Interpretations of the classical electron radius

The classical electron radius: $$\begin{equation} r_{\text{e}} = \frac{1}{4 \pi \epsilon _0}\frac{e^2}{m c^2} \simeq 2.82 \, \text{fm} \label{eq:redef} \end{equation}$$

is usually defined in terms of the electrostatic energy of a charged sphere. The sphere radius is chosen such that, when the total charge equals the elementary one, the energy equals the rest mass of the electron (up to a numerical prefactor).

Two charges

Clearly, this physical picture is completely unrealistic; its only merit is providing a mnemonic for $r_{\text{e}}$ as the connection between the relativistic concept of rest mass (or energy) and the electrostatic interaction. It is however too complicated to be useful: the simplest way of expressing this connection is to say that:

The electrostatic energy of two elementary charges in vacuum at a distance $r_{\text{e}}$ is $m c^2$. 

This is much easier to remember and yields directly $\eqref{eq:redef}$, without any additional prefactor, but $r_{\text{e}}$ is now seen as the distance between two electrons, instead of the size of one particle. To retrieve this feature, we will look next at:

Thomson scattering

Consider a free electron submitted to an electromagnetic plane wave, with $E_0$ the amplitude of the electric field. The electric field scattered by the induced dipole can be written as [1]:

$$\begin{equation} E_{\text{s}}(\mathbf{r}) = -E_0\frac{r_{\text{e}}}{r}\exp(i k r) \cos \psi \label{eq:Thomson} \end{equation}$$

where $\psi$ is the polarization angle. While the incident wave has the same amplitude everywhere, the scattered wave is spherical, so that $E_{\text{s}}$ diverges at the origin and decreases towards zero at infinity. For what radius $r$ does one have $|E_{\text{s}}|=|E_0|$? Neglecting the phase and the polarization factor, we see that this occurs exactly at $r=r_{\text{e}}$. A perfectly reflecting surface placed at this position would yield a scattered field with the same amplitude. We can then say that:

$r_{\text{e}}$ is the size of a perfectly reflecting sphere that would have the same scattering as a free electron. 

This conclusion illustrates the concept of scattering cross-section as an "effective surface" seen by the incoming field. For the electron, based on the intuitive argument above we expect this surface to be of the order of $\pi r_{\text{e}}^2$. This is indeed the case, up to a prefactor of 8/3.

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[1] Jens Als-Nielsen and Des McMorrow, Elements of Modern X-ray Physics (2nd ed.), Wiley 2011 (appendix B).