### Boltzmann Entropy

This semester, I’ve been teaching a Physics for non-Science majors (mostly Business School students) class.

Towards the end of the semester, we turned to Thermodynamics and, in particular, the subject of Entropy. The textbook had a discussion of ideal gases and of heat engines and whatnot. But, somewhere along the line, they made a totally mysterious leap to Boltzmann’s definition of Entropy. As important as Boltzmann’s insight is, it was presented in a fashion totally disconnected from Thermodynamics, or anything else that came before.

So, equipped with the Ideal Gas Law, and a little *baby* kinetic theory, I decided to see if I could present the argument leading to Boltzmann’s definition. I think I mostly succeeded. Herewith is a, somewhat fancied-up, version of the argument.

We start with Clausius’s definition^{1} of the entropy

the First Law of Thermodynamics

(where $W= p d V$) and the Ideal Gas Law

where $U$ is the internal energy of the gas and $\alpha = 3/2$ for a monatomic ideal gas.

Let’s consider an isothermal process. Since $T=\text{const}$, $d U =0$ and $Q = N k T \ln(V_f/V_i)$ Comparing this with (1), we conclude that

where $f(T)$ is some volume-independent function of the temperature.

Repeating the same analysis for an adiabatic process, $Q=0$, and hence $d U = \alpha N k d T = -W = -\frac{N k T}{V} d V$ or

Since $d S=0$, we can solve for the previously unknown function $f(T)$

where the constant is independent of both $V$ and $T$.

This is (almost) the answer we are after. But it behoves us to pause and note that it has a very suggestive interpretation. We don’t know where any particular gas molecule is located. But we *do* know that it must be somewhere within the volume $V$. Similarly, we don’t know what the velocity of any particular gas molecule is. But baby kinetic theory^{2} tells us that
$v_{\text{RMS}} \propto T^{1/2}$
So $T^{3/2}$ is (proportional to) the volume in “velocity space” in which we expect to find the molecule of gas and $V T^{3/2}$ is the volume in “phase space” for a single molecule. It represents, in other words, our lack of knowledge of the state of that gas molecule. For $N$ molecules, the volume in phase space is ${(V T^{3/2})}^N$, which is what appears as the argument of the logarithm in (6).

So the Boltzmann entropy is $k$ times the natural logarithm of the volume in phase space of the system.

That’s about as far as I got in my lecture, but one can go a little further. (6) is *wrong* because it isn’t extensive. If we take two container of the same gas, at the same temperature and pressure, we should find that the total entropy $S= S_1 + S_2$. Instead, with (6), we find

${(S- S_1 - S_2)}^{\text{naïve}} = N k \ln(N) - N_1 k \ln(N_1) - N_2 k \ln(N_2)$

where $N=N_1+N_2$.

But this discrepancy is easy to fix. The quantity that behaves extensively is

where $c_{1,2}$ are constants. Using Stirling’s formula, for large $N$, we can then write this as

That is, we should treat the gas molecules as *identical particles*, and take $k$ times logarithm of the volume in phase space, where we’ve *modded out* by the permutations of the $N$ particles.

Despite having had to gloss over a couple of steps where a little calculus was required, I’m rather proud of this “elementary” derivation. I don’t think I’ve seen anything even remotely resembling a satisfactory explanation in any of the elementary textbooks (even the calculus-based ones).

^{1} The course was, by no means, calculus-based. Expressions like “$d X$” mean “a small change in $X$.” So (1) was read as: add a small amount of heat, $Q$ to the system at a temperature $T$, and you get a small change in the entropy, $d S = \tfrac{Q}{T}$. As a result, I had to cheat in a couple of steps in the derivation. But these weren’t terribly big cheats.

^{2} We’d previously argued for this, on the basis of a simple model, in which molecules, whose average kinetic energy is $\tfrac{1}{2}m v_{\text{RMS}}^2= \tfrac{3}{2} k T$, collide elastically with the walls of the container. This simple-minded model reproduces the pressure, $p$, predicted by the Ideal Gas Law.

## Re: Boltzmann Entropy

Nice. It also shows the falsity of the oft-made claim that quantum mechanics is needed to get the correct entropy formula.