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+---
+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.