# Spherical Coordinates

1. Sep 3, 2004

I'm trying to find the line element in spherical coordinates as well as a velocity element. I know that they are (ds)^2=(dr)^2+r^2(sin(theta))^2(dtheta)^2+r^2(dphi)^2 and sqrt[(dr/dt)^2+r^2(sin theta)^2(dtheta/dt)^2+r^2(dphi/dt)^2].

I know that this should be a quick and easy problem, but I simply can not figure it out. I would really appreciate some help on this one.

2. Sep 3, 2004

### Tide

Start from Cartesian coordinates in which $ds^2 = dx^2+dy^2+dz^2$ then calculate the differentials dx, dy and dz using:

$$x = r \sin \theta \cos \phi$$
$$y = r \sin \theta \sin \phi$$
$$z = r \cos \theta$$

Substitute for dx, dy and dz in $ds^2 = dx^2+dy^2+dz^2$ and after a bit of algebra you should get the desired result.

3. Sep 3, 2004

### robphy

To find the velocity vector, write the position vector as $r(t) \hat r$.
Then $v(t)=\frac{d}{dt} \left[ r(t) \hat r \right]$.
Use the product rule.
You'll have to compute $\frac{d}{dt} \hat r$,
where $$\hat r= \sin\theta\cos\phi \hat\imath + \sin\theta\sin\phi \hat\jmath + \cos\theta \hat k$$.

To simplify what you get, you might find it useful to know that
$$\hat \theta= \cos\theta\cos\phi \hat\imath + \cos\theta\sin\phi \hat\jmath - \sin\theta \hat k$$
and $$\hat \phi= -\sin\phi \hat\imath + \cos\phi \hat\jmath$$

You can derive these expressions for the spherical-polar unit vectors if you calculate the vectorial element
$$d \vec s = (dx)\hat \imath + (dy)\hat \jmath + (dz)\hat k$$
using Tide's expressions for x, y, and z. [The strategy is to group the terms in $dr$, $d\theta$, and $d\phi$.]