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## The continuity equation for phase space density

Posted by peeterjoot on February 5, 2013

Thinking back to the origin of the 3D continuity equation from fluid mechanics, we used the geometrical argument that any change to the mass in a volume had to leave through the boundary. This was

\begin{aligned}\frac{\partial }{\partial t} \int_V \rho dV = -\int_{\partial V} \left( \rho \mathbf{u} \right) \cdot \hat{\mathbf{n}} dA= -\int_{V} \boldsymbol{\nabla} \cdot \left( \rho \mathbf{u} \right) dV.\end{aligned} \hspace{\stretch{1}}(1.0.1)

We used Green’s theorem above, allowing us to write, provided the volume is fixed

\begin{aligned}0 =\int_V \left( \frac{\partial {\rho}}{\partial t} + \boldsymbol{\nabla} \cdot \left( \rho \mathbf{u} \right) \right)dV,\end{aligned} \hspace{\stretch{1}}(1.0.2)

or

\begin{aligned}0 =\frac{\partial {\rho}}{\partial t} + \boldsymbol{\nabla} \cdot \left( \rho \mathbf{u} \right).\end{aligned} \hspace{\stretch{1}}(1.0.3)

Consider the following phase space picture (Fig1).  The time evolution of any individual particle (or set of particles that lie in the same element of phase space) is directed in the direction $(\dot{x}, \dot{p})$. So, the phase space density leaving through the surface is in proportion to the normal component of $\mathbf{j} = \rho (\dot{q}, \dot{p})$ (red in the figure).

Fig1: Phase space current

With this geometrical picture in mind, the $6N$ dimensional phase space equivalent of 1.0.1, and a basis $\mathbf{e}_{i_{\alpha}}$ for the positions $q_{i_\alpha}$ and $\mathbf{f}_{i_\alpha}$ for the momentum components $p_{i_\alpha}$ is then

\begin{aligned}\frac{\partial }{\partial t} \int_{V_{6N}} \rho d^{3N} q d^{3N} p &= -\int_{\partial V_{6N}} \rho \sum_{i_\alpha} \left( \mathbf{e}_{i_\alpha} \dot{q}_{i_\alpha} + \mathbf{f}_{i_\alpha} \dot{p}_{i_\alpha} \right) \cdot \hat{\mathbf{n}} dA \\ &= -\int_{V_{6N}} d^{3N} q d^{3N} p \sum_{i_\alpha} \left( \frac{ \partial \left( \rho \dot{q}_{i_\alpha} \right) }{\partial q_{i_\alpha} } + \frac{ \partial \left( \rho \dot{p}_{i_\alpha} \right) }{\partial p_{i_\alpha} } \right)\end{aligned} \hspace{\stretch{1}}(1.0.5)

Here $dA$ is the surface area element for the phase space and $\hat{\mathbf{n}}$ is the unit normal to this surface. We have to assume the existance of a divergence theorem for the $6N$ dimensional space.

We can now regroup, and find for the integrand

\begin{aligned} 0 = \frac{\partial \rho}{\partial t} + \sum_{i_\alpha} \left( \frac{\partial \left( \rho \dot{q}_{i_\alpha} \right) }{ \partial q_{i_\alpha} } + \frac{\partial \left( \rho \dot{p}_{i_\alpha} \right) }{ \partial p_{i_\alpha} } \right),\end{aligned} \hspace{\stretch{1}}(1.0.5)

which is the \underlineAndIndex{continuity equation}. The assumptions that we have to make are that the flow of the density in phase space through the surface is proportional to the projection of the vector $\rho (\dot{q}_{i_\alpha}, \dot{p}_{i_\alpha})$, and then use the same old arguments (extended to a $6N$ dimensional space) as we did for the continuity equation for 3D masses.