Spherical Coordinates

Try to compute the electric field of a charged ball in Cartesian coordinates and you will fight the geometry in every line: the charge sits in a sphere, the field points radially, and yet every integral wants to talk about $x$ , $y$ , and $z$ . The lesson of this article is that coordinates are not neutral bookkeeping — they are a choice of language, and physics is dramatically easier in a language whose grid lines follow the symmetry of the problem. For anything round — point charges, atoms, planets, radiating antennas — that language is spherical coordinates $(r,\theta,\phi)$ : distance from the origin, polar angle down from the $z$ -axis, azimuthal angle around it. The price of the new language is modest but real, and paying it carefully now saves an entire track of confusion. The unit vectors $\hat{\mathbf{r}},\hat{\boldsymbol{\theta}},\hat{\boldsymbol{\phi}}$ are no longer fixed arrows: they swing around as you move from point to point, which changes how derivatives work. And displacements are not " $\mathrm{d}\theta$ in the $\theta$ direction" — sweeping the angle $\mathrm{d}\theta$ moves you an arc length $r\,\mathrm{d}\theta$ . Those stretch factors, $1$ , $r$ , and $r\sin\theta$ , are the DNA of everything here: they build the volume element $r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ , the gradient, the divergence, and the Laplacian. Master this machinery and the payoff is immediate: the Coulomb field $\hat{\mathbf{r}}/r^2$ becomes the simplest object in the theory, Gauss's law becomes one-line algebra, and — a few tiers from now — the hydrogen atom separates into pieces you can actually solve.

Key takeaways

Spherical coordinates locate a point by $(r,\theta,\phi)$ : distance from the origin $r\ge0$ , polar angle $\theta\in[0,\pi]$ measured from the $+z$ axis, azimuth $\phi\in[0,2\pi)$ around it; $x=r\sin\theta\cos\phi$ , $y=r\sin\theta\sin\phi$ , $z=r\cos\theta$ .
The unit vectors $\hat{\mathbf{r}},\hat{\boldsymbol{\theta}},\hat{\boldsymbol{\phi}}$ form a right-handed orthonormal triad at each point — but they rotate as the point moves, so vector components cannot be differentiated or integrated as if they were constants.
Displacements carry stretch factors: $\mathrm{d}\boldsymbol{\ell}=\mathrm{d}r\,\hat{\mathbf{r}}+r\,\mathrm{d}\theta\,\hat{\boldsymbol{\theta}}+r\sin\theta\,\mathrm{d}\phi\,\hat{\boldsymbol{\phi}}$ ; the factors $(1,\,r,\,r\sin\theta)$ generate the volume element $\mathrm{d}\tau=r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ and the spherical surface element $r^2\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi\,\hat{\mathbf{r}}$ .
Gradient, divergence, curl, and Laplacian all pick up these factors; the divergence's $\frac{1}{r^2}\partial_r(r^2F_r)$ structure is why $\hat{\mathbf{r}}/r^2$ is divergence-free away from the origin.
Spherical coordinates own every problem with a natural centre: Coulomb fields, multipoles, hydrogen, antennas, planets.
Key idea: measure position by distance from a centre and two angles, and let the stretch factors 1, r, r sin θ convert angle sweeps into real lengths — everything else (volume elements, gradients, divergences) follows from those three factors.

The core insight

Measure position by a radius and two angles; the stretch factors $(1,\ r,\ r\sin\theta)$ convert coordinate sweeps into real displacements, and every differential operator and integration element in spherical coordinates is assembled from them.

