Does an object falling radially in a gravitational field of a planet emit gravitational waves as seen by an observer on the surface of the planet?

What about an observer falling along with the object? Are any gravitational waves detected by them?

My question is similar to this one, but for gravitational radiation instead of electromagnetic radiation.


1 Answer 1


This is an interesting question, because the electromagnetic version has been studied extensively and has been controversial. Any arguments about the electromagnetic version should probably carry over, at least qualitatively, to the gravitational case. However, the gravitational version doesn't threaten the equivalence principle, which is presumably why you don't see it discussed as much.

It seems pretty obvious to me that an observer at infinity will see radiation. After all, we detect gravitational waves from inspiraling binary systems, and the fact that this example is radial shouldn't matter. (IIRC one way of stating the criterion for radiation is that the mass quadrupole moment has to have a nonvanishing third time derivative, which is certainly true in this thought experiment.)

For an observer comoving with one body or the other, maybe a little bit more delicacy is required.

In the discussion of the electromagnetic case, one of the fundamental issues is how to define a radiation field. It isn't an unambiguously defined thing, and this is why you can get apparently contradictory answers to the question of whether falling charges radiate -- both "yes" and "no" can be correct, depending on your definition of a radiation field.

Because the system is not a black hole binary, the motion of the two objects is at nonrelativistic speeds, and therefore the wavelength of the gravitational waves is large compared to the size of the system. This means that two observers falling along with the two objects will both be in the near-field region, and will both be essentially observing the same wave. So I guess we have to ask whether there's some unambiguous way to define what is or is not a radiation field, purely locally, in the near-field region. I'm not sure that there is.

It is certainly going to be true that each observer will see their own object accelerate. This is not a violation of the equivalence principle, because the EP only predicts universality of free-fall motion in the case of test particles. In that limit, radiation effects vanish. I think the only issue is whether those observers would describe these anomalous accelerations as being due to a gravitational wave. You could try to resolve this ambiguity by saying, OK, will LIGO detect a gravitational wave in this situation? The problem then is that LIGO is sensitive to all sorts of gravitational effects, e.g., lunar tides, that we wouldn't classify as gravitational waves.

  • $\begingroup$ Regarding your first paragraph: Is it therefore correct to say that, similarly to EM radiation, gravitational radiation is not a generally invariant concept? (That EM radiation is not an invariant concept was, in brief, the conclusion of Rohrlich's work in the 60s) $\endgroup$
    – Povel
    Aug 18, 2019 at 20:32
  • $\begingroup$ Regarding your last paragraph: By saying that each observer will see their own object accelerate you mean coordinate acceleration, right? An observer standing on one of the two bodies would detect a proper acceleration (its weight), but surely it wouldn't be able to tell that itself and the body it is standing on are falling toward the other body, because that is coordinate acceleration, which is not detected by an accelerometer, correct? $\endgroup$
    – Povel
    Aug 18, 2019 at 20:42
  • $\begingroup$ By saying that each observer will see their own object accelerate you mean coordinate acceleration, right? No, it's a proper acceleration. They're close to the body, so it's a local measurement. $\endgroup$
    – user4552
    Aug 19, 2019 at 13:25

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