# Momentum probability density change in colisions (Drude Model)

1. Apr 3, 2013

### benf.stokes

1. The problem statement, all variables and given/known data

A particle suffers elastic colisions with scattering centers with a probability of colision per unit time $\lambda$. After a colision the particle is in a direction caracterized by a solid angle $d\Omega$ with probability $\omega(\theta) d\Omega$, that depends only on the angle between the inicial direction $\vec{{p}'}$ and the final direction $\vec{p}$. Assume only elastic colisions, $p={p}'$

a) Obtain the folowing equation of movement for the density of probability $f(\vec{p},t)$:

$\frac{\partial f(\vec{p},t)}{\partial t}=-\lambda\cdot f(\vec{p},t)+\lambda \cdot \int d{\Omega}' \cdot \omega(\theta)f(\vec{{p}'},t)$

Where the integration is over the solid angle of $\vec{{p}'} (d{\Omega}'=sin{\theta}' d{\theta}'d{\phi}'$)

b)Show that the equation of movement of the average momentum is:

$\frac{\partial <\vec{p}>}{\partial t}=-\frac{<\vec{p}>}{\tau_{tr}}$

where $\tau_{tr}$, the transport time is defined by:

$\frac{1}{\tau_{tr}}=\lambda \int d\Omega (1-cos \;\theta) \omega(\theta)$

c)Deduce the Drude condutivity with this model.

2. The attempt at a solution

a) I can arrive at the given expression, so no problems here

b) I start by writing:

$<\vec{p}>=\int d^3p \; f(\vec{p},t)\cdot \vec{p}$

And I derive this expressions with respect to t, getting from a) that:

$\frac{\partial <\vec{p}>}{\partial t}=-\lambda\cdot <\vec{p}>+\lambda \cdot \int d^3 p \; \vec{p} \int d{\Omega}' \cdot \omega(\theta) f(\vec{{p}'},t)$

And from here I don't what to do, is there something that I'm missing or does it need some kind of trick?

c) Can't do it without doing b) :p

Last edited: Apr 3, 2013