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I Absolute Energy in General Relativity?

  1. Jan 26, 2017 #1
    Let us assume that a spacetime has a global time-displacement symmetry described by a timelike Killing vector field ##\xi^\mu=(1,0,0,0)##.

    Further assume that we have a particle with four-momentum ##P^\mu=m\ V^\mu##.

    It is well-known that a stationary observer with four-velocity ##U^\mu## measures the energy of the particle, ##E_{obs}##, relative to himself as
    $$E_{obs}=U_\mu V^\mu.$$
    My question: Is the absolute energy of the particle given by ##E=\xi_\mu V^\mu##?

    I use the word "absolute" as ##\xi^\mu## is a global symmetry of the spacetime so that I would have thought that ##E## should be the energy of the particle relative to the spacetime itself.

    PS I'm sorry if I've asked similar questions in the past. I feel that I have formulated my question more succinctly in this post.
     
    Last edited: Jan 26, 2017
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  3. Jan 26, 2017 #2

    ShayanJ

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    As far as I understand your question, you're just asking us whether you can call some quantity a particular name. How is that physics?
     
  4. Jan 26, 2017 #3

    pervect

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    I've never seen the phrase "absolute energy" used in any paper. So the short answer would be "probably not", unless you can find some peer-reviewed paper that uses that term. Then you could justify the term by referring to the peer reviewed paper. You'd probably be called on to do so - the term has some unfortunate implications.

    For a longer answer, let's assume that the particle you're talking about with the 4-momentum ##P^{\mu}## is a test particle, Then assuming there are no external forces other than gravity acting on it, it will follow a geodesic. Since it follows a geodesic, the quantity ##\xi_{\mu} P^{\mu}## will be a constant of motion. See for instance Wald's appendix on Killing Vectors, the inner product of the Killing vector with a tangent vector to a geodesic is constant along the geodesic, and that's basically the mathematical justification for why this quantity is a constant of motion for our test particle.

    One of my texts (MTW) calls this constant of motion of a test particle "the energy at infinity", which is the term I usually use. I don't believe Wald uses this terminology, I don't recall him giving this quantity a name offhand.

    Note that ##\xi_\mu = g_{\mu\nu} \xi^{\nu}##, and if we choose coordinates such that we have ##\xi^\mu## = (1,0,0,0) and ##g_{\mu\nu}## = diag(-1,1,1,1) at infinity, then ##\xi_\mu## is (1,0,0,0) at infinity.

    What you can not do is take a collection of test particles and add together the "energies at infinity" to get a total energy. Basically, while the test particle mass doesn't have to be zero, it has to be small compared to the mass associated with the static space-time.

    What is this mass (or if you prefer energy) associated with the static space-time? It's known as the Komar mass. There's a formula given for it on Wald on pg 289, Wald does not use the term "Komar mass", IIRC, but does say that
    where ##\epsilon## are the Levi-civita symbols (or was it tensors? I get the two confused), and ##\xi## is the Killing vector. I won't attempt to justify the formula here, I'll refer you to the original text. I expect that it might will require some considerable study and additional questions if you seriously want to follow up on this point. I'm not sure how interested you are in the idea, though it seems potentially relevant to your question.

    It's handy to give this particular concept of mass applicable to static or stationary space-times a name. Wald didn't name this concept of mass, at least not IIRC, but it's generally known as the Komar mass. I believe it's also mentioned in a later section that the concept also applies to stationary space-times, not just static space-times.

    Note that the Komarr mass it's a mass associated with an entire space-time. There are well known problems with trying to localize the mass of a space-time and assign it to particular locations. I'd dig up a reference on why this is a problem, but this post has gone on long enough already - hopefully it's not gone on too long, though I suspect I've presented too much information at one gulp.

    But though I feel I've probably presented too much info to quickly, I don't see anyting I want to cut out.
     
  5. Jan 26, 2017 #4

    PeterDonis

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    This is actually the energy per unit mass. The energy is ##E_{obs} = U_\mu P^\mu##. In many contexts energy per unit mass is more useful, which is why you often see that form (and why the "per unit mass" is often omitted for brevity, on the assumption that the reader can put it back in based on the context); but it's important to understand the distinction.

