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Blackbody Radiation (Rayleigh-Jean) Equipartion of Energy Question

  1. Jul 2, 2007 #1


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    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. jcsd
  3. Jul 2, 2007 #2
    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).
  4. Jul 2, 2007 #3


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    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.
  5. Jul 2, 2007 #4


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    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?
  6. Sep 25, 2009 #5
    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.
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