In this wikipedia article it is described how a beam of light, with its locally constant speed, can travel "faster than light". That is to say it travels a distance, which, from a special relativistic point of view, is surprisingly big.

I wonder if a gravitational wave on such a curved spacetime (of which the wave is actually part of) behaves equally.

Does a gravitational wave also ride on expanding spacetime, just as light does? Do the nonlinearities of gravitation-gravitation interaction influence the propagation of a wave (like e.g. a plasma) such that light and gravity are effectively not equally fast?

If I want to send a fast signal in this expanding universe scenario, in what fashion do I decide to I send it?

  • $\begingroup$ Related: physics.stackexchange.com/q/12049/2451 $\endgroup$ – Qmechanic Apr 21 '12 at 0:01
  • $\begingroup$ I assume the real question you have is if there is any back-reaction. To be more specific, one might expect that in the limit of small gravitational waves on a static background, those waves would propagate with speed $c$. I can only say that in GR it does --- for details see any of the textbooks. I suppose from dimensional analysis one would expect nothing different as long as the wavelength was above Plank scale and below the radius of curvature of the underlying static background. $\endgroup$ – genneth Apr 21 '12 at 0:47
  • $\begingroup$ @genneth: Yes, that's exactly my question. (I can't even see how the question can be read otherwise, the answers explain points regarding locally constant speed and faster than light even if that's both pointed out in the first line of my post. In the question section I emphasise the gravitation-gravitation interaction and give the plasma analog.) I don't see yet how you conclude from dimensional analysis to a pure linear propagation, though. $\endgroup$ – Nikolaj-K Apr 21 '12 at 10:18
  • $\begingroup$ You don't get linear dispersion from dimensional analysis (at least I don't) --- you go to full GR and calculate the propagation of small ripples on a background --- this can be found in textbooks. The question of in what regimes this is a good approximation is/can be answered by dimensional analysis. $\endgroup$ – genneth Apr 21 '12 at 10:28
  • $\begingroup$ +++ Just as a comment, I don't consider this question answered. $\endgroup$ – Nikolaj-K Aug 8 '12 at 18:22

The speed of light is only locally invariant. For example if you watch a beam of light falling onto a black hole event horizon you'll find it slows down, and if you waited an infinite time you'd see it stop at the event horizon. Whether the speed of light is really changing depends on your point of view. I'd say it's just your choice of co-ordinates that makes it look as if it's changing, but opinions will differ.

Anyhow, the "faster than $c$" motion referred to in your link is similar in that it's a matter of geometry and the co-ordinates you're using. Since it's just geometry gravity waves will be affected in the same way as light, as indeed will everything that propagtes in space including objects with a non-zero rest mass.


It's important to stress, once again, that nothing is actually traveling faster than light in a literal sense. It's just that the universe expands, so that object that sent the signal is, measuring with today's ruler, farther away than the speed of light times the age of the universe. You can't receive signals from objects too far away, though - we can only communicate with objects inside our horizon.

All of the above reasoning also applies to gravitational wave signals, which also travel at the speed of light.

  • $\begingroup$ Yeah, I though parenthesis and the frases "That is to say" and "from a special relativistical point of view" would emphasise that enought. In any case thanks for the answer, although I'm acutally surprised that there are supposed to be no backcoupling effects due to non-linearity. $\endgroup$ – Nikolaj-K Apr 21 '12 at 0:29
  • $\begingroup$ @NickKidman the non-linearity doesn't matter because the gravitational waves that travel long distances are small perturbations to the overall metric. $\endgroup$ – kleingordon Apr 22 '12 at 6:45

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