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

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

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

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.

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

role="math" localid="1655961512676" $\overline{z}\sin \varphi =-y{\sin }^2\varphi +z\cos \varphi \sin \varphi .........\left(4\right)$

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

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

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

role="math" localid="1655964470265" $$\begin{gathered} \frac{\partial \overline{v_y}}{\partial y}=\frac{\partial }{\partial y}(v_y\cos \varphi +v_z\sin \varphi ) \\ =\frac{\partial \overline{v_y}}{\partial y}\cos \varphi +\frac{\partial v_z}{\partial y}\sin \varphi \\ =\left(\frac{\partial v_y}{\partial y}\frac{\partial y}{\partial y}+\frac{\partial v_y}{\partial y}\frac{\partial z}{\partial y}\right)\cos \varphi +\left(\frac{\partial v_z}{\partial y}\frac{\partial y}{\partial y}+\frac{\partial v_z}{\partial z}\frac{\partial z}{\partial y}\right)\sin \varphi \end{gathered}$$

Substitute $\sin \varphi$for role="math" localid="1655965188625" $\frac{\partial z}{\partial \overline{y}},\cos \varphi$for $\frac{\partial z}{\partial \overline{z}},\cos \varphi$for $\frac{\partial y}{\partial \overline{y}}$ and $-\sin \varphi$ for role="math" localid="1655965364288" $\frac{\partial y}{\partial \overline{z}}$ 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

role="math" localid="1655964873819" $$\begin{gathered} \frac{\partial \overline{v_y}}{\partial z}=\frac{\partial v_y}{\partial y}(-v_y\sin \varphi +v_z\cos \varphi ) \\ =\frac{\partial v_y}{\partial \overline{z}}\sin \varphi +\frac{\partial v_z}{\partial \overline{z}}\cos \varphi \\ =\left(\frac{\partial v_y}{\partial y}\frac{\partial y}{\partial \overline{yz}}+\frac{\partial v_y}{\partial z}\frac{\partial z}{\partial \overline{z}}\right)\sin \varphi +\left(\frac{\partial v_z}{\partial y}\frac{\partial y}{\partial \overline{z}}+\frac{\partial v_z}{\partial z}\frac{\partial z}{\partial \overline{z}}\right)\cos \varphi \end{gathered}$$

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 role="math" localid="1655964045847" $-\left(\frac{\partial v_y}{\partial y}(-\sin \varphi )+\frac{\partial v_y}{\partial z}\cos \varphi \right)\sin \varphi +\left(\frac{\partial v_z}{\partial y}(-\sin \varphi )+\frac{\partial v_z}{\partial z}\cos \varphi \right)\cos \left(\varphi \right)$ 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 $\frac{\partial \overline{v_y}}{\partial z}+\frac{\partial \overline{v_y}}{\partial y}$

role="math" localid="1655963767096" $$\begin{gathered} \frac{\partial \overline{v_y}}{\partial z}+\frac{\partial \overline{v_y}}{\partial y}=\left[\begin{array}{cc}\left(\frac{\partial v_y}{\partial y}(-\sin \varphi )+\frac{\partial v_y}{\partial z}\cos \varphi \right)\sin \varphi +\left(\frac{\partial v_z}{\partial y}(-\sin \varphi )+\frac{\partial v_z}{\partial z}\cos \varphi \right)\cos \left(\varphi \right)& \\ +\left(\frac{\partial v_y}{\partial y}\cos \varphi +\frac{\partial v_y}{\partial z}\sin \varphi \right)\cos \varphi +\left(\frac{\partial v_z}{\partial y}(\cos \varphi )+\frac{\partial v_z}{\partial z}\sin \varphi \right)\sin \varphi & \end{array}\right] \\ =\frac{\partial v_y}{\partial y}({\cos }^2\varphi +{\sin }^2\varphi )+\frac{\partial v_y}{\partial z}\left({\cos }^2\varphi +{\sin }^2\varphi \right) \\ =\frac{\partial v_y}{\partial y}\left(1\right)+\frac{\partial v_y}{\partial z}\left(1\right) \\ =\frac{\partial v_y}{\partial y}+\frac{\partial v_y}{\partial z} \end{gathered}$$

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 $\frac{\partial \overline{v_y}}{\partial z}+\frac{\partial \overline{v_y}}{\partial y}=\frac{\partial \overline{v_y}}{\partial y}+\frac{\partial \overline{v_y}}{\partial z}$