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A General relativity -- Proof of energy measured by observer

  1. Apr 13, 2017 #1
    I want to prove that ##E = -g_{\mu \nu}u^\mu p^\nu## is the energy measured by an observer with velocity ##u^\mu## of an object with momentum ##p^\mu##. My reasoning is that in special relativity we know that ##\gamma m = E##. We can transform to coordinates where ##u'^\mu = (1,\vec{0})##. Since the measurement is happening locally to the observer, the space appears flat, and ##g = \eta## and hence ##dt/d\tau = \gamma##.
    $$
    -g_{\mu \nu}u^\mu p^\nu = -\eta_{\mu \nu}u'^\mu p'^\nu =
    -(1,\vec{0})\cdot(-\gamma m, \gamma \vec{v}) = \gamma m = E
    $$
    My question. Is this kind of reasoning legal? Is it enough to argue that since the measurement happens locally, we can take the space to be flat? The reason I ask is because it would surprise me since pretty much any kind of measurement I can think of is a local concept.
     
    Last edited: Apr 13, 2017
  2. jcsd
  3. Apr 13, 2017 #2

    PeroK

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    I think that's reasonable, although I don't think you need to introduce ##\gamma## into the equation. In the observer's LIRF, the energy of a particle is the first component of its four-velocity. In this LIRF, the observer's four-velocity is, as you say, ##(1, \vec{0})##. Therefore, the measured energy is the inner product of these two four velocities.

    The inner product is coordinate-independent (assuming that it is defined using the metric appropriate to the coordinates in each case), hence the result follows.
     
  4. Apr 13, 2017 #3
    Thanks!
     
  5. Apr 13, 2017 #4

    PeterDonis

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    You do if you want to get the right answer. :wink:

    Not quite; it's the first component of its 4-momentum. That's where the factor of ##m## in the correct answer ##\gamma m## comes from. If it were the first component of the object's 4-velocity, that would just be ##\gamma##, with no factor of ##m##.
     
  6. Apr 13, 2017 #5

    PeterDonis

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    Yes. (With the proviso that it's spacetime that we take to be flat, not space.)

    Plenty of quantities of interest do not involve purely local measurements. For example, the redshift of distant objects due to the expansion of the universe.
     
  7. Apr 13, 2017 #6
    What do you mean? Is it wrong to use the gamma? I was taught that we obtain GR locally. Thanks
     
  8. Apr 13, 2017 #7

    PeterDonis

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    No, it's right. I was responding to PeroK who appeared to be implying that you didn't need to include ##\gamma## in the equation; you do.
     
  9. Apr 14, 2017 #8

    PeroK

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    Yes, of course, four-momentum rather than just the four-velocity.

    The point I was making is that:

    ##-\mathbf{p}.\mathbf{u_{obs}}##

    In invariant in any coordinate system, including in the LIRF, where we know that it equals the particle's observed energy. It's irrelevant to the argument that ##E= \gamma m## in this reference frame.

    In other words, you could write more simply:

    $$
    -g_{\mu \nu}u^\mu p^\nu = -\eta_{\mu \nu}u'^\mu p'^\nu =
    -(1,\vec{0})\cdot(E, \vec{p}) = E
    $$
     
  10. Apr 14, 2017 #9

    PeterDonis

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    Ah, I see. Yes, this is correct.
     
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