Q14P - Introduction to Electrodynamics

Question

Suppose that f is a function of two variables (y and z) only. Show that the gradient ${}^{\mathbf{\nabla }\mathbf{f}\mathbf{ }\mathbf{=}\mathbf{(}\mathbf{\partial }\mathbf{f}\mathbf{/}\mathbf{\partial }\mathbf{y}\mathbf{)}\widehat{\mathbf{ }\mathbf{y}}\mathbf{ }\mathbf{(}\mathbf{\partial }\mathbf{f}\mathbf{/}\mathbf{\partial }\mathbf{z}\mathbf{)}\mathbf{ }\hat{\mathbf{z}}}$ transforms as a vector under rotations, Eq 1.29. [Hint: ${}^{\mathbf{(}\mathbf{\partial }\mathbf{f}\mathbf{/}\mathbf{\partial }\overline{\mathbf{y}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{\partial }\mathbf{f}\mathbf{/}\mathbf{\partial }\mathbf{y}\mathbf{)}\mathbf{(}\mathbf{\partial }\mathbf{f}\mathbf{/}\mathbf{\partial }\overline{\mathbf{y}}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{\partial }\mathbf{f}\mathbf{/}\mathbf{\partial }\mathbf{z}\mathbf{)}\mathbf{(}\mathbf{\partial }\mathbf{z}\mathbf{/}\mathbf{\partial }\overline{\mathbf{y}}\mathbf{)}\mathbf{,}}$and the analogous formula for ${}^{\mathbf{\partial }\mathbf{f}\mathbf{/}\mathbf{\partial }\overline{\mathbf{z}}}$. We know that ${}^{\overline{\mathbf{y}}\mathbf{ }\mathbf{=}\mathbf{ycos\varphi }\mathbf{ }\mathbf{+}\mathbf{zsin\varphi }}$ and ${}^{\overline{\mathbf{z}}\mathbf{ }\mathbf{=}\mathbf{-}\mathbf{ycos}\mathbf{\varnothing }\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{zcos}\mathbf{\varnothing }\mathbf{;}}$”solve” these equations for y and z (as functions of ${}^{\overline{\mathbf{y}}}$ and ${}^{\overline{\mathbf{z}}}$ (as functions of ${}^{\overline{\mathbf{y}}}$ and ${}^{\overline{\mathbf{z}}}$), and compute the needed derivatives ${}^{\mathbf{\partial }\mathbf{f}\mathbf{/}\mathbf{\partial }{\overline{\mathbf{y}}}_{\mathbf{,}}\mathbf{ }\mathbf{\partial }\mathbf{z}\mathbf{/}\mathbf{\partial }\overline{\mathbf{y}}\mathbf{ }}$, etc]

Step-by-Step Solution

Answer

The matrix ${}^{\left( \overline{\nabla f} \right)=\left[\begin{array}{c}\cos \varnothing \sin \varnothing \\ -\sin \varnothing \cos \varnothing \end{array}\right]=(\nabla f)}$proves that’s ${}^{\nabla \mathrm{f}}$ is used to transform the vector under the rotation.

Step 1: Write the expression for the coordinates.

Consider the change in coordinates is ${}^{\mathbf{f}\mathbf{ }\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{\to }\mathbf{f}\mathbf{ }\mathbf{(}{\overline{\mathbf{y}}}_{\mathbf{,}\mathbf{ }}\overline{\mathbf{z}}\mathbf{ }\mathbf{)}}$.

$$\begin{gathered} \overline{y}=y \cos \varphi +z {\sin }^2\varphi \\ \overline{z} \cos \varphi =-y \sin \varphi +z\cos \varphi \end{gathered}$$

Here, the variables of the function are y and z. The shifted coordinates are ${}^{\overline{\mathrm{y}}}$ and ${}^{\overline{\mathrm{z}}}$ are the changed coordinates. The angle of rotation is ${}^{\varnothing }$.

Rewrite the equations as,

$$\begin{gathered} \overline{y} \sin \varphi =y \cos \varphi \sin \varphi +z{\sin }^2\varphi \\ \overline{z} \cos \varphi =-y \sin \varphi \cos \varphi +z {\cos }^2\varphi \end{gathered}$$

Add the equation for the two coordinates as,

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

Rewrite the coordinates as,

$$\begin{gathered} \overline{y} \cos \varphi =y {\cos }^2\varphi +z \sin \varphi \cos \varphi \\ \overline{z} \sin \varphi =-y \sin \varphi \cos \varphi +z \cos \varphi \sin \varphi \end{gathered}$$

Subtract the two equations as,

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

Determine the partial derivatives of the equation (1).

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

Determine the partial derivatives of the equation (2).

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

Step 2: Determine the proof that ∇ f transform as the vector under rotation.

Write the expression for the gradient of the function with respect to y.

${\left(\overline{\nabla f}\right)}_y=\frac{\partial f}{\partial y}.\frac{\partial y}{\partial \overline{y}}+\frac{\partial f}{\partial z}.\frac{\partial z}{\partial \overline{y}}$

Write the equation for the gradient in terms of the partial derivative as,

${\left(\overline{\nabla f}\right)}_y={\left(\nabla f\right)}_y \cos \varphi +{\left(\nabla f\right)}_z \sin \varphi$

Write the expression for the gradient in terms of the z.

$$\begin{gathered} {\left(\overline{\nabla f}\right)}_z=\frac{\partial f}{\partial z}.\frac{\partial y}{\partial \overline{z}}+\frac{\partial \mathrm{f}}{\partial z}.\frac{\partial z}{\partial \overline{z}} \\ ={\left(\overline{\nabla f}\right)}_y-\sin \varphi +{\left(\overline{\nabla f}\right)}_z\cos \varphi \end{gathered}$$

Write the expression for the gradient in the matrix form.

$\left(\overline{\nabla f}\right)=\left[\begin{array}{cc}\cos \varphi & \sin \varphi \\ -\sin \varphi & \cos \varphi \end{array}\right]\left(\nabla f\right)$

Thus, the matrix $\left(\overline{\nabla f}\right)=\left[\begin{array}{cc}\cos \varphi & \sin \varphi \\ -\sin \varphi & \cos \varphi \end{array}\right]\left(\nabla f\right)$ proves that’s the  $\nabla f$is used to transform the vector under the rotation.