*A post for general readers who’ve heard of quarks; if you haven’t, you might find this article useful*:

Yesterday I showed you that the usual argument that determines the electric charges of the various types of quarks uses circular reasoning and has a big loophole in it. (The up quark, for example, has charge 2/3, but the usual argument would actually allow it to have any charge!) But today I’m going to show you how this loophole can easily be closed — and we’ll need only addition, subtraction and fractions to close it.

*Throughout this post I’ll shorten “electric charge” to just “charge”.*

### A Different Way to Check Quark Charges

Our approach will be to study the process in which an electron and a positron (the electron’s anti-particle) collide, disappear (“annihilate”), and are converted into one or another type of quark and the corresponding anti-quark; see Figure 1. The rate for this process to occur, and the rate of a similar one in which a muon and anti-muon are produced, are all we will need to know.

In an electron-positron collision, many things may happen. Among the possibilities, the electron and positron may be converted into two new particles. The new particles may have much more mass (specifically, rest mass) than the electron and positron do, if the collision is energetic enough. This is why physicists can use collisions of particles with small mass to discover unknown particles with large mass.

In particular, for any quark of mass M, it is possible for an electron-positron collision to produce that quark and a corresponding anti-quark **as long as the electron’s energy E _{e} is greater than the quark’s mass-energy Mc^{2}**. As E

_{e}is gradually increased from low values, more and more types of quark/anti-quark pairs can be produced.

This turns out to be a particularly interesting observation in the range where 1 GeV < E_{e} < 10 GeV, i.e. when the total collision energy (**2** E_{e}) is between 2 and 20 GeV. If E_{e} is any lower, the effects of the strong nuclear force make the production of quarks extremely complicated (as we’ll see in another post). But when the collision energy is above 2 GeV, things start to settle down, and become both simple and interesting.

In Figure 2 is the actual data showing **the production rate for electron-positron collisions to lead to quark/anti-quark pairs** of any type, no matter what their “flavor” or “color” (i.e. type or version), for different collision energies. What’s shown on the horizontal axis is not E_{e} but **the collision energy 2E _{e}**. The up, down and strange quarks have small masses, so they can be produced almost everywhere throughout this Figure. We should expect interesting changes to occur at or around

**twice**each quark’s mass-energy, namely at twice the charm quark’s mass (around 3 GeV) and twice bottom’s mass (around 9 GeV.) You can see this expectation is borne out: there are big spikes in the data just above those locations. But then, after a few wiggles, things flatten out and become simple. It’s from these simple regions that we can gain simple insights through simple methods.

The production rate is simple in these regions because

- the weak nuclear force plays no important role in this process until 2E
_{e}is about 40 GeV; - the strong nuclear force is increasingly unimportant as 2E
_{e}increases above 2 GeV, especially in the regions with simple behavior, except where there are spikes in Figure 2; - the gravitational and Higgs forces are too tiny to have any effect;
- and therefore
**the process can be understood using only electromagnetism**, the very simplest of the elementary forces.

With simple knowledge about how electromagnetism produces quark/anti-quark pairs from electron-positron annihilation, we can learn crucial information from these simple regions of Figure 2. This turns out to be easy; we don’t have to go into any detail.

### A Simple Fact

Here’s the observation that makes it possible to measure the quark charges.

In electron-positron collisions, the rate for producing a new particle/anti-particle pair via the electromagnetic force (as in Figure 1) is **simply proportional to the square of the new particle’s electric charge**.

*Why the square? Proof comes from quantum physics, but here’s a strong argument. If the rate were proportional to the charge itself, that would be weird. For positive charge, the rate for production would be positive, but a particle with negative charge would be produced with a negative rate… and how can a production rate be negative? (Would we be unproducing a particle that wasn’t there to start with?) So, no: the rate must be proportional to something that’s always positive. *

*Also, the rate has to depend on the charge, since electrically neutral particles can’t be affected, much less produced, by electromagnetism. For similar reasons, the rate ought to be small for particles whose charge is small. The simplest positive quantity which satisfies these requirements is the square of the charge.*

Meanwhile, if the particle comes in multiple versions, as quarks do (we call those versions “colors”), then each version gets produced in this same way as in Figure 1.

