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- Thread starter ensabah6
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- #2

mathman

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Since pions are made from quarks, presumably there are partners made from the quark partners.

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With *minimal* susy every particle has a superpartner (ignoring susy singlets). With extended susy there are more partners for a given particle. If we just consider minimal right now, your question is about a composite particle of quarks (and gluons), which obey fermi-dirac statistics. The super-partners of spin-1/2 particles obey bose-einstein statistics since they are either scalar or vector bosons. Therefore, there aren't smeson partners of the mesons. However, the partners of bosons such as the photon are fermions so that you *could* form composites of these; but superpartners have the same quantum numbers (other than spin) as their partners. The partners of photons will not carry charge. Consider instead the gluinos (spin-1/2 partners of gluons). They obey fermi-dirac statistics and carry color charge, though they are in the adjoint representation like their gauge vector partners. Group theory tells us that two spin-1/2 particles in the adjoint representation of SU(3) (the color group) can form a color-singlet and therefore can occur as a composite. Similar arguments hold for SU(n) gauge multiplets in general: Two spin-1/2 gauginos in the adjoint representation of SU(n) can form a neutral composite. You can move on to composites of more than 2 fermions, checking whether there is a singlet in the product of representations.

Extended susy doesn't really change the situation, except each particle has more superpartners.

- #4

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The super-partners of spin-1/2 particles obey bose-einstein statistics since they are either scalar or vector bosons. Therefore, there aren't smeson partners of the mesons. However, the partners of bosons such as the photon are fermions so that you *could* form composites of these; but superpartners have the same quantum numbers (other than spin) as their partners.

That would imply that the lightest squark is a free stable particle with fractional charge. I'd expect that to be an easily detectable signature in cosmic ray detectors.

- #5

Vanadium 50

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That would imply that the lightest squark is a free stable particle with fractional charge. I'd expect that to be an easily detectable signature in cosmic ray detectors.

Why? The lightest squark can decay via [itex]\tilde{q} \rightarrow q + \tilde{\chi}}^2[/itex].

- #6

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Let me give another answer to the original question while I'm at it.

Since the original question is about a super-pion (meson), we can ask: In the effective field theory in which pions appear, are there superpartners for them? They would have to be effective fields with spin-1/2 and the same quantum numbers otherwise. The only other elementary particles to do this in the underlying QCD would be by using gluinos (partners of the gluons) to form spin-1/2 composites. But since gluinos are electrically neutral (like gluons) these can't be superpartners of the pions (anyway, we shouldn't have expected that composites from different supermultiplets would be superpartners).

- #7

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Why? The lightest squark can decay via [itex]\tilde{q} \rightarrow q + \tilde{\chi}}^2[/itex].

If the lightest squark is free to begin with, then it's either prohibited to decay into the regular quark by QCD, or it decays into a free quark, which have never been observed (I'm not quite certain on this).

Since the original question is about a super-pion (meson), we can ask: In the effective field theory in which pions appear, are there superpartners for them? They would have to be effective fields with spin-1/2 and the same quantum numbers otherwise.

I don't think that was the question... The question was whether superpartners of quarks form composites such as mesons and baryons. Your answer implied that they can't, therefore they are free, therefore see my objection above.

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Hadronization never really occurs for a single quark, due to local conservation of color it always involves all initial "jets". If we produce squarks, we will produce them in pairs, and the hadronization for their decay quarks will involve presumably all "jets" as well.If the lightest squark is free to begin with, then it's either prohibited to decay into the regular quark by QCD, or it decays into a free quark, which have never been observed (I'm not quite certain on this).

- #9

Vanadium 50

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If the lightest squark is free to begin with, then it's either prohibited to decay into the regular quark by QCD, or it decays into a free quark, which have never been observed (I'm not quite certain on this).

The lightest squark is not free; it will always be confined. (It may decay before it hadronizes, like the top quark, of course, if it's lifetime is small enough. It would bind with an antiquark to produce what is conventionally called an R-hadron: [itex]\tilde{q}\overline{q}[/itex].

It will either decay as I described above, if something called R-parity is conserved (essentially, requiring a conservation of "supersymmetricness") or if it is violated via [itex]\tilde{q} \rightarrow qq\overline{q}[/itex].

- #10

arivero

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The answers in the thread, combined, reflect the orthodoxy. Now I sustain, since 2006 or so, an heterodox view:

So, if we produce the mesons from global SU(5) flavour, there are 24 mesons. Of these 12 are neutral, 6 have charge +1, and the other 6 have charge -1. So it is not only that the mass of the pion is near of the muon; it also happen that the number of degrees of freedom of the leptons is the same that the number of mesons, charge by charge. It smells SUSY.

A priory, it is not trivial that the same thing is going to happen with quarks. There is no interaction changing leptons to quarks, so the fact of being able to order the three generations of leptons in an SU(5)_f representation, and survive it to SU(3)_f x SU(2)_f , does not imply that we are going to be able to do the same with quarks. I could be wrong here, and it could be that I am missing some obvious -say- argument.

Still, a naive counting for quark shows that we have 6 diquarks for -2/3, 6 diquarks for +1/3, and, ahem, 3 diquarks from +4/3. So we have again the three generations of quarks plus three generations of an exotic object which can not be arranged to partner a dirac fermion. It seems the partner of a majorana, but it is not neutral.

