# 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: $$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int_{E<H<E+dE} d\Gamma$$ which also solves the problem of zero integral and the dimensionality issue (it is now a dimensionless quantity) but the downside now is that it feels to me as if it is not well defined in this fashion because I defined the PDF by some infinitesimal qunatity. I know that there is a formal consturction of the infinitesimals but I'm not familiar with it.

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

• dE should not be taken to be infinitesimal, it's a finite but small quantity that defines a coarse graining scale. The whole point of statistical physics is to describe microscopic degrees of freedom statistically while keeping track of only a small number of variables that characterize the macroscopic properties of the system. – Count Iblis Jul 17 '16 at 0:40

## 2 Answers

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

$$\tilde \Omega(N,V,E) = \sum_{\text{hypercubes}: E<H<E+E_0} \frac{\delta x^{3N} \delta p^{3N}}{ h^{3N}} \simeq \frac{1}{h^{3N}} \int_{E<H<E+E_0} d^{3N}p \ d^{3N}q$$

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

$$\int_{E<H<E+E_0} d^{3N}p \ d^{3N}q \simeq E_0 \int d^{3N} p \ d^{3N} q \ \delta (H(\{q,p\})-E)$$

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

$$V_{shell}=\int_{R<\| \vec x \|<R+\epsilon} d^D x = V_D(R+\epsilon) - V_D(R)$$

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

$$V_{shell} =\int_{R<\| \vec x \|<R+\epsilon} d^D x = \frac{ \pi^{D/2}} {\Gamma(D/2+1)} \left( (R+\epsilon)^D - R^D \right) = \frac{ \pi^{D/2} R^D} {\Gamma(D/2+1)} \left[\left(1+\frac{\epsilon}{R}\right)^D - 1 \right]$$

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

$$\int_{E<\sum p_i^2/2m<E+E_0} d^{3N}p \ d^{3N}q = V^N \int_{E<\sum p_i^2/2m<E+E_0} d^{3N}p$$

and this integral is exactly an hyperspherical shell.

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

• Thank you for your great answer! although I have some questions: $\\$ 1) you said "The correct PDF is therefore $P(\{q,p\})=\frac{E_0}{N! h^{3N}} \ \delta (H(\{q,p\})-E)$" shouldn't I technically incorporate the factors $\frac{1}{N! h^{3N}}$ in the infinitesimal volume $d\Gamma$ of phase space when I perform the integration rather then in the PDF? 2) just to make sure, the solution to the problem I mentioned in my second definition of the PDF is to take $E_0$ instead of $dE$ where $E_0$ is some energy smaller than the resolution of our energy measurment? – dorsh605 Jul 17 '16 at 22:33
• 3) another thing that bother me is that you say that the integral $\int d^{3N} p \ d^{3N} q \ \delta (H(\{q,p\})-E)$ equals to the surface of the constraint $H(\{q,p\})=E$ and I think it's not accurate to say this – dorsh605 Jul 17 '16 at 22:33
• @dorsh605 1) You can incorporate the factor $1/N!$ wherever you want, but it has to appear somewhere because otherwise you would be counting twice the states in which the position in phase space of two particles is exchanged. 2) $dE=E_0$ must be a finite quantity, this is why I prefer not to use the notation $dE$ (some author, like Huang, use $\Delta E$). Actually it doesn't matter what value the constant has, as long as it is $\ll E$. But it is nice to think of it as the resolution of our instrument (not smaller than that). – valerio Jul 18 '16 at 7:52
• @dorsh605 3) I'm not saying that the integral equals the surface. If you read carefully, I'm saying that the integral equals the surface times the thickness, because I am integrating on a shell. This is like when you integrate some function $f(\vec r)$ between $R$ and $R+\Delta R$ in spherical coordinates and say that the integral is approximately equal to $f(R) \Delta R$. I'll add this example to the answer to clarify. – valerio Jul 18 '16 at 7:55
• @dorsh605 $f(r)$ has two roots, $\pm R$. I did the calculation with your method and got $4 \pi R$, so looks like we were both wrong. Anyway this is expected because we are doing this in 3 dimensions but as I said that approximation is expected to be good in an high-dimensional space. What I mean is that there is a great difference between $R$ and $R^2$ but there is a small difference between $R^{99}$ and $R^{100}$... – valerio Jul 18 '16 at 8:47

I think the key here is that you're misunderstanding these integrals.

Let's look at the following integral you wrote:

$$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int_{H=E} d\Gamma$$

And let's look at a common example when H is a function of p and q. Now this definition becomes:

$$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int_{H(p,q)=E} d\Gamma(p,q)$$

Can you see more clearly now that we are integrating along a constraint? This integral only adds up values in which the integration parameters are constrained to equal to energy. Can you see in this specific example that the resultant integral is not always zero? You say this general integral is always zero, but this is not true. Your redefinitions are not therefore necessary (and, by the way, your redefinition is actually the exact same definition, you just rewrote a piecewise function in a different form using a dirac delta).

We can rewrite this integral along a constraint using dirac delta functions:

$$\int_{H(p,q)=E} d\Gamma =\int \delta(H(q,p)-E) dq dp$$

When we write the LHS we mean the RHS. This is a mathematical definition. This is correct and complete, and there's no need for extra rigor here.

You were also considering the following:

$$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int_{E<H<E+deltaE} d\Gamma$$

This is a completely different physical system then the one you are considering before. Here you are allowing the energies of the hamiltonian from before to have some finite "thickness" (So E+$\Delta$E not E+dE). Now for normalization you can handle this with a second integral to add up each of these allowed energies.

Technically you are correct that in the limit where deltaE goes to zero this becomes the first definition, so in a sense it's more "general." But defining your state that could be described by the first definition by using this second definition in the limit where deltaE goes to zero is in now way more rigorous than just using the first.

Also you mentioned the units of the first integral aren't unity but this is not true.

Units of hbar:

Energytime = kg(m/s) (m/s)*s = p m (and since you're integrating over p and x you'll recover units of momentum times position)

• about the first part. I think I understoood how to perform this integration and I don't agree that $\int_{H(p,q)=E} d\Gamma =\int \delta(H(q,p)-E) dq dp$. think of integrating on the constraint of a sphere which is 2D while the integration is carried out in a 3-dimensional space the set of points of the constraint are then a set of measure zero, thus, the integral will be zero. that's why I have to put the dirac delta. – dorsh605 Jul 17 '16 at 6:29
• about the dimensions of the definition of omega with the dirac delta. you didn't mention the dimension of the delta which is one over the dimension of it's argument, hence, one over energy. – dorsh605 Jul 17 '16 at 6:29
• I'm still convinced you don't understand the initial definition. $\int_{H(p,q)=E} d\Gamma(p,q)$ is a path integral, not a volume integral. You can reexpress it as a volume integral with a dirac delta function. – Steven Sagona Jul 17 '16 at 7:09
• why do you say it is a path integral? $d\Gamma$ represents an infinitesimal volume in phase space and the constraint is just the domain of integration in phase space. – dorsh605 Jul 17 '16 at 7:16
• it doesn't make sense that it's a path integral. we got this expression from the demand that $P$ would be normalized. $P$ is defined on all of phase space so the normalization condition is that the volume integral of $P$ over all phase space will be equal to $1$ and from there we got the definition of $\Omega$ – dorsh605 Jul 17 '16 at 7:24