Q17P - Introduction to Electrodynamics

Question

In two dimensions, show that the divergence transforms as a scalar under rotations. [Hint: Use Eq. 1.29 to determine ${}^{\overline{{\mathbf{v}}_{\mathbf{y}}}}$ and${}^{{\overline{\mathbf{v}}}_{\mathbf{z}}}$ and the method of Prob. 1.14 to calculate the derivatives. Your aim is to show that  

$\frac{\mathbf{\partial }\overline{{\mathbf{v}}_{\mathbf{y}}}}{\mathbf{\partial }\mathbf{y}}\mathbf{+}\frac{\mathbf{\partial }\overline{{\mathbf{v}}_{\mathbf{y}}}}{\mathbf{\partial }\mathbf{z}}\mathbf{=}\frac{\mathbf{\partial }{\mathbf{v}}_{\mathbf{ }\mathbf{y}}}{\mathbf{\partial }\mathbf{y}}\mathbf{+}\frac{\mathbf{\partial }{\mathbf{v}}_{\mathbf{z}}}{\mathbf{\partial }\mathbf{z}}$

Step-by-Step Solution

Answer

It is shown that$\frac{\partial \overline{{\mathrm{v}}_{\mathrm{y}}}}{\partial z}+\frac{\partial \overline{{\mathrm{v}}_{\mathrm{y}}}}{\partial y}=\frac{\partial {\mathrm{v}}_{ \mathrm{y}}}{\partial \mathrm{y}}+\frac{\partial {\mathrm{v}}_y}{\partial \mathrm{z}}$. The divergence transforms as a scalar under rotation, in two dimensions as$\frac{\partial \overline{{\mathrm{v}}_{\mathrm{y}}}}{\partial z}+\frac{\partial \overline{{\mathrm{v}}_{\mathrm{y}}}}{\partial y}=\frac{\partial {\mathrm{v}}_{ \mathrm{y}}}{\partial \mathrm{y}}+\frac{\partial {\mathrm{v}}_y}{\partial \mathrm{z}}$

Step 1: Define the given information.

It is to be shown that the divergence transforms as a scalar under rotation, in two dimensions as $\frac{\partial \overline{{\mathrm{v}}_{\mathrm{y}}}}{\partial z}+\frac{\partial \overline{{\mathrm{v}}_{\mathrm{y}}}}{\partial y}=\frac{\partial {\mathrm{v}}_{ \mathrm{y}}}{\partial \mathrm{y}}+\frac{\partial {\mathrm{v}}_y}{\partial \mathrm{z}}$

Step 2: Define a vector.

The vector point function is a function which possess both magnitude and direction, the mathematical representation of a vector point function is as follows:

$\mathbf{a}\mathbf{=}{\mathbf{a}}_{\mathbf{x}}\mathbf{i}\mathbf{+}{\mathbf{a}}_{\mathbf{y}}\mathbf{j}\mathbf{+}{\mathbf{a}}_{\mathbf{z}}\mathbf{k}$

Here,${}^{\mathbf{ax}\mathbf{,}\mathbf{ay}\mathbf{,}\mathbf{xz}}$ are the components of vector function${}^{\mathbf{a}}$ in ${}^{\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}}$ plane respectively.

Step 3: Set up relation between two set of dimensions.

Write the value of the ${}^{\overline{\mathbf{y}}}$ and ${}^{\overline{\mathbf{z}}}$, as

 $$\begin{gathered} \overline{y}=y\cos \varphi +z\sin \varphi .......\left(1\right) \\ \overline{z}=-y\sin \varphi +z\cos \varphi .......\left(2\right) \end{gathered}$$

Multiply  on both side of equation (1) as

$\overline{z}\cos \varphi =y{\cos }^2\varphi +z\sin \varphi \cos \varphi .........\left(3\right)$

Multiply${}^{\sin \varnothing }$ on both side of equation (2) as

$\overline{z}\sin \varphi =-y{\sin }^2\varphi +z\cos \varphi \sin \varphi .........\left(4\right)$

Subtract equation (4) from equation (3)

