91Ó°ÊÓ

The spherical coordinates are defined in terms of $r,\mathrm{θ},\mathrm{ϕ}$, where r is the distance from origin, $\mathrm{θ}$is the pole angle and $\mathrm{ϕ}$ is the azimuthal angle.

The spherical coordinates can be drawn as,

The scalar potentials is $v=\frac{{\displaystyle \stackrel{âˆ\S }{r}}}{{\displaystyle r^2}}$ and the position vector is $\stackrel{→}{r}=x \text{i}+y \text{j}+z \text{k}$. The unit vector in the direction of $\stackrel{→}{r}$, is obtained as,

$\begin{array}{rcl}\stackrel{âˆ\S }{r}& =& \frac{\stackrel{→}{r}}{\left|r\right|}\\ & =& \frac{x \text{i}+y \text{j}+z \text{k}}{\left|\sqrt{x^2+y^2+z^2}\right|}\end{array}$

The spherical coordinates of the system is defined as,

$x=\left(r\sin \mathrm{θ}\right)\cos \mathrm{ϕ}$

$y=\left(r\sin \mathrm{θ}\right)\sin \mathrm{ϕ}$

$z=r\cos \mathrm{θ}$

Substitute $\left(r\sin \mathrm{θ}\right)\cos \mathrm{ϕ}$for x , $\left(r\sin \mathrm{θ}\right)\sin \mathrm{ϕ}$for y and $r\cos \mathrm{θ}$ for z into

$\stackrel{→}{r}=x \text{i}+y \text{j}+z \text{k}$

$\begin{array}{rcl}\stackrel{→}{r}& =& x \text{i}+y \text{j}+z \text{k}\\ & =& \left(r\sin \mathrm{θ}\right)\cos \mathrm{ϕ} \text{i}+\left(r\sin \mathrm{θ}\right)\sin \mathrm{ϕ} \text{j}+r\cos \mathrm{θ} k\\ & & \end{array}$

The unit vector $\stackrel{âˆ\S }{r}$is obtained as$\stackrel{âˆ\S }{r}=\sin \mathrm{θ}\cos \mathrm{Ï•} \stackrel{âˆ\S }{x}+\sin \mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}+\cos \mathrm{θ}\stackrel{âˆ\S }{z}$

The infinitesimal displacement along the direction$\mathrm{θ}$, is obtained as ${\stackrel{→}{dl}}_{\mathrm{θ}}=rd\mathrm{θ}\stackrel{âˆ\S }{\mathrm{θ}}$ ……. (3)

The infinitesimal displacement along the direction $\mathrm{θ}$, in terms of Cartesian coordinates is written as,

${\stackrel{→}{dl}}_{\mathrm{θ}}=dx\stackrel{âˆ\S }{x}+dy\stackrel{âˆ\S }{y}+dz\stackrel{âˆ\S }{x}$

As $x=\left(r\sin \mathrm{θ}\right)\cos \mathrm{ϕ}$ , $y=\left(r\sin \mathrm{θ}\right)\sin \mathrm{ϕ}$, $z=r\cos \mathrm{θ}$, infinitesimal displacement along the direction $\mathrm{θ}$, can be written as,

${\stackrel{→}{dl}}_{\mathrm{θ}}=\left(\left(r\sin \mathrm{θ}\right)\cos \mathrm{Ï•}\right)\stackrel{âˆ\S }{x}+\left(\left(r\sin \mathrm{θ}\right)\sin \mathrm{Ï•}\right)\stackrel{âˆ\S }{y}+\left(r\cos \mathrm{θ}\right)\stackrel{âˆ\S }{z}$

From equation (3), ${\stackrel{→}{dl}}_{\mathrm{θ}}=rd\mathrm{θ}\stackrel{âˆ\S }{\mathrm{θ}}$

$rd\mathrm{θ} \stackrel{âˆ\S }{\mathrm{θ}}=\left(\left(r\sin \mathrm{θ}\right)\cos \mathrm{Ï•}\right)\stackrel{âˆ\S }{x}+\left(\left(r\sin \mathrm{θ}\right)\sin \mathrm{Ï•}\right)\stackrel{âˆ\S }{y}+\left(r\cos \mathrm{θ}\right)\stackrel{âˆ\S }{z}$

The infinitesimal displacement along the direction $\mathrm{θ}$, is obtained as

${\stackrel{→}{dl}}_{\mathrm{Ï•}}=r\sin \mathrm{θ}d\mathrm{Ï•}\stackrel{âˆ\S }{\mathrm{Ï•}}$ ……. (3)

The infinitesimal displacement along the direction $\mathrm{θ}$, in terms of Cartesian coordinates is written as,

