Talk:Proton radius puzzle

Does the Proton actually have a radius?

If we refer to the proton 'radius' are we not assuming that it is in essence a point (spherical) mass ? But if one considers the extremely high electrostatic force (relative to "mass") acting on the electron, it becomes evident that there is some kind of motion in which the electron accelerates very rapidly towards the speed of light. Under such condition the electron's mass effectively 'oscillates' - it changes constantly from mass to energy and back again. And one can only imagine that the proton is subject to a similar form of 'oscillation'. In which case there is neither fixed mass nor fixed radius - only some kind of 'averaged' value(s) the determination of which may be susceptible to whichever form of experiment is used in the measurement thereof. It is the classical conundrum of quantum mechanics at work - measurement affects the values to be measured!

Yes, which is why the article notes the following,

"The experiment does not directly measure the physical size of the radius, but rather an energy difference between two separate energy levels of the atom, known as the Lamb shift, which is sensitive to the radius."

It is measuring the energy itself directly. Regardless though, if the measurements being conducted are done so in the same way, they should consistently return the same values. The fact that they don't in this case is why there is cause for confusion. Silverseren^C 20:10, 21 August 2017 (UTC)

What exactly do physicists mean when they talk about the proton's "charge radius" ? And how does the value of the proton radius affect the value of the Rydberg constant. If I look at the formula for the Rydberg constant:

$R_{\infty }={\frac {m_{\text{e}}e^{4}}{8{\varepsilon _{0}}^{2}h^{3}c}}=1.097\;373\;156\;8508(65)\times 10^{7}\,{\text{m}}^{-1},$

... it doesn't contain r at all unless it is re-written in this form:

$R_{\infty }={\frac {\alpha ^{2}m_{\text{e}}c}{4\pi \hbar }}={\frac {\alpha ^{2}}{2\lambda _{\text{e}}}}={\frac {\alpha }{4\pi a_{0}}}$

and then the radius referred to is the Bohr radius which I thought was an antiquated relic of a long since discarded model of the hydrogen atom. For that matter the Rydberg constant itself is a 'relic' of the Bohr model so I don't understand why it remains topical in modern Physics other than that it fascinates by virtue of containing so many fundamental constants linked together.

We seem to be trying to quantify a somewhat nebulous "radius" linked to a somewhat ghostly "constant" which apparently is one of the most precisely determined of all quantities in Physics ??? I am probably out of depth here but in general the modern formulation of the "Rydberg formula" is the Dirac formula and the constant outside the brackets (or rather on top of a somewhat complicated square root) is simply:

$m_{e}c^{2}$

With allowance being made for the fact that we need to use the electron's reduced mass in the equation. The only other important constant in the Dirac formula (for the Hydrogen spectrum) is the fine structure constant.

Neil Parker (talk) 17:37, 22 August 2017 (UTC)

The root-mean-square charge radius $R$ is defined so that $<R^{2}>={\frac {\int _{r}\rho (r)r^{2}{\rm {d}}{\mathbf {r} }}{\int _{r}\rho (r){\rm {d}}{\mathbf {r} }}}$ where $\rho (r)$ is the charge density. It's defined with $r^{2}$ because we measure it as a cross-section. 202.140.199.194 (talk) 08:18, 5 June 2020 (UTC)

Update to Measurements

The entry for Proton Radius Puzzle probably needs an update to include the new test that was just released a few days ago.

"Abstract

The surprising discrepancy between results from different methods for measuring the proton charge radius is referred to as the proton radius puzzle. In particular, measurements using electrons seem to lead to a different radius compared with those using muons. Here, a direct measurement of the n = 2 Lamb shift of atomic hydrogen is presented. Our measurement determines the proton radius to be rp = 0.833 femtometers, with an uncertainty of ±0.010 femtometers. This electron-based measurement of rp agrees with that obtained from the analogous muon-based Lamb shift measurement but is not consistent with the larger radius that was obtained from the averaging of previous electron-based measurements."

https://science.sciencemag.org/content/365/6457/1007

Science 06 Sep 2019: Vol. 365, Issue 6457, pp. 1007-1012 DOI: 10.1126/science.aau7807

WikiStumps (talk) 23:29, 18 September 2019 (UTC)

Proton is not a sphere.

How can the radius of a proton be measured when it is internally composed of a minimum of three particles and additionally the charge is not uniformly distributed? 5.173.162.11 (talk) 19:58, 30 July 2023 (UTC)

A possible solution

In 2021 I discovered an equation that the proton charge radius equals two times the proton Compton wavelength divided by pi. However, I never published this because a few months later I found out that two Danish physicists discovered this equation first in their paper "On proton charge radius definition" in 2019. So they were the first, obviously then me discovering this equation independently of them is irrelevant. Either way, this equation is in great agreement with the smaller value from the muonic hydrogen experiments. 109.92.60.120 (talk) 14:07, 23 September 2023 (UTC)

Proton radius size

First I agree that proton itself as well as neutrons are not necessary spherical particles however the effect that quarks gluonicly make one toward other either attractive strong force or repulsive the same type quarks become smaller with distance from a quark. Also the protons and neutrons are not too close one to another in the atom they are in what we would measure a size of a protium 0.8 femtometers however they are not closer in atoms one to another then 3 femtometers 3.6 in some measurements. However the proton charge radius is additive depending on the amount of protons and neutrons and its charges from amount of quarks in them. So the charge radius of the proton and of the atom is the size of the atom more protons and neutrons the bigger the size of the charge area where waves get captured and turned into what people call electrons although there are no electrons but trapped electric charge around the nucleus. Also as I said the neutrons add some amount of charge trapping capability although less then their - neutrino and a wave charged particles in other words protons. — Preceding unsigned comment added by Markoshiva (talk • contribs) 19:35, 19 October 2024 (UTC)