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Is energy of gravitational waves negative?

  1. Jan 10, 2014 #1

    Demystifier

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    Owing to the Hamiltonian constraint in general relativity, the total energy is zero. (To avoid ambiguity, here energy is DEFINED as the corresponding part of the ADM Hamiltonian.) Since matter has positive energy, it implies that gravitational field has negative energy.

    Does it mean that gravitational waves also have negative ADM energy? If no, then why not? If yes, then does it mean that detection of gravitational waves should manifest as a REDUCTION of the detector energy?
     
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  3. Jan 10, 2014 #2

    Simon Bridge

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  4. Jan 10, 2014 #3

    PeterDonis

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    But this is not the same as what is usually referred to as the "ADM energy" of an asymptotically flat spacetime (I realize that the ADM Hamiltonian itself is not limited to such spacetimes). The ADM energy of Schwarzschild spacetime, for example, is equal to the mass ##M## that appears in the metric. See further comments below.

    For particular interpretations of those terms, yes. I can think of two:

    (1) The one you give, which, for example, I have seen used to describe the universe as a whole: the matter in the universe has positive energy which is exactly balanced by the negative energy in the gravitational field, for a total energy of zero. This works because the universe is a closed system, so the Hamiltonian constraint has an obvious physical application.

    (2) If you evaluate the Komar mass of an asymptotically flat spacetime, such as the exterior + interior Schwarzschild solution describing a static star, you find that it is *smaller* than what you would get if you just summed up the stress-energy in the obvious way. The difference is standardly described as the system having negative gravitational binding energy; the process of "assembling" the system from masses at infinity releases energy, so that the final system has less total energy than the initial system did. (If we wanted to be precise, we could quantify this as the difference between the ADM energy and the Bondi energy of the spacetime as a whole, similar to what we do with gravitational wave energy--see below.)

    However, there are other possible interpretations. See below.

    In an asymptotically flat spacetime, the energy emitted by gravitational waves can be defined as the difference between the ADM energy (using the definition I referred to above, *not* the "ADM Hamiltonian" definition) and the Bondi energy. (This actually works for any type of radiation that escapes to infinity; see above.) For matter satisfying the usual energy conditions, there are theorems showing that ##E_{ADM} >= E_{Bondi} >= 0## for any asymptotically flat spacetime. This means that all three energies--the ADM energy, the Bondi energy, and the energy emitted by gravitational waves--are nonnegative. (And, of course, if we predict what will happen if the gravitational waves, if present, are absorbed by a detector, we predict that the detector's energy will increase, not decrease.)

    So IMO it comes down to how you interpret the term "energy". There is no one unique interpretation of that term.
     
  5. Jan 10, 2014 #4

    PAllen

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    Since this is discussed in relation to SC interior + exterior, I think it is incorrect. There is no radiation reaching infinity for SC exterior. Therefore Bondi energy = ADM energy.
     
  6. Jan 10, 2014 #5

    PeterDonis

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    Yes, you're right: a perfectly spherically symmetric collapse, with a Schwarzschild vacuum exterior, can't emit any radiation. But in such a case the "gravitational binding energy" would be zero, which isn't physically realistic; the system wouldn't actually be bound.

    What I was actually envisioning (which I should have been clearer about) was a collapse that *can* emit radiation, so that it starts out with a *non* vacuum exterior, containing outgoing radiation, as well as matter collapsing from "infinity" (or at least some very large radius) to a stable bound gravitating mass which ends up spherically symmetric. We can suppose, for purposes of illustration, that this radiation can be modeled as a region of outgoing spherically symmetric null dust (similar to the outgoing Vaidya metric), although obviously this is still highly idealized.

    Once the collapse is complete, the central object stops emitting radiation, so there will be some "last outgoing null surface" to the future of which the exterior of the central object is the Schwarzschild vacuum solution. The only difference from the "standard" SC exterior + interior model would be that at some finite but very large (and increasing) value of ##r##, the exterior would not be vacuum; there would be a spherically symmetric, expanding region of null dust. (To clarify, I don't envision this being gravitational radiation; I envision it being EM radiation emitted during the collapse process.) Because it is spherically symmetric, it would not affect the metric inside.

    This idealized spacetime would have an ADM and Bondi mass that were not the same; the ADM mass would be the initial mass of the system, and the Bondi mass (at least when evaluated far enough along future null infinity, to the future of the last outgoing null surface of radiation) would be the final mass of the central object, minus the energy in the emitted radiation. (If we evaluated the Bondi mass in the region of future null infinity where the radiation was arriving, it would be a smooth function of the outgoing null coordinate ##u## along future null infinity, decreasing from the ADM mass to the final value of the Bondi mass.)
     
  7. Jan 10, 2014 #6

    PAllen

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    Agreed (with respect to Peter's last post).

    Of course, I think Komar mass (as you referenced earlier) shows binding energy much more directly. Locally measured mass of each piece (in a static, bound system) contributes less to Komar mass the more total Komar mass there is. This looks at the static picture after 'everything else' has happened.
     
  8. Jan 10, 2014 #7

    PeterDonis

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    Yes, agreed. For a SC interior + exterior spacetime with nothing else present, Komar mass = ADM mass, both of which are less than the "naive" integral of the SET over a spacelike slice, indicating negative gravitational binding energy. And for the spacetime with outgoing radiation present that I described, Komar mass = ADM mass before the collapse starts, and Komar mass = Bondi mass after the collapse has finished and all radiation has been emitted (note that in this case we only do the Komar integral over a compact region enclosing the object, excluding the radiation), and in both cases the mass is less than the "naive" integral of the SET over the object.
     
  9. Jan 13, 2014 #8

    Demystifier

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    Thank you all for your answers. However, as neither of the answers completely satisfied me, I have found my own answer. To get it, I have used Weinberg's "Gravitation and Cosmology", Secs. 7.6 and 10.3. Let me summarize my findings.

    To simplify notation, let me use units in which the gravitational constant is 1/8 pi. Then Einstein equation can be written as
    G=-T
    where G is the Einstein tensor, T is the matter energy-momentum tensor, and the tensor indices are suppressed. The 00-component of this can be written as the Hamiltonian constraint
    H=G_00+T_00=0
    Here T_00 cannot be negative, so G_00 can be thought of as gravitational energy which cannot be positive.

    However, when one considers gravitational waves, it is common first to make a decomposition of the metric
    g=eta+h
    where eta is the flat (Minkowski) background and h is the "gravitational field", which is usually small compared to eta. When G is expanded in the powers of h, it turns out that the terms linear in h do not vanish, i.e.
    G=G_1+G_2+...
    where G_1 is linear in h, G_2 is quadratic in h, etc. On the other hand, in analogy with other field theories in Minkowski spacetime, one wants to have energy-momentum of the "field" h which does not have terms linear in h. Therefore, one defines the energy-momentum of gravity not as G, but as
    t=G-G_1
    where the term linear in h is subtracted. In this way, even though G_00 cannot be positive, it turns out that energy t_00 is positive.

    Hence, the energy of the gravitational wave is positive because it is defined such that the dominant negative contribution linear in h is subtracted from it.
     
    Last edited: Jan 13, 2014
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