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Math Help - verify that gamma = 1 + p^2/[(m^2)(c^2)]

  1. #1
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    verify that gamma = 1 + p^2/[(m^2)(c^2)]

    Verify that 1/sqrt(1-(v^2)/(c^2)) = 1 + p^2/[(m^2)(c^2)]

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    OK, so on the right side I changed p^2 into (m^2)(v^2) and cancelled out the m^2 in the numerator and denominator. This leaves me with:

    1 + (v^2)/(c^2) = 1/sqrt(1-(v^2)/(c^2))

    This seems tantalizingly close, but I have no clue what to do next.

    Any guidance would be appreciated.

    Yes, this is a homework problem, but homework isn't turned in. I'd still like to know how to do it though.

    ------------
    By the way, how do you guys type math? I'd like to make it more readable.
    Last edited by Pewł; September 9th 2009 at 09:39 PM.
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  2. #2
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    Are you sure you're not supposed to be using the relativistic definition of momentum here: p = \gamma mv . After all, you are dealing with the lorentz factor here.

    Also, it looks like your right hand expression turns out to be \gamma ^2 . Are you sure you typed in the question correctly?
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    Thanks for the prompt response.

    I thought of using the relativistic definition, but it seems the nonrelativistic one got me closer. When using p = gamma mv, I get 1+gamma^2-gamma^4 for the right hand side. Can't factor that . . . or can I?

    Yes, I typed it as it appears in the book. I'll look for a list of errata, though.
    I'm taking gamma to be one over the square root of (1-(v^2)/(c^2))
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    Although your two expressions look similar, they are not equal. Just plug in some values of v and you'll find out for yourself.

    As for your problem, start with the RHS:
    \begin{aligned} 1 + \frac{p^2}{m^2c^2} &= 1 + \frac{{\color{red}\gamma ^2} {m^2} v^2}{m^2 c^2} \\ & = 1 + {\color{red}\frac{1}{1 - \frac{v^2}{c^2}}} \cdot \frac{v^2}{c^2} \\ & = 1 + \frac{v^2}{c^2 - v^2}\end{aligned}

    Keep going with this. Combine the fractions and simplify. You should end up with \gamma ^2.
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    Yes, I do end up getting gamma squared. After combining the fractions, I multiplied top and bottom by 1/(c^2).

    So I guess it was a bad problem then. International editions do have a lot of errors (probably intentional).
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