5
$\begingroup$

How did the metric evolve inside the event horizons of the black holes whose merger caused the GW150914 signal?

In principle the Schwarzchild metric of a non-rotating black hole is known inside the event horizon, although the analogous Kerr solution for rotating black holes seems to have unphysical properties in this region. Is it possible to at least simulate the dynamics of the metric inside the event horizon during a black hole merger, and get a meaningful answer? If so, what happens and what would an observer inside the event horizon see? If not, why not?

The main inspiration for the question is my semi-Newtonian intuition that once the event horizons merge, the two singularities would rapidly orbit each other inside the event horizon, and eventually crash into each other due to emission of gravitational waves (which of course must remain trapped inside the event horizon). I highly doubt that this intuition is correct. Can general relativity give us a better answer?

$\endgroup$
  • 2
    $\begingroup$ I rather strongly dislike the existing answers which say "we can't say anything". There are numerical GR simulations of the inspiral and the ringdown, which presumably do involve the spacetime regions inside the event horizons. (If not, an answer to this question would be "the existing simulations actually do stop at the event horizon".) The question explicitly asks about what happens in those regions (at least, in the existing simulations), so there is no reason why there shouldn't be an answer. $\endgroup$ – Emilio Pisanty Feb 15 '16 at 21:58
  • $\begingroup$ @EmilioPisanty I agree, I am also unsatisfied with the current answers. However, I expect the simulations that we have seen actually do stop at the event horizon. I can't see why they would need to include the black hole interiors since they're causally disconnected from the spacetime region of interest. But I feel like the inside should be affected by the outside (e.g. things can fall in), so this seems on the face of it to be a well-posed question. $\endgroup$ – Mark Mitchison Feb 15 '16 at 22:26
  • 1
    $\begingroup$ Once the horizons merge no information is getting out besides what could already be seen in the shapes of the horizons and the surrounding spacetime. That said, @Emilio is right in that simulations go inside the horizon (apparent, because one cannot calculate actual horizons without knowing the full spacetime). They don't go all the way to the singularity, but they do go slightly inside. This way, one doesn't have to worry about what numerical boundary conditions to impose on the domain. $\endgroup$ – user10851 Feb 15 '16 at 23:04
  • 1
    $\begingroup$ Thanks @ChrisWhite, this is the closest I've got so far to an answer. Since apparently the metric can be simulated a little bit inside the horizons, presumably it can also be simulated a long way inside (staying away from the singularity of course)? What can one say qualitatively about what the metric does inside the horizons during the merger, e.g. the point when the horizons touch and coalesce? $\endgroup$ – Mark Mitchison Feb 15 '16 at 23:10
  • $\begingroup$ The causal disconnection is true, but I would expect simulations to have trouble finding the boundary since nothing much happens there locally; it mostly feels like a flappy boundary condition that would be hard to handle numerically. It makes an interesting separate question, though. $\endgroup$ – Emilio Pisanty Feb 15 '16 at 23:21
1
$\begingroup$

This might not literally be what you are aiming for, but everything we see or hear is from events before the horizon forms. And the gravitational waves are also emitted before the horizon forms.

Which means there are events in those stars and between those stars that happen after the waves we've seen are emitted. All the way in the center of the star we called the first black hole, and all the way in the center of the star we call the second black hole and in any part of the space in between.

So we do know what that looks like. And we can solve for it. And you do see the two stars orbiting each other. No singularities. No event horizons. Just time dilated signals coming out towards us.

So its "inside" in the sense that there are no holes or missing parts. It's just that the pre horizon events are very time dilated so a little bit of them covers a lot of our time. And at least those are the most scientific parts.

$\endgroup$
0
$\begingroup$

I was just thinking about this particular question, and I think I came up with an answer you probably won't like: it doesn't matter. Unfortunately for us (or fortunately?), anything that happens inside an event horizon is completely unknowable to the outside. While we can have a picture of the metric inside the event horizon for some distance, we should expect that the classical approximation breaks down the further we get. The fact that there is a singularity at the center of the black hole in GR indicates not that there is a region of infinite density but that GR as a whole breaks down on that scale.

Also, note that the gravitational waves emitted in the recent merger that we detected are not emitted from inside the event horizon, but from outside of the horizon - otherwise, they wouldn't escape, as you mentioned.

However, one paper I read described the shape of the photon sphere of the merger of (maximally charged) black holes, which has an analytic solution. I would expect (but I haven't done the calculation, so I'm not quite certain) that the event horizon takes on a form similar to that of the photon sphere (but smaller) during the actual merger event. Once the event horizon becomes that of a sphere, what happens on the interior is no longer relevant - everything outside the event horizon cannot tell the difference between a black hole with two singularities or a black hole with one singularity, so for all intents and purposes, we can consider it as having a single singularity.

One question that is possibly worth investigating, however, is exactly how the event horizons combine and the dynamics there - what happens when they overlap, or can they even overlap? Further, an interesting continuation of your question is, if you have two black holes (of equal mass) about to collide with impact parameter zero, and place a point particle exactly between the two bodies, what happens to that point particle? Clearly, the black holes do collide - however, from a distant observer's perspective, the particle can never pass the event horizon of either black hole. So does it get squeezed out? Spread out equally over both horizons? That's more to consider.

