How does one show that $\beta$ is the same for different substances in thermal equilibrium?

Question

In the section regarding quantum statistical mechanics, Griffiths uses the method of Lagrange multipliers to calculate the most probable energy configuration $(N_1,N_2,\dots)$, where $Q(N_1,N_2,\dots)$ is the number of ways to achieve that particular configuration, subject to the constraints of fixed total number of particles and energy. He uses the symbols $\alpha$ and $\beta$ to set up the Lagrange multiplier function $$ G = \ln Q + \alpha\left( N - \sum_{n=1}^\infty N_n \right) + \beta\left( E - \sum_{n=1}^\infty E_nN_n \right). $$ Plugging in the formula for $Q$ for distinguishable particles, and carrying out the method, he obtains $$ N_n = d_n e^{-(\alpha + \beta E_n)}. $$ As a specific example, he considers distinguishable particles in the 3D infinite square well. Solving for $\beta$ from the constraints, he gets $$ E = \frac{3N}{2\beta}, $$ which is reminiscent of the formula for temperature $$ \frac E N = \frac 32 k_B T, $$ from which it follows that $$ \beta = \frac 1{k_B T}. $$ Griffiths claims that this relationship between $\beta$ and $T$ is true in general, and says that in order to show this, one would have to show that the value of $\beta$ is the same for different substances in thermal equilibrium with each other. How does one show this?

Answer 1

After doing some research on my own, I came across this wikipedia page on the "thermodynamic beta", which seems to be the same $\beta$ that I referred to in the question. In the statistical interpretation section, the article says that $\beta$ is a numerical quantity relating two macroscopic systems in equilibrium, continuing to show that when two systems are at thermal equilibrium, $$ \frac{\mathrm d \ln \Omega_1}{\mathrm d E_1} = \frac{\mathrm d \ln \Omega_2}{\mathrm d E_2}. $$ This motivates the following definition for $\beta$ for a particle system: $$ \beta \equiv \frac{\mathrm d \ln \Omega}{\mathrm dE}. $$ Now I just have to show that this definition is congruent with or follows from our use of $\beta$ as a Lagrange multiplier in this case. Given that $$ E = \sum_{n=1}^\infty E_n N_n, $$ properties of differentials give $$ \mathrm d E = \sum_{n=1}^\infty E_n \mathrm d N_n. $$ Similarly, $$ \mathrm d \ln Q = \sum_{n=1}^\infty \frac{\partial \ln Q}{\partial N_n} \mathrm dN_n. $$ The method of Lagrange multipliers sets $\partial_{N_n} G = 0$ for all $N_n$. Hence $$ \frac{\partial \ln Q}{\partial N_n} - \alpha - \beta E_n = 0 \implies \mathrm d \ln Q = \sum_{n=1}^\infty (\alpha + \beta E_n ) \mathrm dN_n. $$ So $$ \frac{\mathrm d \ln Q }{\mathrm d E} = \frac{ \sum (\alpha + \beta E_n ) \mathrm dN_n}{\displaystyle\sum E_n \mathrm d N_n} = \alpha \frac{ \sum\mathrm dN_n}{ \sum E_n\mathrm dN_n} + \beta. $$ Now, since $E_n$ strictly increases as $n$ increases, and the summation is carried out to infinity, I am inclined to believe that the fraction above should be zero, which just leaves $$ \frac{\mathrm d \ln Q }{\mathrm d E} = \beta. $$ I know this proof isn't rigorous at all, so if I made any mistakes, please let me know!