$y=\overline{y}\cos \varphi -\overline{z}\sin \varphi$

Differentiate above equation with respect to y and z as

$\frac{\partial y}{\partial \overline{y}}=cos\varphi$

$\frac{\partial y}{\partial \overline{z}}=-\sin \varphi$

Multiply on both side of equation (1) as

$\overline{y}\sin \varphi =y\mathrm{cossin}\varphi +z^2\sin \varphi .....\left(5\right)$

Multiply${}^{\cos \varphi }$ on both side of equation (2) as

$\overline{z}\cos \varphi =-\overline{y}\sin \varphi \cos \varphi +\overline{z}\cos \varphi .....\left(6\right)$

Add equation (5) and equation (6)

$z=\overline{y}\sin \varphi +\overline{z}\cos \varphi$

Differentiate above equation with respect to y and z as

$$\begin{gathered} \frac{\partial z}{\partial \overline{y}}=\sin \varphi \\ \frac{\partial z}{\partial \overline{z}}=\cos \varphi \end{gathered}$$

Step 4: Prove the required equation.

The position of ${}^{x,y,z}$ axis with respect to ${}^{\overline{x},\overline{y},\overline{z}}$ can be represented via matrix as

$\left[\frac{\overline{v_y}}{v_z}\right]\left[\begin{array}{cc}\cos \varphi & \sin \varphi \\ -\sin \varphi & \cos \varphi \end{array}\right]\left[\begin{array}{c}{v}_y\\ v_z\end{array}\right]$

Expand above matrix as

$\frac{\overline{v_y}}{v_z}\begin{array}{cc}=v_{y } \cos \varphi +v_z\sin \varphi & \\ =-v_y \sin \varphi + v_z\cos \varphi & \end{array}$

Partially differentiate${}^{\overline{v_y}}$ with respect to ${}^{\overline{\partial y}}$ using chain rule, as

$$\begin{gathered} \frac{\partial \overline{v_y}}{\partial y}=\frac{\partial }{\partial y}(v_y\cos \varphi +v_z\sin \varphi ) \\ =\frac{\partial \overline{v_y}}{\partial y}\cos \varphi +\frac{\partial v_z}{\partial y}\sin \varphi \\ =\left(\frac{\partial v_y}{\partial y}\frac{\partial y}{\partial y}+\frac{\partial v_y}{\partial y}\frac{\partial z}{\partial y}\right)\cos \varphi +\left(\frac{\partial v_z}{\partial y}\frac{\partial y}{\partial y}+\frac{\partial v_z}{\partial z}\frac{\partial z}{\partial y}\right)\sin \varphi \end{gathered}$$

Substitute  $\sin \varphi$ for $\frac{\partial z}{\partial \overline{y}},\cos \varphi$ for $\frac{\partial z}{\partial \overline{z}},\cos \varphi$ for  $\frac{\partial y}{\partial \overline{y}}$ and  $-\sin \varphi$ for $\frac{\partial y}{\partial \overline{z}}$  int above expression.

$\frac{\partial \overline{v_y}}{\partial y}=\left(\frac{\partial v_y}{\partial y}\cos \varphi +\frac{\partial v_y}{\partial z}\sin \varphi \right)\cos \varphi +\left(\frac{\partial v_z}{\partial y}\cos \varphi +\frac{\partial v_z}{\partial z}\sin \varphi \right)\sin \varphi$
 

Partially differentiate $\overline{v_z}$  with resect to $\overline{\partial z}$ ,using chain rule, as

  $$\begin{gathered} \frac{\partial \overline{v_y}}{\partial z}=\frac{\partial v_y}{\partial y}(-v_y\sin \varphi +v_z\cos \varphi ) \\ =\frac{\partial v_y}{\partial \overline{z}}\sin \varphi +\frac{\partial v_z}{\partial \overline{z}}\cos \varphi \\ =\left(\frac{\partial v_y}{\partial y}\frac{\partial y}{\partial \overline{yz}}+\frac{\partial v_y}{\partial z}\frac{\partial z}{\partial \overline{z}}\right)\sin \varphi +\left(\frac{\partial v_z}{\partial y}\frac{\partial y}{\partial \overline{z}}+\frac{\partial v_z}{\partial z}\frac{\partial z}{\partial \overline{z}}\right)\cos \varphi \end{gathered}$$

Substitute  $\sin \varphi$ for $\frac{\partial z}{\partial \overline{y}} ,\cos \varphi$ for $\frac{\partial z}{\partial \overline{z}} ,\cos \varphi$ for $\frac{\partial y}{\partial \overline{y}} ,\cos \varphi$ and $-\sin \varphi for \frac{\partial y}{\partial \overline{z}}$  for   into above expression.

