# Is there any evidence for a 4th family of particles?

1. Dec 18, 2003

### davelee

In the standard model, there are three families of matter particles, headed up by the electron, tau and muon, and their associated neutrinos, and quarks. Is there any reason to suspect the existence of a 4th or even 5th+ series of matter particles? (which, if they existed, would probably be nearly impossible to produce.)

In a similar vein, does anyone know if the tau/tau neutrino/top/bottom quark family predicted to exist, or was one of them found by "accident" first?

Finally, I don't quite understand the technical details, but it appears that neutrinos may actually be an oscillating superposition of electron, tau and muon neutrinos. Is it possible that something similar could happen with the electron, tau, muon, or is the neutrino oscillation specific to how it interacts only via the weak force?

2. Dec 19, 2003

### jcsd

If you want to divide everything up into massive leptons, neutrinoes and quarks, then yes you do find there is another family of particles that aren't included: gauge bosons which are force carriers like the photon or gluon.

I cvan't rember, but I'm pretty sure that the top quark was predicted by the standard model rather than being detected.

In the standard model electrons, tauons and muons can be thought of as the same particle in different states as their main differnce is obviously their mass. I'm not quite sure how the mechanism works but I'm pretty certain that for neutrinoes there is an uncertainty relationship between their mass and their flavor, but I've never heard of a simlair uncertainty relationship between the flavour and the mass of the massive leptons.

3. Dec 19, 2003

### GRQC

The gauge bosons are not generally considered a "family" of particles in the same sense as the three flavor/generations of leptons:

(e,nu_e), (mu, nu_mu), (tau, nu_tau)
(u, d) (c, s) (t, b)

The Standard Model also predicts the Higgs boson, which is believed to contribute to the origins of mass (what's called symmetry breaking). This has yet to be discovered, but they're currently looking for it at CERN.

There are in theory more families of leptons. Many papers will discuss a fourth "sterile" neutrino flavor to solve their problems (e.g. with neutrino oscillations). Whether these are for real or not is up in the air.

Pretty much all three quarks above charm were predicted and later discovered. The (somewhat shakey) evidence for the top quark was published in about 1996 from the FNAL-D0 and CDF groups.

Neutrinos can undergo flavor oscillations because they have definite states in two "spaces" (mass eigenstates and flavor eigenstates). Bottom line is that the can oscillated into a different type of neutrino when passing through dense matter (hence the presumed origin of the solar neutrino problem). Leptons aren't believed to display the same type of oscillation behavior (but quarks do).

The possibility also exists that neutrinos can have gravitational eigenstates as well. That is, each family of leptons could couple to gravity differently (and violate the equivalence principle).

4. Dec 19, 2003

### marcus

hello GRQC, and welcome here at PF
habitual visits to arXiv
hope you post lots of good links thereto

marcus

5. Dec 19, 2003

### mormonator_rm

6. Dec 22, 2003

### suyver

I seem to remember that there are only three families of quarks: u,d ; c,s ; t,b. I vaguely remember that the linewidth of the Z boson can only be fitted from the standard model if one assumes three massive quark families. Does anybody know if I am remembering this right?

7. Dec 22, 2003

### chroot

Staff Emeritus
Yeah, there is some experimental evidence that there are exactly three generations of matter. I'll have to look it up tomorrow.

- Warren

8. Dec 22, 2003

### suyver

Same here...

9. Dec 23, 2003

### suyver

OK, I looked it up.

At the LEP the following reaction has been observed:

$$e^+ + e^- \rightarrow Z^0 \rightarrow \nu_i + \bar{\nu}_i$$

At center-of-maas energies comparable to the $Z^0$ mass, one observes the so-called $Z^0$ resonance. This means that the cross-section for the formation of the $Z^0$ increases for energies close to the $Z^0$ mass. Of course, at even higher energies, the cross-section decreases again.

