## Saturday, November 3, 2012

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