23 January 2014

Interpretations of the classical electron radius

r_{\text{e}} = \frac{1}{4 \pi \epsilon _0}\frac{e^2}{m c^2} \simeq 2.82 \, \text{fm}
\label{eq:redef}

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]:

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

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.

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