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General relativity prevents light from escaping a black hole, but does it also apply to gravitational waves?

Qmechanic
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jhourback
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  • 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. – Gyro Gearloose Feb 27 '16 at 22:42
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    Possible duplicate of How does gravity escape a black hole? – Asher Feb 27 '16 at 22:47
  • @GyroGearloose The same for me. But nor I can not support my view. – dominecf Feb 27 '16 at 22:48
  • @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. – Gyro Gearloose Feb 27 '16 at 22:56
  • @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.) – Asher Feb 27 '16 at 23:03
  • @Asher this exactly what I think (but don't know for sure). My heuristic tells me that there is a conceptual difference between static gravity and change of gravity ... but I'm far from having even a heuristic concept how change ( as a function of time) could be embedded in space-time, where time is an integral part and not an available parameter. – Gyro Gearloose Feb 27 '16 at 23:09
  • @GyroGearloose Curvature evolves according to the Einstein Equation. Just as electromagnetic fields evolve according to the Maxwell Equation. You don't have to label things as this kind of field or that kind of field, you just put the fields in and solve for the evolution. – Timaeus Feb 28 '16 at 01:26

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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.

Community
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Yukterez
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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.

Timaeus
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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.

kpv
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  • 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. – Timaeus Feb 28 '16 at 01:24
  • @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. – kpv Feb 28 '16 at 01:35
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    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. – Timaeus Feb 28 '16 at 01:41
  • @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. – kpv Feb 28 '16 at 01:47
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    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. – Timaeus Feb 28 '16 at 01:52
  • You could use the Einstein Equation actively to solve a Cauchy problem. Or you could use it more passively to take a whole manifold and then give a thumbs up or a thumbs down about whether it satisfies the Einstein Equation or not. The manifold and the metric and the family of observers make the manifold. It does stop and ask something what it is before declaring an event to be in the last light cone or not. It doesn't tell different waves that they are or aren't in the past light cone depending on what thing is at that event. The event itself (the point on the manifold) is in the cone or isnt – Timaeus Feb 28 '16 at 01:53
  • I agree that it may not be possible for the GW to be generated inside EH. But we do not know what goes on in there. And by that logic, the question itself may becomes meaningless because we would never know whether the GW was generated inside, or outside EH. Another example - speed of light slows down near EH, If everything that applies to light cone, also applied to gravity, then speed of gravity would also slow down near EH. Which I think is not true. By that logic, gravity (like light) would not be able to make out of EH making EH itself ineffective. – kpv Feb 28 '16 at 02:05
  • If it was generated outside, then by definition we would know. Obviously you haven't bothered to read any questions and answers about how gravity works. As a ball of gas collapses it leaves the spacetime outside the new location of the gas in a curved state, and things react to that curvature. It doesn't broadcast what is going on inside. You really need to learn the basics of relativity. The metric evolves according to the Einstein Equation, that's what is going on and you have manifolds with stress-energy tensors and metric tensors. – Timaeus Feb 28 '16 at 02:11
  • OK, in terms of tensors, metric, and manifolds - aren't the gravitational waves, ripples in the tensor/metric etc., or ripples in their evolution? I am calling the evolved combination of all these as gravity (or curvature). If that evolved combo ripples inside event horizon, it would not restrict its own ripple from escaping. It can restrict everything else, but not itself. If still does not make sense, one of us must be missing something. – kpv Feb 28 '16 at 02:47
  • If the final evolution inside EH is not defined, or known, then its ripple remains undefined as well, implying that creation of GW inside EH is not defined. GW are ripple in space (or its curvature). Is that curvature defined inside EH? – kpv Feb 28 '16 at 02:55