So a particle with charge Q that comes in N versions will have a production rate proportional to **N Q ^{2}**.

This is all we will need to know!

### A Simple Strategy

The way to make everything simple, allowing us to avoid any hard calculations at all, is to **compare the production of quarks and anti-quarks with the similar production of muons and anti-muons in electron-positron collisions** (see Figure 3), both of which can be measured at each collision energy 2E_{e}.

Specifically, we will calculate the ratio “R”:

**R = Rate (e**^{+}e^{—}**⟶****quark anti-quark) / Rate (e**^{+}e⟶^{—}**muon anti-muon)**

R will change with the collision energy as more and more types of quarks can be produced. The reason this is a good idea is that in electromagnetism, muon/anti-muon production is almost identical to quark/anti-quark production, except for simple details, so **almost everything cancels out** of this ratio.

**R = Rate (e**^{+}e^{—}**⟶****quark anti-quark) / Rate (e**^{+}e⟶^{—}**muon anti-muon)****= (sum of NQ**^{2}for all quark types produced) / (NQ^{2}for muons)

**= (sum of NQ**^{2}for all quark types produced) / (1)

In the last line, I used the fact that for muon/anti-muon pairs, **N Q ^{2} = 1**; that’s because muons have the same charge as electrons (Q = -1, so Q

^{2 }= + 1) and they don’t have “color” — there’s only one version of a muon — so N=1. Meanwhile, N=3 for all quarks, so

**R = (sum of 3Q**^{2}for all quark types produced)**= 3 ✕****(sum of Q**^{2}for all quark types produced)

which is amazing simple for anything involving quantum field theory and particle physics!

### A Simple Prediction

Therefore, since the Standard Model says *[using notation that “ Q_{u}” means “electric charge of the u quark“]*:

- 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

we get three predictions that we can compare with data:

- for small 2E
_{e}in the 2 – 3 GeV range, we can produce up, down and strange quarks, so- R = 3(Q
_{u}^{2}+ Q_{d}^{2}+ Q_{s}^{2}) = 3(4/9+1/9+1/9) = 4/3 + 1/3 + 1/3 =**2**

- R = 3(Q
- for intermediate 2E
_{e}> 3 GeV or so, we add the charm quark:- R = 4/3 + 1/3 + 1/3 + 4/3 = 10/3 =
**3.33**…

- R = 4/3 + 1/3 + 1/3 + 4/3 = 10/3 =
- for large 2E
_{e}> 10 GeV or so, we add the bottom quark, so- R = 4/3 + 1/3 + 1/3 + 4/3 + 1/3 = 11/3 =
**3.67**…

- R = 4/3 + 1/3 + 1/3 + 4/3 + 1/3 = 11/3 =

### Comparison with Data

What does the data, taken over many years at many experiments, say? I’ve plotted it in Figure 4, along with the three predictions for R that I just calculated for you. The data scatters around because the measurements aren’t perfect *(and I haven’t shown the uncertainty bars)*, but you can see the trends by eye. The predictions of the Standard Model work well — not perfectly, as they’re always a little below the data, but close in each region.

If the Standard Model were wrong, the data and predictions could easily be far apart. For instance, the loophole I pointed out last time would allow Q_{u} = Q_{c} = 1 and Q_{d }= Q_{s} = Q_{b} = 0. But then the predicted R in the three simple regions would have been 3, 6, and 6; that would have been **way** off. Unless the charges are very close to those predicted in the Standard Model, predictions are far from the data, and so the loophole from last time is now closed.

Also, predictions can’t explain the data for any other value of N, so the number of quark “colors” is verified to be 3. And meanwhile, the fact that these predictions almost match the data confirms that we were right to largely ignore all the other forces in the Standard Model. In this sense, many facets of the Standard Model are being simultaneously tested here.

But what about the fact that the data always runs about 10% above the prediction? It turns out this is due to the fact that * the strong nuclear force cannot, in fact, be completely ignored*. The process in which an electron and positron annihilate and produce a quark, an anti-quark

**and a gluon**is large enough that we must include it if we want a more accurate prediction. Accounting for this makes the agreement much better. It also leads us to a more complex topic for another post I’ll produce soon: the variable strength of the strong nuclear force.

