# Microscopic transport equations

1. Jul 18, 2007

### jdstokes

Hi,

I'm working my way through Klimontovich's `statistical mechanics' and I'm having trouble with his derivation of the microscopic transport equation:

$\frac{\partial N(x,t)}{\partial t} + \mathbf{v} \frac{\partial N}{\partial \mathbf{r}} + \mathbf{F}^M(\mathbf{r},t)\frac{\partial N (x,t)}{\partial \mathbf{p}} = 0$ ... (1)

where $N(x,t) = \sum_{i=1}^N \delta (\mathbf{r} - \mathbf{r}_i(t))\delta(\mathbf{p} - \mathbf{p}_i(t)) = \sum_{i=1}^N \delta (x-x_i(t))$.

Klimontovich appears to be using the chain rule

$\frac{\partial N}{ \partial t} = \sum_{i=1}^N \left( \mathbf{v}_i \frac{\partial}{\partial \mathbf{r}_i} + \mathbf{F}^M(\mathbf{r}_i,t) \frac{\partial}{\partial \mathbf{p}_i}\right) \delta (x - x_i(t))$

where $F^M(\mathbf{r}_i,t) = d\mathbf{p}_i/dt$ is the microscopic force.

He then claims that since, e.g., $(\partial/\partial \mathbf{r}_i)\delta (\mathbf{r}-\mathbf{r}_i)=-(\partial/\partial \mathbf{r})\delta(\mathbf{r} - \mathbf{r}_i)$,

the above expression simplifies to

$-\mathbf{v} \frac{\partial N}{\partial \mathbf{r}} -\mathbf{F}^M(\mathbf{r},t)\frac{\partial N (x,t)}{\partial \mathbf{p}}$.

But I don't see this. In order to make this simplification we must have

$(\mathbf{v}_i\partial/\partial \mathbf{r}_i)\delta (\mathbf{r}-\mathbf{r}_i)=-(\mathbf{v}\partial/\partial \mathbf{r})\delta(\mathbf{r} - \mathbf{r}_i)$,

which is a little bit difficult to believe IMO.

Can anyone shed some light on this please.

Thanks

Note:
$X \in \mathbb{R}^{6N}$
$x \in \mathbb{R}^{6}$
$\mathbf{r},\mathbf{p} \in \mathbb{R}^{3}$

Last edited: Jul 18, 2007