Spherical Polar Coordinates

Spherical polar coordinates are the orthogonal curvilinear system built for problems with a centre of symmetry, and they earn that role by trading three Cartesian lengths for one length and two angles: the distance $r$ from the origin, the polar (colatitude) angle $\theta$ that the position vector makes with a chosen axis, and the azimuth $\phi$ of its shadow in the equatorial plane. This is the physicists' convention of Arfken–Weber, Hassani, Stone–Goldbart and Byron–Fuller; many mathematics texts swap the names of $\theta$ and $\phi$ , so the convention must be stated before a single integral is written. The reward for the change of variables is that a sphere, a cone and a half-plane — the surfaces of constant $r$ , $\theta$ and $\phi$ — meet everywhere at right angles, so the system is orthogonal and every piece of vector calculus can be assembled from just three numbers. Those three numbers are the scale factors $h_r=1$ , $h_\theta=r$ , $h_\phi=r\sin\theta$ , and this article's whole claim is that they are all you memorise. From them the line, surface and volume elements, the gradient, divergence, curl and Laplacian, and the Jacobian $r^2\sin\theta$ all follow by substitution into the general orthogonal-curvilinear formulae. The chart is global except on a measure-zero set — the origin $r=0$ and the polar axis $\theta=0,\pi$ — where the azimuth loses its meaning; these are coordinate singularities, not singularities of space, and they become boundary and regularity conditions when solving differential equations. The payoff is enormous: the Coulomb and Newtonian potentials depend on $r$ alone, the multipole expansion sorts sources by angular shape, and separating Laplace's equation here produces the spherical harmonics $Y_\ell^m(\theta,\phi)$ that underlie atomic structure, gravitational geodesy and the cosmic microwave background.

Key takeaways

Spherical coordinates $(r,\theta,\phi)$ replace $(x,y,z)$ by a radius $r\ge0$ , a polar angle $\theta\in[0,\pi]$ from the $z$ -axis, and an azimuth $\phi\in[0,2\pi)$ in the $xy$ -plane.
The system is orthogonal with scale factors $h_r=1$ , $h_\theta=r$ , $h_\phi=r\sin\theta$ — the physical distance moved per unit change of each coordinate.
Every element and operator follows from those factors: $\mathrm{d}V=r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ , $\mathrm{d}\Omega=\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi$ , and the Jacobian $r^2\sin\theta$ .
The radial Laplacian $\tfrac{1}{r^2}\partial_r(r^2\partial_r)$ plus the angular operator whose eigenvalue is $-\ell(\ell+1)$ governs every central-force and radiation problem.
The chart is singular at $r=0$ and on the polar axis $\theta=0,\pi$ , where the $\phi$ -circle shrinks to a point — a defect of the coordinates, not of space.
Key idea: once you know the three scale factors $h_r=1,\ h_\theta=r,\ h_\phi=r\sin\theta$ , every differential operator and integration element in spherical coordinates follows mechanically from the general orthogonal-curvilinear formulae — there is nothing more to memorise.

Where this lives today

The coordinate system that Legendre used for the shape of the Earth is still the working language of anything with a centre or a sky. In cosmology the temperature of the cosmic microwave background is a function on the celestial sphere, and it is expanded in spherical harmonics $T(\theta,\phi)=\sum_{\ell m}a_{\ell m}Y_\ell^m$ ; the angular power spectrum $C_\ell=\langle|a_{\ell m}|^2\rangle$ measured by WMAP and Planck shows its first acoustic peak near $\ell\approx220$ (angular scale $\sim1^{\circ}$ ), and the position of that peak is one of the sharpest measurements of the geometry of the universe. In geodesy the Earth's gravitational potential is tabulated as a spherical-harmonic model — EGM2008 runs to degree and order $2190$ — and every GPS fix and satellite orbit is corrected against it. In atomic and molecular physics the orbitals you draw are $|Y_\ell^m|$ : the $s$ , $p$ , $d$ shapes are the $\ell=0,1,2$ harmonics, and the whole periodic table is the bookkeeping of the quantum numbers $(n,\ell,m)$ that separation in these coordinates hands you. And in engineering the same $\sin\theta$ that vanishes on the axis sets the doughnut radiation pattern of a dipole antenna and the differential cross-section $\mathrm d\sigma/\mathrm d\Omega$ that every scattering experiment reports.

The physical picture

Stand at the origin and point at a star. You need exactly three numbers to fix where it is: how far away it is ( $r$ ), how far its direction tips away from straight up ( $\theta$ , measured down from the pole), and which way you have turned to face it ( $\phi$ , the compass bearing). That is the whole content of $(r,\theta,\phi)$ — one radial reach and two swings. The first figure fixes the geometry and the transformation to Cartesian coordinates, and it quietly carries the one relation everything else leans on: the distance from the polar axis is $\rho=r\sin\theta$ , which is why the azimuthal circle has radius $r\sin\theta$ and not $r$ .

