# 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?

• So lnQ is entropy, the alpha part is a chemical potential and the beta part is the inner energy. It's not fully true "in general", that beta is the same in both substances in thermal equilibrium, as thermal equilibrium in the classical sense is an oversimplification. I recommend you to read about Tsallis equilibrium, that takes into account the situation where there are long range correlations in the system. In such a case for example beta needs to be multiplied with a factor C(T) , taking into account some effects (the same factor appears in the definition of the entropy). Aug 7, 2022 at 8:44

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!