# Cauchy distribution instead of Coulomb law?

A recent question by alexv - and his comment that the answer will eventually be used in Gravity modeling - has triggered the following in my mind. It's about Electric modeling instead of Gravity modeling, but the inverse square law is similar.

The energy density in the electric field of a point charge $q$ is given by $\frac{1}{2} \epsilon_0 E^2$ where the electric field $E$ at a distance $r$ is: $$E = \frac{q}{4 \pi \epsilon_0 r^2}$$ The total energy in the field is thus given by the integral: $$U = \int_0^\infty\frac{1}{2}\epsilon_0 \left(\frac{q}{4\pi\epsilon_0 r^2} \right)^2 4 \pi r^2 dr \ = \frac{q^2}{8 \pi \epsilon_0} \int_0^\infty \frac{dr}{r^2} = \frac{q^2}{8 \pi \epsilon_0} \left[ \frac{1}{r} \right]_0^\infty = \infty$$ Hence there is an infinite outcome for de self energy of the electron. As is well known, this is quite a serious problem in classical electrodynamics.
It is shown below how this problem can be resolved by sort of renormalization, as understood by this author. Replace $r^2$ by $(r^2+\sigma^2)$ where $\sigma$ is interpreted as the "size" of the electron. Such a Cauchy distribution is in agreement with Coulomb's law at reasonable distances from the origin.
At the same time the singularity at the origin is effectively removed, because we have the following extremely simple result: $$U = \frac{q^2}{8 \pi \epsilon_0} \int_0^\infty \frac{r^2\,dr}{(r^2+\sigma^2)^2} = \frac{q^2}{8 \pi \epsilon_0} \frac{1}{4}\frac{\pi}{\sigma} = \frac{q^2}{32\,\epsilon_0\sigma}$$ The outcome must be equal to $m_0 c^2$, hence we calculate for the electron radius: $$\sigma = \frac{q^2}{32\,\epsilon_0 m_0 c^2} = \frac{\pi}{8} r_e$$ Where $r_e$ is the classical electron radius: Classical electron radius
I'm wondering if an approach as shown here been tried before. And if not, why not?

• Which physical principles did you base this on? – TBrendle Oct 25 '13 at 17:57
• (1) The self-energy of the electron must be finite. (2) The difference between this and Coulomb's law can not be measured. I'm not sure about (2) however. – Han de Bruijn Oct 25 '13 at 17:59
• So you're just proposing a tweaking of the classical electric field law because it looks prettier when you integrate it out? – TBrendle Oct 25 '13 at 18:01
• So what if you can't (physically) tell the difference? – Han de Bruijn Oct 25 '13 at 18:04
• The justification is a finite outcome. The model "that has worked for hundreds of years" gives an infinite mass for the electron, contradicting your claim. While, on the contrary, it has been rigorously tested that electrons have finite mass. – Han de Bruijn Oct 25 '13 at 18:15

The field law you are trying to replace derives directly from Coulomb's law that the electric force between charges is proportional to the product of their charges divided by the square of the distance between them. Coulomb's law didn't drop from the sky. There are good empirical and even mathematical reasons for adopting it (which you can read in any undergraduate physics textbook), and there have been 200 years of rigorous experiments that have confirmed its truth within its domain of applicability, which, remember, is to calculate the force between charged particles.

Yes, the classical model breaks down at the edges of its applicability. You are not the first person to notice this. In fact, there was quite a hubub when this observation was new. The modern point of view isn't that equations need to be "fixed", it's that they are only applicable in certain contexts. And your example of trying to find the size of an electron, is simply an example of where the model fails to give intelligible results. The classical model assumes point particles. You can't use that tool to do what you're trying to do. And why would you? If you want to measure the size of an electron you would do an experiment, not fool around with the laws on paper. Physics is a science. Instead of inserting ad hoc tweaks into already well-proven empirical theories, modern physics has developed new theories that extend rather than replace classical laws. You should study a modern treatment before proposing "fixes" to the classical laws.

Moreover, you state that the difference between the original law and your fix is undetectable. If true, then your improvement improves absolutely nothing in a modeling context.

Edit: I understand now that you're not asking a good-faith question but rather trying to advance a new physical law and support it against criticisms. Good luck with that here on math stack exchange. And no, your suggestion is not new. It is the most obvious way to remove the singularity and was one of the first to be suggested, historically. It turns out much more fundamental "fixes" are required if you want to model physics at that small a scale.

• Again, can the difference between my model and Coulomb's law ever be measured? I will immediately capitulate if such is the case. – Han de Bruijn Oct 25 '13 at 18:20
• Science does not deal with things that can't be measured! I could add to our theory of the solar system, a teapot orbiting mars. Would you ever be able to prove it's not there? I'm trying to answer your question about why physicists wouldn't use your kind of improvement. I'm not denying it could be an interesting mathematical exercise. – TBrendle Oct 25 '13 at 18:24
• Except me .. And Richard P. Feynman in The Feynman Lectures on Physics volume II chapter 28 on Electromagnetic Mass. – Han de Bruijn Oct 25 '13 at 18:50
• I should have said: the difference between the original law and my fix is undetectable at large distances. The main point is that it gives a FINITE outcome for the electron mass. And I think it's unwise to consider the classical electrodynamics context as being outdated. – Han de Bruijn Oct 25 '13 at 19:06
• I think he is trying to find a law that works for actual particles. If you claim that the old law only works for point particles, then it never works. If you claim it works for electrons, you claim electrons have infinite mass, contradicting observations of finite mass electrons. – PyRulez Oct 25 '13 at 19:46