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Work-energy theorem integral

  1. Jul 31, 2014 #1
    I'm having trouble with an integral involved in deriving the work-energy theorem


    1. The problem statement, all variables and given/known data

    I'm trying to get from ∫mv/√(1-v^2/c^2)dv to -mc^2(1-v^2/c^2).


    2. Relevant equations



    3. The attempt at a solution

    I start out by putting gamma on top to yield: ∫mv(1-v^2/c^2)^-1/2, then I square everything to get rid of the square root term and end up with: ∫m^2 v^2 - m^2 v^2 (v^2/c^2)dv, and then to get rid of the fractions I end up with: ∫c^2 m^2 v^2 - m^2 v^4dv, which, when I try to integrate, gets me no where close to the answer. What am I doing wrong? Should I be doing a U-substitution or chain rule?
     
  2. jcsd
  3. Jul 31, 2014 #2

    Simon Bridge

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    Here, let me help,
    $$\int \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\;dv \rightarrow -mc^2\left(1-\frac{v^2}{c^2}\right) $$... you are trying to work the maths from the LHS to the RHS.
    ... yes :)

    try ##u=v/c## to start with.
     
  4. Aug 1, 2014 #3
    Ok, now I see.. If I take U=1-v^2/c^2, then I get dU/dx=-2v/c^2, and then just follow the protocol, and sure enough, there's the right answer.

    Thanks Simon!
     
  5. Aug 1, 2014 #4

    Simon Bridge

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    Well done - you usually want to look for some sort of substitution for the bit that gives you problems.
    Don't be frightened to try several different ones.

    Since you will be doing a lot of this, it is best practice to learn LaTeX :)
     
  6. Aug 2, 2014 #5

    vanhees71

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    To be fair one should say that the original statement is not correct. There's a square root missing in the given result of the integral!
     
  7. Aug 2, 2014 #6

    HallsofIvy

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    Also, since the integral on the left is an indefinite integral, there should be an added "constant of integration". What you are trying to prove simply isn't true!
     
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