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Deriving formula for Relativistic Kinetic Energy

  1. May 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Hey all,
    I am encountering a problem with a derivation of the formula [tex] K_{ineticEnergy}=mc^2-m_{0}c^2[/tex] as it is described by my textbook. I need someone to explain to me how the author changes the integral and the upper limit of it in the final part. I'll now give you the equations.



    2. Relevant equations
    The author states that the formula can be derived straight from the definition of kinetic energy
    [tex] K=\int^{s}_{0} Fds [/tex] and using Newton's 2nd law [tex] F= \frac{d(mu)}{dt}[/tex] we get
    [tex]K=\int^{s}_{0}\frac{d(mu)}{dt}ds=\int^{mu}_{0}ud(mu)=\int^{u}_{0}ud(\frac{m_{0}u}{\sqrt{1-\frac{u^2}{c^2}}}) [/tex]
    and goes on in a way I can now follow.



    3. The attempt at a solution
    If someone could tell me the steps the author ommits that would be terrific. I've found that wikipedia's derivation of the formula doesn't have such confusing parts.
     
  2. jcsd
  3. May 17, 2012 #2
    No steps are omitted. One step relates to the quantity [itex]\frac{ds}{dt}[/itex] in the first integral, and the second step is simply writing the relativistic mass in terms of the rest mass.
     
  4. May 17, 2012 #3
    Ok but there are some things I'm not sure about:

    1) why does the upper limit change from s to mu, simply because we swap ds/dt for u and use d(mu) for the variable of integration, it just doesn't sound too strict mathematically to me.

    2) why does m vanish in the final upper limit just because we use relativistic mass?
     
  5. May 17, 2012 #4
    Integrating [itex]d(mu)[/itex] from one position to another position doesn't make sense, so the integral is written in terms of going from one momentum to another. There's nothing mathematically flawed about changing the variable of integration, and therefore changing the limits of the integration. Of course, this is one of those instances where just writing [itex]u = ds/dt[/itex] can get you in trouble, if you don't think about what else must change to accompany that if you're doing that operation inside an integral.

    For the same reason. The integration variable is no longer [itex]mu[/itex]; the integral depends only on [itex]u[/itex] at this point.
     
  6. May 17, 2012 #5
    I see, I think I get it , thanks!
     
  7. May 17, 2012 #6
    If you're at all ever confused about this sort of thing, never hesitate to do a formal variable substitution like you would for a traditional integral. The author is being lax, like physicists typically are, and assuming that the integral of [itex]ds[/itex] from the initial position to the final position is equivalent to the integral of [itex]d(mu)[/itex] from the initial momentum to the final momentum, since it is the same motion that is being integrated over in either case. This is of course obvious to the physicist, but it rightfully does make the mathematician shudder :)
     
  8. May 17, 2012 #7
    Seeing as I'm studying Physics, this may be a bad sign
     
  9. May 17, 2012 #8
    I didn't mean it like that all! It's only obvious to the physicist after you've had some practice with it (which you're now starting to get). One of the great things about being an undergrad is that you do it the hard way now, so that you can justify doing it the easy way for the rest of your career :)
     
  10. May 17, 2012 #9
    This is Introductory physics? :O
     
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