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Simple derivation of Casimir Force

  1. May 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Derive the Casimir Force on each plate, for a two parallel plate system (L x L), separated at a distance of 'a' apart.

    The solution was found in en.wikipedia.org/wiki/Casimir_effect#Derivation_of_Casimir_effect_assuming_zeta-regularization. (sorry I couldn't include link yet). Now my question is how did the (2∏)^2 came out in the integral for <E>,


    I would think it as the constant from fourier transform but I was unable to prove that. Any idea how did that thing pop up of nowhere?
  2. jcsd
  3. May 23, 2012 #2
    oh this question is moot. It's basically multiplication of the DOS.

    Thanks anyway
  4. May 23, 2012 #3
    Assume a large hypercubic box in d dimensions of length L. Impose periodic boundary conditions (PBCs) on any function:
    \psi(x_1 + L, x_2, \ldots, x_d) = \psi(x_1, x_2 + L, \ldots, x_d) = \ldots = \psi(x_1, x_2, \ldots, x_d + L)
    Then, we can expand the function in multidimensional Fourier series:
    \psi(\mathbf{x}) = \sum_{\mathbf{k}}{c_{\mathbf{k}} \, e^{i \mathbf{k} \cdot \mathbf{x}}}
    \mathbf{k} = \frac{2\pi}{L} \langle n_1, n_2, \ldots, n_d \rangle
    is a multidimensional wave vector that can take on discrete values.

    In an interval [itex](k_i, k_i + dk_i)[/itex] of the ith component, there are
    dn_i = \frac{L}{2\pi} \, dk_i
    To find the total number of states within an infinitesimal volume of k space
    dn = \mathrm{\Pi}_{i = 1}^{d}{dn_{i}} = \frac{L^{d}}{(2\pi)^{d}) \, d^{d}k
    So, the famous factor [itex]L^{d}/(2\pi)^{d}[/itex] gives the density of states in k space.
  5. May 24, 2012 #4
    Ah. Yeah Ive forgot about the DOS.
    Thanks for the detailed explaination!
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