General relativity prevents light from escaping a black hole, but does it also apply to gravitational waves?
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$\begingroup$ I would say "no" short-reasoning that nothing carrying mass/energy can overcome the gravitational forces defining a black hole. ... But then, I would not be surprised if my reasoning would miss something. $\endgroup$– Gyro GearlooseCommented Feb 27, 2016 at 22:42
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5$\begingroup$ Possible duplicate of How does gravity escape a black hole? $\endgroup$– AsherCommented Feb 27, 2016 at 22:47
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$\begingroup$ @GyroGearloose The same for me. But nor I can not support my view. $\endgroup$– dominecfCommented Feb 27, 2016 at 22:48
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$\begingroup$ @Asher I don't think it is a duplicate because the question you linked asks about a static property, the mass of the black hole. Maybe the answers over there cover some part of this question, too. They are worth reading anyway. $\endgroup$– Gyro GearlooseCommented Feb 27, 2016 at 22:56
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$\begingroup$ @GyroGearloose The top answer there points out that what we call gravity is really an effect of the local spacetime curvature, not the spacetime curvature somewhere else... though now that I think about it, waves do conceptually "generate" somewhere and "propagate" to somewhere else, so I won't contest your disagreement. Regarding your first comment, "nothing carrying mass/energy can overcome the gravitational forces defining a black hole," how would gravity effect a graviton? (Rhetorically; I doubt we have any answer for that.) $\endgroup$– AsherCommented Feb 27, 2016 at 23:03
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3 Answers
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Even if all the mass behind the horizon would magically disappear in one moment, you would not notice that from outside the horizon.
One could assume that in this case there was no mass left that could bend spacetime anymore and the gravitational field would disappear with c, but on the other side you have to take time dilation into account:
From the perspective of the outside observer everything that makes up the black hole is stacked up at the horizon and asymptotically approaches it as time goes to infinity, simply because the factor for the time dilation approaches 0 as an object approaches the horizon.
Therefore, what happens inside a black hole at a given proper time of an infalling observer does not even have a corresponding coordinate time on the outside of the black hole (mathematically an imaginary one, but technically after infinity), because from that perspective it takes an infinite amount of time to even get near the horizon, not to mention behind it.
If from our perspective there is nothing behind the horizon there is nothing that could create gravitational waves behind the horizon. In other words: whatever happens behind the horizon, outside the horizon it has not happened yet.
So the answear would be no, you can not send gravitational waves from the inside of a black hole to the outside, simply because you haven't even yet been falling though the horizon in the system of an outside observer. From his perspective you are always outside the black hole and never go through the horizon until infinity.
Leonard Susskind explains this here and here, and John Rennie mentions it in this post.
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Any gravitational waves emitted inside the event horizon fail to make out of the event horizon because they travel at lightspeed. And a lightspeed signal from inside stays inside.
Therefore, any waves emitted inside the event horizon are never observed on the outside of the event horizon.
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If a gravitational wave can be created inside event horizon, I do not see any reason why it would not escape from it. It may not be possible to create a GW inside event horizon, but that is a different question.
When we say nothing can escape event horizon, the reason for that is enormous gravity beyond event horizon. This argument likely will not apply to gravity which is cause for the argument itself.
Remember, gravity causes black hole, not the other way. Because, gravity was there before black hole had formed, not that gravity appeared after black hole formed. Gravity controls black hole, not that black hole controls gravity.
GR itself predicts GW, not sure if it specifies location of origin of it.
Also, if a GW could not escape EH, and suppose, it can enter EH, then that would imply a GW can be absorbed by a BH. If it can not even enter, then that would imply GW can be reflected off EH. These would be follow up questions.
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$\begingroup$ A gravitational wave can't escape an event horizon because a GW travels at lightspeed and the inside if an event horizon is (by definition) causally disconnected from the outside in the sense that the past light cone of every event on the outside fails to intersect the inside at all. $\endgroup$– TimaeusCommented Feb 28, 2016 at 1:24
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$\begingroup$ @Timaeus : the cause/reason/basis for that definition is gravity itself. I do not think gravity is going to restrict itself (or its own wave) from escaping. $\endgroup$– kpvCommented Feb 28, 2016 at 1:35
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2$\begingroup$ I have no idea what you are saying. An event can be outside the past light cone of another event even without gravity. You can have a family of observers (such as hyperbolic observers in SR) that have a horizon such that the past light cone of every event of the observers fails to intersect some region of spacetime. We literally defined the event horizon in terms of being outside the past light cones of a family of observers. It's merely the definition. Like saying some integers are odd when they fail to be two times an integer. It is merely a definition. It is not deep at all. $\endgroup$– TimaeusCommented Feb 28, 2016 at 1:41
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$\begingroup$ @Timaeus: Yes, I understand it. What I am saying is because the definition itself is in terms of 2, by definition (without testing) 2 is even . You may be missing the basic reason for all the definitions regarding EH. That reason is none other than gravity as far as I know. $\endgroup$– kpvCommented Feb 28, 2016 at 1:47
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2$\begingroup$ No, someone could hand you a manifold and a metric and define what a past light cone is then they could label a family of observers and ask you for the event horizon for that family. You could compute the past light cone of every event for the family of observers and then take the union of that set, and then take the boundary, and that's the event horizon. You could do that without studying GR or knowing what gravity is. It's a function of the metric and the manifold and the definition. Purely a function of that. All the Einstein Equation does is tell the metric how to evolve. $\endgroup$– TimaeusCommented Feb 28, 2016 at 1:52