thermodynamics - Is temperature a Lagrange multiplier of energy?

Someone told me that temperature is a Lagrange multiplier of energy... Is this true? Why?

Answer

In brief, this comes from the Gibbs formalism, where, at equilibrium, the probability of a microstate is given by $p = \frac{1}{Z}\exp(-\beta E)$ where E is the energy of the microstate and Z is a partition function. Here, $\beta$ is the Lagrange multiplier, which can be seen to be $\beta = 1/T$. This set of lecture notes from the University of Edinburgh might be a useful reference to read. Note that we can only define temperature at (local) equilibrium.