The spherical polar coordinates represent the coordinates of points on the surface of a sphere in a co-variant form. The coordinates of the point $P$ in this system is represented by the radial vector $r$ which is the distance from the origin to the point, the polar or zenith angle $\theta$ which is the angle the radial vector makes with respect to the z axis and the azimuth or longitudinal angle $\phi$ which is which is the normal polar coordinate in the x − y plane as shown below in the figure.

These co-ordinates are related to the rectangular coordinates x, y, and z through

$$x=r\sin \theta cos\varphi$$

$$y=r\sin \theta \sin \varphi$$

$$z=r\cos \theta$$

Spherical polar coordinates in terms of Cartesian coordinates are

$$r=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$$

$$\tan \theta =\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}}{z}$$

$$\tan \varphi =\frac{y}{z}$$

Unit vectors in spherical coordinate system

**1.** $\hat {e_r}$ : points towards the $r$ axis that is in the direction of vector $\vec r$ along which only coordinate $r$ changes. We can also write$\vec r= r \hat {e_r}$.

**2.** $\hat {e_\theta}$: Unit vector $\hat {e_\theta}$ is tangent at $P$ to circle $SPT$ . Displacement along this circle only changes coordinate.

**3.** $\hat {e_\varphi}$: This unit vector is also tangent at $P$ if circle under consideration is $PP’$ produced by rotation of $OP$ along z-axis. Displacement along this circle only changes $\varphi$.

The general differential displacement for any particle P in spherical polar coordinate is

$$d\vec{r}=dr\widehat{{{e}_{r}}}+rd\theta \widehat{{{e}_{\theta }}}+rsin\theta d\varphi \widehat{{{e}_{\varphi }}}$$

The unit vectors $\hat {e_r}$ ,$\hat {e_\theta}$ and $\hat {e_\varphi}$ can be expressed in terms of $\hat i$ , $\hat j$ and $\hat k$ :

$$\hat {{e_r}} = \sin \theta cos\varphi \hat i + \sin \theta sin\varphi \hat j + cos\theta \hat k$$

$$\hat {{e_\theta }} = \cos \theta cos\varphi \hat i + \cos \theta sin\varphi \hat j – \sin \theta \hat k$$

$$\hat {{e_\varphi }} = – \sin \varphi \hat i + \cos \varphi \hat j $$

The unit vectors $\left( \widehat{{{e}_{r}}},\widehat{{{e}_{\theta }}},\widehat{{{e}_{\varphi }}} \right)$ , unlike $\left( \hat{i},\hat{j},\hat{k} \right)$ are not constant vectors but change in direction as co-ordinates $\theta$ and $\varphi$ change. At each point they constitute an orthogonal right handed co-ordinate system, that is we have

$${{\hat{e}}_{r}}\cdot {{\hat{e}}_{\varphi }}={{\hat{e}}_{r}}\cdot {{\hat{e}}_{\theta }}={{\hat{e}}_{\theta }}\cdot {{\hat{e}}_{\varphi }}=0$$

\[{{\hat{e}}_{r}}\times {{\hat{e}}_{\theta }}={{\hat{e}}_{\varphi }}\]

\[{{\hat{e}}_{\theta }}\times {{\hat{e}}_{\varphi }}={{\hat{e}}_{r}}\]

\[{{\hat{e}}_{\varphi }}\times {{\hat{e}}_{r}}={{\hat{e}}_{\theta }}\]

## Solved Problems on Spherical Polar co-ordinates

**Question 1**

How to Find Kinetic energy in terms of \((r,\theta ,\phi )\)

**Solution 1**

We have to find the kinetic energy in terms of \((r,\theta ,\phi )\) that is in terms of spherical c0-ordinates. Kinetic energy in terms of Cartesian co-ordinates is

\(T = \frac{1}{2}m\left( {{{\dot x}^2} + {{\dot y}^2} + {{\dot z}^2}} \right)\) (1)

Where \(\dot x,\dot y{\rm{ and }}\dot z\) are derivatives of z, y and z with respect to time.