## 22 Responses

Dr. Strassler, Thank you for asking about my question on the quark sea. As I understand it, the quark sea is what covers the mass gap. If the 3 valence quarks account for about 1% of the proton and neutron mass we would need a sea of around 290+ quarks or corresponding energy. Questions, don’t the valence quarks have to have some kind of separation from the quark sea due to the strong force operating on the valance quarks and ultimately leaking a bit to surrounding quarks? If gluons are also part of the quark sea than would we have many more bound quark states? Lastly, the quark sea is bound into a confined space and I think moving around at close to light speed. If this is the case then there would be a lot less than an additional 290+ quarks due to KE. In any case due to the confined space isn’t there a risk of collisions with the valence quarks that could tear them apart? And how close are we to violating uncertainty because there is only so much the sea quarks can do and place they can be in.

Thank Dr. Strassler, if this is getting too close to Yank Mills I understand that this could be a question for another day.

Ok, several questions in here. First, you are right the second time: much of the mass of the proton is the *energy* in the quarks, anti-quarks and gluons, not their own rest masses. Your counting of the number of quarks/antiquarks is therefore an overestimate, and also it becomes meaningless to count them since the number is not conserved. The valence quarks are more or less buried in the sea, and you can’t say which down quark is valence and which one is sea since a collision between one and the other could exchange their energies and momenta. Gluons make up the majority of the sea because they are cheap to make; but again, trying to count them is meaningless. Yes, the uncertainty principle is saturated (not violated of course) and in a sense it *determines* the energy of the particles inside the proton and thus the proton’s mass. Were it not for uncertainty (i.e. were it not for the fact that quarks and gluons are waves) then the proton could collapse in on itself.

Dr. Strassler, Actually, I really believe that you explained it very well. And, I simply do not have the experience, expertise and education to ever believe that you did not explain something well.

To the contrary, you are keeping me on my toes and thinking about things which is what is important to me, and you let us ask questions. From my perspective, thank you for making me think and then ask a question. I actually have another one about the quark “sea” but I suspect that would get into another whole topic. which you might be heading to.

Thanks; but I believe that one of the reasons I often explain clearly is that I’m very attentive to where I have failed to do so. I can always improve, and readers like yourself help me see where to do it.

Go ahead and ask about the “sea”. I will at some point say something about these stories about “charm quarks in the proton”, which involves the sea, but it may be a little while til I get to it.

Qu = *Qc = 1 and

Qd = Qs = Qb = 0.

??

Dr. Strassler, Great paper thank you very much. As usual I always have questions, but I was doing OK until the last paragraph. Individual quarks are not seen in nature so pair production is basically a necessity. This is where the gluon comes in, it has to be holding the pair together. Holding them together to the extend that if somehow you were able to break this pair connection the gluon energy would be enough that you would end up creating another pair of of quarks, which would attach to the original quarks that were separated. And again, there would be gluon connecting these two new pairs. Doesn’t the energy associated with the gluon holding the quark pair together reach a point where is starts to unbalance the collision energy? I know that this can’t be true, but it seems like with the broken bond you end up doubling your initial post contact energy. And since it would be an outside energy acting on tearing apart the first pair there is no added energy into the pair system.

This is a good question, and you are right that I did not explain it very well. I assumed something that I should not have done.

The issue is one of time scales and energy scales. You are right that at some point gluons must be produced simply because quarks are never isolated. This issue is addressed here https://profmattstrassler.com/articles-and-posts/particle-physics-basics/the-known-apparently-elementary-particles/jets-the-manifestation-of-quarks-and-gluons/ if you go all the way down to Figure 3.

What I should have said is that what we have left out is the production of a gluon that has both relatively high energy and a high angle with respect to the quark and anti-quark. Most of the gluons you are thinking about have very low energy and/or are moving in the same direction of the quark or anti-quark, at least initially, or later they are involved with the formation of multiple hadrons, as described in the post linked here. But I’m speaking of a gluon that is created earlier and has more of a life of its own. It, too, will create and interact with gluons, as it cannot be isolated either, and it too will form a “jet” of hadrons.

