There is some confidence that electron is a perfect point e.g. to simplify QFT calculations. However, searching for experimental evidence (stack), Wikipedia article only points argument based on g-factor being close to 2: Dehmelt's 1988 paper extrapolating from proton and triton behavior that RMS (root mean square) radius for particles composed of 3 fermions should be $\approx g-2$:

enter image description here

Using more than two points for fitting this parabola it wouldn't look so great, e.g. neutron (udd) has $g\approx-3.8$ and $<r^2_n>\approx -0.1 fm^2$.

And while classically $g$-factor is said to be 1 for rotating object, it is for assuming equal mass and charge density ($\rho_m\propto\rho_q$). Generally we can classically get any $g$ by modifying charge-mass distribution:

$$g=\frac{2m}{q} \frac{\mu}{L}=\frac{2m}{q} \frac{\int AdI}{\omega I}=\frac{2m}{q} \frac{\int \pi r^2 \rho_q(r)\frac{\omega}{2\pi} dr}{\omega I}= \frac{m}{q}\frac{\int \rho_q(r) r^2 dr}{\int \rho_m(r) r^2 dr}$$

Another argument for point nature of electron is tiny cross-section, so let's look at it for electron-positron collisions:

enter image description here

Beside some bumps corresponding to resonances, we see a linear trend in this log-log plot: $\approx 10^{-6}$ mb for 10GeVs (5GeV per lepton), $\approx 10^{-4}$ mb for 1GeV. The 1GeV case means $\gamma\approx 1000$, which is also in Lorentz contraction: geometrically means $\gamma$ times reduction of size, hence $\gamma^2$ times reduction of cross-section - exactly as in this line on log-log scale plot.

More proper explanation is that it is for collision - transforming to frame of reference where one particle rests, we get $\gamma\to\approx \gamma^2$. This asymptotic $\sigma \propto 1/E^2$ behavior in colliders is well known (e.g. (10) here) - wanting size of resting electron, we need to take it from GeVs to E=511keVs.

Extrapolating this line (no resonances) to resting electron ($\gamma=1$), we get $\approx 100$ mb, corresponding to $\approx 2$ fm radius.

From the other side we know that two EM photons having 2 x 511keV energy can create electron-positron pair, hence energy conservation doesn't allow electric field of electron to exceed 511keV energy, what requires some its deformation in femtometer scale from $E\propto 1/r^2$:

$$\int_{1.4fm}^\infty \frac{1}{2} |E|^2 4\pi r^2 dr\approx 511keV$$

Could anybody elaborate on concluding upper bound for electron radius from g-factor itself, or point different experimental boundary?

Does it forbid electron's parton structure: being "composed of three smaller fermions" as Dehmelt writes? Does it also forbid some deformation/regularization of electric field to a finite energy?

  • There are theoretical constraints, not just experimental ones. Some relevant search terms are "preon" and "confinement problem". Experimentally, I think the bound should be no worse than $hc/E$, where $E$ is the the energy scale probed by the LHC, so about $10^{-18}$ m. (There may be a lower experimental bound coming from high-energy cosmic rays, or from high-precision measurements.) – Ben Crowell Apr 2 at 19:37
  • @BenCrowell, thank you - I have looked at preon, but regarding electron radius it only mentions Dehmelt's paper (above) - criticizing electron composed of three fermions. Regarding probing in high energy, we should have in mind that there would be involved Lorentz contraction of hypothetical size - to get boundary for resting electron, we should extrapolate it to gamma=1, which for electron-positron collisions discussed above suggest fm-scale for resting electron (?) – Jarek Duda Apr 2 at 20:49
  • @annav, it only looks at g-factor, for which Dehmelt's original argument used parabola fitted to two points (top plot above) against electron being built of 3 smaller fermions - is this argument proper? – Jarek Duda Sep 11 at 11:40
  • it calculates g from an α , that fits the data for the magneton, and then estimates a radius. What convinces me is that calcualting the magneton "If the electron is composed of constituent particles bound together by some unknown attraction then we would expect that the standard model formula displayed above would not accurately predict the measured magnetic moment. Antiprotons and protons, for example, are not at all well described by this equation. As is well known, this is because antiprotons and protons are not the point particles with no size that are assumed in deriving the formula.' – anna v Sep 11 at 12:00

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