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Derivation of relativistic E_kin

  1. Oct 9, 2013 #1
    In my textbook there is a derivation of relativistic kinetic energy starting from an integral of the force applied over the distance required to take the particle's speed from 0 to v.
    There's one stage of the derivation I don't understand on mathematical grounds, which is going from:

    $$E_k = \int_0^{v} u\ d\left( \frac{mu}{\sqrt{1-(u^2/c^2)}}\right )$$
    To the next line, which is:
    $$E_k = \left [ \frac{mu^2}{\sqrt{1-(u^2/c^2)}}\right]_0^v - \int_0^v \frac{mu}{\sqrt{1-(u^2/c^2)}}\ du$$
    I guess they have used integration by parts, but how do you get the change of integrating variable? I only really know the rule ##\int f'g dx = fg - \int fg' dx## and can't seem to make it fit here.
    Thanks in advance
     
  2. jcsd
  3. Oct 9, 2013 #2

    Simon Bridge

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    What was the original variable being integrated over?
    What was substituted in it's place?
     
  4. Oct 9, 2013 #3

    phyzguy

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    Take:
    [tex] f' = d (\frac{mu}{\sqrt{1-(u^2/c^2)}}), g = u[/tex]
     
  5. Oct 9, 2013 #4
    Ah, thanks phyzguy...the penny has dropped.
     
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