Blackbody Radiation (Rayleigh-Jean) Equipartion of Energy Question

1. Jul 2, 2007

uart

There's one thing that has always puzzled me about the derivation of the old classical Rayleigh-Jean Law of blackbody radiation. I understand how they calculate the density of modes in the cavity however I don't see why they assign an "equipartition" energy of kT per mode instead of 1/2 kT as is used for example in the kinetic theory of gases to find the heat capacity of gases.

I'm pretty sure the correct "equipartiion" is 1/2 kT per mode and my hunch is that in the blackbody radiation derivation they use kT to allow for independent vertical/horizontal polarizations, though I've never seen this explicitly stated. Or is it 1/2 kT for the E field and another 1/2 kT for the B field, can someone please enlighten me as to the precise reason.

Thanks. :)

2. Jul 2, 2007

K.J.Healey

I don't know the answer to this specific question, though I will add that most likely when its a multiple of 1/2kT its due to another degree of freedom (as you have stated).

3. Jul 2, 2007

olgranpappy

You treat each mode as a harmonic oscillator contributing (1/2)kT. Each mode can be labelled by its polarization and its wave-vector. The polarization takes two values for each value of the wave-vector--that's where the factor of two comes from. I.e., when you sum over modes in this case the sum over polarizations is trivial and just gives a factor of two.

4. Jul 2, 2007

uart

So my hunch about polarization was correct. Thanks for confirming this guys.

One more question. Am I correct in thinking that the polarization must be the same for all three components of the wave vector. Otherwise you'd get a density of modes proportional to the 5th power of frequency instead of the correct result which is a 2nd power in freq {equiv to (-7)th power in Lambda instead of the correct (-4)th power of Lambda}. My electromagnetic is a bit rusty but I think it makes sense that all three components must have the same polarization - right?

5. Sep 25, 2009

Kuwait

polirization is the wrong answer. The degrees of freedom come from both the kinetic and the potential part of the oscillator with each part contributing one half.