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Relativistic Momentum Help With Equation Reduction

  1. Aug 17, 2011 #1
    Hi, so basically have been looking at http://en.wikibooks.org/wiki/Special_Relativity:_Dynamics#Momentum" and working my way through the maths for myself. However I have hit this point and can't get past it:

    \begin{align}
    u = \frac{v - u}{1-\frac{uv}{c^2}}
    \end{align}
    Which should be able to become:
    \begin{align}
    u = \frac{c^2}{v(1-\sqrt{1-\frac{v^2}{c^2}})}
    \end{align}
    I have tried and tried but can't seem to get it to work. Can anyone help me out on this one please?!

    Thanks
    James
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Aug 17, 2011 #2

    G01

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    Homework Helper
    Gold Member

    That wiki entry is not being clear with it's order of operations!

    What you should be getting is:

    [tex] u = \frac{c^2}{v}(1-\sqrt{1-\frac{v^2}{c^2}})[/tex]
     
    Last edited by a moderator: Apr 26, 2017
  4. Aug 17, 2011 #3
    Hi G01, thanks for your reply. Have tried aiming for that equation instead and still can't get there. I have tried all sorts of different approaches, such as getting it in the form of a quadratic or dividing the top and bottom of the original fraction by c^2, but I just end up nowhere. Could you perhaps give me a hand in the right direction, maybe the first step or two - but don't make it too easy for me!
     
  5. Aug 17, 2011 #4

    jtbell

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    Staff: Mentor

    Show us exactly what you've tried, and where you get stuck, and tell us why you're stuck.
     
  6. Aug 17, 2011 #5
    Ok, have shut my computer down now and am replying on my phone. I will post my attempts tomorrow.
    Thanks,
    James
     
  7. Aug 17, 2011 #6

    PAllen

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    Science Advisor
    Gold Member

    Well, brute force use of the quadratic formula leads directly to the desired answer.
     
  8. Aug 19, 2011 #7
    Haha, went back over my general quadratic attempt after PAllen's suggestion and realised that I had made a rather fundamental cross multiplication error! Fixed that and got to the desired equation. Lovely!

    Last question, the equation that I get is:
    \begin{align}
    u = \frac{c^2}{v}(1\pm\sqrt{1-\frac{v^2}{c^2}})
    \end{align}
    I assume that we choose the negative version of the equation, because the positive version would yield a value for u which is greater than c?

    Cheers
     
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