Figure

A point located by radius r, polar angle θ from the z axis, and azimuth φ in the equatorial plane, with x = r sinθ cosφ, y = r sinθ sinφ, z = r cosθ and projected axis distance ρ = r sinθ

Now ask the operational question the scale factors answer: nudge one coordinate by a tiny amount and how far in metres do you actually travel? Nudging $r$ moves you a plain $\mathrm{d}r$ , so $h_r=1$ . Nudging $\theta$ sweeps an arc of radius $r$ , so $r\,\mathrm{d}\theta$ and $h_\theta=r$ . Nudging $\phi$ sweeps a circle whose radius is the distance from the axis, $\rho=r\sin\theta$ , so $r\sin\theta\,\mathrm{d}\phi$ and $h_\phi=r\sin\theta$ . Stack the three perpendicular displacements into a little box and its volume is the product of the edges — that is the Jacobian $r^2\sin\theta$ , and the second figure is that box. It is the punchline of the whole system: the same $\sin\theta$ that measures the shrinking $\phi$ -circle is what makes the volume element vanish smoothly onto the polar axis.

Figure

The curvilinear volume element is a box with edges dr, r dθ and r sinθ dφ, so dV = r² sinθ dr dθ dφ — the azimuthal edge r sinθ dφ shrinks to zero on the polar axis

The last thing the picture makes visible is that the unit vectors are not fixed arrows. The triad $\{\hat{\mathbf e}_r,\hat{\mathbf e}_\theta,\hat{\mathbf e}_\phi\}$ rides along with the point like the local up, north and east on a globe: at two different places they point in genuinely different directions. That is harmless for a scalar field, but the moment you differentiate a vector field the basis itself has to be differentiated too, and those derivatives are exactly where the extra $r^2$ and $\sin\theta$ terms in the divergence and curl come from.

The core equations

The starting point is the change of variables itself: three trigonometric statements that plant the point at radius $r$ , tilt it by $\theta$ off the axis, and spin it by $\phi$ around the axis. Reading them off the first figure, the height is $z=r\cos\theta$ and the in-plane shadow of length $r\sin\theta$ resolves into $x$ and $y$ .

x=r\sin\theta\cos\phi,\quad y=r\sin\theta\sin\phi,\quad z=r\cos\theta \qquad (1)

where: $r$ — radial distance from the origin, $r\ge0$ ; $\theta$ — polar angle from the $+z$ axis, $0\le\theta\le\pi$ ; $\phi$ — azimuthal angle in the $xy$ -plane from the $+x$ axis, $0\le\phi<2\pi$ . Reading: fix $r$ and you are on a sphere; fix $\theta$ and you are on a cone; fix $\phi$ and you are on a half-plane hinged on the axis. Inverting the transformation recovers the coordinates from a Cartesian point, and the only subtlety is the azimuth, which needs the quadrant-aware two-argument arctangent so that points in the third and fourth quadrants are not folded onto the first.

r=\sqrt{x^2+y^2+z^2},\quad \theta=\arccos\!\frac{z}{r},\quad \phi=\operatorname{atan2}(y,x) \qquad (2)

where: $\operatorname{atan2}(y,x)$ — the two-argument arctangent returning $\phi$ in the correct quadrant of $[0,2\pi)$ (or $(-\pi,\pi]$ ). Reading: $r$ and $\theta$ are single-valued off the origin, but $\phi$ is undefined on the whole polar axis where $x=y=0$ . Everything downstream is bookkeeping on three scale factors, defined as the lengths of the tangent vectors $\partial\mathbf r/\partial q_i$ . Derivation A extracts them; here they are as the object to remember.

h_r=\left|\frac{\partial\mathbf r}{\partial r}\right|=1,\qquad h_\theta=\left|\frac{\partial\mathbf r}{\partial\theta}\right|=r,\qquad h_\phi=\left|\frac{\partial\mathbf r}{\partial\phi}\right|=r\sin\theta \qquad (3)

where: $h_i$ — physical distance moved per unit change of coordinate $q_i$ . Reading: $h_r$ is pure length (an angle-free coordinate), $h_\theta$ carries one factor of $r$ (an arc), and $h_\phi$ carries $r\sin\theta$ (a circle about the axis). The infinitesimal displacement is then the sum of the three orthogonal steps, and squaring it gives the metric — the Pythagorean rule in curved coordinates, with each coordinate weighted by its scale factor squared.

