Question

Construct a vector function that has zero divergence and zero curl everywhere. (A constant will do the job, of course, but make it something a little more interesting than that!)

Short answer

For the vector $\mathrm{v}=\frac{1}{{\mathrm{r}}^2}\hat{\mathrm{r}}$the divergence and curl is 0 everywhere.

Step-by-step solution

Define the divergence in spherical coordinates

A vector function that has zero divergence and zero curl everywhere has to be obtained.

Define the divergence in spherical coordinates

The integral of derivative of a function $\mathit{f}\mathbf{}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}\mathbf{)}$ over an open surface area is equal to the volume integral of the function, $\mathbf{\int }\mathbf{(}\mathbf{\nabla }\mathbf{.}\mathbf{v}\mathbf{)}\mathbf{.}\mathbf{d}\mathbf{\tau }\mathbf{=}{\mathbf{\oint }}_{\mathbf{s}}\mathbf{v}\mathbf{-}\mathbf{d}\mathbf{a}$.

The divergence of vector function $F(r,\theta ,\varphi )$ in spherical coordinates is

$\nabla F(r,\theta ,\varphi )=\frac{1}{r^2}\frac{\partial \left(r^2{F}_1\right)}{\partial r}+\frac{1}{r\sin \theta }\frac{\partial \left(\sin \theta F_2\right)}{\partial \theta }+\frac{1}{r\sin \theta }\frac{\partial F_3}{\partial \varphi }$

Here, $r,\theta ,\varphi$ are the spherical coordinates.

Step: 3 Compute divergence of the function.

Let the required function is $\mathrm{v}=\frac{1}{{\mathrm{r}}^2}\hat{\mathrm{r}}$and the del operator is defined as

$\nabla =\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k$. The divergence of vector v is computed as follows:

$$\begin{gathered} \nabla .v=\frac{1}{r}\frac{\partial \left(r^2{v}_r\right)}{\partial r}+\frac{1}{r\sin \theta }\frac{\partial \left(\sin \theta v_{\theta }\right)}{\partial \theta }+\frac{1}{r\sin \theta }\frac{\partial \left(v_{\phi }\right)}{\partial \phi } \\ =\frac{1}{r}\frac{\partial \left(r^2{\displaystyle \frac{1}{r^2}}\right)}{\partial r}+\frac{1}{r\sin \theta }\frac{\partial \left(\sin \theta \right(0\left)\right)}{\partial \theta }+\frac{1}{r\sin \theta }\frac{\partial \left(0\right)}{\partial \phi } \\ =\frac{1}{r}\frac{\partial \left(1\right)}{\partial r}+0+0 \\ =0 \end{gathered}$$

Compute the curl of vector v.

The curl of vector v is calculated as follows:

$$\begin{gathered} \nabla .v=\left[\begin{array}{cc}\frac{1}{r\sin \theta }& \left[\frac{\partial \left(\sin \theta v_{\theta }\right)}{\partial \theta }-\frac{\partial \left(v_{\phi }\right)}{\partial \phi }\right]\hat{r}+\frac{1}{r}\left[\frac{1}{\sin \theta }\frac{\partial \left(v_r\right)}{\partial \phi }-\frac{\partial \left(r{v}_{\phi }\right)}{\partial r}\right]\hat{\theta }\\ & +\frac{1}{r}\left[\frac{\partial \left(r{v}_{\phi }\right)}{\partial r}-\frac{\left(r{v}_r\right)}{\partial \theta }\right]\hat{\varphi }\end{array}\right] \\ =\left[\begin{array}{cc}\frac{1}{r\sin \theta }& \left[\frac{\partial \left(\sin \theta \right(0\left)\right)}{\partial \theta }-\frac{\partial \left(0\right)}{\partial \phi }\right]\hat{r}+\frac{1}{r}\left[\frac{1}{\sin \theta }\frac{\partial \left({\displaystyle \frac{1}{r^2}}\right)}{\partial \phi }-\frac{\partial \left(r\right(0\left)\right)}{\partial r}\right]\hat{\theta }\\ & +\frac{1}{r}\left[\frac{\partial \left(r\right(0\left)\right)}{\partial r}-\frac{\left(r\left({\displaystyle \frac{1}{r^2}}\right)\right)}{\partial \theta }\right]\hat{\varphi }\end{array}\right] \\ =0 \end{gathered}$$

Therefore, curl of v is $\nabla \times v=0$.

Hence, for the vector $\mathrm{v}=\frac{1}{{\mathrm{r}}^2}\hat{\mathrm{r}}$the divergence and curl is 0 everywhere