Peeter Joot's (OLD) Blog.

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Liouville’s theorem questions on density and current

Posted by peeterjoot on February 5, 2013

[Click here for a PDF of this post with nicer formatting and figures if the post had any]

Liouville’s theorem questions on density and current

In the midterm we were asked to state and prove Liouville’s theorem. I couldn’t remember the proof, having only a recollection that it had something to do with the continuity equation

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \frac{\partial {j}}{\partial {x}},\end{aligned} \hspace{\stretch{1}}(1.1.1)

but unfortunately couldn’t remember what the j was. Looking up the proof, it’s actually really simple, just the application of chain rule for a function \rho that’s presumed to be a function of time, position and momentum variables. It didn’t appear to me that this proof has anything to do with any sort of notion of density, so I posed the following questions.

Context

The core of the proof can be distilled to one dimension, removing all the indexes that obfuscate what’s being one. For that case, application of the chain rule to a function \rho(t, x, p), we have

\begin{aligned}\frac{d{{\rho}}}{dt} &= \frac{\partial {\rho}}{\partial {t}} + \frac{\partial {x}}{\partial {t}} \frac{\partial {\rho}}{\partial {x}} + \frac{\partial {p}}{\partial {t}} \frac{\partial {\rho}}{\partial {p}} \\ &= \frac{\partial {\rho}}{\partial {t}} + \dot{x} \frac{\partial {\rho}}{\partial {x}} + \dot{p} \frac{\partial {\rho}}{\partial {p}} \\ &= \frac{\partial {\rho}}{\partial {t}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {x}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {p}} - \rho \left(  \frac{\partial {\dot{x}}}{\partial {x}} + \frac{\partial {\dot{p}}}{\partial {p}}  \right) \\ &= \frac{\partial {\rho}}{\partial {t}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {x}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {p}} - \rho \underbrace{\left(\frac{\partial {}}{\partial {x}} \left(  \frac{\partial {H}}{\partial {p}}  \right)+\frac{\partial {}}{\partial {p}} \left(  -\frac{\partial {H}}{\partial {x}}  \right)\right)}_{= 0}\end{aligned} \hspace{\stretch{1}}(1.1.2)

Wrong interpretation

From this I’d thought that the theorem was about steady states. If we do have a steady state, where d\rho/dt = 0 we have

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {x}} + \frac{\partial {\left( \dot{p} \rho \right)}}{\partial {p}}.\end{aligned} \hspace{\stretch{1}}(1.1.3)

That would answer the question of what the current is, it’s this tuple

\begin{aligned}\mathbf{j} = \rho (\dot{x}, \dot{p}),\end{aligned} \hspace{\stretch{1}}(1.1.4)

so if we introduce a “phase space” gradient

\begin{aligned}\boldsymbol{\nabla} = \left(  \frac{\partial {}}{\partial {x}}, \frac{\partial {}}{\partial {p}}  \right)\end{aligned} \hspace{\stretch{1}}(1.1.5)

we’ve got something that looks like a continuity equation

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \boldsymbol{\nabla} \cdot \mathbf{j}.\end{aligned} \hspace{\stretch{1}}(1.1.6)

Given this misinterpretation of the theorem, I had the following two questions

  • This function \rho appears to be pretty much arbitrary. I don’t see how this connects to any notion of density?
  • If we pick a specific Hamiltonian, say the 1D SHO, what physical interpretation do we have for this “current” \mathbf{j}?

The clarification

Asking about this, the response was “Actually, equation 1.1.3 has to be assumed for the proof. This equation holds if \rho is the phase space density and since the pair in 1.1.4 is the current density in phase space. The theorem then states that d\rho/dt = 0 whether or not one is in the steady state. This means even as the system is evolving in time, if we sit on a particular phase space point and follow it around as it evolves, the density in our neighborhood will be a constant.”

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