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Relativistic Momentum derivative respect to velocity

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    show the d(γmu) = m(1-(u^2/c^2))^(-3/2)

    2. Relevant equations

    C = constant, m = constant, γ= (1-(u^2/c^2))^(-1/2)

    3. The attempt at a solution

    So in calculating d(γmu)/du, I thought I would take out the m first since it is a constant and then just add it in later.

    m * d(γu)/du

    So I just have to calculate what γ is. γ= (1-(u^2/c^2))^(-1/2), so by chain rule the derivative of this is (-1/2)((1-(u^2/c^2))^(-3/2) * -2u/c^2. Then I just multiply this by m and I get
    m(u/c^2)(1-(u^2/c^2))^(-3/2).

    This differs from the correct solution by the factor (u/c^2) which i got from doing the chain rule and calculating the derivative of the inside with respect to u. If someone could please show me the error in my ways, that would be greatly appreciated. Thank you.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 7, 2011 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Is it \gamma * m * u ? If so, you must use the product rule first, then the chain rule.
     
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