# If the electron is point like, then what is the significance of the classical radius of the electron?

What is the physical meaning/significance of the classical radius of the electron if we know from experiments that the electron is point like?

Is there similarly a classical radius of the photon? The W and Z bosons?

-

The classical electron radius is a length scale at which the classical self-energy of the electron completely accounts for the mass. It tells you where the classical theory of a pointlike electron breaks down.

The compton wavelength tells you where quantum mechanics takes over. The ratio of the compton wavelength to the classical electron radius is the reciprocal of the fine-structure constant, and the fine structure constant tells you the strength of the successive quantum corrections.

So the classical electron radius is telling you small the compton wavelength of the electron can be given a fixed mass of the electron (so the charge is changing), before the quantum theory would be as bad as the classical theory. QED is well in the safe region, having a classical electron radius much smaller than the Compton wavelength, and is therefore well described by a quantum field theory.

This argument suggests that the theory of a massive electron whose charge is so big that it's Compton wavelength is smaller than its classical radius is inconsistent. This is the limit of large fine-structure constant, in which the theory of quantum electrodynamics is believed to be inconsistent, because of the Landau triviality issue.

-

Let me note that the cross-section of Thomson scattering of low-frequency electromagnetic radiation on a free electron is of the same order of magnitude as the classical radius of electron squared (loosely speaking).

-

That wiki article itself provides the answer:

In simple terms, the classical electron radius is roughly the size the electron would need to have for its mass to be completely due to its electrostatic potential energy - not taking quantum mechanics into account.

So since the the photon is massless, and uncharged (it doesn't interact with itself), its "classical radius" would be zero. The W bosons are charged and could similarly be given classical radii, but the Z boson is neutral and could not. In fact that article also gives the formula for finding the classical radius of any charged particle: $$r=\frac{1}{4\pi\epsilon_0}\frac{q^2}{mc^2}$$ where $q$ is the charge and $m$ is the mass.

Edit in response to Revo's comment: Care to clarify what it is about the quoted section that you don't understand?

The electrostatic interaction between charged particles contributes to their potential energy, e.g. the repulsion between like-charged particles is implemented by their "wanting" to move farther apart in order to decrease their potential energy. So if one were to assemble a sphere of charge, it would cost energy to hold it together since the same-charged parts repel each other. The bigger the sphere is, the less energy it takes because the charge is allowed to be more spread out, as it wants. This electrostatic energy is then viewed as the "source" of the electron's mass via Einstein's mass-energy equivalence relation. So given the electron's charge, one can solve for the size of the sphere necessary to get the correct electrostatic energy that is equivalent to the electron's mass.

-
I have already read the wiki article, and I did not understand it. It is not clear to me. That is why I asked the question here. –  Revo Aug 17 '12 at 23:16
Good answer. I would only add that $r_\mathrm{e}$ has probably stuck around because that combination of quantities often appears in formulas, and so it helps make some equations easier to read and intuit. –  Chris White Aug 18 '12 at 1:22

The point electron can be modeled by allowing the point charge to revolve in a Compton wavelength orbit. Multiplying the rotational frequency by h, you obtain the mass energy. (Nothing new about that, Compton knew that). However, if you multiply the frequency by the charge, you get the current. Multiply the current by the area and you get the Bohr magneton, identically. This is all done in 6 easy equations.

-

## protected by Qmechanic♦Jan 15 '14 at 23:24

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.