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Deriving power from the relativistic energy equation

  1. Sep 27, 2011 #1
    1. The problem statement, all variables and given/known data
    I recently finished a test that asks you to derive
    [itex]Power = \frac{dE}{dt} = F \times v[/itex]
    from the energy equation:
    [itex]E^2 = E_{0}^2 + (pc)^2[/itex]

    2. Relevant equations
    [itex]Power = \frac{dE}{dt} = F \times v[/itex]
    [itex]E^2 = E_{0}^2 + (pc)^2[/itex]
    [itex]p = \gamma m v[/itex]

    3. The attempt at a solution
    I got there in kind of a messy way but I would like to know how I could have more cleanly shown how to put it together. Here's the way I got to it:
    [itex]2E \frac{dE}{dt} = 0 + 2pc \frac{dp}{dt}[/itex]
    [itex]2(\gamma mc^2) \frac{dE}{dt} = 0 + 2 \gamma mvc \frac{dp}{dt}[/itex]
    [itex]\frac{dp}{dt} = F \stackrel{and\rightarrow}{} \frac{dx}{dt} = v [/itex]
    therefore [itex]\frac{dE}{dt} = P = Fv[/itex]

    Of course, I also realize I may have bungled this, so corrections or at least references to the rules would also be much appreciated. Please, weigh in.
     
  2. jcsd
  3. Sep 27, 2011 #2

    vela

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    Your notation is a little confusing because both the force F and the velocity v are vectors, so using the symbol [itex]\times[/itex] looks like you're taking the cross product of the two. What you want to show is[tex]\frac{dE}{dt} = \vec{F}\cdot\vec{v}[/tex]Use the fact that [itex]\vec{p}^2 = \vec{p}\cdot\vec{p}[/itex] and just basically do what you did, and you'll get that result.
     
  4. Sep 27, 2011 #3
    Sorry about the notation confusion. Still getting used to using LateX.

    The lightbulb came on with that comparison with the momentum squared. Thank you very much!
     
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