Is a diffuse scatterer necessarily a Lambertian emitter?

[Philip Koeck]

TL;DR Summary

Do all diffuse scatterers also emit according to Lambert's cosine law.

The title contains the whole question.

I would argue that the answer is yes, but I'm wondering what other people have to say about this?

 

[sophiecentaur]

I'd agree because there is no specular element in this ideal process. But Raleigh scattering is not omnidirectional nor cosine.
I try to avoid to much 'classification' in these matters because you can waste more time worrying about it without increasing understanding; it's only words.

 

[sophiecentaur]

@Philip Koeck There was a thread about this some while ago. There seemed to be some confused thinking (including mine) but I came to the conclusion that the term `'lambertian" can really only refer to an absorbing cavity with a hole of area A. The absorption of energy(from one direction) will depend on the area of the hole, presented to the incoming flux. That's got to correspond to the Cosine of the angle from the normal. But that wouldn't apply to a general scatterer. IMO, the term "lambertian" just makes life harder. You can get a 'correct' answer by appropriately integrating the incident energy.

 

[Philip Koeck]

@Philip Koeck There was a thread about this some while ago. There seemed to be some confused thinking (including mine) but I came to the conclusion that the term `'lambertian" can really only refer to an absorbing cavity with a hole of area A. The absorption of energy(from one direction) will depend on the area of the hole, presented to the incoming flux. That's got to correspond to the Cosine of the angle from the normal. But that wouldn't apply to a general scatterer. IMO, the term "lambertian" just makes life harder. You can get a 'correct' answer by appropriately integrating the incident energy.

I would say all absorption is necessarily Lambertian, but it's not usually called that as far as I can see.
Absorption is proportional to the cosine of the incident angle purely for geometric reasons so it's so obvious that it doesn't really need to have a special name I would think.
This is of course true for absorption by a solid surface as well, not only for absorption by a hole in a cavity.

That emission and scattering can be Lambertian is not so obvious in my opinion.

For emission from a hole in a cavity it follows from geometry if the radiation inside the cavity is isotropic and homogeneous. So there the mechanism is clear.

For emission from a solid surface the mechanism that leads to Lambertian emission is not clear to me.

What is clear is that a black body surface, just like hole in the wall of a cavity, has to emit according to the cosine law in order to allow radiation equilibrium between surfaces of equal temperature.

We can go a step further and consider surfaces that absorb some of the incoming radiation and scatter some of it with a cosine dependence (i.e. perfectly diffuse scattering), but don't reflect like a mirror and don't transmit any radiation.
I think we can conclude that such a surface has to be a Lambertian emitter as well even though it's not a black body.

 

[WernerQH]

There's an extensive theory of "Radiative Transfer", see e.g. Chandrasekhar's monograph by that title. It turns out that Lambert's law provides at best an approximate description of radiation escaping from "diffuse" surfaces (walls or clouds). Even for isotropic scattering the escaping radiation has a more complicated angular dependence, if you look at the hairy calculations in Chandrasekhar's book.

I was curious about the brightness of the moon, assuming "Lambertian" characteristics. It turned out that, as a Lambertian emitter, full moon should be brighter than half moon by only a factor of $\pi$, whereas the observed difference is almost a factor of 10 (more than two magnitudes). Of course, the moon's surface is anything but "smooth".

 

[Philip Koeck]

... Even for isotropic scattering the escaping radiation has a more complicated angular dependence, ...

This sounds like a contradiction to me.
The only way scattering (or emission) can be isotropic as measured by an observer is by being Lambertian, which means it follows a simple cosine. That's how I understand it, anyway.

 

[WernerQH]

The only way scattering (or emission) can be isotropic as measured by an observer is by being Lambertian, which means it follows a simple cosine. That's how I understand it, anyway.

