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Microscopic transport equations

  1. Jul 18, 2007 #1
    Hi,

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

    [itex]\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[/itex] ... (1)

    where [itex]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))[/itex].

    Klimontovich appears to be using the chain rule

    [itex] \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))[/itex]

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

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

    the above expression simplifies to

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

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

    [itex](\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)[/itex],

    which is a little bit difficult to believe IMO.

    Can anyone shed some light on this please.

    Thanks

    Note:
    [itex] X \in \mathbb{R}^{6N}[/itex]
    [itex] x \in \mathbb{R}^{6}[/itex]
    [itex] \mathbf{r},\mathbf{p} \in \mathbb{R}^{3}[/itex]
     
    Last edited: Jul 18, 2007
  2. jcsd
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