I'm a 2nd year physics undergraduate and recently I've volunteered to give a short presentation on the Sackur-Tetrode equation derivation and its use at removing the Gibbs paradox. I've looked on the Internet and in some books and found many methods for deriving this result. The one I'm going with is the derivation from the book Thermodynamics and Statistical Mechanics by W. Greiner. However, there are some things I'm still not certain about (I dared to expand the argument a little) and in order to seem like I know what I'm talking about, I need some clarification. Here's more or less how I see it going:

In order to obtain an expression for the entropy of an ideal gas with $N$ (identical) particles of mass $m$ occupying a volume $V$ and with internal energy $E$, we use Boltzmann's entropy formula: $$S(N,V,E)=k\ln \Omega(N,V,E).$$

This reduces the problem to finding $\Omega(N,V,E)$, the number of microstates of our system which realise the macrostate $(N,V,E)$.

A microstate of the system gives information about the entire system at a microscopic level, that is, it specifies the exact momentum and position of each constituent of the system. There are $N$ particles and each one has three directions of position and of momentum, therefore our system has $6N$ degrees of freedom.

We imagine a $6N$ dimensional space, called the phase space, where each axis corresponds to a degree of freedom or the system. Each point in this space corresponds to a specific microstate: $(r_x^{(1)},r_y^{(1)},r_z^{(1)}, p_x^{(1)}, p_y^{(1)}, p_z^{(1)},...,r_x^{(N)},r_y^{(N)},r_z^{(N)}, p_x^{(N)}, p_y^{(N)}, p_z^{(N)})$.

The microstates which correspond to our particular macrostate $(N,V,E)$ form some sort of high-dimensional hypersurface. The "volume" of this surface, $\phi(N,V,E)$, would give us a quantity related to the number of microstates. Of course in phase space, "volumes" of surfaces have dimension of action raised to some power, and since we're interested in a number, the idea is to introduce some sort of elementary bit of "volume", $\phi_0$ and then divide to produce a number $\frac{\phi(N,V,E)}{\phi_0}$ which we claim to be the number of relevant microstates.

Fortunately, it physically makes sense for such an elementary "volume" to exist, since from the uncertainty principle we know that a point in phase space really cannot be specified with infinite precision, and that there is a quantum of action, which is nice because our "volumes" have dimension of action to some power.

We already used the fact that we have $N$ particles. Now it's time to take our other givens into account. Let's start with $E$. Classicaly, for an isolated system of noninteracting particles, the total energy is purely kinetic and given by the sum of the kinetic energies of each particle:

$$E=\sum_{k=1}^{N} \frac{ [p_{x}^{(k)} ]^2 +[p_{y}^{(k)} ]^2 +[p_{z}^{(k)} ]^2 }{2m}$$

That gives a bound for our momenta coordinates. The position coordinates can be likewise bounded using our last given: the volume $V$. We may assume that the system occupies a cube of length $l$, then for every $k=1,2,...N$ we have:

$$0\le r_{x}^{(k)} \le l ; \quad 0\le r_{y}^{(k)} \le l ; \quad 0\le r_{z}^{(k)} \le l ; $$

This seems like cheating a bit (why assume a particular shape of the container), but the assumption can be later dropped because in further calculation we don't really care about the bounds on the position coordinates, thanks to the fact that the total energy is independent of them. So this works for arbitrary $V$.

The two previous equations specify the desired "volume" in our phase space. The dimension of this object is $6N-1$ because we only have one constraint, namely the energy being constant. To calculate its "volume" we need to consider the wonderful looking integral:

$$\phi(N,V,E)=|A|=\int_{A} dr_{x}^{(1) }dr_{y}^{(1)} dr_{z}^{(1)} dp_{x}^{(1) }dp_{y}^{(1)} dp_{z}^{(1)} ... dr_{x}^{(N) }dr_{y}^{(N)} dr_{z}^{(N)} dp_{x}^{(N) }dp_{y}^{(N)} dp_{z}^{(N)}$$

Where $A$ is the set given by the previous equations. We can simplify this notation by writing

$$\phi(N,V,E)=|A|=\int_{A} d^{3N}r \ d^{3N}p$$

Notice that the position coordinates are independent of momentum coordinates, so really, the integral factors:

$$\phi(N,V,E)=|A|=\int_{A_r} d^{3N}r \int_{A_p} d^{3N}p$$

Where $A_r$ is the set given by the position conditions and $A_p$ by the momentum conditions. The first factor is simply $V^N$, or $l^{3N}$. The second factor is slightly trickier to calculate, but it's just the surface of a 3N dimensional sphere with radius $\sqrt{2mE}$. This already implies that the units will be $\mbox{[momentum]}^{3N-1}$, and so the units of $|A|$ are $\mbox{[momentum]}^{3N-1} \mbox{[length]}^{3N}$.

