Derivation of relativistic E_kin

  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
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What was the original variable being integrated over?
What was substituted in it's place?
 
  • #3
phyzguy
Science Advisor
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Take:
[tex] f' = d (\frac{mu}{\sqrt{1-(u^2/c^2)}}), g = u[/tex]
 
  • #4
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Ah, thanks phyzguy...the penny has dropped.
 

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