# How can a curl lead to delta distributions?

My professor defined the following vector field:

$$\vec{A}=\left(\frac{-gy}{r(r+z)},\frac{gx}{r(r+1)},0\right)$$

($g$ being a constant, $r=\sqrt{x^2+y^2+z^2}$ and told us to derive the 3-dimensional curl:

$$\mathrm{curl}\vec{A}=\vec\nabla\times\vec{A}=g\frac{\vec{r}}{r^3}+4\pi\delta(x)\delta(y)\theta(-z)$$

where $\theta$ is the Heaviside function, and $\delta$ a Dirac distribution.

I have absolutely no clue how to derive that. I have never seen a curl leading to Dirac delta distributions.

Any hints are welcome.

• Are you sure about the $r+1$ in the denominator? Furthermore the curl would have $g$ as a factor. – Christian Blatter Aug 21 '16 at 13:05

## 1 Answer

Take the function $$f(x,y,z) = \frac{x}{\sqrt{x^2+y^2+z^2}}$$ And say we want to find $\frac{\partial f}{\partial x}$. If either of $y$ or $z$ are nonzero, then this is straightforward, as the function is differentiable; you get $\frac{y^2+z^2}{r^3}$. However, if $y = z = 0$, then $$f(x, 0, 0) = \frac{x}{\sqrt{x^2}} = \textrm{sign}(x)$$ Strictly speaking, thus function is not differentiable at $x = 0$. However, we can informally say $$\frac{d}{dx} \textrm{sign}(x) = 2\delta(x)$$ because $$\int_a^b 2\delta(x) dx = \textrm{sign}(b)-\textrm{sign}(b)$$ since both the LHS and RHS are equal to $2$ when $a < 0 < b$, and $0$ when $a < b < 0$ or $0 < a < b$. Notice that $2$ is the "surface area" of a unit $0$-sphere; similar to how you have a $4\pi$ in your epression, the surface area of a unit $2$-sphere. This isn't a coincidence.

Anyway, putting these together, we see that $$\frac{\partial f}{\partial x} = \frac{y^2+z^2}{r^3} + 2\delta(x) \chi(x,y,z)$$ where $\chi$ is the indicator function for the set $\{(x,0,0) : x\in\Bbb{R}\}$.

I hope this makes more sense now. Again, everything here was completely informal/heuristic.