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Relativistic Energy Derivation math problem

  1. Jul 10, 2013 #1

    In a derivation of relativistic energy (in Physics for Scientists and Engineers, 5th edition, Serway and Beichner) they use a method of integration by substitution:

    Given that


    and relativistic momentum is given by


    [itex]W=∫F dx=∫\frac{dp}{dt} dx[/itex]

    which is fine then they say


    (This is about halfway through the derivation.)
    its that last step that i dont understand. why does the denominator term now have a 3/2 index? and why is du/dt given like that? Any help would be appreciated.
    Last edited: Jul 10, 2013
  2. jcsd
  3. Jul 10, 2013 #2


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    is there a missing square-root symbol?
  4. Jul 10, 2013 #3


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    (d/dt)(mv(1-v2)-1/2) = m(1-v2)-1/2dv/dt -1/2(-2v)(mv)(1-v2)-3/2)dv/dt
    [first term is from d/dt'ing the v in the numerator, second is from d/dt'ing the denominator]

    = m(1-v2 + v2)(1-v2)-3/2dv/dt = m(1-v2)-3/2dv/dt
  5. Jul 10, 2013 #4
    Sorry but that's not very clear. I have trouble reading non formatted math like that. As a second year physics student in uni, should I be able to do that differentiation by hand?
  6. Jul 10, 2013 #5


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    Try looking up "chain rule". Wiki has a definition, it might not be the easiest to read. Note that you want to find (d/dt) p(t) which is equal to (d/dt) p(v(t)), and that m is assumed to be constant (not a function of time).
  7. Jul 10, 2013 #6
    Sorry I use u instead of v for velocity

    In plain English the reason you get an exponent of -3/2 is because γ(u) is an exponent of -1/2, and what's the derivative of x-1/2? It's (-1/2)x-3/2.

    I actually just posted a similar question regarding this topic, but with the time component rather than space.
    Last edited: Jul 10, 2013
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