2.3.6 Spherical coordinates

Cartesian coordinates are perfect for boxes and planes, and the cylindrical coordinates of Section 2.3.4 are perfect for wires and pipes. Many of the most important problems of electromagnetics, however, are organised around a single point: the field of a point charge, of a charged sphere, of a small antenna seen from far away. In such problems the natural questions are "how far from the centre?" and "in which direction?". Spherical coordinates answer exactly these two questions.

You already use spherical coordinates whenever you look at a map. A place on the Earth is given by two angles, its latitude (how far north or south of the equator) and its longitude (how far east or west of a reference meridian). Together with the distance from the centre of the Earth, these two angles fix the point completely. Spherical coordinates are the same idea, with two small changes of convention that we explain below (Figure 2.3.16).

We build the spherical system in the same way as the cylindrical one: coordinates, position vector, unit vectors and scale factors, and differential elements. We then compare the three systems of the course, learn to convert vector components between them, and finally use the new elements to compute areas, volumes and fluxes over spheres.

Definition of the coordinates

A point $\mathrm{P}$ in space is located by three numbers $(r,\theta ,\varphi )$, see Figure 2.3.18(a):

• the radial distance $r$ is the distance from the origin O to P, $0\le r<\infty$,
• the polar angle (zenith angle) $\theta$ is the angle between the positive $z$ axis and the line OP, $0\le \theta \le \pi$, and
• the azimuthal angle $\varphi$ is the angle, measured in the $xy$-plane from the positive $x$ axis towards the positive $y$ axis, between the $x$ axis and the projection of OP onto the $xy$-plane, $0\le \varphi <2\pi$. It is the same angle $\varphi$ as in cylindrical coordinates.

Note the ranges carefully: $\varphi$ sweeps a full turn, but $\theta$ only runs from the "north pole" ($\theta =0$) to the "south pole" ($\theta =\pi$). This is enough: once we know how far down from the north pole the point is ($\theta$) and on which side of the $z$ axis it lies ($\varphi$), every direction has been reached exactly once.

The geographic picture

Figure 2.3.16 makes the analogy with the globe precise. Geographers measure the latitude from the equator (from $90^{\circ }$S to $90^{\circ }$N), whereas we measure $\theta$ from the north pole, from $0$ to $180^{\circ }$, so $\theta =90^{\circ }-\text{latitude}$ for points in the northern hemisphere and $\theta =90^{\circ }+\text{(southern latitude)}$ in the southern one. Geographers measure the longitude from $0$ to $180^{\circ }$ east or west, while we measure $\varphi$ eastwards only, from $0$ to $360^{\circ }$ (in radians $0\le \varphi <2\pi$). A circle of latitude (a parallel) is a curve on which $\theta$ is constant, and a meridian is a half circle from pole to pole on which $\varphi$ is constant. We shall use the words "parallel" and "meridian" in exactly this sense.

Coordinate surfaces

Each coordinate, held constant, defines a coordinate surface (Figures 2.3.17 and 2.3.18(b)):

• $r=\text{const}$ is a sphere centred at the origin,
• $\theta =\text{const}$ is a circular cone with its apex at the origin and the $z$ axis as its axis (for $\theta =\pi /2$ the cone opens up into the $xy$-plane, and for $\theta >\pi /2$ it opens downwards), and
• $\varphi =\text{const}$ is a half-plane hinged on the $z$ axis.

The point P is the single common point of one sphere, one cone and one half-plane. These three surfaces meet at right angles to each other at every point, which is why the spherical system, like the Cartesian and the cylindrical ones, is an orthogonal coordinate system. The intersection of two coordinate surfaces is a coordinate curve: the sphere and the cone meet in a parallel (along which only $\varphi$ varies), the sphere and the half-plane in a meridian (only $\theta$ varies), and the cone and the half-plane in a ray from the origin (only $r$ varies).

Drag $r$, $\theta$ and $\varphi$ and watch the sphere, the cone and the half-plane through P move, and see how the three unit vectors $a_r$, $a_{\theta }$, $a_{\varphi }$ turn with the point while staying mutually perpendicular.

