Four-Quark Hadron confirmed by CERN

My "high horse" is truth and not posting beyond my competence. In that I have never said you invented TQFT, I have limited myself to saying you lack the prerequisites to draw inferences from it and that use it more like a juju totem than the field of physics-inspired mathematics that it is. I welcome you to join me on the "high horse" that Clint Eastwood recommended:

 

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Phooey. The juju is peddling ephemeral patterns in the entrails as particles when they ain't. Those so-called tetraquarks last for no time flat. Then they decay into kaons and pions which don't last long either. Yes, charged pions last better than neutral pions, but what do they decay to? Muons which themselves decay, electrons, which can be annihilated to gamma photons, neutrinos which travel at c or thereabouts, and gamma photons. And where have the quarks gone? Poof, they aren't there any more. So don't you give me juju, rpenner. Not when we've never seen a free quark and gluons are virtual.

 

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Empty assertion. Quarks are Dirac fields, electronlike particles, and they overlap, like electrons in an atom.

The stability of the proton is due to it being the lowest-mass baryon and the lack of baryon-number-violating interactions, not some gibberish about knots.

A pentaquark is likely to be as stable as a tetraquark.
Tetraquark = 2 mesons = 2 ordinary quarks + 2 antiquarks
Pentaquark = meson + baryon = 4 ordinary quarks + 1 antiquark

Even if a pentaquark is stable against strong interactions, it will still be unstable under weak interactions, just like a meson.

Instability is totally irrelevant.
I repeat.
Instability is totally irrelevant.

Quarks have antiquarks, and mesons are composed of an ordinary quark and an antiquark. Many mesons can decay by annihilation of those two particles, meaning that those quarks are then gone.

 

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Only we've never seen a free quark. We've seen plenty of electrons, but never a free quark.

Baloney and you know it. How about telling us about the lowest-mass meson? Baryon number, pah. Talk about a non-answer.

A proton is a triquark. It's 3 ordinary quarks. You can't justify your assertion at all, because the pentaquark might be composed of ordinary quarks too.

What, you mean the weak interactions that are allegedly mediated by a particle weighing in at 80GeV or 91GeV? Like the weak interactions of Beta decay? You know, where a 939MeV neutron decays into a proton electron and antineutrino?

It is crucial.

They're always gone. Like I said, we've never seen a free quark. And everybody who knows about TQFT knows why.

 

Irrelevant. In fact, there's a whole theory about it: color confinement.

Me: The stability of the proton is due to it being the lowest-mass baryon and the lack of baryon-number-violating interactions, not some gibberish about knots.

Baryon-number conservation is not baloney. What about the lowest-mass meson?

Except that that would be contrary to quantum chromodynamics and color confinement, both of which have been very well-tested.

Me: pentaquarks being unstable under weak interactions

Yes. What did you think it is?

Me: instability is irrelevant

What tripe.

Except that you seem to be the only one who knows about this alleged solution to color confinement.

 

Enough, lpetrich.

 

Says the guy who has never used TQFT other than as a stage prop in his flim-flam show.

 

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hilarious.

 

As to a pentaquark being all ordinary quarks, that is contrary to color confinement. An all-ordinary-quark pentaquark would also have fractional electric charge, and there are very strong upper limits on that. Checking on Particle Data Group, it did not have much to say about searches for free quarks, other than to refer to reviews of searches of free fractionally-charged particles. Searching elsewhere, I found this 2009 review: Searches for Fractionally Charged Particles - Article-1.pdf There is absolutely zero observational evidence of fractionally charged free particles. Zero.

What does this tell us about the quark content of free particles?

It means that they must always be made up of combinations of quarks whose charges add up to integer charge. Ordinary quarks have charge (integer) - 1/3, and their antiparticles charge (integer) + 1/3. It's rather easy to show that one comes up with this sum rule:
(# ordinary quarks) - (# antiquarks) = multiple of 3

It is easy to show that an all-ordinary-quark pentaquark violates it.

Let's see what the possible symmetry groups are for quark color. The mesons don't tell us all that much, since they are quark-antiquark states, but the baryons are very revealing, since they are 3-quark states. There are no free quarks or free 2-quark states or free 4-quark states, so there must be something special about 3 that does not apply to 1, 2, or 4. If baryons are colored states, then that makes them not much different from other combinations of quarks, which may also be colored. But if baryons are colorless states and diquarks cannot be, we get a constraint on the possible symmetry group of color. No colorless diquarks means that quark color must be in a complex representation of the color symmetry group. Real and pseudoreal ones would allow colorless diquarks. That limits the possible symmetry groups to SU(n) for n >= 3, SO(4n+2), E6, and their products. Since baryons must be antisymmetric in color to compensate for their symmetry in spin and flavor, that means that we must consider which representations will have antisymmetric cubes that are colorless. Of these ones, only SU(3) remains. The group G2 has a rep that implies 7 colors, and that rep has an colorless antisymmetric cube, but it also has a colorless symmetric square, and we don't observe free diquarks.

So we conclude that the QCD symmetry group is SU(3), and that all free-particle states are QCD scalars/singlets. That successfully accounts for the aforementioned quark sum rule.

There is independent evidence of quarks having three colors. Electron-positron annihilation into hadrons in e+e- collider experiments. One can calculate the rate of particle production from e+e- -> virtual photon or Z -> quarks or muons; the muons are for comparison. Treating quarks as electronlike particles with the appropriate charges, particles that make hadrons, one finds that the hadron rate is 3 times what one would expect from colorless quarks.

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Gluons are the QCD analog of photons, and they have similar field equations. However, gluons have 8 color states, (color * anticolor) - colorless = 3*3-1, and they interact with each other. So an important test of QCD is interactions of gluons with quarks and with each other. One has to disentangle the strengths of these interactions from experimental results, usually jets of hadrons from high-energy collisions or decays of heavy-quark hadrons. One finds that gluons indeed have 8 color states and that gluon-gluon vs. quark-gluon interactions have the relative values predicted from SU(3). An interesting bit of evidence is the running of the QCD equivalent of the electromagnetic fine-structure constant, alpha-s. It's mainly measured from quark-gluon interactions, but its running is due to gluon-gluon interactions, so one gets a good test of (gluon-gluon)/(quark-gluon). Its value is 0.3 for an interaction energy of 2 GeV, 0.18 for 10 GeV, and 0.11 for 100 GeV, in good agreement with what one predicts from QCD.

As one can see, alpha-s gets very big for small energies, and one thus has to use lattice gauge theory to do calculations. One divides space-time up into a grid, and one then randomly generates field configurations, weighted by the theory's Lagrangian, a sort of summary of the equations of motion. It takes an enormous amount of number-crunching to get usable results, but it's been possible to get the proton's mass to within a few percent relative to some other quantities.

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So QCD makes lots of successful predictions, and we can thus count on it to predict that there will be no Farsight pentaquarks.