Turning summations into integrals in the thermodynamic limit

[Gabriel Maia]

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

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

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

$\sum_{\vec{k}} \rightarrow \frac{\displaystyle V}{\displaystyle (2\pi)}\int_{0}^{\infty}\,4\pi\,k^{2}dk$Could you explain how is this change from the summation to the integral is done?

Thank you very much.

 

[king vitamin]

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