\mathrm{d}\mathbf s=\mathrm{d}r\,\hat{\mathbf e}_r+r\,\mathrm{d}\theta\,\hat{\mathbf e}_\theta+r\sin\theta\,\mathrm{d}\phi\,\hat{\mathbf e}_\phi,\qquad \mathrm{d}s^2=\mathrm{d}r^2+r^2\mathrm{d}\theta^2+r^2\sin^2\theta\,\mathrm{d}\phi^2 \qquad (4)

where: $\hat{\mathbf e}_i$ — the position-dependent orthonormal basis along increasing $r,\theta,\phi$ . Reading: no cross terms appear ( $\mathrm{d}r\,\mathrm{d}\theta$ , etc. are absent) precisely because the system is orthogonal. Because the three steps are mutually perpendicular, the volume element is simply the product of the edge lengths, and the pure-angle part is the solid-angle element that integrates to $4\pi$ over the whole sky.

\mathrm{d}V=h_r h_\theta h_\phi\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi=r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi,\qquad \mathrm{d}\Omega=\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi \qquad (5)

where: $\mathrm{d}V$ — volume element; $\mathrm{d}\Omega$ — solid-angle element with $\int\mathrm{d}\Omega=4\pi$ . Reading: the $r^2$ makes shells at large radius fat, and the $\sin\theta$ makes them vanish onto the axis. The gradient divides each partial derivative by the corresponding scale factor, converting a coordinate-rate of change into a spatial-rate of change with the right units.

\nabla f=\frac{\partial f}{\partial r}\hat{\mathbf e}_r+\frac{1}{r}\frac{\partial f}{\partial\theta}\hat{\mathbf e}_\theta+\frac{1}{r\sin\theta}\frac{\partial f}{\partial\phi}\hat{\mathbf e}_\phi \qquad (6)

where: $f$ — a scalar field $f(r,\theta,\phi)$ . Reading: the $1/r$ and $1/(r\sin\theta)$ are exactly $1/h_\theta$ and $1/h_\phi$ — a far-away point needs a large angular change to move a metre, so its angular gradient is small. The divergence and Laplacian pick up the moving-basis terms as metric weights inside the derivatives; the radial Laplacian $\tfrac{1}{r^2}\partial_r(r^2\partial_r)$ is the operator that governs central-force and radiation problems, and Derivation C shows how the two angular pieces assemble into the operator behind the spherical harmonics.

\nabla\cdot\mathbf A=\frac{1}{r^2}\frac{\partial(r^2 A_r)}{\partial r}+\frac{1}{r\sin\theta}\frac{\partial(\sin\theta\,A_\theta)}{\partial\theta}+\frac{1}{r\sin\theta}\frac{\partial A_\phi}{\partial\phi},\qquad \nabla^2 f=\frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial f}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial f}{\partial\theta}\right)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2 f}{\partial\phi^2} \qquad (7)

where: $A_i$ — physical components of $\mathbf A$ along $\hat{\mathbf e}_i$ ; $\nabla^2=\nabla\cdot\nabla$ . Reading: the $r^2$ and $\sin\theta$ inside the derivatives are the fingerprints of the curved basis — drop them and Gauss's law and the hydrogen atom both come out wrong.

Derivation AScale factors from the position vector

The entire toolkit of spherical vector calculus rests on three numbers, and this derivation extracts them from first principles by differentiating the position vector rather than by quoting a table. A scale factor $h_i$ answers a purely geometric question — when the coordinate $q_i$ is nudged by $\mathrm{d}q_i$ while the others are held fixed, how many metres does the point actually move? — and the answer is the length of the tangent vector $\partial\mathbf r/\partial q_i$ . Once $h_r,h_\theta,h_\phi$ are in hand, every line, surface and volume element and every differential operator follows by substitution, which is the sense in which they are the only thing worth memorising.

Step 1. Write the position vector from equation (1), $\mathbf r=r\sin\theta\cos\phi\,\hat{\mathbf x}+r\sin\theta\sin\phi\,\hat{\mathbf y}+r\cos\theta\,\hat{\mathbf z}$ , and differentiate with respect to $r$ at fixed angles: $\partial\mathbf r/\partial r=(\sin\theta\cos\phi,\ \sin\theta\sin\phi,\ \cos\theta)$ . Its length is $\sqrt{\sin^2\theta(\cos^2\phi+\sin^2\phi)+\cos^2\theta}=\sqrt{\sin^2\theta+\cos^2\theta}=1$ , so $h_r=1$ . Why: moving out along the radius is ordinary Cartesian motion, so a unit change of $r$ is a unit change of length — the radial coordinate carries no distortion.

Step 2. Differentiate with respect to $\theta$ : $\partial\mathbf r/\partial\theta=r(\cos\theta\cos\phi,\ \cos\theta\sin\phi,\ -\sin\theta)$ , whose length is $r\sqrt{\cos^2\theta+\sin^2\theta}=r$ , so $h_\theta=r$ . Why: a change in the polar angle sweeps the point along a great-circle arc of radius $r$ , and arc length is radius times angle — the further out, the more metres a given angular nudge buys.

