How to calculate $\vec\nabla\times\frac{\vec{r}}{r^3}$ when $r$ can be $0$?

Question

I already know that $\vec\nabla\cdot\frac{\vec r}{r^3} = 4\pi\delta(\vec r)$. In that case, we calculate it directly when $r\neq 0$ and use Gauss' Law to prove it is equivalent to $4\pi\delta(\vec r)$. So I wonder how to calculate $\vec\nabla\times\frac{\vec r}{r^3}$ if we don't rule out the condition that $r$ may be zero?

Answer 1

Thanks to @jim. I add an answer.

Here,

$$\vec \nabla \times ({\vec r\over r^3})=\det{\begin{bmatrix}\hat a_x&\hat a_y&\hat a_z\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ {x\over r^3} & {y\over r^3} & {z\over r^3} \end{bmatrix}}$$

we know that

$$\frac{\partial}{\partial x}({y\over r^3})=y\frac{\partial}{\partial x}({1\over r^3})=y\frac{x}{r^3}$$

$$\frac{\partial}{\partial y}({x\over r^3})=x\frac{\partial}{\partial y}({1\over r^3})=x\frac{y}{r^3}$$

So, each of the cross multiplications used for the calculation of determinant give an equal results and the difference between them is zero. Hence, the entire determinant is zero:

$$\vec \nabla \times ({\vec r\over r^3})=\vec 0$$

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I am an electrical engineer. If we look at it an a practical way, an electric field has this form

$$E={q \over 4 \pi \epsilon_0}{ \vec r \over r^3}$$

when the charge $q$ is located in the origin.

To find the charge density at each location, we need divergence:

$$\nabla.E={\rho \over \epsilon_0}$$

where $\rho$ is the density of charge.

In this case

$$\nabla.E={q \over 4 \pi \epsilon_0} 4 \pi \delta(\vec r)={q \delta(\vec r) \over \epsilon_0}$$ and $q \delta(\vec r)$ is equal to $\rho$

The stationary electric fields (unlike magnetic fields) are conservative. It means if a charge moves to another location and returns back to its original location regardless of the path, the applied potential energy is zero. In another term:

$$\vec \nabla \times E=\vec 0$$

Since:

$$\vec \nabla \times {\vec r\over r^3}=\vec 0$$

It is interesting to know that any vector field can be decomposed into two parts:

$$\vec V(x,y,z)=\vec D(x,y,z)+\vec C(x,y,z)$$

where $\vec \nabla \times \vec D = 0$ and $\nabla . C = 0$

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Update

In addition, ${\vec r\over r^3}$ is gradient of a scalar field:

$${\vec r\over r^3}=\nabla {-1\over r}$$

Curl of a gradient of any scalar field is zero, thus:

$$\vec\nabla\times {\vec r\over r^3}=\vec\nabla\times \vec\nabla {-1\over r}=\vec 0$$