  • One of us is missing something. You are failing to understand how GR works. There is nothing in the equations that label things and treat them differently. If you prefer to to define an event horizon as a surface that nothing escapes (even gravitational waves) then you'll finally grasp this is just a consequence of a definition. You'll get the spirit of the correct answer. Then you'll just have to learn how gravity works. Which is that the metric evolves according to the Einstein Equation like how the electromagnetic field evolves according to the Maxwell Equation. Study the SR example – Timaeus Feb 28 '16 at 03:35
  • I do not fully understand GR math, but I understand the concept enough to answer this question. Let me try one last time. GR works based upon certain properties of empty space which may include tensors, metrics etc. etc. Value of these properties and amount of mass, decide where the EH is. Space itself is nothing other than these properties. Therefore, a ripple in space actually, has to be a ripple in the value of these properties. This ripple in values of the properties will alter the EH itself. I again say, if GW can be created inside EH, then they can also escape it. – kpv Feb 28 '16 at 06:06
  • You are just wrong, seriously wrong on many many levels. Curvature isn't caused by mass. The event horizon is determined by the full manifold and its global properties and people doing simulations actually sometimes can't even find it accurately because of its dependence on global structure. The EH is not I repeat not anything other than the boundary between things that can affect the outside and things that never will. It is not some special distance from some mass. You do not understand enough to answer this question correctly. You need to learn the definition of an EH. – Timaeus Feb 28 '16 at 06:57
  • @Timaeus: Adding to last comments - the GW (space ripple) that passes by, how does it carry energy? It does so in the form of changed values of space properties, which return to their normal/natural value after the wave has passed. This ripple in value of the properties is not going to be stopped by anything because it ripples the behavior of everything on its way including black holes. If you substitute different values of these properties in GR, GR will give different results. These different results can ripple through black hole without violating the equations. – kpv Feb 28 '16 at 07:00
  • The EH is actually defined as the boundary between things that can get out and things that can't. Read a proper definition. Anything that gets out must by definition have originated outside the event horizon. This isn't deep or controversial in the slightest, it's the stupidest tautology ever. Since we defined the event horizon as the boundary between things that can get out and things that can't the boundary depends on the entire global structure (future, past, the whole thing) and by definition nothing ever ever ever ever ever ever ever ever gets out. It's literally that simple. – Timaeus Feb 28 '16 at 07:05
  • And, you are just making stuff up. That's literally offensive. – Timaeus Feb 28 '16 at 07:06
  • Yes, I made-up the energy part. My apologies on that. Is there an explanation on how (in what form) they carry energy, please point me to a resource. As I understand, space is set of certain property values, nothing else and to carry energy, only these properties/values can be used, because, there is nothing else available to use. – kpv Feb 28 '16 at 07:11
  • Gravity waves are ripples in spacetime, so don't forget the time component of spacetime - time stands still at the horizon, and if the ripples are produced inside the horizon the time outside the black hole has already passed infinity. So to be observed from outside the black hole the ripples would need to travel not only infinitely far back in time, but also on a perpendicular time axis, which they don't. – Yukterez Feb 28 '16 at 07:15
  • May be not a good analogy, but time stands still for a photon as well, but it still reaches us. – kpv Feb 28 '16 at 07:19
  • A photon has no proper time. That does not mean zero, it simply has none, not even zero. It also has no rest frame. You also must distinguish between relative time dilation due motion through space and absolute time dilation due the metric of spacetime when you have gravity. If a photon gets created at the horizon it never reaches you, because it does not even have time to swing on it's frequency so technically it is not even a photon, it does not exist. – Yukterez Feb 28 '16 at 07:20
  • @kpv The event horizon is defined as the boundary between events that can reach us and those that can't. So if there were some waves that reached us, we define the event horizon to be located somewhere where they don't have to cross it. If something crossed it has to cross in since we define the EH to exclude the things that never get to us. This is a semantic argument where you simply don't know the proper definitions. There isn't even any physics in this conversation. Just you not knowing the definitions of the words you use and saying things that are completely wrong – Timaeus Feb 28 '16 at 07:24
  • @СимонТыран: May I say, gravitational time dilation also happens due to these property values of space and so, their ripple can still make through (otherwise) stand still time? – kpv Feb 28 '16 at 07:25
  • @kpv Please stop making things up. – Timaeus Feb 28 '16 at 07:27
  • OK, let us conclude this thread here. Thank you both for your insight. – kpv Feb 28 '16 at 07:29
  • @Timaeus: A last question - If nothing could come out of the EH, how do we explain the big bang. – kpv Mar 01 '16 at 20:33