Step 3. Differentiate with respect to $\phi$ : $\partial\mathbf r/\partial\phi=r\sin\theta(-\sin\phi,\ \cos\phi,\ 0)$ , whose length is $r\sin\theta$ , so $h_\phi=r\sin\theta$ . Why: spinning about the axis moves the point on a circle whose radius is not $r$ but the distance from the axis, $\rho=r\sin\theta$ — which is why this factor, and not $r$ , controls the azimuthal geometry.

Step 4. Check orthogonality by dotting the three tangent vectors pairwise; each product vanishes, e.g. $(\partial\mathbf r/\partial r)\cdot(\partial\mathbf r/\partial\theta)=r(\sin\theta\cos\theta\cos^2\phi+\sin\theta\cos\theta\sin^2\phi-\sin\theta\cos\theta)=0$ . Why: orthogonality is what licenses every later shortcut — it lets the metric be diagonal, the volume element be a plain product $h_r h_\theta h_\phi$ , and the operators be assembled term by term with no cross-coupling.

Go deeper · Derivation B · 1 minThe volume element and the Jacobian r squared sinθ

Derivation BThe volume element and the Jacobian r squared sinθ

The factor $r^2\sin\theta$ that decorates every spherical volume integral is the Jacobian determinant of the map $(r,\theta,\phi)\mapsto(x,y,z)$ , and this derivation computes it two independent ways so that the answer is beyond doubt. The first way is brute force — assemble the matrix of partial derivatives and take its determinant — and the second is the orthogonal shortcut in which the determinant collapses to the product of the scale factors. That the two agree is not a coincidence but a theorem about orthogonal systems, and it doubles as the cleanest check that the scale factors of Derivation A were found correctly.

Step 1. Assemble the Jacobian matrix whose columns are the tangent vectors of Derivation A,

J=\begin{pmatrix}\sin\theta\cos\phi & r\cos\theta\cos\phi & -r\sin\theta\sin\phi\\ \sin\theta\sin\phi & r\cos\theta\sin\phi & r\sin\theta\cos\phi\\ \cos\theta & -r\sin\theta & 0\end{pmatrix}.

Why: the change-of-variables theorem replaces $\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$ by $|\det J|\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ , so the whole task is to evaluate this one determinant.

Step 2. Expand along the bottom row, whose zero entry kills the third cofactor: $\det J=\cos\theta\,C_{31}+(-r\sin\theta)\,C_{32}$ , with $C_{31}=r^2\sin\theta\cos\theta$ and $C_{32}=-\,r\sin^2\theta$ . Why: expanding on the row with a zero is the least-arithmetic route, and each surviving cofactor is a $2\times2$ minor that simplifies immediately with $\cos^2\phi+\sin^2\phi=1$ .

Step 3. Combine the two terms: $\det J=r^2\sin\theta\cos^2\theta+r^2\sin^3\theta=r^2\sin\theta(\cos^2\theta+\sin^2\theta)=r^2\sin\theta$ , so $\mathrm{d}V=r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ . Why: the Pythagorean identity in $\theta$ does the final collapse, and the result is non-negative on $\theta\in[0,\pi]$ so no absolute value is needed.

Step 4. Cross-check with the orthogonal shortcut $\det J=h_r h_\theta h_\phi=1\cdot r\cdot r\sin\theta=r^2\sin\theta$ , and confirm the global consequence $\int_0^{2\pi}\!\int_0^{\pi}\!\int_0^{R}r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi=\tfrac{4}{3}\pi R^3$ . Why: recovering the sphere volume from the element is the acid test — if the Jacobian were wrong, this elementary integral would not close on the schoolbook answer. *Both routes agree: direct cofactor expansion and the orthogonal shortcut $h_r h_\theta h_\phi$ both give $\det J=r^2\sin\theta$ , and the shell integral $\int_0^R r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ returns $\tfrac{4}{3}\pi R^3$ , confirming the Jacobian is correct.*

Go deeper · Derivation C · 1 minThe spherical Laplacian and the angular operator

Derivation CThe spherical Laplacian and the angular operator

The Laplacian is the operator that turns spherical coordinates from a convenience into a necessity, because it is the one that separates for central problems, and this derivation builds it by substitution into the universal orthogonal-curvilinear formula rather than by grinding through second derivatives of the basis vectors. The general rule packages all the moving-basis bookkeeping into a single template, so the work is purely algebraic — insert the three scale factors, cancel, and read off the standard form. The final step names the angular remainder as the operator whose eigenfunctions are the spherical harmonics, which is the door to atomic and multipole physics.

