- #1
PerUlven
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Homework Statement
On http://www.chem.arizona.edu/~salzmanr/480b/statt02/statt02.html 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.
Homework 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]
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.
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