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 $$ \frac{\varepsilon}{(r^2+\varepsilon)^{5/2}}~\to~ \frac{4\pi}{3}\delta^3(\vec{r}) \quad\text{for}\quad\varepsilon\to 0^+. \tag{3}$$


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