be hard for a perfect smooth black body surface to do so, too, especially at close to ambient temperatures in the atmosphere where convective processes always are at hand and mostly dominate the energy transfer between matter.”

Hans is “hand-waving” here. The total emissivity of an object depends on the object’s emissivity at each wavelength. When one integrates the power output over all wavelengths at particular temperature, one can calculate the total emissivity. It isn’t a hard problem once you know the emissivity at all wavelengths. (Scientists have been studying in the laboratory how the optical properties of gases change with temperature, pressure and wavelength for decades, so these values are known.) Convection, of course, has nothing to do with the physics of radiation. Energy flux by other mechanisms (conduction and convection) occurs completely independently of radiation. It’s ludicrous to postulate that any particular molecule on a surface “decides” whether or not to emit (or absorb) a photon based on whether other molecules on the surface are gaining or losing energy by conduction or convection or radiation. All three processes occur independently.

The behavior of radiation as it passes through matter is described by the Schwartzschild eqn:

dI/ds = -I*n*o.ds + B(lamba,T)*n*o.ds – scattering

Blackbody radiation, the need for an emissivity term when scattering is involved, TOA emission from the earth and many other phenomena can be derived from this equation. The S-B equation itself comes from integrating the Planck function, B(lamba,T), over all wavelengths.

A “cavity in thermal equilibrium” is a description of how the Planck function was first derived by a semi-classical approach. Instead, picture molecules in equilibrium “photon gas” whose energy distribution (which is the same the Planck function except for geometric factors) is derived from Bose-Einstein statistics for massless particles. The probability of a photon with the right amount of energy (E = hv) being near enough to be absorbed and excite a molecule is proportional to the Planck function (and the density of molecules and their absorption cross-section). For a system at equilibrium, emission equals absorption and depends on the same factors. Thus, the emission term in the Schwartzschild equation is B(lamba,T)*n*o.ds

74. Frank says:

Hans: If you still confused by Nick Stock’s explanation, consider two parallel planes radiating across a narrow gap. All of the energy emitted upward by the lower plane is absorbed by the upper plane. Now imagine the planes have been tipped so that they are perpendicular. Now none of the energy will be absorbed. For two planes at an angle, the absorption will depend on the cosine of the angle between them and the origin of the Lambert cosine law.

Think of your sphere as a collection of tiny planes at different angles to each other before integrating.

75. […] been arguing over at Tallbloke’s. It’s one of those posts where a sceptic does an elementary analysis and makes elementary […]

76. Richard says:

If I put a superconducting black body inside another black body then they will both achieve the same equilbrium temperature. Anything else is just crazy.

77. DocMartyn says:

Reed Coray, thank you for pointing me to Planck’s blackbody radiation law. I had assumed that cross sectional area ratios would allow me to estimate the back radiation to inner core and inner surface; but obviously not.
I would still like to know the temperature of my inner Pu/Pb sphere at steady state.

78. tchannon says:

Nick Stokes makes an elementary mistake. Conflating sceptic with something else, suggesting he has an agenda.

Maybe what he ought to have written in public: experts in a field are less likely to make elementary mistakes, the obvious which ought not to need saying. Why did he?

I point out there are rife mistakes in science from out of expertise.

79. Nick Stokes says:

tchannon,
“experts in a field are less likely to make elementary mistakes”

It’s not the making of a mistake, it’s what happens afterward. I’ve made the same mistake as Hans – forgetting about the cos thing and getting a discrepancy. But I didn’t then conclude that electromagnetic radiation theory and thermodynamics might be all wrong. I knew that I must have got it wrong, and went over it and over until I could find the error.

80. tchannon says:

As a subject the whole thing is worth a long discussion, pervades most of life. This is perhaps a topic for an article, would take quite some time to write.

Even recognising mistakes, where that word is also an euphemism, is just a small part of things. The environment, culture, upbringing, maturity, loads of things come into it, including of course making other mistakes. Loss of face can be writ big, blame culture too.

Did you understand what I meant by an agenda?

I can’t comment on the lambert thing, haven’t been following, or that is what I assume it is. What to do if there is a definite problem is difficult at the best of times, in recorded public is even more tricky. My inclination is de-escalate, not necessarily what others do. This is not weakness, there are usually better ways.

If the whole thing had been formal and from authority, different matter. If anyone has learnt, good, if it is just minor rough and tumble of safe discussion, fine.