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- #11

arivero

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Still, a naive counting for quark shows...

A detail. If you do not like naive counting, you can check Slansky

http://ccdb4fs.kek.jp/cgi-bin/img/reduced_gif?198102061+201+265 [Broken]

(Note we are using flavour groups. It is coincidental that they look as gauge groups in a GUT. In any case, SU(5) GUT proceeds differently, via the 10 and the 5)

[tex]SU_5 \supset SU_2 \times SU_3 \times U_1 [/tex]

[tex]15=(3,1)(6)+(2,3)(1)+(1,6)(-4)[/tex]

[tex]24=(1,1)(0)+(3,1)(0)+(2,3)(-5)+(2, \bar 3)(5)+(1,8)(0)[/tex]

For leptons, all of them, the 24 is produced from 5x(\bar 5)=1+24. For antiquarks, we produce the 15 from 5x5=10+15, and same for quarks.

Note that the difference in U(1) charge is the same, -5-0= -4-1, thus the connection between this U(1) label, q, and electric charge, Q, is Q= 1/5 (q + C), where C is 0 for leptons and, hmm, 2/3 for quarks. So C is proportional to barion number.

It is very tempting to think that the scheme can be promoted to a theory of strings, with groups acting in the extremes of an open string. Some lore (Marcus-Sagnotti) on world-sheet fermions says than the group is promoted from SU(5) to SU(2^5) and then the physical meaning, as flavours, is lost. Besides, this 2^5 happens to coincide with 2^D/2, coming from the total dimension of space. So it is not easy to sketch a coherent research path across this panorama.

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- #12

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it also happen that the number of degrees of freedom of the leptons is the same that the number of mesons, charge by charge

Numbers seem to coincide. To complete the argument it would be nice to provide some kind of mapping between 24 leptons to 24 spin-zero mesons.

You have six negatively-charged leptons:

- left and right electron (0.5 MeV)

- left and right muon (140 MeV)

- left and right tau (1.8 GeV)

On the other hand, you have six negatively charged spin zero mesons from your SU(5):

[tex]\pi^- (d\bar{u})[/tex] 140 MeV

[tex]D^- (d\bar{c})[/tex] 1.9 GeV

[tex]K^- (s\bar{u})[/tex] 500 MeV

[tex]D^-_s (s\bar{c})[/tex] 2 GeV

[tex]B^- (b\bar{u})[/tex] 5.3 GeV

[tex]B^-_c (b\bar{c})[/tex] 6.3 GeV

Should there be some kind of natural one-to-one mapping between these?

- #13

arivero

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Should there be some kind of natural one-to-one mapping between these?

Good question. If there is SUSY, it is obviously broken. I have been thinking about the one-to-one map, but I would like to hear suggestions from the audience.

- #14

arivero

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1) if we restaurate the SU(2) "up flavour" symmetry, letting c->u, we have three masses:

[itex]

\pi^- (d\bar{u}) D^- (d\bar{c})

[/itex] 139.5 MeV

[itex]

K^- (s\bar{u})D^-_s (s\bar{c})

[/itex] 493.7 MeV

[itex]

B^- (b\bar{u}) B^-_c (b\bar{c})

[/itex] 5279 MeV

2) If we restaurate SU(3) "down flavour", letting s,b -> d, we have two masses

[itex]

\pi^- (d\bar{u}) K^- (s\bar{u}) B^- (b\bar{u})

[/itex] 139.5 MeV

[itex]

D^- (d\bar{c}) D^-_s (s\bar{c}) B^-_c (b\bar{c})

[/itex] 1870 MeV

1 and 2) In both cases, we need to compare with the masses of the leptons: electron, muon, tau, respectively 0.511, 105.66 and 1777 Mev

In the case 2) we have two hits and we could think of some combination to justify the masslessness of electron. In the case 1) I wonder if we could have still some pattern coming from spontaneus susy breaking.

Brannen (http://www.brannenworks.com/koidehadrons.pdf ) could point out that the triplet (0, 139.5, 1870) still meets Koide's. And I point out that the original derivation of Koide's was in a model of composite particles.

- #15

arivero

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it would be nice to provide some kind of mapping

Well, from all the above, I think the map should be:

tau --> the pion and eta8 mixing of [itex]

D^- D^-_s B^-_c

[/itex]

mu -> the pion and eta8 mixing of [itex]

\pi^- K^- B^-

[/itex]

electron -> the eta1 mixing of [itex]

D^- D^-_s B^-_c

[/itex] and the eta1 mixing of [itex]

\pi^- K^- B^-

[/itex]

The mass term of the mixing should be a "democratic" one generating two zero eigenvalues (for pion and eta8) and an negative eigenvalue (for eta1) proportional somehow to the mass of the "upside" quark (u or c), some kind of "binding energy".

Furthermore, there is probably hint of chirality: the leptonic release of a W- particle should correspond to a change from "antiup" to "antidown" or to a change from "down" to "up". In the later case we can reach four neutral states (the (1,1) and the (3,1) in post #11) In the former case, we can reach eight neutral states (the (1,8)(0)), Nature could use chirality to solve this asymmetry. Perhaps.

- #16

arivero

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I have upload part of the arguments of this thread to arXiv:0910.4793 for book-keeping :-)

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