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There exist a calculation about electromagnetic mass: $$m_\mathrm{em} = \int {1\over 2}E^2 \, dV = \int\limits_{r_e}^\infty \frac{1}{2} \left( {q\over 4\pi r^2} \right)^2 4\pi r^2 \, dr = {q^2 \over 8\pi r_e}$$

Reducing $r_e$ we get infinite mass. Now, let us suppose that the electric field of an electron is finite and the integral is from $r_e$ to some $r_o$. Under this circumstance, is it allowed to reduce $r_e$ to zero? And what happens when the interval of an integral changes from infinity to a constant number?

HolgerFiedler
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1 Answers1

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Your integral with limit $\int_{r_e}^{r_o}$ will produce $\frac{q^2}{8\pi r_e}-\frac{q^2}{8\pi r_o}$ (in fact this follows from $\int_{r_e}^{r_o}=\int_{r_e}^\infty-\int_{r_o}^\infty$), so it does make sense to replace the infinite upper limit with a finite one (which would however feel arbitrary), but even then you cannot replace the lower limti with $0$.

Hagen von Eitzen
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  • Something seems obviously for you, but not for me. I hope that a finite upper limit does not allow to reduce the lower limit to zero, but how to justify this? – HolgerFiedler May 21 '15 at 18:08