Equilibrium between objects at different temperatures?

[hutchphd]

This is indeed much more interesting than one (me, in particular) might expect at first glance. The etendu theorem is really a statement about a sort of "optical free energy" (my colloquial term), considering both energy and entropy. The theorem's requirement that etendu cannot passively decrease is the second law of thermodynamics for a beam of light. Attempts to show it using energy conservation alone cannot be viable. Thanks for the question @Philip Koeck

 

[Philip Koeck]

A quick update on whether or not radiation from a black body is isotropic.

I've read a bit about Lambert's law in an old book (Optical Physics by Max Garbuny) and he mentions that a glowing sphere looks like a flat disk, which means that radiation from the surface of a black body is actually isotropic.
If radiation was more intense in the direction orthogonal to the surface the sphere would look like a disk that fades out towards its edge.

The brightness (or radiance), however, is proportional to the cosine of the angle (measured from the direction orthogonal to the surface). (This is Lambert's law.)
This is simply because brightness is given as power per solid angle and per emitting area.

 

[DrClaude]

Here is how Walter Greiner presents it in Quantum Mechanics: An Introduction:

 

[Philip Koeck]

Here is how Walter Greiner presents it in Quantum Mechanics: An Introduction:
View attachment 331919
View attachment 331920
View attachment 331921
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View attachment 331923
View attachment 331924

Greiner and Garbuny use the word radiance for different things, I think.

Anyway, if I read Greiner correctly, a detector of fixed area placed inside a hot black body cavity will detect the same power no matter where it is placed and in what direction it is facing.

Does that sound right?

That's in principle what I assumed in my calculation from a previous post.

 

[Philip Koeck]

Greiner and Garbuny use the word radiance for different things, I think.

Anyway, if I read Greiner correctly, a detector of fixed area placed inside a hot black body cavity will detect the same power no matter where it is placed and in what direction it is facing.

Does that sound right?

That's in principle what I assumed in my calculation from a previous post.

I think I have to correct that.
Right now I believe I assumed that the radiation coming from a given surface element of fixed size on the cavity is isotropic, and that is not right. That should actually follow Lambert's law.
That would change my derivation quite a lot.
I have to think a bit more about that.

 

[Philip Koeck]

I think I have to correct that.
Right now I believe I assumed that the radiation coming from a given surface element of fixed size on the cavity is isotropic, and that is not right. That should actually follow Lambert's law.
That would change my derivation quite a lot.
I have to think a bit more about that.

Good news: If I take into account that radiation power per solid angle emitted from a fixed surface element is proportional to cos φ, where φ is the angle measured from the direction perpendicular to the surface, then the fraction f from post 30 is actually given by sin²a and therefore T_S=T_C.

A connected question: Does Lambert's law follow only from the second law or is there some other way to show it?

 

[hutchphd]

You need to specify a particular derivation if you desire meaningful commentary. This is very difficult to follow. There are a number of ways to state Lambert's Law(s) for instance.

 

[hutchphd]

A connected question: Does Lambert's law follow only from the second law or is there some other way to show it?

I invoke the Anna Karenina principle here and direct you to a paper by Zhang

 

[Philip Koeck]

You need to specify a particular derivation if you desire meaningful commentary. This is very difficult to follow. There are a number of ways to state Lambert's Law(s) for instance.

To start with here's the calculation I discussed in post 30 with the correction to a brightness that's proportional to cos φ.
Sorry for the quality of the image. It looks good when I open the file on my computer, but it get's fuzzy when I upload it.

This text shows that this angular dependence of radiation (what I call Lambert's law) is in agreement with the second law.

I'll send another, more direct, argument for Lambert's law when I get round to it.

 

[vanhees71]

In the restframe of the black-body radiation it's homogeneous and isotropic. If it were not, there'd be heat flow from some parts of the container walls to others through radiation until thermal equilibrium is reached.

Lambert's cosine Law follows from this. It's however not about the distribution of black-body radiation within the cavity, which is homogeneous and isotropic in its rest frame, but about the radiation of a black body to the outside. It describes the energy flow of radiation from a surface, and the cosine is that of the angle between the emitter's surface normal and the direction of observation. The Wikipedia is pretty good, as far as I can see:

https://en.wikipedia.org/wiki/Lambert's_cosine_law

 

[Philip Koeck]

In the restframe of the black-body radiation it's homogeneous and isotropic. If it were not, there'd be heat flow from some parts of the container walls to others through radiation until thermal equilibrium is reached.

