Can Black Holes Have Equal Spin and Behave as Indistinguishable Fermions?

[snorkack]

Well, do they?
Also: black holes, unlike elementary particles, have continuous distribution of rest mass, because they are free to absorb photons and gravitons of arbitrary mass, and kinetic energy of particles they capture.

If two Kerr or Newman black holes have equal spin, which happens to be a half-integer one, how close do their rest masses have to be for them to behave as indistinguishable fermions?

 

[rootone]

A black hole is black.
It absorbs all photons at every wavelength and emits none.

 

[newjerseyrunner]

None that you could see. Theoretically, black holes emit Hawking radiation, the wavelength of which is determined by the size of the black hole. According to wikipedia, the wavelength of the black hole's black body radiation is 16 times it's schwarzschild radius. So for the supermassive black hole at the center of our galaxy would radiate with a wavelength of 200million kilometers. So extremely low energy radio.

The smaller the black hole, the hotter it gets, so when a micro black hole evaporates, it emits a gamma ray. There is a conceivable size that for a moment a black hole might emit a photon in the visible spectrum.

 

[mfb]

A black hole with a mass of about 10¹⁹ kg would have its peak emission in the visible light range. At that mass, its remaining lifetime is roughly 1 second. Which also means its brightness would be comparable to the whole Milky Way at 10³⁶ W.
Well, obviously it would be visible before that, but with stronger infrared emission (compared to visible light) then.

It is unclear how black holes would fit to the quantum-mechanical framework. Their mass could be quantized in some way - we just don't know.

 

[Gaz]

Regardless of it being black and absorbing all light that passes its event horizon it should create a gravitational lens though right? Making it look like a bubble in space ?

 

[rootone]

Yes it should create a gravitational lens.
Actual gravitational lenses have been observed, but as far as I know they are attributed to large galaxy clusters rather than black holes though.

 

[mfb]

The "focal length" of such a black hole would be very short I think.
Deflection of light by black holes has been observed indirectly - in binary systems or even just to describe the accretion disk it is an important part for the description of the system.

 

[PeterDonis]

If two Kerr or Newman black holes have equal spin, which happens to be a half-integer one, how close do their rest masses have to be for them to behave as indistinguishable fermions?

You're mixing models. The Kerr-Newman black hole geometry is classical; there's no such thing as a "fermion" (or "boson" for that matter) in this model. A Kerr-Newman hole can have a continuous infinity of possible masses, and each one is different.

As mfb noted, we don't currently have a quantum theory of black holes, because we don't have a quantum theory of gravity. So even if we revise your question so it doesn't assume a classical model that isn't applicable, your question is not answerable at our current state of knowledge.

 

[snorkack]

If a quark falls inside the event horizon of a black hole, does the black hole acquire a colour which enables the now coloured black hole to be a source of a gluon chain and pull in the rest of the quarks and gluons of the hadron?

 

[newjerseyrunner]

You are confusing color with color charge. Color is a component of photons and only photons, it's it's frequency. Color charge is a component of quarks and gluons and has nothing to do with what it actually looks like.

As for pulling in quarks, I don't think so. I don't think quarks actually have a defined position within a hadron. So if you have a meson, I don't think you can be know that quark is on one side of the event horizon and one is on the other side. Don't quote me on that though, not totally sure. Definitely sure about the difference between color and color charge though.

 

[mfb]

Would probably need a quantum theory of gravity for a better answer. In practice, there is no way a part of a hadron could remain outside.

 

[Vanadium 50]

A black hole is a color singlet. Since there are no free quarks, a single quark cannot fall into one.

 

[vanhees71]

Well, theoretically you can have a black hole with color (in the sense of the charge of the strong interaction), and funnily enough GR + a nonabelian gauge theory has black holes with hairs, which is different from GR + Maxwell (abelian gauge theory), where black holes have no hairs according to a famous theorem discovered by Wheeler.

https://en.wikipedia.org/wiki/No-hair_theorem

 

[PeterDonis]

GR + a nonabelian gauge theory has black holes with hairs, which is different from GR + Maxwell (abelian gauge theory), where black holes have no hairs

The original "no hair" theorem didn't say no hair, period; it said no hair other than mass, charge, and angular momentum. Charge, of course, comes from GR + Maxwell, so, strictly speaking, GR + Maxwell does have hair. The hair in GR + Yang-Mills is basically the non-abelian gauge theory version of the GR + Maxwell hair; but AFAIK this Yang-Mills hair is not color charge, exactly, it's actually more like a "magnetic" charge than an "electric" charge.

