3
$\begingroup$

In a physics problem I came across a function which seems to be $0$ over all space except in $\vec{0}$ where it is undefined. However the function represents a charge density so I have strong reasons to believe that it takes the form of some kind of dirac-like distribution. My idea was therefore to apply a test function and see what I could get out of it. The function is the following in spherical coordinates (with $p$ and $k$ two constants) :

$$\varrho(r, \theta, \phi)=\frac{1}{4\pi}\nabla\cdot\underbrace{\left[2\left(\frac{1}{r^2}-\frac{ik}{r}\right)\frac{e^{ikr}}{r}p\cos(\theta)\vec{u}_r+\left(\frac{1}{r^2}-\frac{ik}{r}-k^2\right)\frac{e^{ikr}}{r}p\sin(\theta)\vec{u}_{\theta}\right]}_{\vec{A}(r, \theta, \phi)}$$

It indeed can be shown using the expression of divergence in spherical coordinates that this zero is everywhere except in $\vec{0}$ where it is undefined. From a physics perspective it is expected that $\varrho = \vec{p}\cdot\nabla\delta^{(3)}$ where $\vec{p}=p\vec{u}_z=p\cos(\theta)\vec{u}_r-p\sin(\theta)\vec{u}_{\theta}$. The idea was therefore to apply a test function $f$ and proove that somehow : $$\int_{\mathbb{R}^3}f\varrho = p\frac{\partial f}{\partial z}\Big|_{r=0} $$ I thought that the fact that $\varrho$ could be expressed as the divergence of another vector field could help me, the idea would be that similarly to showing that $\nabla\cdot\left(\frac{\vec{r}}{r^3}\right) = 4\pi\delta^{(3)}(\vec{r})$ I could show that $\nabla\cdot\vec{A}=4\pi\vec{p}\cdot\nabla\delta^{(3)}$. I tried using Green's first identity : $$\int_Vf\nabla\cdot\vec{A}\;dV = \int_{\partial V}f\vec{A}\cdot\vec{dS} - \int_V\nabla f\cdot\vec{A}\; dV$$ but I couldn't quite make it work. Could someone help me to proove it please ?

$\endgroup$
1
  • $\begingroup$ The electric dipole is treated in my Phys.SE answer here. $\endgroup$
    – Qmechanic
    Commented Mar 7, 2022 at 9:11

1 Answer 1

1
$\begingroup$

The meaning of the equation $$\operatorname{div}\left(\frac{\boldsymbol r}{r^3}\right)=4\pi~\delta(\boldsymbol r)$$ Is that, for any $\epsilon > 0$, $$\int\limits_{\mathbb{B}(0,\epsilon)}\operatorname{div}\left(\frac{\boldsymbol r}{r^3}\right)\mathrm d^3 \boldsymbol r=4\pi$$ Owing to the definition of the Dirac delta $$\int\limits_{\mathbb{B}(0,\epsilon)}\delta(\boldsymbol r)\mathrm d^3\boldsymbol r=1$$ This is easy to show. Due to the divergence theorem, $$\int\limits_{\mathbb{B}(0,\epsilon)}\operatorname{div}\left(\frac{\boldsymbol r}{r^3}\right)\mathrm d^3 \boldsymbol r=\int\limits_{\partial\mathbb{B}(0,\epsilon)}\frac{\boldsymbol {r}}{r^3}\cdot \hat{\boldsymbol r}~ \mathrm d^2 \boldsymbol r\\ =\int\limits_{\partial\mathbb{B}(0,\epsilon)}\frac{r\hat{\boldsymbol {r}}}{r^3}\cdot \hat{\boldsymbol r}~ \mathrm d^2 \boldsymbol r \\ =\int\limits_0^\pi\int\limits_0^{2\pi}\frac{1}{\epsilon^2}~\epsilon^2\sin\phi ~\mathrm d\theta\mathrm d\phi \\ =4\pi.$$ QED.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .