# 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?

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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.

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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

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).

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