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Does object lose kinetic energy due to emitting gravitational waves?

  1. Dec 16, 2014 #1
    I always thought when there is no force acting on an object it moves at constant speed but every object that moves generates gravitational waves what causes that object to lose energy. Does it mean that object loses kinetic energy and slows down and after some (very long) time will stop.
     
  2. jcsd
  3. Dec 16, 2014 #2

    PAllen

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    Basically, yes. In GR, all orbits are, in principle, unstable and inspiral occurs until merger. Except for massive, compact objects orbiting closely, the effect is too small and too slow to ever be observed, in practice. Ultimately, this is similar to the classical EM prediction that co-orbiting opposite charges will spiral together emitting EM radiation, except that for GR the power of the radiation is miniscule and the time scale of inspiral is huge (for ordinary stars and planets).
     
  4. Dec 16, 2014 #3

    Vanadium 50

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    At constant velocity an object will not emit gravitational waves. There will be a frame where it is at rest, and it obviously cannot lose energy in that frame. It needs to be accelerated (such as in an orbit, where the speed is constant, but the velocity is not)
     
  5. Dec 16, 2014 #4
    There are still 2 things I don't understand about it
    1.Does not moving object lose energy due to gravitation field it creates?
    2.What if gravitation force caused by gravitation field is the force that causes object to move and thus lose energy? If first statement is false wouldn't it mean the energy "is created" which is considered as impossible?
     
  6. Dec 16, 2014 #5

    A.T.

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    No

    With two objects the system has potential energy, which is converted into kinetic energy.
     
  7. Dec 16, 2014 #6
    but when object 1 creates gravitational field and don't lose any energy to does it what causes object 2 constant acceleration gaining kinetic energy and vice versa isn't it like "gravitation constantly creates kinetic energy" ?
     
  8. Dec 16, 2014 #7

    A.T.

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    No, see post #5.
     
  9. Dec 16, 2014 #8

    PeterDonis

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    Actually, even that isn't enough; if the acceleration is constant (here we mean coordinate acceleration, since it is present for an object in a free-fall orbit), as for example in a perfectly circular orbit, no gravitational waves are emitted. The acceleration has to be changing with time (which for any real orbit it will be, since no real orbit will be perfectly circular).
     
  10. Dec 16, 2014 #9

    PAllen

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    Isn't the direction of coordinate acceleration changing for a circular orbit? So you are claiming that doesn't matter? Note that the quadrupole moment is changing in time for a circular orbit, and has nonzero second derivative (at least in cartesion coordinates). Certainly the case of two equal masses co-orbiting in a circle is a canonic example of GW production. If, in the limit, of infinite mass ratio there is no GW, this seems rather trivial because, in some sense, the second body doesn't really exist.
     
  11. Dec 16, 2014 #10

    PeterDonis

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    Not for GW emission, no.

    IIRC, the GW emission depends on the third time derivative of the quadrupole moment, not the second. The third derivative is zero for a circular orbit (of a test body--the mass ratio issue you bring up is a valid point).
     
    Last edited: Dec 16, 2014
  12. Dec 16, 2014 #11

    PAllen

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    See:

    http://preposterousuniverse.com/grnotes/grnotes-six.pdf [Broken]
    p. 158 on

    The formula is for second derivative, not third. Further, exact circular orbit is discussed on p. 159 as producing GW. The test body case is s a non-sequiter not because 'exact symmetry' can't be achieved but because exact test body for finite larger mass, means the test body doesn't exist.
     
    Last edited by a moderator: May 7, 2017
  13. Dec 16, 2014 #12

    PeterDonis

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    The formula for the metric perturbation ##h_{\mu \nu}## depends on the second derivative. But the formula for radiated power, which is what we've been discussing, depends on the third derivative. See equation 6.105.

    Regarding the circular orbit, yes, you are correct; I was mis-remembering how the third time derivative of the quadrupole moment behaves. I agree now that any two bodies with finite mass orbiting each other will have a nonzero third time derivative of their quadrupole moment, even if both orbits are exactly circular. Sorry for the confusion on my part.
     
  14. Dec 16, 2014 #13

    Vanadium 50

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    That's true. I felt that the OP's difficulty was probably not that the problem wasn't complicated enough, though. If you feel that adding more levels of complication will help, more power to you.
     
  15. Dec 16, 2014 #14

    PAllen

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    I've always seen the first formula as the one that defines the 'existence' of GW (the formula Carroll give goes back to Einstein (1916)!). Carroll says:

    "The gravitational wave produced by an isolated nonrelativistic object is therefore proportional
    to the second derivative of the quadrupole moment of the energy density"

    However, I would agree that a perturbation carrying no energy is not 'real radiation', and for that you need 3d derivative to be nonzero.
     
  16. Dec 16, 2014 #15

    PeterDonis

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    Touche. :eek:
     
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