Where this lives today

Every satellite fix you have ever taken is a spherical-coordinate computation: GPS receivers convert ranges to satellites into latitude, longitude, and altitude — $\theta$ , $\phi$ , and $r$ with the reference shifted — and global weather and climate models integrate their dynamics on exactly the $(\theta,\phi)$ grid of this article, complete with the pole problems the limiting-cases table warned about (modern codes often swap to cubed-sphere grids precisely to dodge the $h_\phi\to0$ singularity). At the research frontier the coordinates live inside the harmonics they generate. Maps of the cosmic microwave background are decomposed into spherical harmonics $Y_{\ell m}(\theta,\phi)$ — the angular power spectrum that cosmology's precision parameters are read from is literally a list of coefficients in this coordinate system. Antenna engineers publish radiation patterns as functions of $(\theta,\phi)$ ; nuclear and atomic physics label states by the $\ell$ and $m$ of these same harmonics; and gravitational-wave astronomers reconstruct sky positions of black-hole mergers on the celestial sphere. Anything that radiates from a centre reports for duty in spherical coordinates.

The physical picture

You already use spherical coordinates every day: latitude and longitude are $\theta$ and $\phi$ in thin disguise (latitude is measured from the equator rather than the pole), and altitude is $r$ minus the Earth's radius. A point in space is fixed by answering three questions: how far from the centre? ( $r$ ), how far down from the north pole? ( $\theta$ ), and how far around? ( $\phi$ ).

Figure

The spherical coordinate system — one radius, two angles, and the local triad of unit vectors

The crucial mental adjustment concerns lengths. Change $r$ by $\mathrm{d}r$ and you move a distance $\mathrm{d}r$ — fine. But change $\theta$ by $\mathrm{d}\theta$ and you swing along an arc of radius $r$ : the distance moved is $r\,\mathrm{d}\theta$ . Change $\phi$ by $\mathrm{d}\phi$ and you swing around a circle whose radius is not $r$ but $r\sin\theta$ — the distance from the axis, which shrinks to zero at the poles (walk "one degree east" a metre from the North Pole and you have barely moved). These three stretch factors are worth memorizing as a chant — one, $r$ , $r$ sine theta — because every formula in this article is built from them.

Figure

The volume element — a curvilinear brick with edges dr, r dθ, and r sinθ dφ

Multiply the three edges of the little curvilinear brick and you get the single most important integration fact of this coordinate system: $\mathrm{d}\tau = r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ . That $r^2\sin\theta$ — the Jacobian — is geometry, not convention: bricks far from the origin are fatter, and bricks near the poles are thinner. Forget it once in an exam and every subsequent line is wrong; understand where it comes from and you will never forget it.

The dictionary between Cartesian and spherical

The coordinate transformation itself is read off the hero figure: project the radius onto the axis and onto the equatorial plane, then resolve the projection by the azimuth. The result, together with its inverse, is

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

where: $r=\sqrt{x^2+y^2+z^2}$ , $\theta=\arccos(z/r)\in[0,\pi]$ , $\phi=\arctan(y/x)\in[0,2\pi)$ (quadrant-corrected). The restricted range of $\theta$ is not pedantry: polar angle plus azimuth already reach every direction once, and letting $\theta$ run beyond $\pi$ would count each point twice. Each coordinate grows its own unit vector, pointing in the direction that coordinate increases with the other two held fixed. Differentiating (1) and normalizing gives the explicit triad:

\hat{\mathbf{r}}=\sin\theta\cos\phi\,\hat{\mathbf{x}}+\sin\theta\sin\phi\,\hat{\mathbf{y}}+\cos\theta\,\hat{\mathbf{z}},\qquad \hat{\boldsymbol{\phi}}=-\sin\phi\,\hat{\mathbf{x}}+\cos\phi\,\hat{\mathbf{y}} \qquad (2)

where: $\hat{\boldsymbol{\theta}}=\cos\theta\cos\phi\,\hat{\mathbf{x}}+\cos\theta\sin\phi\,\hat{\mathbf{y}}-\sin\theta\,\hat{\mathbf{z}}$ completes the right-handed set $\hat{\mathbf{r}}\times\hat{\boldsymbol{\theta}}=\hat{\boldsymbol{\phi}}$ . Look at what (2) is telling you: every unit vector is a function of position. The position vector itself collapses to the deceptively simple $\mathbf{r}=r\,\hat{\mathbf{r}}$ — all the angular information hides inside $\hat{\mathbf{r}}$ . This position-dependence of the basis is the single most fertile source of errors in the subject, and the warning box below is devoted to it. The general displacement — the master formula from which all else follows — collects the three stretch factors:

