Spherical coordinates system (Spherical polar coordinates)

This article is about Spherical Polar coordinates and is aimed for First-year physics students and also for those appearing for exams like JAM/GATE etc. Spherical coordinates system (or Spherical polar coordinates) are very convenient in those problems of physics where there no preferred direction and the force in the problem is spherically symmetrical for example Coulomb’s Law due to point charge and the gravitational force due to a point mass.

Rectangular to spherical coordinates

rectangular to spherical coordinates
Figure 1:- Relationship between spherical polar coordinates and the rectangular Cartesian coordinates.
The spherical polar coordinates represent the coordinates of points on the surface of a sphere in a covariant 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}



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. $t$ is
\(\begin{array}{l}\dot x = \frac{{dx}}{{dt}} = \dot r\sin \theta \cos \phi + r\cos \theta \cos \phi \dot \theta – r\sin \theta \sin \phi \dot \phi \\\dot y = \frac{{dy}}{{dt}} = \dot r\sin \theta \sin \phi + r\cos \theta \sin \phi \dot \theta + r\sin \theta \cos \phi \dot \phi \\\dot z = \frac{{dz}}{{dt}} = \dot r\cos \theta – r\sin \theta \dot \theta \end{array}\)
\(\dot r = \frac{{dr}}{{dt}},\dot \theta = \frac{{d\theta }}{{dt}},\dot \phi = \frac{{d\phi }}{{dt}}\)
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 \(\phi \).
\(T = \frac{1}{2}m({\dot r^2} + {r^2}{\dot \theta ^2} + {r^2}{\sin ^2}\theta {\dot \phi ^2})\)

spherical coordinates unit vectors

1. $\hat {r} $: points towards the $r$ axis that is in the direction of the vector $\vec r$ along which only coordinate $r$ changes. We can also write$\vec r= r \hat {r}$.
2. $\hat {\theta}$: Unit vector $\hat {\theta}$ is tangent at $P$ to circle $SPT$ . Displacement along this circle only changes coordinate.
3. $\hat {\varphi}$: This unit vector is also tangent at $P$ if circle under consideration is $PP’$ produced by rotation of $OP$ along the z-axis. Displacement along this circle only changes $\varphi$.
The general differential displacement for any particle P in spherical polar coordinate is
$$d\vec{r}=dr\hat{r}+rd\theta \hat{\theta }+r\sin\theta d\varphi \hat{\varphi}$$
The unit vectors $\hat {r}$ ,$\hat {\theta}$ and $\hat {\varphi}$ can be expressed in terms of $\hat i$ , $\hat j$ and $\hat k$ :
\hat {r} = \sin \theta \cos\varphi \hat i + \sin \theta \sin\varphi \hat j + \cos\theta \hat k \\
\hat {\theta } = \cos \theta cos\varphi \hat i + \cos \theta \sin\varphi \hat j – \sin \theta \hat k \\
\hat {\varphi } = – \sin \varphi \hat i + \cos \varphi \hat j
The unit vectors $\left( \widehat{r},\widehat{\theta },\widehat{\varphi } \right)$ , 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{r}\cdot \hat{\varphi}&=\hat{r}\cdot \hat{\theta}=\hat{\theta}\cdot \hat{\varphi}=0 \\
\hat{r}\times \hat{\theta} &=\hat{\varphi} \\
\hat{\theta }\times \hat{\varphi }&=\hat{r} \\
\hat{\varphi}\times \hat{r}&=\hat{\theta}

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