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

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

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

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

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

$$\frac{\partial \langle\vec{p}\rangle}{\partial t}=-\frac{\langle\vec{p}\rangle}{\tau_\rm{tr}}$$

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

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

Attempt at a solution

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

b) I start by writing:

$$\langle\vec{p}\rangle =\int \mathrm d^3p \; f(\vec{p},t)\cdot \vec{p}$$

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

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

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

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Sorry, I don't have the answer for now either but I am at least confused by the notations. Is $\theta$ in expression a) the same as the one in the expression of $\tau_{tr}$?

Is $\omega(\theta)$ in fact $\omega(\theta|\theta')=\omega(|\theta-\theta'|)$?

Also, to be consistent, I think that is average $\langle \vec{p}\rangle$ can be written as

$\langle \vec{p} \rangle = \int d \Omega \:f(\vec{p},t)\vec{p}$

since your probability space is simply that of orientations for your vectors.

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It's okay, just having a discussion about the problem should be pretty helpful. Thanks! Regarding your remarks: yeah, i agree about the writing of $<\vec{p}>$; The $\theta$ in $\omega(\theta)$ was also giving me some confusion but I'm pretty sure it is $\omega(\left |\theta-{\theta}' \right|)$ as you claim (it is what makes the most sense). – benfstokes Apr 4 '13 at 13:30

One can perform the $\textbf{p}$ integral before the $\Omega$ integral in the last equation you wrote: $$ \int d{\Omega} \int d^3 p \; \vec{p}\, \omega(\theta) f(\vec{{p}'},t), $$ where I removed primes in angular variables because they are integrated over anyway.

Here, $\textbf{p} = R(\theta,\phi) \textbf{p}^{\prime}$, with some appropriate rotation matrix $R(\theta,\phi)$, and with this, you can express $\textbf{p}$ and $d^{3}p$ in terms of $\textbf{p}^{\prime}$ and $d^{3}p^{\prime}$ . Then, the rest will follow.

I think this may be a sufficient hint, but I'll provide more details if you want.

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Don't understand why I didn't realize that this question is almost a year old. – higgsss Jan 31 '14 at 13:12

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