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Number of photons in a mode (in a blackbody)

  1. Feb 7, 2012 #1
    # of photons in a mode (in a blackbody)

    I've been going over a proof again concerning the thermal average number of photons in a mode of a cubical blackbody of length 'L' at equilibrium with a reservoir of temperature 'tau' (has a small hole in it, etc. etc. etc.) The logic went as usual, determine what the energy levels of the system are, sum up the Boltzmann factors to create the partition function, then do the sum over all states of the quantity in question (the number of photons, here denoted 's') times the probability of finding the system in that state.

    A quick definition, by "mode" I mean different waveform possible to any one photon in the cavity, so each mode has its own unique 'ω' value.

    The derivation goes as follows:

    [itex]\large \epsilon_{s} = s\hbar\omega[/itex]
    Here 's' denotes number of photons in a mode, so this is the energy of having 's' photons in a mode of frequency 'ω'. Note that 's' is not the integer multiple 'n' that would be used to denote an exited level of one photon in the box (ie. integer multiples of 'n' would denote different modes or waveforms that the photon can be in IE. [itex]\omega = \frac{n\pi c}{L}[/itex]).

    [itex]\large Z = \sum_{s=0}^\infty e^\frac{-s\hbar\omega}{\tau}[/itex]
    Summation over the Boltzmann factors for all possible number of photons in a particular mode of frequency 'ω'.

    [itex]\large P(s) = \frac{e^\frac{-s\hbar\omega}{\tau}}{Z}[/itex]
    P(s) is the probability of finding the system in state 's'.

    [itex]\large <s> = \sum_{s=0}^\infty sP(s)[/itex]
    <s> is the then the thermal average number of photons in a mode.

    So.... My problem is that I don't understand why the initial sum to create the partition function wasn't a sum over ALL modes and ALL possible number of photons that could be in each mode (it should be a doulbe sum and not a single sum). Was each mode or possible wavefunction configuration considered as an isolated system at equilibrium with all other other modes? It should be that 'ω' can also very and create different energy levels in the system as well as the number of photons in each mode of frequency 'ω' (which in the specific case of the photon in a box of length 'L' should be [itex]\omega = \frac{n\pi c}{L}[/itex]. Why was the thermal average number of photons in a mode of frequency 'ω' independent of the other modes that existed inside the cavity? I guess I'm confused about what classifies as a system? If you can arbitrarily break the partition function apart and not sum all "states" it shouldn't give the correct answer. The photons could have been in any mode.

    Sorry if this seems long and drawn out, I just want to make sure everyone is seeing the exact derivation I've seen using the same syntax/variables/etc. If the variables aren't clearly defined things can get confusing really quick especially with the distinction of variables like 's' and 'n' and words like "mode" and "state".

    Any light shed would be greatly appreciated, this has been bothering me, I didn't think I was this confused by this.

    (This is just verbatim from Thermal Physics by Kittel and Kroemer 2nd. edition btw)
     
  2. jcsd
  3. May 29, 2012 #2
    Re: # of photons in a mode (in a blackbody)

    Hmmm, I feel slightly embarrassed now that I see that this wasn't a topic that people responded to, but I'll give it one more try and bump it.
     
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