$\frac{\partial \overline{v_y}}{\partial z}=-\left(\frac{\partial v_y}{\partial y}(-\sin \varphi )+\frac{\partial v_y}{\partial z}\cos \varphi \right)\sin \varphi +\left(\frac{\partial v_z}{\partial y}(-\sin \varphi )+\frac{\partial v_z}{\partial z}\cos \varphi \right)\cos \left(\varphi \right)$
 

Add the expression for$\frac{\partial \overline{v_y}}{\partial z}and \frac{\partial \overline{v_y}}{\partial y}as,\frac{\partial \overline{v_y}}{\partial z}+\frac{\partial \overline{v_y}}{\partial y}$ 

Substitute $-\left(\frac{\partial v_y}{\partial y}(-\sin \varphi )+\frac{\partial v_y}{\partial z}\cos \varphi \right)\sin \varphi +\left(\frac{\partial v_z}{\partial y}(-\sin \varphi )+\frac{\partial v_z}{\partial z}\cos \varphi \right)\cos \left(\varphi \right)$  for $\frac{\partial \overline{v_y}}{\partial z}$  , and$\left(\frac{\partial v_y}{\partial y}\cos \varphi +\frac{\partial v_y}{\partial z}\sin \varphi \right)\cos \varphi +\left(\frac{\partial v_z}{\partial y}(\cos \varphi )+\frac{\partial v_z}{\partial z}\sin \varphi \right)\sin \varphi$   for  $\frac{\partial \overline{v_y}}{\partial y}$ into $\frac{\partial \overline{v_y}}{\partial z}+\frac{\partial \overline{v_y}}{\partial y}$ 

$$\begin{gathered} \frac{\partial \overline{v_y}}{\partial z}+\frac{\partial \overline{v_y}}{\partial y}=\left[\begin{array}{cc}\left(\frac{\partial v_y}{\partial y}(-\sin \varphi )+\frac{\partial v_y}{\partial z}\cos \varphi \right)\sin \varphi +\left(\frac{\partial v_z}{\partial y}(-\sin \varphi )+\frac{\partial v_z}{\partial z}\cos \varphi \right)\cos \left(\varphi \right)& \\ +\left(\frac{\partial v_y}{\partial y}\cos \varphi +\frac{\partial v_y}{\partial z}\sin \varphi \right)\cos \varphi +\left(\frac{\partial v_z}{\partial y}(\cos \varphi )+\frac{\partial v_z}{\partial z}\sin \varphi \right)\sin \varphi & \end{array}\right] \\ =\frac{\partial v_y}{\partial y}({\cos }^2\varphi +{\sin }^2\varphi )+\frac{\partial v_y}{\partial z}\left({\cos }^2\varphi +{\sin }^2\varphi \right) \\ =\frac{\partial v_y}{\partial y}\left(1\right)+\frac{\partial v_y}{\partial z}\left(1\right) \\ =\frac{\partial v_y}{\partial y} +\frac{\partial v_y}{\partial z} \end{gathered}$$ 

 

It is obtained that $\frac{\partial \overline{v_y}}{\partial z}+\frac{\partial \overline{v_y}}{\partial y}=\frac{\partial v_y}{\partial y}+\frac{\partial v_y}{\partial z}$ . 

 

Thus, the divergence transforms as a scalar under rotation, in two dimensions as $\frac{\partial \overline{v_y}}{\partial z}+\frac{\partial \overline{v_y}}{\partial y}=\frac{\partial \overline{v_y}}{\partial y}+\frac{\partial \overline{v_y}}{\partial z}$