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Alice is radially free-falling towards a black hole. 1000 years later, Bob is following her, also free-falling. Will he meet Alice at the event horizon?

The problem is that from the point of view of a far-away observer, Alice will not reach the event horizon within a finite time, and so will Bob. That means that from the point of view of a far-away observer, both events (event 1: Alice reaches the event horizon at coordinate xy, event 2: Bob reaches the event horizon at coordinate xy) are not distinguishable. Does that imply that Bob will reach the event horizon simultaneously with Alice? Is there simultaneity in infinity? Will they meet?

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    $\begingroup$ "That means that from the point of view of a far-away observer, both events (event 1: Alice reaches the event horizon at coordinate xy, event 2: Bob reaches the event horizon at coordinate xy) are not distinguishable". This is an assertion without an argument. The fact that the assertion happens to be false is beside the point. $\endgroup$
    – WillO
    Commented Dec 20, 2021 at 17:38

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No, they do not meet.

From their points of view, they cross the event horizon and meet the singularity at some finite proper time later.

They don't meet from the point of view of an external observer either. They both get asymptotically closer to the event horizon, but Bob is always further from the horizon than Alice.

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  • $\begingroup$ That's all fine, but: What exactly tells you that Alice reaches the event horizon before Bob? The external observer is not able to tell us because both events happen outside of his time measurement, in infinity. $\endgroup$
    – Moonraker
    Commented Dec 20, 2021 at 19:30
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    $\begingroup$ @Moonraker GR tells us that. You can call it a prediction of GR if you like. I can't answer your question in any other framework. Neither Alice or Bob reach the event horizon from the perspective of a distant observer. $\endgroup$
    – ProfRob
    Commented Dec 20, 2021 at 20:45
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From some comments by the OP,

An "intersection" would require a look beyond the EH, I am only claiming that both worldlines "meet" i.e. "touch" at the EH [...] the worldlines of Alice and Bob are approaching each other - spatially - when approaching a point at finite distance on the event horizon, and we know that they will effectively reach the EH within finite time.

This is a misunderstanding. Consider ordinary $(1+1)$-dimensional Minkowski spacetime with line element $\mathrm ds^2 = -\mathrm dt^2 + \mathrm dx^2$, and consider two objects $a$ and $b$ following the trajectories $x_a(t) = t$ and $x_b(t) = t+1$. Their trajectories look like this:

enter image description here

Presumably you agree that (i) these worldlines never touch and (ii) nothing particularly special is happening at $t=1$. However, now I will perform a rather odd change of coordinates, in which I define $u = x e^{-1/(1-t)}$ for all $t<1$. On the patch of spacetime defined by $t<1$ and $|x|<(1-t)^2$, this is a perfectly valid choice of spatial coordinate, and yields the line element $$\mathrm ds^2 = -\left(1- \frac{x^2}{(1-t)^4}\right) \mathrm dt^2 + \frac{2xe^{1/(1-t)}}{(1-t)^2} \mathrm dx \mathrm dt + e^{2/(1-t)} \mathrm du^2$$

If we now look at $u_a(t)$ and $u_b(t)$, we see something interesting - namely that the two trajectories appear to be converging at $t=1$.

enter image description here

This seems to fly in the face of the claim made above. The trick is that this apparent convergence is not real, and is an artifact of our strange choice in coordinates.

If we had started by using $(u,t)$-coordinates then this would perhaps not be so obvious. However, note that the metric components diverge as $t\rightarrow 1$, which is indicative of a coordinate singularity; on the other hand, coordinate-independent scalars (such as the Ricci scalar, which is obviously $0$ here) do not diverge at $t=1$. Furthermore, observe that the fact that the coordinates are getting closer together does not mean that the objects are; indeed at any fixed $t$, we see that $$\mathrm ds = e^{1/(1-t)}\mathrm du= dx$$ and so the distance between $a$ and $b$ is the same for all $t<1$, despite the fact that their $u$-coordinates are getting closer together.


My intention here was to demonstrate that we must be careful how we interpret statements made at the level of coordinate charts. In the Swarzschild spacetime, the standard Swarzschild coordinates $(t,r,\theta,\phi)$ have a coordinate singularity as $r\rightarrow r_s$, and exhibits the same symptoms (limited domain of definition, divergent metric components but well-defined curvature, etc) at the event horizon as my toy model did at $t=1$. Choosing a coordinate chart which isn't singular at $r_s$ - such as the Eddington-Finkelstein chart - makes it clear that the worldlines of two infalling objects will not intersect at the event horizon.

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  • $\begingroup$ Thank you, the last paragraph is a well-done résumé of your point of view which I will keep in mind. However, you still insist on some "intersection" which I don't claim, and you did not mention the Kruskal metric where the event horizon is the simultaneity line t = infinity. $\endgroup$
    – Moonraker
    Commented Dec 21, 2021 at 20:18
  • $\begingroup$ @Moonraker I can't imagine what you mean by "They will meet!" if not that the worldlines intersect at the horizon. The events "Alice crosses the horizon" and "Bob crosses the horizon" do not coincide. As far as the Kruskal coordinates go, the event horizon is the set of events $X= \pm T$, and distinct, timelike, radial geodesics cross the horizon at different points. I'm not sure what else there is to say, but if you can elaborate on your question I can try. $\endgroup$
    – J. Murray
    Commented Dec 21, 2021 at 20:53
  • $\begingroup$ See my new answer, I hope it is more precise now. $\endgroup$
    – Moonraker
    Commented Dec 25, 2021 at 16:32
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Alice is radially free-falling towards a black hole. 1000 years later, Bob is following her, also free-falling. Will he meet Alice at the event horizon?

No, they will not meet at the event horizon. Nor will they meet below the event horizon. In a coordinate-independent sense their worldlines do not intersect.

Does that imply that Bob will reach the event horizon simultaneously with Alice?

This part of the question is not well-posed. The coordinate chart assumed in this section is the Schwarzschild coordinates. The event horizon is not part of that coordinate chart. So asking what happens at the event horizon in those coordinates is ill posed, it is asking about the behavior of a function outside of its domain.

However, in general a coordinate chart is a one-to-one mapping between events in spacetime and points in $\mathbb{R}^4$. So if two worldlines do not share any events then they also do not share any coordinate points. In other words, if they do not meet in the coordinate independent sense described above then they do not meet in any possible coordinate chart.

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