statistical mechanics - number of states in microcanonical ensemble

I have a problem with the definition of $\Omega(E,V,N)$ — the number of microstates with $V$, $N$ and energy $E$. It starts with the definition of the PDF. If one defines the PDF as follows:

$P(\{q_i,p_i\})=\dfrac{1}{\Omega(E,V,N)}$ if $H(\{q_i,p_i\})=E$ and $P(\{q_i,p_i\})=0$ otherwise.

In this case it follows that in order for $P$ to be normalized we should demand that:

$$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int_{H=E} d\Gamma$$ but this integral is zero because the domain of integration is a set of measure zero. So as I saw, there are two options. the first is to redefine the PDF as: $$P(\{q_i,p_i\})=\frac{1}{\Omega(E,V,N)}\cdot \delta(H(\{q_i,p_i\})-E)$$ and then we get: $$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int \delta(H(\{q_i,p_i\})-E) d\Gamma$$ which solves the problem of zero integral but now it has the dimension of $\dfrac{1}{[E]}$ which is problematic if I want to consider $\log \Omega(E,V,N)$.

The second option is to redefine the PDF as:

$P(\{q_i,p_i\})=\dfrac{1}{\Omega(E,V,N)}$ if $H(\{q_i,p_i\})\in[E,E+dE]$ and $P(\{q_i,p_i\})=0$ otherwise

and then we get: $$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int_{E

can someone please shed some light on which of these 2 definitions are "better"?

Answer

The dimension issue is solved easily by defining the probability density function(PDF) as

$$P(\{q,p\})=\frac{E_0}{h^{3N}} \ \delta (H(\{q,p\})-E)$$

where $E_0$ is an arbitrary constant which will not affect any thermodynamic quantity or equilibrium property.

Actually, this definition is incomplete. We have to take into account the indistinguishability of particles, or we will encounter Gibbs' paradox. The correct PDF is therefore

$$P(\{q,p\})=\frac{E_0}{N! h^{3N}} \ \delta (H(\{q,p\})-E)$$

Also, it can be shown that the two definition of the PDF that you mention are almost equivalent:

Imagine to divide the phase space in hypercubes of volume $\delta x^{3N} \delta p^{3N} = h^{3N}$, and let's say that you can only determine the energy if each microstate to be within $E$ and $E+E_0$. We can then define a coarse-grained partition funcion

$$\tilde \Omega(N,V,E) = \sum_{\text{hypercubes}: E

Now, since we are integrating over a shell, we can approximate this integral as the hypersurface $H=E$ times the thickness $E_0$, obtaining

$$\tilde \Omega(N,V,E) \simeq \frac{E_0}{h^{3N}} \int d^{3N} p \ d^{3N} q \ \delta (H(\{q,p\})-E) = \Omega(N,V,E)$$

(The indistinguishability must be taken into account separately based on combinatorial arguments and is not included in the previous discussion).

The approximation

$$\int_{E

is good when $E_0$ is small. To see that this is plausible, consider the particular case of the integral over an hyperspherical shell:

$$V_{shell}=\int_{R<\| \vec x \|

where $V_D (R)$ is the volume of the hypersphere

$$V_D(R) = \frac{R^D \pi^{D/2}} {\Gamma(D/2+1)}$$

We then have

$$V_{shell} =\int_{R<\| \vec x \|

we do the following approximation for $\epsilon \ll R$

$$\left(1+\frac \epsilon R \right)^D \simeq 1 + D \frac{\epsilon}{R}$$

from which

$$V_{shell} \simeq \frac{ \pi^{D/2} R^{D-1}} {\Gamma(D/2+1)} \ \epsilon D$$

Now, since

$$\Gamma(D/2+1) = \frac D 2 \Gamma\left(\frac D 2 \right)$$

we have

$$V_{shell} \simeq \epsilon \frac{2 \pi^{D/2} R^{D-1}}{\Gamma(D/2)} = \epsilon S_D(R)$$

where $S_D(R)$ is the surface area of the hypersphere.

Fun fact: if we are dealing with non-interacting particles, i.e.

$$H=\sum_i^N \frac{p_i^2}{2m}$$

we have

$$\int_{E<\sum p_i^2/2m

and this integral is exactly an hyperspherical shell.

References: M. E. Tuckerman, Statistical Mechanics: Theory and Molecular Simulation