Charge distribution of a proton

Question

I've been studying high-school level physics and I noticed that protons are composed of up-up-down quarks. It is known that fields can be non-uniform due to geometries: Earth's gravity field is not uniform due to different mass distributions (mountains, continents, oceans, etc.). Up and down quarks also have nonidentical charges: 2/3 and -1/3 respectively.

Does this mean that the electrostatic field around a proton is non-uniform in the same manner as the non-uniformity of Earth's gravity field?

Answer 1

If quarks had been little point charges with definite positions, then their electric fields would have added up to a non-uniform electric field. But due to quantum mechanics their positions are indeterminate: they are in a sense spread out so the total field is the sum of all possible configurations. So that makes the overall field look like a (slightly diffuse) point charge.

...Except that this is not 100% correct. The quarks have probability distributions that don't exactly even out perfectly, especially since the proton has spin. This gives it a tiny dipole moment.

Answer 2

This is a great question. Of the answers you've gotten so far, the one by Anders Sandberg is wrong, and the one by Sebastiano doesn't actually address the question.

The proton is the simplest nucleus: one proton and zero neutrons. In general it's very common for nuclei to be nonspherical, like an American football. When this happens, the usual way to detect it and measure the amount is through a quantity known as the electric quadrupole moment. Following your analogy, the earth does have a gravitational quadrupole moment because its rotation causes it to be flattened at the poles (the opposite of the type of deformation usually seen in nuclei). This causes the gravitational field close to the earth to be significantly different than the field of a pointlike mass.

However, not all nuclei have a nonvanishing electric quadrupole moment, and the proton is one of the ones that doesn't. This turns out to be because its spin is 1/2, whereas only particles with a spin of at least 2 can have a quadrupole moment. I don't know of a good elementary explanation for this, but here is a technical one that would be above the head of a high school student, sorry :-) https://archive.org/details/in.ernet.dli.2015.502596/page/n49/mode/2up

The answer by Anders Sandberg points out that the proton has a nonvanishing magnetic dipole moment. That's true, but it isn't a correct analogy of the type you're asking about. A perfectly uniform sphere of charge can still have a nonvanishing magnetic dipole moment.

Answer 3

disclaimer: This post is about a proton electric dipole moment, but it might be the case that the proton electric quadrapole moment is known to be non-zero..

I don't know the physics of quarks and gluons, but I know some basic electric theory and some experimental physics.

The potential from some source charge near the origin can be written as $$ V(r, \theta, \phi) = \sum_{j=1}^{\infty}\sum_{\ell=0}^{\infty}\sum_{m=-\ell}^{+\ell} \frac{D_{\ell, j}^m}{r^j} Y_{\ell}^m(\theta, \phi) $$ By symmetry, if the charge distribution $\rho(r, \theta, \phi)$ is spherically symmetric then the only terms $D_{\ell, j}^m$ with $\ell, m=0$ will contribute. These are the monopole terms. The first order deviation from spherical symmetry would be the inclusion of a dipolar terms with $\ell=1$.

"The storage ring proton EDM experiment" describes an experiment looking for an electric dipole moment of the proton. I do not know how this experiment works, but I can copy and paste this figure and its caption.

This figure shows that we have never observe an electric dipole moment of the proton, but that it has been constrained to be $d<10^{-24}\text{ e}\cdot\text{cm}$.

Here's a comparison I have to put that number into perspective. In atoms the radius of the electron cloud is the Bohr radius $\approx a_0 \approx 50 \text{ pm}$. When electrons in atoms are driven between the $S$ and $P$ orbitals there is an associated dipole moment of a scale $d\approx e\cdot a_0$. That is, the length scale of the dipole moment is as large as the radius of the charge distribution.

Apparently the charge radius of the proton is $\approx 1 \text{ fm} = 10^{-13}\text{ cm}$. The length scale of the proton dipole moment has been constrained be 11 orders of magnitude smaller than this. What that tells me as a naive experimentalist is that if the proton is made of multiple components then, for whatever reason, those components occupy orbitals that are HIGHLY spherical.