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Formula for energy in GR

  1. Nov 25, 2008 #1
    If a stationary observer throws a rock out in the radial direction in a Schwarzschild geometry, what is the relationship between the energy of the rock and its speed in the observer's frame?

    I'm a bit confused because the book seems to say it's [tex]E = \frac{m}{\sqrt{1-V^2}}[/tex], but I thought that only applied to inertial frames in special relativity.

  2. jcsd
  3. Nov 25, 2008 #2


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    That formula only makes sense if it is assumed that [itex]c=1[/itex], otherwise it could be written

    [tex]E = \frac{mc^2}{\sqrt{1-V^2/c^2}}[/tex]

    which I don't recognise. Over to someone else.
  4. Nov 25, 2008 #3

    George Jones

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    If, with respect to an orthonormal frame on the observer's worldline, the rock has speed [itex]V[/itex] on release, then, at release and with respect to this frame, the total energy of the rock is given by the same expression as in special relativity.
  5. Nov 25, 2008 #4
    Thanks for the replies.

    Is that because at release the positions of the observer and the rock coincide and since the spacetime is tangentially flat, the formula applies as in special relativity?

    So is it true that if the rock's position was different from the observer's, the formula wouldn't apply?
  6. Nov 28, 2008 #5
    Can someone tell me why the formula applies in this situation?
  7. Nov 28, 2008 #6


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    "Energy", especially "potential Energy" is a bit tricky, because for the latter there is not always a straightforward definition in GR.
    So if you use local standards to measure the velocity of the stone, said formula holds. Energy is not conserved because you use a different frame for every point of the stone's worldline.
    If you use a single (static) frame, like the Schwarzschild metric, there is a conserved quantity which one can identif with energy. If [tex]E_{kin}[/tex] is the kinetic energy as defined above, [tex]E_{kin}\sqrt{1-\frac{2M}{r}}[/tex] is the total energy, a constant.
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