What would happen if a black hole disappeared? [closed]

Question

Imagine if a black hole disappeared. Would spacetime act like a rubber band and propel objects that used to be caught in its gravitational field outwards - i.e. some kind of space time explosion? How fast would spacetime start to flatten out? Or, would planets, stars, etc just move off in a straight line as if nothing happened? Also, if the position of a black hole changed due to a quantum effect, would a similar "space time explosion" occur?

Answer 1

This question has pushed the rubber sheet analogy beyond the event horizon [rim shot]. As pointed out in the comments, we can't answer what physics says will happen when we ignore the laws of physics, however briefly. Nevertheless, it is somewhat of a tradition in relativity; for example, with the question, "What happens if we could teleport/communicate instantly over long distances." (What happens is that you realize, instant teleportation is a frame dependent thing, and you have trouble asking the question. But you may gain insight into the relativity of simultaneity).

So what happens if a black hole disappears? The 1st problem is: what disappears? A black hole is made out of spacetime. Like Oakland, California: there is no there there.

The Schwarzschild metric:

$$ -c^2d\tau^2 = -\big(1-\frac{r_s}r\big)c^2dt^2 +(1-\frac{r_s}r\big)^{-1}dr^2 + r^2(d\theta^2+\sin^2\theta d\phi^2) $$

is a vacuum solution: there is no $T_{\mu\nu}$ to create local non-zero $G_{\mu\nu}$, and there is no $T_{\mu\nu}$ to take away.

As the comments suggest, we could just take away the matter...cut it out. I think that is what the singularity has already done. The baryons, leptons, dark matter, and photons that crossed the event horizon are gone. Their mass has vanished in the singularity and are no longer part of our spacetime manifold. Quantum gravity and a solution to the blackhole information paradox notwithstanding, all that's left of them is spacetime curvature.

I'm trying to imagine cutting the singularity out, but I don't know what that means. It is a cut out.

OK, so maybe, and this is the non-physical, and perhaps non-mathematical, part: let's imagine suddenly adding an "opposite" singularity that cancels it out. Here we are relying on:

$$ (+\infty) + (-\infty) = 0 $$

Not a confidence booster, but it's essential the question.

Now as $r\rightarrow 0$, instead of diverging to infinity, we jump back to zero. Does a single point of $g_{\mu\nu} = \eta_{\mu\nu} $ do anything? The Schwarzschild metric still solves the Einstein field equations at $r>0$. It fails at $r=0$, but it always failed at $r=0$. That's why there's a singularity.

OK, let's at another canceling infinity, to cancel the Schwarzschild metric:

$$ (+\infty) + 2(-\infty) = -\infty $$

Now we've gone into a dark forest. The curvature is going to cancel the blackhole (assuming arguendo, this is non-physical). But there is a problem: like Las Vegas (NV), what happens in the event horizon stays in the event horizon.

This isn't about light can't escape, this is about the future. As we all know, when objects fall into a blackhole, we never see them cross the horizon. They slowly fade away, slow down, and freeze above the horizon. The observer falling in crosses the horizon and see all of our future occur above him in an ever shrinking circle light, until he reaches his future: the singularity.

For those inside the horizon, the singularity is a time in their very near future. For us, on the outside, it's in our infinite future. That's how the boundary got its name: the event horizon. Canceling the singularity surely can't propagate backwards in time? (That was the same problem with teleportation).

So: "idk" is it. Even if we could do the experiment, we'd never know the outcome. Hopefully some insight into the nature of blackholes was gained.