Step 1. Start from the orthogonal template $\nabla^2 f=\dfrac{1}{h_r h_\theta h_\phi}\sum_{i}\dfrac{\partial}{\partial q_i}\!\left(\dfrac{h_r h_\theta h_\phi}{h_i^2}\dfrac{\partial f}{\partial q_i}\right)$ with $h_r h_\theta h_\phi=r^2\sin\theta$ . Why: this template is where all the $r^2$ and $\sin\theta$ weights are generated automatically, so nothing about the rotating triad has to be tracked by hand.

Step 2. Radial term ( $h_r=1$ ): $\dfrac{1}{r^2\sin\theta}\partial_r\!\left(\dfrac{r^2\sin\theta}{1}\partial_r f\right)=\dfrac{1}{r^2}\partial_r(r^2\partial_r f)$ , the $\sin\theta$ cancelling because it does not depend on $r$ . Why: this is the piece that survives for any spherically symmetric field and reduces the problem to a single radial ODE.

Step 3. Angular terms ( $h_\theta^2=r^2$ , $h_\phi^2=r^2\sin^2\theta$ ): $\dfrac{1}{r^2\sin\theta}\partial_\theta(\sin\theta\,\partial_\theta f)+\dfrac{1}{r^2\sin^2\theta}\partial_\phi^2 f$ , so altogether $\nabla^2 f=\dfrac{1}{r^2}\partial_r(r^2\partial_r f)+\dfrac{1}{r^2}\,\Lambda f$ with $\Lambda=\dfrac{1}{\sin\theta}\partial_\theta(\sin\theta\,\partial_\theta)+\dfrac{1}{\sin^2\theta}\partial_\phi^2$ . Why: isolating the $1/r^2$ times a pure-angle operator $\Lambda$ is exactly the structure that makes the equation separable into a radial factor and an angular factor.

Step 4. Identify $\Lambda$ : its eigenfunctions are the spherical harmonics $Y_\ell^m(\theta,\phi)$ with $\Lambda Y_\ell^m=-\ell(\ell+1)Y_\ell^m$ , so $-\hbar^2\Lambda$ is the squared orbital angular momentum $\hat L^2$ . Why: the eigenvalue $-\ell(\ell+1)$ becomes the centrifugal term $\ell(\ell+1)/r^2$ in the radial equation, which is how the coordinate geometry hands quantum mechanics its angular-momentum quantum number. *As a check, apply the Laplacian to $f=1/r$ : only the radial term contributes, and it gives $\nabla^2(1/r)=0$ for $r\ne0$ — confirming the harmonicity of the Coulomb potential.*

Limiting cases

Spherical coordinates degrade gracefully: in special regimes the machinery collapses onto simpler, more familiar systems, and each collapse is a check on the general formulae.

Case	Result	Meaning
Spherical symmetry	$f=f(r):\ \ \nabla^2 f=\frac{1}{r^2}\frac{\mathrm d}{\mathrm dr}\!\left(r^2\frac{\mathrm df}{\mathrm dr}\right)$	the angular terms vanish and the PDE becomes a single radial ODE — the Coulomb and Newtonian potentials, and every ground-state problem
Equatorial plane	$\theta=\tfrac{\pi}{2}:\ \ h_\phi=r,\ (r,\phi)\to$ plane polar	setting the colatitude to $90^{\circ}$ recovers plane polar coordinates, with $\sin\theta=1$
Cylindrical link	$\rho=r\sin\theta,\ z=r\cos\theta$	identifying the axial distance $\rho$ ties spherical to cylindrical coordinates term by term
Near the axis	$\theta\to0:\ \ \mathrm d\Omega=\sin\theta\,\mathrm d\theta\,\mathrm d\phi\to\theta\,\mathrm d\theta\,\mathrm d\phi$	the solid-angle patch shrinks linearly in $\theta$ because the $\phi$ -circle has radius $r\sin\theta\to0$
Far field	$r\to\infty:\ \ h_\theta,h_\phi\to\infty$	a fixed angular step spans an ever-larger distance — the regime of radiation fields and the outgoing $1/r$ wave

Common trapThinking the unit vectors are constants, or that the axis is a hole in space

