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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?

Qmechanic
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Javier
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  • In general, I do not see super big problems in using one or the other. On the other hand, looking at mathematical consistency, did you have a look at *canonical variables*? – mikuszefski May 16 '17 at 15:21
  • Regarding the super big problems, that's the point of my question: if it will never make a difference, I would like to know why, or see some kind of proof or argument. And I know what canonical variables are, but in the context of mechanics you can use them or you can use the Lagrangian and everything will be fine. Is it the same here? – Javier May 16 '17 at 15:26
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    @Javier notice that you rescaled Z by dropping a factor $m$. if so you have to put it somewhere else, because if not your probabilities will not normalize to 1. –  May 16 '17 at 15:56
  • Isn't this just the fact that stat mech should be quantum, and the semiclassical limit implicitly taken here says that quantum states correspond to fixed volumes of phase space? – knzhou May 16 '17 at 16:00
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    Related: https://physics.stackexchange.com/q/51534/2451 , https://physics.stackexchange.com/q/31534/2451 and links therein. – Qmechanic May 16 '17 at 17:47
  • Since you are using $p=m\dot{q}$ the second $Z$ would be $Z = m \int dq\ d\dot{q}\ e^{-\beta E(\dot q, q)}$ and integrating you would obtain $Z = L \sqrt{2\pi m k T}$ in both. – juanrga Jul 17 '18 at 16:22

1 Answers1

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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.

dmckee --- ex-moderator kitten
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  • For atoms in a gas momentum and velocity only differ by a multiplicative constant. If volume in phase space is conserved, then it is also conserved in configuration space. –  May 16 '17 at 19:13
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    @WillyBillyWilliams Consider what happens when the system includes objects of more than one mass. Even if we restrict ourselves to ideal gases, thre is no unique multiplier. – dmckee --- ex-moderator kitten May 17 '17 at 15:52