# Derivation of Planck's Law

1. Dec 19, 2011

### skippy1729

Can Planck's law be derived from QED and statistical mechanics? Any references appreciated.

Skippy

2. Dec 19, 2011

### dextercioby

Re: Derivation of {lanck's Law

From QED I would say no. It is derived from the statistics of the free Bose gas.

3. Dec 19, 2011

### Demystifier

Re: Derivation of {lanck's Law

Yes it can. See
J. J. Sakurai, Advanced Quantum Mechanics,
the end of Sec. 2.4 (pages 46-47).

4. Dec 19, 2011

### skippy1729

Re: Derivation of {lanck's Law

Really? In two pages? Sorry if I am skeptical but I am already into my 2014 book budget and I live hundreds of miles from a real library.

I am looking for something that starts with fundamental emission, absorbtion and scattering processes with minimal "hand waving". Is Sakuri going to do this? I tried google books and Amazon but they don't have a preview.

Thanks, Skippy

5. Dec 20, 2011

### Demystifier

Re: Derivation of {lanck's Law

Well, before these two pages Sakurai makes other derivations on emission and absorption which he uses also for other purposes. So, the full derivation takes more than two pages. But it's all there in the book.

I send you additional info in a PM.

6. Dec 21, 2011

### skippy1729

In Sakurai's derivation he says: "Suppose the radiation field is enclosed by "black" walls which are made up of atoms and are capable of absorbing and re-emitting photons of any frequency." There is nothing about how the "almost continuous" classical spectrum arises from the discrete spectrum of atomic transitions. In other words how does the spectrum smear out or thermalize? It must be some type of inelastic scattering process with the neutral particles of the walls or the contained gas. I can't seem to find any explanation of this in terms of fundamental processes.

Skippy

7. Dec 21, 2011

### Demystifier

Here is one mechanism I am aware of. Atoms are not at rest, but have a velocity due to the thermal motions. The spectrum of kinetic energies due to the atom motion is continuous. The total energy is the sum of this kinetic energy and the usual discrete energy of electron levels. Of course, this effect is larger at higher temperatures.

8. Dec 21, 2011

### skippy1729

Of course, neutral atoms in a gas or solid will eventually thermalize due to motions and collisions. But what is the mechanism whereby a hot lump of coal will glow and emit photons with frequencies not in the discrete spectrum of carbon? This is what puzzles me.

Skippy

9. Dec 21, 2011

### dextercioby

After checking Sakurai's argument, I keep my statement above in post #2 and say that Sakurai doesn't derive it from QED, but from the theory of the quantized electromagnetic field.

10. Dec 22, 2011

### Ken G

I believe the answer to that is that the carbon atoms are strongly perturbed by their environment, so the energies of emission are not restricted to those of isolated carbon.

11. Dec 22, 2011

### tom.stoer

Only isolated atoms have a discerete spectrum; in a solid the discrete of the energy levels are split into a quasi-continuum of levels, a so-called band; in a hot gas or plasma the spectrum is contuous due to recoil and doppler broadening.

Regarding QED, emission, absorbtion and scattering processes: the funny thing with statistical mechanics is that you don't need these details; the macrsoscopic behavior of a free bose gas is rather independent, therefore it works with nearly all hot lumps of matter

12. Dec 22, 2011

### vanhees71

The modern derivation of Planck's Law is to assume to have a massless non-interacting gas of photons. It's one of the most simple calculations in finite-temperature field theory. You may quantize the free electromagnetic field either by completely fixing the gauge, using the radiation-gauge conditions,

$$A^0=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$

Then you only describe physical photons (two transverse spacelike polarization modes, e.g., helicity eigenmodes).

Then you calculate the partition sum in the rest frame of the heat bath.

$$Z[\beta]=\mathrm{Tr} [\exp(-\beta \hat{H})].$$

This leads to the thermodynamic quantities of a massless ideal gas with the weight 2 for the 2 polarization states.

13. Dec 22, 2011

### skippy1729

Thanks to all for the constructive replies.

So, the atoms of the black walls of a cavity will have will have smeared energy levels due to the fact that they are not governed by a free Hamiltonian. They can emit and absorb photons from an almost continuous spectrum.

Suppose the cavity is filled with a low density hydrogen gas at a temperature lower than the walls of the cavity. There will be some conduction at the surface of the cavity wall from the occasional contact with the hydrogen atoms and a resulting transfer of momentum. There will be some absorption and emmision of energy by the hydrogen at energies of its discrete spectrum which will not have a net corresponding momentum transfer and thus no heating effect. So is the heating of the neutral hydrogen gas solely due to conduction at the walls and convection to the interior? Is radiation involved at all in reaching the equilibrium state? I am assuming that the individual hydrogen atoms in a sufficiently rarefied gas will retain their discrete spectrum.

Skippy

14. Dec 23, 2011