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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. It seems to me that the expression $\int {1\over 2}E^2 \, dV$ is infinite too for infinite volume V. Please refute or confirm.

HolgerFiedler
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  • If the integral were infinite in value (divergent) then the electric field would have infinite energy. So as a practical matter--and to make the physics sensible--we somehow 'regularize' $E$ such that it has compact support or other vanishes sufficiently quickly. – Simon S May 22 '15 at 20:18
  • @SimonS I am out of my depth physics-wise, but mathematically I observe the integral diverges here not because the integrand fails to decay sufficiently as $r\to\infty$ but rather because of its pole at $r=0$. – anon May 22 '15 at 20:20
  • No, the integral of $1/r^2$ only diverges at the lower limit, $r=0$; it converges at the upper limit, $r=\infty$. – mjqxxxx May 22 '15 at 20:21
  • @anon, oops I misspoke, edited. Thanks. – Simon S May 22 '15 at 20:21
  • @SimonS No chance because the declaration of an infinity of the electric field of an electron is untouchable in physics. – HolgerFiedler May 22 '15 at 20:25
  • @anon I'm about the first integral. Please don't look only on it (if possible). – HolgerFiedler May 22 '15 at 20:27
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    The $r_e$ here is known as the classical electron radius. Its value is about $2.8179 \times 10^{-15}$m. Since it is smaller than the Compton wavelength of electron which is about $2.4263 \times 10^{-12}$m. There is no issue that $r_e$ becomes too small. Before that, quantum mechanics will take control of the physics. – achille hui May 22 '15 at 20:29
  • @achillehui I'm not about $r_e$. I'm asking about the first integral and does it is infinite for infinite volume. – HolgerFiedler May 22 '15 at 20:38
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    It is finite, the integrand decay as $\frac{1}{r^2}$ and the integral converge nicely. – achille hui May 22 '15 at 20:39
  • @achillehui To give a statement for $\int {1\over 2}E^2 , dV$ only is not possible? – HolgerFiedler May 22 '15 at 20:45
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    By Gauss law, the electric field $E$ associated with any localized set of charges need to falls off as least as fast as $r^{-2}$. Since the volume element is proportional to $r^2 dr$, the integrand of your integral for any localized set of charges is always $O(r^{-2})$. The assertion is not limited to electric field for a single point charge. – achille hui May 22 '15 at 20:53
  • @achillehui Thanks, I get it now. – HolgerFiedler May 23 '15 at 05:48

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