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Homework Help: Integration simplification

  1. May 29, 2012 #1
    I simply don't understand this conversion. I've only worked with du, dx, dy, etc. before. How do you change a d(γmu) into some kind of du?
    (This is in the contest of relativity, although this info is not essential i think)

    Thanks! :)
     

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  3. May 29, 2012 #2

    SammyS

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    attachment.php?attachmentid=47775&d=1338348275.jpg


    What is [itex]\displaystyle \frac{d}{du}\left(\frac{mu}{\sqrt{1-u^2/c^2}}\right)\ ?[/itex]
     
  4. May 29, 2012 #3


    Let [tex]x = \frac{{mu}}{{\sqrt {1 - \frac{{{u^2}}}{{c_{0}^2}}} }}[/tex]

    Then solve for u.
     
  5. May 29, 2012 #4
    Figured it out thanks! :))
     
  6. May 29, 2012 #5
    Wait dimension10, how do you solve it your way, after getting u = x/sqrt(m^2+x^2/c^2)?
     
  7. May 29, 2012 #6
    [tex]\begin{array}{l}
    {x^2} = \frac{{{m^2}{u^2}}}{{1 - \frac{{{u^2}}}{{{c_0}^2}}}}\\
    {x^2} = \frac{{{m^2}}}{{\frac{1}{{{u^2}}} - \frac{1}{{{c_0}^2}}}}\\
    \frac{{{m^2}}}{{{x^2}}} + \frac{1}{{c_0^2}} = \frac{1}{{{u^2}}}\\
    {u^2} = \frac{1}{{\frac{{{m^2}}}{{{x^2}}} + \frac{1}{{c_0^2}}}}\\
    {u^2} = \frac{{c_0^2{x^2}}}{{{m^2}c_0^2 + {x^2}}}\\
    u = \frac{{{c_0}x}}{{\sqrt {{m^2}c_0^2 + {x^2}} }}\\
    ...\\
    \int {\frac{{{c_0}x}}{{\sqrt {{m^2}c_0^2 + {x^2}} }}{\rm{d}}x} = \int {m{{\left( {1 - \frac{{{x^2}}}{{c_0^2}}} \right)}^{ - \frac{3}{2}}}x{\rm{ d}}u} = \int {m{{\left( {1 - \frac{{{u^2}}}{{c_0^2}}} \right)}^{ - \frac{3}{2}}}u{\rm{ d}}u}
    \end{array}[/tex]
     
  8. May 29, 2012 #7
    Thanks a lot!
     
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