Undergrad Equilibrium between objects at different temperatures?

[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
$$P(t)=\int_A \mathrm{d}^2 \vec{f} \cdot \vec{S}(t,\vec{x}).$$

 

[Philip Koeck]

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
$$P(t)=\int_A \mathrm{d}^2 \vec{f} \cdot \vec{S}(t,\vec{x}).$$

I completely agree with that, but I can't see how to apply that to emission from a black body. The surface of a black body gives off radiation from its surface. No energy is flowing through the surface as far as I can see.

I'd really like to focus the discussion on thermal emission. So no scattering or reflection.
Unfortunately the Wiki shared by Sophie seems to jump between these topics and many sources only discuss scattering, as pointed out by Sophie.

What it is clear, I think is the following:

A black body is a Lambertian emitter and it always looks flat no matter what shape it is.
For example a sphere looks like a disk. (See several statements in the Wiki and other sources.)

This means (to me) that the current per solid angle and projected surface area is independent of θ.

The above means that the current per solid angle and actual surface area is proportional to cos(θ).

I believe the above statement is Lambert's law for emission, but I might be wrong.

Based on the above statement I argued in post 63 that Lambert's law is not due to the projection being smaller than the actual surface by a factor cos(θ), but that it has to come from the properties of the "elementary emitters", whatever they might be.

I'd really like to know what people think of this argument

An afterthought:
If I don't assume that the "elementary sources" sit on the surface, but rather in a small volume just below the surface, then I would get the cos-dependence even for isotropic emission (on average) from the individual sources.
Maybe that's a better model.

 

[sophiecentaur]

it has to come from the properties of the "elementary emitters", whatever they might be.

I see you've built in an 'unknown' into that statement and I don't think that's the best way to try for an explanation. (valiant attempt tho')

Personally, I don't have a problem with using the experience of the measurer to define a phenomenon. Whatever you measure radiation with will have an aperture and the number of photons admitted through this aperture from the emitting area will follow a cosine law.

This will apply to any size of emitting surface and the same argument also applies to the cos law 'spreading' effect when a tilted surface is illuminated by a plane wave front (from infinity). There's a distinct difference between receiving illumination from a single distant source and emitting radiation in all directions from a diffuse source. That difference provides a bit of a cognitive dissonance, particularly when we think too hard about the situation.

 

[Philip Koeck]

... the number of photons admitted through this aperture from the emitting area will follow a cosine law.

I don't think that's consistent with what I read in various sources.

According to the Wiki and a book I have (Max Garbunny) a sphere (for example) that is a Lambertian emitter will look flat, like a disk, to an observer. To me that means that the photon current passing through an aperture and reaching the detector is independent of the orientation of the surface.
That's just an experimental finding!

 

[vanhees71]

I completely agree with that, but I can't see how to apply that to emission from a black body. The surface of a black body gives off radiation from its surface. No energy is flowing through the surface as far as I can see.

I'd really like to focus the discussion on thermal emission. So no scattering or reflection.
Unfortunately the Wiki shared by Sophie seems to jump between these topics and many sources only discuss scattering, as pointed out by Sophie.

What it is clear, I think is the following:

A black body is a Lambertian emitter and it always looks flat no matter what shape it is.
For example a sphere looks like a disk. (See several statements in the Wiki and other sources.)

This means (to me) that the current per solid angle and projected surface area is independent of θ.

The above means that the current per solid angle and actual surface area is proportional to cos(θ).

I believe the above statement is Lambert's law for emission, but I might be wrong.

Based on the above statement I argued in post 63 that Lambert's law is not due to the projection being smaller than the actual surface by a factor cos(θ), but that it has to come from the properties of the "elementary emitters", whatever they might be.

I'd really like to know what people think of this argument

An afterthought:
If I don't assume that the "elementary sources" sit on the surface, but rather in a small volume just below the surface, then I would get the cos-dependence even for isotropic emission (on average) from the individual sources.
Maybe that's a better model.

It's thermal radiation. So from any surface element (in its rest frame) there's simply a radially symmetric emission from each surface element of the emitting surface.

 

[Philip Koeck]

It's thermal radiation. So from any surface element (in its rest frame) there's simply a radially symmetric emission from each surface element of the emitting surface.

In that case, if the "emitting units" sit on the surface and emit isotropically, I refer back to my post 63.

 

[sophiecentaur]

That's just an experimental finding!

All theories stand by being confirmed by measurement(?).

In that case, if the "emitting units" sit on the surface and emit isotropically, I refer back to my post 63.

It may help is you assume that there must be something wrong with your argument and perception. That is often the biggest hurdle. I think there is a lot of sloppy use of terms about Mr Lambert and the cosine law is often taken (wrongly) out of context. It's only since reading around with an open mind that I am happy with it.

