Quantum tunneling into a black hole

Question

The currently accepted answer to Throwing a micro black hole into the sun: does it collapse into a black hole or does it result in a supernova? states that a small black hole of mass approximately $10^9$ kg cannot accrete matter quickly, because radiation pressure would push particles away more strongly than gravity would pull them in. The argument given is that the micro black hole would exceed the Eddington limit. In other words, for a proton near the event horizon, the outward force due to radiation is

$$\frac{ L_\text{Hawking}\sigma_\text{Thomson} }{Ac} = \frac{\hbar c^6 \cdot \sigma_\text{Thomson} \cdot c^4}{M^2 15360 \pi G^2 \cdot 16\pi M^2 G^2 \cdot c} \approx 3 \times 10^{12} \,\text{N},$$

while (using the Newtonian approximation for simplicity), the gravitational force is

$$\frac{GMm_p}{R_\text{Schwarzschild}^2} = \frac{GMm_p}{(2GM/c^2)^2} = \frac{m_p c^4}{4GM} \approx 5 \times 10^7 \,\text{N}.$$

Thus, the proton is pushed away from the black hole rather than pulled towards it.

But since the proton's wave function is nonzero inside the black hole, isn't there a significant chance that the proton will simply tunnel into the black hole and never return?

I'm asking this question purely out of academic interest. I realize that it's irrelevant to the correctness of the answer linked above, since tunneling would presumably only affect a tiny number of protons near the event horizon. I also understand that giving a full answer would probably require a theory of quantum gravity. However, is there anything that can be said from semiclassical gravity alone?

Answer 1

First of all, the first equation used in the question is not applicable for the black hole mass under consideration. The temperature of Hawking radiation for $M=10^9 \,\text{kg}$ exceeds $10\,\text{GeV}$, therefore there would be a lot more particles being radiated besides photons: electrons, muons, pions, … and their corresponding antiparticles and since the number of species of particles would be large the overall luminosity would be dozens of times greater than the value in the OP. Also, cross sections of interactions with protons for some of the species would also be larger than for photons.

And you are right, it is possible that a proton with a nonzero wavefunction around the black hole horizon would tunnel into this blackhole. It's just the probability of such tunneling would be very small if we take somewhat realistic values for the conditions near this black hole.

Imagine a wavepacket (which would be the wavefunction of a proton) travelling toward the black hole. If we ignore for a moment Hawking radiation, the most likely outcome would be simply an elastic scattering with proton flying away without losing any energy. At the same time there would be a small but nonzero probability that the proton would be absorbed by a black hole directly. More likely however (but much smaller than probability of elastic scattering), if the kinetic energy of a proton is small enough, would be a probability for a proton to emit a photon and enter into a bound state around the black hole (much like an electron in a hydrogen atom). Once in a bound state proton could be directly absorbed by black hole horizon or could drop into a lower bound state by emitting another photon, and so forth. As the energy of bound states of proton lowers, the probability of absorption by a black hole in a unit of time becomes greater until the proton gets absorbed.

If we now include the quanta of Hawking radiation into consideration, then at each stage of the process outlined above we must include the probability that a proton will interact with a Hawking particle. Since these particles are flying away from the black hole, any interaction with them would almost certainly prevent the proton from being absorbed by the black hole. So by the time the original wavepacket of a proton will have the chance to enter the vicinity of a black hole where it could be absorbed or enter a bound state, it would be irradiated be so many quanta of Hawking radiation that it would be almost completely dispersed. Even if the bound state is formed (and remember, this is more probable outcome than direct absorption) it does not guarantee that the proton will be absorbed, since it is much more likely that the proton would become unbound (“ionized”) by Hawking radiation before it could be absorbed.