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It appears that it is impossible to detect the event horizon from a local experiment. Is this a consequence that General Relativity always reduces locally to a Lorentz invariant theory?

Or in other words, is this statement (the inability to detect an event horizon by any local experiment) true in all modified gravity theories that reduce to special relativity locally?

Níckolas Alves
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konstle
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  • You can detect the event horizon in some special cases. For example, if you are in a radial free fall to a Schwarzschild black hole and you are looking exactly back, then you’d see the stars redshifted exactly twice when you are at the horizon. – safesphere Mar 28 '22 at 06:42

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Yes, it is.

I'll try to avoid using too much terminology, since that seems to not be quite the point in here. If you wish, you can check, e.g., Chap. 12 of Wald's General Relativity for further mathematical detail and precision.

By definition, in a black hole spacetime, the event horizon is the topological boundary of the black hole region. "Black hole region" means, roughly, the region of spacetime which does not influence causally any of the observers that go to infinity in infinite time. More technically, the black hole region $\mathcal{B}$ of a spacetime $\mathcal{M}$ is defined as $\mathcal{B} = \mathcal{M} \setminus J^{-}(\mathcal{I}^+)$, where $\mathcal{I}^+$ denotes future null infinity. The event horizon is the boundary of this region: it splits the region of spacetime that respects this from the region of spacetime in which information can reach at least one of those observers.

I tried to emphasize the keywords in these definitions. If you have an event in spacetime that can causally influence at least one of those observers going to infinity, then that event is not in the black hole. Notice, then, that you need information about the entire future of the spacetime in order to figure out whether that one particular event you're considering can affect at least one observer. If it does, not in the black hole. If it doesn't, it is inside the black hole.

Notice then that for you to figure out where the black hole (and hence the event horizon) is, you must have global information, as exemplified in the accepted answer of the question you referred to.

All of these facts depend on the geometry of spacetime, but none of them make reference to the explicit field equations you are assuming. As long as you are dealing with a metric theory of gravity, you will run in this due to the very definition of event horizon.

To close the answer with an analogy, the issue is similar with if I said that someone's "half-life" is the instant right in the middle of someone's life. For example, if you die on your 40 year birthday, your half-life would be when you're 20, and so on. Is doesn't matter how you live, what you do, how long you live, your cause of death, etc, you'll never be able to tell when your half-life is until you have global information (in this case, until you die). That is not because of some local property of what you chose to do on some day, or not even on how life works (whether you're a carbon-based organism or something weirder, etc), but because the very definition I gave of half-life requires knowledge of your time of death.

Níckolas Alves
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  • A strong argument can be made that the spacetime inside the horizon is a mathematical abstraction that does not exist in reality. In this case the worldline of a falling observer ends at the horizon and his matter is converted to the gravitational energy of the black hole. Similarly, matter must lose all information (convert to the gravitational energy) at the horizon for the Hawking radiation to exist. In either case the observer dies at the horizon. Can his death be considered a local detection of the horizon? (Good answer +1) – safesphere Mar 28 '22 at 07:01