This might be a bit more general question about how to figure out what is the appropriate (delta) expression in singular points, but e.g. for the dipole, we can derive its potential by a taylor approiximation:
However, this was derived from $$\int d^3r' \rho(\vec{r'})-\vec{r'}\nabla\frac{1}{r},$$ for which the above potential expression is valid only for $r\neq 0$.
However, the field is said to be derived from the above potential (which seems to be true only for $r\neq 0$), and it is
The first term can be derived by calculating $E=-\nabla\Phi$ when assuming $r\neq 0$, but for the second element it's unclear how to derive it.
As this basically a result of $\Delta \frac{1}{r}$ I can see why $4\pi\delta(\vec{r})$ is involved, but how does one derive this exact term? and why how come the field is a sum of a delta function at 0, and a function the isn't defined/converges at 0?


Hint: Formally one should introduce testfunctions to deal with distributions. Another more physical approach is to regularize the dipole potential $$ \Phi_{\varepsilon}~=~ \frac{\vec{p}\cdot\vec{r}}{(r^2+\varepsilon)^{3/2}}, \tag{1} $$ similar to my Phys.SE answer here. The regularized dipole potential $\Phi_{\varepsilon}\in C^{\infty}(\mathbb{R}^3)$ is infinitely many times differentiable. The regularized electric field then becomes: $$ \vec{E}_{\varepsilon}~=~-\vec{\nabla}\Phi_{\varepsilon} ~=~\frac{3(\vec{p}\cdot\vec{r})\vec{r}-r^2\vec{p} }{(r^2+\varepsilon)^{5/2}} - \vec{p}\frac{\varepsilon}{(r^2+\varepsilon)^{5/2}}. \tag{2}$$ Next it is straightforward to check that the last term is a regularized 3D Dirac delta distribution $$ \frac{\varepsilon}{(r^2+\varepsilon)^{5/2}}~\to~ \frac{4\pi}{3}\delta^3(\vec{r}) \quad\text{for}\quad\varepsilon\to 0^+. \tag{3}$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.