Q63P - Introduction to Electrodynamics

Question

(a) Find the divergence of the function

v=r^r$\mathit{v}\mathbf{=}\frac{\hat{\mathbf{r}}}{\mathbf{r}}$

v=r^rFirst compute it directly, as in Eq. 1.84. Test your result using the divergence theorem, as in Eq. 1.85. Is there a delta function at the origin, as there was for $\frac{\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}$? What is the general formula for the divergence of ${\mathbf{r}}^{\mathbf{n}} \hat{\mathbf{r}}$ ? [Answer: $\mathbf{\nabla }\mathbf{.}\mathbf{(}{\mathbf{r}}^{\mathbf{n}}\mathbf{ }\hat{\mathbf{r}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}{\mathbf{r}}^{\mathbf{n}\mathbf{-}\mathbf{1}}$] unless $\mathbf{n}\mathbf{=}\mathbf{-}\mathbf{2}$ , in which case it is $\mathbf{4}{\mathbf{\pi \delta }}^{\mathbf{3}}$for$\mathbf{n}\mathbf{<}\mathbf{2}$ the divergence is ill-defined at the origin.]

(b) Find the curl of ${\mathbf{r}}^{\mathbf{n}}\mathbf{ }\hat{\mathbf{r}}$ . Test your conclusion using Prob. 1.61b. [Answer:$\mathbf{\nabla }\mathbf{\times }\mathbf{(}{\mathbf{r}}^{\mathbf{n}}\mathbf{ }\hat{\mathbf{r}}\mathbf{ }\mathbf{)}\mathbf{=}\mathbf{0}$]

Step-by-Step Solution

Answer

(a) The divergence of the radial vector is$\nabla .v=\frac{1}{r^2}$ . No delta function is involved at the origin in the surface integral and is dependent on the radius R.The left and right side of gauss divergence theorem for the function  is $\frac{\hat{r}}{r}$satisfied. The general formula for the divergence of the radial vector $r^n \hat{r}$ is $\nabla .v=\left(n+2\right)r^{n-1}$

(b)The curl of the vector $v=r^n \hat{r}$ is 0.Stokes theorem is verified for the function $r^n \hat{r}$

Step 1: Describe the given information

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

$\nabla v (r,\theta ,\varphi )=\frac{1}{r^2}\frac{\partial r^2{v}_r}{\partial r}+\frac{1}{r \sin \theta }\frac{\partial \left(\sin \theta v_{\theta }\right)}{\partial \theta }+\frac{1}{r \sin \theta }\frac{\partial v_{\varphi }}{\partial \varphi }$

 

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

Step 2: 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, $\mathbf{\int }\mathbf{(}\mathbf{\nabla }\mathbf{.}\mathit{v}\mathbf{)}\mathbf{ }\mathbf{.}\mathit{d}\mathit{\tau }\mathbf{=}{\mathbf{\oint }}_{\mathbf{s}}\mathit{v}\mathbf{.}\mathit{d}\mathit{a}\mathbf{.}$.

 

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

$\nabla v (r,\theta ,\varphi )=\frac{1}{r^2}\frac{\partial r^2{v}_r}{\partial r}+\frac{1}{r \sin \theta }\frac{\partial \left(\sin \theta v_{\theta }\right)}{\partial \theta }+\frac{1}{r \sin \theta }\frac{\partial v_{\varphi }}{\partial \varphi }$

 

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

Step: 3 Verify Gauss divergence theorem for the function v in part (a)

The given function is $v=\frac{\hat{r}}{r}$ . On comparing it with the vector, $\mathrm{v}={\mathrm{v}}_{\mathrm{r}} \hat{\mathrm{r}}+{\mathrm{v}}_{\mathrm{\theta }} \hat{\mathrm{\theta }}+{\mathrm{v}}_{\mathrm{\varphi }}\hat{\mathrm{\varphi }}$ the components are written as

$$\begin{gathered} v_r=\frac{1}{r} \\ {v}_{\theta }=0 \\ {v}_{\varphi }=0 \end{gathered}$$

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

$$\begin{gathered} \nabla .\mathrm{v}=\frac{1}{{\mathrm{r}}^2}\frac{\partial \left({\mathrm{r}}^2{\mathrm{v}}_{\mathrm{r}}\right)}{\partial \mathrm{r}}+\frac{1}{\mathrm{r} \sin \mathrm{\theta }}\frac{\partial \left(\mathrm{sin\theta }\right({\mathrm{v}}_{\mathrm{\theta }}\left)\right)}{\partial \mathrm{\theta }}+\frac{1}{\mathrm{r} \mathrm{sin\theta }}\frac{\partial \left({\mathrm{v}}_{\mathrm{\varphi }}\right)}{\partial \mathrm{\varphi }} \\ =\frac{1}{{\mathrm{r}}^2}\frac{\partial \left({\mathrm{r}}^2\left(\frac{1}{\mathrm{r}}\right)\right)}{\partial \mathrm{r}}+\frac{1}{\mathrm{r} \mathrm{sin\theta }}\frac{\partial \left(\mathrm{sin\theta }\right(0)}{\partial \mathrm{\theta }}+\frac{1}{\mathrm{r} \mathrm{sin\theta }}\frac{\partial \left(0\right)}{\partial \mathrm{\varphi }} \\ =\frac{1}{{\mathrm{r}}^2}\frac{\partial ( \mathrm{r} )}{\partial \mathrm{r}}+0+0 \\ =\frac{1}{{\mathrm{r}}^2} \end{gathered}$$

Thus, the divergence of the radial vector is $\nabla .v=\frac{1}{r^2}$.