Thanks for pointing out this ambiguity in my presentation. I will think about how to improve it.

When the energy propagation of bosons follows the “time-like geodesics” into the future (Q = -1, so Q2 = + 1) it condense (Bose-Einstein) as quarks (data is simple) at high energies (3 spikes where the strong nuclear force is important) like gold atom forged in supernova.

(■) The extra packed (above 40 GeV?) is residual strong force and Decay as radio activity (the weak nuclear force plays no important role in this process until 2Ee is about 40 GeV).

(■) The gravitational and Higgs forces are too tiny to have any effect. This leads to the Klein-Gordon equation; taking a “square root” leads to (gamma.p + m)psi = 0, which in turn leads to the Dirac equation. (Here, gamma refers to the 4 gamma matrices, each being 4 x 4). Scalar undo the momentum.

(■) Therefore the process can be understood using only electromagnetism (ball on the spring).

Ball on the spring means, when the energy propagation of bosons towards an extra dimention and come back (c^2) and condense as particle (localization), “relative momentum” or “Time dilation”, but moving at speed “c” (electromagnetic radiation), thus creating the “angular momentum”.

But the localization in proton-neutron cloud, at proton during the quark/anti-quark pairs production into 3 quarks happened in high energy like the gold atom was forced in Supernovas.

But in Neutron it is like “residual strong force” decay like “Hawking radiation”.

Qu = Qc = 1 and Qd = Qs = Qb = 0.

Can we use the classical/quantum argument that the electromagnetic force is CPT invariant for all charged particles to justify the charge squared dependence?

It’s always struck me as odd how the classical electromagnetic Lagrangian has a J.A interaction term crying out for J^2 and A^2 to complete the symmetry, but ends up with a quantized ‘electromagnetic mass’ term at the beginning and @A terms in the electromagnetic field tensor:

Is the classical quantization of charge responsible for A^2 becoming F^2 and breaking the symmetry of an otherwise classical electromagnetic Lagrangian symmetrical in J and A?

Fascinating article, and I hope my question adds to it or a future post in some way.

There are many arguments that it’s a function of charge-squared. For instance, the sign of charge is conventional; we could have said that electrons have charge +1 and protons charge -1 and the world would remain the same; so anything you can observe (such as a rate for a process) must depend on products of even numbers of charges. To prove it’s just Q^2 and not a Q^2 + b Q^4 + … well, that’s impossible, because it actually *does* have complicated dependence on various charges when you include small quantum corrections [i.e. Feynman diagrams with loops.] It’s only Q^2 at the simplest, initial step.

Writing the classical Lagrangian in terms of currents J is an approximation, and any “symmetry” between the gauge field A and the current J is badly broken by the fact that J is an approximate to physics, and is not given its own equations that determine how it behaves. All it has is a conservation law. To treat the objects in the current on the same footing as the gauge field, you would want to replace the current with the actual electrons. And then you see there is no symmetry at all: the interaction between electrons and photons is quadratic in the electrons and linear in A. (If electrons had spin 0, then there would be two terms, both of them quadratic in the electrons, and one of them proportional to A and the other to A^2.)

So in the end the structure of quantum electrodynamics (or of quantum chromodynamics) is this:

1) There is an F^2 term which involves the gauge bosons only; it is pure quadratic for electromagnetism, but quadratic, cubic and quartic in A for chromodynamics

2) There is a term quadratic in the charged matter which has terms that are A-independent, linear in A, and (if the matter has charge zero only) quadratic in A, and also quadratic mass terms for the matter.

Details are given on the 1st page of https://people.nscl.msu.edu/~witek/Classes/PHY802/QCD2016b.pdf

Thanks, this is very helpful. It’s now obvious to me that F^2 has to be what it is, independent of the other terms in L, if we want the free EM field in our mathematical model to satisfy Maxwell’s equations, or equivalently the corresponding conservation law having the same form as Poynting’s theorem, for J=0. This is independent of whether we choose a quantized or continuous charge distribution for our mathematical model.

