Question

a) Show that$\mathbf{\nabla }\mathbf{\cdot }\mathbf{(}{\mathit{e}}_{\mathbf{r}}\mathbf{/}{\mathit{r}}^{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{0}$for$\mathit{r}\mathbf{>}\mathbf{0}$.

(b) Show that$\mathbf{\nabla }\mathbf{(}\mathbf{1}\mathbf{/}\mathit{r}\mathbf{)}\mathbf{=}\mathbf{-}{\mathit{e}}_{\mathbf{r}}\mathbf{/}{\mathit{r}}^{\mathbf{2}}$.

Short answer

(a) The solution is $\frac{2F\left(r\right)}{r}+r{F}^{\text{'}\text{'}}\left(r\right)$

(b) The solution is$-e_r/r^2$

Step-by-step solution

Given information

The given expressions is

(a)

The vectors$e_r/r^2$ and $e_{r}F\left(r\right)$.

(b)

The function$f\left(r\right)=1/r$.

Formula of Spherical Coordinates

The divergence of a vector in spherical coordinates is given as

$\mathbf{\nabla }\mathbf{\cdot }\mathit{V}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\mathbf{\left(}{\mathit{r}}^{\mathbf{2}}{\mathit{V}}_{\mathbf{r}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{r}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{\theta }}\mathbf{\left(}{\mathit{V}}_{\mathbf{\theta }}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{r}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{\theta }}\mathbf{\left(}{\mathit{V}}_{\mathbf{\varphi }}\mathbf{\right)}\mathbf{.}$

The gradient of a function in spherical coordinates is given as

$\mathbf{\nabla }\mathit{f}\mathbf{=}{\mathit{e}}_{\mathbf{r}}\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{r}}\mathbf{+}{\mathit{e}}_{\mathbf{o}}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{\theta }}\mathbf{+}{\mathit{e}}_{\mathbf{\varphi }}\frac{\mathbf{1}}{\mathbf{r}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\varphi }}\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{\theta }}$

Show the given function ∇⋅(er/r2)=0

Solve the equation

$$\begin{gathered} \nabla .\left(e_r/r^2\right)=\nabla .\left(1/r^2.e_r+0.e_0+0.e_{\varphi }\right) \\ =\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2.1/r^2\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(0.\sin \theta \right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(0\right) \\ =\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2.1/r^2\right)+0+0 \\ =\frac{1}{r^2}\frac{\partial }{\partial r}\left(1\right) \end{gathered}$$

On further calculating,

$$\begin{gathered} \nabla .\left(e_r/r^2\right)=\frac{1}{r^2}.0 \\ =0 \end{gathered}$$

Thus,$\nabla \cdot (e_r/r^2)=0$ for r>0.

And,

$$\begin{gathered} \nabla .\left[e_{r}F\left(r\right)\right]=\nabla .\left(F\left(r\right)e_r+0.e_{\theta }+0.e_{\varphi }\right) \\ =\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2.F\left(r\right)\right)+\frac{1}{rs\mathrm{in}\theta }\frac{\partial }{\partial \theta }\left(0.\sin \theta \right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(0\right) \\ =\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2.F\left(r\right)\right)+0+0 \\ =\frac{1}{r^2}\left[2rF\left(r\right)+r^2{F}^{\text{'}\text{'}}\left(r\right)\right] \\ =\frac{2F\left(r\right)}{r}+r{F}^{\text{'}\text{'}}\left(r\right) \end{gathered}$$

Which is a function ofr only.

Show the given function

Solve the equation

$$\begin{gathered} \nabla \left(1/r\right)=e_r\frac{\partial }{\partial r}\left(1/r\right)+e_0\frac{1}{r}\frac{\partial }{\partial \theta }\left(1/r\right)+e_{\varphi }\frac{1}{r\sin \varphi }\frac{\partial }{\partial \varphi }\left(1/r\right) \\ =e_r\left(-\frac{1}{r^2}\right)+e_0\frac{1}{r}.0+e_{\varphi }\frac{1}{r\sin \varphi }.0 \\ =e_r\left(-\frac{1}{r^2}\right)+0+0 \\ =-e_r/r^2 \end{gathered}$$

Thus, the solution is $-e_r/r^2$.