Two errors dominate. The first is treating $\hat{\mathbf e}_r,\hat{\mathbf e}_\theta,\hat{\mathbf e}_\phi$ like the fixed Cartesian $\hat{\mathbf x},\hat{\mathbf y},\hat{\mathbf z}$ and dropping their derivatives; but $\partial_\theta\hat{\mathbf e}_r=\hat{\mathbf e}_\theta$ , $\partial_\phi\hat{\mathbf e}_r=\sin\theta\,\hat{\mathbf e}_\phi$ and $\partial_\theta\hat{\mathbf e}_\theta=-\hat{\mathbf e}_r$ are nonzero, and it is precisely those derivatives that generate the $r^2$ and $\sin\theta$ factors inside the divergence and curl. Forget them and you compute $\nabla\cdot\mathbf A$ as a bare sum of partials and get the wrong answer for even the simplest radial field. The second error is reading the coordinate singularity at $r=0$ and on the axis $\theta=0,\pi$ as a defect of space. It is not: a physically smooth field passes straight through the pole. Only the chart degenerates there — the azimuth $\phi$ labels a circle that has shrunk to a point — which is why one imposes finiteness at the origin and single-valuedness in $\phi$ as boundary conditions, and never as physics. (A third, cheaper trap: forgetting the $\sin\theta$ in $\mathrm dV$ or $\mathrm d\Omega$ entirely, which quietly breaks every normalisation.)

Worked examples

1 · Moment of inertia of a uniform solid sphere (fully worked)

A solid sphere of mass $M$ , radius $R$ and uniform density $\rho$ spins about a diameter. Find its moment of inertia, then evaluate it for an Earth-sized uniform body.

Step 1. The moment of inertia about the $z$ -axis is $I=\int(x^2+y^2)\,\mathrm dm=\rho\int(r\sin\theta)^2\,\mathrm dV$ , since the distance from the axis is $\rho_\perp=r\sin\theta$ . Insert $\mathrm dV=r^2\sin\theta\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi$ , so the integrand becomes $r^2\sin^2\theta\cdot r^2\sin\theta=r^4\sin^3\theta$ .

Step 2. The integral factorises because the limits are constant: $I=\rho\left(\int_0^R r^4\,\mathrm dr\right)\!\left(\int_0^\pi\sin^3\theta\,\mathrm d\theta\right)\!\left(\int_0^{2\pi}\mathrm d\phi\right)$ . Evaluate each piece: $\int_0^R r^4\,\mathrm dr=\tfrac{R^5}{5}$ , $\int_0^\pi\sin^3\theta\,\mathrm d\theta=\big[-\cos\theta+\tfrac13\cos^3\theta\big]_0^\pi=\tfrac43$ , and $\int_0^{2\pi}\mathrm d\phi=2\pi$ . Hence $I=\rho\cdot\tfrac{R^5}{5}\cdot\tfrac43\cdot2\pi=\tfrac{8\pi}{15}\rho R^5$ .

Step 3. Eliminate the density with $M=\rho\cdot\tfrac43\pi R^3$ , i.e. $\rho=\dfrac{3M}{4\pi R^3}$ , giving $I=\dfrac{8\pi}{15}\cdot\dfrac{3M}{4\pi R^3}\cdot R^5=\dfrac{2}{5}MR^2$ . So $I=\tfrac{2}{5}MR^2$ for any uniform sphere, independent of its size or density.

Step 4. Numbers: for $M=5.97\times10^{24}\ \mathrm{kg}$ and $R=6.37\times10^6\ \mathrm{m}$ , $I=0.4\times5.97\times10^{24}\times(6.37\times10^6)^2\approx\mathbf{9.7\times10^{37}\ kg\,m^2}$ . The measured Earth value is about $8.0\times10^{37}\ \mathrm{kg\,m^2}$ , i.e. a coefficient near $0.33$ rather than $0.40$ — a direct readout that Earth's mass is concentrated toward its dense iron core rather than spread uniformly.

2 · Solid angle of a cone, and the sky as 4π

What solid angle does a right circular cone of half-angle $\alpha$ (apex at the origin, axis along $z$ ) subtend? Evaluate it for $\alpha=30^{\circ}$ and for the Moon.

Step 1. Integrate the solid-angle element over the cap the cone cuts on the unit sphere: $\Omega=\int_0^{2\pi}\!\int_0^{\alpha}\sin\theta\,\mathrm d\theta\,\mathrm d\phi=2\pi\big[-\cos\theta\big]_0^{\alpha}$ , so $\Omega=2\pi(1-\cos\alpha)$ .

Step 2. Sanity-check the endpoints: $\alpha=90^{\circ}$ gives $\Omega=2\pi$ (a hemisphere) and $\alpha=180^{\circ}$ gives $\Omega=4\pi$ (the whole sky), both correct. For $\alpha=30^{\circ}$ , $\cos30^{\circ}=0.86603$ , so $\Omega=2\pi(1-0.86603)=2\pi(0.13397)=\mathbf{0.842\ sr}$ , which is $\Omega/4\pi=6.70\%$ of the full sphere.

