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