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

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