Microscopic transport equations

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

 

[Jhero]

Thank you for reaching out for help with your understanding of Klimontovich's derivation of the microscopic transport equation. As a fellow scientist, I understand the importance of fully understanding the concepts and equations we work with.

After reviewing Klimontovich's derivation and the forum post you referenced, I can offer some clarification on the simplification you are questioning. The key to understanding this simplification lies in the definition of the Dirac delta function and its properties.

As you mentioned, we have N(x,t) = \sum_{i=1}^N \delta (\mathbf{r} - \mathbf{r}_i(t))\delta(\mathbf{p} - \mathbf{p}_i(t)). The Dirac delta function is defined as \delta(\mathbf{x}) = 0 for \mathbf{x} \neq 0 and \int_{-\infty}^{\infty} \delta(\mathbf{x})d\mathbf{x} = 1. This means that when we take the derivative of N with respect to \mathbf{r}, we can treat the delta function as a constant with respect to \mathbf{r}. This is why we can simplify (\partial/\partial \mathbf{r}_i)\delta (\mathbf{r}-\mathbf{r}_i) to -(\partial/\partial \mathbf{r})\delta(\mathbf{r} - \mathbf{r}_i).

In other words, when we take the derivative of N with respect to \mathbf{r}, we are essentially taking the derivative of a sum of constants (the delta functions) and we can simply bring the derivative inside the sum. This is a common property of the Dirac delta function and is often used in mathematical derivations.

I hope this helps to clarify the simplification in Klimontovich's derivation. If you have any further questions or concerns, please don't hesitate to ask. Good luck with your studies!