A relation between Gibbs Entropy and Boltzmann Entropy

Question

We have Gibbs Entropy which is defined as $S_G(U,V,N)=k_B \ln \Omega_0$, where $\Omega_0$ is the number of microscopic states with energy equal or smaller than $U$. Note that this definition differs from the definition of Boltzmann Entropy, $S_B(U,V,\delta U, N) = k_B \ln \Omega$, where $\Omega = (\partial \Omega_0 / \partial U)\delta U$. Consider the different definitions of tempratures: $$T_B = (\partial S_B / \partial U)^{-1}$$ $$T_G = (\partial S_G / \partial U)^{-1}$$

I want to show that: $$T_B/T_G = 1/(1+k_B/C)$$ where $C = (\partial T_G /\partial E)^{-1}$ is the total heat capacity associated with $T_G$.

I am not sure how to show it, after some algebraic manipulations I got to: $$T_B/T_G = \frac{\Omega / \Omega_0}{(\partial \Omega \partial U) / (\partial \Omega_0 / \partial U)} = $$ $$ = \frac{\delta U (\partial \Omega_0 / \partial U)/\Omega_0}{\partial^2 \Omega_0 /\partial U^2\delta U / (\partial \Omega_0 /\partial U)} = $$ $$ = \frac{(\partial \Omega_0 / \partial U)^2 1/ \Omega_0}{\partial^2 \Omega_0 / \partial U^2}$$

And $$C^{-1} = \frac{-1}{(\partial S_G / \partial U)^2} \frac{\partial^2 S_G}{\partial U \partial E}$$

I am stuck, does someone have a detailed reference where this is shown or knows how to show it?

Answer 1

Okay, let primes be partial derivatives with respect to $U$ and $\sigma = S_B$ and $S = S_G$ and let me redefine $\Omega$ as $\Omega_0$ so that I don't have to keep writing these subscript-zeros on every $\Omega_0$ (in other words I will not refer to your $\Omega$ explicitly at all in this answer). Then you have the corresponding temperature definitions $$\begin{array}{rcl} \tau^{-1} &= \sigma' =& k ~\Omega'' /\Omega',\\ T^{-1} &= S' =& k ~\Omega'/\Omega. \end{array} $$ You have defined $$C^{-1} = T' = (1/S')' = \frac{-S''}{(S')^2} = -\frac{k \big[\Omega''/\Omega ~-~(\Omega')^2/\Omega^2\big]}{k^2 (\Omega')^2 / \Omega^2}.$$ Seems pretty straightforward to me, though I can see why you might have been put off by the complexity of the symbols, especially if you failed to unite conceptually $E$ and $U$ in your expressions. After cancellation above you should get:$$k/C = 1 - \frac{\Omega'' ~\Omega}{(\Omega')^2} = 1 - \frac{T}{\tau}.$$The only variation I'm seeing from your expression is that I get a $\tau/T = 1/(1 - k/C)$ expression, and I don't see a way to fix that difference in our signs; that minus sign seems morally destined to be there unless I'm missing something like $E$ is some energy flow into a system and therefore $E = - U$ and $C^{-1} = - T'$ or so.