I am taking a course in mathematics for physicists and in one of the lectures my professor has written the following paragraph:
"given the transformation $q1 = q1(x,y,z)$, $q2 = q2(x,y,z)$, $q3 = q3(x,y,z)$ and $x = x(q1,q2,q3)$, $y = y(q1,q2,q3)$, $z = z(q1,q2,q3)$, we can use $(q1,q2,q3)$ as general coordinates that span $\mathbb R^3$: $\vec r (q1,q2,q3): \mathbb R^3 \rightarrow \mathbb R^3$"
What is the meaning of $\vec r (q1,q2,q3): \mathbb R^3 \rightarrow \mathbb R^3$? could you give an example?
I found an example in older lecture notes of another professor, which was using spherical coordinates $(\rho,\theta,\phi)$ for $(q1,q2,q3)$ and cartesian coordinates:
$\vec r (\rho,\theta,\phi): \mathbb R^3 \rightarrow \mathbb R^3 ; \vec r (\rho,\theta,\phi) = ( \rho sin(\theta)cos(\phi),\rho sin(\theta)sin(\phi),\rho cos(\theta))$. is it correct?
Thank you in advance!
I suppose that your professor is assuming $$ \overrightarrow{r} = \begin{bmatrix} x(q_1, q_2, q_3)\\ y(q_1, q_2, q_3)\\ z(q_1, q_2, q_3)\\ \end{bmatrix} $$ so $\overrightarrow{r}$ defines a bijection $\overrightarrow{r} : \mathbb{R}^3 \to \mathbb{R}^3$ which is diffeomorphic if the functions $x, y, z : \mathbb{R}^3 \to \mathbb{R}$ and $q_1, q_2, q_3 : \mathbb{R}^3 \to \mathbb{R}$ are smooth. Functions like these are often called transformations, or coordinate change by physicists.
As example you can take $$ \overrightarrow{r} = \begin{bmatrix} 2 q_2 + q_1\\ q_3^3\\ q_1 - q_2\\ \end{bmatrix} $$
This can be easily checked to be bijective and smooth, but not diffeomorphic! Moreover notice that $\mathbb{R}^3$ is not special at all: the same argument can be repeated in any dimension.
The idea behind spherical coordinates is the same, however in this case the function $$\overrightarrow{r}= \begin{bmatrix} \rho(q_1, q_2, q_3)\\ \theta(q_1, q_2, q_3)\\ \phi(q_1, q_2, q_3)\\ \end{bmatrix} $$ is not diffeomorphic anymore, so spherical coordinates are, in some way, "singular". As excercise you could think about this fact and check points where things go wrong.