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- Thread starter bgfnfgh
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In summary: No, the third derivative is zero for a circular orbit (of a test body--the mass ratio issue you bring up is a valid point).

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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).bgfnfgh said:

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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?

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Nobgfnfgh said:1.Does not moving object lose energy due to gravitation field it creates?

With two objects the system has potential energy, which is converted into kinetic energy.bgfnfgh said: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?

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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" ?A.T. said:With two objects the system has potential energy, which is converted into kinetic energy.

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No, see post #5.bgfnfgh said:sn't it like "gravitation constantly creates kinetic energy" ?

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Vanadium 50 said:It needs to be accelerated

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).

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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.PeterDonis said: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).

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PAllen said:Isn't the direction of coordinate acceleration changing for a circular orbit? So you are claiming that doesn't matter?

Not for GW emission, no.

PAllen said:Note that the quadrupole moment is changing in time for a circular orbit, and has nonzero second derivative (at least in cartesion coordinates).

IIRC, the GW emission depends on the

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See:PeterDonis said:Not for GW emission, no.

IIRC, the GW emission depends on thethirdtime 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).

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.

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PAllen said:The formula is for second derivative, not third.

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.

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PeterDonis said: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).

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.

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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:PeterDonis said: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.

"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.

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Vanadium 50 said:If you feel that adding more levels of complication will help, more power to you.

Touche.

Gravitational waves are ripples in the fabric of space-time that are produced when two massive objects, such as black holes or neutron stars, orbit each other. As these objects move, they emit energy in the form of gravitational waves, which can cause a loss of kinetic energy in the system.

No, the loss of kinetic energy due to gravitational waves is a gradual process and depends on the mass and velocity of the objects involved. In most cases, the objects will continue to have some kinetic energy even after emitting gravitational waves.

The emission of gravitational waves causes a decrease in the kinetic energy of an object, which in turn can lead to a change in its velocity and orbit. This effect is most noticeable in systems with extremely massive or rapidly moving objects.

No, the emission of gravitational waves and the resulting loss of kinetic energy are only significant in extremely massive and energetic systems, such as those involving black holes or neutron stars. In everyday situations, the effects of gravitational waves are too small to be measured.

At the moment, it is not possible to harness the energy from gravitational waves as a source of power. The amount of energy emitted in the form of gravitational waves is extremely small and difficult to detect, making it impractical for use as an energy source. However, research is ongoing to develop technologies that may one day make it possible to capture and use this energy.

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