# What is the tangent vector representing rotation?

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?

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.

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!