# Planck Law without a cavity

1. Aug 5, 2011

### RGann

It is perfectly possible to derive, for instance, the Maxwell distribution of speeds in a heuristic way with only two things:

1. The Gibbs hypothesis, which says that to count the number of states at speed v you take all the points in phase space between v and v+dv in velocity space. This makes the "number of ways", g, proportional to v2.
2. The Boltzmann factor

Then you can write the distribution of speeds by $$P(v) = A g(E) P(E) = A v^2 e^{-m v^2/2 k T},$$ and normalize the distribution. The idea, of course, is that there are more ways to have a given speed v than a smaller one, but the probability goes down from the Boltzmann factor, the competition of which gives the MB distribution. This is not rigorous, but it's intuitively satisfying.

I wonder whether the same is not possible for the blackbody spectrum (Planck's Law). The typical proof given depends on the body being in a cavity, so that the modes can be counted. This then allows for the calculation of the properties of a photon gas in equilibrium with the object, so that when you poke a hole in the cavity, the photon gas leaks out and you see the spectrum.

Needless to say, I find this proof distasteful for conceptual reasons. The object radiating is not in a cavity, and I find it hard to believe that this construct is the only way of doing it. Moreover, I don't think that students have a hard time with discrete energy levels of an atom. So I wonder if an argument couldn't be constructed, with a toy model, as follows:

In a substance (of any phase), where the atoms are in equilibrium, the way the equilibrium is mediated (e.g. phonons) can give rise to promotion of electrons in the atoms to higher energy levels. Presuming that the outer electrons are in equilibrium with the substance, the average level they can be promoted to is indicated by the Boltzmann factor, e^{-E/k T} where the energy is given by n h f. Then, the average energy of the excited atom would be
$$E_\text{avg.} = \frac{h f}{e^{h f/ k T}-1}.$$ (This can be computed by finding the partition function, which is a geometric series, and either differentiating -ln Z with respect to beta, or finding the probability by (1/Z) times the sum of E e^{-E/kT}.) These electrons then decay to the ground state giving rise to a spectrum of photons.

You can see on http://en.wikipedia.org/wiki/Planck%27s_law" [Broken], where they reach the same. However, the density of states, g, is now needed to compute the internal energy. I'm having some kind of mental disconnect. Without use of the cavity, how does one say what the density of states is? Is it possible to state it in terms of the components of the wave vector?

The ultimate answer is the intensity (energy radiated per time per area), which is
$$I(f,T) = \frac{2 h f^3 }{c^2} \frac{1}{e^{h f/kT}-1}.$$
I have scoured the web, Am. J. Phys, Found. Phys., and not found any treatment that gives one a feel for this problem. Any ideas?

Last edited by a moderator: May 5, 2017
2. Aug 6, 2011

### vanhees71

This is a quantum problem and thus perhaps my answer is not welcome here. Then one of the adminstrators may move it to the quantum-physics section. To stress it: This problem cannot be treated in classical physics. That's why quantum theory has been born with Planck's finding of the black-body-radiation distribution.

The density of states should indeed first be counted in a finite quantization volume. For simplicity's sake, take a cubic box with periodic boundary conditions. A complete set of single-photon states is then given by the momentum eigenstates. The possible momentum values are

$$\vec{p}=\frac{2 \pi}{L} \vec{n}, \quad \vec{n} \in \mathbb{Z}^3.$$

Now make the box very large (against the considered typical wavelengths of the problem). Then in a momentum-space-volume element $\mathrm{d}^3 \vec{p}$ you have

$$L^3 \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3}$$

momentum states. Since the photon is a massless particle with spin 1 it has two helicity states per momentum state, and thus the correct phase-space distribution for photons in the thermodynamical limit reads

$$f(\vec{p})=\frac{2}{(2 \pi)^3} \frac{1}{\exp(|\vec{p}|/T)-1},$$

because the dispersion relation reads $E=|\vec{p}|$.

I used natural units here, i.e., $\hbar=k=1$.

3. Aug 6, 2011

### RGann

Thanks for the reply. I think this forum is the closest, since it's the only one labeled for thermodynamics.

I suppose the argument being made is that confining a gas of photons to a solid necessarily forces periodic boundary conditions, thereby constraining momentum so that standing waves can be formed. The fact that I find it distasteful does not mean that it isn't true.

4. Aug 6, 2011

### RedX

Have you tried using a Debye solid? A solid is naturally confined to itself.

5. Aug 7, 2011

### RGann

To my recollection, I've never seen a treatment of Debye theory that did not also invoke periodic boundary conditions to set the phonon wavelength with nodes at the boundaries. In many respects, the two are not very different. That's too bad, because invoking periodic boundary conditions in a problem tends to get weird looks from students---and justifiably so.

6. Aug 7, 2011

### RedX

Well it's been awhile since I looked at thermodynamics, but I thought a Debye solid didn't require periodic boundary conditions, just vanishing at the endpoints.

So you have a solid, and the frequencies it can vibrate at correspond to wavelengths that are 2L/n for positive integers n and length L of the solid. So if the solid can only vibrate at those frequencies, then the electromagnetic radiation can only occur at those frequencies, and then you can get the energy density, since 2L/n was the same way to derive the energy density in the cavity.

The problem though would be that frequencies of vibrations using the speed of sound at a given wavelength are different than using the speed of light on the same wavelengths.

7. Aug 7, 2011

### RGann

I agree that this is sound. Functionally, periodic BCs and vanishing at the boundary are equivalent. At that point, though, you've resorted to treating the spectrum as being due to a particle in a box. That is, without talking about a photon gas confined to a volume, you can't really get anywhere. And for whatever reason, I guess because I'm a dumb experimentalist, that doesn't seem like it should be necessary.