# Rutherford scattering question

1. Aug 31, 2006

### schattenjaeger

http://img168.imageshack.us/my.php?image=fundn7.png

d(chi)/dt =l/(mr^2), I don't see how to get there

if you use the definition of angular momentum, you get l=r x mv(vectors where appropriate)

so you can write v as v=dr/dt*r + rdx/dt*x where x is the symbol chi(close enough)and that r and x at the end of the two expressions are unit vectors

So you can write the magnitude of l=|mr*v| where v is as above. Then I'm stumped

2. Sep 1, 2006

### lightgrav

Only the tangential part of v contributes to L ... it is ANGULAR momentum.
the radial part has r x (dr/dt)r_hat = 0 .
So, with the angular part, as you say, L = r m r dtheta/dt
=> dtheta/dt = L/mr^2 .

3. Sep 1, 2006

### schattenjaeger

Okokok I think you cleared it up

So to write it out completely, L=mr x (dr/dtr_hat+rdx/dtx_hat)

There's a distributive property for the cross product, right? So you get mr x dr/dtr_hat, and r and r_hat are parallel so it's 0, and you're left with L=mr x r dx/dtx_hat, the magnitude of which is mr^2dx/dt?