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Casimir Effect, no idea

  1. Oct 24, 2013 #1
    Hello,

    I am attempting to repeat the math found on page 4 of this paper using the Euler-maclaurin summation formula. How would I incorporate the conversion factor because I can not figure it out for the life in me!

    Thank you!

    http://www.hep.caltech.edu/~phys199/lectures/lect5_6_cas.pdf
     
    Last edited: Oct 24, 2013
  2. jcsd
  3. Oct 25, 2013 #2

    fzero

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    Using the conversion factor means that we can write

    $$\nu = -\lim_{\epsilon\rightarrow 0} \frac{d}{d\epsilon} e^{-\epsilon \nu}.$$

    Note that I take ##\epsilon\rightarrow 0## instead of ##\infty## like in the text. I suspect the ##\infty## is a typo.

    This conversion factor is useful, because we can then write the sum as

    $$ \sum_{\nu = 1}^\infty \nu = -\lim_{\epsilon\rightarrow 0} \frac{d}{d\epsilon} \sum_{\nu = 1}^\infty e^{-\epsilon \nu}.$$

    Since ##e^{-\epsilon} < 1##, we can recognize this as a geometric series and do the sum. Expanding the result in ##\epsilon## will leave a divergent term, a finite term, and terms that vanish as ##\epsilon\rightarrow 0##. Introducing ## e^{-\epsilon \nu}## into the integral gives a result that cancels the divergent term in the sum. We are then left with a finite result when we compute the difference between the sum and integral.

    Alternatively, one can use the Euler-Maclaurin formula, which is also referred to in the text.
     
  4. Oct 25, 2013 #3
    Thank you, but the text states that they used the Euler-Macluarin and the conversion factor. Could this have been a typo?
     
  5. Oct 25, 2013 #4

    fzero

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    You only need one of the methods. Give it a try.
     
  6. Oct 25, 2013 #5
    oh ok thank you. I will let you know how it turns out asap!
     
  7. Oct 25, 2013 #6

    TSny

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    For using the Euler-Maclaurin formula, the hint might be that you should use ##e^{-\epsilon \nu}## as a "convergence" factor (rather than a "conversion" factor). Then let ##\epsilon \rightarrow 0## (not ∞, as fzero has already noted).

    Thus, consider the argument of the sum or integral to be ##\nu e^{-\epsilon \nu}##. Without the convergence factor, you run into trouble for ##\nu \rightarrow \infty## in the Euler-Maclaurin formula.
     
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