*A post for general readers who’ve heard of quarks; if you haven’t, try reading here:*

The universe has six types of quarks, some of which are found in protons and neutrons, and thus throughout all ordinary material. For no good reasons, we call them **up, down, strange, charm, bottom **and** top**. Today and tomorrow I want to show you how we know their electric charges, even though we can’t measure them directly. The only math we’ll need is addition, subtraction, and fractions.

This also intersects with my most recent post in this series on the Standard Model, which explained how we know that each type of quark comes in three “colors”, or versions — each one a type of strong nuclear charge akin to electric charge.

Today we’ll review the usual lore that you can find in any book or on any website, but we’ll see that there’s a big loophole in the lore that we need to close. Tomorrow we’ll use a clever method to close that loophole and verify the lore is really true.

### The Lore for Protons and Neutrons

Physicists usually define electric charge so that

**the proton has electric charge +1****the electron has charge -1,****the neutron has charge 0**(i.e. electrically neutral, hence its name).

*[Throughout the remainder of this post, I’ll abbreviate “electric charge” as simply “ charge“.]*

As for the six types of quarks, the lore is that their charges are *[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

But how do we know this?

Initial evidence comes from the proton and neutron. The 1960’s view of these particles, which is still often found on websites, is that a proton is made from two up quarks and a down quark (uud) and nothing else, while a neutron is the same as a proton with one up quark replaced with a down quark.

Their charges just come from the charges of the quarks they contain, added together:

- proton (uud): Q
_{u}+ Q_{u}+ Q_{d}= 2 ✕ (2/3) + (-1/3) = 4/3 – 1/3 = 1 - neutron (udd): Q
_{u}+ Q_{d}+ Q_{d}= (2/3) + 2 ✕ (-1/3) = 2/3 – 2/3 = 0

But in the 1970’s, people understood that (as I’ve explained here) really a proton’s made from

- 2 up quarks + 1 down quark + many quark-antiquark pairs + many gluons

where the first three quarks are called the “**valence quarks**“, and the remaining particles are called the “**sea**“. The sea, however, carries no charge:

- gluons are neutral, and
- in each quark/anti-quark pair the charges always cancel.

That means **the proton’s charge arises just from the charge of its valence quarks**, and so the algebra we did a moment ago still gives the right answer for the proton’s charge.

*(Despite recent press articles, the proton does not have a charm quark inside it. It very occasionally has virtual charm quark/charm anti-quark pairs, whose net charge is zero, contributing to the sea. These quark/anti-quarks pairs must be virtual since otherwise their total mass would be much larger than the proton’s. This issue is far more subtle than I can cover here, and it’s admittedly quite difficult for science journalists to get it right. I’ll come back to this in a future post.)*

### The Loophole in the Lore

More generally, a “baryon” is any particle with **three valence quarks and a sea** of gluons and quark/anti-quark pairs. There are lots of them. For example, the Δ^{++} (“Delta-plus-plus”) has three valence up quarks (uuu) and so its charge is 3Q_{u} = 2. The Ω^{–} (“Omega-minus”) has three strange quarks (sss) and so its charge is 3Q_{s} = -1. The Λ (“Lambda”), with (uds) valence quarks, has charge Q_{u} + Q_{d} + Q_{s} = 0. And so on.

The baryon charges all work out correctly and agree with experiment — of course. That’s why what I just told you about the quark charges is the usual lore. But the reasoning is circular. We assumed we knew what was in the proton and neutron and other baryons, and checked this was consistent. But what if we were wrong? For instance, what if, in every baryon, there were not only three valence quarks and a sea but also one additional valence particle X, perhaps essential for the baryon to form, with charge Q_{X}? Then the proton is not (uud) but (uudX); the neutron is (uddX), the Ω^{–} is (sssX), and so on.

With a little algebra you can check that **whatever Q _{X} you choose, there is some choice of quark charges that still works**. For instance, suppose Q

_{X}=-1; then if you take Q

_{u}= +1 and Q

_{d}= Q

_{s}= 0, you get

- proton (uudX) : 2Q
_{u}+ Q_{d}+ Q_{X}= 2 ✕ 1 + 0 + (-1) = 1 - Ω
^{–}(sssX): 3Q_{s}+ Q_{X}= 3 ✕ 0 + (-1) = -1

which are correct. If instead you take Q_{X} = +1, then you need Q_{u} = +1/3 and Q_{d} = Q_{s} = -2/3

- proton (uudX) : 2Q
_{u}+ Q_{d}+ Q_{X}= 2 ✕ (1/3) + (-2/3) + 1 = 2/3 – 2/3 + 1 = 1 - Ω
^{–}(sssX): 3Q_{s}+ Q_{X}= 3 ✕ (-2/3) + 1 = -2 + 1 = -1

And so on for all the baryons. It always works as long as you take take Q_{u} = (2+Q_{X})/3 and Q_{d} = Q_{s} = Q_{u}-1.

