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Calculation of Casimir Effect

  1. Dec 3, 2009 #1
    Hello I am trying to work out the Casimir force via the Abel-Plana equation. I have been following the derivation in http://arxiv.org/abs/quant-ph/0106045.

    Specifically I can't figure out for the life of me how the author is going from the first part of equation 2.33 to the second part of it. I am trying to figure out if it something simple I am just over looking.

    In my calculations the last term of the first equality has a 1/2 that refuses to go away. This prevents regularization using the Abel-Plana equation. The only way I can see around it, is to assume the integration over dk3 goes from -infinity to infinity and convert that to twice the integral of 0 to infinity. In which case one has to ask how does negative wave numbers make any sense?

    I am typing this really late as I have spent all day on this, I will try to put up some equations tomorrow if I have time. But everything that is relevant is in the arXiv paper.
     
  2. jcsd
  3. Dec 3, 2009 #2

    diazona

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    wow :surprised that's a long paper...
    That seems quite reasonable to me. Typically when you see an integral written without limits in a paper or book on quantum theory, it's implied that the integral is over the whole applicable region, which for a 1-dimensional integral is usually [itex]-\infty[/itex] to [itex]+\infty[/itex].

    The negative wavenumbers represent waves that are traveling in the negative direction. You're familiar with the expression
    [tex]e^{i(k x - \omega t)}[/tex]
    for a complex wave, right? For [itex]k > 0[/itex] (and [itex]\omega > 0[/itex]), the wave's velocity is positive, as you can see if you use the stationary phase condition,
    [tex]k x - \omega t = k(x - v t) = \phi_0[/tex]
    (As t increases, x must also increase to keep the combination (x - vt) a constant) But if you change to [itex]k < 0[/itex], it's the same kind of wave just moving in the negative direction. I seem to remember seeing a picture somewhere on Wikipedia that would show this rather well, but I can't find it now.
     
  4. Dec 4, 2009 #3
    Yes I realized that too earlier today but then why is that same thing not also done with the dk1 and dk2 integrals? Also if you look a bit further up at the example of the simple massless scalar field in 1D the integral over dk (no subscript) this is equation 2.17, is explicitly from 0 to infinity. Of course this whole thing just keeps on getting more confusing because a bit further up equation 2.14, k does go from -infinity to infinity!

    As of today the best explanation I can come up with is since in equation 2.29 the discrete sum is taken from -infinity and to infinity to account for two polarizations of photons (this being the 3D parallel plates example now rather then the 1D Scalar field) this is equivalent to multiplying a discrete sum from 0 to infinity by 2 which is the 1D scalar field energy of between the plates - equation 2.11 (the fact that the latter starts from 1 and the former starts from 0 can be rectified by subtracting out of the n=0 term).

    Therefore by the same logic comparing equations 2.32 and 2.16 which are the free space energies of the vacuum without boundary conditions in 3D and 1D respectively one should multiply 2.32 also by 2 in order to account for photon polarization. This will resolve the issue. But then why would they not say that explicitly, but sneak it in in the middle of a damn derivation. Which anyone who didn't sit down with a pencil and tried working out the math themselves could have easily missed.
     
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