In this article "Reflections on Maxwell’s Treatise", Section 4.2, it says:

He replaces $\mathbf{m}$ with a volume element of magnetization $\mathbf{M}\ dV$ , integrates over $V$ , and lets the same integral define the magnetic potential inside as well as outside the magnetization to get in Art. $385$ equation $8$ (of Maxwell's treatise):

$\displaystyle \psi_m (\mathbf {r})=\dfrac{1}{4 \pi} \int_{V'} \mathbf{M(r')}.\nabla' \left( \dfrac{1}{\left| \mathbf{r}-\mathbf{r'} \right|} \right) dV' \tag{22}$

Why is the singularity in equation $(22)$ not taken into account?

Shouldn't we use the following improper integral instead of equation $(22)$:

$$\displaystyle \psi_m (\mathbf {r})=\dfrac{1}{4 \pi} \lim \limits_{\delta V' \to 0} \int_{V'-\delta V'} \mathbf{M(r')}.\nabla' \left( \dfrac{1}{\left| \mathbf{r}-\mathbf{r'} \right|} \right) dV' $$

How shall we show this integral converges (without using the spherical volume element) using Cartesian coordiante system?

  • $\begingroup$ What do you mean by ignored? If it doesn't make the improper integral divergent, there's nothing to ignore. $\endgroup$ – Ruslan Mar 5 at 10:48
  • $\begingroup$ Have you checked if it really is a singularity if you take into account the suppressing factor in the volume element? $\endgroup$ – Qmechanic Mar 5 at 10:52
  • $\begingroup$ @Qmechanic: I think using spherical element, it would be removable singularity??? $\endgroup$ – N.G.Tyson Mar 5 at 10:54
  • $\begingroup$ @Ruslan: How can we show the improper integral $\displaystyle \psi_m (\mathbf {r})=\dfrac{1}{4 \pi} \lim \limits_{\delta V' \to 0} \int_{V'-\delta V'} \mathbf{M(r')}.\nabla' \left( \dfrac{1}{\left| \mathbf{r}-\mathbf{r'} \right|} \right) dV'$ converges? $\endgroup$ – N.G.Tyson Mar 5 at 10:58
  • 2
    $\begingroup$ Also posted (under a different user id?) at math.stackexchange.com/q/3135817 $\endgroup$ – user197851 Mar 5 at 11:51

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