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Blackbody radiation

  1. Aug 11, 2014 #1
    Hello,

    in Schwartz's QFT-book it says that:
    "Classically, a box of size L supports standing electromagnetic waves with angular frequencies [itex]\omega_n = \frac{2\pi}{L}\left|\vec{n}\right|c [/itex] (...)"

    I wonder if the factor 2 is really correct, I only get this factor 2 if I suppose that eg. for the wavelength in x-direction the possible values are [itex]L = n\lambda_{x,n}[/itex], however I would expect that [itex]L = n\lambda_{x,n}/2[/itex].
     
  2. jcsd
  3. Aug 29, 2014 #2
  4. Aug 30, 2014 #3

    vanhees71

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    Well, the answer is formally correct, but the issue is more subtle! Periodic boundary conditions are nice, because you can use them without mathematical quibbles in quantum theory, because there you still have momentum operators, which is not the case for rigid boundary conditions. The calculation nevertheless goes through as stated, because you only need the Hamiltonian, which is still well defined also with rigid boundary conditions.

    Also physically it's more appropriate to use periodic boundary conditions, because you treat the black-body radiation in the sense of the canonical ensemble, i.e., you put a finite volume box into the much thermal radiation field contained in a very much larger volume and consider the energy conservation on average within the subvolume (there are enery fluctuations, because photons can leave and enter the subvolume). The maximum-entropy principle then justifies the use of the canonical distribution function. Under the constraint of a given mean energy the unique Statistical Operator fulfilling the maximum-entropy principle is the canonical distribution
    [tex]\hat{\rho}=\frac{1}{Z} \exp(-\beta \hat{H}), \quad Z=\mathrm{Tr} \exp(-\beta \hat{H}).[/tex]
    BTW: This posting should appear in the quantum-theory forum, because blackbody radiation cannot be understood in terms of classical physics. In fact this problem let Planck to discover quantum theory (after about a decade of work on the problem!).
     
  5. Aug 31, 2014 #4
    There is a reason why I posted it in the classical physics section. The books starts with the "wrong"/classical argument. That's also why in my first post it says "Classically, a box of size L supports...". And that's probably also the reason why I thought of rigid boundary conditions, because this is what makes sense to me in a classical sense.

    Also I don't know what you mean by "into the much thermal radiation field".
     
  6. Aug 31, 2014 #5

    vanhees71

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    I meant "...into the much larger volume of the radiation field..."
     
  7. Aug 31, 2014 #6

    atyy

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    I assume PBCs are like putting the system on a torus? Does one really get different results in the canonical ensemble if the bath is on a torus or just another much bigger box?
     
  8. Aug 31, 2014 #7

    vanhees71

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    I don't think that one gets a different answer when taking the thermodynamic limit, i.e., going from the finite volume with discrete momenta to a "very large volume" and taking the limit to momentum integrals. I only have my quibble with the interpretation of particles in a box with rigid boundary conditions, because then there are, strictly speaking, no momentum operators anymore.

    That's why I also would use the example of a Schroedinger particle in a box potential with infinitely high walls (which shows that this example is unrealistic anyway) only to demonstrate the important difference between hermitean and self-adjoint operators.
     
  9. Aug 31, 2014 #8

    atyy

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    I guess there are some strange cases (not the EM field) where the ground state degeneracy does depend on the topology of the manifold. I'm not sure how this plays out in thermodynamics. http://arxiv.org/abs/cond-mat/9711223
     
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