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## Main Question or Discussion Point

What are miniature black holes at a quantum level? What does string theory say?

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What are miniature black holes at a quantum level? What does string theory say?

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Haelfix

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Decay channels for one.

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Can you expand upon this please?Decay channels for one.

If some new particle shows up that the standard model didn't predict, then how can you constrain the decay channels?

Or is it something really clear, like blackholes don't conserve color or lepton number or something?

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Chronos

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Vanadium 50

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Vanadium 50

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None have ever been observed.Micro black holes are made all the time at particle accelerator labs like Fermi-Lab

Except that they don't. Perhaps you are thinking of RHIC.where they accelerate gold atoms to super-high speeds

That is correct.there are no worries of the earth being sucked up by the micro black holes that are constantly made

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Thank you for the explanation.

Can that reasoning be used to draw conclusions about the currently known elementary particles?

For example: Can the fact that a muon DOESN'T decay like a blackhole be taken as evidence that it is not a point particle?

And can the 'no hair' theorems be used to claim blackholes

If color isn't conserved, does this mean an up - anti-up pair could form near a black hole horizon and a single quark could escape? The black hole would be essentially neutral color, and so a single quark would be floating free trying to hadronize, but since color is conserved outside the black hole, it wouldn't be able to complete this process.

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Vanadium 50

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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.For example: Can the fact that a muon DOESN'T decay like a blackhole be taken as evidence that it is not a point particle?

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.And can the 'no hair' theorems be used to claim blackholesonlyconserve energy, angular momentum, and electric charge ... so it doesn't conserve say color, lepton number, weak charge, etc. ?

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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.

So it seems like it should be evidence of

Much was learned when the question of why an electron "orbitting" a proton is stable. It seems like

If anyone can suggest some good articles, please do let me know.

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.

Can someone that knows the details here comment about this?

It seems like the community agrees that for an astrophysical sized blackhole, Hawking radiation predicted by the standard model would NOT conserve lepton number for an outside observer. Yet every interaction in the standard model conserves lepton number. Conversely though, electric charge IS still conserved for an outside observer.

The problem is I don't know the details of these derivations, so on the face it seems arbitrary to me which quantities a black hole will still appear to conserve for an outside observer. Again, it seems plausible that color or weak-charge is conserved by the blackhole... but is it? How about other quantities we consider conserved in quantum mechanics?

If anyone is familiar with the literature on this, can someone point the way to a good review article about the spectra of blackhole evaporation?

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But what makes you think a muon is a black hole?But a black hole that small should decay much quicker than 2 microseconds.

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.

Third, the appropriate metric for a charged object is

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.It seems like the community agrees that for an astrophysical sized blackhole, Hawking radiation predicted by the standard model would NOT conserve lepton number for an outside observer.

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Haelfix

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However whatever it is that completes the unification paradigm probably *will* have their quantum numbers conserved (or at least some sort of holographic dual), that is, so long as quantum mechanics stays valid and the information loss problem is wrong as expected.

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I second the notion that a muon is not a black hole, hence does not behave as such.

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But as you point out, theThere is the sense that if you feed a bunch of protons to a blackhole, according to semiclassical reasoning you probably won't conserve say total baryon number, since a bh will radiate democratically.

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It's the other way around, isn't it? A muon doesn't behave like a black hole, so therefore it's something else.I second the notion that a muon is not a black hole, hence does not behave as such.

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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 blackhole 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 blackhole, Hawking radiation predicted by the standard model would NOT conserve lepton number for an outside observer.

A quick google search brought up this source, which uses the same logic Haelfix mentioned to argue lepton number would not be conserved:

Black Holes (authors Jean-Pierre Luminet, Alison Bullough, Andrew King)

... 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 blackhole. 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] onlyequalnumbers 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.

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No, I am not asking that at all. (For as already mentioned, particle decay is NOT like black-hole radiation.)

I'm sorry to repeat myself, but my questions were:

A black hole with the mass of an elementary particle should evaporate much quicker than the time scale of other interactions.

So it seems like this should be able to be combined with experimental evidence to give indications of

I also asked if the "no hair" theorems can be used to say electric charge and angular momentum are conserved in a black hole, but not any other quantum numbers? If not, then what quantum numbers are conserved?

-------------------------------------

The thrust of the question is this:

Currently in the standard model we consider elementary particles to be point particles (we can discuss creation and annihilation operators labelled by space-time coordinates). In semi-classical GR, a "point" mass is a black hole and evaporates violating lepton number. Even as weak as gravity is, the semi-classical result shows that the time scale for decay via gravity interactions is much much faster than decay via any other couplings. The overarching question is: What can we learn from this?

It seems to me that it suggests elementary particles are not point particles (our starting assumption was wrong) or that gravity interactions become much weaker on short scales (we learn from how the semi-classical application went wrong). If the case is the later, then how heavy of 'elementary particles' we can create and not see them decay through evaporation gives a limit on the scale at which gravity becomes essentially (assymptotically) free.

==================================

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.

Second, you seem to be missing the point of the questions here. I am not extrapolating experimental evidence to claim there is no structure experimentally down to the 'swarzchild radius'. If we take the particle to be pointlike, we get conclusions that don't match experiment. So this seems to suggest particles are NOT pointlike (maybe stringy, or as in LQC there is a finite minimum volume), or indicates gravity (which is already a 'weak' force) becomes even weaker on short scales.

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 isnotthe 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 possess horizons.

So let us instead focus on spin 0 uncharged particles. If we find a Higgs particle at the LHC (which seems very likely), experimentalists are expecting the most probable decay mode to be a bottom anti-bottom pair. Yet if it is a point particle, the time scale for gavitational evaporation will be much quicker (extrapolating the 'strength' of gravity coupling to be the same down to that scale).

So my question could be reformulated:

If they find the Higgs, and the decay matches standard model predictions, what can be concluded about the particle "size" or limits on gavitational strength interaction? i.e.

It seems to me we can learn a lot from this.

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