How to Find Kinetic energy in terms of spherical co-ordinates

Question

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

Solution

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

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