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Relativistic Moment F = dp/dt

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data
    Newton's second law is given by [tex]\vec{F}[/tex]=[tex]\frac{d\vec{p}}{dt}[/tex]. If the force is always parallel to the velocity, show that [tex]\vec{F}[/tex]=[tex]\gamma^{3}m\vec{a}[/tex].


    2. Relevant equations
    [tex]p=\gamma m\vec{v}[/tex]
    [tex]\frac{d\vec{v}}{dt}=\vec{a}[/tex]
    [tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    3. The attempt at a solution
    I figure all you need to do would be to find d/dt of p. After a couple tries, I'm still unable to get the result that it provides.
     
  2. jcsd
  3. Sep 13, 2007 #2

    Dick

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    Yes, that's what you do, and it's straightforward enough. No real tricks. Maybe you could show us an attempt?
     
  4. Sep 13, 2007 #3
    [​IMG]

    Here's a scan of what I did... at the very bottom I was then lost.
     
  5. Sep 13, 2007 #4

    Dick

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    You missed a factor of dv/dt in the first term when you did the chain rule. So factoring out all of the common terms (gamma*m*dv/dt), you only need to show that (v^2/c^2)/(1-v^2/c^2)+1=gamma^2. You are almost there.
     
  6. Sep 13, 2007 #5
    Thanks for your help... I didn't realize that when I would take the derivative of -v^2/c^2 that it would be (-2v/c^2)(dv/dt). I think that's what you were getting at. I was able to get the answer that way :)
     
  7. Sep 13, 2007 #6

    Dick

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    That's what I was getting at.
     
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