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Rutherford scattering question

  1. Aug 31, 2006 #1

    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. jcsd
  3. Sep 1, 2006 #2


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    Homework Helper

    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 .
  4. Sep 1, 2006 #3
    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?
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