I am a Mathematics student preparing myself for the exam in biological models. A part of the lectures was about statistical mechanics. After a brief introduction (hamiltonian, microcanonical ensamble) we proceeded to the definition of the entropy and its properties. There was some part I did not understand then and reading a textbook (Kerson Huang - "Statistical Mechanics", 6.2, p. 133) I am still confused.

Why should we take the values $$\bar E_1, \bar E_2 $$ that maximise the function $$ \Gamma(E) = \Gamma(\bar E_1)\Gamma(\bar E_2) $$ and why does the following $$ S(E,V) = S(\bar E_1,V_1)+S(\bar E_2,V_2)+O(\log N)$$ prove the extensive property of the entropy? For me it proves only the special case with energies that maximise the function.

I saw a similiar approach in prof. David Tong's lecture notes on his website, but I was not able to understand them. I think I am missing some fundamental assumption or do not understand the gamma function properly. I do not have strong background in physics, so would appreciate a detailed answer.


2 Answers 2


Before being able to exchange energy the two systems will have definite energies $E_1$ and $E_2$, and so definite entropies $S_1(E_1)$ and $S_2(E_2)$. This derivation doesn't show that the entropy of the final 1+2 system is $\bf S_1(E_1)+S_2(E_2)$. In fact, it can't show this is the case because it's not true (why?).

What the derivation does prove is that

  1. when two different systems are allowed to exchange energy then at equilibrium the two systems will reach definite energies $\bar{E_1}$ and $\bar{E_2}$, that may be different from the initial energies (but satisfy the condition $E_1+E_2=\bar{E_1}+\bar{E_2}$)

  2. the entropy of the system 1+2 at equilibrium is precisely equal to the sum of the entropies of the two single systems, when they have the equilibrium energies $\bar{E_1}$ and $\bar{E_2}$

  3. this entropy is the maximum possibile entropy that the 1+2 system can have

What is meant by the extensiveness of entropy is what is addressed by point number 2.


$\bar E_1, \bar E_2 $ are the most probable values. You can think about that in different ways, they are the most probable values because they maximize $ \Gamma(E) = \Gamma(\bar E_1)\Gamma(\bar E_2) $ and therefore they maximize the entropy, according to the Principle of Maximum Entropy

If you understand why maximizing $\Gamma$ you maximize the entropy you can skip the next paragraph(2 questions with answers), but Since you said you don't understand the $\Gamma$ function I'll try to add some notions and relate them to the question.

  • So what is $\Gamma(E)$?

    It is the number of microstates available to the system, or if you want a more formal definition, it is the phase space volume occupied by all the reference points of the system. So it is basically a volume in phase space which represents how many microstates system can be in. Note that all these microstates correspond to the same physical macrostate.

  • How does that relate to the question?

    It does considering that the entropy of the system is the $\ln \Gamma(E)$, or more precisely, it is the logarithm of $\frac {\Gamma}{Planck \, constant }$. In this way you obtain a pure number inside the logarithm. In classical statistical mechanics you can't prove that you have to divide for the Planck constant. In this framework you just divide for a constant of the dimension of an action (phase space volume), the constant is then fixed by quantum statistical mechanics.

That said, I repeat that those values of the energy $\bar E_1, \bar E_2 $ are the most probable values, so Kerson & Huang's derivation proves the extensive property of entropy for all the systems that respects the principle of maximum entropy, that is as far as i know, every known system.


I don't remember the details of their proof, but I remember that to be carried out, that proof needed an important assumption, which if I'm correct is implicit in their textbook, that is for a typical system, the number of microstates $\Gamma(E)$ pertaining to any macrostate that departs even slightly from the most probable one is “orders of magnitude” smaller than the number pertaining to the latter, which is $\Gamma(\bar E)$. Without this you couldn't do some mathematical passages in the proof.$


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