Is it possible to consistently say that a particle falls into a black hole in quantum mechanics?

This question is closely connected to this old question of mine. The reason for this new one is that I realize part of the issue with that old question is that its main point - the idea of throwing a particle into a black hole - seems very ill formulated. Let me elaborate.

Classical Discussion

Consider the Schwarzschild spacetime and suppose a massless particle is sent radially on the direction of the horizon. Considering the exterior region of the black hole with its usual Schwarzschild coordinates the metric becomes $$g=-f(r)dt^2+f(r)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2),\quad f(r)=1-\frac{2M}{r}.$$

The particle being thrown will have a trajectory $$\gamma(\lambda)$$ obeying the geodesic equation. We may solve the equation by demanding the curve be radial and ingoing. Doing so we find that $$r$$ itself works as an affine parameter and that $$t(r)$$ reads $$t(r)=-\left(\lambda+2M\ln\frac{|r-2M|}{2M}\right)+C.$$

As $$r\to 2M$$ we get $$t(r)\to +\infty$$. In this sense: a Schwarzschild observer never sees the particle crossing $$r = 2M$$.

We may get out of this, though, by noticing that $$r = 2M$$ is approached in finite affine parameter of the geodesic. In that sense we may say that the particle does indeed crosses the horizon.

If the particle were massive this last point would mean that in finite propertime it crosses the horizon, so it certainly does even if the Schwarzschild observer can't see this.

Quantum Discussion

Now let's take quantum mechanics into consideration. I want to consider the particle as quantum of a field. So let us take a massless Klein-Gordon field $$\phi$$.

How do we formulate and discuss the idea of "a quantum of $$\phi$$ falling through the horizon"? Honestly I have no idea and I see lots of problems:

1. The idea of particles is ill-defined in curved spacetimes because (1) distinct observers disagree on what particles are, and (2) the background may create particles. In that sense, even if we say that in the far past an observer has thrown a particle in the direction of the horizon, talking about the particle falling in seems ill-defined from start.

2. The particle no longer has a well-defined geodesic with an affine parameter. In the classical case even though the Schwarzschild observer could not acknowledge the particle falling into the black hole, the fact that along its geodesic the horizon is crossed in finite affine parameter shows that indeed it has gone into the hole. Here we can't do this analysis.

Taking (1) and (2) into account, it seems we could never speak of a quantum mechanical particle falling into a black hole. I see no way of "the particle acknowledging it" like with the affine parameter thing, and I see no way of taking one observer to acknowledge it as well.

This is tremendously puzzling to me. First because it obviously makes sense to me that things may fall into a black hole. Second because I see in several occasions people talking about particles falling into black holes as if it were the most common thing to consider. I won't get one extensive list on this, but take this paper, in the conclusions they say:

Our results may have implications for the black hole information loss problem. Virtually all discussions of information loss in the black hole context rely on the possibility of localizing particles - from throwing a particle into a black hole to keeping information localized.

So, how on Earth can we talk about throwing a quantum particle into a black hole in the context of QFT on which we see particles as quanta of fields? How can this be turn into a well defined idea?

• "We may get out of this, though, by noticing that $r=2M$ is approached in finite affine parameter of the geodesic. In that sense we may say that the particle does indeed crosses the horizon." - This is a circular logic: "$r$ crosses the point of $r=2M$, because $r$ crosses this point". It is absolutely illogical and unphysical. Plus it has nothing to do with time. Here is the proper logic: (1) In no reference frame can this particle be observed to cross the horizon. (2) The Schwarzschild coordinates describe the entire physical manifold. The event of crossing does not exist in this manifold. Commented Oct 25, 2019 at 5:33
• "If the particle were massive this last point would mean that in finite propertime it crosses the horizon, so it certainly does even if the Schwarzschild observer can't see this." - Another logical flaw, universally common. At the singularity, what happens to the proper time? It is finite, yet it ends. So there is no obligation for the proper time to continue just because it is finite. The proper time at the horizon is finite and it ends there. Period. In no valid application of differential geometry can the horizon be crossed. This is mathematically forbidden despite any human intuition. Commented Oct 25, 2019 at 5:53
• @safesphere: According to your "logic" flat space in polar coordinates and flat space in Cartesian coordinates are different manifolds. Please actually pick up a book about differential geometry and learn about atlases and coordinate charts. Commented Oct 25, 2019 at 6:11
• @mmeent "flat space in polar coordinates and flat space in Cartesian coordinates are different manifolds" - You said it, I didn't - it is your "logic", not mine. You misinterpret what I've said. I also find your comment unfriendly, a shame for a physics website. Commented Oct 25, 2019 at 7:03
• @safesphere "This is a circular logic: r crosses the point of r=2M, because r crosses this point". you forgot the part where it is checked that r can be used as affine parameter. Commented Oct 25, 2019 at 12:31

A lot of issues with quantum field theory in curved spacetime can be finessed by instead looking at the classical limit involving many quanta: wave propagation in curved spacetime. You imagine a plane wave of, say, a classical scalar field impinging on a black hole and study what fraction of the wave’s energy is absorbed by the hole and how much is scattered in various directions. For an example of this kind of analysis, see this review.

In this sense: a Schwarzschild observer never sees the particle crossing r=2M.

Your reasoning includes a wrong conclusion, because the proper time of the infalling particle goes beyond the end of our time (!)

For this, you must be aware of the difference between "be seen by an outside observer" and "simultaneity from the point of view of an outside observer"

One instructive mean to show what happens around a black hole is the Kruskal metric. In the following diagram it is important to notice that the simultaneity lines of an outside observer are the radial lines t=1, t=2 etc.

As an example, particle A is infalling, and particle B is an outside observer whose worldline remains outside the event horizon. Following the radial lines you will see that according to the Kruskal spacetime diagram of an outside observer, the spacetime position of B will never be simultaneous with the point where A is crossing the event horizon.

A different question is if B will s e e A crossing the event horizon. The answer is not provided by the radial Kruskal lines but by the arrows which are simulating communication at speed of light between A and B. And you will see that B never will see A crossing the event horizon.

In conclusion, the outside observer does not only never see the particle falling in, but there is also no simultaneity. That means that, if it would be possible for a particle to cross the event horizon, this would happen after the end of our time, that means after the end of the (outside) universe.

• Good point. The event of crossing the horizon does not exist in any slice of physical time of a black hole spacetime. Commented Oct 25, 2019 at 6:17
• "In conclusion, the outside observer does not only never see the particle falling in, but there is also no simultaneity. That means that, if it would be possible for a particle to cross the event horizon, this would happen after the end of our time, that means after the end of the (outside) universe." This is because by that time in the outside observer's frame the BH has already evaporated. This is a contradiction, because the particle never crosses the horizon, but it evaporates too. Commented Oct 25, 2019 at 20:45