The actual derivation of the surface of an $n$-dimensional sphere of radius $r$ will be left out of this, and we just quote the result: $S_n = \frac{2 \pi ^{n/2}}{\Gamma (n/2)}r^{n-1}$

Anyway, we have obtained a solution for our "volume" $|A|$:

$$\phi(N,V,E)=|A|=2 V^N \frac{\pi ^{3N/2}}{\Gamma(3N/2)}\sqrt{2mE}^{3N-1} \quad \mbox{[momentum]}^{3N-1} \mbox{[length]}^{3N}$$

Here I have a big problem. As said before, the number of states is $\frac{\phi(N,V,E)}{\phi_0}$. I would like for $\phi_0$ to be something like $h^{3N}$ or $h^{3N-1}$ but this gives wrong units! Ignoring this for the moment and just taking $\phi_0 = h^{3N-1}$ here's the rest of the derivation:

We thus obtain an expression for the number of microstates:

$$\Omega(N,V,E) = 2 V^N \frac{\pi^{3N/2}}{\Gamma(3N/2)} \left( \frac{\sqrt{2mE}}{h} \right) ^{3N-1}$$

Taking the log of this and using appropriate approximations etc.(I skip this here but won't during the presentation ofc.) we arrive at:

$$ S(N,V,E) = kN \left( \frac{3}{2} + \ln V \left( \frac{4 \pi m E}{h^2} \right) ^{\frac{3}{2}} \right)$$

This unfortunately leads to the Gibbs paradox. The reason for this is that we've assumed the particles to be disinguishable when in reality they are not. In order to account for this, we need to introduce the Gibbs correction factor into the total amount of states $\Omega$:

$$\Omega_C(N,V,E) = \frac{1}{N!} \Omega(N,V,E) $$

Dividing by the amount of permutations accounts for indistinguishability. Going through the same derivation leads to the proper formula now. Thus the paradox is fixed.

I would love to hear some feedback about this presentation and particularly with the problem I had with $\phi_0$. Additionally, the introduction of the Gibbs correction factor seems kind of shady to me, I read somewhere that it's again just an approximation and that proper counting would produce a more complicated result than just diving by the factorial, but once again the error is negligible.

  • 1
    $\begingroup$ Be careful with the volume of phase space, the number of states with energy E is zero! (in classical physics). You have to consider the number of states with energy between $E$ and $E+dE$, in the thermodynamic limit you can forget about $dE$ $\endgroup$
    – Ikiperu
    May 22, 2013 at 13:31
  • $\begingroup$ Please elaborate on that if you could (see my comments to the answer below). Is there actually any point trying to do this derivation classically? $\endgroup$ May 22, 2013 at 14:57

1 Answer 1


The integral carries units of $[momentum]^{3N}[space]^{3N}$ which is exactley the same as $1/h^{3N}$, so this is the factor you need for making the phase space volume dimensionless. I don't understand why you say that it has to be $[momentum]^{3N-1}$, just look directly at the integral.

Furthermore, and maybe this is the main problem, the phase space VOLUME ist given bei the volume of a 3N-dimensional sphere times $V^{3N}$, not only the surface. Now we are able to calculate the number of microstates which obey the constraint $H<E$:

$\Omega=\phi(E+dE)-\phi(E)\approx\frac{\partial\phi}{dE}dE $ which is, obviously, dimensionless again. In your derivation, I can not see any difference between $\phi$ and $\Omega$ except the wrong dimensions of $\phi$. Here at this point the power of E decreases.

I think this solves both your problems. I have to admit that I don't have the Greiner book here, so I don't know what he states. This is how I learned it.

  • $\begingroup$ I asked a question about the integral on math SE earlier: math.stackexchange.com/a/398746/75576 , and that's where I got the units from. Because the thing I'm calculating is a 6N-1 dimensional object (or so I thought - now I'm confused again) $\endgroup$ May 22, 2013 at 14:19
  • $\begingroup$ This is what I tried to point out: $\phi$ is the whole volume of phase space, so it has to have units of l^3N*p^3N. $\Omega$ is the "surface" of this volume, or more exactly, the volume of an infinitesimal shell. $\endgroup$
    – Noldig
    May 22, 2013 at 14:24
  • $\begingroup$ Also, why did you say that $\frac{d \phi}{dE}dE$ is dimensionless? Doesn't it have dimension of $\phi$, i.e. phase space volume? $\endgroup$ May 22, 2013 at 14:34
  • $\begingroup$ No. Thanks to Stirling's approximation for large N we find $S=k\ln\Omega\approx k\ln\phi$ hence $\phi$ has to be dimensionless as well. The thing is, you need the factor $1/h^{3N}$ already for $\phi$. The difference between $\phi$ and $\Omega$ is NOT only the correction of units. $\endgroup$
    – Noldig
    May 22, 2013 at 14:38
  • $\begingroup$ But you just said $\phi$ has to have units of action^3N in the first comment? I don't get it. Let's say I define it anew: $\phi(E) = \frac{\mbox{Volume of phase space s.t.} H(r,p) \le E}{h^{3N}}$, i.e. the number of states with energy less or equal $E$. Then I see that given a finite resolution, $\Delta E$, I can think of the number of states with energy $E\le H(r,p) \le E+\Delta E$, which is $\phi(E+\Delta E) - \phi(E) \approx \frac{d \phi}{dE} \Delta E$. So then, classicaly, it boils down to the fact that the amount of states of a system with energy precisely $E$ must be zero... $\endgroup$ May 22, 2013 at 14:47

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