Rotations around the x and y axes of stereographic sphere

[IanBerkman]

Homework Statement

Show that the equations
$$ \delta \phi = \cot \theta \cot \phi \delta \theta, \quad \delta \phi =- \cot \theta \tan \phi \delta \theta$$
represent rotations around the x and y axes respectively of a stereographic sphere.
Both these two rotations map the sphere on itself and map any geodesic line on another geodesic line.

Homework Equations

The sphere is a momentum sphere with radius $p_0$. Assuming $\theta$ is the angle from the z-axis and $\phi$ the angle from the x-axis. We obtain the stereographic projection in the $p_x,p_y$-plane by
$$
\textbf{p} = p_0\cot(\theta/2)(\cos\phi, \sin\phi, 0)$$

The Attempt at a Solution

The meaning of $\delta$ is not clearly stated and I used it as a deviation (similar to $\Delta$).
An infinitesimal small deviation gives for the x-rotation
$$d \phi = \cot \theta \cot \phi d\theta$$
Integrating gives
$$\cos\phi = a\csc\theta$$
Where $a$ is the constant of the integration.
Intuitively I think the answer of it being a rotation around the x-axis is hidden in this equation, but I fail to see it.

Furthermore, I need to express the new stereographic project of the momentum under a rotation.
Differentiating w.r.t. $\phi$ gives
$$ \sin \phi = a \cot\theta\csc\theta \frac{d\theta}{d\phi} =
a\csc\theta\tan\phi$$
Substituting these equations of $\sin\phi$ and $\cos\phi$ into the expression $\textbf{p}$ at the top gives
$$
\textbf{p}=ap_0\cot(\theta/2)\csc\theta(1,\tan\phi,0)$$

The problem is, I do not know if I do this correct or entirely wrong. I want to hear your opinion about this.

Thanks in advance,

Ian

 

[IanBerkman]

Substituting the $\cos\phi$ term in the spherical coordinates gives an constant for $x$. Furthermore, I obtain $y=rc\tan\phi$ and $z=rc\tan\phi\sec\phi$. This gives $y^2+z^2\neq constant$, which is not possible.