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Turning summations into integrals in the thermodynamic limit

  1. Jul 2, 2017 #1
    Hi. I'm reading a solution to a problem concerning a gas of photons. In the solution, the energy of the gas is given as

    [itex]E=2\sum_{\vec{k}} \frac{\displaystyle \epsilon_{\vec{k}}}{\displaystyle \exp[\beta\epsilon_{\vec{k}}]+1}[/itex]


    where [itex]\epsilon_{\vec{k}}[/itex] is one photon's energy. It is said then that in the thermodynamic limit we have

    [itex]\sum_{\vec{k}} \rightarrow \frac{\displaystyle V}{\displaystyle (2\pi)}\int_{0}^{\infty}\,4\pi\,k^{2}dk[/itex]


    Could you explain how is this change from the summation to the integral is done?

    Thank you very much.
     
  2. jcsd
  3. Jul 5, 2017 #2
    I'll just do the 1D case for simplicity. In a finite 1D box with periodic BCs, the momentum [itex]k[/itex] is really defined as
    [tex]
    k = k(n) = \frac{2 \pi n}{L}
    [/tex]
    where [itex]n[/itex] runs over all of the integers. Then the sum is really over these integers
    [tex]
    \sum_{k} f(k) = \sum_{n = \infty}^{\infty} f(k(n))
    [/tex]
    The thermodynamic limit can be though of as approximating this sum as an integral
    [tex]
    \sum_{n = \infty}^{\infty} f(k(n)) \rightarrow \int_{-\infty}^{\infty} dn f(k(n))
    [/tex]
    However, since [itex]k[/itex] has a more natural physical interpretation, we prefer to change variables. Then using the fact that [itex]dn = \frac{L}{2 \pi} dk[/itex], we obtain
    [tex]
    \int_{-\infty}^{\infty} dn f(k(n)) = \frac{L}{2 \pi} \int_{-\infty}^{\infty} dk f(k)
    [/tex]
    In [itex]d[/itex] dimensions, the factor is [itex]V/(2 \pi)^d[/itex] instead.
     
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