Do they know how many types of quarks exist?

Abidal Sala

Do physicists know exactly how many types of quarks exist? does this allow them to predict some particles should exist like Higgs Boson?

phyzguy

It is thought that the 6 quarks that we know about are all that there are. The strongest constraint that I know of is that of the "invisible" Z boson decays. This is where the Z-boson decays to a pair of neutrinos. The width of this decay strongly suggests that there are only three families of neutrinos, and since the leptons come in families with the quarks, that there are only three families of quarks.

phinds

You might find it informative to look up the Standard Model. Most links will give a comprehensive list of all known particles including those phzguy discussed

tom.stoer

In addition to the number of quarks the number of leptons and their electro-weak charges are known. That means that when a sevenths quark is found there must be an 8th quark plus a heavy electron-partner (like the myon) plus an additional neutrino (to complete the fourth generation) as well; an incomplete generation causes quantum anomalies i.e. the model would be inconsistent mathematically.

mfb

4th generation particles (in addition to the known 3 generations) are one of the things the LHC looks for. However, this 4th generation would need some strange properties, like a very heavy neutrino.
Using invisible Z decays, the number of light neutrinos was determined to be 2.98 +- 0.07(stat) +- 0.07(sys), in good agreement with 3 (publication).

seerongo

Maybe this thread is a bit old, but here goes.

What would the discovery of a 4th generation of quarks do to the Standard Model? It's my understanding (non-physicist) that three generations fully satisfies SM. Is there even room for a 4th? And, if a 4th generation of quarks is found, would that mean that there must be a 4th for all the other leptons as well?

lpetrich

The additional generations of quarks would not be alone. There would have to be some additional particles to make the Triangle Anomaly cancel. I'd mentioned that in an earlier thread: Empirical tests of quark charges, any info?

The triangle anomaly is the result of a chiral-fermion loop that gauge particles couple to. Chirality = handedness (left vs. right). The anomaly only disappears if their couplings to the gauge particles cancel. That's essentially
Tr({T_a,T_b,T_c}) (left - right)
T = gauge operator on the fermions for particles a, b, c

The unbroken Standard Model has gauge symmetry SU(3)_QCD * SU(2)_WIS * U(1)_WHC
QCD = quantum chromodynamics
WIS = weak isospin I
WHC = weak hypercharge Y

QCD multiplets can be denoted by their Lie-algebra highest-weight vectors: (w0,w1). Their multiplicity is (1/2)(1+w0)(1+w1)(2+w0+w1), and their operator powers are
1: 0
2: (3*w0 + 3*w1 + w0² + w0*w1 + w1²)
3: (1/2) * (w0 - w1) * (9 + 9*w0 + 9*w1 + 2*w0² + 5*w0*w1 + 2*w1²)

Weak isospin behaves like 3D angular momentum, with multiplicity 2I+1 and operator powers 1, 2, 3: 0, I(I+1), 0

Weak hypercharge behaves like electric charge - it's the average charge of a multiplet - and operator powers 1, 2, 3: Y, Y², Y³

Electric charge Q = I₃ + Y, where I₃ = -I, -I+1, ..., I-1, I

The anomalies that don't automatically cancel:
(QCD)³
(QCD)² * (WHC)
(WIS)² * (WHC)
(WHC)³
(WHC)

Let's see how the Standard-Model chiral fermions stack up. Each generation is:
Left up+dn: (3 = (1,0), 1/2, 1/6) -- 6 * (10, 4*(1/6), (3/4)*(1/6), 1/216, 1/6)
Right up: (3* = (0,1), 0, -2/3) -- 3 * (-10, 4*(-2/3), 0, -8/27, -2/3)
Right dn: (3* = (0,1), 0, 1/3) -- 3 * (-10, 4*(1/3), 0, 1/27, 1/3)
Left nu+el: (1 = (0,0), 1/2, -1/2) -- 2 * (0, 0, (3/4)*(-1/2), -1/8, -1/2)
Right nu: (1 = (0,0), 0, 0) -- 1 * (0, 0, 0, 0, 0)
Right el: (1 = (0,0), 0, 1) -- 1 * (0, 0, (3/4)*1, 1, 1)

Quarks alone: (0, 0, 3/4, -3/4, 0)
Leptons alone: (0, 0, -3/4, 3/4, 0)
Both: (0, 0, 0, 0, 0) - cancellation!

