# Divergent self-energy of point charges in Classical Electrodynamics

Assuming the electron to be a classical point particle, if one calculates the self-energy one finds $$U=-\frac{e^2}{8\pi\epsilon_0r}$$ which diverges as $r\rightarrow 0$. Therefore, the measured mass of the electron should be $$m_{0e}+U/c^2=m_e.$$

Now, $m_{0e}$ is the mass of the electron in absence of its electric field (which is therefore unobservable because its electric field cannot be switched off) and $m_e$ is the measured electron mass.

Why is it a problem that in classical electrodynamics, the self-energy of a point electron diverge? The divergence in $U$ may be absorbed in $m_{0e}$ which is unobservable as often done in renormalization.

This is presented as a problem in textbooks because often one equates $m_ec^2=U$, and forget about $m_{e0}$. But I don't see a reason why $m_{e0}$ should be neglected. See page 28 here.

• Divergences are always a problem, whether we can reabsorb them into unobservable parameters or not. Just because we can hide the divergence into a counter-term doesn't mean that there is not a problem. In classical mechanics, and in quantum mechanics, a divergence always make the theory ill-defined. (note that "rigorous" treatments of QFT always study the regularised theory, say, in dimensional regulatisation, in which case there are no divergences at all, and the theory is "well" defined, at least in a perturbative sense). Dec 25 '16 at 19:43
• @AccidentalFourierTransform Assuming a cut-off in length (by assuming new physics exists at short distance), one can take a similar regularization scheme here, too. Right?
– SRS
Dec 25 '16 at 20:08
• @SRS Yes but that kind of regularization would not be Lorentz (ie Relativistic) invariant.
– user139175
Dec 27 '16 at 9:03

• Angular momentum density is given by $\vec{r}\times \vec{N}/c$, where $\vec{N}$ is the Poynting vector. Torque would be the rate of change of that, which is net non-zero when magnetic and rotation axes are misaligned. Dec 24 '19 at 11:52