jonhswon
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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.jonhswon said:I meant thermal radiation pattern emitted from a surface being lambertian is natural. :)
Philip Koeck said:By some ingenious optical device all the radiation from body 1 is focused onto body 2 and vice versa.
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.Philip Koeck said:Me too.
Maybe the whole approach of comparing outgoing and incoming radiation is wrong, but why?
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.sophiecentaur said: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.
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.Philip Koeck said:It seems that the cosine-dependence is actually essential for the 2nd law to hold.
So far the only argument for this cos-dependence (Lambert's law) I've seen is that it's required by the second law.sophiecentaur said: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.
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?sophiecentaur said: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
Maybe you could explain this once more. What is the vertical plane and what do you mean by λ/r4 above the ground?sophiecentaur said:A cos pattern in the vertical plane is what you get from a horizontal transmitting dipole, 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.Philip Koeck said:Maybe you could explain this once more. What is the vertical plane and what do you mean by λ/r4 above the ground?
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.sophiecentaur said: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.
Maybe I'm misunderstanding, but let's discuss based on the Wiki article.sophiecentaur said: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.)
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.Philip Koeck said:Obviously this number doesn't change when you look at the surface from a different direction.
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).sophiecentaur said: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'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)Philip Koeck said:2. The photon current per solid angle of emission and projected surface area of the emitter is proportional to cos(θ).
At least for thermal emission Lambert's law implies that an observer sees the same brightness at every angle to the surface.sophiecentaur said:the limbs look bright where Lambert would suggest they should be dark.
So far I've been assuming that option 1 is correct. The way I read the Wikipedia article it supports that too.Philip Koeck said: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).
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.vanhees71 said: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 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')Philip Koeck said:it has to come from the properties of the "elementary emitters", whatever they might be.
I don't think that's consistent with what I read in various sources.sophiecentaur said:... the number of photons admitted through this aperture from the emitting area will follow a cosine law.
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 said: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.
In that case, if the "emitting units" sit on the surface and emit isotropically, I refer back to my post 63.vanhees71 said: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.
All theories stand by being confirmed by measurement(?).Philip Koeck said:That's just an experimental finding!
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.Philip Koeck said:In that case, if the "emitting units" sit on the surface and emit isotropically, I refer back to my post 63.
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).sophiecentaur said: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.
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.Philip Koeck said:What I was thinking of was an actual surface made of "black body material", not a hole.
sophiecentaur said: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.
sophiecentaur said:It may be annoying but is there anywhere else to go?
Are we basically down to the nature of a lambertian surface? How's this idea?Philip Koeck said:In a way the second law is a result but not a cause.
Was this post aimed at a different thread?vanhees71 said: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.vanhees71 said:I still do not understand, where your problem is. The energy flow of electromagnetic waves is described by the Poynting vector. That's it!
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.Philip Koeck said:I think it is meant for this thread.
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.Philip Koeck said: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,
That's definitely an interesting idea!sophiecentaur said: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.
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 said:That's definitely an interesting idea!
Maybe one shouldn't expect the same mechanism for Lambertian scatterers and emitters.sophiecentaur said: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.
'Imagining' is not really a basis for a theory. I often fall into this trap too.Philip Koeck said:For emission I could imagine that the cosine law emerges even for atomically flat surfaces.
Do you have a reference or some web resource that discusses this?sophiecentaur said: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.
The Maxwell Demon was a thought experiment to demonstrate how heat could be made to flow 'uphill' but only by violating the second law of thermodynamics. That basic principle must hold in the scenario of the OP. Try a search on violating the second law.Philip Koeck said:Do you have a reference or some web resource that discusses this?
Sounds interesting.
I'm not looking for any loophole, quite on the contrary.sophiecentaur said:The Maxwell Demon was a thought experiment to demonstrate how heat could be made to flow 'uphill' but only by violating the second law of thermodynamics. That basic principle must hold in the scenario of the OP. Try a search on violating the second law.
Trying to find a 'loophole' in this by involving lambertian surfaces is hardly worth doing because the lambertian cosine law is only a concept and the second law doesn't hang on the characteristic of a particular surface. We're sort of chasing our tails here, I think.
Here's the thought experiment:sophiecentaur said:... the lambertian cosine law is only a concept and the second law doesn't hang on the characteristic of a particular surface. We're sort of chasing our tails here, I think.
Can you give me a derivation of that? It isn't obviously the case so it needs to be justified, I think.Philip Koeck said:A is larger than B by a factor 1/cosθ.
I realize that I was sloppy, but it should hold approximately for narrow cones.sophiecentaur said:Can you give me a derivation of that? It isn't obviously the case so it needs to be justified, I think.
Wouldn't you expect a "squared" in there, as the areas are involved? I really don't follow your argument and where's it going? You invent a complicated, approximate model and neither you nor I have the Maths to solve it.Philip Koeck said:I realize that I was sloppy, but it should hold approximately for narrow cones.
Why should there be a "squared"? The tilting only introduces a change normal to the tilt axis so the ratio between the tilted area and the corresponding untilted area becomes 1/cosθ.sophiecentaur said:Wouldn't you expect a "squared" in there, as the areas are involved? I really don't follow your argument and where's it going? You invent a complicated, approximate model and neither you nor I have the Maths to solve it.
The result of both models must be that the second law works but the spheres must be less involved than the reflecting cones. Also, you have to remember that equilibrium will occur with or without lambertian surfaces or you could have a Maxwell Demon by just choosing appropriate surfaces.Philip Koeck said:exact, but a bit more involved.
I'm quite sure (though not 100 %) that for black bodies to be in thermal equilibrium at equal temperatures they have to be lambertian emitters.sophiecentaur said:The result of both models must be that the second law works but the spheres must be less involved than the reflecting cones. Also, you have to remember that equilibrium will occur with or without lambertian surfaces or you could have a Maxwell Demon by just choosing appropriate surfaces.
I'm beginning to lose interest in this thread because nothing new or helpful has turned up for quite a while. Shall we call it a day?