The Standard Model More Deeply: Lessons on the Strong Nuclear Force from Quark Electric Charges

For readers who want to go a bit deeper into details (though I suggest you read last week’s posts for general readers first [post 1, post 2]):

Last week, using just addition and subtraction of fractions, we saw that the ratio of production rates

• R = Rate (e⁺ e^— ⟶ quark anti-quark) / Rate (e⁺ e^— ⟶ muon anti-muon)

(where e^— stands for “electron” and e⁺ for “positron”) can be used to verify the electric charges of the quarks of nature. [In this post I’ll usually drop the word “electric” from “electric charge”.] Specifically, the ratio R, at different energies, is both sensitive to and consistent with the Standard Model of particle physics, not only confirming the quarks’ charges but also the fact that they come in three “colors”. (About colors, you can read recent posts here, here and here.)

To keep the previous posts short, I didn’t give evidence that the data agrees only with the Standard Model; I’ll start today by doing that. But I did point out that the data doesn’t quite match the simple prediction. You can see that in the figure below, repeated from last time; it shows the data (black dots) lies close to the predictions (the solid lines) but generally lies a few percent above them. Why is this? The answer: we neglected a small but noticeable effect from the strong nuclear force. Not only does accounting for this effect fix the problem, it allows us to get a rough measure of the strength of the strong nuclear force. From these considerations we can learn several immensely important facts about nature, as we’ll see today and in the next post.

Checking that the Data Really Verifies the Standard Model

Figure 1 shows data roughly agrees with the Standard Model prediction, in which quarks come in N=3 colors and have charges

• Up, Charm, Top (u,c,t): Q_u = Q_c = Q_t = 2/3
• Down, Strange, Bottom (d,s,b): Q_d = Q_s = Q_b = -1/3

where “Q_u” means “charge of the u quark.” But maybe it agrees with lots of other possibilities too? Are there other choices of charges and/or N that would work just as well?

If we assume N=3, but allow Q_u to vary (always keeping Q_u = Q_c = Q_t and Q_d = Q_s = Q_b = Q_u -1 , as is required by the charges of protons, neutrons and other “baryons”), predictions for R are shown in Figure 2. Blue, green and red curves correspond to the three regions in Figure 1: low energy (2 – 3 GeV), medium energy (5 – 10 GeV) and high energy (11-20 GeV). The location where Q_u = Q_c = Q_t = 2/3 is marked with a vertical black line, and the predictions of the Standard Model for the values of R in the three regions where we predicted it are shown with blue, green and red stars. Meanwhile, from Figure 1 the values of R from the three regions can be estimated from the data; these are plotted as thick horizontal dashed lines. That the stars (the Standard Model prediction) lie nearly on top of the dashed lines (the data) means the Standard Model is consistent with nature. But you can also see that there is no other value of Q_u where the predictions (the three curved lines) match the data (the thick dashed lines). So if N=3, then Q_u at (or very close to) 2/3 is really the only acceptable option.

Things don’t work for other values of N, either. For N=4 and any Q_u, the prediction for R is always too big. For N=2, prediction would almost work for Q_u = 1, near where the blue curve equals 2, but the red and green prediction are identical there and equal to 4, which clearly disagrees with the data.

Understanding the Remaining Discrepancy

Despite the near-agreement in Figure 1, the remaining discrepancy is troubling. Data is always above the prediction in each of the three regions. Did we leave something out?

In the predictions we made for R, we assumed that only electromagnetism is important and that all other forces can be neglected. But this is not quite a fair assumption. What we’ve left out, conceptually, is the possibility that when an electron and positron annihilate, they become a quark, a corresponding anti-quark and a gluon, as in Figure 4. Specifically, this is a gluon which carries a substantial fraction of the available energy and whose direction makes a wide angle with the quark and anti-quark directions. (Gluons that move along the quark or anti-quark directions, or have low energy, must be treated differently; they are the ones responsible for jets and quark confinement, and are implicitly already accounted for — a very long story.)

The effect of emitting an energetic, wide-angle gluon from a quark or anti-quark can be calculated, and slightly increases the rate for production of “hadrons” (particles containing quarks, anti-quarks and gluons) from electron-positron collisions. Specifically, it increases R by a factor

• (1 + α_s / π)

where α_s is the characteristic strength (or “coupling”) of the strong nuclear force, and π is the usual quantity from math class. If we simply accept this result without question, we can see from Figure 1 that data and prediction would agree much better if α_s were about 0.2 to 0.3, so that (1 + α_s / π) would be larger than 1 by about 5% to 10%. We can even view the discrepancy as our first measurement of α_s, admittedly an imprecise one.

The Strength of a Force

What does this “strength” mean? There’s an analogous quantity in electromagnetism, often denoted “α” or “α_em” , and it is measured to be about 1/137.04 = 0.00730, at least in familiar contexts. This small but mighty number arises when we write the law for electric forces — Coulomb’s law — which tells us the force F between two objects of charge Q₁ and Q₂ that are a distance r apart. In first-year physics textbooks you’ll see this written as

• F = k Q₁ Q₂ / r²

where k is Coulomb’s constant and the charges are in units called “Coulombs”; for instance, the electron’s charge is -e, where e is about 1.6 x 10^-19 Coulombs.

But professional physicists write this differently, using the fundamental constants ħ (Planck’s constant) and c (the cosmic speed limit):

• F = α ħc Q₁ Q₂ / r²

Here the charges are pure numbers: the electron’s charge is -1, for instance. The “e” from first-year physics, with units of Coulombs, has been absorbed into α, which is now itself a pure number, namely 1/137.04. Since the force F is proportional to α, we can say that α sets the strength of all electric forces.

Although α is often called the electromagnetic coupling constant (or, historically, the “fine structure constant”, referring to its effect on atomic energy levels), it is not in fact constant. At distances shorter than a trillionth of a meter, Coulomb’s law is slightly wrong, and we can understand the cause as distance-dependence in α itself. This change in the electromagnetic coupling arises from quantum effects which make empty space polarizable. We’ll get back to this next time.

For the strong nuclear force, there’s an analogous law to Coulomb’s law that governs the force between a quark and an anti-quark at a fixed distance r. It would read

• F = α_s ħc (4/3) / r²

if α_s were constant. But in many contexts it is a bad approximation to treat α_s as constant, and we’ll return to this next time. For now, though, we’ll just say that the data from Figure 1 suggests that, at least in the range of energies around 1 — 20 GeV, α_s is somewhere around 0.2 to 0.3.

Did we miss something?

But let’s finish for today by answering an obvious question. Muons, unlike quarks, can’t radiate gluons — to do so would require the action of the strong nuclear force, to which the muons are immune — but they can certainly radiate photons via electromagnetism. If, in calculating the ratio R, we have to include electron+positron ⟶ quark + anti-quark + gluon in the numerator to make things work, shouldn’t we, for consistency, have to account for electron+positron ⟶ muon + anti-muon + photon in the denominator? And in fact, what about electron+positron ⟶ quark + anti-quark + photon in the numerator?

The point is that effects involving photons, on either the numerator or denominator of R, are too small to worry about. For instance, the effect on the denominator of R of photon emission is of the size

(1 + α_em / π) = 1.002

a shift of two tenths of a percent, far too small to see in Figure 1. The only important effect that we left out comes from the strong nuclear force, precisely because it is strong, i.e. because α_s >> α_em . We’ll see more examples of this next time.