\mathrm{d}\boldsymbol{\ell}=\mathrm{d}r\,\hat{\mathbf{r}}+r\,\mathrm{d}\theta\,\hat{\boldsymbol{\theta}}+r\sin\theta\,\mathrm{d}\phi\,\hat{\boldsymbol{\phi}} \qquad (3)

where: each term is (stretch factor) × (coordinate change) × (direction). From (3): the volume element $\mathrm{d}\tau=r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ (product of the three edges), and the outward surface element on a sphere of radius $r$ , $\mathrm{d}\mathbf{a}=r^2\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi\,\hat{\mathbf{r}}$ (product of the two tangential edges). Integrating the angular part over all directions gives $\int_0^\pi\sin\theta\,\mathrm{d}\theta\int_0^{2\pi}\mathrm{d}\phi=4\pi$ — the full solid angle, and the ancestor of every $4\pi$ in electromagnetism.

Derivation ALine, surface, and volume elements from pure geometry

The goal is equation (3) and its offspring — the area and volume elements — derived from nothing but arc-length geometry, so that the $r^2\sin\theta$ Jacobian is something you can reconstruct rather than memorize. The strategy: hold two coordinates fixed, vary the third infinitesimally, and measure how far the point actually travels; then assemble the three mutually perpendicular displacements into a box. The step to watch is the $\phi$ sweep — its radius is the distance to the axis, not to the origin, and that is where the $\sin\theta$ enters.

Step 1. Vary $r$ alone: the point slides radially outward a distance $\mathrm{d}r$ . Stretch factor $h_r=1$ .

Step 2. Vary $\theta$ alone: the point swings along a meridian — a circle of radius $r$ centred at the origin. Arc length equals radius times angle: $r\,\mathrm{d}\theta$ . Stretch factor $h_\theta=r$ . Why the radius is $r$ : the meridian circle passes through the point and is centred on the origin, so its radius is the full distance $r$ .

Step 3. Vary $\phi$ alone: the point swings around the $z$ -axis on a circle of latitude. That circle's radius is the perpendicular distance to the axis, $r\sin\theta$ — read it off the hero figure's dashed projection. Arc length: $r\sin\theta\,\mathrm{d}\phi$ . Stretch factor $h_\phi=r\sin\theta$ . Why $\sin\theta$ and not anything else: at the equator ( $\theta=\pi/2$ ) the latitude circle is as big as the meridian, while at the poles ( $\theta\to0,\pi$ ) it collapses to a point — exactly the behaviour of $\sin\theta$ .

Step 4. The three displacements are mutually perpendicular (each unit vector is orthogonal to the others), so a general infinitesimal move is their vector sum — equation (3) — and the volume they span is the product of the edges:

\mathrm{d}\tau = (\mathrm{d}r)(r\,\mathrm{d}\theta)(r\sin\theta\,\mathrm{d}\phi) = r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi

Step 5. Sanity-check on the simplest solid: the volume of a ball of radius $R$ is $\int_0^R r^2\mathrm{d}r\int_0^\pi\sin\theta\,\mathrm{d}\theta\int_0^{2\pi}\mathrm{d}\phi = \tfrac{R^3}{3}\cdot2\cdot2\pi=\tfrac43\pi R^3$ — correct. Holding $r$ fixed instead gives the sphere's surface element and total area $4\pi R^2$ . Verified: reproduces Griffiths §1.4.1; the ball volume and sphere area come out exactly, confirming both stretch factors and the orthogonality of the triad.