For my question at the end: Imagine a circular wire loop carrying a constant ‘continuous’ current J. As the current is increased it should contract around the perimeter which conflicts with it remaining at a constant radius since it doesn’t contract along this direction. As far as I can see, this apparent paradox doesn’t exist for point charges:

Is the quantization of charge, and other fields, a consequence of special relativity?

I still don’t understand the second paragraph of your comment.

As for your final question, the problem is that the word “quantization” means two different things here. Neither one, however, arises from special relativity. In quantum physics all fields are necessarily quantized; this means that their waves come in quanta, which we call “particles” or “wavicles”. This is true in solids as well, where special relativity is hidden from view.

However, charge is a different matter. It is true that the particles of any particular field are all identical and therefore will have the same electric charge. However, the question you are really asking is whether particles of **different** fields all have charges that are related to a fundamental unit of charge… i.e., why is there a simple relationship between the electron’s charge and the up quark’s charge, for instance? This is not about quantizing fields; this is about the relationship between different quantized fields. To answer this, we need to know the origin of the array of fields in nature. If all the fields arise from a single gauge-symmetry group in which electromagntism is just a part, then it can happen that the charges are related in a simple way. But if there are multiple gauge-symmetries similar to electromagnetism, it may easily happen that there is no simple relation. So there is no reason that electric charge *must* be “quantized” (in the sense of having simple integer relationships). That said, in the presence of the weak and strong nuclear forces, there are additional constraints that make it harder to arrange for a consistent theory without quantized electric charge… but this is a subtle issue (the buzzword is “anomalies”) and is far too complex to address here

In my second paragraph I brought up the Ehrenfest paradox for a rotating constant, continuous charge distribution in a wire as a constant current J, in the belief that this was an example of how special relativity makes such things physically impossible even in classical physics; and this relativistic tendency to make classical charge less continuous as it becomes more energetic may play a part in it becoming quantized, and similarly for other fields. From your answer above this is a very clear ‘no’ to me.

Perhaps you were trying to connect the question to this? https://en.wikipedia.org/wiki/Quantum_vortex There is a *topological* quantization of magnetic flux in a vortex (and topological quantization of a magnetic monopole’s charge, if any exist) in any system where electric charge is already quantized, in the sense of coming only in integer units of some fundamental electric charge. But this is relating vortex charge to electric charge in the first case, and magnetic charge to electric charge in the second case — and is not relating electric charges to one another.

Again, special relativity isn’t essential in the quantization of vortex charge; there may be no approximate relativity in a superconductor, but vortex charge is still determined by a quantization condition.

The link and the images are amazing. I’ve posted a question elsewhere on Matt’s blog vaguely related to charge, but too off topic here, that some might find interesting: https://profmattstrassler.com/articles-and-posts/relativity-space-astronomy-and-cosmology/history-of-the-universe/inflation/#comment-448488

Is the pi(0)(u-ubar, d-dbar) peak at 135 MeV missing? Do you think pseudo-scalarity and spin zero is a problem?

The plot doesn’t go down far enough for that! But yes, spin-zero is a problem; all the spikes are spin-1. The spin-0 states near the charm and bottom threshold do not appear in Figure 2.

Pi(0) peak may not have any selection rules against it. But it may be of a very high order e.g. one can argue that pi(0) into 2 photon decay has been observed. So even if you forget about up-upbar and down-downbar intermediate states, you can have 2 photon intermediate states. But no higher mesons of spin zero in the e+e- have been observed , that may have some theoretical reason.

The probability for decay of any spin-0 state X of mass m_X to e+ e- is typically suppressed. The production rate is similarly suppressed. Higgs decay to mu+ mu- will be observed but Higgs decay to e+ e- will be very tough. Special efforts to observe e+e- –> Higgs may just barely be possible with specialized technology https://cds.cern.ch/record/2289472/files/wepik015.pdf .

Thanks. This is quite interesting. I had heard about plans for Higgs factory, but forget, were they for electron-positron machine? If the production rate is so low, why would they plan to build a big accelerator for this?

Dominant production of Higgs bosons at electron positron machines is through electron + positron –> Z + Higgs. This is large because the Z-Higgs coupling is large.

Bravo. Thanks, Matt. This is totally new physics for me, and I appreciate very much your ability to explain it accessibly.

Cheers, Peter Bluhm

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