Position vector and the relations with Cartesian coordinates

To relate $(r,\theta ,\varphi )$ to $(x,y,z)$ we project the position vector $r$ of P (length $r$) twice, as shown in Figure 2.3.19.

1. In the vertical half-plane that contains P (panel a), the projection of $r$ onto the $z$ axis is $r\cos \theta$, and its projection onto the $xy$-plane has length $r\sin \theta$. This horizontal length is precisely the cylindrical radial distance $\rho$ of Section 2.3.4.
2. In the $xy$-plane (panel b), the horizontal projection of length $r\sin \theta$ makes the angle $\varphi$ with the $x$ axis, so it splits into $r\sin \theta \cos \varphi$ along $x$ and $r\sin \theta \sin \varphi$ along $y$.

Hence

and the position vector of P reads

Inverting Eq. (2.3.30),

The first relation is Pythagoras' theorem in three dimensions. The arccos function returns values in $[0,\pi ]$, which is exactly the range of $\theta$, so the polar angle is uniquely fixed by $z/r$. For the azimuth this is not so: $\arccos (x/r\sin \theta )$ alone cannot distinguish $\varphi$ from $2\pi -\varphi$, and $\arcsin (y/r\sin \theta )$ alone cannot distinguish $\varphi$ from $\pi -\varphi$. The azimuth is the angle whose cosine and sine are given by Eq. (2.3.32). Equivalently, $\tan \varphi =y/x$ with the quadrant chosen according to the signs of $x$ and $y$. Since $r\sin \theta =\sqrt{x^2+y^2}=\rho$, these are the same relations as for the cylindrical $\varphi$. On the $z$ axis ($\sin \theta =0$) the azimuth is undefined, just as $\theta$ is undefined at the origin. Such exceptional points are harmless in integrals.

Unit vectors and scale factors

In Section 2.3.2 we learnt the recipe that produces the unit vectors and the scale factors of any orthogonal coordinate system $(u_1,u_2,u_3)$: differentiate the position vector with respect to one coordinate while keeping the other two fixed,

The vector $\partial r/\partial u_i$ is tangent to the curve traced out by P when only $u_i$ changes. Its direction is the unit vector $a_i$, and its length $h_i$ tells us how many metres P moves per unit change of $u_i$. Let us apply the recipe to Eq. (2.3.31), one coordinate at a time.

Radial direction

Only $r$ changes, so we differentiate Eq. (2.3.31) with $\theta$ and $\varphi$ held fixed:

Increasing $r$ moves P straight away from the origin, one metre per metre: $h_r=1$.

Polar direction

Only $\theta$ changes:

Increasing $\theta$ moves P "southwards" along a meridian, a circle of radius $r$, so a change $d\theta$ (in radians) is an arc of length $rd\theta$, hence $h_{\theta }=r$.

Azimuthal direction

Only $\varphi$ changes:

Increasing $\varphi$ moves P "eastwards" along a parallel (a circle of latitude) whose radius is the distance $r\sin \theta$ from the $z$ axis. Hence $h_{\varphi }=r\sin \theta$ (recall that $\sin \theta \ge 0$ for $0\le \theta \le \pi$).

It is worth checking that these vectors really form an orthonormal, right-handed triple. Each has unit length by construction. Their mutual dot products vanish, e.g.

and similarly $a_r\cdot a_{\varphi }=a_{\theta }\cdot a_{\varphi }=0$. A short computation of the cross product gives

so the order $(r,\theta ,\varphi )$ is right-handed, just like $(x,y,z)$ and $(\rho ,\varphi ,z)$. The order matters: in formulas such as the curl determinant of Section 2.3.10, the coordinates must always be listed as $(r,\theta ,\varphi )$.

Comparing Eq. (2.3.34) with Eq. (2.3.31) we notice that $r=r{a}_r$: in spherical coordinates the position vector has a single component. The information about the direction of P, i.e. about $\theta$ and $\varphi$, is hidden inside the unit vector $a_r$ itself. This brings us to the most important practical difference between Cartesian and curvilinear coordinates.

