Picture an electron falling into an event horizon, so far from what I read gravity is either not a force or it is extremely weak therefore it doesn't cause wavefunction to collapse unless the strength of gravity is very strong like just after Big Bang or near the event horizon so is that true? Is this perhaps also how Hawking radiation is produced?
1 Answer
This question can be addressed either in a more QM-like manner up to a discussion involving Quantum Field Theory in Curved Spacetime. I'll break this answer in a few different sections due to this.
Quantum Mechanical Answer
There is no need to have a wavefunction collapse even if the interaction is strong. Notice that, for example, the electron in a hydrogen atom doesn't has its wavefunction to collapse just because it is interacting with the electromagnetic field. The wavefunction is different in the presence of electromagnetic or gravitational effects, but this does not mean it collapses.
Quantum Field Theory in Curved Spacetime Answer
It doesn't really make sense to talk about an electron close to a black hole. To define what we mean by a particle, we need to have some sort of natural notion of time that allows us to pick solutions of the wave equation ruling the evolution of the particle we're interested in and open these solutions with a Fourier transform to distinguish the modes with positive and negative frequencies (the former are associated with particles, the latter with antiparticles). This can be done in Minkowski spacetime because one can simply use the notion of time translation symmetry provided by the Poincaré group. However, there is no natural way of doing this close to a black hole. In other words, close to a black hole, it doesn't make sense to talk about particles. We can only talk about quantum fields, which is in fact all we need to do Physics.
Furthermore, the physical state of a quantum field is a global notion. The presence of the event horizon won't change its properties nor lead to any sort of collapse. Notice also that it is a bit more difficult to interpret Quantum Mechanics in terms of a wavefunction collapse once Relativity is taken into account (the collapse is instantaneous, but Relativity doesn't allow for instantaneous spread of information).
Hawking Radiation
The origin of Hawking radiation has nothing to do with wavefunction collapse. Instead, it is a consequence of what I mentioned that the definition of particles depends on a preferred notion of time, and in different moments one can have different preferred notions of time. I'll sketch the main ideas in here.
Suppose you have an observer looking at a star from a large distance. There is a natural notion of time because the spacetime can be taken to be stationary (at least in the far past). The star suddenly collapses (there is no time translation symmetry during this process, so we have no natural notion of time nor notion of particles) and turns into a black hole, which eventually stabilizes. Now we once again have a stationary spacetime and a natural notion of time. However, this is not the same notion of time we had to begin with, and as a consequence we have a different notion of particles. The quantum state we used to call vacuum when the star was stable we now perceive as being a thermal state filled with particles emanating from the black hole. It is all a consequence of the fact that the notion of particle is not a fundamental concept in physics. This does not lead to any paradoxes, because the state of the quantum field is independent of observer.
I have also discussed these issues about the Hawking effect in a couple of other posts. See, e.g., this, and this.
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$\begingroup$ “There would only be a collapse if there was a measurement” - What exactly is “measurement”? $\endgroup$ Jan 28, 2022 at 7:27
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$\begingroup$ @safesphere I edited that phrase out to avoid turning this question into something about the measurement problem $\endgroup$ Jan 28, 2022 at 7:30