homework and exercises - Is there a quick way of finding the kinetic energy on spherical coordinates?

Assume a particle in 3D euclidean space. Its kinetic energy: $$T = \frac{1}{2}m\left(\dot x^2 + \dot y^2 + \dot z^2\right)$$

I need to change to spherical coordinates and find its kinetic energy: $$T = \frac{1}{2}m\left(\dot r^2 + r^2\dot\theta^2 + r^2\sin^2\theta\dot\phi^2\right)$$

Its well known that: $$x = r\sin\theta\cos\phi \\ y = r\sin\theta\sin\phi \\ z = r\cos\theta$$

A way of doing it is taking the time derivatives, arriving with $3+3+2=8$ different terms with some squares, then open it arriving at $6+6+3 = 12$ different terms majority of them with 4 sine or cossine multiplications. Then to cancel out some terms somehow to arrive in this neat $3$-term expression for kinetic energy in spherical coordinates. In short, a lot of work just to arrive in a simple expression.

Here is my question: Is there a shorter way? Or even better: is there an effortless way?

Answer

There is an effortless way, if you accept geometrical reasoning.

You know, that $T = \frac 1 2 m \vec v^2 = \frac 1 2 m \lvert \vec v \rvert^2$. Furthermore, spherical coordinates are orthogonal, therefore you can just write:

$$\lvert \vec v \rvert = \sqrt{v_\phi^2 + v_\theta^2 + v_r^2}$$

Geometrically, one easily finds: $v_r = \dot r$, $v_\theta = r \dot \theta$ and $v_\phi = r \sin(\theta) \dot \phi$.

And thus the result:

$$\lvert\vec v\rvert = \sqrt{\dot r^2 + r^2 \dot \theta^2 + r^2 \sin^2(\theta) \dot \phi^2}.$$