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Homework Help: Need help with Special Relativity: Force and Energy

  1. Jun 20, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm trying to show that

    [tex]\frac{dE}{dT}[/tex]= F x V where F is force and V is velocity

    Can someone please help me out?


    2. Relevant equations

    F=[tex]\frac{dP}{dT}[/tex]=M x V x [tex]\gamma[/tex]


    All of the other equations from special relativity (contraction, dilation, energy) are probably important as well.

    3. The attempt at a solution

    Honestly, I have no idea where to start. Can someone please point me in the right direction. I'm a high school junior and I have very little experience with relativity. I understand that I should show some work, but I truly don't have any but I have been trying for a while.

    Please help me get started!
     
  2. jcsd
  3. Jun 20, 2010 #2

    vela

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    You should know the formula for the relativistic energy E of an object. Try differentiating that with respect to time and see what you get.
     
  4. Jun 20, 2010 #3
    The formulas for relativistic energy don't involve time (t) as far as I can tell.
    I'm not sure how taking the derivative of that would help unless I'm
    overlooking something.
     
  5. Jun 20, 2010 #4

    vela

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    It doesn't depend explicitly on time, but the energy depends on [itex]\gamma[/itex], which, in turn, depends on velocity, which does change with time.
     
  6. Jun 20, 2010 #5
    gosh, now I'm even more stuck, thanks anyway.

    so you meant that I should use [tex]\frac{x}{t}[/tex] instead of v?

    well after doing that, and taking the derivative, I can see no way for it to manipulate in F x V.
     
  7. Jun 20, 2010 #6

    vela

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    No, I think that'll make it more complicated. Keep it in terms of just the velocity v.

    Start with this first. You know that

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

    What do you get if you differentiate it with respect to time, remembering that v is a function of time?
     
    Last edited: Jun 20, 2010
  8. Jun 20, 2010 #7
    ok, this is what I have

    [tex]\frac{dE}{dt}[/tex]=[tex]mc^{2}\frac{-\gamma^{3}v}{c^{2}}\frac{dv}{dt}[/tex]= F x V

    am I doing this correctly? when equate the equation above with F x V (where F = gamma*mass*acceleration*v)
    i get that -[tex]\gamma^{3}=\gamma[/tex] so I get the feeling I messed up somewhere...

    I really appreciate your help!
     
    Last edited: Jun 20, 2010
  9. Jun 20, 2010 #8

    vela

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    Hmm, not sure how you came up with that. Can you show more steps?
     
  10. Jun 20, 2010 #9
    Sure,

    So here's the derivative of gamma
    [tex] \dot {\gamma} = \frac{d}{dt} \left( 1- \frac{v^2}{c^2} \right) ^{-1/2} = \left( \frac{-1}{2} \right) \left( \frac{-2v \dot {v} }{c^2} \right) \left( 1- \frac{v^2}{c^2} \right) ^{-3/2} = \gamma ^3 \left( \frac{v \dot {v}}{c^2} \right) [/tex]

    hence
    [tex]\frac{dE}{dt}=mc^{2}\frac{-\gamma^{3}v}{c^{2}}\frac{dv}{dt}[/tex]

    because E = mc[tex]^{2}\gamma[/tex].

    [tex]F\bulletV=m\gamma\frac{dv}{dt}v = \frac{dE}{dt}=mc^{2}\frac{\gamma^{3}v}{c^{2}}\frac{dv}{dt}[/tex]

    and then I simplified this
     
    Last edited: Jun 20, 2010
  11. Jun 20, 2010 #10

    vela

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    You're getting there. You need to be a bit more careful when calculating F. You know that the momentum is given by [itex]p=\gamma mv[/itex]. When you differentiate it, you need to use the product rule.
     
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