The cosine law applies to incident radiation; it works at the Earth's poles and at the equator. PV cells are mounted with a tilt because of the cos law.
When you observe the Moon, it appears of uniform brightness despite the fact that it's a sphere because you can see a bigger area of poorly illuminated (but isotropic) moon surface at the edges. That all makes sense and agrees with our intuition.

For an emitter you assume a black body, which can be treated as a hole in a large sphere with radiation coming out of it and NOT a set of isotropic emitters on the surface. Wiki goes on to say:

"The observer at angle θ to the normal will be seeing the scene through the same aperture of area dA0 (still corresponding to a dΩ wedge) and from this oblique vantage the area element dA is foreshortened and will subtend a (solid) angle of dΩ0 cos(θ). "

So then you have the same situation as that of an illuminated patch and the angle relative to the normal introduces "foreshortening". I just realised where the apparent paradox lies. Whaddya think?

 

[Philip Koeck]

For an emitter you assume a black body, which can be treated as a hole in a large sphere with radiation coming out of it and NOT a set of isotropic emitters on the surface.

Here I completely agree with you. If we think of a black body as a hole in a cavity and assume that the radiation inside the cavity is isotropic and homogeneous then the radiation coming out from the hole will follow the cos-law simply because the effective area of the hole will be proportional to cos(θ).
That's completely in line with what vanhees71 wrote too.
Here we actually have radiation that comes from inside the cavity and traverses a hole that is tilted by θ with respect to the radiation.

What I was thinking of was an actual surface made of "black body material", not a hole.
In that case I don't see that geometry alone will lead to the cosine law if the elementary emitters are in the surface. If the emitters were in a thicker layer below the surface the situation would be different. In fact it would be a bit like the cavity with a hole in it.

 

[sophiecentaur]

What I was thinking of was an actual surface made of "black body material", not a hole.

I appreciate that but the evidence is that the cosine law has been found, by experiment, to apply. That implies that a lambertian emitting surface behaves as if it were a hole in a surface with a large (enough) volume beneath it with all the walls at the temperature of the surface.
1. We can't argue against the evidence of cosine law behaviour
2. This behaviour for emission is the reciprocal of an absorbing surface and there is a very plausible argument for cosine behaviour in emission.
3. I have a feeling that having different laws for absorption and emission would involve some basic failure of thermodynamics laws - and would require a Maxwell Demon to explain it.

It may be annoying but is there anywhere else to go?

 

[Philip Koeck]

I appreciate that but the evidence is that the cosine law has been found, by experiment, to apply. That implies that a lambertian emitting surface behaves as if it were a hole in a surface with a large (enough) volume beneath it with all the walls at the temperature of the surface.
1. We can't argue against the evidence of cosine law behaviour
2. This behaviour for emission is the reciprocal of an absorbing surface and there is a very plausible argument for cosine behaviour in emission.
3. I have a feeling that having different laws for absorption and emission would involve some basic failure of thermodynamics laws - and would require a Maxwell Demon to explain it.

That's sort of what I've found so far.
Experimental evidence shows that the cosine law applies and the cosine-law is necessary for the 2nd law.
Just as you say, if absorption is proportional to cos(θ) then emission must be too. Otherwise you can't get equilibrium at equal temperatures.

That's similar to what I found in my calculation for a specific case (sphere in a spherical cavity) discussed in an earlier post (55?). The sphere and cavity can only be in equilibrium at equal temperatures if the cosine law for emission holds.

So, in summary, what we have is that the cosine law has to hold for emission in order for the 2nd law to hold.

It may be annoying but is there anywhere else to go?

I think there might be.

The second law doesn't really explain why something happens, at least not in the sense that it provides a mechanism.
In a way the second law is a result but not a cause.

In the case of the cavity with hot inner surfaces and a hole in the wall where radiation escapes the cosine law for emission (and absorption) follows from pure geometry. That's a very simple "mechanistic" explanation.

For a solid surface I don't see that, at least not if the radiation sources are isotropic and only located on the surface. (see earlier post 63).

Maybe we can conclude that even in the case of a solid surface thermal radiation doesn't come from the surface but from deeper inside the solid.
Then the cosine-law would make perfect sense even if the "radiation inside the solid" (whatever that might be) is isotropic, or, if you prefer, each elementary source is isotropic.

 

[sophiecentaur]

Ok but we have yet to find an example of the second law failing. There would be incredible knock-ins from that. So your hoped for model would need to survive a very stringent test.

I do agree that I get brain pain when I try to imagine what must happen to EM waves to make them appear to emerge from a virtual image at infinity (or even less).

 

[sophiecentaur]

In a way the second law is a result but not a cause.

Are we basically down to the nature of a lambertian surface? How's this idea?

It's definitely not flat because it needs to scatter equally in all directions. That would affect the ray paths from all points on the surface. More and more surface peaks would obscure the view from a low angle. Because of the equilibrium condition all over the surface there would be no change in temperature where the scattered rays hit the obscuring peaks so the net emission in a given direction would depend only on the rays that get through without being absorbed by peaks. (It's the thermal equilibrium over the surface that's the clever bit.)

