0
$\begingroup$

This question already has an answer here:

What happens to the event horizons of each black hole as the two identical non-rotating black holes collide head on?

At a point exactly between them, is the curvature zero (do the individual curvatures due to each black hole cancel out)?

Does it mean that when the holes begin to merge (event horizons touch) the event horizon of each black hole recedes away because the curvatures cancel out?

Does that mean I would then be able to peer inside a black hole? [tantamount to information now escaping from the original insides of the black holes]

$\endgroup$

marked as duplicate by PM 2Ring, GiorgioP, HDE 226868, Kyle Kanos, user191954 May 2 at 4:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I also think that it's misleading to refer to the event horizons in this context, per discussion in other questions on this topic. $\endgroup$ – Jerry Schirmer May 1 at 16:21
1
$\begingroup$

At a point exactly between them, is the curvature zero (do the individual curvatures due to each black hole cancel out)?

No. GR is a nonlinear theory, so curvatures don't add in any simple way. Even if they did, the behavior of the Riemann tensor under a parity transformation along one spatial axis is not just a sign flip.

Does it mean that when the holes begin to merge (event horizons touch) the event horizon of each black hole recedes away because the curvatures cancel out?

There is no direct link between curvature at a point (which is a local property) and the question of whether that point is on the horizon (which is a global property).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.