Lambert's cosine Law follows from this. It's however not about the distribution of black-body radiation within the cavity, which is homogeneous and isotropic in its rest frame, but about the radiation of a black body to the outside. It describes the energy flow of radiation from a surface, and the cosine is that of the angle between the emitter's surface normal and the direction of observation. The Wikipedia is pretty good, as far as I can see:

https://en.wikipedia.org/wiki/Lambert's_cosine_law

I'll skip my own thought experiment, since it's essentially the same as what's discussed in the Wikipedia page shared by vanHees71 and also in the sections from Greiner shared by DrClaude (plus some other sources).

The argument for the cosine law is that it's necessary to make radiation isotropic and homogeneous and that, in turn, is necessary for two bodies (or parts of the same surface) to be in equlibrium at equal temperatures.

So you could say that the second law requires Lambert's law.

Is there any other way to show that the cosine law has to hold?

 

[vanhees71]

That there must be this cosine in the calculation of how much radiation energy flows through a surface is very general. It's just the definition of the "energy-current density". As for any current density, the em. field energy going through an arbitrary given surface per unit time is
$$\int_S \mathrm{d}^2 \vec{f} \cdot \vec{S},$$
where $\vec{S} \propto \vec{E} \times \vec{B}$ is the Poynting vector. The cosine is hidden, of course, in the scalar product between the Poynting vector and the surface-normal vector under the intgral.

 

[Philip Koeck]

That there must be this cosine in the calculation of how much radiation energy flows through a surface is very general. It's just the definition of the "energy-current density". As for any current density, the em. field energy going through an arbitrary given surface per unit time is
$$\int_S \mathrm{d}^2 \vec{f} \cdot \vec{S},$$
where $\vec{S} \propto \vec{E} \times \vec{B}$ is the Poynting vector. The cosine is hidden, of course, in the scalar product between the Poynting vector and the surface-normal vector under the intgral.

I'm not sure whether that's really what I'm looking for.

Clearly there must be a cosine factor for radiation that flows through a surface, but what about radiation that's emitted from a surface.

Why is the power per solid angle and emitting surface not independent of the angle?

 

[jonhswon]

I think thermal radiation lambertian emission pattern is how nature works, and so is the related law of thermo. Is there a need to establish which one gives birth to the other? They just work together.

Isotropic pattern is unrealistic/wrong, but combined with such a wrong view factor that in the end it yields correct result, it works. Like (-1)*(-1)=1.

edit: added the word "thermal radiation"

 

[Philip Koeck]

I think lambertian emission pattern is how nature works, and so is the related law of thermo. Is there a need to establish which one gives birth to the other? They just work together.

Radiation from an electric dipole, for example, is not lambertian.

I would say things that happen in nature are in agreement with the 2nd law, but that doesn't mean they are caused by the 2nd law.

Heat is never completely converted to work (without also producing waste heat) simply because this extremely unlikely. The 2nd law simply states that such an unlikely event will usually not happen. It doesn't cause it not to happen.

I think some other reason for Lambert's law would be interesting.
At the moment we only have: "It has to be like that in order to agree with the 2nd law."

 

[jonhswon]

I meant thermal radiation pattern emitted from a surface being lambertian is natural. :)

 

[Philip Koeck]

I meant thermal radiation pattern emitted from a surface being lambertian is natural. :)

Yes, I know you meant that, but I'm wondering what causes it to be lambertian.

The 2nd law doesn't qualify as a cause of things. It only tells you what cannot happen, but it doesn't actually do a good job predicting what will happen.

You can think of the following process:
You place a lump of metal on a metal surface. Then you add heat to the metal lump and suddenly the lump starts gliding along the surface and some waste heat is transferred to the metal surface.

As long as the amount of waste heat is large enough the 2nd law has no problem with this process, but that doesn't mean it will happen.

In this example you see that 2nd law doesn't tell you what's going to happen, simply because it has nothing to do with mechanisms.

All things that happen do so in agreement with the 2nd law, but not because of it.

That's my view anyway.

 

[sophiecentaur]

By some ingenious optical device all the radiation from body 1 is focused onto body 2 and vice versa.

Me too.

