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The question is in the title, so basically you drop a mass into a non rotating black hole, and then the distribution of mass inside the black hole will not be spherically symmetric (assume the mass did not reach the singularity yet). Will this information affect the gravitational field surrounding the black hole outside the horizon? or will this field be always spherically symmetric?

My confusion is that we cannot know anything from inside the horizon (other than total mass), but an asymmetric distribution of matter should result in an asymmetric gravitational field. So what is wrong in my reasoning?

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  • $\begingroup$ Presumably if the black hole's mass was asymmetric the event horizon wouldn't be spherical? $\endgroup$
    – Alan Birtles
    Commented Nov 11, 2020 at 15:58

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As the falling object approaches the horizon from outside, the situation is already not spherically symmetric so neither is the spacetime. In a reference frame in which the object passes the horizon in a finite time, the spacetime outside remains non-spherically symmetric after the object passes the horizon, though the asymmetry will be dynamic and will tend to radiate away until eventually the whole thing settles to a symmetric shape (in the absence of angular momentum). At all times the spacetime at any given place is influenced only by matter and signals which can arrive there, in the future light cone of other events. So it is not influenced by events beyond the horizon.

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  • $\begingroup$ So, to be more specific, a black hole with a non uniform distribution of mass inside (of course this will last a short time before that mass reaches the singularity) has a spherically symmetric gravitational field? $\endgroup$
    – user65081
    Commented Nov 10, 2020 at 22:43
  • $\begingroup$ No, it will not be symmetric but the asymmetry will radiate away. And you need to be a bit careful about what region of spacetime you are referring to when you use the phrase "a black hole with X inside" because the region beyond the horizon is dynamic and cannot avoid being dynamic, and nothing outside is in the future light cone of stuff inside. $\endgroup$
    – Andrew Steane
    Commented Nov 10, 2020 at 22:49
  • $\begingroup$ but does not the last statement contradict the first? if it is not symmetric, so you can guess mass distribution inside, correct?, on the other hand no information can escape, so how can you calculate the distribution based on the asymmetry? $\endgroup$
    – user65081
    Commented Nov 10, 2020 at 23:01
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    $\begingroup$ An observed asymmetry outside does not mean you deduce anything about inside, it means you deduce that the events outside that caused the shape of spacetime there were not symmetric. $\endgroup$
    – Andrew Steane
    Commented Nov 10, 2020 at 23:12
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    $\begingroup$ @Wolphramjonny The drone can influence the region of spacetime in its future light cone. Once the drone has passed the horizon, its future lightcone is all inside the horizon. So your last statement is correct. $\endgroup$
    – Andrew Steane
    Commented Nov 11, 2020 at 8:34

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