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---
title: "Spherical coordinates"
sort_title: "Spherical coordinates"
date: 2021-03-04
categories:
- Mathematics
- Physics
layout: "concept"
---
**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 coordinate system is
an [orthogonal curvilinear system](/know/concept/orthogonal-curvilinear-coordinates/),
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 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:
$$\begin{aligned}
\boxed{
\nabla f
= \vu{e}_r \pdv{f}{r}
+ \vu{e}_\theta \frac{1}{r} \pdv{f}{\theta} + \mathbf{e}_\varphi \frac{1}{r \sin\theta} \pdv{f}{\varphi}
}
\end{aligned}$$
$$\begin{aligned}
\boxed{
\nabla \cdot \vb{V}
= \frac{1}{r^2} \pdv{(r^2 V_r)}{r}
+ \frac{1}{r \sin\theta} \pdv{(\sin\theta V_\theta)}{\theta}
+ \frac{1}{r \sin\theta} \pdv{V_\varphi}{\varphi}
}
\end{aligned}$$
$$\begin{aligned}
\boxed{
\nabla^2 f
= \frac{1}{r^2} \pdv{}{r}\Big( r^2 \pdv{f}{r} \Big)
+ \frac{1}{r^2 \sin\theta} \pdv{}{\theta}\Big( \sin\theta \pdv{f}{\theta} \Big)
+ \frac{1}{r^2 \sin^2(\theta)} \pdvn{2}{f}{\varphi}
}
\end{aligned}$$
$$\begin{aligned}
\boxed{
\begin{aligned}
\nabla \times \vb{V}
&= \frac{\vu{e}_r}{r \sin\theta} \Big( \pdv{(\sin\theta V_\varphi)}{\theta} - \pdv{V_\theta}{\varphi} \Big)
\\
&+ \frac{\vu{e}_\theta}{r} \Big( \frac{1}{\sin\theta} \pdv{V_r}{\varphi} - \pdv{(r V_\varphi)}{r} \Big)
\\
&+ \frac{\vu{e}_\varphi}{r} \Big( \pdv{(r V_\theta)}{r} - \pdv{V_r}{\theta} \Big)
\end{aligned}
}
\end{aligned}$$
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 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.
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