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Is there a relativistic version of the work-energy theorem?

  1. Feb 25, 2012 #1


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    The work-energy theorem is stated here on Wikipedia. On the same page it says "regardless of the choice of reference frame, the work energy theorem remains valid and the work done on the object is equal to the change in kinetic energy."

    I am wondering if there is a relativistic version of the work-energy theorem.


  2. jcsd
  3. Feb 25, 2012 #2
    No. The classical version applies to relativity too. But the result is different.
  4. Feb 25, 2012 #3
    [itex]W=\Delta E_k=(\gamma -1)mc^2[/itex]

    Just integrate force over some distance, starting from rest:

    [tex]W=\int Fdx=\int \frac{dp~dx}{dt}=\int vdp=vp-\int pdv=\gamma mv^2-\int \gamma mvdv[/tex]

    What you end up with is:

    [itex]W=\gamma mc^2+C[/itex]

    with some constant of integration C. If you set v=0, you take W=0 and γ=1, therefore C=-mc2. So what you get is:

    [itex]W=\gamma mc^2-mc^2=(\gamma -1)mc^2=E_k[/itex]
  5. Feb 25, 2012 #4


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    In general, [itex]W=\int F dx[/itex], but the Lorentz transformations tell you the relationship between dx in the two different inertial frames as [itex]dx = {{dx'}\over{\sqrt{1-{{v^2}\over{c^2}}}}}[/itex].
  6. Feb 25, 2012 #5
    Two different inertial frames?
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