What is a magnetic radius (e.g. of a hadron)? And how does this differ from the charge radius?

  • $\begingroup$ By magnetic radius, do you mean something different from gyroradius? $\endgroup$
    – Kyle Kanos
    Jan 8, 2018 at 3:17
  • $\begingroup$ Yes. That is why I posted here. Google kept giving a lot of bad results, with "gyroradius" being the first one. $\endgroup$ Jan 8, 2018 at 3:36

1 Answer 1


The proton is the most well measured hadron: they are plentiful and available in cryo-liquid form for electron scattering experiments. Moreover, the charge radius can be estimated via hyper-fine splitting measurements. The neutron is a bit more difficult, as there are no free neutron targets.

The charge radius is the RMS radius, $r$, of the charge density, $\rho(r)$, and is 0.84-0.88 fm.

The magnetic radius of the proton is the RMS radius of the magnetic moment distribution, and is very nearly 0.78 fm.

A detailed comparison of the two is a major undertaking (on the scale of SE answer) involving comparisons of statistical and systematic errors involved in various experimental techniques and theoretical models.

Edited by Nicole Sharp with data from https://pdg.lbl.gov/

  • $\begingroup$ Thanks. I was wondering if it might be related to magnetic moment. So it is basically the average amount of space that the magnetic moment is localized to. $\endgroup$ Jan 8, 2018 at 3:25
  • 1
    $\begingroup$ The Particle Data Group actually gives a smaller magnetic radius (0.78 fm) for the proton than its charge radius (0.84-0.88 fm). $\endgroup$ Jan 8, 2018 at 3:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.