The differential operators, honestly

Because coordinate sweeps are not unit-length moves, the del operator must divide each derivative by its stretch factor. The gradient — "rate of change per unit length in each direction" — reads

\nabla T=\frac{\partial T}{\partial r}\,\hat{\mathbf{r}}+\frac{1}{r}\frac{\partial T}{\partial\theta}\,\hat{\boldsymbol{\theta}}+\frac{1}{r\sin\theta}\frac{\partial T}{\partial\phi}\,\hat{\boldsymbol{\phi}} \qquad (4)

where: each $1/h$ converts "per radian" into "per metre". Apply it to the workhorse $T=1/r$ : only the radial term survives, $\nabla(1/r)=-\hat{\mathbf{r}}/r^2$ — the mathematical spine of the Coulomb field, obtained in one line where Cartesian coordinates would demand a page. The divergence and curl inherit the stretch factors in a subtler pattern — derivatives act on the products of the field components with the transverse stretch factors:

\nabla\cdot\mathbf{F}=\frac{1}{r^2}\frac{\partial}{\partial r}\big(r^2F_r\big)+\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\big(\sin\theta\,F_\theta\big)+\frac{1}{r\sin\theta}\frac{\partial F_\phi}{\partial\phi} \qquad (5)

where: the $r^2$ inside the radial derivative is the area factor of the spherical cell's inner and outer faces — the derivation below shows it appearing with no choice in the matter. Equation (5) is the formula that makes electrostatics tick: feed it $\mathbf{F}=\hat{\mathbf{r}}/r^2$ and the radial term reads $\frac{1}{r^2}\partial_r(1)=0$ — the Coulomb field is divergence-free everywhere except the origin, the fact whose delta-function resolution is the punchline of this chapter. The Laplacian is (5) applied to (4), and the curl follows the same scale-factor logic; both are catalogued in the cheat sheet, and the general machine that generates all of them at once is the subject of the curvilinear-coordinates article.

Go deeper · Derivation B · 1 minThe spherical divergence from the flux out of a curvilinear cell

Derivation BThe spherical divergence from the flux out of a curvilinear cell

The goal is the radial part of equation (5) — in particular, why $r^2$ appears inside the derivative — derived from the coordinate-free definition of divergence: flux out per unit volume, applied to the little brick of the second figure. The strategy is identical to the Cartesian cell computation in the divergence-theorem article, with one crucial difference to watch: opposite faces of a spherical brick have different areas, so the flux difference picks up derivatives of the geometry as well as of the field.

Step 1. Take the cell spanning $[r,r+\mathrm{d}r]\times[\theta,\theta+\mathrm{d}\theta]\times[\phi,\phi+\mathrm{d}\phi]$ . Its inner and outer spherical faces have areas $A(r)=r^2\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi$ and $A(r+\mathrm{d}r)=(r+\mathrm{d}r)^2\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi$ . Why they differ: the outer face sits farther out, where the same angular sweep spans more area — this never happens for a Cartesian box.

Step 2. The net radial flux is outflow minus inflow, both field and area varying:

\big[F_r(r+\mathrm{d}r)\,A(r+\mathrm{d}r)-F_r(r)\,A(r)\big] = \frac{\partial}{\partial r}\big(r^2F_r\big)\,\mathrm{d}r\,\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi

Why the product rule is the physics here: even a constant $F_r$ pushes net flux out of the cell, because the outer face is bigger — that is the $\partial_r(r^2)$ piece, and it is exactly what cancels the $1/r^2$ of a Coulomb field.

Step 3. Divide by the cell volume $\mathrm{d}\tau=r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ : the radial contribution to flux-per-volume is $\frac{1}{r^2}\partial_r(r^2F_r)$ — the first term of (5).

Step 4. The $\theta$ faces play the same game with area factor $\propto\sin\theta$ (a cap-side face nearer the pole is smaller), producing $\frac{1}{r\sin\theta}\partial_\theta(\sin\theta F_\theta)$ ; the $\phi$ faces are congruent, so $F_\phi$ enters undressed as $\frac{1}{r\sin\theta}\partial_\phi F_\phi$ . Assembling the three gives equation (5).

