Do Protons Have a Surface?

In searching for the radius of a proton, the Wikipedia page "Proton" states "Because protons are not fundamental particles, they possess a measurable size;...." As my physics is pre-quark, I am confused. I ask:

1. Is this statement generally accepted, suggesting that quarks are not measureable in size?
2. If a proton has a radius, does it necessarily have a surface, and, in particular, a surface that might be described by the thermodynamics of surfaces?
• Possible duplicates: physics.stackexchange.com/q/234979/50583, physics.stackexchange.com/q/119732/50583, physics.stackexchange.com/q/277565/50583 (yes, they're about electrons and not protons, but the answers there explain what "point-like" means for a particle (and "has a measureable size" in your question means "is not point-like") Dec 5, 2021 at 0:36
• 1.Quarks have no size (substructure) to our knowledge, to date. 2. The whole point about a composite QM size is that it is fuzzy. Protons have a surface to the extent a cloud or a swarm of bees has a surface. You really mean to go there? Dec 5, 2021 at 1:08

Here's what the facts is.

At relatively low energies, an incident beam of electrons will scatter off a proton just as if were a hard sphere of approx. 1x10^-15 meter in diameter.

However, as you turn up the energy of the incident electron beam, the electrons begin to penetrate that sphere and instead of scattering coherently off its (apparently) "spherical" surface, they penetrate the interior of the proton and instead start scattering off the 3 point-like quarks inside that sphere.

Particle physicists describe scattering behavior by means of something called the scattering function, and experiments from which the scattering function of a proton are obtained as a function of the incident probe (electron) energy clearly show a transition from scattering off a single sphere of finite size to scattering off of 3 pointlike constituents as the beam energy is increased.

This work was done at SLAC in the late 1960's and early 1970's and constituted the first solid evidence that hadrons were composed of quarks.

• I have an urge to correct the grammar of your first sentence, but I can’t tell whether the error was intentional. Dec 5, 2021 at 5:41
• @gilbert, the line is taken from the lyrics of a once-popular song by the Steve Miller Band. It was intended as a sly reference to a period in my own life when I was a struggling musician. -NN Dec 5, 2021 at 18:55
• Thanks for the clear answer that even includes historical references. I'd give it a double "like" if I could. I'm mistrustful of the term "point-like" as well as "singularity" because they seem like words that cloud some other reality. Is there is a mathematical or QM description of these words (or are they concepts?). I'm comfortable with the words as postulates. As to earlier comments, it depends on what is is. Dec 6, 2021 at 18:27
• @IncredibleII, thank you for you kind comments. Yes, the quantum mechanics specialists have ways of defining "point-like" in terms of scattering formulas and also have ways of dealing with the associated math which one would otherwise expect to have infinities in it. Singularities are even harder to get a handle on and would recommend you pose these questions here- perhaps one of the true experts can help! -NN Dec 6, 2021 at 18:44

they possess a measurable size

...

1. Is this statement generally accepted,

The statement describes what is observed experimentally. The main interactions in keeping the proton as a particle with a given mass composed of quarks are the electromagnetic and the strong, whose balance, one attractive (strong ) one repulsive (electromagnetic) define the bound state of a proton, quantum mechanically . This is modeled in various ways and the models are validated with new experiments. Quantum mechanics is essential for modeling mathematically the microcosm of particles.

suggesting that quarks are not measurable in size?

It does not suggest that, though as far as QM modeling goes, all elementary particles, including quarks, are axiomatically assumed to be point particles.

1. If a proton has a radius, does it necessarily have a surface, and, in particular, a surface that might be described by the thermodynamics of surfaces?

One cannot apply classical mechanics theories, as thermodynamic surfaces, to quantum mechanical entities. Quantum mechanics is a probabilistic theory, giving the probability of interaction. In the case of the proton, when it interacts with a charged particle the spill over charges of the quarks inside will interact with it , either attractively or repulsively . For a group/gas of protons, they will interact repulsively to each other when at low temperatures, and this repulsion will define statistically a surface for the proton. Depending on their energy the interaction starts seeing the individual quark-quark scatters, as at the energies of the LHC

In wikipedia:

Because protons are not fundamental particles, they possess a measurable size; the root mean square charge radius of a proton is about 0.84–0.87 fm

This radius defines a type of surface but it certainly has nothing to do with thermodynamic classical concepts.

• QM is probabilistic and thermodynamics is statistical. I might claim thermodynamics is also probablistic. I don't see the difference. Dec 6, 2021 at 18:34
• @IncredibleII thermodynamics and its probabilities emerge from the classcial mechanics equations of particle interactions, where the (x,y,z,t) of each particle follows newtonian mechanics and has a specific trajectory and then a large number of particles interacting can be described with probability distributions. . Quantum mechanics does not describe individual particle trajectories but gives the probability for a given particle to be at (x,y,z,t) according to the wavefunction of the system under study , the $Ψ*Ψ$. Dec 6, 2021 at 18:40