${\stackrel{→}{dl}}_{\mathrm{θ}}=dx \stackrel{âˆ\S }{x}+dy\stackrel{âˆ\S }{y}+dz\stackrel{âˆ\S }{z}$

As $x=\left(r\sin \mathrm{θ}\right)\cos \mathrm{ϕ}$, $y=\left(r\sin \mathrm{θ}\right)\sin \mathrm{ϕ}$, $z=r\cos \mathrm{θ}$, infinitesimal displacement along the direction $\mathrm{θ}$, can be written as,

${\stackrel{→}{dl}}_{\mathrm{θ}}=\left(\left(r\sin \mathrm{θ}\right)\cos \mathrm{Ï•}\right)\stackrel{âˆ\S }{x}+\left(\left(r\sin \mathrm{θ}\right)\sin \mathrm{Ï•}\right)\stackrel{âˆ\S }{y} +\left(r\cos \mathrm{θ}\right)\stackrel{âˆ\S }{z}$

From equation (3), ${\stackrel{→}{dl}}_{\mathrm{Ï•}}=r\sin \mathrm{θ}d\mathrm{Ï•}\stackrel{âˆ\S }{\mathrm{Ï•}}$

$\begin{array}{rcl}r\sin \mathrm{θ}d\mathrm{Ï•}\stackrel{âˆ\S }{\mathrm{Ï•}}& =& -\left(\left(r\sin \mathrm{θ}\right)\cos \mathrm{Ï•}\right)\stackrel{âˆ\S }{x}+\left(\left(r\sin \mathrm{θ}\right)\sin \mathrm{Ï•}\right)\stackrel{âˆ\S }{y}\\ \stackrel{âˆ\S }{\mathrm{Ï•}}& =&   -\sin \mathrm{Ï•}\stackrel{âˆ\S }{x}+\cos \mathrm{Ï•} \stackrel{âˆ\S }{y}\end{array}$

The product of $\stackrel{âˆ\S }{r}.\stackrel{âˆ\S }{r}$, is calculated as,

$\begin{array}{rcl}\stackrel{âˆ\S }{r}.\stackrel{âˆ\S }{r}& =& {\sin }^2\mathrm{θ}\left({\cos }^2\mathrm{Ï•}+{\sin }^2\mathrm{Ï•}\right)+{\cos }^2\mathrm{θ}\\ & =& {\sin }^2\mathrm{θ}+{\cos }^2\mathrm{θ}\\ & =& 1\\ & & \end{array}$

Multiply the vectors $\stackrel{âˆ\S }{\mathrm{θ}}$ and $\stackrel{âˆ\S }{\mathrm{Ï•}}$

$\begin{array}{rcl}\stackrel{âˆ\S }{\mathrm{θ}}.\stackrel{âˆ\S }{\mathrm{Ï•}}& =& -\cos \mathrm{θ}\sin \mathrm{Ï•}\cos \mathrm{Ï•}+\cos \mathrm{θ}\sin \mathrm{Ï•}\cos \mathrm{Ï•}\\ & =& 0\\ & & \end{array}$

As$x=\left(r\sin \mathrm{θ}\right)\cos \mathrm{ϕ}$, $y=\left(r\sin \mathrm{θ}\right)\sin \mathrm{ϕ}$, $z=r\cos \mathrm{θ}$, the position vector

$\stackrel{âˆ\S }{r}=\left(r\sin \mathrm{θ}\right)\cos  \stackrel{âˆ\S }{x}+\sin \mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}+\cos \mathrm{θ}\stackrel{âˆ\S }{z}$

Multiply above equation by $\sin \mathrm{θ}$on both sides,

$\sin \mathrm{θ}\stackrel{âˆ\S }{r}={\sin }^2\mathrm{θ}\cos \stackrel{âˆ\S }{x} +{\sin }^2\mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}+\sin \mathrm{θ}\cos \mathrm{θ} \stackrel{âˆ\S }{z}$ ……. (1)

Now the theta vector is $\stackrel{âˆ\S }{\mathrm{θ}}=\cos \mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{x}+\cos \mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}-\sin \mathrm{θ}\stackrel{âˆ\S }{z}$

Multiply above equation by $\cos \mathrm{Ï•}$on both sides,

$\cos \mathrm{θ}\stackrel{âˆ\S }{\mathrm{θ}}={\cos }^2\mathrm{θ}\cos \mathrm{Ï•} \stackrel{âˆ\S }{x}+{\cos }^2\mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}-\sin \mathrm{θ}\cos \mathrm{θ} \stackrel{âˆ\S }{z}$ ……. (2)

