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I find in the liuterature (e.g. Landau & Lifshitz [1]) that the entropy in a microcanonical ensemble is given as: $S = k_B \log(\Omega),$ where $\Omega$ is the mutiplicity of microstates (Landau uses the notation $\Delta \Gamma = \Omega$), i.e., the entropy is given by the boltzman constant times the logarithm of the the number of different microstates compatible with a given constraint (e.g. energy in a given energy shell).

On the other hand, I find that [2] $S = k_B \log(W)$, where $W$ is the available phase space volume.

My conclusion from these two definitions is that the phase space volume $W$ and the multiplicity $\Omega$ are related by $\Omega = W/\hbar^n$ in $n$ dimensions, i.e., quantizing phase space volume to count the number of available states. Is this conclusion correct, and if it is, then I wonder where the $\hbar^n$ goes, since the two definitions of the entropy would have different units?

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  • $\begingroup$ The usual relation of multiplicity and phase volume is $\Omega = W/h^{3N}$, with the original Planck constant $h$ = 6.6e-34 Js. Also usually the multiplicity number is denoted $W$, and phase volume $\Omega$ (opposite to your definition). $\endgroup$ Feb 3 at 19:55

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In classical mechanics, the states of a system live on a smooth manifold: $\vec{p}$ and $\vec{q}$ are continuous variables for each particle that can take an infinite number of values, even after you ensure that your total energy is $E$ in the microcanonical ensemble. This means you can't really "count" the number of states—that's like asking to count the number of points on the sphere. What we can do is calculate the phase space volume available to the system at a particular energy, which is $$ V(E) = \epsilon \int d^{3N} \vec{p} \, d^{3N} \vec{q} \, \delta(E - H(\{\vec{p}_i, \vec{q}_i\})), $$ where $\epsilon$ determines the thickness of the $6N$-dimensional hypersphere whose volume we're calculating. This is a dimensionful quantity, with the same dimensions as $q^{3N} p^{3N}$ or $M^{3N} T^{-3N}L^{6N}$. The purely classical explanation of what happens next is that we divide $V(E)$ by an dimensionful constant, suggestively named $h^{3N}$, that cancels the dimension of $V$ (if you're being careful, you should not take logarithms of dimensionful quantities) and allows us to state that $\Omega \equiv V(E)/h^{3N}$ is the "number of states." If we take the 3rd law of thermodynamics as an axiom (putting aside the idea of residual entropy for a moment), we can fix the value of $h$ to ensure that $S=k\log \Omega$ goes to zero at the minimum value of $E$, which corresponds to zero temperature in the canonical ensemble.

Of course, the real world is described by quantum mechanics, where the available states are quantized, and it just so happens that the $h$ that makes classical statmech give you zero entropy at zero temperature is Planck's constant, the defining scale for quantum phenomena to become important. If you treat your system quantumly, you can just count up the states and say that that number is $\Omega$, and if you take the hot, large limit, you will get a result that agrees with the classical picture. There was lots of speculation about the relationship between $h$-the-blackbody-constant and $h$-the-quantum-of-action around the time that quantum mechanics was being formulated, but I don't know the history well enough to tell that story properly.

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  • $\begingroup$ Your $V(E)$ has dimensions $q^{3N}p^{3N}/E$, so $M^{3N-1}T^{2-3N}L^{6N-2}$, because delta distribution has dimensions of inverse energy. The correct definition of $V$, giving phase volume numerically and with proper units of $q^{3N}p^{3N}$, has additional constant factor $\epsilon$ with units of energy. This factor defines thickness of the $6N$-dimensional hypershell of energy $E$ whose phase volume we get. $\endgroup$ Feb 3 at 20:08
  • $\begingroup$ Ah I'm sorry, you're right! I'll change it. $\endgroup$ Feb 3 at 20:23
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    $\begingroup$ Beware, although $\epsilon$ determines thickness of the hypershell in the $6N$ dimensional phase space, it is not it, because lengths in phase space are not measured in units of energy. $\endgroup$ Feb 3 at 22:09

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