If you created an annular ring of compact matter, delicately balancing it and continuously adding matter to the ring in a stable orbit around a dormant black hole, could you peer beyond the theoretical event horizon if you were an observer between the ring and the event horizon?

If the external ring was dense enough, you might even end up with Lagrange points as you get closer to the black hole. As you add matter, the ring distorts space in front of it and decays gravitationally into the black hole, shifting the gravitation force of space between the ring and the core. I created a little visualization to show the effect. Between the outer and inner gravitational attractors, you get flat(er?) space. Ds space around the black hole

  • $\begingroup$ The horizon will move toward the ring. The more mass you add to the ring, the closer the horizon will move to the ring. $\endgroup$
    – safesphere
    Commented Apr 19 at 9:48
  • $\begingroup$ I think only for a distant observer. For a local observer around the ring, the ring pulls on objects just like the black hole does. $\endgroup$
    – Travis R
    Commented Apr 19 at 11:10
  • $\begingroup$ If you replace the ring with a hollow spherical shell around a black hole, then this setup has an analytical solution per the Birkhoff theorem. There is no gravitational attraction to the shell on the inside. Same as in the a Newton Shell theorem. $\endgroup$
    – safesphere
    Commented Apr 20 at 2:14
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    $\begingroup$ This would be invalidated in this special case, especially given the... intransitivity of the black hole. Birkhoff works because regardless of where you are inside the sphere, there are counter particles everywhere in the sphere balancing them. But in this case, even though its the 2D analog of Birkhoff its at least partially blocked by the black hole. But this whole thing does make you wonder if you can feel the gravity of a massive object on the other side of the black hole... You would think the black hole would create a complete filter of gravity for whatever is behind it. $\endgroup$
    – Travis R
    Commented Apr 20 at 23:30
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    $\begingroup$ Thinking of gravity in terms of acceleration (attraction) often yields wrong conclusions. A better way is thinking in terms of potential. Then acceleration is the rate of change (gradient) of the potential. The potential inside a shell is constant, but not zero. The potential around a black hole changes with distance. The horizon is where the potential reaches the critical value. So if you put a black hole inside a shell, their potential add up, so the critical value is reached at a larger distance from the black hole. This means the horizon becomes larger and closer to the shell. $\endgroup$
    – safesphere
    Commented Apr 21 at 17:35

1 Answer 1


could you peer beyond the theoretical event horizon [...]?

No. You don't even need to check the details of the spacetime to answer this question, in fact. The definition of a black hole is (roughly) "the region of spacetime that can't be seen by any observer that goes infinitely far within infinite time". The condition of "infinitely far in infinite time" is meant only to assure that you are considering the observers that do not fall in the black hole. The event horizon is then the (topological) boundary of the black hole.

This means that, by definition of a black hole, you cannot look inside of it without entering the black hole. If somehow you could peek inside the event horizon without entering the black hole, then, by definition, that would not be the event horizon.

By definition of a black hole there is no situation in which you can look inside the black hole and come back. The only way to look inside it is by falling in.


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