In a view of a remote observer, an object falling into a black hole is "hanging" at the horizon (slowly falling with a deceleration). Around this moment, the event horizon expands for some reason that is beyond the scope of this question (e.g. the black hole merges with another one or whatever). The new horizon is larger than the initial distance to the object. Logically, there are two possibilities of what can happen:

  1. The object ends up inside the black hole "swallowed" by the expanded event horizon.

  2. The object moves farther away from the black hole and remains "handing" at the expanded horizon at a larger distance from the center than before.

Which of these two options is correct?


In a comment below John Rennie states:

In a black hole merger no horizons exist either before or after the merger (in the frame of the distant observer).

This is because time slows down infinitely as matter approaches the "apparent horizon" (the radius of where the horizon would be), but the actual horizon never forms. (Let the experts decide if this is correct.)

As a clarification, this is irrelevant to my question. The question is if the object ends up inside or outside the horizon regardless of whether the horizon is real or "apparent".

  • $\begingroup$ Susskind has a good model for the different perspectives, using fish in a pond (and I since found out about the gravity/fluid correspondance making it more compelling!) The Universe As A Hologram youtu.be/2DIl3Hfh9tY The distant universe is being 'swallowed' by an event horizon, as the vast distances turn time-like seperatiin into space-like seperation. The fact no light from there could ever reach us does not affect them at all- $\endgroup$ – CriglCragl Jul 19 '18 at 23:46

There is a mechanism by which matter remains sticked to the event horizon, and not at the constant distance from the BH center.

The mechanism is called frame-dragging. Near the BH surface frame-dragging has enormous power. For instance, there is so-called ergosphere, which is the volume surrounding the event horizon, in which any body has to rotate in the same direction as the BH does. If the BH moves, so do all bodies close enough to the event horizon.

This effect happens with all massive bodies, and has been experimentally detected around Earth, but only in the case of BH frame-dragging can be stronger than pure gravitational attraction.

In fact even the very slow-down of infalling matter in Schwarzschield coordinates can be seen as an effect of frame-dragging near a massive body.

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A quick introductory note: from the comments it's clear the OP is thinking of a black hole merger. My answer was written before I realised this so it assumes the black hole is growing outwards by the gradual accretion of matter. My argument wouldn't be applicable to black hole mergers.

There isn't an analytic solution for the situation you describe, but there is a related model system that we use to get an idea of what happens. This is the Oppenheimer-Snyder metric. The OS metric describes a sphere of uniform density, pressureless gas collapsing under its own gravity. Real stars are neither uniform nor pressureless so the OS metric can at best give us a guide as to the main features of the collapse, but let's go with it and see what happens.

In the rest frame of an observer on the surface of the collapsing ball the event horizon first appears at the centre of the sphere and grows outwards towards the observer as the ball collapses. The event horizon passes the observer at the moment when the radius of the sphere is equal to its Schwarzschild radius.

The relevance to your question is that we can consider the observer on the surface as your object hanging at the horizon. The collapse causes the horizon to grow past and engulf the observer as you describe in your question. The OS metric allows us to calculate the time at which this happens both in the rest frame of the infalling observer and in the frame of the observer far from the collapsing sphere.

The equations we need are given in this article on the GR Wiki. We use a timelike parameter $T$ - note that this is not the proper time of any observer, just a parameter. Then the proper time of the infalling observer is given by:

$$ t' = \frac{A(0)}{2}\left(T + \sin T\right) \tag{1} $$

The parameter $A$ is the scale factor describing the collapse of the sphere i.e. $A$ starts at a finite value at the beginning of the collapse and decreases to zero at the moment the singularity forms. $A$ is given by:

$$ A(T) = \frac{A(0)}{2}\left(1 + \cos T\right) \tag{2} $$

and $A(0)$ is related to the initial radius of the sphere $r_0$ as measured by the observer far from the black hole by:

$$ A(0) = \sqrt{\frac{r_0{}^3}{r_s}} $$

The scale factor $A$ falls to zero when $T=\pi$, so collapse occurs over the range from $T=0$ to $T=\pi$. Substituting $T=\pi$ in equation (1) gives a finite answer so the collapse completes in a finite time as measured by our observer sitting on the surface of the sphere. That means:

The observer on the surface of the sphere observes themselves to pass the growing event horizon in a finite time

