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Math Help - Theory OF Relativity

  1. #1
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    Unhappy Theory OF Relativity

    Im having trouble with this homework problem, Its just that the formua throws me off, and I cant figure out what is asking.

    #7. According to the Theory of Relativity the energy E of a body with mass "m" is given as a function of its speed "v" by



    Where c is constant, the speed of light.

    A. Assuming v< c, expand E as a series in (V/C) to the second nonzero term.

    B. Explain why (v/c) is very small, then E can be well approximated by
    E=1/2 mv^2
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  2. #2
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    Hi

    For \frac{v}{c} << 1

    E = mc^2\left[\left(1-\frac{v^2}{c^2}\right)^{-\frac12}-1\right] = mc^2\left(1+\frac12\:\frac{v^2}{c^2}-1\right) = \frac12\:mv^2
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  3. #3
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    Can you explain this please. Would that be number A.Thank you
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  4. #4
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    When a is negligible with respect to 1 you can expand \left(1+a\right)^n as 1+na at 1st order

    Here a=-\frac{v^2}{c^2} and n = -\frac12
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  5. #5
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    So the second non zero term would be 1/2 mv^2?... As for be (v/c) is very small because V is always less than C, automatically thats less than 1.. it further becomes small when u square it..
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  6. #6
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    Yes. If \frac{v}{c} << 1 then \frac{v^2}{c^2} << 1 and \left(1-\frac{v^2}{c^2}\right)^{-\frac12} \simeq 1+\frac12\:\frac{v^2}{c^2}
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  7. #7
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    Ok So I get part B of this question Thank You. Im still having a little trouble expanding The series (v/c) to the second non zero term as in part A.
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