Maybe the whole approach of comparing outgoing and incoming radiation is wrong, but why?

I think the apparent paradox is to do with the assumption that the two bodies only exchange energy. I thought about a perfect ellipsoid (already discussed) and my conclusion was that a single object at one focus would only be radiating and receiving energy from itself. That would be an equilibrium situation. Put a minute object at the other focus. It only takes up a small part of the image of the larger object so it cannot interact with it all and its equilibrium temperature would be reached when the energy received equalled the energy emitted. The small object would be an exact equivalent of a small part (same area) of the large object. Stefan's law says this will happen when the added object is at the same temperature as the original object. Stefan's law assumes / implies that the object under discussion only radiates outwards and ignores the above situation.

The same argument applies for a small object inside a big one. The effective area of the inside of the big one is actually the same as the area of the small one inside. All the energy that's unaccounted for is passing between the parts of the internal face of the outside one.

 

[Philip Koeck]

The same argument applies for a small object inside a big one. The effective area of the inside of the big one is actually the same as the area of the small one inside. All the energy that's unaccounted for is passing between the parts of the internal face of the outside one.

In post 44 I shared a (visually rather fuzzy) calculation for a spherical black object (arbitrary diameter as long as it fits) in the center of a spherical black cavity.
The result is that the two surfaces will be in equilibrium at equal temperature if (and, I believe, only if) radiation given off from a surface is proportional to cos a, where a is the angle with respect to the surface's normal.
Without the cosine dependence I don't get equilibrium at equal temperatures.
It seems that the cosine-dependence is actually essential for the 2nd law to hold.

 

[sophiecentaur]

It seems that the cosine-dependence is actually essential for the 2nd law to hold.

That's interesting. If the object inside is reduced in size to be just a tiny (infinitessimal) 'probe', you'd expect it to measure the temperature of the inside face of the big hole. If the internal object / probe were almost to fit entirely within the big object then, again, it would be at the same temperature.
Choose an intermediate size and could there be any argument that the inside object could be at a higher or lower temperature and what sort of curve would result as the inner object expanded or contracted?
That implies (and cross checks?) that your cos dependence would have to apply.

 

[Philip Koeck]

That's interesting. If the object inside is reduced in size to be just a tiny (infinitessimal) 'probe', you'd expect it to measure the temperature of the inside face of the big hole. If the internal object / probe were almost to fit entirely within the big object then, again, it would be at the same temperature.
Choose an intermediate size and could there be any argument that the inside object could be at a higher or lower temperature and what sort of curve would result as the inner object expanded or contracted?
That implies (and cross checks?) that your cos dependence would have to apply.

So far the only argument for this cos-dependence (Lambert's law) I've seen is that it's required by the second law.
People have also argued that it's natural, but I can only see that for absorption, not for emission, unless I again say it's required by the second law.

What I'm looking for now is some other explanation for Lambert's law, for example based on the emission properties of optical phonons.

 

[sophiecentaur]

A cos pattern in the vertical plane is what you get from a horizontal transmitting dipole, λ/r4 above the ground. Thinking in terms of EM theory, it seems to me that currents flowing randomly across the surface of a radiator (in all directions) could produce a radiation distribution which follows Lambert's Law. The maxes and nulls in the horizontal plane would average out to half.

Is that crazy, do you think?

 

[Philip Koeck]

A cos pattern in the vertical plane is what you get from a horizontal transmitting dipole, λ/r4 above the ground. Thinking in terms of EM theory, it seems to me that currents flowing randomly across the surface of a radiator (in all directions) could produce a radiation distribution which follows Lambert's Law. The maxes and nulls in the horizontal plane would average out to half.

Is that crazy, do you think?View attachment 332745

Just to clarify. Is the model that there are currents and maybe dipole oscillation mainly in the plane of the surface, but otherwise in random orientations?
This would mean that each "donut" of radiation has a maximum in the direction normal to the plain.
Now these donuts actually have a sin²-profile, but due to the random in-plane orientations this could average out to a cos-profile.
Could work!

I'm not quite sure how you would get zero for a direction parallel to the surface, though.

What does the sin²-donut give when you rotationally average it?

Just to explain why I'm discussing this: It would be interesting that the cos-behaviour of black-body radiation follows from some other physics and then actually agrees with the second law.
That would be like other physics "adapting" to the 2nd law, in a way.