More generally, *you can get all the baryon charges right by taking any Q _{u} you like and demanding*

- Up, Charm, Top (u,c,t):
**Q**_{u}= Q_{c}= Q_{t} - Down, Strange, Bottom (d,s,b):
**Q**_{d}= Q_{s}= Q_{b}= Q_{u}– 1

* as long as you allow for the possibility of additional particles like X that are common to all baryons.* (

*It must always be that Q*

_{d}= Q_{u}-1, because the replacement of one u quark with a d quark turns a proton of charge 1 to a neutron of charge 0. Similarly a neutron (udd) has the same charge as a Lambda (uds), so Q_{d}must equal Q_{s}. The presence of additional particles such as X can’t change these fundamental facts.)How, then, can we exclude the possibility of additional charged valence particles such as the X? I’ll show you tomorrow how **we can measure the quark charges**. Along the way we will also double-check that each type of quark comes in three “colors” [i..e. versions of strong nuclear charge.] Stay tuned!

Excellent, informative articles. Especially to a person who just retired and now finds time to try and understand better the wonders of physics. Thank You professor. God give strength to continue.

Nice article, thanks. Plus I love it whenever anyone mentions the Delta plus-plus. To repurpose the famous Rabi quote: “Who ordered _that?_

Well, once you order a quark with charge 2/3 and a force with 3 colors, aren’t you guaranteed a particle with charge 2? So you should ask who ordered those ingredients.

Nicely stated! Confined 1/3 is weird, but 2/3 is even weirder, and Delta++ is the proof.

Weird is in the eye of the beholder; in SU(5) or SO(10) grand unification, -1/3 and 2/3 are guaranteed.

Last I checked, about 20 years ago, SU(5) failed. Anything new?

That’s a separate question, not for today. The point I’m making is that we know plenty of contexts in which 2/3 is not weird.

You, as well as most populizers, say “The particles in the proton and neutron move around at near light-speed, constantly colliding and changing.” Perhaps its just popularization, a figure of speech. But in reality, since a proton is a “ground state” of whatever is in it, isn’t it necessarily static. That means its a static linear combination of all those different possible “wavefunctions” with different constituents.

Since I’m formally a chemist, I’d compare this to the two Kekule structures of Benzene. Those bonds in Benzene certainly don’t move around! (The only “different constituents” in Benzene are all those virtual photons.) If I’m missing something here, does it not have to be fact that the proton is clearly relativistic?

The ground state of hydrogen is static, and the electron’s wave function around the proton is therefore static, but nevertheless, the electron’s average velocity-squared (p^2/m^2) in that state can be measured (or calculated) and it is certainly not zero. Equivalently we can determine the electron’s kinetic energy (p^2/2m) and learn the same thing.

In the same sense, the average velocity-squared of the types of particles in the proton could be measured, at least in principle, and would be found very close to c. Said another way, their energies are generally large compared to their E=mc^2 mass-energy.

Does that answer your question? Even in static wave functions, particles’ kinetic energies can be relativistic and average velocity-squares can be high.

Your answer to my question is not exactly my point … of course the momentum of the particles, be it in a proton or a benzene molecule are non-zero. By static I mean a single eigenstate of the composite system. For a proton, a non-static state would be a linear combination of the usual ground state, and a state where one valence quark (d) was in a 2s or 3s state Would this be a N(1440) or N(1710)?

If such a linear combination could be made, the probability of it decaying would oscillate versus time at a frequency given by the energy (1710 or 1440) – 938. That’s what I mean by nonstatic.

Certainly the system can be excited, as with any composite system. But you cannot view this as a set of nearly-free particles in one-particle orbitals, as you do for electrons in simple atoms., and try to identify one or another quark state within the excited proton. In fact you’d probably do just as well to use the string theory picture of baryons, where the gluons are actually dominant degrees of freedom and the quarks just follow along — meaning that the N(1440) would perhaps be viewed as an excited string bound to two non-excited strings by a D-brane. (If those words mean nothing to you, read https://profmattstrassler.com/2018/02/05/a-brilliant-light-disappears-over-the-horizon-in-memory-of-joe-polchinski/ .)

In any case, there is a spin excitation which gives you the Delta resonances: https://en.wikipedia.org/wiki/Delta_baryon — these are spin 3/2 vs the proton’s spin-1/2. The excitations called “N” all have spin-1/2 and are probably better thought of as spatial resonances of some kind, but as far as I know you can’t think of them as due to a valence quark being in an excited state. These excited resonances are hard to compute because we have limited analytic methods, while numerical strategies are limited when particles are unstable. They can perhaps be estimated by working with the number of colors going to infinity (rather than just 3), a trick which generally works much better than you’d expect, but it’s still not trivial since the number of valence quarks goes to infinity in this limit.

