*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 Q_{1} and Q_{2} that are a distance **r** apart. In first-year physics textbooks you’ll see this written as

- F = k Q
_{1}Q_{2}/ r^{2}

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 Q
_{1}Q_{2}/ r^{2}

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) / r^{2}

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.

>… “effects involving photons … are too small to worry about.”

Independently of electroweak unification, doesn’t this magnitudes data argue powerfully against any attempt at U(3) inclusion of the photon as the ninth oddball, infinite-range gluon?

Yes it does. But it was harder to explain so I didn’t lead with it.

The data is where it all starts, though, so that works great. Thanks!

Argh. Matt Strassler, I like your data argument for about two orders of magnitude difference in alpha scales within nucleons, I really do. It helps me understand the genesis of the SU(3) model.

I tried to convince myself that it gets rid of the Glashow mnemonic cube. But of course, it doesn’t, else Glashow would never have mentioned it. In the end, the alpha-ratio argument is a scaling factor that visually flattens the Glashow cube U(3)-like relationships without altering them in any fundamental way.

Let me flip the question around in case I’m missing the obvious: What is the Standard Model explanation for Glashow’s cubes? Surely the fact that all fundamental fermions form lovely T3 isospin pairs across the corners of two U(3) Glashow cubes has been noticed and explained?

I’ve looked but have not had much luck.

The cube is a “charge lattice”, and the whole thing fits inside SO(10), the grand unified group. This can be partially explained within the Standard Model (but not completely) by appealing to consistency conditions called “anomalies”. The full structure of Standard Model matter is not explained within the Standard Model; that’s one of several reasons that physicists are frustrated with the Standard Model and were hoping for some clues from the Large Hadron Collider.

Color charges (mass?) are Conjugative localized quantization like qbits?

It “appears” same for Proton and Electron (mutually repulsive), like…

the Sun and the Moon “appears” in same size..!?

sorry… mutually attractive force …

Nice review. Two questions. (1) Does one find experimental value of R (q,qbar) by summing cross sections of all hadron production upto those quark masses? (2) Theoretical values by one photon exchange? But then there will be rhos, omegas etc. exchanges as you mentioned last time. In that case there may be double counting since these particles are also made from quarks-antiquarks.

(1) Although it’s not easy to build a detector which can identify any particular hadron in terms of whether it’s a pion or a kaon or a proton, it’s not hard to build one that can distinguish hadrons from other types of particles. So it’s a relatively (!) easy and natural experiment to study electron-positron collisions and simply count how many collisions result in some hadrons being produced. The denominator is also easy to measure because muons and anti-muons are easy to identify experimentally.

(2) Photons, rhos and omegas “mix”. All low interactions between electrons/positrons and quarks/anti-quarks would cease, including those from rhos and omegas, if the electromagnetic interaction were shut off completely. So the photon is always essential at all energy scales… the rho is only produced through its mixing with the photon… and it’s *you* who are double-counting 🙂 .

Thanks Matt. I did some phenomenological work in 60s and 70s. Those were bubble chamber days. Then switched to NMR biophysics and quantum optics. I have not kept up much with either theoretical or experimental work for how current detectors work. So please excuse my ignorant questions! I understand mostly (1). For (2) I would still like to understand if the diagram e+-e- to one photon in s-channel and then it going to quarks is the principle contribution to R (theory). Thanks again. I find your blogs very educational.

dimensions of (h_bar c) is (J.m), so why didn’t you write

F= (alpha*h_bar*c) (Q_1 Q_2)/r^2 instead of

F= alpha/(h_bar*c) (Q_1 Q_2)/r^2 ?

Thanks — an error of transposition and inattention. Fixed now.