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Relativistic Kinetic Energy Derivation

  1. Feb 13, 2016 #1
    1. The problem statement, all variables and given/known data

    This problem comes from an intermediate step in the textbook's derivation of relativistic energy. It states that

    [tex]E_k\:=\:\int _0^u\frac{d\left(\gamma mu\right)}{dt}dx[/tex]

    then leaves the following intermediate calculation as an exercise to the reader:

    Show that

    [tex]d\left(\gamma mu\right)=\frac{m}{\left(1-\frac{u^2}{c^2}\right)^{\frac{3}{2}}\:}du [/tex]

    2. Relevant equations

    Where

    [tex]\gamma \:=\:\frac{1}{\sqrt{1-\frac{u^2}{c^2}}} [/tex]


    3. The attempt at a solution:


    I tried using the product rule

    [tex]\frac{d}{du}\left(\gamma mu\right)=m\:\frac{d}{du}\left(\gamma u\right)=\:m\left(\gamma 'u+u'\gamma \right)=\:m\left[u\:\frac{d}{du}\left(1-\frac{u^2}{c^2}\right)^{-\frac{1}{2}}+\left(1-\frac{u^2}{c^2}\right)^{-\frac{1}{2}}\right][/tex]

    [tex] =m\left[\frac{u^2}{c^2}\left(1-\frac{u^2}{c^2}\right)^{-\frac{3}{2}}+\left(1+\frac{u^2}{c^2}\right)^{-\frac{1}{2}}\right] [/tex]

    Anyone know what I'm doing wrong?
     
  2. jcsd
  3. Feb 13, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Everything looks good so far. Keep going. :oldsmile:

    [Oops, I do see a typo error in your last equation. Check the signs inside the parentheses.]
     
  4. Feb 13, 2016 #3
    Thank you!

    I figured it out. I can't believe I did not see that this entire time.
     
    Last edited: Feb 13, 2016
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