You must distinguish the microscopic scattering events from the overall radiation pattern that emerges as a result of many radiating electrons (atoms, molecules) distributed in the medium. It is the far field that can be measured by an observer, and it is usually far from isotropic. Visible light is not in equilibrium with a white wall. (Only light in the far infrared perhaps.) A white wall is usually illuminated from a particular direction. The radiation hitting the wall is highly anisotropic, while the reflected radiation is nearly isotropic (in the half-space) if you consider the intensity, but has the $\cos \theta$-dependence considering the flux.

 

[sophiecentaur]

That emission and scattering can be Lambertian is not so obvious in my opinion.

The emission from a hole in a plane surface is, I think, 'obviously' proportional to the projected area, which follows the cos function. But a flat matt surface will have a profile which will let you down for large angles of reflection. It's hard to understand the equilibrium requirement is hard to understand and would have to involve the conductivity of the material. Think in terms of the surface of Mercury which has huge differences in temperature over the surface. This effect has to (?) result in non uniform surface temperature distribution in the various slopes of the surface texture.

Perhaps it's best to assume Lambert works and not be too bothered about why or how. (That's not 'good' Physics, though)

 

[Philip Koeck]

The emission from a hole in a plane surface is, I think, 'obviously' proportional to the projected area, which follows the cos function.

I completely agree.

But a flat matt surface will have a profile which will let you down for large angles of reflection. It's hard to understand the equilibrium requirement is hard to understand and would have to involve the conductivity of the material. Think in terms of the surface of Mercury which has huge differences in temperature over the surface. This effect has to (?) result in non uniform surface temperature distribution in the various slopes of the surface texture.

My thinking is that all surfaces that are in equilibrium with each other have to have the same temperature no matter whether conduction is involved or not.
Consider a cavity made of a material with (almost) zero thermal conductivity.
All surfaces inside this cavity should arrive at the same temperature due to radiation alone no matter what the shape of this cavity is.
This should be true for all surfaces, not only for black bodies.
Do you agree?

PS: The various parts of Mercury are not in equilibrium with each other I would say.

 

[sophiecentaur]

PS: The various parts of Mercury are not in equilibrium with each other I would say.

I'm inclined to agree but the 'system' (Sun Mercury Deep Space) is arguably in equilibrium because the temperatures are constant - assume that Mercury's rotation and orbit are in sync.

All surfaces inside this cavity should arrive at the same temperature due to radiation alone no matter what the shape of this cavity is.

But would that apply to a surface, rather than a cavity? I don't think it would - from the Mercury model.

 

[Philip Koeck]

I'm inclined to agree but the 'system' (Sun Mercury Deep Space) is arguably in equilibrium because the temperatures are constant - assume that Mercury's rotation and orbit are in sync.

There's a net flow of energy from the sun, so that's not thermodynamic equilibrium.
It's more like a stationary or steady state. The temperatures have settled to constant values in accordance with the flow of heat.
Something like that.

But would that apply to a surface, rather than a cavity? I don't think it would - from the Mercury model.

The inside of the cavity is a surface.
It's an arbitrary cavity so it can also contain objects floating inside it.
The point of using a cavity in the argument is simply to create a closed system.

I would argue that in any closed system the temperature has to become the same everywhere.

 

[sophiecentaur]

The point of using a cavity in the argument is simple to create a closed system.

I get that but a suitable cavity needs to have no significant specular reflections (as with a surface). The specification seems to a bit recursive. That's fine as long as any prediction assumes a long timescale and overall equilibrium.
Bit it's clearly near enough for jazz in most cases.

 

[Philip Koeck]

I get that but a suitable cavity needs to have no significant specular reflections (as with a surface). The specification seems to a bit recursive. That's fine as long as any prediction assumes a long timescale and overall equilibrium.
Bit it's clearly near enough for jazz in most cases.

I don't understand everything you are saying above.

Anyway, I believe all surfaces in a cavity that's isolated from the environment have to have the same temperature when the cavity has settled into equilibrium.
It shouldn't matter whether the surfaces only absorb and emit radiation or whether they reflect or scatter it, or any combination of these properties.
If you can achieve temperature gradients simply by choosing a clever combination of surface properties you're immediately in trouble with the second law.