So to cancel anomalies, there must be some additional particles to go along with those additional quarks, and additional leptons would fit without trouble.

seerongo

Thanks for the explanation, also the link to the other thread. I'd always heard that any new generation would have to involve the other leptons, but didn't know why. Is there anything else other than anomaly cancellation, which would be satisfied by adding the additional particles, that might make SM choke on a 4th generation?

What's the consensus, if there is one? Good chance for a 4th generation, or probably not, or "no way"?

mfb

If you ask scientists directly involved in the searches or some theoretic physicists working in this area, they will probably tell you that chances are good to find a 4th generation.
But apart from that, I think the usual expectation is that there are just 3 generations. The Z decays are very convincing. 3 neutrino generations in the range of meV and a 4th generation heavier by at least 12 orders of magnitude?
I would say "problably not".

seerongo

Wow, I would think they would be extremely unstable. Any models to suggest decay times on such things?

mfb

What would be unstable?
A 4th-generation neutrino would be stable, as long as it is lighter than the 4th-generation equivalent of the electron. I don't know how neutrino mixing would look like...

seerongo

Okay. I guess I was just thinking of the electron - muon - tau, each generation more unstable than the other, but maybe that doesn't necessarily follow.

Vanadium 50

Where do you get this?

Vanadium 50

The SM, by definition, has three generations.

If you try and put in a (complete - as described above, an incomplete one breaks the theory) 4th generation and nothing else you will run afoul of indirect measurements. (Including the Higgs). So you need to either make this quark behave differently than the others, or you need to add in some additional new physics to undo what the new quark has done to these measurements.

lpetrich

Another limit is from the strength of the coupling to the Higgs particle. That cannot be greater than something around 1, or else one violates unitarity limits. One can get more particles coming out than going in. From [1007.0043] Higgs Mass Constraints on a Fourth Family: Upper and Lower Limits on CKM Mixing, the unitarity limit is about 500 GeV. The LHC's detectors ought to see lots of evidence of such particles -- if they exist.

Another theoretical argument is from Grand Unified Theories - colored particles must be accompanied by colorless ones; quarklike ones by leptonlike ones. That does not constrain the number of generations, but instead the content of each generation.

Georgi-Glashow SU(5):
1 (0000) L = (R nu)*
5 (1000) R = (L nu+el)* + (R dn)
10 (0100) L = (L up+dn) + (R up)* + (R el)*
10* (0010) R = (L up+dn)* + (R up) + (R el)
5* (0001) L = (L nu+el) + (R dn)*
1 (0000) R = (R nu)

Pati-Salam SU(4)*SU(2)*SU(2):
8 (4 = (100), 1/2, 0) L = (L up+dn) + (L nu+el)
8 (4* = (001), 0, 1/2) L = (R up)* + (R dn)* + (R el)* + (R nu)*
8 (4* = (001), 1/2, 0) R = (L up+dn)* + (L nu+el)*
8 (4 = (100), 0, 1/2) R = (R up) + (R dn) + (R el) + (R nu)

Both the GG and the PS multiplets have quarks and leptons side by side.

Those theories still have the problem of triangle anomalies not automatically canceling, but the simplest superset of both of them does have automatic cancellation: Fritzsch-Minkowski SO(10). It includes each generation of elementary-fermion multiplets in a 16 spinor multiplet, quarks and leptons side by side.

mfb

It is based on my observation that scientists working in area X tend to be confident to measure something interesting in area X - at least more confident than others. That is nothing wrong, and it does not imply causality in any direction. It is just something you should keep in mind if a theoretician tells you "my model is the best and LHC will confirm it".

AdrianTheRock

Why? What would prevent it decaying into a lower generation particle, given that we know the charged weak current mixes generations?