Cartesian co-ordinates x, y, z in terms of \(r,\theta \) and \(\phi \) are

\(\begin{array}{l}x = r\sin \theta \cos \phi \\y = r\sin \theta \sin \phi \\z = r\cos \theta \end{array}\)

Now derivatives of x, y and z w.r.t. t are

\(\begin{array}{l}\dot x = \frac{{dx}}{{dt}} = \dot r\sin \theta \cos \phi + r\cos \theta \cos \phi \dot \theta – r\sin \theta \sin \phi \dot \phi \\\dot y = \frac{{dy}}{{dt}} = \dot r\sin \theta \sin \phi + r\cos \theta \sin \phi \dot \theta + r\sin \theta \cos \phi \dot \phi \\\dot z = \frac{{dz}}{{dt}} = \dot r\cos \theta – r\sin \theta \dot \theta \end{array}\)

Where

\(\dot r = \frac{{dr}}{{dt}},\dot \theta = \frac{{d\theta }}{{dt}},\dot \phi = \frac{{d\phi }}{{dt}}\)

means all \(r,\theta \) and \(\phi \) changes with time as the particle moves or changes its position with time.

Now calculate for

\({\dot x^2},{\dot y^2},{\dot z^2}\)

and add them. After adding them we get

\({(\dot x)^2} + {(\dot y)^2} + {(\dot z)^2} = {\dot r^2} + {r^2}{\dot \theta ^2} + {r^2}{\sin ^2}\theta {\dot \phi ^2}\)

Putting this value of \({(\dot x)^2} + {(\dot y)^2} + {(\dot z)^2}\)in equation 1 we get kinetic energy of particle or system in terms of \(r,\theta \) and \(\phi \).

Hence

\(T = \frac{1}{2}m({\dot r^2} + {r^2}{\dot \theta ^2} + {r^2}{\sin ^2}\theta {\dot \phi ^2})\)

## Motion in two dimensions

Let us denote the two dimensional plane as XY Plane as shown below in the figure.

The position vector of the particle is given by

$\vec{r}=x\hat{i}+y\hat{j}$

as $Z=0$ for particle in motion.

In two dimension case the xy plane corresponds to the situation where $\vec{r}=\rho$. The polar angle $\theta$ given in figure 1 is then $\frac{\pi}{2}$ for the particle and spherical polar coordinates $(r, \theta, \phi)$ then reduced to circular polar coordinates $(r, \phi);\theta=90^0$.

Now position of the particle P in terms of radial distance $r$ from origin O and $\phi$ is given by

$r=\sqrt{x^2+y^2}$

$x=rcos\varphi$

$tan\varphi=\frac{x}{y}$

$y=rsin\varphi$

We already have displacement vector $d\vec{r}$ in three dimensionn that is

\[d\vec{r}=dr\widehat{{{e}_{r}}}+rd\theta \widehat{{{e}_{\theta }}}+rsin\theta d\varphi \widehat{{{e}_{\varphi }}}\]

The displacement vector $d\vec r$ in circular polar coordinates can be obtained by putting $\thata = 90^0$ and $d\thata = 0$ in above equation. So , we get

\[d\vec{r}=dr\widehat{{{e}_{r}}}+rd\varphi \widehat{{{e}_{\varphi }}}\]

where unit vectors $\hat {{e_r}}$ and $\hat {{e_\varphi }}$ are given by

$$\hat {{e_r}} = cos\varphi \hat i + sin\varphi \hat j $$

$$\hat {{e_\varphi }} = – \sin \varphi \hat i + \cos \varphi \hat j $$

##### Reference books

1. Mechanics Paperback – 2007 by P.K. Srivastava

2. MECHANICS Paperback – 14 Jul 2003 by H Hans (Author), S Puri (Author)

##### Reference material for further reading

**1.** Refer this link to determine the spherical unit vectors in terms of Cartesian coordinates

http://web.physics.ucsb.edu/~fratus/phys103/Disc/disc_notes_3_pdf.pdf

**2.** http://planetmath.org/unitvectorsincurvilinearcoordinates to know more about curvilinear cordinates

**3.** Find extra stuff at this link http://mathworld.wolfram.com/SphericalCoordinates.html

Definitely there is much much more information to be added to this topic. I’ll do it some other time.