Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In the standard model of particle physics, there are three generations of quarks (up/down, strange/charm, and top/bottom), along with three generations of leptons (electron, muon, and tau). All of these particles have been observed experimentally, and we don't seem to have seen anything new along these lines. A priori, this doesn't eliminate the possibility of a fourth generation, but the physicists I've spoken to do not think additional generations are likely.

Question: What sort of theoretical or experimental reasons do we have for this limitation?

One reason I heard from my officemate is that we haven't seen new neutrinos. Neutrinos seem to be light enough that if another generation's neutrino is too heavy to be detected, then the corresponding quarks would be massive enough that new physics might interfere with their existence. This suggests the question: is there a general rule relating neutrino masses to quark masses, or would an exceptionally heavy neutrino just look bizarre but otherwise be okay with our current state of knowledge?

Another reason I've heard involves the Yukawa coupling between quarks and the Higgs field. Apparently, if quark masses get much beyond the top quark mass, the coupling gets strong enough that QCD fails to accurately describe the resulting theory. My wild guess is that this really means perturbative expansions in Feynman diagrams don't even pretend to converge, but that it may not necessarily eliminate alternative techniques like lattice QCD (about which I know nothing).

Additional reasons would be greatly appreciated, and any words or references (the more mathy the better) that would help to illuminate the previous paragraphs would be nice.

share|improve this question
2  
Excellent question! I remember reading something about this that I'll see if I can look up if nobody else gets there first. –  David Z Dec 18 '10 at 22:38
    
I agree, good question. And I also remember reading an answer to this question some time ago. –  Noldorin Dec 19 '10 at 2:04

5 Answers 5

up vote 22 down vote accepted

There are very good experimental limits on light neutrinos that have the same electroweak couplings as the neutrinos in the first 3 generations from the measured width of the $Z$ boson. Here light means $m_\nu < m_Z/2$. Note this does not involve direct detection of neutrinos, it is an indirect measurement based on the calculation of the $Z$ width given the number of light neutrinos. Here's the PDG citation:

http://pdg.lbl.gov/2010/listings/rpp2010-list-number-neutrino-types.pdf

There is also a cosmological bound on the number of neutrino generations coming from production of Helium during big-bang nucleosynthesis. This is discussed in "The Early Universe" by Kolb and Turner although I am sure there are now more up to date reviews. This bound is around 3 or 4.

There is no direct relationship between quark and neutrino masses, although you can derive possible relations by embedding the Standard Model in various GUTS such as those based on $SO(10)$ or $E_6$. The most straightforward explanation in such models of why neutrinos are light is called the see-saw mechanism

http://en.wikipedia.org/wiki/Seesaw_mechanism

and leads to neutrinos masses $m_\nu \sim m_q^2/M$ where $M$ is some large mass scale on the order of $10^{11} ~GeV$ associated with the vacuum expectation value of some Higgs field that plays a role in breaking the GUT symmetry down to $SU(3) \times SU(2) \times U(1)$. If the same mechanism is at play for additional generations one would expect the neutrinos to be lighter than $M_Z$ even if the quarks are quite heavy. Also, as you mentioned, if you try to make fourth or higher generations very heavy you have to increase the Yukawa coupling to the point that you are outside the range of perturbation theory. These are rough theoretical explanations and the full story is much more complicated but the combination of the excellent experimental limits, cosmological bounds and theoretical expectations makes most people skeptical of further generations. Sorry this wasn't mathier.

share|improve this answer
    
Thank you; this was very informative (no need to apologize). –  Scott Carnahan Dec 19 '10 at 9:08

Theoretical reasons for three generations.

Traditional. A. Anything less than 3 generations could not introduce CP violation into heavy quark decay. This actually lead to the prediction of the bottom and top quarks.

GUT/String theory B. The biggest special lie group is E8, this happens to nicely split into three copies of E6, producing 3 generations.

share|improve this answer
1  
-1: This is nonsense. Generations are not copies of the gauge group, E8 does not split into any more than 1 copy of E6, certainly not in standard string compactifications, you don't add up dimensions to figure out how Lie groups split, because some generators disappear completely during the breaking. The generation number is determined by Fermionic zero modes on the compactification manifold, not by the Lie Group. –  Ron Maimon Nov 27 '11 at 7:25
1  
The 248-dimensional adjoint representation of E8 transforms under SU(3)×E6 as: (8,1) + (1,78) + (3,27) + (\overline{3},\overline{27}). –  Dr BDO Adams Nov 28 '11 at 1:00
    
What you wrote is the standard embedding of SU(3)xE6 in E8 to reproduce an E6 GUT. Notice that there is only one copy of E6, not three. –  Ron Maimon Nov 28 '11 at 2:34
    
One copy of the forces the 78 adjoint version, but free copies of the fermions the 27-multpet. Having the forces and fermions in the same group, means it has to be a supersymmetric theory, with E8_bosons * E8_fermions. –  Dr BDO Adams Dec 30 '11 at 8:54

My research involves a geometric model of spin-1/2 particles, though the discussion of the three generations is beyond the scope of my thesis. However, if I can figure out how to mention this speculation in the Future Work section at the end of my thesis, I will probably do so.

