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  Experimental boundaries for size of electron?

+ 2 like - 0 dislike
1721 views

There is some confidence that electron is a perfect point e.g. to simplify QFT calculations. However, searching for experimental evidence, e.g. 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 $\approx g-2$:

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:

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 $E=2\cdot 511$keVs.

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?

asked Jan 10, 2020 in Open problems by Jarek (10 points) [ no revision ]

My paper here might be helpful to you.

Dear Vladimir, briefly looking I don't see answer to the question of focus here. Also for others, knowing such experimental evidence for electron size, please briefly specify: which experiment, how it provides boundary for electron size, and how large this boundary is.

Yes, my paper answers your question. In fact there are different definitions of size. If you mean the elastic picture, the size of a free electron is infinity.

1 Answer

+ 1 like - 0 dislike

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

See the item ''Are electrons pointlike/structureless?'' from Chapter B2: Photons and Electrons of my Theoetical Physics FAQ. The experimental upper bound of the electron charge radius seems to be ~ 10^{-16} cm. (a 1982 figure); the 2019 listing of the particle data group doesn't give a value.

''Does it forbid electron's parton structure: being "composed of three smaller fermions"''

No. This depends on the assumed properties and interactions of the Preons making up the electron.

''Does it also forbid some deformation/regularization of electric field to a finite energy?''

Yes, since there the form factors enter, which are not those of a point particle.

answered Jan 10, 2020 by Arnold Neumaier (15,787 points) [ revision history ]

Could you maybe elaborate on this 10^-16 cm bound? In your webpage there is a paper with damaged link and missing thesis.

Regarding possibility of deformation of electric field of point charge to finite energy you say is forbidden, we can create electron-positron from 2x511keV EM energy - so where does the remaining of this infinite energy of electric field comes from?

Also, running coupling says that alpha deforms e.g. to ~1/127 for 90GeV collisions - would such deformation be needed for perfect point charges?

@Jarek: I corrected the link. I don't know more about the matter than is listed in my FAQ. In any case, the electron is definitely not a point, only pointlike.

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