Question about an "exact" distribution function

[dRic2]

Suppose I have an exact microscopic distribution function in phase-space defined as a sum of delta-functions, i.e
$$F( \mathbf x, \mathbf v, t) = \sum_{i} \delta( \mathbf x - \mathbf x_i ) \delta (\mathbf v - \mathbf v_i )$$
Can I conclude that, in absence of creation/destruction of particles,
$$ \frac {dF( \mathbf x, \mathbf v, t)}{dt} = \frac {\partial F} {\partial t} + \mathbf v \cdot \frac {\partial F} {\partial \mathbf x} + \mathbf a \cdot \frac {\partial F} {\partial \mathbf v} = 0$$
where $\mathbf a = \frac {d \mathbf v}{dt}$ Note that here no statistical averaging is involved.

I found that this is "trivially" derived from phase-space conservation, but although I get a feeling for it, I don't see it so clearly.

Thanks
Ric

 

[dRic2]

Maybe something like this?

Consider a volume $\tau$ in the case space. Then $\frac {dN \tau}{dy} = 0$ because the total number of particles can not change. Then differentiation of the product yields
$\frac {dN}{dt}\tau = -N \frac {d\tau}{dt} = 0$ because Liouville's theorem tells that phase space behaves as an incompressible fluid, so the volume element must be unchanged.

 

[vanhees71]

In the OP the most important thing is missing in the intial equation, namely the (ensemble) average over the phase-space trajectories $(x_j(t),v_j(t))$ (though I'd prefer $p_j$ rather then $v_j$).

 

[dRic2]

My question is what happens if I do *not* perform an ensemble average.

This is not statistical mechanics just yet, it's just a question about the *exact* equation of motion for a system of particles

 

[vanhees71]

Then I don't understand what you are aiming at. Without an averaging the left-hand-side of the equation is not a phase-space distribution function.

 

[dRic2]

https://farside.ph.utexas.edu/teaching/plasma/lectures1/node28.html

If you have time, please read the first page. In particular I'm referring to eq 169