As explained in many books, for the microscopic justification of the second law of thermodynamics (lets formulate it as the total entropy takes maximum among all possible exchanges of two systems), you don't have to enter the realm of the canonical ensamble.

Where is the microcanonical phase space density $$\varrho=const. \ \ \ \ \text{if}\ \ \ \ E<H<E+\Delta,$$ used in the computation of the phase space which comes from the composition of two other systems?

Let $E=E_1+E_2$. For fixed energies $E_1$ and $E_2$, the new volume is given by $\Gamma(E)=\Gamma(E_1)\Gamma(E_2)$ and at the intermediate point where one considers all the possible energy exchanges, one writes $\Gamma(E)=\sum_{\epsilon}\Gamma(E_1+\epsilon)\Gamma(E_2-\epsilon)$.

I don't see how the construction of the composed phase space is computationally influenced by $\varrho$, and it also seems to me that this composed space would be computable without stating $\varrho$ explicitly. It's a volume after all, it shold just be the product in any case.

Furthermore, is $\varrho$ involved in the derivation that the maximum among the possible composed phase volumes is very sharp? (Is there a general derivation?)

  • $\begingroup$ I think one should also consider the number of particles exchanges. $\endgroup$
    – jjcale
    Jul 24, 2012 at 23:42
  • $\begingroup$ @jjcale: One certainly can (I don't know if one has to, though). $\endgroup$
    – Nikolaj-K
    Jul 25, 2012 at 7:04

1 Answer 1


The "microcanonical ensemble" is just saying each of the $\Gamma(E)$ states is equally likely. When you multiply the volumes at energy $\pm\epsilon$ and add over $\epsilon$, you are using the microcanonical ensemble assumption that all the states are equally likely, to get that the probability is the volume.

As for the sharpness, this is from the thermodynamic observation that one of the systems 1 or 2 is very big, so that it has a value of $\partial_U S $, which tells you how the volume changes with energy. This volume changes in a way proportional to the macroscopic size, since the S is extensive, and there are Avogadro's number of particles.

  • $\begingroup$ Okay so in the first paragraph you say that it justifies the ' "geometric" volume equals propability'-statement, right? In the second paragraph, is the "has a volume" sentence complete. What has this value you mention - it would be $\frac{\Gamma'(E)}{\Gamma(E)}$, wouldn't it? $\endgroup$
    – Nikolaj-K
    Jul 24, 2012 at 21:49
  • $\begingroup$ @NickKidman: I don't find an incomplete sentence, nor the phrase 'has a volume'. I don't know what you mean by the ratio of the volumes. If you calculate in one example, like an ideal gas or a harmonic solid, you will no longer be confused on these things. $\endgroup$
    – Ron Maimon
    Jul 24, 2012 at 22:01
  • $\begingroup$ "has a volume" was ments to be has a value. The ration is the derivative of $S(E)=log(\Gamma(E))$. $\endgroup$
    – Nikolaj-K
    Jul 24, 2012 at 22:55

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