Add equations (1) and (2) as,

$\begin{array}{rcl}\sin \mathrm{θ}\stackrel{âˆ\S }{r}+\cos \mathrm{θ}\stackrel{âˆ\S }{\mathrm{θ}}& =& {\sin }^2\mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{x}+{\sin }^2\mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}+\sin \mathrm{θ}\cos \mathrm{θ}\stackrel{âˆ\S }{z}+{\cos }^2\mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{x}+{\cos }^2\mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}-\sin \mathrm{θ}\cos \mathrm{θ}\stackrel{âˆ\S }{z}\\ & =& {\sin }^2\mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{x}+{\sin }^2\mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}+{\cos }^2\mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{x}+{\cos }^2\mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}\\ & =& \left({\sin }^2\mathrm{θ}+{\cos }^2\mathrm{θ}\stackrel{âˆ\S }{x}\right)\cos \mathrm{Ï•}\stackrel{}{x+\left({\sin }^2\mathrm{θ}+{\cos }^2\mathrm{θ}\right)\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}}\\ & =& \cos \mathrm{Ï•}\stackrel{âˆ\S }{x}+\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}\end{array}$

solve further as,

$\stackrel{âˆ\S }{\mathrm{Ï•}}=-\sin \mathrm{Ï•}\stackrel{âˆ\S }{x}+\cos \mathrm{Ï•}\stackrel{âˆ\S }{y}$

Multiply $\sin \mathrm{θ}\stackrel{âˆ\S }{r}+\cos \mathrm{θ}\stackrel{âˆ\S }{\mathrm{θ}}=\cos \mathrm{Ï•}\stackrel{âˆ\S }{x}+\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}$ by $\cos \mathrm{Ï•}$on both sides,

$\sin \mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{r}+\cos \mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{\mathrm{θ}}={\cos }^2\mathrm{Ï•}\stackrel{âˆ\S }{x}+\sin \mathrm{Ï•}\cos \mathrm{Ï•}\stackrel{âˆ\S }{y}$ ……. (3)

Multiply $\stackrel{âˆ\S }{\mathrm{Ï•}}=-\sin \mathrm{Ï•}\stackrel{âˆ\S }{x}+\cos \mathrm{Ï•}\stackrel{âˆ\S }{y}$by $\sin \mathrm{Ï•}$on both sides,

$\sin \mathrm{Ï•}\stackrel{âˆ\S }{\mathrm{Ï•}}=-{\sin }^2\mathrm{Ï•}\stackrel{âˆ\S }{x}+\cos \mathrm{Ï•}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}$ ……. (4)

Subtract equation (4) from equation (3).

$\begin{array}{rcl}\sin \mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{r}+\cos \mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{\mathrm{θ}}-\sin \mathrm{Ï•}\stackrel{âˆ\S }{\mathrm{Ï•}}& =& {\cos }^2\mathrm{Ï•}\stackrel{âˆ\S }{x}+\sin \mathrm{Ï•}\cos \mathrm{Ï•}\stackrel{âˆ\S }{y}-\sin \mathrm{Ï•}\cos \mathrm{Ï•}\stackrel{âˆ\S }{y}+{\sin }^2\mathrm{Ï•}\stackrel{âˆ\S }{x}\\ & =& {\cos }^2\mathrm{Ï•}\stackrel{âˆ\S }{x}+{\sin }^2\mathrm{Ï•}\stackrel{âˆ\S }{x}\\ & =& \stackrel{âˆ\S }{x}\end{array}$

Thus, $\stackrel{âˆ\S }{x}=\sin \mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{r}+\cos \mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{\mathrm{θ}}-\sin \mathrm{Ï•}\stackrel{âˆ\S }{\mathrm{Ï•}}$

Multiply $\sin \mathrm{θ}\stackrel{âˆ\S }{r}+\cos \mathrm{θ}\stackrel{âˆ\S }{\mathrm{θ}}=\cos \mathrm{Ï•}\stackrel{âˆ\S }{x} +\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}$by $\sin \mathrm{Ï•}$ on both sides,

$\sin \mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{r}+\cos \mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{\mathrm{θ}}=\cos \mathrm{Ï•}\sin \mathrm{Ï•} \stackrel{âˆ\S }{x}+{\sin }^2\mathrm{Ï•}\stackrel{âˆ\S }{y}$ ……. (5)

Multiply $\stackrel{âˆ\S }{\mathrm{Ï•}}=-\sin \mathrm{Ï•} \stackrel{âˆ\S }{x}+\cos \mathrm{Ï•}\stackrel{âˆ\S }{y}$ by $\cos \mathrm{Ï•}$ on both sides,

$\cos \mathrm{Ï•}\stackrel{âˆ\S }{\mathrm{Ï•}}=-\sin \mathrm{Ï•} \cos \mathrm{Ï•}\stackrel{âˆ\S }{x}+{\cos }^2\mathrm{Ï•}\stackrel{âˆ\S }{y}$ ……. (5)

Add equation (5) and (6).

