Q41P - Introduction to Electrodynamics

Question

Compute the gradient and Laplacian of the function$\mathit{T}\mathbf{=}\mathit{r}\mathbf{ }\mathbf{(}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\mathbf{ }\mathit{\theta }\mathit{c}\mathit{o}\mathit{s}\mathbf{ }\mathit{\varphi }\mathbf{)}$. Check the Laplacian by converting T to Cartesian coordinates and using Eq. 1.42. Test the gradient theorem for this function, using the path shown in Fig. 1.41, from (0, 0, 0) to (0, 0, 2).

                                     

Step-by-Step Solution

Answer

According to gradient theorem,.The Laplacian of the function T is 0.

Step 1: Define the dirac delta function

The function   is defined as$T=r (\cos \theta +\sin \theta \cos \varphi )$  . Differentiate the function  with resect to $r, \theta , \varphi$  as follows:

$$\begin{gathered} \frac{\partial T}{\partial r}=\cos \theta +\sin \theta \cos \varphi \\ \frac{\partial T}{\partial \theta }=r(-\sin \theta +\cos \theta \cos \varphi ) \\ \frac{\partial T}{\partial \varphi }=r(-\sin \theta \sin \varphi ) \end{gathered}$$

Step: 2 Compute gradient of the function T

The gradient of function  is defined as ,

The given function is   and the del operatoe is defined as  $\nabla =\frac{\partial }{\partial \mathcal{x}}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k$.

 

The gradient of function   is computed as follows:

$$\begin{gathered} \nabla T=\frac{\partial T}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial T}{\partial \theta }\hat{\theta }+\frac{1}{r\sin \theta }\frac{\partial T}{\partial \varphi }\hat{\varphi } \\ =\left[\begin{array}{c}(\cos \theta +\sin \theta \cos \theta \cos \varphi )\hat{r}+\frac{r}{r}\left(-\sin \theta +\cos \theta \cos \varphi \right)\hat{\theta }+\\ \frac{1}{r \sin \theta }\left(r(-\sin \theta \sin \varphi \right))\hat{\varphi }\end{array}\right] \\ =(\cos \theta +\sin \theta \cos \varphi )\hat{r}(-\sin \theta +\cos \theta \cos \varphi )\hat{\theta }-\sin \varphi \hat{\varphi } \end{gathered}$$

Step 3: Compute the Laplacian of function T

The Laplacian of function T is computed as: 

Solve further as,

Therefore the Laplacian of function T is 0.

Step 4: Draw the path described

The path described in the given information is drawn as follows:                                                          

Step 5: Compute the gradint of the function along path (i)

Along plath (i) ,$\theta =\frac{\pi }{2},\varphi =0$. The differential length vector becomes$\stackrel{\rightharpoonup }{dl}=dr r$  .

The gradient of vector T , along the path (i) is computed as: 

 $$\begin{gathered} \int \nabla \mathrm{T} \mathrm{dl}=\underset{0}{\overset{2}{\int }} \left({\mathrm{r}}^{2 }(\mathrm{cos\theta }+\mathrm{sin\theta cos\varphi }\right)\hat{\mathrm{r}}+(-\mathrm{sin\theta }+\mathrm{cos\theta cos\varphi })\hat{\mathrm{\theta }}-\mathrm{sin\varphi }\hat{\mathrm{\varphi }})\mathrm{dr} \hat{\mathrm{r}} \\ =\underset{0}{\overset{2}{\int }} (\mathrm{cos\theta }+\mathrm{sin\theta cos\varphi })\mathrm{dr} \\ =\underset{0}{\overset{2}{\int }} \left(\cos \frac{\mathrm{\pi }}{2}+\sin \frac{\mathrm{\pi }}{2}\cos 0\right)\mathrm{dr} \\ =2 \end{gathered}$$

Step 6: Compute the gradint of the function along path (ii)

Along plath (ii) , $\theta =\frac{\pi }{2},\varphi =0,r=2$. The differential length vector becomes $\stackrel{\rightharpoonup }{dl}=r \sin \theta d\varphi \hat{\varphi }$ , which can be simplified as,

 $$\begin{gathered} \stackrel{\rightharpoonup }{dl}=2\sin \frac{\pi }{2}d\varphi \hat{\varphi } \\ =2d\varphi \hat{\varphi } \end{gathered}$$

 

.The gradient of vector T , along the path (i) is computed as: 

$$\begin{gathered} \int \nabla \mathrm{T} \mathrm{dl}=\underset{0}{\overset{\pi /2}{\int }} \left({\mathrm{r}}^{2 }(\mathrm{cos\theta }+\mathrm{sin\theta cos\varphi }\right)\hat{\mathrm{r}}+(-\mathrm{sin\theta }+\mathrm{cos\theta cos\varphi })\hat{\mathrm{\theta }}-\mathrm{sin\varphi }\hat{\mathrm{\varphi }})2d\varphi \hat{\varphi } \\ =\underset{0}{\overset{\pi /2}{\int }}\sin \varphi d\varphi \\ ={{\left(2\cos \varphi \right)}_0}^{\pi /2} \\ =-2 \end{gathered}$$

Step 7: Compute the gradint of the function along path (iii)

Along plath (iii) , $\theta =\frac{\pi }{2}to0,\varphi =\frac{\pi }{2},r=2$  . The differential length vector becomes $\stackrel{\rightharpoonup }{dl}=rd\theta \hat{\theta }$ , which can be simplified as,

 $$\begin{gathered} \stackrel{\rightharpoonup }{dl}=rd\theta \hat{\theta } \\ =2d\theta \hat{\theta } \end{gathered}$$

The gradient of vector T , along the path (iii) is computed as: 

$$\begin{gathered} \int \nabla \mathrm{T} \mathrm{dl}=\underset{\pi /2}{\overset{0}{\int }} \left({\mathrm{r}}^{2 }(\mathrm{cos\theta }+\mathrm{sin\theta cos\varphi }\right)\hat{\mathrm{r}}+(-\mathrm{sin\theta }+\mathrm{cos\theta cos\varphi })\hat{\mathrm{\theta }}-\mathrm{sin\varphi }\hat{\mathrm{\varphi }})2d\varphi \hat{\varphi } \\ =\underset{0}{\overset{\pi /2}{\int }}\sin \theta d\theta \\ ={{\left(2\cos \varphi \right)}^0}_{\pi /2} \\ =2 \end{gathered}$$

 

 The gradient is the sum of all the gradient along the paths traced, that is

 $$\begin{gathered} \underset{a}{\overset{b}{\int }} \nabla T dl=2-2+2 \\ =2 \end{gathered}$$

Thus according to gradient theorem, the line integral of the gradient of a function is equal to difference of the function value at the destination point from the function value at the source point, as shown below:

$$\begin{gathered} \underset{a}{\overset{b}{\int }} \nabla T dl=t\left(b\right)-t\left(a\right) \\ =t(0,0,2)-t(0,0,0) \\ =2 \end{gathered}$$