What is the tangent vector representing rotation?

Question

I am reading Mathematics for physics: A guided tour for graduate students by Michael Stone. On the page 379, the book says

The surface of the unit sphere is a manifold...We may label its points with spherical polar coordinates, $\theta$ measuring the co-latitude and $\phi$ measuring the longitude...In this coordinate basis, the tangent vector representing the velocity field due to a rigid rotation of one radian per second about the $z$-axis is $$V_z=\partial_\phi$$

I think it is like saying that, when rotating through $\alpha$ rad about $z$-axis, a point $(1,\theta,\phi)$ transforms to $(1,\theta,\phi+\alpha)$. But I cannot understand the next line, it says

Similarly, $$V_x=-\sin \phi \partial_\theta - \cot \theta \cos \phi \partial_\phi$$

I try to think about how a point transforms after rotation but failed. Has anyone read this book or can anyone help me to understand this line?

Answer 1

I don't read the book, but I suggest some way to understand it.

Suppose $\mathbf{\omega}$ is an angular velocity of given system. Then $\mathbf{V} = \omega \times \mathbf{r}$ for $\mathbf{r}$, the coordinate of given points on the sphere. Note that, $\mathbf{V}$ is an element of tangent space. You may know this is common mathematical tool to represent the vector field on manifolds.

Now, $(\partial_x, \partial_y, \partial_z)$ be the commonly given cartesian basis vector of $\mathbb{R}^3$ and $(\partial_r, \partial_\theta, \partial_\phi)$ be the basis of tangent space (of sphere) with normal vector $\partial_r$. Then they have a relation as follow.

$$ \partial_x = \sin(\theta)\cos(\phi) \partial_r + \cos(\theta)\cos(\phi) \partial_\theta - \frac{\sin(\phi)}{\sin(\theta)}\partial_\phi\\ \partial_y = \sin(\theta)\sin(\phi) \partial_r + \cos(\theta)\sin(\phi) \partial_\theta + \frac{\cos(\phi)}{\sin(\theta)} \partial_\phi\\ \partial_z = \cos(\theta) \partial_r - \sin(\theta)\partial_\theta $$ When the case $r = 1$. (Because the sphere is an unit sphere)

In the case $\omega = \partial_z$,

Then, by calculation, we could get $$ \mathbf{V}_z = x \partial_y - y \partial_x = \partial_\phi $$

In a similar sense, if $\omega = \partial_x$, We may get, $$ \mathbf{V}_x = y \partial_z - z \partial_y = -\sin(\phi)\partial_\theta - \cos(\phi) \cot(\theta) \partial_\phi $$ what we want.

Answer 2

Thanks for reading our book!

ChoMedit's solution is correct. If you rotate with unit angular velocity about the $x$-axis, then after time $\delta t$ then the point with spherical coordinates $(\theta,\phi)$ will shift as $$ (\theta,\phi) \to (\theta-\delta t \sin\phi , \phi-\delta t \cot\theta \cos\phi ). $$ It takes some trig to figure this out geometrically, so we used the $z$ axis as the main example and hoped you would simply believe us for the other two axes!