 

[snorkack]

A black hole is a color singlet. Since there are no free quarks, a single quark cannot fall into one.

So if a hadron passes event horizon so that one quark winds out just inside the horizon, what does the rest of the hadron do?

 

[ChrisVer]

So if a hadron passes event horizon so that one quark winds out just inside the horizon, what does the rest of the hadron do?

even at the case quarks would matter, I guess it would split into several hadrons...

 

[snorkack]

even at the case quarks would matter, I guess it would split into several hadrons...

The problem is that any number of hadrons would have to be white. Which the part of the hadron left outside event horizon is not.

 

[Vanadium 50]

So if a magnet passes event horizon so that one pole winds out just inside the horizon, do I have a monopole?

 

[snorkack]

So if a magnet passes event horizon so that one pole winds out just inside the horizon, do I have a monopole?

Magnetic poles are not quite so assigned to specific particles.
Since a black hole can have angular momentum and electric charge, but no other hair, what happens to the dipole magnetic field of an electron when it falls in a black hole? Looks like the whole field with all field lines falls in the hole leaving nothing outside, right?

 

[Khashishi]

Black holes have a temperature and are the perfect black body emitter. The temperature and therefore color depends on its size. Oops newjerseyrunner already covered it.

 

[ChrisVer]

The problem is that any number of hadrons would have to be white.

We say colorless, rather than white... again things are not colors, but rather a quantum number or charge...

Which the part of the hadron left outside event horizon is not.

What I referred to is called hadronization. If you, by any means, try to separate the quarks of the hadron, you give the system potential energy. At some point the system will have enough energy to rather produce two (or more) hadrons than keep adding to its potential energy. The final hadrons are still color-neutral.
Since at the horizon I don't see any reason for quantum gravitational effects to somehow alter what we know of particle physics (in particular QCD), I am assuming that what I say is legit.

 

[nikkkom]

I believe macroscopic objects can have color charge (my favorite example is a macroscopic lump of quark-gluon plasma freezing out), but they are hardly realizable in practice, and even if they are, you'd get a quite mundane macroscopic picture: say, if you have two lumps of matter, each with one unit of color charge ("one of proton's quarks magically teleports from one rock to a nearby one"), they will just be attracted with a force of about 100 Newtons. Considering that this force affects nucleons, you'd end up with a quick sequence of pions generated between these two objects, balancing the changes. (Energy to create pions comes from the fact that a state with color-charges separated by a large distance has higher energy than pions' rest mass).

 

[PeterDonis]

I believe macroscopic objects can have color charge

Not according to our best current theory--at least, not at temperatures that are achievable in any of our current or foreseeable experiments.

one of proton's quarks magically teleports from one rock to a nearby one

This violates the laws of physics.

pions

The actual force carrier for the strong interaction is the gluon, not the pion. The pion theory is an approximation only, and since pions are colorless, color can't be transferred from one object to another via pion exchange anyway.

 

[PeterDonis]

At some point the system will have enough energy to rather produce two (or more) hadrons than keep adding to its potential energy.

And just to give some numbers, this point is a separation of the same order of magnitude as the size of a proton. So on any size scale larger than that one would not be able to observe free quarks.

 

[nikkkom]

Not according to our best current theory--at least, not at temperatures that are achievable in any of our current or foreseeable experiments.

I give you a ring of bulk-color-neutral quark-gluon plasma 1 light year in diameter. It freezes out and forms hadrons. How can it form hadrons in a way that all quarks everywhere along the ring "pair" perfectly into colorless triples? Knowing how to do that correctly requires quarks to have instantaneous communications.

 

[PeterDonis]

I give you a ring of bulk-color-neutral quark-gluon plasma 1 light year in diameter.

By "bulk color-neutral" you mean, I assume, that the entire plasma, globally, is color neutral, but individual local regions of it need not be, correct?

Assuming this is the case, the plasma must, of course, be at a high enough temperature that quarks are not confined. Otherwise it would be impossible to have any local regions that were not color neutral.

It freezes out and forms hadrons. How can it form hadrons in a way that all quarks everywhere along the ring "pair" perfectly into colorless triples?

Because the strong interaction between the quarks gets stronger as the temperature drops. As it gets stronger, the interaction between quarks drives them to distribute themselves through the plasma in a way that is locally color neutral. (Note that, as the interaction gets stronger, this process can involve new quark-antiquark pairs being created locally; it does not necessarily require quarks to move large distances through the plasma.) Then they form bound hadrons as the temperature drops still lower.

