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Question
I wish to construct a general 3D revolved, orthogonal, curvilinear co-ordinate system, where the axis of revolution is coincident with the Cartesian $z$-axis. If the first two co-ordinates are $\alpha_1$ and $\alpha_2$, and the angle of revolution is taken to be $\phi$, the 3D co-ordinate system can be represented in Cartesian co-ordinates as:
\begin{align} x &= r(\alpha_1,\alpha_2)\cos(\phi)\\ y &= r(\alpha_1,\alpha_2)\sin(\phi)\\ z &= z(\alpha_1,\alpha_2) \end{align}
The above definition seems logically consistent to me, as when $\phi=0$, $x=r$, and $y=0$ (position vector lies in the $x-z$ plane). Moreover, we can look at various revolved co-ordinate systems such as cylindrical, spherical, toroidal, etc... which have this structure.
However, I believe the above system is under-constrained; that is, If I have chosen two co-ordinate definitions ($x$ and $y$) the third should be fully constrained to satisfy orthoganality, and I should somehow be able to find $z=z(r(\alpha_1,\alpha_2))$. So the core of my question is: What is the relationship between $z$ and $r$?
What I've tried
The base vectors are given by:
\begin{align} \mathbf{g}_1 &= \frac{\dd r}{\dd \alpha_1} \cos(\phi) \hat{\mathbf{i}} + \frac{\dd r}{\dd \alpha_1} \sin(\phi) \hat{\mathbf{j}} + \frac{\dd z}{\dd \alpha_1} \hat{\mathbf{k}} \\ \mathbf{g}_2 &= \frac{\dd r}{\dd \alpha_2} \cos(\phi) \hat{\mathbf{i}} + \frac{\dd r}{\dd \alpha_2} \sin(\phi) \hat{\mathbf{j}} + \frac{\dd z}{\dd \alpha_2} \hat{\mathbf{k}} \\ \mathbf{g}_3 &= -z\sin(\phi)\hat{\mathbf{i}} + z\cos(\phi)\hat{\mathbf{j}} \end{align}
The scale factors are given by:
\begin{align} h_1^2 &= \left(\frac{\dd r}{\dd \alpha_1}\right)^2 + \left(\frac{\dd z}{\dd \alpha_1}\right)^2 \\ h_2^2 &= \left(\frac{\dd r}{\dd \alpha_2}\right)^2 + \left(\frac{\dd z}{\dd \alpha_2}\right)^2 \\ h_3^2 &= r^2 \end{align}
To find this relationship I have tried to enforce orthonormality by writing:
\begin{gather} \mathbf{g}_1 \cdot \mathbf{g}_2 = 0 \\ \frac{\dd r}{\dd \alpha_1}\frac{\dd r}{\dd \alpha_2} + \frac{\dd z}{\dd \alpha_1}\frac{\dd z}{\dd \alpha_2} = 0 \label{eq:dot}\tag{1} \end{gather}
I have also written the cross-product between unit vectors:
\begin{equation} \hat{\mathbf{e}}_1 \times\hat{\mathbf{e}}_2 = \hat{\mathbf{e}}_3 \end{equation}
This results in two linearly dependent equations, the first is given by:
\begin{equation} \frac{1}{h_1h_2}\left(\frac{\dd r}{\dd \alpha_1}\frac{\dd z}{\dd \alpha_2} - \frac{\dd r}{\dd \alpha_2}\frac{\dd z}{\dd \alpha_1}\right) = -\frac{r}{h_3} = -1 \label{eq:cross}\tag{2} \end{equation}
After some algebra, it is possible to show that equation \eqref{eq:cross} collapses to equation \eqref{eq:dot}, which suggests any choice of $z$ is valid so long as equation \eqref{eq:dot} is satisfied. I'm not sure where the additional constraint comes from. All the 3D revolved co-ordinate systems I've looked at have a simple expression for $z$ which seems very closely related to $r$. Could I be missing something much simpler that will lead to an obvious relationship?
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While pondering this question I have since realized the fact that $r$ and $z$ must themselves form a 2D co-ordinate system (it seems really obvious in hindsight). Because any 2D orthogonal co-ordinate system can be generated using a holomorphic function, the functions $z$ and $r$ must satisfy the Cauchy-Riemann equations:
\begin{equation} \frac{\dd z}{\dd \alpha_1} = \frac{\dd r}{\dd \alpha_2} \quad\quad \frac{\dd z}{\dd \alpha_2} = -\frac{\dd r}{\dd \alpha_1} \label{eq:CR}\tag{3} \end{equation}
These are the relationships between $z$ and $r$ I was looking for. In addition, equations \eqref{eq:CR} satisfy equation (1) from the question.
Furthermore, the scale factors can all be written in terms of $r$ only:
\begin{align} h_1^2 &= h_2^2 = \left(\frac{\dd r}{\dd \alpha_1}\right)^2 + \left(\frac{\dd r}{\dd \alpha_2}\right)^2 \\ h_3^2 &= r^2 \end{align}