I'm puzzled because the gravitational waves do warp the spacetime locally. They do it continuously as they propagate. So the photons travelling in tandem with these waves must be constantly following a curved path and hence must be travelling more distance to reach earth than 'normal' photons from the same source that travel without these disturbances. So do they take more time to reach earth?


In vacuum, that is, in absence of matter and electromagnetic interactions, photons and gravity waves follow the same geodesics (if gravitons are actually massless... Let us assume it for we have no strong evidences for the contrary). In their propagation photons are not affected by the ripples and the stretches in space-time caused by the waves, because they are travelling along with them. There are no front crossings.

You can figure this out looking at the photons as a surfer riding a wave; if he moves along with the wave, that is, he's at rest with respect to the wave, and he doesn't perform any trick or strange path, the space he covers is exactly the same as if the sea were calm. He doesn't measure the slope and the curves in the surface of the water, which would constitute and increment in the path length respect to when the sea is calm. In conclusion, neglecting the electromagnetic interaction of the photons with the interstellar medium, photons travelling in tandem with the gravity waves, and those leaving later from the same source, take exactly the same time to reach us.

EDIT: consider also that the stretching effects of gravitational waves are only transverse (at least in General Relativity) to the propagation direction of the waves. Thus, a light ray travelling in their same direction, won't be effected by any effect of dilatation/compression of the space-time. This is indeed the working principle of laser interferometric detectors: measure the interference of two orthogonal light beams. GWs propagating along the $z$ direction and their transversal effects.

  • $\begingroup$ Thanks for answering. However I do not understand both points. I see what you mean here. But I don't see how a travelling spacetime ripple is different from a bump in spacetime formed by a star. In the second case it seems to curve the spacetime. So do you mean to say that light travelling along this curved path really doesn't travel more distance? $\endgroup$ Feb 13 '16 at 12:54
  • $\begingroup$ Regarding the transverse aspect, do you mean to say that if the wave is coming from south to north, then instead of the possibility that the space is compressed in north-south direction, it is the possibility that the g.wave rotates the light beam travelling in the east-west arm to make it out of sync with the other one's polarization to produce the interference? The description given here (ligo.caltech.edu/page/what-is-interferometer) doesn't mention about this transverse polarization aspect. Did they make it too sugary for public? $\endgroup$ Feb 13 '16 at 13:01
  • $\begingroup$ Hi! As regards the first question, you see (high frequency) ripples in the space-time fabric because you are at rest with the source and the waves are travelling towards you at the speed of light. Instead, light rays are travelling in tandem with the gravity waves, so they are at rest with them and they don't see the effects of such ripples. I think that the surfer analogy could help. According to his point of view, at rest with the wave, he is on a slope (the front of the wave) and he doesn't experience the waving movement that would experience a boat bound at the harbour. $\endgroup$ Feb 13 '16 at 15:44
  • $\begingroup$ And the LIGO link you pointed to does in fact tell the same as regards the transverse deformations the waves produce. In the picture at the top of the page ('Basic schematic...') the wave is moving from top to bottom in a direction perpendicular to the plane of the detector. Imagine such plane defined by the two arms of the interferometer, and name them $x$ and $y$. Then the wave that moves towards with the $z$ direction produce ripples in the space(-time) of the ($x,y$) plane, that is the one of the detector. $\endgroup$ Feb 13 '16 at 15:52
  • 1
    $\begingroup$ Re "if gravitons are actually massless": one of the results from the LIGO detection is a very strict upper limit to the mass of the graviton ($< 10^{-22}$ eV), actually more stringent than the current limit on the mass of the photon. $\endgroup$
    – rob
    Feb 13 '16 at 16:50

You are right - it would take longer for light (and gravity and anything else) to make the trip during the event than after. But the effect is tiny, as gravitational waves shrink and expand space in virtually equal amounts - and in the linear regime - which the waves are in for virtually the whole trip, the compression and expansion is perpendicular to the direction of motion.

It is only because of the non - linear behaviour of general relativity that there is a very slight overall increase in time. I don't know what the amount of time would be, but something like 99% of the extra delay would occur in say the first million km from the event. So even if the effect was say 1% over the first million km, that would amount to only an extra (light takes 3 seconds to go a million km) 0.03 seconds. Then virtually nothing for the next billion years of travel. Those numbers are for illustration only - I'm sure someone has done the calculation somewhere.

After a million km of travel, the gravitational waves are well into the linear regime and the effect turns from a small effect into a completely ignorable tiny amount.

So 1.3 billion years for light travel without the event, and 1.3billion years + 0.03 seconds for light travelling in the event.

Please note that the 0.03 seconds is nothing more than a complete guess, I would like to see the real number.

Gravitational waves will only look like the neat linear fully transverse waves as one gets out of the intensive emission region.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.