I can't help but marvel at the coincidence of the number three for generations as well as for dimensions of space (where the inertial reference frame fixes the time dimension related to the spatial dimensions). If spin was treated as an oscillation (not just an "intrinsic angular momentum"), then higher-generational particles could have more complicated modes of oscillation: second- and third-generation particles could have two- and three- dimensional spin modes, respectively. If spin was somehow related to mass (which the magnetic dipole moment seems to say it is), then the greater masses of the higher-generational particles could be explained by these higher-dimensional oscillations. Somehow. :)

I am only putting this idea out because I don't suspect I will have the chance to investigate it myself in a more thorough manner. But who knows, maybe I will, and maybe your comments on the idea will help me hone it. Or maybe someone else will take it and run with it, which is fine with me as long as I am mentioned in the credits somewhere. ;)

share|improve this answer
    
I am upvoting because I think this is an idea that we all have enjoyed to speculate with when we were youngsters, and we digg for quaternions and clifford algebras ans alternative vector products... It is bold of you to tell it explicitly. –  arivero Sep 17 '11 at 1:09
    
This idea is not original--- the idea that the muon is an oscillation excitation is as old as the muon. It is not the best model available today, because the quarks and leptons are fundamental. –  Ron Maimon Nov 27 '11 at 7:33
    
+1 for relating mass to higher oscillation modes in three dimensions. –  Stefan Bischof Mar 18 '13 at 23:21

If instead of "why do we...", you had asked "why do I...", speculative answers could be considered too. In 25 years, I have thought some ones; perhaps some people would like to add more, here as community wiki (doesn't generate rep) or in the comments if rep is less than 100.

  • Lock between colour and flavour.
  • mass matrix would need to be 3x3 for some reason.
  • mass matrix is 3x3 as a minimum to violate CP (but it could be more, then)
  • mass matrix is 3x3 in order to use the involved lengths in some discretization of the calculus of derivatives up to second order. Related to the ambiguity of choosing an ordering when quantising terms such as $xp$.
  • three generations come from the the relationship between bosonic strings, with 24 transversal directions, and superstrings, with 8. Related to Leech lattice, heterotic strings, etc.
  • three generations without the neutrinos are 84 helicities, or three generations with Right and Left neutrinos but excluding the top quark are also 84 helicities. This is also the number of components of the source of the 11D Membrane, of M-theory fame.
  • three generations is the only solution for my (@arivero) petty theory, the sBootstrap, to work with leptons besides quarks. And even only with quarks, any other solution is uglier.
share|improve this answer

One part of the answer to this question is that the neutrinos are Majorana particles (or Weyl--- the two are the same in 4d), which can only acquire mass from nonrenormalizable corrections. The neutrinos do not have a right handed partner in an accessible energy range. If there is such a partner, it is very very heavy. So this means that they have to be exactly massless if the standard model is exactly renormalizable.

The interactions that give neutrinos mass are two-Higgs two-Lepton scattering events in the standard model Lagrangian, where the term is $HHLL$ with the SU(2) indices of each H contracted with an L. This term gives neutrino masses, but is dimension 5, so is suppressed by the natural energy scale, which is 1016 GeV, the GUT scale. This gives the measured neutrino masses. This term also rules out a low energy Planck scale.

If you have another generation, the next neutrino would have to be light, just because of this suppression. There is no way to couple the Higgs to the next neutrino much stronger than the other three. There are only 3 light neutrinos, as revealed by the Z width, BBN, as others said.

share|improve this answer
5  
It is unclear whether or not Neutrinos are Majorana particles or Dirac particles. It is more elegant if they are the former, but it is an open question in particle physics. –  Columbia Sep 10 '11 at 22:14
1  
@Columbia: Neutrinos are Majorana in the standard model, and they are certainly Majorana in real life, although I agree that experimentally it is an open question. Sterile neutrinos can have any mass you like, they are not stabilized to be zero mass by a gauge charge, so they require a ridiculous fine tuning to be TeV mass, let alone eV mass. Barring any evidence that they are there, this possibility should be excluded a-priori. –  Ron Maimon Sep 11 '11 at 1:52
1  
@Ron Maimon: excluding anything that is possible a priori is ridiculous, and has historically led to all sorts of insanity, like the energists rejecting Boltzmann, or Einstein's war on Quantum Mechanics. Be open to and consider all possible options. Especially since large-mass sterile neutrinos make at least a plausible dark matter candidate. –  Jerry Schirmer Sep 14 '11 at 20:52
1  
@Ron: None of that makes it WRONG, just unlikely. A priori arguemnts are just that. No one wanted to believe in GSW theory because it was achiral. No one wanted to believe in U(3) theory for the strong interaction because it was nonperturbative in the low energy limit. Reality bats last. If someone is doing something wrong, call them out on it. If you find something unlikely, call them out on it. But don't run around being smug and better than everyone by confusing the two, and then hide behind technical language to intimidate others. –  Jerry Schirmer Sep 17 '11 at 14:16
3  
Um, I don't think many people besides Lubos would argue that string theory is accepted true science. I know that the string theory professors at my graduate department certainly didn't. –  Jerry Schirmer Sep 17 '11 at 18:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.