I know, there have been multiple questions posed about sphere coordinates, but I state that mine is a little different:
originating from a manifold given by M = $\{(x,y,z)\in \mathbb{R}^3 \vert x^2+y^2+z^2=1,\quad z\geq0\}$ that describes a unit sphere u can find the simple parametrization:
$\psi_1(x,y)=\left(\begin{array}{cc}x\\y\\\sqrt{1-x^2-z^2}\end{array}\right)$
or with spheric coordinates:
$\psi_2(r,\varphi)=\left(\begin{array}{cc}r\,\cos(\varphi)\\r\,\sin(\varphi)\\\sqrt{1-r^2}\end{array}\right) \quad \text{where}\quad r\in[0,1], \varphi \in[0,2\pi]$
Now, I wonder how this should equal to the usual sphere coordinates:
$\psi_3(\varphi,\theta)=\left(\begin{array}{cc}r\,\cos(\varphi)\,r\,\sin(\theta)\\r\,\sin(\varphi)\,r\,\sin(\theta)\\r\,\cos(\theta)\end{array}\right)\quad \text{where}\quad r=1, \varphi \in[0,2\pi], \theta\in[0,\frac{\pi}{2}]$
especially because the parametrization depends on different parameter.
How are they equal to each other. Or are they at all?
Best Answer
You can describe the set $\mathbb{R}^{3}$ using 3 different geometries, these are Cartesian (usual), cylindrical and spherical geometry. Among these geometries, the way of describing a point in space changes, but they are equivalent, since there are bijective transformations to go from one to the other.
For example: From cylindrical to Cartesian coordinates \begin{eqnarray*} f: \mathbb{R}^3 & \to & \mathbb{R}^3\\ (\rho,\phi,z) & \mapsto & f(\rho,\phi,z)=(\rho\cos(\phi),\rho\sin(\phi),z)=(x,y,z) \end{eqnarray*} this function is a biyection for $\phi\in[0,2\pi)$, $\rho\geq 0$ and $z\in\mathbb{R}$.
The above allows writing different regions through parameterizations that may be easier.