Q37P - Introduction to Electrodynamics

Question

Question: Find formulas for  $\mathit{r}\mathbf{,}\mathit{\theta }\mathbf{,}\mathit{\varphi }$ in terms of x, y, z (the inverse, in other words, of Eq. 1.62)

Step-by-Step Solution

Answer

The formula of $r$is obtained to be equal to $r=\sqrt{x^2+y^2+z^2}$ . The formula for $\theta$ is obtained as $\theta ={\cos }^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)$ and the value of $\varphi$ is obtained as $\varphi ={\tan }^{-1}\left(\frac{y}{x}\right)$ .

Step 1: Define the spherical coordinates

The spherical coordinates are defined in terms of $\mathit{r}\mathbf{,}\mathit{\theta }\mathbf{,}\mathit{\varphi }$ , where r is the distance from origin, $\mathit{\theta }$ is the polar angle and $\mathit{\varphi }$ is the azimuthal angle.

 

The spherical coordinates is drawn as, 

 

 

In the triangle OMP, the angle M is $90°$ and the length of OP is r . From the figure, it can be written as $OM=r \sin \theta$ . Also from the figure, we can write as,

 $$\begin{gathered} x=\left(r \sin \theta \right)\cos \varphi \\ y=\left(r \sin \theta \right)\sin \varphi \\ r=r \cos \theta \end{gathered}$$

Substitute $\left(r \sin \theta \right)\cos \varphi$ for $x$ ,  $\left(r \sin \theta \right)\sin \varphi$ for $y$ and $r \cos \theta$ for  into $x^2+y^2+z^2$ .

$$\begin{gathered} x^2+y^2+z^2={\left[(r \sin \theta )\cos \theta \right]}^2+{\left[(r \sin \theta )\sin \theta \right]}^2+{\left[(r \cos \theta )\right]}^2 \\ =r^3\left[{\sin }^2\theta ({\cos }^2\varphi +{\sin }^2\theta )+\left({\cos }^2\varphi \right)\right] \\ =r^2\left[{\sin }^2\theta \left(1\right)+{\cos }^2\varphi \right] \\ =r^2 \end{gathered}$$

Thus, the formula of $r$ is obtained to be equal to $r=\sqrt{x^2+y^2+z^2}$ .

Step: 2 Obtain the formula θ for ϕ

The formula for z is $z=r \cos \theta$  , which can be rearranged as $\theta ={\cos }^{-1}\left(\frac{z}{r}\right)$ . 

Substitute $r=\sqrt{x^2+y^2+z^2}$  for $r$  into $\theta ={\cos }^{-1}\left(\frac{z}{r}\right)$ .

 $\theta ={\cos }^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)$

 

Thus formula for $\theta$ is obtained as $\theta ={\cos }^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)$ .

 

Now, Divide the formula of $y=(r \sin \theta )\sin \varphi \mathrm{by} x=(r \sin \theta )\cos \varphi \mathrm{as},$

 $$\begin{gathered} \frac{y}{x}=\frac{(r \sin \theta )\sin \varphi }{\mathbf{(}r sin\theta \mathbf{)}cos\varphi } \\ =\frac{ \sin \varphi }{\cos \varphi } \\ =\tan \varphi \end{gathered}$$

  

Therefore, $\frac{y}{x}=\tan \varphi$  , then value of $\varphi$ can be obtained as,

 $$\begin{gathered} \frac{y}{x}=\tan \varphi \\ \varphi ={\tan }^{-1}\left(\frac{y}{x}\right) \end{gathered}$$

Therefore, the value of $\varphi$ is obtained as  $\varphi ={\tan }^{-1}\left(\frac{y}{x}\right)$ .