# Show that, for two large systems in thermal contact, the number $\Omega^{0}(E^{0},E_1)$ can be expressed as a Gaussian in the variable $E_1$

This problem below is from the book "Statistical Mechanics" by Pathria. The author defined the number of microstates of a system with two subsystems exchanging energy as: $$\Omega_1(E_1) \Omega_2(E_2) = \Omega_1(E_1) \Omega_2(E^{0}-E_1) = \Omega^{0}(E^{0},E_1)$$

Show that, for two large systems in thermal contact, the number $\Omega^{0}(E^{0},E_1)$ can be expressed as a Gaussian in the variable $E_1$.

Here is my attempt:

I tried to work backwards in this problem, in the hopes that I could get some insights by assuming that the function $\Omega^{0}(E^{0},E_1)$ takes the form: $$\Omega^{0}(E^{0},E_1) = a e^{-b E_{1}^2 /2}$$

If that were true, we would have: $$\ln \Omega^{0}(E^{0},E_1) = \ln [a e^{-b E_{1}^2 /2}] = c -\frac{b E_1^2}{2}$$ Where $c = \ln a$.

But I cannot say anything about the form of the function $\Omega^{0}(E^{0},E_1)$. However, I do know that: $$\left(\frac{\partial S}{E_1} \right)_{V,N} = \frac{1}{T}$$

So since $S = k \ln \Omega^{0}(E^{0},E_1)$, combining the last two equations I found: $$\frac{\partial S}{E_1} = -bkE_1 \Rightarrow T = -\frac{1}{bkE_1}$$

It is possible to work backwards, but in order to prove the result stated in the problem I would have to assume that $T = -1/bkE_1$; and this looks quite odd to me.

Could someone clarify this issue?

## 1 Answer

Suggestion:

It is not hard to taylor expand to get

$$\ln \Omega_1(E_1) = \ln \Omega_1( \bar E_1) + \beta_1 (\bar E_1) (E_1 - \bar E_1) + \gamma_1 (E_1 - \bar E_1)^2 + \dots$$ $$\ln \Omega_2(E_2) = \ln \Omega_2( \bar E_2) + \beta_2 (\bar E_2) (E_2 - \bar E_2) + \gamma_2(E_2 - \bar E_2)^2 + \dots$$ with $E_2 - \bar E_2 = -(E_1 - \bar E_1)$, so $$\ln \Omega_1(E_1) \Omega_2 (E_2) = \ln \Omega_1( \bar E_1)\Omega_2( \bar E_2) + (\beta_1 - \beta_2)(E_1 - \bar E_1) + (\gamma_1 + \gamma_2) (E_1 - \bar E_1)^2 + \dots$$ At thermodynamic equilibrium this function is maximized wrt $(E_1 - \bar E_1)$, so the linear term vanishes.

• any ideas why the higher terms will vanish? I estimated they will be proportional to $V$, (with their extensive properties) but no idea if anything goes wrong. Mar 13 at 12:07
• So I haven't touched this stuff in ages but you can use thermodynamic formulae to show that the higher order term is proportional to 1/Cv, the heat capacity of the system, which increases rapidly with number of DOF. Quick search yielded mcgreevy.physics.ucsd.edu/s12/lecture-notes/chapter06.pdf Mar 15 at 10:26
• It isn't exactly a tight argument. I think it's more informative to take a combinatorial approach with the derivation. The calculation I gave is just a quick way to get an analytic expression, but it depends on other machinery. Mar 15 at 10:28