Knowing how to do that correctly requires quarks to have instantaneous communications.

No, it doesn't. If the plasma is 1 light-year in diameter, it could take at least a year for it to undergo the process, if it were necessary for quarks on opposite sides of the plasma to interact (which it might not be, see above). There is no requirement that the process be instantaneous.

I should emphasize, btw, that quark confinement and hadronization are not very well understood, nor is the transition from free quarks to confined quarks as the temperature drops. What I've described above is heuristic.

 

[nikkkom]

Because the strong interaction between the quarks gets stronger as the temperature drops. As it gets stronger, the interaction between quarks drives them to distribute themselves through the plasma in a way that is locally color neutral.

Without instantaneous communication, this can't be done so that plasma is totally color-neutral everywhere. There will be "leftovers" which can't be "paired up" (tripled up?) correctly.

An example from electromagnetism. Here's hydrogen plasma (protons and electrons):

...pepepepe...<HUGE DISTANCE>....pepepepepepepepepe...

Now pair them up, _perfectly_ (no leftovers) into neutral atoms. Without communication between two spatially-separated regions shown.

At the left, should it be

...[pe][pe][pe][pe]...

or
...p][ep][ep][ep][e... ?

Both are "locally correctly paired", sure, but only one will match the pairing chosen in the "right" region. Else, you end up having one unpaired charged particle between these two regions. Clearly, globally correct pairing *can't be known* without looking at what's going on in the "right" region.

If the plasma is 1 light-year in diameter, it will take at least a year for it to undergo the process. There is no requirement that the process be instantaneous.

Plasma in my example is not a ball 1 ly across. It's a thin ring (say, 1 proton diameter) 1 ly in circumference, kept at the temperature when it's still plasma. Then the machinery which keeps it hot is switched off along entire ring at once. (I chose ring because in this configuration there are no "ends" and there can't be argument "it will decay from ends towards the middle").

 

[PeterDonis]

There will be "leftovers" which can't be "paired up" (tripled up?) correctly.

Not if extra quark-antiquark pairs are created as needed. Or if pairings (triplings) switch as necessary to adjust (see below).

Both are "locally correctly paired", sure, but only one will match the pairing chosen in the "right" region.

And that means the pairings will switch around as needed to match up the regions. There's no need for an electron or proton to travel all the way from one end to the other.

For example, if at some point you have one chain ...pepepe trying to match up with another chain epepep..., the two electrons at the ends will repel each other but attract the protons next along, so a swap will be induced, which will then propagate along one of the two chains until it meets another mismatch, at which point it will fix that mismatch up.

globally correct pairing *can't be known* without looking at what's going on in the "right" region.

There's no need for the globally correct pairing to be known all at once. It can be "figured out", locally, by the process described above. (Bear in mind, also, that these are not really classical point particles but quantum fields, and the process of condensation described in quantum field language is not at all the same. So, as I said before, all this classical-style description is just heuristic anyway.)

It's a thin ring (say, 1 proton diameter) 1 ly in circumference, kept at the temperature when it's still plasma. Then the machinery which keeps it hot is switched off along entire ring at once.

Ah, I see, this is a deliberately constructed experiment, not a "natural" configuration. So the experiment must have started from a condensed state that was locally color neutral everywhere, and it must have somehow converted it into the high-temperature plasma state where there are local regions which are not color neutral. Whatever process did that, it has an inverse, which takes the plasma state back to a condensed, locally color neutral state. If that were not the case, there would be no way for your experiment to create the plasma state you describe in the first place.

 

[vanhees71]

This is a very hard problem in contemporary heavy-ion physics, and one must admit that there's so far not a completely satisfactory answer to the hadronization problem yet, i.e., how to dynamically describe the hadronization of the rapidly evolving fireballs of partonic matter created in ultra-relativistic heavy-ion collisions at CERN (SPS and LHC) and RHIC.

 

[ChrisVer]

I don't know, what would be the average speed [or energy] of the proton once it reaches the Schwarzschild's radius?
I mean should you consider what is going on there as "ultra relativistic" or in other words is the region there allowing for free quarks or for confined?

 

[mfb]

Ultrarelativistic in which frame? Confinement does not depend on the speed of protons.

 

[ChrisVer]

Confinement does not depend on the speed of protons.

it has to depend on the energy or temperature, if the QGP can exist...

 

[mfb]

Energy in the center of mass frame, sure. An isolated proton is never a quark-gluon-plasma, no matter how fast it is relative to you. Why should your reference frame matter at all? It is arbitrary.