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Why does Planck's constant appear in classical statistical mechanics.

I gather a constant appears in because we would like to count classical states in phase space and so therefore we have to separate phase space into "boxes" such that $$\delta q_i\delta p_i=h_0$$

Is a good reason as to why this constant is equal to $\hbar$?


marked as duplicate by ACuriousMind, Kyle Kanos, Rob Jeffries, user10851, John Rennie May 5 '15 at 7:02

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    $\begingroup$ Well, the reason comes from QM. But the important thing in classical statistical mechanics is that as long as $\hbar$ is small, results are independent of the actual value of $\hbar$ $\endgroup$ – Ihle May 4 '15 at 12:10

$\hbar$ does not need to appear in classical statistical mechanics. You are free to replace it with any quantity with units of angular momentum, say $\hbar_{\mathrm{C}}$. As long as this is choosen smaller than the size you can experimentally probe (i.e., as long as you don't ask questions of the theory that contain structure on this length scale or below) then the theory's predictions will be independent of $\hbar_{\mathrm{C}}$.


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