---
title: "Cylindrical parabolic coordinates"
firstLetter: "C"
publishDate: 2021-03-04
categories:
- Mathematics
- Physics

date: 2021-03-04T15:07:46+01:00
draft: false
markup: pandoc
---

# Cylindrical parabolic coordinates

**Cylindrical parabolic coordinates** are a coordinate system
that describes a point in space using three coordinates $(\sigma, \tau, z)$.
The $z$-axis is unchanged from the Cartesian system,
hence it is called a *cylindrical* system.
In the $z$-isoplane, however, confocal parabolas are used.
These coordinates can be converted to the Cartesian $(x, y, z)$ as follows:

$$\begin{aligned}
    \boxed{
        x = \frac{1}{2} (\tau^2 - \sigma^2 )
        \qquad
        y = \sigma \tau
        \qquad
        z = z
    }
\end{aligned}$$

Converting the other way is a bit trickier.
It can be done by solving the following equations,
and potentially involves some fiddling with signs:

$$\begin{aligned}
    2 x
    = \frac{y^2}{\sigma^2} - \sigma^2
    \qquad \quad
    2 x
    = - \frac{y^2}{\tau^2} + \tau^2
\end{aligned}$$

Cylindrical parabolic coordinates form an orthogonal
[curvilinear](/know/concept/curvilinear-coordinates/) system,
so we would like to find its scale factors $h_\sigma$, $h_\tau$ and $h_z$.
The differentials of the Cartesian coordinates are as follows:

$$\begin{aligned}
    \dd{x} = - \sigma \dd{\sigma} + \tau \dd{\tau}
    \qquad
    \dd{y} = \tau \dd{\sigma} + \sigma \dd{\tau}
    \qquad
    \dd{z} = \dd{z}
\end{aligned}$$

We calculate the line segment $\dd{\ell}^2$,
skipping many terms thanks to orthogonality:

$$\begin{aligned}
    \dd{\ell}^2
    &= (\sigma^2 + \tau^2) \:\dd{\sigma}^2 + (\tau^2 + \sigma^2) \:\dd{\tau}^2 + \dd{z}^2
\end{aligned}$$

From this, we can directly read off the scale factors $h_\sigma^2$, $h_\tau^2$ and $h_z^2$,
which turn out to be:

$$\begin{aligned}
    \boxed{
        h_\sigma = \sqrt{\sigma^2 + \tau^2}
        \qquad
        h_\tau = \sqrt{\sigma^2 + \tau^2}
        \qquad
        h_z = 1
    }
\end{aligned}$$

With these scale factors, we can use
the general formulae for orthogonal curvilinear coordinates
to easily to convert things from the Cartesian system.
The basis vectors are:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \vu{e}_\sigma
            &= \frac{- \sigma}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_x + \frac{\tau}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_y
            \\
            \vu{e}_\tau
            &= \frac{\tau}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_x + \frac{\sigma}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_y
            \\
            \vu{e}_z
            &= \vu{e}_z
        \end{aligned}
    }
\end{aligned}$$

The basic vector operations (gradient, divergence, Laplacian and curl) are given by:

$$\begin{aligned}
    \boxed{
        \nabla f
        = \frac{\mathbf{e}_\sigma}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\sigma}
        + \frac{\mathbf{e}_\tau}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\tau}
        + \mathbf{e}_z \pdv{f}{z}
    }
\end{aligned}$$

$$\begin{aligned}
    \boxed{
        \nabla \cdot \mathbf{V}
        = \frac{1}{\sigma^2 + \tau^2}
        \Big( \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{d\sigma} + \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{d\tau} \Big) + \pdv{V_z}{z}
    }
\end{aligned}$$

$$\begin{aligned}
    \boxed{
        \nabla^2 f
        = \frac{1}{\sigma^2 + \tau^2} \Big( \pdv[2]{f}{\sigma} + \pdv[2]{f}{\tau} \Big) + \pdv[2]{f}{z}
    }
\end{aligned}$$

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \nabla \times \mathbf{V}
            &= \mathbf{e}_\sigma \Big( \frac{\mathbf{e}_1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\tau} - \pdv{V_\tau}{z} \Big)
            \\
            &+ \mathbf{e}_\tau \Big( \pdv{V_\sigma}{z} - \frac{1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\sigma} \Big)
            \\
            &+ \frac{\mathbf{e}_z}{\sigma^2 + \tau^2}
            \Big( \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{\sigma} - \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{\tau} \Big)
        \end{aligned}
    }
\end{aligned}$$

The differential element of volume $\dd{V}$
in cylindrical parabolic coordinates is given by:

$$\begin{aligned}
    \boxed{
        \dd{V} = (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau} \dd{z}
    }
\end{aligned}$$

The differential elements of the isosurfaces are as follows,
where $\dd{S_\sigma}$ is the $\sigma$-isosurface, etc.:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \dd{S_\sigma} &= \sqrt{\sigma^2 + \tau^2} \dd{\tau} \dd{z}
            \\
            \dd{S_\tau} &= \sqrt{\sigma^2 + \tau^2} \dd{\sigma} \dd{z}
            \\
            \dd{S_z} &= (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau}
        \end{aligned}
    }
\end{aligned}$$

The normal element $\dd{\vu{S}}$ of a surface and
the tangent element $\dd{\vu{\ell}}$ of a curve are respectively:

$$\begin{aligned}
    \boxed{
        \dd{\vu{S}}
        = \mathbf{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\tau} \dd{z}
        + \mathbf{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\sigma} \dd{z}
        + \mathbf{e}_z (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau}
    }
\end{aligned}$$

$$\begin{aligned}
    \boxed{
        \dd{\vu{\ell}}
        = \mathbf{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\sigma}
        + \mathbf{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\tau}
        + \mathbf{e}_z \dd{z}
    }
\end{aligned}$$


## References
1.  M.L. Boas,
    *Mathematical methods in the physical sciences*, 2nd edition,
    Wiley.