The question asks about the view of a remote observer. If we take the observer out to infinity then the observer's coordinates are the Schwarzschild coordinates and specifically the observer's time is the Schwarzschild $t$ coordinate. For this observer the time at the surface of the sphere, i.e. at the position of the infalling observer, is given by the rather ugly expression:

$$ t = r_s\left( \ln\left( \frac{\sqrt{r_0/r_s - 1} + \tan(T/2)}{\sqrt{r_0/r_s - 1} - \tan(T/2)} \right) + \sqrt{r_0/r_s - 1}\left( T + \frac{r_0}{2r_s}(T + \sin T)\right) \right) \tag{3} $$

Although this is rather involved we need only note that the right hand side goes to infinity when the denominator in the log term goes to zero i.e. when:

$$ \sqrt{\frac{r_0}{r_s} - 1} = \tan\left(\frac{T}{2}\right) \tag{4} $$

and this is the moment when the radius of the sphere is equal to the Schwarzschild radius (as measured by the observer at infinity). So the conclusion is:

The observer far from the sphere observes the observer on the surface of the sphere to take an infinite time to pass the point $r=r_s$

So if you are prepared to accept the above as an acceptable model for the situation you describe then neither of the two options you present is correct. For the distant observer no event horizon ever forms and the the infalling observer takes an infinite time to pass the point $r = r_s$ where the horizon would form given infinite time.

I would guess you are thinking of an established black hole with a horizon at $r=r_s$, and what happens if this horizon grows (maybe because a load of mass is dumped into the black hole). The problem is that this is an unphysical situation as for the distant observer an event horizon takes infinite time to form. So the experiment could never be done. The calculation I've described (given the limitations of the OS metric) illustrates what would actually happen.

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  • $\begingroup$ Thank you John! I appreciate this information, as it is very useful to me in general, but it seems unrelated to my question. A collision of two black holes has been numerically modeled to take a rather short finite time for a remote observer. Thus my question describes a very much physical situation that your answer unfortunately does not address. $\endgroup$ – safesphere Jul 7 '18 at 7:53
  • $\begingroup$ In a black hole merger calculation we start with two Kerr geometries i.e. we start with event horizons present even though those horizons would take an infinite time to form, and therefore don't actually exist. Then the calculation looks at the evolution of those horizons with time. So the calculation is unphysical. In a black hole merger no horizons exist either before or after the merger (in the frame of the distant observer). $\endgroup$ – John Rennie Jul 7 '18 at 8:52
  • $\begingroup$ This is irrelevant to my question. In the view of a remote observer, even if there is no horizon, there is an "apparent horizon", at which the infalling matter is "frozen" in time. When two black holes merge, the resulting apparent horizon is larger, even if the actual horizon still does not exist. The gravitational waveforms of the merger have been experimentally confirmed to take finite time. My question is if an object at the horizon (doesn't matter apparent or real) would move further away from the center. $\endgroup$ – safesphere Jul 7 '18 at 13:33
  • $\begingroup$ @safesphere let me see if I understand you correctly. Suppose we take two merging black holes, one with our test object at a distance $\epsilon$ from the horizon (in the distant observer frame) and moving slowly as a result. The question is whether after the merger the distance of the object from the merged horizon, $\epsilon'$ is greater or less than epsilon, and if less whether $\epsilon \lt 0$. Yes? $\endgroup$ – John Rennie Jul 7 '18 at 14:41
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    $\begingroup$ Only if $\epsilon$ after the merger is positive or negative. In other words, if the object that initially is very close to the original horizon ends up inside or outside the final horizon. Thanks! $\endgroup$ – safesphere Jul 7 '18 at 15:12

As it is usually defined, the event horizon of a black hole is the imaginary surface which nothing can escape to infinity from. You cannot actually identify the event horizon of a black hole without taking into account everything that will happen in the indefinite future. So if you add more mass to a black hole at time $t$, the position of the horizon at all times previous to $t$ gets recalculated, and expands slightly. This recalculation alters the position at times much less than $t$ infinitesimally, though; it changes it just enough to let a ray of light moving outward at the horizon escape at infinite time.

Unlike a ray of light escaping, however, an object falling into the black hole will not move outward when the horizon expands due to new mass falling in. Its calculated position will retroactively move outward infinitesimally when the mass is added at time $t$, but it will still be well within the new horizon, assuming it is significantly larger than the old horizon.

Note that this analysis is classical, and does not take into account the quantum mechanics of black holes. If firewalls exist, and are taken into account, the answer could be completely different.

