Willmutt reminded me of a paper I saw in the morning,
At that time, the particle was conjectured to be so heavily bound that it would be a tachyon, \(m^2\lt 0\). I actually think that composite tachyons can't exist in tachyon-free theories, can they? (You better believe that such a tachyonic particle is impossible because such a man-made Cosmos-eating tachyonic toplet would be even worse than an Earth-eating strangelet LOL.)
The zodiac, a similarly strange bound state of 12 particles.
Unlike my numerologically driven weakly bound states of new particles, they propose that the particle could be a heavily bound state of 12 top quarks in total.
More precisely, they say that there should be 6 top quarks and 6 top antiquarks in the beast. The number 6 is preferred because all \(2\times 3 = 6\) arrangements of the spin-and-color are represented – both for quarks and antiquarks. So this complete list could potentially make a particle that is as stable as the atom of helium; or the helium-4 nucleus (the alpha-particle). The whole low-lying "shell" is occupied in all these cases!
The binding energy could come from the exchange of the virtual Higgs quanta. Note that for the odd messenger spins, \(J=1,3,5,\dots\), i.e. for electromagnetism, the like charges repel. For the \(J=2\) gravity, the like charges (positive masses) attract. For \(J=0\), the like charges must attract, too. A closer analysis of the signs in the Dirac fermionic bilinears implies that the opposite sources of the Higgs field actually attract as well – so the "sign of the top quark" is ignored. An ironic side effect of this rule is that when a top quark-antiquark pair is created, the total field they produce jumps discontinuously. But unlike the electric charge, the "charge sourcing the Higgs field" isn't conserved, so this jump isn't contradicting anything.
Twelve top quarks have the mass of \(12\times 173\GeV=2076\GeV\) so you need the interaction energy \(-1326\GeV\) to get down to \(750\GeV\). There are \(12\times 11/ 2\times 1=66\) pairs of "tops" (or antitops) in the proposed bound state. If each of them contributes \(-20\GeV\) in average, you will be fine. But do they contribute \(-20\GeV\) in such bounds states? Cannot someone just calculate these things, e.g. some lattice QCD methods? Cannot one see this \(-20\GeV\) in the toponium?
Both authors claim that \(pp\to SS\) where \(S\) is their 12-particle bound state has the cross section of 0.2 pb and 2 pb at \(8\TeV\) and \(13\TeV\), respectively, which seem good enough. The dominant decay modes should be (in this order) \(S\to t\bar t,gg,hh,W^+W^-,ZZ,\) and \(\gamma\gamma\). Given the low status of the diphoton, that doesn't look too good, does it? It is pretty hard to imagine how this complicated beast decays at all – twelve particles have to be liquidated almost simultaneously. That only occurs in some very high order, doesn't it? I am actually surprised by the high production cross section for the same reason.
But the simplicity makes the proposal attractive even if the absence of the Beyond the Standard Model physics could be disappointing at the end.
Production and Decay of \(750\GeV\) state of 6 top and 6 anti top quarksby two experienced physicists, Froggatt and (co-father of string theory) Nielsen, that proposes that the \(750\GeV\) cernette could be real – and it could be a part of the Standard Model. They've been talking about the bound state (now proposed to be the cernette) since 2003.
At that time, the particle was conjectured to be so heavily bound that it would be a tachyon, \(m^2\lt 0\). I actually think that composite tachyons can't exist in tachyon-free theories, can they? (You better believe that such a tachyonic particle is impossible because such a man-made Cosmos-eating tachyonic toplet would be even worse than an Earth-eating strangelet LOL.)
The zodiac, a similarly strange bound state of 12 particles.
Unlike my numerologically driven weakly bound states of new particles, they propose that the particle could be a heavily bound state of 12 top quarks in total.
More precisely, they say that there should be 6 top quarks and 6 top antiquarks in the beast. The number 6 is preferred because all \(2\times 3 = 6\) arrangements of the spin-and-color are represented – both for quarks and antiquarks. So this complete list could potentially make a particle that is as stable as the atom of helium; or the helium-4 nucleus (the alpha-particle). The whole low-lying "shell" is occupied in all these cases!
The binding energy could come from the exchange of the virtual Higgs quanta. Note that for the odd messenger spins, \(J=1,3,5,\dots\), i.e. for electromagnetism, the like charges repel. For the \(J=2\) gravity, the like charges (positive masses) attract. For \(J=0\), the like charges must attract, too. A closer analysis of the signs in the Dirac fermionic bilinears implies that the opposite sources of the Higgs field actually attract as well – so the "sign of the top quark" is ignored. An ironic side effect of this rule is that when a top quark-antiquark pair is created, the total field they produce jumps discontinuously. But unlike the electric charge, the "charge sourcing the Higgs field" isn't conserved, so this jump isn't contradicting anything.
Twelve top quarks have the mass of \(12\times 173\GeV=2076\GeV\) so you need the interaction energy \(-1326\GeV\) to get down to \(750\GeV\). There are \(12\times 11/ 2\times 1=66\) pairs of "tops" (or antitops) in the proposed bound state. If each of them contributes \(-20\GeV\) in average, you will be fine. But do they contribute \(-20\GeV\) in such bounds states? Cannot someone just calculate these things, e.g. some lattice QCD methods? Cannot one see this \(-20\GeV\) in the toponium?
Both authors claim that \(pp\to SS\) where \(S\) is their 12-particle bound state has the cross section of 0.2 pb and 2 pb at \(8\TeV\) and \(13\TeV\), respectively, which seem good enough. The dominant decay modes should be (in this order) \(S\to t\bar t,gg,hh,W^+W^-,ZZ,\) and \(\gamma\gamma\). Given the low status of the diphoton, that doesn't look too good, does it? It is pretty hard to imagine how this complicated beast decays at all – twelve particles have to be liquidated almost simultaneously. That only occurs in some very high order, doesn't it? I am actually surprised by the high production cross section for the same reason.
But the simplicity makes the proposal attractive even if the absence of the Beyond the Standard Model physics could be disappointing at the end.
Cernette: a bound state of 12 top quarks?
![Cernette: a bound state of 12 top quarks?]()
Reviewed by MCH
on
May 13, 2016
Rating:
No comments: