Why do we use momenta and not velocities in the partition function of statistical mechanics?

Question

The question is in the title, what follows is an example.

Consider for example a free classical particle in a 1D box of length $L$ in contact with a thermal reservoir. According to Boltzmann the probability that it is in a state of energy $E$ is $\exp(-E/kT)$, so we calculate the partition function by integrating over all states:

$$Z = \int dq\ dp\ e^{-\beta H(p, q)} = \int dq\ dp\ e^{-\frac{\beta}{2m} p^2} = L \sqrt{2\pi m k T}$$

Now, I don't see why we couldn't label states by $(q, \dot q)$ instead of $(q, p)$; if we did that we would get

$$Z = \int dq\ d\dot{q}\ e^{-\beta E(\dot q, q)} = \int dq\ d\dot{q}\ e^{-\frac{\beta m}{2} \dot{q}^2} = L \sqrt{\frac{2\pi k T}{m}}$$

which is different. Now, in this case the difference doesn't matter since $F = -kT \log Z$ and an additive constant just shifts the energy. Will this always be the case? Or is there a system for which the two partition functions have a nontrivial difference, and in that case, why do we use $p$ instead of $\dot q$ as an integration variable?

Answer 1

The basic issue is that the Louisville Theorem is defined in terms of the Hamiltonian phase space, not the Lagrangian configuration space.

And why is that? Because the things that the traditional partition functions measure are volumes in phase space and those stay constant over the time-evolution of the system. On the other hand there is no theorem that says volumes in configuration space stay constant in time, so a critical ingredient is missing when you try to build a statistical mechanics in the Lagrangian framework.