I've been reviewing these course notes on the microcanonical ensemble, and there is something I'm not quite understanding. McGreevy goes through the classic example of two gases connected by a diathermal wall. He makes the statement that the probability of an arbitrary energy of the first box, $E_1$, is:

$$p(E_1)=\frac{\text{volume of accessible phase space consistent with }E_1}{\text{volume of accessible phase space}}=\frac{\Omega_1(E_1)\Omega_2(E-E_1)}{\Omega(E)} \tag{1}$$


$$E=E_1+E_2=\text{constant for the total isolated system}$$

This is odd to me, because under the exact same construction in Pathria, Statistical Mechanics, we can enumerate the total number of microstates by multiplying $\Omega_1$ by $\Omega_2$: $$\Omega(E)=\Omega(E_1)\Omega(E_2) = \Omega(E_1)\Omega(E-E_1) \tag{2}$$

Plugging $(2)$ into $(1)$ we would get $p(E_1)=1$. This seems like a nonsensical result. We haven't invoked the equilibrium condition yet, but if we take the partial derivative of the logarithm of $(1)$ w.r.t. $E_1$,

$$\frac{\partial \ln p(E_1)}{\partial E_1} = \frac{\partial}{\partial E_1}1 = 0$$

We end up with something that is mathematically equivalent to the equilibrium condition. As a point of comparison, Pathria defines the equilibrium condition to be when the total number of microstates is maximized with respect to either energy,

$$\frac{\partial \Omega(E)}{\partial E_1} = \frac{\partial \Omega_1(E_1)\Omega_2(E-E_1)}{\partial E_1} = 0$$

I just wonder if it physically makes sense. How can we assert $p(E_1)=1$ prior to establishing equilibrium?

If all of this is valid, then we should be able to say the following:

  • We can only ascribe the definition $\Omega=\Omega_1 \Omega_2$ at equilibrium

  • At equilibrium the probability of either $E_1$ or $E_2$ "collapses" to 1

This reasoning seems circular to me. Perhaps I am missing something simple.


1 Answer 1


In equation (1), the denominator is the total number of accessible states for a given total energy $E$. But your equation (2) only tells us the number of states permitted for a given energy $E_1$ of a partition of your system (with fixed $E$). So in equation (1), the denominator isn't (your) $\Omega(E)$.

Indeed, it's $\Omega_{T}(E)$, the sum over ALL possible configurations for a given $E$ (whatever $E_1$ could be), so you have:


With $E$ fixed of course.

  • 1
    $\begingroup$ This makes sense. There's a slight notational difference between Pathria and the link I provided, and it tripped me up. Thanks for your answer. $\endgroup$
    – michael b
    Oct 10, 2021 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.