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Relativistic Energy- well a basic algebraic simplification

  1. Sep 27, 2005 #1
    a little rough on this simplification, is this correct?
    solving for "u"

    E = mc^2 / (root)(1-u^2/c^2)

    (root)(1-u^2/c^2) = mc^2/E

    1 - u^2/c^2 = (mc^2/E)^2

    1 = (mc^2/E)^2 + u^2/c^2

    c^2 = (mc^2/E)^2 (c^2) + u^2 -> (sq root everything)

    u = c - mc^3/E

    pretty bad i don't know this basic simplification, but it'll all come back quickly enuf

    thanks for the help :smile:

    Last edited: Sep 27, 2005
  2. jcsd
  3. Sep 28, 2005 #2
    c^2 = (mc^2/E)^2 (c^2) + u^2 -> (sq root everything)

    u = c - mc^3/E

    Your mistake occurs on the transition between these two lines

    when you take the root of a function which has two terms added together, it is not equivalent to the root of each one added together

    ie root (a^2 + b^2) is not equal to (a + b)

    so you should really have
    u = root ((mc^2/E)^2(c^2) + u^2)
  4. Sep 28, 2005 #3
    alright thank you :smile:

    but do you mean

    u = root ((mc^2/E)^2(c^2) + c^2) but actually u = root (c^2 - (mc^2/E)^2(c^2) ) :confused:
  5. Sep 28, 2005 #4
    ah sorry, yes thats what I meant to type

    also you could further make it so that

    u = root (c^2(1 - (mc^2/E)^2))
    u = c*root(1 - (mc^2/E)^2)
  6. Sep 28, 2005 #5
    great! thanks a lot for the help :smile:
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