---
title: "Cylindrical polar coordinates"
firstLetter: "C"
publishDate: 2021-07-26
categories:
- Mathematics
- Physics

date: 2021-07-26T16:08:46+02:00
draft: false
markup: pandoc
---

# Cylindrical polar coordinates

**Cylindrical polar coordinates** are an extension of polar coordinates to 3D,
which describes the location of a point in space
using the coordinates $(r, \varphi, z)$.
The $z$-axis is unchanged from Cartesian coordinates,
hence it is called a *cylindrical* system.

Cartesian coordinates $(x, y, z)$
and the cylindrical system $(r, \varphi, z)$ are related by:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            x &= r \cos\varphi \\
            y &= r \sin\varphi \\
            z &= z
        \end{aligned}
    }
\end{aligned}$$

Conversely, a point given in $(x, y, z)$
can be converted to $(r, \varphi, z)$
using these formulae:

$$\begin{aligned}
    \boxed{
        r = \sqrt{x^2 + y^2}
        \qquad
        \varphi = \mathtt{atan2}(y, x)
        \qquad
        z = z
    }
\end{aligned}$$

The cylindrical polar coordinates form an orthogonal
[curvilinear system](/know/concept/curvilinear-coordinates/),
whose scale factors $h_r$, $h_\varphi$ and $h_z$ we want to find.
To do so, we calculate the differentials of the Cartesian coordinates:

$$\begin{aligned}
    \dd{x} = \dd{r} \cos\varphi - \dd{\varphi} r \sin\varphi
    \qquad
    \dd{y} = \dd{r} \sin\varphi + \dd{\varphi} r \cos\varphi
    \qquad
    \dd{z} = \dd{z}
\end{aligned}$$

And then we calculate the line element $\dd{\ell}^2$,
skipping many terms thanks to orthogonality,

$$\begin{aligned}
    \dd{\ell}^2
    &= \dd{r}^2 \big( \cos^2(\varphi) + \sin^2(\varphi) \big)
    + \dd{\varphi}^2 \big( r^2 \sin^2(\varphi) + r^2 \cos^2(\varphi) \big)
    + \dd{z}^2
    \\
    &= \dd{r}^2 + r^2 \: \dd{\varphi}^2 + \dd{z}^2
\end{aligned}$$

Finally, we can simply read off
the squares of the desired scale factors
$h_r^2$, $h_\varphi^2$ and $h_z^2$:

$$\begin{aligned}
    \boxed{
        h_r = 1
        \qquad
        h_\varphi = r
        \qquad
        h_z = 1
    }
\end{aligned}$$

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

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \vu{e}_r
            &= \cos\varphi \:\vu{e}_x + \sin\varphi \:\vu{e}_y
            \\
            \vu{e}_\varphi
            &= - \sin\varphi \:\vu{e}_x + \cos\varphi \:\vu{e}_y
            \\
            \vu{e}_z
            &= \vu{e}_z
        \end{aligned}
    }
\end{aligned}$$

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

$$\begin{aligned}
    \boxed{
        \nabla f
        = \vu{e}_r \pdv{f}{r}
        + \vu{e}_\varphi \frac{1}{r} \pdv{f}{\varphi}
        + \mathbf{e}_z \pdv{f}{z}
    }
\end{aligned}$$

$$\begin{aligned}
    \boxed{
        \nabla \cdot \vb{V}
        = \frac{1}{r} \pdv{(r V_r)}{r}
        + \frac{1}{r} \pdv{V_\varphi}{\varphi}
        + \pdv{V_z}{z}
    }
\end{aligned}$$

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

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \nabla \times \vb{V}
            &= \vu{e}_r \Big( \frac{1}{r} \pdv{V_z}{\varphi} - \pdv{V_\varphi}{z} \Big)
            \\
            &+ \vu{e}_\varphi \Big( \pdv{V_r}{z} - \pdv{V_z}{r} \Big)
            \\
            &+ \frac{\vu{e}_z}{r} \Big( \pdv{(r V_\varphi)}{r} - \pdv{V_r}{\varphi} \Big)
        \end{aligned}
    }
\end{aligned}$$

The differential element of volume $\dd{V}$
takes the following form:

$$\begin{aligned}
    \boxed{
        \dd{V}
        = r \dd{r} \dd{\varphi} \dd{z}
    }
\end{aligned}$$

So, for example, an integral over all of space is converted like so:

$$\begin{aligned}
    \iiint_{-\infty}^\infty f(x, y, z) \dd{V}
    = \int_{-\infty}^{\infty} \int_0^{2\pi} \int_0^\infty f(r, \varphi, z) \: r \dd{r} \dd{\varphi} \dd{z}
\end{aligned}$$

The isosurface elements are as follows, where $S_r$ is a surface at constant $r$, etc.:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \dd{S}_r = r \dd{\varphi} \dd{z}
            \qquad
            \dd{S}_\varphi = \dd{r} \dd{z}
            \qquad
            \dd{S}_z = r \dd{r} \dd{\varphi}
        \end{aligned}
    }
\end{aligned}$$

Similarly, the normal vector element $\dd{\vu{S}}$ for an arbitrary surface is given by:

$$\begin{aligned}
    \boxed{
        \dd{\vu{S}}
        = \vu{e}_r \: r \dd{\varphi} \dd{z}
        + \vu{e}_\varphi \dd{r} \dd{z}
        + \vu{e}_z \: r \dd{r} \dd{\varphi}
    }
\end{aligned}$$

And finally, the tangent vector element $\dd{\vu{\ell}}$ of a given curve is as follows:

$$\begin{aligned}
    \boxed{
        \dd{\vu{\ell}}
        = \vu{e}_r \dd{r}
        + \vu{e}_\varphi \: r \dd{\varphi}
        + \vu{e}_z \dd{z}
    }
\end{aligned}$$


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