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Does the producing process of gravitational waves produce gravitation?

  1. Jan 2, 2012 #1
    Gravitational waves are generated when the mass quadrupole moment changes in time.
    We also know motion of mass contributes to its gravitation. Does the producing process of gravitational waves, which involves mass in accelerated motion, produce gravitation as well? If so, is it of less, equal or more magnitude than the gravitational waves being generated?
     
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  3. Jan 2, 2012 #2

    Mentz114

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    I think it must. For instance a rotating asymmetrical mass would radiate, and the energy flux and momentum would contribute to the gravitational field through the energy-momentum tensor.

    The energy of the GWs would be a tiny fraction of the rest + kinetic energies involved. I can't back this up rigorously, but we'd be detecting GWs if it were otherwise.
     
  4. Jan 2, 2012 #3

    bcrowell

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    Sure. Anything that makes a non-oscillatory contribution to the stress-energy tensor is going to contribute to the Newtonian far field.

    It's less simply because radiation fields fall off more slowly than static fields. It's the same idea as in E&M. In E&M, the electric field of a point charge falls off as 1/r2, but the far field of a spherical wave falls off as 1/r (because the energy density goes like 1/r2, and the field goes like the square root of the energy density). It's the same idea with gravity, but with different exponents because the field equations have one more derivative. The static Newtonian field (Petrov type D) has a curvature that falls off like 1/r3, while the distant radiation field (Petrov type N) has curvature like 1/r.

    This is a different question than what I interpret the OP as having asked. The energy of the waves builds up over time as more and more energy heads out into space. In a decaying system like the Hulse-Taylor pulsar, the energy in the waves will eventually be comparable to the initial KE+PE of the system, since it's basically radiating away what it originally had. If the final state is a black hole, I don't know what fraction of the initial rest mass is radiated away. Maybe it depends on how close the final black hole is to the limiting angular momentum?
     
    Last edited: Jan 2, 2012
  5. Jan 3, 2012 #4

    tom.stoer

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    I think it helps to look at the electromagnetic field first.

    Starting with Maxwell's equation it's unclear where the static Coulomb field with its 1/r behavior comes from. Usually one looks at the electrostatic case and solves the Poisson equation for a delta-charge and then for a general charge density as an integral over delta charges. One can show that Maxwell's theory allowes for a formulation where this static Coulomb field contributes (as one piece of the general interaction involving currents, el.-mag. fields etc.) even in the general case (i.e. w/o referring to electrostatics).

    Even if GR is much more complicated there is a "weak field limit" where such a reformulation is again possible.

    This explains the following "paradoxical situation": a black whole does not allow for any dynamical entity to come out of the event horizon. Now how does gravity (or curvature) carried by gravitational waves comes out of the whole??? The answer is that it doesn't!!! The gravitational field outside the event horizon is not 'produced' by the black hole but it's the static relic (1/r in a certain approximation) which was already there before the star collapsed to black hole).
     
  6. Jan 3, 2012 #5

    timmdeeg

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    One can't expect graviational waves anyway, as there is no mass quadrupole moment variation in time.
    As to the static relic notion, yes that's an usual interpretation. It might also not be wrong to say that as long as the black hole's mass is part of the universe, it must show it's gravitational field.
     
  7. Jan 3, 2012 #6

    tom.stoer

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    You are right; in the same sense electromagnetic waves can't explain the static Coulomb potential b/c there is no monopole radiation which would violate the conservation of electric charge.

    Again I fully agree; the main question I wanted to answer was "how gravity can exist even w/o gravitational waves"; perhaps this was a bit off-topic.
     
  8. Jan 3, 2012 #7

    timmdeeg

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    According to John Baez's General Relativity Tutorial there is Weyl curvature (no Ricci curvature) due to gravitational waves propagating in otherwise flat spacetime. I wonder what curvature means in the context of the Weyl tensor, besides being responsible for tidal forces. To my understanding there is no curvature in the x-y plane in a sense of expecting e.g. shapiro delay. Any comment to this is appreciated.
     
  9. Jan 3, 2012 #8

    Mentz114

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    The closest thing we've got to a gravitational wave solution is the pp-wave metric, which is written in Brinkmann coordinates (u,v,y,z). The Ricci curvature is zero so all the curvature is Weyl. The most interesting thing for me is the tidal tensor which ( in the y-z directions) is

    [tex]\pmatrix{ A & B\cr B & -A}[/tex]

    where A and B are functions of u only. It seems the GWs cause a tidal effect which is opposite in the y and z directions. If B is non-zero there is polarization.

    This 'solution' is a purely mathematical construct but certainly looks like transverse wave of Weyl curvature propagating at c.
     
  10. Jan 4, 2012 #9

    timmdeeg

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    Ok, supposed there is Weyl curvature only, then as shown in Figure 1.1 http://books.google.de/books?id=W-J5TUBEy-AC&lpg=PP1&dq=gravitational%20waves&pg=PA11#v=onepage&q&f=false [Broken] the latter oscillates between negative and positive values.

    But in which plane do the light ray triangles actually show spacetime curvature? In the plane of propagation (a) as shown in the Fig. or in the transverse plane (b), what can't be shown, but could be meant?

    Thinking of the phrase 'plane wave solution' it's hard to imagine (b).

    Any comments are very welcome.
     
    Last edited by a moderator: May 5, 2017
  11. Jan 4, 2012 #10

    Mentz114

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    Fig. 1.2 shows the plane perpendicular to the motion. If you drop an L-shaped interferometer into the circle, then presumably the arms will change in length.
     
  12. Jan 5, 2012 #11

    timmdeeg

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    Agreed, masses in the transverse plane of the gravitational wave are accelerated, like the mirrors in an interferometer.

    But the question regards the curvature. Let us consider triangle measurements performed by lightrays in the tranverse plane. The time needed for that is very short compared to the period of the wave, which anyway can be thought arbitralily long. By doing this repeatedly (like snapshots) one obtains the sum of the angles and thus the curvature as a function of the phase of the wave.

    What would be the result? Would the curvature oscillate between concave and convex as Fig. 1.1 shows? But how then would this fit to the fact that the Ricci curvature vanishes according to John Baez?
     
  13. Jan 5, 2012 #12

    Mentz114

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    Yes, I think it will.


    It fits. There is no Ricci curvature associated with GWs. Ricci curvature is caused by the EMT, Weyl curvature isn't.

    Remember that the Riemann curvature tensor may be decomposed into Ricci and Weyl components.
     
  14. Jan 6, 2012 #13

    timmdeeg

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    Ok, thank you. So, supposed there is infact spacetime curvature in the transverse plane of the gravitational wave, would you agree, that the laser interferometer measurement should be obscured by effects like Shapiro-delay or gravitational red/blue shifting?

    Of course such effects which light experiences by traveling through curved spacetime are tiny, the displacement of the mirrors however too. Unfortunately I was unable to find anything regarding this matter in the web.
     
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