$\begin{array}{rcl}\sin \mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{r}+\cos \mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{\mathrm{θ}}+\cos \mathrm{Ï•}\stackrel{âˆ\S }{\mathrm{Ï•}}& =& \cos \mathrm{Ï•}\sin \mathrm{Ï•}\stackrel{âˆ\S }{x}+{\sin }^2\mathrm{Ï•}\stackrel{âˆ\S }{y}-\sin \mathrm{Ï•} \cos \mathrm{Ï•}\stackrel{âˆ\S }{x}+{\cos }^2\mathrm{Ï•}\stackrel{âˆ\S }{y}\\ & =& {\sin }^2\mathrm{Ï•}\stackrel{âˆ\S }{y}+{\cos }^2\mathrm{Ï•}\stackrel{âˆ\S }{y}\\ & =& \stackrel{âˆ\S }{y}\end{array}$

Thus, $\stackrel{âˆ\S }{y}=\begin{array}{rcl}& & \sin \end{array}\begin{array}{rcl}& & \mathrm{θ}\end{array}\begin{array}{rcl}& & \sin \end{array}\begin{array}{rcl}& & \mathrm{Ï•}\end{array}\begin{array}{rcl}& & \stackrel{âˆ\S }{r}\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & \cos \end{array}\begin{array}{rcl}& & \mathrm{θ}\end{array}\begin{array}{rcl}& & \sin \end{array}\begin{array}{rcl}& & \mathrm{Ï•}\end{array}\begin{array}{rcl}& & \stackrel{âˆ\S }{\mathrm{θ}}\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & \cos \end{array}\begin{array}{rcl}& & \mathrm{Ï•}\end{array}\begin{array}{rcl}& & \stackrel{âˆ\S }{\mathrm{Ï•}}\end{array}$

.

As $x=\left(r\sin \mathrm{θ}\right)\cos \mathrm{ϕ}$, $y=\left(r\sin \mathrm{θ}\right)\sin \mathrm{ϕ}$, $z=r\cos \mathrm{θ}$, the position vector is

$\stackrel{}{r}=\left(r\sin \mathrm{θ}\right)\cos \stackrel{âˆ\S }{x}+\sin \mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}+\cos \mathrm{θ}\stackrel{âˆ\S }{z}$

Multiply above equation by $\cos \mathrm{θ}$on both sides,

$\cos \mathrm{θ}\stackrel{âˆ\S }{r}=\sin \mathrm{θ}\cos \mathrm{θ}\cos \stackrel{âˆ\S }{x}+\sin \mathrm{θ}\cos \mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}+{\cos }^2\mathrm{θ}\stackrel{âˆ\S }{z}$ ……. (6)

Now the theta vector is $\stackrel{âˆ\S }{\mathrm{θ}}==\cos \mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{x}+\cos \mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}-\sin \mathrm{θ} \stackrel{âˆ\S }{z}$

Multiply above equation by $\sin \mathrm{θ}$on both sides,

$\sin \mathrm{θ}\stackrel{âˆ\S }{\mathrm{θ}}=\sin \mathrm{θ}\cos \mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{x}+\sin \mathrm{θ}\cos \mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}-{\sin }^2\mathrm{θ}\stackrel{âˆ\S }{z}$ ……. (7)

Subtract equation (7) from equation (8) as,

$\begin{array}{rcl}\cos \mathrm{θ}\stackrel{âˆ\S }{r}-\sin \mathrm{θ}\stackrel{âˆ\S }{\mathrm{θ}}& =& \left[\sin \mathrm{θ}\cos \mathrm{θ}\cos \stackrel{âˆ\S }{x}+\sin \mathrm{θ}\cos \mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}+{\cos }^2\mathrm{θ}\stackrel{âˆ\S }{z}-\sin \mathrm{θ}\cos \mathrm{θ}\cos \mathrm{Ï•}\stackrel{âˆ\S }{x}-\sin \mathrm{θ}\cos \mathrm{θ}\sin \mathrm{Ï•}\stackrel{âˆ\S }{y}+{\sin }^2\mathrm{θ}\stackrel{âˆ\S }{z}\right]\\ & =& {\cos }^2\mathrm{θ} \stackrel{âˆ\S }{z}+{\sin }^2\mathrm{θ} \stackrel{âˆ\S }{z}\\ & =& \stackrel{âˆ\S }{z}\\ & & \end{array}$

Thus, $\stackrel{âˆ\S }{z}=\cos \mathrm{θ}\stackrel{âˆ\S }{r}-\sin \mathrm{θ}\stackrel{âˆ\S }{\mathrm{θ}}$