About thermodynamic beta First of all, I'm not majoring this part and just finding some answer about Arrhenius rate equation: $$ v \propto \exp\left(-\frac{E_a}{kT}\right) .$$
To derive this relation, I started with Maxwell-Boltzmann distribution. When there are $N=\sum_j n_j $ particles with energy distribution $$ E_{tot} = \sum_j \epsilon_j n_j $$, the disorder number becomes $$\Omega = \frac{N!}{n_1 ! n_2 ! n_3 ! \ldots}. $$ To  find when to get the maximum $\Omega$, I used Lagrange multiplyer. Therefore, not considering degeneracy, I got this:$$\textrm{d}(\ln\Omega)=\sum_j\left(\ln \frac{1}{n_j} +\alpha -\beta\epsilon_j\right)\textrm{d}n_j$$
To make $\textrm{d}(\ln\Omega)$ be zero, I obtained $$ n(\epsilon)=e^\alpha e^{-\beta\epsilon}.$$
But, the question is how we know that $\beta = \frac{1}{k_B T}$? Somewhere i found that by using $\beta = \frac{\textrm{d}\ln\Omega}{\textrm{d}E}$ and the equation of entropy it is possible. However, I assumed that total energy $E$ is constant so I think $\textrm{d}E$ should be 0. What's the problem in my way? 
 A: In the derivation of the Boltzmann distribution, one finds the maximum of the function
\begin{equation}\
\ln \Omega(\{n_{i}\}) = \ln\!\Big[\Big(\sum_{i} n_{i}\Big)!\Big]-\sum_{i}\ln n_{i}!
\end{equation}
under the constraints
\begin{equation}
\sum_{i} n_{i} = N,\quad \sum_{i} n_{i}\epsilon_{i} = E,
\end{equation}
where the corresponding Lagrange multipliers are respectively called $\alpha$ and $\beta$. The constrained maximum of $\ln \Omega$ occurring at $n_{i} = \exp(-\alpha-\beta\epsilon_{i})$ (i.e., the Boltzmann distribution) is a function of $N$ and $E$. Let's denote it as $\ln \tilde{\Omega}(N,E)$. A theorem on constrained stationary point (which I will state and prove below) guarantees that
\begin{equation}
\beta = \frac{\partial \ln\tilde{\Omega}}{\partial E}.
\end{equation}


Theorem. Suppose that a stationary point of a function $f(\{x_{i}\})$ under a set of constraints $\big\{g_{\alpha}(\{x_{i}\}) = c_{\alpha}\big\}$ occurs at $\big\{\tilde{x}_{i}(\{c_{\alpha}\})\big\}$. Define $\tilde{f}(\{c_{\alpha}\})\equiv f\Big(\Big\{\tilde{x}_{i}\big(\{c_{\alpha}\}\big)\Big\}\Big)$. Then,
  \begin{equation}
\frac{\partial\tilde{f}}{\partial c_{\alpha}} = \lambda_{\alpha},
\end{equation}
  where $\lambda_{\alpha}$ is the Lagrange multiplier for the $\alpha$-th constraint, i.e.,
  \begin{equation}
\left[\frac{\partial f}{\partial x_{i}} - \sum_{\alpha} \lambda_{\alpha}\frac{\partial g_{\alpha}}{\partial x_{i}}\right]_{x_{i}=\tilde{x}_{i}} = 0.
\end{equation}
Proof.
  \begin{equation}
\begin{split}
\frac{\partial \tilde{f}}{\partial c_{\alpha}} &= \sum_{i}\frac{\partial f}{\partial x_{i}}\Bigg|_{x_{i}=\tilde{x}_{i}} \frac{\partial \tilde{x}_{i}}{\partial c_{\alpha}} = \sum_{i} \sum_{\beta}\lambda_{\beta}\frac{\partial g_{\beta}}{\partial x_{i}}\Bigg|_{x_{i}=\tilde{x}_{i}} \frac{\partial \tilde{x}_{i}}{\partial c_{\alpha}}\\
&=\sum_{\beta}\lambda_{\beta}\frac{\partial g_{\beta}\Big(\Big\{\tilde{x}_{i}\big(\{c_{\gamma}\}\big)\Big\}\Big)}{\partial c_{\alpha}} = \sum_{\beta}\lambda_{\beta} \frac{\partial c_{\beta}}{\partial c_{\alpha}} =\sum_{\beta}\lambda_{\beta}\delta_{\alpha\beta}\\
&= \lambda_{\alpha},
\end{split}
\end{equation}
  where the first and third equalities follow from the chain rule.

