Take the 2-minute tour ×
Astronomy Stack Exchange is a question and answer site for astronomers and astrophysicists. It's 100% free, no registration required.

Gravity is sometimes described as a curvature in space-time. Due to relativity, doesn't this imply that gravity doesn't propagate?

A thought experiment. If you are moving toward a black hole, this could also be considered as if the black hole is moving toward you. If a black hole was moving toward you at the speed of light, and gravity propagates at the speed of light, you would experience something similar to a supersonic blast, except its a gravity blast, and it would allow you to remain unaffected until you arrived at the very centre of the black hole?

If you were able to turn around, just before you arrive at the center, and travel away from the black hole at the speed of light you would never notice it, even though you would be travelling at the same velocity as the black hole in the same direction?

share|improve this question
    
Gravity is sometimes described as a curvature in space-time. Due to relativity, doesn't this imply that gravity doesn't propagate? –  Marc Jan 10 at 6:17

2 Answers 2

Gravity is sometimes described as a curvature in space-time. Due to relativity, doesn't this imply that gravity doesn't propagate?

There's a fairly precise sense in which gravity propagates: if you have a spacetime and you perturb it a bit, then you can think of the new spacetime as the old "background" spacetime with a small change on top of it. Then it makes sense to discuss the speed at which this change propagates.

But in the general case, the speed of propagation of gravity has no particular meaning. This is sensible: to talk about speed, you need some standard to compare it to, and hence a background spacetime. But gravity is nonlinear, so to have an objective answer to that question, the changes to that background need to be either small or tightly constrained.

If a black hole was moving toward you at the speed of light ... [would you] experience something similar to a supersonic blast, except its a gravity blast, ...

Not analogously to a supersonic (so here, superluminal?) one, no.

For an electric charge, the ultra-relativistic limit of its electromagnetic field is a plane wave that's impulsive, an infinitely-thin Dirac delta profile. This travels at the speed of light. The gravitational analogue of this (for a Schwarzschild black hole) is the Aichelburg-Sexl ultraboost, which is a spacetime that's axially symmetric and everywhere flat except for an impulsive plane wave.

... and it would allow you to remain unaffected until you arrived at the very centre of the black hole?

No.

If you were able to turn around, just before you arrive at the center, and travel away from the black hole at the speed of light you would never notice it, even though you would be travelling at the same velocity as the black hole in the same direction?

That's the same thing as saying you're stationary above the black hole. Hence, you would notice it because you'd need to be accelerated to stay stationary. Plus, if you're turning around after passing the horizon, it's far too late.

share|improve this answer

Black holes distort the geometry of spacetime. So you need to take care about that.

Changes of gravity propagate with the speed of light. That's thought to occur for accelerated mass, e.g. a binary of black holes.

A second fundamental point: Speeds don't add up in an additive way, but in a subadditive way: speed of light + speed of light = speed of light.

If you or the black hole travel with the speed of light, you resp. the black hole needs to be massless at rest, otherwise your relativistic mass, or that of the black hole would be infinite.

Photons are massless at rest, so you're actually talking of light, or something which behaves rather similar to light in many aspects.

So at the end you're talking of light directed to a black hole and reflected by a mirror inside the black hole.

The light can get inside the black hole, but it cannot leave the hole after being reflected from the mirror due to the high curvature of spacetime below the event horizon.

If you are travelling towards a binary of black holes, you get blue-shifted changes of gravity.

share|improve this answer
    
I don't get that gravity can change, without adding energy or mass. Adding energy or mass, implies taking it from something. I get that due to exponential gravity, binary black holes would display as a pulsating amount of gravity, almost like a wave, but not that it would be blueshifted due to doppler. Perhaps due to time dilation. How else can it create changes in amount of gravity? And subadditive implies constant velocity of 0 for everything, but that makes sort of sense if you are moving through time as well as space, but that may be not understanding subadditive. –  frodeborli Jan 9 at 18:56
    
Subadditivity: en.wikipedia.org/wiki/Subadditivity, it means that two velocity don't fully add up to the sum. Adding velocity to the speed of light doesn't increase the velocity: It remains the speed of light. –  Gerald Jan 9 at 19:36
    
The oscillation of gravitational waves is blue-shifted for an observer approching the black hole binary. This means, the frequency of the waves looks increased. –  Gerald Jan 9 at 19:40
    
Details about the relativistic addition of velocities, e.g. here: marxists.org/reference/archive/einstein/works/1910s/relative/… –  Gerald Jan 9 at 19:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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