    The usual name for the quantity ##\xi_\mu P^\mu## (see above for why I put ##P^\mu## instead of ##V^\mu##) is "energy at infinity", as pervect said. That name is based on the assumption (which is valid for all known cases of practical interest) that the norm of ##\xi^\mu## goes to ##1## at spatial infinity.

    The problem with this is that "spacetime itself" doesn't measure anything, so "energy relative to spacetime itself" makes no sense.
     
  6. Jan 27, 2017 #5

    Ibix

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    Killing vectors are picked out by a symmetry of spacetime. So, am I right in thinking that the OP has done something comparable to choosing to work in co-moving coordinates in cosmology? The invariants can have a meaning beyond just being a measurement by some random observer. For example ##\xi_\mu P^\mu## isn't just the energy measured by some observer as ##V_\mu P^\mu## would be, it is "energy at infinity". But this doesn't make the outcome more "absolute" than any other.
     
  7. Jan 27, 2017 #6
    1. Can I say that ##\xi_\mu P^\mu## is the energy of the particle with respect to the notion of time associated with ##\xi^\mu##?
    2. Can I say that ##\xi_\mu P^\mu## is the energy of the particle as measured by a freely-falling observer at rest in the local inertial frame of the particle?
     
    Last edited: Jan 27, 2017
  8. Jan 27, 2017 #7

    PeterDonis

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    Sort of. ##\xi^\mu## defines a notion of "time translations", i.e., it defines a family of curves, its integral curves, along each of which the spacetime geometry is unchanged. ##\xi^\mu P^\mu## can be thought of as energy with respect to the notion of time translations defined by ##\xi^\mu##, but since the particle will not, in general, be moving along an integral curve of ##\xi^\mu##, the particle itself will not see an unchanging spacetime geometry along its worldline. This also appears as the fact that, with respect to observers who are moving along the integral curves of ##\xi^\mu## with 4-velocity ##U^\mu = \xi^\mu / \sqrt{\xi^\mu \xi_\mu}##, the particle's locally measured energy ##U_\mu P^\mu## will change. (Globally, this can be viewed as a change in the way the particle's energy at infinity is split between kinetic and potential energy.)

    No; that is just the particle's rest mass, i.e., ##P_\mu P^\mu##.
     
  9. Jan 27, 2017 #8
    Assume that a particle travels on a geodesic path ##x^\nu(\lambda)## with four-momentum ##P^\mu=m\ V^\mu##.
    The covariant derivative of ##\xi_\nu P^\nu## along the path is given by
    \begin{eqnarray*}
    \frac{D}{d\lambda}(\xi_\nu P^\nu) &=& V^\mu \nabla_\mu (\xi_\nu P^\nu) \\
    &=& m \ V^\mu \nabla_\mu (\xi_\nu V^\nu) \\
    &=& m \ \xi_\nu V^\mu \nabla_\mu V^\nu + m \ V^\mu V^\nu \nabla_\mu \xi_\nu \\
    &=& m \ V^\mu V^\nu \nabla_{(\mu}\xi_{\nu)} \\
    &=& \frac{1}{2}\ m\ V^\mu V^\nu L_\xi g_{\mu\nu} \\
    &=& 0.
    \end{eqnarray*}
    Since the path is geodesic the first term in the third line vanishes according to the geodesic equation ##V^\mu \nabla_\mu V^\nu=0##.
    Therefore a particle moving on a geodesic does seem to just see an unchanging spacetime geometry along an integral curve of ##\xi^\mu## as we are left with the Lie derivative of the metric in the ##\xi## direction, ##L_\xi g_{\mu\nu}##, which vanishes.
     
  10. Jan 27, 2017 #9

    PeterDonis

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    No, that's not what you calculated. You calculated that the energy at infinity is a constant of the motion, which is true. But that is not the same as the spacetime geometry being constant along a geodesic. To address that question, you would need to calculate the derivative along the geodesic of some invariant that describes the spacetime geometry. For example, you could try the Ricci scalar. If you try, you will see that the derivative along the geodesic of such an invariant does not vanish.

    This says that the spacetime geometry is constant along an integral curve of ##\xi^\mu##. But the geodesics you are talking about are not integral curves of ##\xi^\mu##.
     
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