The unit vectors depend on position

The Cartesian unit vectors point the same way everywhere. The spherical unit vectors do not: $a_r$ at the north pole is $a_z$, while $a_r$ on the positive $x$ axis is $a_x$. On the globe, $a_r$ is "up" (towards the sky), $a_{\theta }$ is "south" and $a_{\varphi }$ is "east". But "up", "south" and "east" in Ankara are different directions in space from "up", "south" and "east" in Tokyo. Figure 2.3.21 shows the triad at several points of a sphere, and Figure 2.3.22(a) shows $a_r$ and $a_{\theta }$ along one meridian. They turn as the point moves.

Differentiating Eqs. (2.3.34) to (2.3.36) and re-expressing the results in the spherical unit vectors gives

For instance, $\partial a_r/\partial \theta =\cos \theta \cos \varphi a_x+\cos \theta \sin \varphi a_y-\sin \theta a_z=a_{\theta }$: tilting the direction southwards turns "up" towards "south". None of the unit vectors changes along the radial direction (moving outwards along a ray does not turn anything), but all of them change with the angles.

Two consequences will accompany us through the whole course.

1. Differentiation. When we differentiate a vector field written in spherical (or cylindrical) components, the unit vectors must be differentiated too. This is why the operator formulas of Section 2.3.10 contain extra factors such as $r^2$ and $\sin \theta$: they are the fingerprints of Eq. (2.3.38).

2. Integration. A unit vector such as $a_r$ may not be taken outside an integral over a region in which it changes direction. For example, integrating $a_r$ over a sphere of radius $b$ centred at the origin does not give $4\pi b^2{a}_r$ (which would not even be well defined: at which point is that $a_r$ evaluated?). Writing $a_r$ in Cartesian form, Eq. (2.3.34), whose unit vectors are constant, and using $dS=b^2\sin \theta d\theta d\varphi$ (see Eq. (2.3.40) below),

   $$\begin{array}{rl}{\oint }_S{a}_{r}dS& =b^2{\int }_0^{2\pi }{\int }_0^{\pi }(\sin \theta \cos \varphi a_x\\ & +\sin \theta \sin \varphi a_y+\cos \theta a_z)\\ & \times \sin \theta d\theta d\varphi =0,\end{array}$$

   because ${\int }_0^{2\pi }\cos \varphi d\varphi ={\int }_0^{2\pi }\sin \varphi d\varphi =0$ and ${\int }_0^{\pi }\cos \theta \sin \theta d\theta =0$. Physically: the outward directions on opposite sides of the sphere cancel. The rule is simple. Before integrating a vector, convert its unit vectors to Cartesian ones. (Scalar integrands, such as a flux $A\cdot dS$, have no such problem.)

Differential elements of length, surface and volume

With the scale factors at hand, the general formula of Section 2.3.2, $d\ell =h_{1}d{u}_1{a}_1+h_{2}d{u}_2{a}_2+h_{3}d{u}_3{a}_3$, gives the differential length

Each term has the dimension of length: $dr$ is a radial step, $rd\theta$ an arc along a meridian of radius $r$, and $r\sin \theta d\varphi$ an arc along a parallel of radius $r\sin \theta$. The angles themselves are dimensionless, so it is the scale factors that carry the metres.

Let the three coordinates increase by $dr$, $d\theta$, $d\varphi$. The point P then sweeps out the small "curved brick" of Figure 2.3.23, bounded by two spheres, two cones and two half-planes. Since the surfaces are mutually orthogonal, its edges $dr$, $rd\theta$ and $r\sin \theta d\varphi$ are mutually perpendicular, and to first order the brick is a rectangular box. Its faces give the differential surface elements (the vector $dS$ is normal to the face, with magnitude equal to its area):

where the sign is chosen according to the required orientation (for a closed surface: outwards). Each area is simply the product of the two edges that lie in that face: on the sphere the edges are $rd\theta$ and $r\sin \theta d\varphi$, and so on. The volume is the product of all three edges,

The factor $r^2\sin \theta$ is the Jacobian of the transformation, Eq. (2.3.30). It expresses two simple geometric facts: bricks of the same $drd\theta d\varphi$ are larger far from the origin (factor $r^2$), and they are thinner near the poles, where the parallels shrink to a point (factor $\sin \theta$). Figure 2.3.24 shows the second fact on a globe.