 

[sophiecentaur]

I don't understand everything you are saying above.

Haha - well said. A fair amount of waffle there.
I think my problem is the 'equilibrium' requirement in practical circumstances. There will be many instances where things are changing too fast to wait for that. Diurnal variations will often be too fast for assumptions of a Lambertian surface but I bet 'they' often reach for the principle when it doesn't apply

There's a sort of assumed authority when someone introduces the term Lambertian and it will often need to be modified for thermal wavelength radiation. For instance, the JWST does a great job of refrigerating itself passively (in addition to the clever heat pumping).

 

[WernerQH]

It shouldn't matter whether the surfaces only absorb and emit radiation or whether they reflect or scatter it, or any combination of these properties.

I thought Lambert's law refers only to the directional dependence of light reflected / scattered from a rough surface illuminated at some (arbitrary) angle. This has little bearing on the approach to thermal equilibrium, because the scattered light has the same frequency as the incident light. The photons carry away as much energy as they had to begin with.

 

[sophiecentaur]

I thought Lambert's law refers only to the directional dependence of light reflected / scattered from a rough surface illuminated at some (arbitrary) angle. This has little bearing on the approach to thermal equilibrium, because the scattered light has the same frequency as the incident light. The photons carry away as much energy as they had to begin with.

What confuses me is where the basic idea of illumination of an area at an angle (cos law) morphs into the net energy absorption. Only after equilibrium is reached will the two be the same. And by then, the net emission will be the same but isotopic.
Am I over thinking this?

 

[WernerQH]

Only after equilibrium is reached will the two be the same. And by then, the net emission will be the same but isotopic.
Am I over thinking this?

Probably. That a photon's probability of passing a surface element $dA$ at an angle $\theta$ is proportional to $\cos \theta$ is consistent with the radiation field being isotropic.

 

[Philip Koeck]

Probably. That a photon's probability of passing a surface element $dA$ at an angle $\theta$ is proportional to $\cos \theta$ is consistent with the radiation field being isotropic.

I would say the probability of a photon hitting an area dA is proportional to cos θ if the radiation is isotropic and homogeneous. This is the "Lambertian law" for absorption, but it's not really called that, I believe, since it's just a result of geometry.

For emission and scattering there's no such simple geometric explanation, as far as I can see.
And, yes, emission can also be Lambertian.
Emission from a black body, for example has to be Lambertian in order to avoid conflict with the second law, but otherwise there's no obvious reason, why it should be Lambertian.

Clearly Lambertian emission or scattering (aka as diffuse scattering) is not isotropic. It decreases with the cosine of the angle from the normal.

However, what an observer sees is isotropic! This, again, is due to geometry. The observer sees an area dA' which is dA cos θ, so to the observer the radiation coming from what he regards as an area dA' is independent of θ.
A typical example for this effect is the appearance of the glowing filament in a light bulb. The filament looks like a flat ribbon although it's actually a cylinder. A glowing sphere looks like a disk (seen face on, of course).

A perfectly diffuse scatterer would also look flat no matter how it is illuminated.

 

[Philip Koeck]

I thought Lambert's law refers only to the directional dependence of light reflected / scattered from a rough surface illuminated at some (arbitrary) angle. This has little bearing on the approach to thermal equilibrium, because the scattered light has the same frequency as the incident light. The photons carry away as much energy as they had to begin with.

This might actually be where I'm going wrong.
So the angular distribution of scattered (and reflected) light has no impact on thermal equilibrium, for example in a cavity.

 

[WernerQH]

Clearly Lambertian emission or scattering (aka as diffuse scattering) is not isotropic. It decreases with the cosine of the angle from the normal.

I think we agree on the physics, and quibble only about terminology. What is the difference between lambertian and isotropic?

Here the intensity falls off as the cosine of the observation angle with respect to the surface normal (Lambert’s law). The radiance (W/m2-sr) is independent of direction.