Step 3. For a distant small disc like the Moon (angular diameter $\approx0.52^{\circ}$ , half-angle $\alpha\approx0.26^{\circ}=4.5\times10^{-3}\ \mathrm{rad}$ ) the small-angle form $1-\cos\alpha\approx\tfrac12\alpha^2$ gives $\Omega\approx2\pi\cdot\tfrac12(4.5\times10^{-3})^2\approx\mathbf{6\times10^{-5}\ sr}$ . The Moon covers about $6\times10^{-5}\ \mathrm{sr}$ out of $4\pi$ — five parts in a million of the sky — which is why "per unit solid angle" is the natural way to state a source's brightness or an antenna's gain.

Physics of the portrait

Spherical coordinates are older than anyone's name attached to them — sailors and astronomers had been giving directions with a distance and two angles for millennia — but their transformation into a calculating engine belongs to the potential theory of the late eighteenth century, and to one man's polynomials in particular.

Origins · 1 minAdrien-Marie Legendre and how this was discovered

Adrien-Marie Legendre

1752–1833 · potential theory, geodesy and the spherical harmonics

Legendre worked at the exact junction this article lives on: the gravitational attraction of not-quite-spherical bodies, computed in spherical coordinates. In his 1782 memoir on the attraction of spheroids he introduced the polynomials $P_\ell(\cos\theta)$ that now bear his name, precisely as the coefficients in the expansion of the Newtonian potential $1/|\mathbf r-\mathbf r'|$ in powers of $r'/r$ — the monopole, dipole and quadrupole terms of the multipole series. Those polynomials are the azimuthally symmetric ( $m=0$ ) spherical harmonics, the angular building blocks that fall out when Laplace's equation is separated in $(r,\theta,\phi)$ ; his contemporary and rival Laplace generalised them to the full $Y_\ell^m(\theta,\phi)$ a few years later. Legendre's fingerprints are all over the machinery a physicist reaches for in this coordinate system: the same functions organise the multipole expansion of a charge distribution, the normal modes of a vibrating sphere, and the shape of the Earth's gravity field in modern geodesy. He was also, with Gauss, a father of the method of least squares and of the systematic study of elliptic integrals — but it is the polynomials born from a problem in spherical coordinates that made his name a fixture of every mathematical-physics course.

History

1782

— Legendre expands the gravitational potential of a spheroid in the polynomials Pℓ(cos θ)

1785

— Laplace writes ∇²V = 0 and its spherical-harmonic solutions for the figures of the planets

1859

— Lamé systematizes orthogonal curvilinear coordinates and their scale factors

1926

— Schrödinger separates the hydrogen atom as ψ = R(r) Yℓᵐ(θ, φ) in spherical coordinates

Going further

Texts. Arfken & Weber, Mathematical Methods for Physicists, 7th ed., §3.10 (spherical polar coordinates) · Hassani, Mathematical Physics, Ch. on curvilinear coordinates and separation · Stone & Goldbart, Mathematics for Physics, §1.3 (orthogonal curvilinear coordinates) · Nakahara, Geometry, Topology and Physics, §7 (metrics and coordinate charts) · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 1 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., §10.9. Key self-consistency checks: $\det J=r^2\sin\theta$ by both direct Jacobian expansion and the product $h_r h_\theta h_\phi$ ; $\int\mathrm{d}\Omega=4\pi$ and the shell integral $\tfrac{4}{3}\pi R^3$ ; $\nabla^2(1/r)=0$ for $r\ne0$ ; the worked examples yield $I=\tfrac{2}{5}MR^2$ and $\Omega=2\pi(1-\cos\alpha)$ . Anchor numbers: Earth $I\approx9.7\times10^{37}$ vs measured $8.0\times10^{37}\ \mathrm{kg\,m^2}$ ; CMB first acoustic peak $\ell\approx220$ ; EGM2008 degree $2190$ ; Moon $\Omega\approx6\times10^{-5}\ \mathrm{sr}$ .

Builds on: Scalar and Vector Fields · The Gradient Operator · Volume Integrals Leads to: The Laplacian Operator · Divergence of a Vector Field · Surface Integrals and Flux · The Divergence (Gauss) Theorem

Ask the tutor: Ask me to separate Laplace's or Schrödinger's equation in these coordinates, or to walk through why the $\sin\theta$ in $\mathrm dV$ is the same one that makes the polar axis a coordinate singularity.

Which is the polar angle, θ or φ — and why does it matter?

In the physics convention used here and in Arfken, Hassani and Stone–Goldbart, $\theta$ is the polar (colatitude) angle measured down from the $z$ -axis ( $0\le\theta\le\pi$ ) and $\phi$ is the azimuth in the equatorial plane ( $0\le\phi<2\pi$ ). Many mathematics and engineering texts (and some software) reverse the two labels, calling the azimuth $\theta$ . It matters because the formulae are not symmetric in the two angles: the volume element carries $\sin\theta$ with $\theta$ the polar angle, and swapping the names silently puts the $\sin$ on the wrong variable, breaking every integral and operator. Always state the convention before you compute.

