gravitational waves are strictly transversal (in the linear regime at least), also their amplitudes are tiny even for cosmic scale events like supernovas or binary black holes (at least far away, maybe we should ask some physicists located a bit closer to the center of the galaxy), but lets put all those facts aside for a second and consider a gravitational source big enough to generate gravitational waves with amplitudes of the order of the galaxy. For instance consider a planar wave like in my mediocre drawing:

$$ h_{\alpha \beta} e^{i (k_{y} y - \omega t)} $$


$$ h_{\alpha \beta} \approx 1 $$

so the perturbation is in the nonlinear regime

gravitational wave reducing distances between stars

i draw two far away objects in three different time slices (this is why they are repeated 3 times), the topmost is the objects without the gravitational wave, the one in the middle represents the objects in the crest of the gravitational wave, and the one in the bottom represents the objects in the valley of the wave.

So, my point is that people would only have to travel an arbitrarily small distance when the wave is on the valley (assuming circular polarization) even if the "normal" distance (i.e: $h_{\mu \nu} = 0$) is several light-years away

Besides being slightly impractical to set up such a mammoth gravitational source, this kind of warp drive is valid from a physical standpoint? Are there any physical limits to gravitational wave amplitudes in such nonlinear regime?

  • 1
    $\begingroup$ @lursher I'm guessing a gravitational field strong enough to shrink a distance by a significant proportion of its beginning value would be stupendously nonlinear: much stronger than describable by weak field Einstein equations. I'm very rusty on GR: I'm guessing that there are nonlinear wave solutions or approximations or known wave behaviours from numerical simulations - and that you are invoking these nonlinear behaviours is this right? $\endgroup$ – WetSavannaAnimal Aug 21 '13 at 1:46
  • $\begingroup$ And how would the would be spacefarer feel in such a strong wave? Could the metric be such that the locally flat regions were big enough to include all of the points in his-her body - so that spaghettifying stresses were not set up? $\endgroup$ – WetSavannaAnimal Aug 21 '13 at 1:50
  • 2
    $\begingroup$ @WetSavannaAnimalakaRodVance: FWIW: you have the plane wave spacetime, which is an exact solution to the vacuum einstein equation: arxiv.org/abs/1203.6173 $\endgroup$ – Jerry Schirmer Sep 19 '13 at 23:28
  • 1
    $\begingroup$ And if you read the linked arxiv article, the authors report a closed null curve in a particualr class of plane wave spacetime. So I'm inclined to say that the answer to lurscher's quesiton is "probably yes", but then you start have to ask questions about generating sufficiently large gravitational waves with the correct waveform using matter satisfying the usual cosmic censorship-style assumptions. $\endgroup$ – Jerry Schirmer Sep 19 '13 at 23:30

I don't think you could use this as a warp drive unless you could collimate the gravity waves. If you consider a spaceship moving at constant velocity through a gravity wave, the ship will be accelerated then decelerated again as the wave passed through but it's average velocity would be unchanged. The only way you could get a net effect from the wave is if you could move from a region of high amplitude to low amplitude within half a cycle of the wave. I can't think of any (plausible) geometry that would allow this. Possibly you could do it very close to a black hole binary, where the gravity wave generation doesn't look like a point source.

  • 2
    $\begingroup$ the gravity wave does not produce any "acceleration" in test particles in the traditional sense of changing net momentum, it is just an oscillation in the metric, so the distance between the far-away objects grows and shrink in a single period by an amount proportional to the wave amplitude. So all objects at every point in the gravitational oscillation are always in free fall. $\endgroup$ – lurscher Jun 22 '12 at 13:06
  • 2
    $\begingroup$ regarding physical plausibility, i agree, it is usually hard to come by with sources of planar waves, and the fact that its gravitational radiation we are talking about does not make it any more realistic. $\endgroup$ – lurscher Jun 22 '12 at 13:20

Here is a paper called Creating spacetime shortcuts with gravitational waveforms

That seems to be close to what you are talking about, but with smaller effects and 'smaller' gravitational wave amplitudes.

What you do is fly your ship so that it only flies through the shrinking part of gravitational waves - so your trip takes place only in the compacted part of some waves which are travelling in a transverse direction to your chosen direction of travel.

So its kinda a big technical stretch - but I think that its actually less audacious than your scheme - it also shows that linear waves will do the kind of effect you are looking for.

As for physical limits on the size of the wave amplitudes you are talking about, remember that the gravitational wave observation of GW150914 had amplitudes close to 1 (well say a tenth) in a region 200 km across or so, and it radiated power at a peak rate of 200 Solar masses per second. So a galaxy sized radiator with amplitudes near 1 would need something like 2 back holes, each the mass of 10 million spiral galaxies merging to get to that power. (I did that 10 million mass calculation rather hastily, feel free to check it. Just make a black hole the size of the Milky Way and see how many solar masses it would take).

  • $\begingroup$ yeah, the required amplitudes are inmense. A back-of-the-napkin estimate suggests at least a million solar masses must be turned into phase-locked gravitational sources, spinning at 3 km/sec, to have a beam of the required amplitude $\endgroup$ – lurscher Mar 26 '16 at 1:07

protected by Community Mar 23 '16 at 21:00

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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