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Show that F = gamma^3*ma

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Newton's second law is given by F = dp/dt. If the force is always parallel to the velocity, show that F = gamma3ma

    2. Relevant equations

    p = gamma*mv
    gamma = 1/(1-v2/c2)1/2

    3. The attempt at a solution

    I really have no clue where to begin. This is what I've done so far, but I don't think I'm even on the right track.

    F = dp/dt = gamma3ma = [1/(1-v2/c2)1/2]3ma
    =1/(1-v2/c2)3/2ma
    =1/(1 - 3v2/c2 + 3v4/c4 - v6/c6)1/2ma

    Thats as far as i get on the right side.

    For the left, all i can manage is dp/dt = d/dx p = d/dx gamma*mv
    This is where I get confused. gamma and m are constant, so does dp/dt = gamma*m*dv/dt?

    Some clarification would be great.

    Thanks!
     
  2. jcsd
  3. Sep 19, 2009 #2
    Are you in Modern Physics and answering this for your Russian Prof?
     
  4. Sep 20, 2009 #3
    You're taking the hard route, you're trying to integrate the given expression, into the derivative. Just take the derivative, and rearrange.

    Some pointers:

    Express [tex]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=({1-\frac{v^2}{c^2}})^{-\tfrac{1}{2}[/tex]

    That should make taking the derivative much more simple.

    Another bit that should help you is: [tex]\frac{d({1-\frac{v^2}{c^2}})}{dt}=\frac{-2v\cdot\dot v}{c^2}[/tex]
     
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