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Math Help - annoying relativity

  1. #1
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    Unhappy annoying relativity

    I am having problems with an equation involving relativity, but my question is directly math related currently the equation is in the form of e=mc^2(blah blah blah), but I need it in the form e=mv^2(blah blah blah).

    the current equation is "E=mc^2((1/(sqrt(1-v^2/c^2)))-1"

    where sqrt(x) signifys the square root of x. please help.

    what I really need is the answer, but I would love to know the steps taken.
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  2. #2
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    Quote Originally Posted by dented42
    I am having problems with an equation involving relativity, but my question is directly math related currently the equation is in the form of e=mc^2(blah blah blah), but I need it in the form e=mv^2(blah blah blah).

    the current equation is "E=mc^2((1/(sqrt(1-v^2/c^2)))-1"

    where sqrt(x) signifys the square root of x. please help.

    what I really need is the answer, but I would love to know the steps taken.
    I do not understand you want to solve for v?
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dented42
    I am having problems with an equation involving relativity, but my question is directly math related currently the equation is in the form of e=mc^2(blah blah blah), but I need it in the form e=mv^2(blah blah blah).

    the current equation is "E=mc^2((1/(sqrt(1-v^2/c^2)))-1"

    where sqrt(x) signifys the square root of x. please help.

    what I really need is the answer, but I would love to know the steps taken.
    What you want to do is expand the gamma function ( \gamma=1/\sqrt{1-v^2/c^2}) in a Taylor series about v=0. So E = mc^2(1+1/1!*\gamma'(v=0)+1/2!*\gamma''(v=0)+...). You should get E=mc^2(1+1/2*v^2/c^2+3/8*v^4/c^4+...).

    -Dan
    Last edited by topsquark; February 14th 2006 at 06:22 AM.
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