 

[Philip Koeck]

A cos pattern in the vertical plane is what you get from a horizontal transmitting dipole, λ/r4 above the ground.

Maybe you could explain this once more. What is the vertical plane and what do you mean by λ/r4 above the ground?

 

[sophiecentaur]

Maybe you could explain this once more. What is the vertical plane and what do you mean by λ/r4 above the ground?

Sorry; a typo. I meant λ/4. Different heights will produce different vertical patterns and my choice of λ/4 was a bit arbitrary but the radiation in the horizontal direction will always be equal if the induced ground image is assumes to be perfect. This link shows patterns of dipoles at different heights. The common feature is a zero elevation. A random selection of independent dipole (uncorrelated signals) will (?) have a max 90 degrees and a null at 0 degrees.
The 'vertical plane' would be the vertical plane through the maximum in the horizontal direction. A large number of dipoles, radiating uncorrelated signals in random azimuths will average out as omnidirectional.

The snag here is that a vertical dipole / monopole will have a maximum at zero elevation. I don't know about the orientation of molecular dipoles on the surface of a solid. Maybe horizontal is a fanciful choice.

 

[Philip Koeck]

Sorry; a typo. I meant λ/4. Different heights will produce different vertical patterns and my choice of λ/4 was a bit arbitrary but the radiation in the horizontal direction will always be equal if the induced ground image is assumes to be perfect. This link shows patterns of dipoles at different heights. The common feature is a zero elevation. A random selection of independent dipole (uncorrelated signals) will (?) have a max 90 degrees and a null at 0 degrees.
The 'vertical plane' would be the vertical plane through the maximum in the horizontal direction. A large number of dipoles, radiating uncorrelated signals in random azimuths will average out as omnidirectional.

The snag here is that a vertical dipole / monopole will have a maximum at zero elevation. I don't know about the orientation of molecular dipoles on the surface of a solid. Maybe horizontal is a fanciful choice.

I think, what is clear is that black body radiation is due to some sort of accelerated charges such as dipole oscillations or varying currents.

If I just think of a model system consisting of oscillating dipoles in a thin layer in vacuum, without a metal surface in the vicinity, then a completely random orientation of these dipoles should actually give an isotropic power per solid angle and emitting surface area (I'll call that brightness).

How can we explain that brightness obeys Lambert's law then?

 

[sophiecentaur]

How can we explain that brightness obeys Lambert's law then?My 'radio ham' approach looks increasingly dodgy. So I revisited the topic from my youth. Wiki seems to suggest that the reason for the cosine law is just a geometrical one and that makes sense to me. An elemental radiator on the surface is omnidirectional but the projection in a direction θ to the normal will be of a radiating area A cosθ. So the flux in that direction will be modified by cosθ.
That seems to me to be a sufficient argument. Am I just being sloppy? (Not for the first time, I'll admit.)

 

[Philip Koeck]

Wiki seems to suggest that the reason for the cosine law is just a geometrical one and that makes sense to me. An elemental radiator on the surface is omnidirectional but the projection in a direction θ to the normal will be of a radiating area A cosθ. So the flux in that direction will be modified by cosθ.
That seems to me to be a sufficient argument. Am I just being sloppy? (Not for the first time, I'll admit.)

Maybe I'm misunderstanding, but let's discuss based on the Wiki article.

I want to limit the discussion to thermal emission from a black surface and refer to figure 1 in the article (shown below) and the text connected to this figure. I'll also use the terminology from the article.

Figure 1 from the Wikipedia article

Here's a passage from this article:
"... the number of photons per second (photon current (my addition)) emitted into the vertical wedge is
I dΩ dA.
The number of photons per second emitted into the wedge at angle θ is I cos(θ) dΩ dA."

To me this means that the photon current emitted by a fixed area dA on the surface into a certain solid angle will be proportional to cos(θ).

The way I see it this is not because the projected area in the emission direction is smaller than dA by a factor cos(θ).

I can offer the following argument for this.
We can picture a certain number of tiny emitters on this surface patch dA. Obviously this number doesn't change when you look at the surface from a different direction.
If each of these emitters radiates isotropically then the sum of all radiation from the surface should also be isotropic if we exclude interference effects.

It is true that the projection of dA is smaller than dA by a factor cos(θ), but the apparent density (in projection) of individual emitters is larger by a factor 1/cos(θ) and the two effects cancel.

