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Writing three-fold summation over n_x, n_y, n_z as single summation over n?

  1. Nov 17, 2011 #1
    1. The problem statement, all variables and given/known data
    On this page they're going from a three-fold summation to a single summation (cubed) (equation 58-59), something like this:

    [itex]
    \sum_{n_x} \sum_{n_y} \sum_{n_z} e^{-a\left(n_x^2 + n_y^2 + n_z^2\right)} = \left( \sum_n e^{-an^2} \right)^3
    [/itex], where [itex]a = \frac{h^2}{8kTmL^2}[/itex]

    Now, I've run into a similar problem, only I have the square root of the n's, like this:

    [itex]
    Z = \sum_{n_x} \sum_{n_y} \sum_{n_z} e^{-c\sqrt{n_x^2 + n_y^2 + n_z^2}}
    [/itex], where [itex]c = \frac{hc}{2kTL}[/itex]

    Could anyone show me how they do the first transition, or some hints to how I should start when trying to do a similar transition on my equation? I'm supposed to show that the sum equals

    [itex]
    Z(\text{high T}) = 8\pi L^3\left(\frac{kT}{hc}\right)^3
    [/itex]

    in the high-temperature limit. I know that I can approximate the sum with an integral, but I need to convert it to a single-sum first.

    2. Relevant equations

    [itex]
    \sum_{n_x} \sum_{n_y} \sum_{n_z} e^{-a\left(n_x^2 + n_y^2 + n_z^2\right)} = \left( \sum_n e^{-an^2} \right)^3 = \left(\int_0^\infty e^{-an^2} dn\right)^3
    [/itex]

    [itex]
    \int_0^\infty e^{-ax}dx = \frac{1}{a}
    [/itex]

    [itex]
    \int_0^\infty e^{-ax^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{a}}
    [/itex]

    3. The attempt at a solution
    I've tried letting [itex]\sqrt{n_x^2+n_y^2+n_z^2} = n[/itex], but then I end up with just [itex]n[/itex] and not [itex]n^2[/itex] in the exponent, and then I don't get the factor [itex]\pi[/itex] when solving the integral (see the two last equations for solutions to the integrals).

    I might be going at this problem all wrong, so I'm open to all suggestions!

    EDIT: Just found the solution myself. I'll show it in detail if anyone wants it, just ask and I'll write it out here.
     
    Last edited: Nov 17, 2011
  2. jcsd
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