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At your usual black hole size, you can simply shoot particles past the event horizon and make it effectively grow. However, once you get to the scales of an atom, your “particles” are larger, or at least comparable in size to this hypothetical nanoscopic black hole.

Ignoring the fact that such a black hole would likely evaporate extremely quickly, would it even be possible for such a black hole to gain mass?

Since we are now working at the scales where matter is better described by waves rather than discrete particles, could a black hole effectively “suck energy” out of this waveform, despite it’s event horizon being smaller than the size of the particle we observe in the macroscopic world, and effectively gain mass? Would extremely energetic gamma rays be a way to feed it?

Or is this question unaswerable since we do not have a verified theory of quantum gravity yet?

Qmechanic
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skytak picus
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2 Answers2

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This problem is correctly described as a "scattering" problem. Your subatomic particle, described by its wave equation, approaches the black hole with some "impact parameter" which corresponds roughly to its closest approach in the absence of the scattering effect. There is a probability that the incident particle is absorbed, and a probability distribution for its final direction if it survives the encounter. The scattering distribution could be thought of as a diffraction pattern if you could arrange a coherent source of scatterers.

You don't have to invoke quantum gravity. A femtometer-scale black hole has a mass of $10^{13}$ kilograms, so the spacetime outside the event horizon shouldn't show any Planck-scale effects.

Because it's continuum relativity, rather than an unknown quantum relativity, wave scattering is not unique to tiny black holes. An astrophysical black hole will diffract radio waves whose wavelengths are long relative to its Schwarzchild radius. Consider the 60Hz radiation emitted by your power lines (50Hz in some parts of the world), with wavelength comparable to the radius of the Earth. Stellar-mass black holes are city-sized, and would have a tiny probability of absorbing an incident photon with this wavelength.

If the momentum transfer with a scattered particle were large enough, you could extract energy from the black hole via "superelastic" scattering. Think of the mostly-correct statement that quarks can't be isolated because the energy required to separate them is more than the energy required to create a quark-antiquark pair, and imagine doing this "stretching" with the tidal interaction near a black hole. This is related to, but not the same as, Hawking radiation: Hawking radiation is the same idea, but without any incident scatterers.

Think of superelastic scattering as a way for the black hole to exchange heat with its environment. It's counterintuitive because a black hole (like any system bound by gravity) has a negative heat capacity, so it gets hotter as heat leaves it.

A nanometer-scale black hole has a Hawking temperature of about $10^{5}$ kelvin. Embedded in cold matter, such a black hole should transfer heat into the medium via superelastic scattering, speeding its Hawking evaporation. In a hot medium, on the other hand, such a black hole should absorb energy, making it more massive, colder, and a better absorber. The "equilibrium," where the black hole and the environment have the same temperature, is not stable; the black hole will, on average, grow or shrink.

rob
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The wavelength of an electron is not the size of the electron. An electron is point like. The wave function (squared) is the distribution of probabilities of where you might observe an electron.

For a proton or neutron, the size is roughly the size of the "orbits" of the point like quarks in it.

You might think of a photon and a pinhole smaller than the wavelength. There is a probability of finding the photon on the other side.

mmesser314
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