Gibbs' Expression for the Entropy [Thermal & Statistical Mechanics I Studied #26]

Now it’s the last one on entropy. I was going to skip this concept, but I’m writing it down just to record that I did try it at least once…. Ha.. seriously, entropy, seriously…. lol

Let’s go!

Building on the concepts we learned before, they say we’re now going to express entropy not thermodynamically, and not by counting the number of ways either, but with probability. Apparently we can do that.

Say~~ there’s a system that can be measured in one of a few macrostates, and for each of those macrostates, let’s assume there really are several microstates that share that same macrostate. (This is the “micro entropy” I’m talking about~)

Then macrostates are possible, and for a specific $i$-th macrostate, let’s say n_i microstates represent that macrostate.

That is, if we say the number of all~~~ the ‘microstates’ that the system can have is $N$,

The n_i of all the macrostates add up: the sum over i of n_i = N

then this holds. So the probability that the $i$-th macrostate shows up can be written down automatically.

If we write it like that,

P(i) = n_i / N (the sum of P(i) = 1 already holds)

And so, using the definition $S = k_B \ln \Omega$ , writing the entropy of the system we set up as $S = k_B \ln N$ gives the entropy that takes every possible microstate into account.

The entropy that only counts the macrostates would be $S = k_B \ln n$ , and figuring out this one is the goal!!! What we can find out by experiment isn’t the microscopic stuff… it’s the macroscopic stuff….

So is this value the truth~~~?!

Nope. In the previous example we got a glimpse of a possible extra entropy, and we said that this possibility, $S + S_{micro}$ (the value with the micro entropy added), would be the total entropy.

Computing $S_{micro}$ makes no sense!!! No sense at all!!! But we did assume that there really are several microstates that share the same macrostate…..

Ha…. since there’s no way to know this guy, the idea is: let’s at least compute its expectation value!

That is,

S_tot = S + S_micro, so k_B ln N = S + k_B Σ P(i) ln n_i. But our goal is to write S in terms of probability, so,

S = k_B ln N − k_B Σ P(i) ln n_i = k_B (ln N − Σ P(i) ln n_i). Σ P(i) = 1, so ln N = Σ ln N · P(i) is totally fine to write. Then S = k_B (Σ ln N · P(i) − Σ P(i) ln n_i) = k_B Σ P(i) (ln N − ln n_i) = k_B Σ P(i) ln (N / n_i)

Since we already said earlier that $P(i) = \frac{n_i}{N}$ ,

S = k_B Σ P(i) ln (N / n_i) = k_B Σ P(i) ln P(i)⁻¹ = −k_B Σ P(i) ln P(i), so S = −k_B Σ P(i) ln P(i)

Like this, entropy can also be written in terms of the probability of being in each macrostate.

This expression for the entropy is called Gibbs’ expression for the entropy, apparently.

Prob 14.8

The probability that a system is in the $i$-th microstate is $P_i = \frac{e^{-\beta E_i}}{Z}$ . E_i is the energy of the $i$-th microstate, and β and Z are constants.

Show that the entropy is given by $\frac{S}{k_B} = \ln Z + \beta U$ . Here $U$ is

U = the sum over i of P_i E_i

the internal energy.

Hmm, Gibbs’ expression for the entropy,

let’s plug the $P_i$ from above into it.

Plugging in P_i: S = −k_B Σ (e^(−βE_i)/Z) ln (e^(−βE_i)/Z) = −k_B Σ (e^(−βE_i)/Z)(ln e^(−βE_i) − ln Z) = −k_B Σ (e^(−βE_i)/Z)(−βE_i − ln Z) = k_B Σ (e^(−βE_i)/Z)(βE_i + ln Z) = k_B Σ ((e^(−βE_i)/Z) βE_i + (e^(−βE_i)/Z) ln Z)

S / k_B = Σ ((e^(−βE_i)/Z) βE_i + (e^(−βE_i)/Z) ln Z)

Got this far. Let’s run toward the answer.

S / k_B = Σ (e^(−βE_i)/Z) βE_i + Σ (e^(−βE_i)/Z) ln Z = Σ P(i) βE_i + Σ P(i) ln Z = β Σ P(i) E_i + ln Z Σ P(i) = βU + ln Z. Yay≈~~QED

Prob 14.8

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