\(\nabla^{2} (1/r)\) behaves as Dirac’s delta
Fig. 3 Small sphere \(S_{\epsilon}\) surrounded by closed surface \(S\) .
At \(\mathbf{r} \neq \mathbf{r}'\) ,
(55) \[ \nabla^{2}\frac{1}{r}
\rightarrow \frac{\partial }{\partial x_{k}} \frac{\partial }{\partial x_{k}} \frac{1}{|\mathbf{r} - \mathbf{r}'|}
= - \left\{ \frac{\delta_{kk}}{|\mathbf{r} - \mathbf{r}'|^{3}} - \frac{3 (x_{k} - x'_{k})^{2}}{|\mathbf{r} - \mathbf{r}'|^{5}} \right\}
= 0\]
Let \(S_{\epsilon}\) be a small sphere of infinitesimal radius \(R\) at \(\mathbf{r}'\) and \(S\) be a surface surrounding volume \(V\) , which includes \(S_{\epsilon}\) (Fig. 3 \(V - V_{\epsilon}\) ,
(56) \[ \iiint_{V - V_{\epsilon}} \nabla^{2} \frac{1}{|\mathbf{r} - \mathbf{r}'|} dV = 0\]
where \(V_{\epsilon}\) is the volume of the small sphere. Applying the divergence theorem yields
(57) \[ \iint_{S + S_{\epsilon}} \mathbf{n} \cdot \nabla \frac{1}{|\mathbf{r} - \mathbf{r}'|} dV
= \iint_{S} \mathbf{n} \cdot \nabla \frac{1}{|\mathbf{r} - \mathbf{r}'|} dS + \iint_{S_{\epsilon}} \mathbf{n} \cdot \nabla \frac{1}{|\mathbf{r} - \mathbf{r}'|} dS = 0\]
Note that \(\mathbf{n}\) in the second term is the unit normal directing the inside of the small sphere, that is, \(\mathbf{n} = - (\mathbf{r} - \mathbf{r}')/|\mathbf{r} - \mathbf{r}'|\) . The second term becomes
(58) \[ \iint_{S_{\epsilon}} \mathbf{n} \cdot \nabla \frac{1}{|\mathbf{r} - \mathbf{r}'|} dS
= - \iint_{S_{\epsilon}} \mathbf{n} \cdot \frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^{3}} dS
= \iint_{S_{\epsilon}} \frac{(\mathbf{r} - \mathbf{r}')\cdot(\mathbf{r} - \mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^{4}} dS
= \iint_{S_{\epsilon}} \frac{dS}{|\mathbf{r} - \mathbf{r}'|^{2}}
= 4 \pi\]
Hence,
(59) \[ \iint_{S} \mathbf{n} \cdot \nabla \frac{1}{|\mathbf{r} - \mathbf{r}'|} dS = - \iint_{S_{\epsilon}} \mathbf{n} \cdot \nabla \frac{1}{|\mathbf{r} - \mathbf{r}'|} dS = - 4 \pi\]
The first equation of the above can however be written as the volume integral for \(V\) :
(60) \[ \iiint_{V} \nabla^{2} \frac{1}{|\mathbf{r} - \mathbf{r}'|} dV = - 4 \pi\]
From Eqs. (56) and (60) , we notice that \(- 4 \pi\) comes only from the infinitesimal sphere \(V_{\epsilon}\) at \(\mathbf{r} - \mathbf{r}'\) , leading to
(61) \[ \iiint_{V} \nabla^{2} \frac{1}{|\mathbf{r} - \mathbf{r}'|} dV = \iiint_{V} - 4 \pi \delta \left( |\mathbf{r} - \mathbf{r}'| \right) dV = - 4 \pi\]
Hence,
(62) \[ \nabla^{2} \frac{1}{|\mathbf{r} - \mathbf{r}'|} = - 4 \pi \delta \left( |\mathbf{r} - \mathbf{r}'| \right)\]