Now comes the point; the $Z^0$ is not stable and it can decay into a neutrino & anti-neutrino: $\nu_i + \bar{\nu}_i$, where $i=e,\mu,\tau$ and maybe a fourth type of neutrino (if it would exist). The width in energy of the $Z^0$ resonance scales with the number of decay-channels that are available. That means that the width of the $Z^0$ resonance is a measure for the number of neutrinos that exist! From the LEP data (all four experiments) is can only be concluded that there are three families of neutrinos.

Note one important point: this argument only works for light neutrinos, as all neutrinos weighing more than half the $Z^0$ mass do not contribute to the resonance and are 'invisible' in this measurement. However, the $Z^0$ weighs about 90 GeV, and the three neutrinos probably weigh about 1 meV or less. So it's quite safe to assume that there is no fourth family.

Merry Christmas,
Freek Suyver.

10. Dec 23, 2003

### Yustas

4th generation quarks are searched for by studying the processes like B-Bbar (D-Dbar, K-Kbar) mixing and flavor-changing neutral current transitions, where 4th generation affects the mixing/decay parameters via loop diagrams. So far, everything is fine without the 4th generation (agrees with theory), so that might mean that there are no 4th generation quarks, or its couplings to the first three generations are tiny...

11. Dec 24, 2003

### mormonator_rm

In other words, do you mean that there should not be a coupling to 4th generation fermions because they are so massive that they could not mix with the first three generations? (i.e. the non-diagonal terms for the 4th generation in the Cabbibo-Kobayashi-Maskawa matrix should drop to zero while the diagonal reaches unity?)

12. Dec 25, 2003

### Yustas

No, what I'm saying is that if 4th gen exists these couplings should be very small (such that V_t'b Vt'd m_t'^2 times usual hadronic stuff for t' contribution to B-Bbar mixing parameter \Delta m_B is smaller then current theoretical and experimental errors), not necessarily zero. So it would still mix with the rest three generation, but the mixing should be tiny...

Another thing is that 4-gen CKM matrix contains four CP-violating phases (compared to one in the 3-generational one), so picture of CP-violation should be much richer (that is, if again those CKM matrix elements are not that small).

Last edited: Jan 15, 2004
13. Jan 15, 2004

### jeff

If you want to delete a post, click edit, click the "delete?" box at the top left of the page and then hit the "delete now" button on the right.

14. Jan 15, 2004

### Yustas

Thanks Jeff!

15. Jan 19, 2004

### arivero

Also, from an experimentalist/empiricist point of view it seems sensible to look for 4th generation quarks even if 4th generation leptons are limited by the Z0 decay width. There is no known particle able to transform between leptons and quarks, so one could find that the number of quark generations is not the same that the one of leptons. (Of course this would spoil very nice symmetries so nobody in the theory side should enjoy it).

16. Jan 19, 2004

### mormonator_rm

If there's a fourth generation of quarks, then there ought to be a fourth generation of leptons, even though they might be suppressed right out of existence. So even if they are not observable, they still ought to be considered. That's just my quasi-theoretical opinion because I like flavour symmetries...

17. Jan 19, 2004

### arivero

:D

Hmm, and of course, I forgot, there is the related issue of cancelation of anomalies. One needs full generations to do it.

18. Jan 19, 2004

### Yustas

In principle, one can avoid anomaly cancellation requirement if 4th generation quarks are not like the ones from the previous three generations (for example, with vector-type couplings, not chiral)...
but it looks ugly to me, even though some people like it.

Of course, 4th generation neutrino might just be heavier than 1/2 mass of the Z... I believe there is a window of masses that are allowed and not ruled out by cosmological considerations...

19. Jan 19, 2004

### Nereid

Staff Emeritus
astronomical/cosmological NO to 4-gen?

It's been bugging me that I recall there were a bunch of good papers that ruled out a 4th generation, based on cosmological models, or COBE observations of the CMB, or ... something! But I just can't find them (or any references to them). Maybe a 4th generation would lead to different primordial nuclide abundances?