 

[Philip Koeck]

I think we agree on the physics, and quibble only about terminology. What is the difference between lambertian and isotropic?

I think there's some confusion in that text.

Isotropic emission should mean that a given surface patch dA emits the same power in all directions.
A Lambertian emitter clearly doesn't.
It's rather important to understand that a Lambertian emitter doesn't emit isotropically.
Otherwise you end up in conflict with the second law.
So it's not just terminology.

However, to an observer the Lambertian emitter looks the same from all angles so in that sense it's isotropic.
That is only the case because the observer sees the projected area dA' rather than the actual area dA as the source of emission.

The example given in the text you quote, the spherically symmetric emitter, is a no-brainer.
Whatever the angular distribution of emission from a surface element on a sphere, the total emission will always be isotropic, simply due to symmetry.

 

[Philip Koeck]

I thought Lambert's law refers only to the directional dependence of light reflected / scattered from a rough surface illuminated at some (arbitrary) angle. This has little bearing on the approach to thermal equilibrium, because the scattered light has the same frequency as the incident light. The photons carry away as much energy as they had to begin with.

I think I can take my argument a step further based on what you say.

A cavity where all surfaces are perfect scatterers and/or reflectors will keep whatever temperature distribution it has if the wall material is a perfect insulator. This is simply because there's no mechanism available for heating or cooling the various surfaces inside the cavity if the surfaces don't absorb or emit.

If the surfaces of the cavity are black bodies, on the other hand, they will all reach the same temperature after a while. In this case it's clear (from several thought experiments all the way back to Kirchoff's and Lambert's own) that emission has to follow Lambert's cosine law in order to be in agreement with the second law.

If we now consider a more general surface material that absorbs, emits, scatters and reflects, but is (essentially) a thermal insulator, then this should mean that emission still has to be Lambertian.
The portion of radiation that is scattered or reflected has no effect on thermal equilibrium, just as you say, and the portion that is absorbed and emitted has to "follow the same rules" as for black bodies.

So I would conclude that all thermal insulators have to be Lambertian emitters.
Could that be correct?

 

[WernerQH]

So I would conclude that all thermal insulators have to be Lambertian emitters.
Could that be correct?

No. Lambert's law has a status quite different from Kirchhoff's. It is not fundamental, and only a useful approximation for some cases (walls, or sheets of paper), but not for others (e.g., window panes or mirrors).

I think there's some confusion in that text.

Palmer's FAQ was intended to dispel confusion. :-)

It's rather important to understand that a Lambertian emitter doesn't emit isotropically.
Otherwise you end up in conflict with the second law.

Do you think Palmer didn't know this? It seems more likely the confusion is yours than Palmer's.

 

[Philip Koeck]

No. Lambert's law has a status quite different from Kirchhoff's. It is not fundamental, and only a useful approximation for some cases (walls, or sheets of paper), but not for others (e.g., window panes or mirrors).

Palmer's FAQ was intended to dispel confusion. :-)

Do you think Palmer didn't know this? It seems more likely the confusion is yours than Palmer's.

Palmer writes: "What is the difference between lambertian and isotropic?

Both terms mean “the same in all directions” and are unfortunately sometimes used interchangeably."

It's pretty clear that Lambertian doesn't mean "the same in all directions". It means "proportional to cos θ".
Isotropic means "the same in all directions". That's why I said that there's some confusion in the text.
Palmer does continue to clarify later in the text, I agree.

I would say it is fundamental that black body emission is Lambertian and not just approximately Lambertian.
It's easy to show that anything else would lead to conflict with the second law.
So in that sense I would classify Lambert's law as a fundamental law and not just an approximation​

 

[Philip Koeck]

No. Lambert's law has a status quite different from Kirchhoff's. It is not fundamental, and only a useful approximation for some cases (walls, or sheets of paper).

I would say a white wall or a white sheet of paper is a Lambertian scatterer and not a Lambertian emitter!
A perfectly white surface doesn't emit at all.