$\endgroup$
  • $\begingroup$ Thanks for the answer Sam, but I don't think this fully answers the question. What happens on the inside of the event horizon certainly cannot affect the outside. But I am asking what happens for observers inside the black hole. It is not true that "While we can have a picture of the metric, it's never observable", because I am free to plunge into a black hole and make observations for myself. (Leaving aside the small matter of my certain death, it's possible in principle.) Just don't expect me to share my scientific results with you later :) $\endgroup$ – Mark Mitchison Feb 14 '16 at 20:36
  • $\begingroup$ Fair point - I was afraid that I wouldn't answer your question in it's entirety! I'll edit that sentence out though, that was too broad. However, I would like to maintain that, as of right now, we can't know what happens at the "center" of a black hole, even in theory, if only because our theory breaks down there! $\endgroup$ – Sam Blitz Feb 14 '16 at 20:48
  • $\begingroup$ I would have to look this up to be completely sure, but my gut feeling on it is that it is relatively unconstrained and is heavily dependant on the recent history of matter that has fallen into the black holes. For example, the Schwarzschild metric is only accurate for black holes that have always existed or, equivalently, for the exterior of a spherically symmetric black hole. The Schwarzschild metric does not tell you what happens on the interior of a black hole that formed a finite time in the past. $\endgroup$ – Graham Reid Feb 15 '16 at 4:52
  • $\begingroup$ A simpler, but illustrative, example is the case of the Vaidya metric which is a metric for (in this case) a collapsing null dust (like massless neutrinos). In this case, you could imagine two shells of matter collapsing inwards with your observer between them. The observer will be outside of the apparent horizon formed by the first shell, but, after a bit of a fall, will be inside the apparent horrizon formed by the second while it is still behind him! Nothing about physics changes for this observer until he hits the singularity! $\endgroup$ – Graham Reid Feb 15 '16 at 4:57
-1
$\begingroup$

A perfect symmetric view without new extensions : after the horizon, there is a privilegied direction. Each BH falls in the other. Nothing escape but the gravitational waves emited in relation with the shape.

From the point of view or each, all happens as if there were a star and a black hole.

$\endgroup$
-5
$\begingroup$

Binary black hole merger viewed from inside the event horizon

At the event horizon the "coordinate" speed of light according to distant observers is zero. So if you were viewing the merger from inside the event horizon, and I was somehow viewing you from where I'm sitting via my bubble of artistic licence, I would say you have no view at all. There's a conflict between this and "the proper speed is c". See Kevin Brown's The Formation and Growth of Black Holes and note the mention of future infinity. As far as I can tell it takes you forever to see anything. You haven't seen it yet, and you never ever will.

How did the metric evolve inside the event horizons of the black holes whose merger caused the GW150914 signal?

I don't know. But the black hole is a black hole for a reason, and it isn't because space is falling down. It's because the "coordinate" speed of light is zero. Which suggests there's no evolution inside the event horizon.

In principle the Schwarzschild metric of a non-rotating black hole is known inside the event horizon

I rather thought that the Schwarzschild metric was only valid outside the event horizon, as per Graham Reid's comment.

although the analogous Kerr solution for rotating black holes seems to have unphysical properties in this region.

Agreed. Take a look at the Einstein digital papers, and you can read this from 1920: "the curvature of light rays occurs only in spaces where the speed of light is spatially variable". He didn't use the word coordinate, which suggests that at the event horizon, the speed of light is zero. Which suggests that the spin rate is zero too.

Is it possible to at least simulate the dynamics of the metric inside the event horizon during a black hole merger, and get a meaningful answer?

Maybe. But my reading of the Einstein digital papers is that there are no dynamics. Rather counterintuitively, the descending photon slows down. See this PhysicsFAQ article by editor Don Koks.

If so, what happens and what would an observer inside the event horizon see?

Maybe there's a way we can have our cake and eat it, in that the expansion of the universe is not limited to the speed of light. But I'm still not hopeful about that observer seeing anything any time soon.

If not, why not?

Because the light doesn't get out. Because gravitational time dilation is infinite. Because the coordinate speed of light is zero.

The main inspiration for the question is my semi-Newtonian intuition that once the event horizons merge, the two singularities would rapidly orbit each other inside the event horizon, and eventually crash into each other due to emission of gravitational waves (which of course must remain trapped inside the event horizon). I highly doubt that this intuition is correct. Can general relativity give us a better answer?

I think it can, but I side with the frozen-star interpretation, and I'm currently in the minority. Again see Kevin Brown's The Formation and Growth of Black Holes: "Incidentally, we should perhaps qualify our dismissal of the "frozen star" interpretation, because it does (arguably) give a serviceable account of phenomena outside the event horizon, at least for an eternal static configuration. Historically the two most common conceptual models for general relativity have been the "geometric interpretation" (as originally conceived by Einstein) and the "field interpretation" (patterned after the quantum field theories of the other fundamental interactions). These two views are operationally equivalent outside event horizons, but they tend to lead to different conceptions of the limit of gravitational collapse. According to the field interpretation, a clock runs increasingly slowly as it approaches the event horizon (due to the strength of the field), and the natural "limit" of this process is that the clock asymptotically approaches "full stop" (i.e., running at a rate of zero). It continues to exist for the rest of time, but it's 'frozen'..."

What this frozen-star interpretation says is that there aren't any point-singularities. Kevin Brown suggests Einstein would have sided with the other more common interpretation, but I don't think he would. When you drop your pencil, it falls down because the speed of light at the floor is less than the speed of light in front of your face. But when you're at the event horizon, the speed of light in front of your face is zero. And it can't go lower than that.

$\endgroup$

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.