  • @kpv if you think the phrase "if nothing can come out of an EH" is ever a way to start a sentence then you simply fail to know what the words mean. So how about an analogy. Imagine a professor that has 100% of your grade based on a paper with a due date and promises to grade any paper submitted before the due date. And the professor puts a time stamp on any paper and if the professor grades a with a certain time stamp, then out of fairness the processor grades any paper submitted earlier. The grade horizon is the time stamp beyond which no papers are graded. – Timaeus Mar 01 '16 at 20:41
  • @kpv You might argue that some papers are special, the papers themselves make arguments (cited by research) that argue why professors should or even must accept late papers. But the fact is that if such a paper is convincing, then the ones submitted before it are also graded. So that paper wasn't submitted after the grade horizon. Because the grade horizon is defined to be the time stamp past which no papers are ever ever ever graded. It's literally the definition and so nothing past that time stamp is ever ever graded becasue if it were then by definition the time stamp cutoff would be after – Timaeus Mar 01 '16 at 20:42
  • @kpv And if the professor lives forever, we don't know where the cutoff is, since we don't know when the professor might change their mind about late papers. And the papers might affect that change of mind. But those papers are by definition not beyond the cutoff. This isn't deep. As for the big bang there are some cosmological models where every timelike path hits a singularity. So in many ways it's like there is no outside becasue you are going to hit a singularity no matter what. Of course singularities and event horizons are totally and completely different as naked singularities show – Timaeus Mar 01 '16 at 20:46
  • @Timaeus: Thanks. Did not know about the naked singularities. However, bringing the term in while discussing big bang indicates that the original singularity is supposed to be a naked one. I understand definition of EH. Event (info) is the last bit we can expect to come out. If last bit does not come out, I said, nothing comes out. That is just how I said, I guess, you knew what I meant. My understanding is - event horizon exists due to gravity itself, otherwise, there would be no event horizon. That is why it is really hard to grasp that EH will apply to gravity, or its wave. Thanks again. – kpv Mar 02 '16 at 00:13
  • @kpv No. The EH isn't "due to gravity" because gravity is local and an event horizon is a global structure. For instance if someone throws one mouth of a traversable wormhole towards a compact object then the region of huge time dilation compared to far away is not an EH because things can escape through the wormhole. An EH is nothing other than the boundary of events that intersect the past cones of the outside. It is not a function of any kind of the local metric or gravity. You've got it all backwards. – Timaeus Mar 02 '16 at 00:18
  • @kpv And there isn't any difficulty seeing an EH apply to gravity when you actually study gravity and how it works. The metric has values and has first time derivatives and the Einstein equations tells you how it evolves. When you have the complete 4d manifold you can then label some events are part of an EH but that's an assignment based on the whole global (including the future) manifold. Gravity is just the metric. The metric is just a field. And like all fields it just evolves and we have equations for how it evolves. It isn't mysterious at all. – Timaeus Mar 02 '16 at 00:24
  • @Timaeus: I am not questioning your explanations, however, I do question my understanding of them - "EH apply to gravity when you actually study gravity". Do you mean "whatever is behind the event horizon" does not have gravitational effect on the stuff outside horizon? There are (tensors, metrics, manifolds, evolution), but eventually the stuff behind EH seems to have g effect on the outside stuff. The g effect also moves with a moving BH. Moving of BH (including EH) itself is an event that can be detected from outside in terms of direction of gravitational acceleration. – kpv Mar 02 '16 at 02:59
  • I agree that the effect of "whatever is behind the event horizon" can not be detected in terms of light. Although it can be figured out by absence of background light, so, I can ignore that. But detection due to gravitation is detection by presence. That is hard to ignore. Even though we can not see, we can tell that the BH moved. I still think gravitational communication remains intact across EH. That communication would apply to any GW as well. – kpv Mar 02 '16 at 03:07
  • @kpv Gravitational interaction is not caused by a long distance communication. The metric evolves according the Einstein equation, and that only depends on what is happening in your local neighborhood. People who do numerical relativity can even let the metric evolve "incorrectly" inside an event horizon and they know it won't affect the metric outside as long as they correctly evolve the metric in a way where it only depends on the local neighborhood. The metric outside a moving black hole changes because it's a second order equation so the metric already has first order time derivatives. – Timaeus Mar 02 '16 at 05:55