I am going through Statistical Mechanics book by Kerson Huang and he defines the entropy as, $$S(E, V) = k_B \log \Gamma(E)$$ where $\Gamma(E)$ is the volume in phase space occupied by the microcanonical ensemble, $$\Gamma(E) = \int_{E<\mathcal{H}(p,q)<E+\Delta}\ d^{3N}p\ d^{3N}q$$ All other books I've studied on Statistical Mechanics define entropy as, $$S = k_B \log\Omega$$ where $\Omega$ is defined as the number of accessible microstates corresponding to the given energy between $E$ and $E+\Delta$. My understanding is that these two definitions will be equivalent to each other only if, $$\Gamma(E) = \int_{E<\mathcal{H}(p,q)<E+\Delta}\ d^{3N}p\ d^{3N}q\ \rho(q, p, t)$$ where $\rho$ is the density function and therefore, $d^{3N}p\ d^{3N}q\ \rho(q, p, t)$ will represent the total number of accessible microstates in the phase space volume $d^{3N}p\ d^{3N}q$. So, please explain how these two seemingly different definitions do not contradict each other. What am I missing?
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- 1$\begingroup$ There is no contradiction: $\rho(q,p)$ is equal to a positive constant for all configurations with energy in the interval and to zero for the other ones. See the "postulate of equal a priori probability", equation (6.7) in the book (second edition). This postulate defines the microcanonical ensemble. $\endgroup$Yvan Velenik– Yvan Velenik2017-03-02 07:10:47 +00:00Commented Mar 2, 2017 at 7:10
- $\begingroup$ @Yvan, if $\rho$ is a constant, then the value of $\Gamma(E)$ would be proportional to the number of accessible microstates, not equal. $\endgroup$sigsegv– sigsegv2017-03-03 17:59:48 +00:00Commented Mar 3, 2017 at 17:59
- $\begingroup$ It does not matter, the thermodynamic entropy is defined only up to a constant. $\endgroup$Yvan Velenik– Yvan Velenik2017-03-03 18:23:35 +00:00Commented Mar 3, 2017 at 18:23
- $\begingroup$ (Moreover, what do you actually mean by "the number of accessible microstates" when discussing a classical system? The energy shell is made of a continuum of distinct microstates.) $\endgroup$Yvan Velenik– Yvan Velenik2017-03-03 18:31:44 +00:00Commented Mar 3, 2017 at 18:31
- $\begingroup$ Number of accessible microstates is the total number of points in the phase space enclosed in the volume $E$ and $E+\Delta$. So, I reasoned like, the higher the volume, the larger the number of points and the number of points should be the density of points times the volume. $\endgroup$sigsegv– sigsegv2017-03-03 18:35:05 +00:00Commented Mar 3, 2017 at 18:35
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1 Answer
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The density $\rho$ would count the number of microstates within the volume $d^{3N}p\,d^{3N}q$ that satisfies the energy constraint $E<\mathcal{H}<E+\Delta$. So you'd actually have:
\begin{align} \Gamma(E) &= \int_{E<\mathcal{H}<E+\Delta} d^{3N}p\,d^{3N}q \\ &= \int \rho\,d^{3N}p\,d^{3N}q \end{align}