Why is there a sinθ in the volume element and not a cosθ?

Because the azimuthal displacement is a circle whose radius is the distance from the polar axis, and that distance is $\rho=r\sin\theta$ . A step $\mathrm d\phi$ therefore moves you $r\sin\theta\,\mathrm d\phi$ metres, giving $h_\phi=r\sin\theta$ , and the volume element is the product $h_r h_\theta h_\phi=r^2\sin\theta$ . The factor must vanish where the $\phi$ -circle shrinks to a point — on the poles $\theta=0,\pi$ — and $\sin\theta$ does exactly that, while $\cos\theta$ would wrongly vanish on the equator and be largest at the poles. So $\sin\theta$ is fixed by the geometry, not chosen.

Is the singularity on the polar axis a real physical problem?

No. The origin and the polar axis are coordinate singularities — places where the chart, not space, breaks down. On the axis the azimuth $\phi$ labels a circle of zero radius, so it becomes meaningless, but a smooth physical field sails straight through the pole with no trouble at all. The practical consequence is only in how you solve equations: you demand that solutions stay finite at $r=0$ and are single-valued as $\phi$ increases by $2\pi$ . Those regularity conditions are what select the physically admissible spherical harmonics and radial functions; they are bookkeeping about the coordinates, never a statement that anything is wrong at those points.

Where do the extra terms in the spherical divergence come from?

From the fact that the basis vectors move. In Cartesian coordinates $\hat{\mathbf x},\hat{\mathbf y},\hat{\mathbf z}$ are the same everywhere, so their derivatives are zero and the divergence is a bare sum of partials. In spherical coordinates $\hat{\mathbf e}_r,\hat{\mathbf e}_\theta,\hat{\mathbf e}_\phi$ swing around as you move, with nonzero derivatives such as $\partial_\theta\hat{\mathbf e}_r=\hat{\mathbf e}_\theta$ . When you differentiate a vector field $\mathbf A=A_r\hat{\mathbf e}_r+\cdots$ , the product rule hits both the components and the basis vectors, and the basis-vector part supplies the $r^2$ and $\sin\theta$ weights you see inside $\frac{1}{r^2}\partial_r(r^2A_r)$ and the angular terms. The orthogonal-curvilinear formula packages all of that automatically.

Why are spherical coordinates the natural home for central-force problems?

Because a central force depends only on $r$ , and in these coordinates that dependence is isolated in a single term. When the potential is $V(r)$ , Laplace's or Schrödinger's equation separates into a radial factor times an angular factor, $\psi=R(r)\,Y_\ell^m(\theta,\phi)$ , and the angular part is solved once and for all by the spherical harmonics. The messy three-dimensional PDE collapses to a one-dimensional radial ODE with a centrifugal term $\ell(\ell+1)/r^2$ . No other coordinate system does this for a general central potential — it is the rotational symmetry of the problem matching the rotational structure of the coordinates, which is why the Kepler orbit and the hydrogen atom are both spherical-coordinate calculations at heart.

What is a solid angle, intuitively, and why does it total 4π?

A solid angle is the two-dimensional analogue of an angle: it measures how much of your field of view a patch of sky fills, in steradians, as the area the patch projects onto a unit sphere. The element is $\mathrm d\Omega=\sin\theta\,\mathrm d\theta\,\mathrm d\phi$ , and integrating over all directions gives $\int_0^{2\pi}\!\int_0^\pi\sin\theta\,\mathrm d\theta\,\mathrm d\phi=2\pi\,[-\cos\theta]_0^\pi=4\pi$ . That $4\pi$ is the total solid angle of the whole sphere, and it is the same $4\pi$ that appears in Gauss's law and in the inverse-square flux, because a point source spreads its field uniformly over all $4\pi$ steradians. A cone of half-angle $\alpha$ subtends $2\pi(1-\cos\alpha)$ of it.

How do spherical coordinates relate to cylindrical and plane polar ones?

They are close cousins tied by two substitutions. Set $\rho=r\sin\theta$ for the distance from the axis and $z=r\cos\theta$ for the height, and spherical coordinates map term by term onto cylindrical coordinates $(\rho,\phi,z)$ . Restrict to the equatorial plane $\theta=\pi/2$ , where $\sin\theta=1$ , and you drop straight to plane polar coordinates $(r,\phi)$ with $h_\phi=r$ . This is why results in the equatorial plane look like the familiar polar ones, and why problems with axial rather than spherical symmetry are usually better done in cylindrical coordinates: the "right" system is the one whose constant-coordinate surfaces match the symmetry of the source or the boundary.