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In Statistical Mechanics, what is the procedure of replacing this summation by the integration given by $$\sum_{\vec k} \rightarrow \frac{V}{(2\pi)^3} \int_{0}^{\infty} 4\pi k^2 dk$$

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  • $\begingroup$ If $|\Delta \vec{k}| \ll |\vec{k}|$ holds for most of the states in the heat bath then we can approximate $\Delta \vec{k} \sim d\vec{k}$ and send the sum to an integral. There is no real procedure involved, you just replace the sum by an integral and send the summand to an integrand. $\endgroup$ Jan 6 '15 at 7:49
  • $\begingroup$ I guess, there should be a method to it. $\endgroup$ Jan 6 '15 at 7:53
  • $\begingroup$ @FenderLesPaul: I think maybe OP wants to understand where the $V$ comes from, etc. This is not completely trivial. Try explaining about the mode density in a box of volume $V$. $\endgroup$
    – DanielSank
    Jan 6 '15 at 9:21
  • $\begingroup$ Essentially a duplicate of physics.stackexchange.com/q/143467/2451 $\endgroup$
    – Qmechanic
    Jan 6 '15 at 9:43
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The answer to your question is already contained inside this answer. Without reiterating the same answer again, essentially there is a conversion from Cartesian to spherical coordinates:

\begin{align} \int\int\int\rho(k)dk_xdk_ydk_z &= \int^k_0\int^{2\pi}_{0}\int^{\pi}_{0}\rho(k)k^2\sin\theta d\theta d\phi dk\\ &= \frac{V}{(2\pi)^3}\int^k_04\pi k^2dk \end{align}

where $k$ is the "radius" of the spherical coordinate system.

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