By the way, non-static states like this can’t really be created practically because the excited states have lifetimes that are extremely short and the mass splittings are very large. Non-static states do appear in a few places in particle physics, such as in B mesons, which have almost equal masses and relatively long lifetimes: https://en.wikipedia.org/wiki/B%E2%80%93Bbar_oscillation

But I still have a doubt: if it is that way that the Proton positive charge is defined, then how is defined the Eletron’s negative electrical charge, that is an elementary particle without quarks and gluons? thanks

Electric charge is defined by the strength and direction of the force created by one particle when it interacts with others. The electron repels other electrons. The proton attracts electrons. Measuring the strength of the attraction and repulsion tells us the electron and proton have opposite charges of equal magnitude. It’s that simple and direct.

A separate question is how we know the quark charges. We cannot so easily measure how individual quarks repel or attract electrons (or other particles) because they are always stuck inside protons, neutrons or other particles. That’s why we need indirect arguments such as I’ve given here and will give tomorrow.

/Weird is in the eye of the beholder; in SU(5) or SO(10) grand unification, -1/3 and 2/3 are guaranteed./

Schrodinger equation is based on Newtonian physics rather than relativistic. It’s just classic kinetic energy + potential. There is no Mass energy,

The schrodinger equation uses the relationship E = p2/2m + V, while the relativistic relation is completely different: E2 = p2 + m2 (c = 1 units).

(■) Proton (uud): Qu + Qu + Qd = 2 ✕ (2/3) + (-1/3) = 4/3 – 1/3 = 1

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).

In “quantum field theory,” time is relative rather than absolute (c = 1, c^2 = 1).

Time is absolute only in non-relativistic quantum physics, commonly known as “quantum mechanics.”

The time evolution operator is unitary, since by the Born rule the norm determines the probability to get a particular result in a measurement, unitarity together with the Born rule guarantees the sum of probabilities is always one.

But “Time dilation” occur when the energy propagation of bosons towards an extra dimention and come back (scalar field energy or c^2) and condense as particle (localization).

So “particle” is not a reality, only “relative momentum”.

Singularity: (■) Neutron (udd): Qu + Qd + Qd = (2/3) + 2 ✕ (-1/3) = 2/3 – 2/3 = 0 .

The ground state is static.

(1) A situation where matter is forced to be compressed to a point (a space-like singularity)

(2) A situation where certain light rays come from a region with infinite curvature (a time-like singularity).

If one follows the “time-like geodesics” into the future, it is impossible for the boundary of the region they form to be generated by the null geodesics from the surface. This means that the boundary must either come from nowhere or the whole future ends at some finite extension into another Universe (boundary to the future).

But.. if an object were squeezed below its own Schwarzschild radius? Then that singularity would be “outside the mass”, and it would mean that GR is breaking down in a region that it shouldn’t.

In falling matter change to energy and then to space, increasing the cosmological constant (#SpaceExpansion). The time to “bounce back (Hawking radiation)” increase until it reaches c^4.

So, if all the null geodesics collide, there is no boundary to the future.

We see a Holographic movie of past (#ColdSpot) zoomed (lot of space inbetween cold spots) and the future (matter falling again into cold spots). ??

Thank you, once again, for a very informative article. Perhaps you will address this in the next article, but is their some reason why the charges of “free” particles (such as the p and n) are always an integer even though they contain confined particles (such as u and d) with fractional charge?

Not a deep reason, no. You can understand it as a mathematical consequence of some symmetries in the Standard Model, but those symmetries might well be accidental. There’s no known principle. (Keep in mind, we could have called the electron charge -3 instead of -1.)

Yes, there is a real natural reason why charge of +/- (1/3 and 2/3 and 3/3) is the rule. Certainly looks like charge is additive? And the reality that yields the (+/-) relation is even more amazing.

Thanks for informing us.

When the particles inside the proton and neutron collide, I’m assuming momentum is conserved. Is there a particular force that ensures the exchange of momentum? Is it some combination of the strong force & electric charge? These posts, about the fundamental workings of the universe, are absolutely fascinating.

Glad you are enjoying the posts!

Forces are not what conserves momentum; the conservation of momentum is even deeper than that. All processes in the universe conserve momentum.

This is a consequence of a fundamental theorem about nature, whose modern form is called “Noether’s theorem”, after its proof by Emmy Noether. One of its consequences is this: in any large region where the laws of nature are independent of location, there will be three quantities conserved in all physical processes. We call these “momentum” (three, because momentum has three components). Similarly, if the laws of nature are independent of time, then there is another conserved quantity, called “energy”; and if they are independent of orientation, there is conserved “angular momentum.”

So yes, momentum is conserved, but all the forces always conserve it. [Gravity can be more complicated when gravity is strong and space is highly curved, but that’s obviously not an issue inside a proton.]