According to the official version, they should remain entangled, but then, according to Mr. Graymer, we should ask the following: if in addition to the 'usual' entanglement there is an entanglement that is especially limited, what will happen? In that case, they would not remain entangled beyond a certain distance, which should depend on some quantum property of the particles considered. The issue is, therefore, to verify if there exists a function with these characteristics that satisfies the Schrödinger equation. Only recently has a solution with these characteristics been found, which will be submitted shortly.

That one falls in the black hole just means that it is following a spacetime curve (geodesic, if it's freely falling, for it could, also, be accelerating, of course), that hits the singularity in finite proper time; while the other definitely isn't. The particle that doesn't hit the singularity isn't free, incidentally.

That they're entangled just means that their state isn't a product state.

So it might be thought that they'll be entangled up to the moment the freely falling particle hits the singularity, that's all; since its evolution after that instant isn't defined at all.

The point is that this moment is at infinite proper time, for the particle, that doesn't hit the singularity; so the particles remain forever entangled, as measured in the outside particle's reference frame.

The two particles don't have a common time in any case. Entanglement simply means that their joint state can't be written as a product of local states, that's all-which means that the bases for each particle, in which their states can be written as linear combinations, can't be globally defined.

For sure 'Quantum entanglement' is the basis for what A Einstein spurned as: 'Spooky action at a distance', or the 'Spooky action' itself, the Zbigniew Motyka response is very good, but my doubt remains that quantum entanglement is supposedly indestructible, eternal, persistent at infinite distances and times, if infinite, same as singularity, were a real thing in our world, not a mathematician's working convention.

Thanks, regards. Salut +

Once more: entanglement means that the probability density of the multiparticle system doesn't factorize into a product over the density of each particle. Nothing more, nothing less.

So it doesn't make sense expressing doubts about this fact. The only way entanglement is affected is when the particles have additional interactions. It suffices to actually try calculating, what can make the terms that prohibit factorization vanish.

And states are defined in phase space, not spacetime. Confusing the two is a common error.

just two? impossible

To Igor: thanks for your remark, new to me. The question was just theoretical; in your same line, I challenge everyone to provide evidence against or for the assumption: 'Singularities do not exist, points exist only in the mathematician's minds, as nothing could be compressed below or beyond the size of strings'

Inside black holes' event horizon, there must be balls, not points.

Salud +

Decoherence may occur even when a black hole is not involved.

The event horizon is a coordinate artifact. Particles that move in a black hole spacetime belong to two classes: those that hit the singularity in finite proper time and those that don't. For neither does the horizon have any meaning.

As mentioned above, the particle that does hit the singularity isn't frozen and the measurements are local to each particle. The particle that doesn't hit the singularity, remains entangled with the one that does for infinite time, since, in its frame, the inner particle never hits the singularity.

Of course it is-it's possible to choose coordinates that are perfectly regular there. It's amazing that this isn't as well known: https://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coordinates

The singularity of the coordinate system at the horizon, when using a certain coordinate system, is nothing more than the singularity of any coordinate system that isn't globally flat. The singularity of the black hole spacetime isn't a coordinate singularity-and this is, also, well-known, since the work of Hawking and Penrose.

The horizon is irrelevant for the observers that don't have the singularity in their future and, also, irrelevant for those that do. (These are the two classes.)

A black hole, that isn't extremal, emits Hawking radiation, since the observer at infinity is, inevitably, accelerating, in order to avoid the singularity-therefore this observer detects a bath at finite temperature.

While not all accelerating observers can avoid the singularity of a non-extemal black hole, only (certain) accelerating observers can avoid it.