In any case, the cosmological consideration would remove any question about a super-massive 4th-gen not being seen in Z-decay.

20. Jan 20, 2004

Staff Emeritus
Given the running coupling of QCD, would sufficiently high mass quarks be unobservable even in principle? And isn't there a renormalization equation limit on the number of fermions?

21. Jan 20, 2004

### arivero

there is some folktale around about the top-antitop boson being already unobservable in principle. I'm not sure about the ground for this, but it could be solid.

22. Jan 20, 2004

### Yustas

Toponium (top-antitop meson) is not observable because top is too heavy to form a bound state. Simplistically speaking, top quark would decay faster via weak interaction than the time needed for "top-antitop system to make one revolution around each other" (note quotes), i.e. boundstate formation time.

As for the QCD constraints, QCD stops being non-abelian (as far as one-loop beta function is concerned) when
$$\frac{g^3}{16 \pi^2} (-11 + 2n_F/3)$$
is greater than zero, i.e. number of flavors greater than 17. I.e. we have a couple of generations to go... so this is not a constraint...

As for the running of QCD coupling -- I'm not sure that the experimental precission is sufficient there, as very heavy quarks would simply decouple at the energies below (at) m_Z. I think people looked at that already and didn't find any useful constraints...

23. Jan 20, 2004

### ahrkron

Staff Emeritus
Can you take out the quotation marks? i.e., can you go a little into the details for this? I've heard this story many times, and I would really like to understand better how it works in the formailsm.

How does time enter into the picture? in general, how do I find how much time is needed to make a bound state?

24. Jan 21, 2004

### Yustas

Sure. Well, it's stil gonna be a bit handwaving. First, let's compute the rate for the eak decauy of top quark (t -> b W). The width would come out to be approximately 2 GeV:
$$\Gamma_t \sim \frac{G_F}{\sqrt{2}} \frac{m_t^3}{8 \pi} \sim 2 ~GeV$$
This implies that the lifetime is
$$\tau \sim 1/\Gamma_t \sim 0.5 GeV^{-1} \sim 3 \times 10^{-25} sec$$

Now comes the handwaving part. Top is heavy, so presumed ttbar boundstate will clearly be in dominated by the Coulomb interaction (and also be nonrelativistic), i.e. ttbar will be moving in a potential
$$V = -\frac{4}{3} \frac{\alpha_s}{r}$$

This means that I can apply what I know from the studies of positronium. In particular, the average velocity of the top quark and radius of the bound state (Bohr's radius):
$$\frac{v}{c} = \frac{4}{3} \alpha_s, ~~ r_0=\frac{3}{2 \alpha_s m_t}$$
(one can be a bit more sophisticated and compute those from the virial theorem or smth similar).

Now is the most handwaving part: time needed to make one revolution is (the system is nonrelativistic)
$$t= \frac{2 \pi r_0}{v}=\frac{9 \pi}{4 \alpha_s^2 m_t} \sim 4 ~GeV^{-1}$$
or 24 * 10^{-25} seconds. In other words, it takes approximately ten times more time for the quarks to complete one revolution that for one of the top quarks to decay.

Of course, this is a handwaving way of showing it. In reality, one acquires another relevant energy scale in this problem, decay width, which effectively cuts off low-energy QCD from the picture. Nothing forbids top quarks to exchange perturbative gluons, so if you are studying, say, electron-positron annihilation into top-antitop, you'd still see the broad peak at the ttbar threshold. But not the sharp spike as in the case of the boundstate...
If you want to continue the discussion, send me an e-mail...

Last edited: Jan 21, 2004
25. Feb 9, 2004

### Ilja

I would like to suggest my interpretation of standard model
fermions:

In geometric terms: hep-th/0310241

If you are not afraid of the "ether"-word, in terms of a
condensed-matter-like interpretation: hep-th/0209167

The point is that above types of interpretation of standard model
fermions require three generations of fermions.