What are miniature black holes at a quantum level? What does string theory say?
Can you expand upon this please?Decay channels for one.
Micro black holes are made all the time at particle accelerator labs like Fermi-Lab
where they accelerate gold atoms to super-high speeds
there are no worries of the Earth being sucked up by the micro black holes that are constantly made
Thank you for the explanation.The decay of a black hole is very "democratic". All particles are produced in these decays, which would be a very striking signature. Additionally, a particle has a definite mass, but black holes will be produced with a range of masses.
For example: Can the fact that a muon DOESN'T decay like a black hole be taken as evidence that it is not a point particle?
And can the 'no hair' theorems be used to claim black holes only conserve energy, angular momentum, and electric charge ... so it doesn't conserve say color, lepton number, weak charge, etc. ?
But a black hole that small should decay much quicker than 2 microseconds.No, it can be taken as evidence that it's not a black hole. More specifically, it can be taken as evidence that it doesn't decay via gravity.
Thank you for your opinion. That sounds quite plausible.Not true either. The "no hair" theorem normally works under the approximation of astrophysical distances, and on these scales only two forces exist: gravity and electromagnetism. So it's proven assuming the strong and weak forces are irrelevant. This approximation is clearly not any good when talking about micro-black holes. I expect that a proper calculation would show that on these scales, color is conserved.
But a black hole that small should decay much quicker than 2 microseconds.
It seems like the community agrees that for an astrophysical sized black hole, Hawking radiation predicted by the standard model would NOT conserve lepton number for an outside observer.
There is the sense that if you feed a bunch of protons to a black hole, according to semiclassical reasoning you probably won't conserve say total baryon number, since a bh will radiate democratically.
Haelfix explained this, but your comment in reply seems to indicate you may have misunderstood or think that Haelfix and I disagree. If we are not in agreement that the current understanding is that black hole evaporation would violate lepton number despite all standard model interactions conserving lepton number, can you please cite a source that says lepton number IS conserved?Why do you think "the community" agrees with this? First, astrophysically sized black holes don't radiate. Secondly, as far as I know everyone agrees that at least statistically, all quantum numbers are conserved. Since this is a thermodynamic process, I don't know how to formulate this in any way except statistically.JustinLevy said:It seems like the community agrees that for an astrophysical sized black hole, Hawking radiation predicted by the standard model would NOT conserve lepton number for an outside observer.
... Each of these has a lepton number which must be conserved during fundamental interactions.
This basic principle of particle physics is cheerfully violated by the quantum black hole. We have already seen that when a black hole is formed or swallows matter it `loses its hair': all information about the particles is lost when they pass through the event horizon. In particular, a black hole formed from baryons (for example the protons and neutrons at the heart of a massive star) does not remember its baryon number. It could just as well have been formed from antibaryons, without our being able to tell the difference. Let us wait patiently. After a time the black hole starts to radiate according to Hawking's mechanism, releasing energy and entropy. Now the fact that the black hole radiates like a black body means that it can emit [on average] only equal numbers of baryons and antibaryons, or leptons and antileptons. In other words, the net baryon number leaving the evaporating black hole is always zero. The evaporation of the black hole violates the rule of conservation of baryon and lepton numbers.
No, I am not asking that at all. (For as already mentioned, particle decay is NOT like black-hole radiation.)Are you asking if particle decay is the same as black-hole radiation, since a point particle can be looked on as micro black-hole?
I don't understand where the self-interacting situation is coming from. I don't think anything I said required self-interaction.If you are thinking that a pointlike self-interacting mass of 106 MeV must be a black hole because r=0 is clearly less than the Schwarzschild radius, you will run into at least three problems. One is that experimentally "pointlike" means "smaller than we have been able to probe", which is somewhere around 10^-18 m, and the Schwarzschild radius of a muon is at least forty orders of magnitude smaller. Extrapolating over forty orders of magnitude is unlikely to be correct.
Second, if you allow a muon to interact with itself, you run into the problem of its electric charge interacting with itself long before you run into the problem of its "gravitational charge" interacting with itself. It takes an infinite amount of energy to cram that much charge into zero space.
That is a VERY good objection. I forgot about those limits. I checked and even for the Z boson which is uncharged, the spin alone easily makes the black hole non-existant 'super-extremal'.Third, the appropriate metric for a charged object is not the Schwarzschild metric. It's the Reissner-Nordstrom metric. Because all elementary particles have charges many orders of magnitude larger than their masses, they don't possesses horizons.