Are three colours needed in particle physics ? November 18, 2007

Posted by dorigo in mathematics, personal, physics, science.

My non-physicist readers will excuse me if this post is above their head… I intend to explain the matter in a more readable way very soon…

In the process of writing about the need for the top quark in the Standard Model,  it occurred to me yesterday that the need for a renormalizable theory of subnuclear interactions – which implies the absence of axial-vector-vector anomalies, and thus a cancellation of all electric charges of the matter fields of the theory – could be combined with the requirement of the existence of charged currents among these fields imposed by the group structure of electroweak interactions, to yield more or less a priori requirements on the number of colors in quantum chromodynamics, the theory of strong interactions which keep atomic nuclei together. Here is an excerpt:

[…]if you were some deity and you were constructing the standard model from scratch, one could argue you would care for renormalizability – it would allow you to compute stuff in a simple mathematical framework. Then, having introduced doublets of quarks and leptons -you need both to construct molecular structures-, you would be facing the choice on the number of colors of quarks (which you need in order to provide stability to nuclear matter, as well as to satisfy the antisymmetric form of the total wavefunction of baryons) and the simultaneous choice of their electric charge.  These are bound to satisfy the rule $\Sigma Q = -1 + N_c \times (Q_u + Q_d)=0$ in order to cancel axial-vector-vector anomalies, as well as the requirement that $Q_u-Q_d=1$ (so that W bosons allow charged current interactions between quarks as well as leptons). You would be hard pressed to choose $N_c$ any different from three if you wanted to build integer-charge mesons and baryons from $q \bar q$ and $qqq$ states! That follows from the fact that the two relations above imply $N_c=\frac{1}{1+2Q_d}$. […] If you know some literature on the topic of building a consistent, alternative version of the SM with different quark charges, I would be glad to get a reference […]

Well, a blog is a wonderful thing, because it sometimes fulfils your wishes! In fact, I was extremely pleased by receiving today an answer to my prayer by a distinguished professor in Stony Brook, Robert Shrock. I was even more pleased to read that the question I had posed to myself was by no means a silly one. Here is Robert’s answer:

Let me respond to your question about quark charges as a function of $N_c$. It is certainly true that $N_c=3$ is special. The condition for the cancellation of anomalies in gauged currents in the standard model is $N_c Y_{Q_L}+Y_{L_L}=0$

where $Y$ denotes weak isospin, and $Q_L=(u,d)_L$ and $L_L=(\nu,e)_L$ denote the left-handed quark and lepton doublets (with notation for the first generation, but the relation holds, of course, for each generation). This yields the relations (e.g., see eqs. (2.15), (2.16) in my paper R. Shrock, Phys. Rev. D53, 6465 (1996) (hep-ph/9512430)): $q_d = q_u-1 = -\frac{1}{2}( 1 + \frac{1}{N_c}(2q_e+1))$

In general, one may consider lepton charges different from the usual ones, which leads to a general set of several classes of quark charges, as displayed in Table 1 of that paper. Note that the relation $q_u-q_d=1$ is independent of $N_c$ and only depends on the $SU(2)_L \times U(1)_Y$ gauge structure with the relation $Q=T_3+(Y/2)$ and the electroweak representation content of the fermions. In section VIII of the paper I discussed some of the phenomenological properties of theories with the various classes of quark charges. For example, for $N_c=5$, if one required the usual lepton charges, then $q_u=3/5$, $q_d=-2/5$ and baryons would be composed of 5 quarks. The proton-like and neutron-like baryons would be given by eqs. (8.3), (8.4), namely $p=uuudd$ and $n=ddduu$ with their usual charges, eqs. (8.5) and (8.6).

The value $N_c=3$ is also special for the idea of grand unification. If one constructs a grand unified theory based on an SO(N) group with $N=4k+2$, $k \ge 2$, then the condition that the SM fermions of a given generation (including an electroweak-singlet neutrino) fit exactly into a spinor representation is (see eq. (9.7)) $2^{N_c+1} = 4(N_c+1)$

which has a solution only for $N_c=3$.

Interestingly, if one considers the $N_c$-extended standard model and studies possible embeddings in a (supersymmetric) GUT, then imposing the conditions of unbroken electromagnetic gauge invariance, asymptotic freedom of color, and three generations of quarks and leptons forces one to choose $N_c=3$. See my recent paper ArXiv:0704.3464, published as Phys. Rev. D76, 055010 (2007).

I am writing this now, but I am feeling guilty, because I should be reading those papers first! The fact is, I want to give them more attention than what I can deserve to them in the control room of the CDF experiment, which is where I am now, with an eye on the keyboard and another on the monitors of the beam, the data taking efficiency, the silicon bias voltages…  And I wanted to give my readers access to those papers as well. So let’s all read about it and get educated on this fascinating issue – I think I will twist Robert’s arm into making a guest post on the matter here! 1. izonei - November 18, 2007

I can’t wait to see the simple version of this. 2. Alejandro Rivero - November 19, 2007

And 3 colors have the simple consequence that two particles have the same charge structure than one antiparticle. With more colors, a pair of particles has a charge that does not appear neither in particles nor antiparticles, and this is puzzling by itself. 3. dorigo - November 19, 2007

That is an excellent point Alejandro. Food for thought.
Cheers,
T. 4. Alejandro Rivero - November 24, 2007

other related question is how does the limit of N_c \to \infty deal with those restrictions. I remember sQCD does some kind of analysis depending on N_f and N_c.

5. Thou shalt have three generations « A Quantum Diaries Survivor - March 25, 2008

[…] larger than that of down quarks. But while the two above “coincidences” are in fact deeply intertwined, the presence of additional fermion generations would make no apparent damage to the overall […]

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