Spherical Polar co-ordinates


This article is about Spherical Polar co-ordinates and is aimed for First year physics students and also for those appearing for exams like JAM/GATE etc.
Spherical Polar co-ordinates

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 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.

Motion of the particle in two dimensional plane.

Figure 2:- Motion of the particle in two dimensional plane.

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.