From this I would conclude that the cos-dependence has to come from the individual emitters (after summing over them).

What do you think?

 

[sophiecentaur]

Obviously this number doesn't change when you look at the surface from a different direction.

This is the nub of the problem. Alternatively, take the point of view of the observer / measuring device. The visible area of the element on the emitting surface will be modified by the angle of view. The element will have maximum area at the normal and the projected area will follow a cos law. Or you can say that the solid angle seen by the observer will be less than on the normal and so the number of photons admitted will be less for non-normal reception.
Whilst I can see that your statement above 'sounds' right it has to be confusing you. Figure 2 on the wiki article explains it and the maths actually shows why.

Edit: Your idea involves a hemisphere of unit radius (centred on the emitting element, showing the equal flux emitted per unit solid angle. Figure 1 on the link shows how the received flux density varies as the sphere due to the projection of the emitting element.

 

[Philip Koeck]

This is the nub of the problem. Alternatively, take the point of view of the observer / measuring device. The visible area of the element on the emitting surface will be modified by the angle of view. The element will have maximum area at the normal and the projected area will follow a cos law. Or you can say that the solid angle seen by the observer will be less than on the normal and so the number of photons admitted will be less for non-normal reception.
Whilst I can see that your statement above 'sounds' right it has to be confusing you. Figure 2 on the wiki article explains it and the maths actually shows why.

I agree that the observer or detector sees isotropic radiation. That's because the observer always sees the same projected area, not the same actual area, and the actual area is bigger than the projected area by a factor 1/cos(θ).

I think the crux of the question is what Lambert's law for emission actually states.

Which of these two statements is correct?

1. The photon current per solid angle of emission and surface area of the emitter is proportional to cos(θ).

2. The photon current per solid angle of emission and projected surface area of the emitter is proportional to cos(θ).

 

[sophiecentaur]

2. The photon current per solid angle of emission and projected surface area of the emitter is proportional to cos(θ).

I'd go for that one because it includes the aperture of the observer. I'd say that the "projected surface area" part is assumed and would only complicate that simple statement. If you were to measure a small emitting area of a large tilted plane with a narrow lens in front of the detector then the apparent projected image would have the same area on the detector. That would give the impression that Lambert doesn't apply. Images of the Moon do not follow Lambert's law; the limbs look bright where Lambert would suggest they should be dark. (Whatever the angle of the Sun's illumination)

The other extreme of things would be to consider a spherical emitter (not a plane) where the detected emissions would be isotropic.

 

[Philip Koeck]

the limbs look bright where Lambert would suggest they should be dark.

At least for thermal emission Lambert's law implies that an observer sees the same brightness at every angle to the surface.
Here's a sentence from the Wiki:
"For example, if the sun were a Lambertian radiator, one would expect to see a constant brightness across the entire solar disc."

Here one has to take into account that an observer sees the projected surface, not the actual one.

 

[Philip Koeck]

Which of these two statements is correct?

1. The photon current per solid angle of emission and surface area of the emitter is proportional to cos(θ).

2. The photon current per solid angle of emission and projected surface area of the emitter is proportional to cos(θ).

So far I've been assuming that option 1 is correct. The way I read the Wikipedia article it supports that too.

There was one online source, though, that stated option 2.

 

[sophiecentaur]

@Philip Koeck
I've been looking round further and the majority of references to Lambert's cosine law involve an ideal diffuse reflector. This link states it succinctly and draws the distinction between the angle of the illumination and the angle of viewing. Illumination of the surface and emission by the surface are equivalent so that all means there is (ha ha should be) no confusion about when to use the cos law and when not.
When you look at a uniform illuminated Lambertian sphere, the edges receive less illumination (cosine law) but the observer sees light from a larger area of the sphere (more isotropic reflectors) in its receiving aperture. That's also a cosine relationship. So the overall appearance is of a uniformly illuminated disc.

The way it's stated often ignores the fact that there could be confusion. But that's the way of many 'explanations' of phenomena.

 

[vanhees71]

What's observed is the energy density or in this case the energy-current density (Poynting vector). Given the Poynting vector $\vec{S}$ the energy flowing through an arbitrarily oriented surface per unit time (power) is given by
$$P(t)=\int_A \mathrm{d}^2 \vec{f} \cdot \vec{S}(t,\vec{x}).$$