---
title: "Spherical coordinates"
firstLetter: "S"
publishDate: 2021-03-04
categories:
- Mathematics
- Physics

date: 2021-03-04T15:05:21+01:00
draft: false
markup: pandoc
---

# Spherical coordinates

**Spherical coordinates** are an extension of polar coordinates to 3D.
The position of a given point in space is described by
three coordinates $(r, \theta, \varphi)$, defined as:

*   $r$: the **radius** or **radial distance**: distance to the origin.
*   $\theta$: the **elevation**, **polar angle** or **colatitude**:
    angle to the positive $z$-axis, or **zenith**, i.e. the "north pole".
*   $\varphi$: the **azimuth**, **azimuthal angle** or **longitude**:
    angle from the positive $x$-axis, typically in the counter-clockwise sense.

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

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

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

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

The spherical basis vectors $\vu{e}_r$, $\vu{e}_\theta$ and $\vu{e}_\varphi$
are expressed in the Cartesian basis like so:

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

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

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

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

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

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

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

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

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

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

$$\begin{aligned}
    \boxed{
        \nabla \cdot \vb{V}
        = \frac{1}{r^2} \pdv{(r^2 V_r)}{r}
        + \frac{1}{r \sin\theta} \pdv{(\sin\theta V_\theta)}{\theta}
        + \frac{1}{r \sin\theta} \pdv{V_\varphi}{\varphi}
    }
\end{aligned}$$

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

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

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

$$\begin{aligned}
    \boxed{
        \dd{V}
        = r^2 \sin\theta \dd{r} \dd{\theta} \dd{\varphi}
    }
\end{aligned}$$

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

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

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

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \dd{S}_r = r^2 \sin\theta \dd{\theta} \dd{\varphi}
            \qquad
            \dd{S}_\theta = r \sin\theta \dd{r} \dd{\varphi}
            \qquad
            \dd{S}_\varphi = r \dd{r} \dd{\theta}
        \end{aligned}
    }
\end{aligned}$$

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

$$\begin{aligned}
    \boxed{
        \dd{\vu{S}}
        = \vu{e}_r \: r^2 \sin\theta \dd{\theta} \dd{\varphi}
        + \vu{e}_\theta \: r \sin\theta \dd{r} \dd{\varphi}
        + \vu{e}_\varphi \: r \dd{r} \dd{\theta}
    }
\end{aligned}$$

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

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


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