The divergence is obtained as $\nabla .v=\frac{1}{r^2}$. The differential volume of the sphere is written as $d\tau =4\pi r^2 dr$ .

 

The surface integral of vector $\stackrel{\rightharpoonup }{v}$, over the volume is computed as:

$$\begin{gathered} \int \left(\nabla v\right).d\tau =\underset{0}{\overset{R}{\int }} \left(\frac{1}{r^2}\right)4\pi r^{2}dr \\ =4\pi \underset{0}{\overset{R}{\int }} dr \\ =4\pi R .............\left(1\right) \end{gathered}$$

From equations (1) and (2), the left and right side of gauss divergence theorem is satisfied.

 

From equation (2), we can conclude that no delta function is involved at the origin in the surface integral and is dependent on the radius R.

 

The another given function is $V=r^n \hat{r}$. On comparing it with the vector,$\mathrm{v}={\mathrm{v}}_{\mathrm{r}} \hat{\mathrm{r}}+{\mathrm{v}}_{\mathrm{\theta }} \hat{\mathrm{\theta }}+{\mathrm{v}}_{\mathrm{\varphi }}\hat{\mathrm{\varphi }}$ the components are written as

$$\begin{gathered} v_r=r^n \\ {v}_{\theta }=0 \\ {v}_{\varphi }=0 \end{gathered}$$

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

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

Solve further as,

$$\begin{gathered} \nabla .v=\frac{1}{r^2}\left(n+2\right)r^{n+2-1} \\ =(n+2)r^{n-1} \end{gathered}$$

Thus, the divergence of the radial vector $r^n \hat{r}$

is $\nabla .\mathrm{v}=(\mathrm{n}+2){\mathrm{r}}^{\mathrm{n}-1}$.

Step 4: Verify stokes theorem for part (b)

The formula of curl of a vector in spherical coordinates is

$\nabla \times v=\left[\begin{array}{c}\frac{1}{r \sin \theta }\left(\frac{\partial (\sin \theta v_{\theta }}{\partial \theta }\right)-\left(\frac{\partial v_{\varphi }}{\partial v_{\varphi }}\right)\widehat{r }\\ +\frac{1}{r}\left(\frac{1}{\sin \theta }\frac{\partial v_{\varphi }}{\partial \varphi }-\frac{\partial \left(r{v}_{\varphi }\right)}{\partial r}\right)\hat{\theta }+\frac{1}{r}\left(\frac{\partial \left(r{v}_{\theta }\right)}{\partial r}-\frac{\partial v_r}{\partial \theta }\right)\hat{\varphi }\end{array}\right]$

The curl of the vector $r^{n }\hat{r}$ is obtained as 

$$\begin{gathered} \nabla \times v=\left[\begin{array}{c}\frac{1}{r \sin \theta }\left(\frac{\partial \left(\sin \theta \right(0\left)\right)}{\partial \theta }\right)-\left(\frac{\partial \left(0\right)}{\partial \varphi }\right)\widehat{r }\\ +\frac{1}{r}\left(\frac{1}{\sin \theta }\frac{\partial \left(r^n\right)}{\partial \varphi }-\frac{\partial (r (0\left)\right)}{\partial r}\right)\hat{\theta }+\frac{1}{r}\left(\frac{\partial (r (0\left)\right)}{\partial r}-\frac{\partial \left(r^n\right)}{\partial \theta }\right)\hat{\varphi }\end{array}\right] \\ =0 \end{gathered}$$

Therefore the curl of the vector $v=r^n \hat{r}$ is 0. Since the curl of the vector $v=r^n \hat{r}$ is 0, that is, $\nabla \times v=0$

 

Substitute 0 for $\nabla \times v$ , into the stokes theorem $\int (\nabla .\mathrm{v}) .\mathrm{d\tau }=-\int \mathrm{v}.\mathrm{da}.$

$$\begin{gathered} \int 0.\mathrm{d\tau }=-\int \mathrm{v}\times \mathrm{da} \\ 0=\int \mathrm{v}\times \mathrm{da} \\ \mathrm{v}\times \mathrm{da}=0 \end{gathered}$$

 This result can also be verified using the fact that v and differential element are in same direction. Thus their cross product is 0, that is $v\times da=0$.

 

Hence stokes theorem is verified.