So you 'make' a lambertian surface and test it by its variation of scattering performance and then you heat it up. It's then the geometry of the rough surface that makes it lambertian for emission too.

AND . . . . as the surface material increases in thermal resistivity, the surface will approach isotropic behaviour. (That bit just came to me)

 

[vanhees71]

I still do not understand, where your problem is. The energy flow of electromagnetic waves is described by the Poynting vector. That's it!

 

[sophiecentaur]

I still do not understand, where your problem is. The energy flow of electromagnetic waves is described by the Poynting vector. That's it!

Was this post aimed at a different thread?

 

[Philip Koeck]

I still do not understand, where your problem is. The energy flow of electromagnetic waves is described by the Poynting vector. That's it!

I think it is meant for this thread.

I assume you are referring to your statement from post 70, which I answered in post 71.

I can understand that reasoning if radiation hits a surface or if radiation escapes through a hole, but I can't understand it if radiation actually originates in a surface.
In post 71 I explain in more detail how I think.

If radiation doesn't originate in the surface but in some finite volume below the surface the situation is just like that for a cavity with a hole in it. In that case I completely buy this reasoning.

 

[sophiecentaur]

I think it is meant for this thread.

The Poynting vector can't help in a situation where many different random sources of EM are littered over a three dimensional ( or even two dimensional) region. Poynting needs E and H vectors (other vectors are available) before it can be defined. It isn't even a diffraction problem. And how does the cosine law emerge from all this? I think we'd need a few more details and a path through solving that maths.

 

[sophiecentaur]

An afterthought:
If I don't assume that the "elementary sources" sit on the surface, but rather in a small volume just below the surface,

Bearing in mind that a lambertian surface has to be 'rough', with unlimited depth, in order to scatter with a cosine law then the emitting surface would have to be the same. If we accept that it's to do with geometry then all that's necessary is to appreciate what happens when radiation from one part of the surface is obscured by another part. That fraction of the 'current' is re-admitted into the surface; the current emitted from the obscuring part of the surface can't be added to by the radiation it obscures because of the temperature equilibrium over the surface.

That doesn't explain the specific cosine nature of the law but it will, 'just by arm waving', have a zero parallel to the surface and have a generically cosine shape. I wouldn't know where to start but I'd bet there's a man, on PF who could.

 

[Philip Koeck]

Bearing in mind that a lambertian surface has to be 'rough', with unlimited depth, in order to scatter with a cosine law then the emitting surface would have to be the same. If we accept that it's to do with geometry then all that's necessary is to appreciate what happens when radiation from one part of the surface is obscured by another part. That fraction of the 'current' is re-admitted into the surface; the current emitted from the obscuring part of the surface can't be added to by the radiation it obscures because of the temperature equilibrium over the surface.

That doesn't explain the specific cosine nature of the law but it will, 'just by arm waving', have a zero parallel to the surface and have a generically cosine shape. I wouldn't know where to start but I'd bet there's a man, on PF who could.

That's definitely an interesting idea!

 

[sophiecentaur]

That's definitely an interesting idea!

It can only ever be an idea imo. No one makes lambertian scatterers but near enough is good enough Even the Moon seems to do a fairly good job and that is multicoloured with rocks and all.

 

[Philip Koeck]

It can only ever be an idea imo. No one makes lambertian scatterers but near enough is good enough Even the Moon seems to do a fairly good job and that is multicoloured with rocks and all.

Maybe one shouldn't expect the same mechanism for Lambertian scatterers and emitters.
A black body is supposed to be a Lambertian emitter, but a black body doesn't scatter at all so it can't be a Lambertian scatterer.

For scattering the surface structure ought to be important for the cosine law.

For emission I could imagine that the cosine law emerges even for atomically flat surfaces.
Maybe the reason for the cosine law is simply that radiation originates below the surface rather than on the surface, so the situation is like that of a cavity with a hole in its wall.

 

[sophiecentaur]

For emission I could imagine that the cosine law emerges even for atomically flat surfaces.

'Imagining' is not really a basis for a theory. I often fall into this trap too.

A thermodynamic argument demands that there must be some reciprocity between scattering and emission. If not then there could be a structure in which the laws are violated. But the lambertian surface and the black body are not real things; they are just models to work with. Trying to explain 'how' either of them could be manufactured is pretty much pointless.

It's interest to note that the nearest thing to a total absorber of light which can be used in TV studios (where black level was important) was a hole, cut in a white board with a void behind which had black velvet lining. The void was a few cm deep and larger than the hole. It was a good enough black level to use in a brightly lit studio from a range of angles. Afaik, no one tried heating the internal surface to measure the emitted thermal spectrum but you could expect the absorbance of the hole would obviously follow a cosine law. (Just throwing that one in of fun.)