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  • $\begingroup$ Thanks Peter, I appreciate the answer. First, could you please provide a link to a reference supporting your conclusions? I would very much like to learn more about this. Secondly, the reasons implied in my question for the object to move outward and stay outside of the expanding horizon were different from the mechanism you described of the future affecting the past. Specifically, in a remote frame, if timelike geodesics cannot cross the lightlike horizon, then they must move outward following the expansion of the horizon. This essentially is a mere result of energy conservation. $\endgroup$ – safesphere Jul 19 '18 at 20:44
  • $\begingroup$ @safesphere: Just to show that the idea of an object moving outward in classical general relativity is rather farfetched, consider the following thought experiment: Suppose you take a planet, and gather enough dust around it so that the dust forms a black hole, and the event horizon forms in the exact middle of the planet? Does the planet suddenly break into pieces and fly outwards with the event horizon? Where does the energy for this come from? The dust is still a very long ways away. $\endgroup$ – Peter Shor Jul 20 '18 at 13:32
  • $\begingroup$ References? I saw 't Hooft give a talk where he considered the scenario in my last comment (although he used a spherical shell of TVs rather than dust) ... I don't know whether he's written it down or not. I'm not going to spend my time looking through all of 't Hooft's papers for something like this, although if anybody finds it, I'd like to know. $\endgroup$ – Peter Shor Jul 20 '18 at 13:36
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    $\begingroup$ Minor note about terminology: what is recalculated is our guess (to the best of our knowledge at a given moment in time) about where the horizon passes and not the actual horizon (which is calculated once for the whole space-time manifold). $\endgroup$ – A.V.S. Aug 2 '18 at 6:46

I don't think that all of these notions of frozen objects are useful in this situation. If the black hole is expanding, then the stack of apparent horizons forms a spacelike surface in the the overall spacetime, and, by construction, this stack of apparent horizons (and at least a neighborhood of their exterior) will be in the interior of the event horizon. No signal from inside the event horizon will ever reach something in the exterior of the event horizon. Asking what you "see" inside of an event horizon is moot.

Any signal that reaches distant infinity cannot be coming from some area sitting infinitesimally outside of the first apparent horizon. Or at least it has to originate at some time sufficently far in the past from the expansion.

And also, remember that while you never see an object enter the black hole, it redshifts as it gets closer and closer to the apparent horizon, and will relatively quickly redshift to such a long frequency that it will become unobservable in any practical sense.

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  • $\begingroup$ Thanks Jerry! I appreciate your insight. "If the black hole is expanding, then the stack of apparent horizons forms a spacelike surface in the the overall spacetime, and, by construction, this stack of apparent horizons (and at least a neighborhood of their exterior) will be in the interior of the event horizon." - What are the details of the construction that puts the apparent horizons on the inside? Also, to clarify, I am not asking what one "sees" inside the event horizon. Finally, I hear your practical arguments, but the answer conceptually defines the spacetime geometry on the inside. $\endgroup$ – safesphere Jul 19 '18 at 22:23
  • $\begingroup$ @safesphere: the event horizon is defined as the inner boundary of everything that makes it to null or timelike infinity. At some point, far after the black hole has hit equilibrium, the event horizon and apparent horizon will coincide. Trace back in null time, and you will eventually hit the point of black hole expansion. The apparent horizons will recede superluminally until the stack of apparent horizons coincide with the early-time equilibrium. So, in the time in between, the event horizon necessarily sits outside the stack of apparent horizons. $\endgroup$ – Jerry Schirmer Jul 20 '18 at 15:28

I don't think it's helpful to talk about objects freezing at the horizon, or about the "view" or frame of reference of a distant observer. An infalling object has zero radial coordinate velocity, in Schwarzschild coordinates, as it passes through the horizon, but this fact is of no physical interest. It is simply a consequence of the misbehavior of the Schwarzschild coordinates at the horizon. General relativity does not have global frames of reference, only local ones.

General relativity, like special relativity, also lacks any preferred notion of simultaneity. Therefore people on earth can't say whether an infalling object has passed through the horizon "now."

The object ends up inside the black hole "swallowed" by the expanded event horizon.

It doesn't matter whether the black hole's mass grows. In either case, the object's world-line passes through the horizon and reaches the singularity in finite proper time.

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    $\begingroup$ I have a hard time seeing how your post relates to my actual question. We observe from a safe distance a spaceship is falling into a black hole. At this moment the black hole merges with another one. The horizon quickly expands, because the combined horizon is larger. While expanding, the horizon either consumes the spaceship or moves it out. In other words, momentarily, either the spaceship distance to the singularity remains about constant (the spaceship is inside) or the spaceship distance to the horizon remains about constant (the spaceship is outside). Two logical choices, which is right? $\endgroup$ – safesphere Jul 29 '18 at 18:25
  • $\begingroup$ If the object can send signals to Earth, it has not passed the horizon when the signals were sent yet. $\endgroup$ – Anixx Jan 4 '19 at 17:17

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