# What is the definition of entropy in microcanonical ensemble?

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?

• 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. Commented Mar 2, 2017 at 7:10
• @Yvan, if $\rho$ is a constant, then the value of $\Gamma(E)$ would be proportional to the number of accessible microstates, not equal. Commented Mar 3, 2017 at 17:59
• It does not matter, the thermodynamic entropy is defined only up to a constant. Commented Mar 3, 2017 at 18:23
• (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.) Commented Mar 3, 2017 at 18:31
• 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. Commented Mar 3, 2017 at 18:35

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: