Is there a quick way of finding the kinetic energy on spherical coordinates?

Question

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 1

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:

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

Answer 2

An important connection to relativity can be made here. Consider the infinitesimal displacement in the Cartesian coordinates: $$ ds^2=dx^2+dy^2+dz^2=dx^a g_{ab}dx^b $$ where $a,b\in\{1,2,3\}$ and $$ dx^a=\left(\begin{array}{c}dx\\dy\\dz\end{array}\right) $$ and $g_{ab}$ the metric, $$ g_{ab}=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right) $$

Since the displacement $ds^2$ should be the same regardless of coordinates, then (via simple geometry), we have $$ ds^2=dr^2+r^2\left(d\theta^2+\sin^2\theta d\phi^2\right)=dx^a g_{ab}dx^b $$ where now the metric takes the form $$ g_{ab}=\left(\begin{array}{ccc}1&0&0\\0&r^2&0\\0&0&r^2\sin^2\theta\end{array}\right) $$ and $$ dx^a=\left(\begin{array}{c}dr\\d\theta\\d\phi\end{array}\right) $$

Thus, you can get the velocities by dividing the displacement $dx^a$ by $dt$, leading to $$ v^av_a=\frac{dx^a}{dt} g_{ab}\frac{dx^b}{dt}\equiv\dot x^ag_{ab}\dot x^b $$ and the expected relation naturally falls out: $$ v^av_a=\begin{cases}\dot x^2+\dot y^2+\dot z^2 \\ \dot r^2+r^2\dot\theta^2+r^2\sin^2\theta\dot\phi^2\end{cases} $$

Answer 3

When you find the total (squared) value of some vector in an orthogonal basis, such as the Cartesian system $(x,y,z)$ or indeed the spherical system $(r,\theta,\phi)$, what you're doing is simply adding the squared values of each component of the vector.

Taking the velocity, let's think about the different components:

1. What is the velocity in the radial direction? That's easy; the radial vector is a straight line, just like any of the basis vectors in the Cartesian system. So the radial velocity is simply $\dot{r}$.
2. How about the azimuthal velocity? Think about the length of an arc on a circle of radius $r$. You know that this length is $r\theta$, and since you're only considering changes in the $\theta$ coordinate, the velocity is just $r\dot{\theta}$.
3. The equatorial component is approached in the same way as part (2), only now you have to account for the fact that the radius of the circle gets smaller as you move away from the equator. How much smaller? Well, you can look at the angle and use 'SOH-CAH-TOA' to convince yourself that the radius of the circle when you're at an azimuthal angle $\theta$ is just $r\sin{\theta}$. So there you have it; the equatorial component of the velocity is $r\sin({\theta})\dot{\phi}$.

Square each of these terms and add them, and that's your total $v^2$ in the spherical coordinate system. Simples!