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Statistical mechanics, confused about an approximation and limits of integration

  1. Sep 25, 2014 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Hello, I tried to solve a problem on my own and then I looked up a solution on the web, and I realize that it seems that I goofed. The problem statement can be found at http://www.hep.fsu.edu/~reina/courses/2012-2013/phy5524/homework/solutions/hw5_sol.pdf [Broken] (Problem 1, part a).


    2. Relevant equations
    Partition function and Helmholt free energy expressions.


    3. The attempt at a solution
    O
    n page 2 of the document, it is written that ##Z_1(T,V)=C\int d \vec R d\vec r \exp \left ( - \frac{\beta K |\vec r|^2}{2} \right )## so far so good. Despite having only 1 "integral sign", it's really 6 integrals to perform.
    Now according to the solution on the web, ##Z_1(T,V)=C V \left ( \frac{2\pi kT}{K} \right )^{3/2}##.
    It means that the solution considers that the triple integral with ##d\vec R## is equal to the volume of the container of the gas, where ##\vec R =\frac{\vec r_1 +\vec r_2}{2}##; I still have to digest this.
    But what really annoys me is the other triple integral, namely ##\int d\vec r \exp \left ( - \frac{\beta K |\vec r|^2}{2} \right )=\left ( \frac{2\pi kT}{K} \right )^{3/2}##
    Where ##\vec r =\vec r_1-\vec r_2##. The result the solution provide implies that the limits of integration are negative and positive infinity... On my draft I considered a cubic box of lengths ##V^{1/3}## so that the result of that triple integral contained the error function (erf).
    And I do not see any argument as to why I could assume that the limits of integrations can be considered as infinite. Is it because the lengths of the sides of the box are huge compared to ##|\vec r|##?
    Ah no... it must be that the integrand is almost 0 for small values of r and onward... okay... and so the approximation is valid. Is this correct?

    The other problem I have with the solution given is when they obtain the Helmholtz free energy as ##F\approx NkT \{ \ln (N) -1 -\ln [Z_1(T,V)] \}## and they just sit on that value.
    But since N is enormous, isn't better to reduce F to approximately ##NkT \{ \ln (N) -\ln [Z_1(T,V)] \}##?
    Why do they keep the "1"... is it because it's multiplied by NkT ? Still that doesn't make sense to me... it's ridiculous small compared to the natural logarithm of N...
     
    Last edited by a moderator: May 7, 2017
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