Does an object approaching a black hole ever cross the combined event horizon of the black hole and itself?

Question

Once you start studying black holes, one of the first things you'll probably hear is that from an outsider's perspective objects falling into the black hole take an infinite time to do so due to time dilation. The object will asymptotically approach, but never reach the event horizon.

As far as I understand the typical description of this, people are referring to the event horizon of the black hole as it was before the approach of the new object. However, my understanding is also that you can calculate a Schwarzschild radius for the total mass $M+m$ (where $M$ is the mass of the black hole and $m$ is the mass of the falling object) which would constitute the radius for the event horizon formed by the combined mass.

This new radius would be only very slightly larger, but positively so. Initially, this new event horizon would be inside the total mass, just like the old one. However, if the assumption here is that the object is approaching the old event horizon, then eventually in finite time, the object would reach the new radius. At that point, the total mass would be inside its Schwarzschild radius, so it would effectively be a black hole of increased mass compared to the original one, while the new event horizon would be exposed outwards.

As such, this would imply that from an outside perspective the falling object would eventually vanish entirely in finite time and the black hole would appear larger.

Some potential flaws in my reasoning that I can see, but I'm not sure about are the following:

1. The total mass of the black hole + the object is not perfectly spherical or symmetrical and therefore the Schwarzschild radius is not necessarily an appropriate way to describe or approximate the situation.
2. The way the mass is distributed inside an event horizon may not matter for the gravitational effects outside it, but it might matter for the effects inside it. Thus the region between the two event horizons does not necessarily behave like the inside of a black hole and could still be visible to outsiders.
3. Even if we account for the new event horizon, the object actually approaches that and not the old horizon, therefore it still only asymptotically approaches it.

If I'm right about any of these doubts I'd appreciate if someone confirmed that, but if there's something else going on that'd be even more interesting.

It's also worth noting that is the strong version of my syllogism where only one object is enough to allow itself to cross the horizon. Even if we conclude that this is not the case, there is a weaker version where we study the effect of multiple objects approaching the black hole. If one assumes a constant inflow of matter (unrealistic, I know, but bear with the hypothetical for a moment) then eventually (given enough time) the area near the event horizon would get dense enough and the expanded horizon wide enough that some, if not all, the approaching matter would end up inside it. If this is correct, then it might also be generalised for cases where the inflow isn't constant, but it's large enough for a long enough time for the effect to kick in.

Answer 1

In the animation below, every colored Kugelblitz shell has an energy equivalent of 1Mc², so together they have a total of 5Mc². The innermost shell will asymptotically, but never quite reach r→2GM/c², while the outermost will converge to r→10GM/c².

The inner shells don't know about the outer shells due to the shell theorem, but the outer ones are affected by the inner ones. So in the frame of a stationary observer far away whose proper time is the Schwarzschild coordinate time t, nothing ever crosses r≤10GM/c².

The innermost shell, representing the core of a collapsed star, does of course end up inside the horizon generated by the outer layers, but still outside the horizon that would be generated if the outer shells were not present:

The gray background doesn't have mass of its own, it just represents the different radii. If the situation were not spherically symmetric it would be a little bit more complicated, but the principle would most likely be the same, although I haven't done the math for the latter case.

Answer 2

No, in the coordinate system of an external observer the infalling object never crosses the event horizon even allowing for the horizon growing outwards to meet the object.

This is a rather arm waving explanation, but I think it's still rigorous. The definition of the horizon is the set of points from which light will never escape to infinity no matter how long we wait. In the case of the infalling object the horizon becomes time dependent, but the definition still applies. That means the light ray that leaves the object as the object crosses the horizon cannot escape to infinity.

This is where it gets arm waving: if the light ray never escapes that must mean for the observer at infinity the time dilation at the horizon is infinite i.e.

$$ \frac{dt}{d\tau}(r = r_s) = \infty $$

where $t$ is the observer's coordinate time and $\tau$ is the proper time. Then for the external observer the elapsed time for the fall to horizon will be something like:

$$ \Delta t = \int_{\tau_0}^{\tau_1} \left(\frac{dt}{d\tau}\right) d\tau $$

where $\tau_0$ and $\tau_1$ are the proper times the object started falling and crossed the horizon. But since $\frac{dt}{d\tau}(\tau_1) = \infty$ the integral diverges and we get an infinite elapsed coordinate time.

I guess the objection is that an infinite derivative does not necessarily mean the integral diverges, but we expect that locally (in space and time) the time dependent metric will look like a Schwarzschild metric and we know the integral diverges for the Schwarzschild metric. Hence we expect it diverges in this case as well.

Answer 3

Once you start studying black holes, one of the first things you'll probably hear is that from an outsider's perspective objects falling into the black hole take an infinite time to do so due to time dilation. The object will asymptotically approach, but never reach the event horizon.

Inprecise statements like this are unfortunately common. They build on concepts that have an intuitive meaning for most people, but do not have a precise well-defined meaning.

The more accurate statement is that an outside observer will never observe anything crossing an event horizon. However, this is also something of a tautology, since the event horizon is itself defined as part of a region of spacetime from which signals can never reach outside observers.

Unlike special relativity, general relativity does not have a unique canonical notion of "the time at event A according to an observer at event B". The best B can do is "the time at which they observe A" (and even that is not necessarily uniquely defined for a far away observer since signals from A to B may follow inequivalent paths and arrive at different times.

So, whether or not object A ever crosses the event horizon according observer B depends on a choice. In general, we can construct both "global times according to B" that will answer in the negative and in the positive. Hyperboloidal slicings are a popular example of the latter, which play and important role in modern numerical relativity.

Answer 4

All of the following statements can be simultaneously true:

1. An observer outside the event horizon will never observe a photon (or any other signal) emitted from inside the event horizon.

2. As an object falls into an event horizon, the time it takes for photons emitted by the falling object to reach an external observed tends to infinity.

3. An observer outside the event horizon can observe photons that appear to have been emitted from a point in space (however you define that) which at some later time will be inside the event horizon.

Statements 1 and 2 together form the basis of the claim that "an external observer never sees anything fall through the event horizon." In particular, statement 1 is basically just a restatement of what the term "event horizon" means, while statement 2 says that an external observer can in principle forever continue to receive a stream of (increasingly faint, sparse and redshifted) photons from the object falling into the event horizon.

(In practice, of course, any light emitted by the falling object will fade out and quickly become impossible for any external observer to detect, even if the last few photons emitted before the object crosses the event horizon could in principle take an arbitrarily long time to reach the external observer.)

However, remember that an event horizon is a boundary in spacetime, not just a line in space. In particular, when we say something happens inside or outside the event horizon, what we mean is that it happens at a point in space that is inside or outside the event horizon at the time when it happens.

In particular, you can observe light emitted from any point in spacetime outside the event horizon (and inside your own past light cone), even if that same spatial location (however you define "same" for this purpose) at some later time might be inside the event horizon.

In other words, it's perfectly possible for you to see an object fall towards (and arbitrarily close to) an event horizon of radius $r$, and then later (indirectly) observe the event horizon expanding to a radius $R > r$. That does not mean that the light from the falling object would've reached you from inside the event horizon, since the object was (by definition!) outside the event horizon when it emitted the light that you saw.

Answer 5

I'm not completely certain of this but I think there's a fundamental flaw here in that real world objects always have size. No object can fly close enough to the black hole to lift the event horizon to cover it without said object having hit the event horizon and not coming back.

The only object without a "size" is another black hole--but I believe in this case when the event horizons touch they're not coming apart again.