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
    Commented Oct 10, 2021 at 15:26

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