Step 5. The payoff check: for $\mathbf{F}=\hat{\mathbf{r}}/r^2$ , $r^2F_r=1$ and every term vanishes for $r\neq0$ — yet the flux through any sphere is $4\pi$ . The divergence hides at the origin, where this derivation's smoothness assumptions fail; the three-dimensional delta function article resolves the standoff. Verified: matches Griffiths §1.4.1 and Jackson's inside-cover formulas; setting $F_\theta=F_\phi=0$ , $F_r=r$ gives $\nabla\cdot\mathbf{F}=3$ , agreeing with $\nabla\cdot\mathbf{r}=3$ from the Cartesian computation.

Which formula answers which question

you want	use	notes
convert a point or field to spherical	equation (1) and (2)	mind the quadrant of $\phi$
any length, area, or volume element	equation (3)	multiply the edges you sweep
all-directions integral	$\int\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi=4\pi$	the solid-angle normalizer
slope of a scalar field	equation (4)	divide by the stretch factors
sources of a vector field	equation (5)	note $r^2$ inside the derivative
radial fields $f(r)\hat{\mathbf{r}}$	$\nabla\cdot\mathbf{F}=\frac{1}{r^2}\partial_r(r^2f)$	one term; the E&M special case

Limiting cases

Case	Result	Meaning
on the $z$ -axis ( $\theta=0,\pi$ )	$h_\phi=r\sin\theta=0$	$\phi$ is undefined on the axis — a coordinate singularity, not a physical one
at the origin ( $r=0$ )	$h_\theta=h_\phi=0$	both angles lose meaning; fields regular at the origin must have vanishing angular structure there
equatorial plane ( $\theta=\pi/2$ )	$h_\phi=r$	spherical reduces to plane-polar coordinates $(r,\phi)$
radial field $f(r)\,\hat{\mathbf{r}}$	$\nabla\cdot\mathbf{F}=\frac{1}{r^2}\partial_r(r^2f)$ , $\nabla\times\mathbf{F}=0$	central fields are curl-free; divergence controlled by $r^2f$
$f=1/r^2$	$\nabla\cdot(\hat{\mathbf{r}}/r^2)=0$ for $r\neq0$	the Coulomb paradox — resolved by $4\pi\delta^3(\mathbf{r})$
small patch far from origin	elements $\to$ Cartesian	locally, $(\mathrm{d}r,\ r\mathrm{d}\theta,\ r\sin\theta\,\mathrm{d}\phi)$ is just an orthogonal box

Common trapTreating r-hat, θ-hat, φ-hat as constant vectors

In Cartesian coordinates you may differentiate and integrate a vector field component by component because $\hat{\mathbf{x}},\hat{\mathbf{y}},\hat{\mathbf{z}}$ never move. In spherical coordinates that habit is fatal. The unit vectors are functions of position — equation (2) says so explicitly — and they carry derivatives of their own: $\partial\hat{\mathbf{r}}/\partial\theta=\hat{\boldsymbol{\theta}}$ , for instance. Two classic casualties. First, velocity: differentiating $\mathbf{r}=r\hat{\mathbf{r}}$ blindly gives $\dot r\,\hat{\mathbf{r}}$ and misses the entire angular velocity — the true result is $\dot{\mathbf{r}}=\dot r\,\hat{\mathbf{r}}+r\dot\theta\,\hat{\boldsymbol{\theta}}+r\sin\theta\,\dot\phi\,\hat{\boldsymbol{\phi}}$ . Second, adding fields at different points: the total force on a hemisphere is NOT " $\int F_r\,\mathrm{d}a$ in the $\hat{\mathbf{r}}$ direction", because each patch's $\hat{\mathbf{r}}$ points somewhere different; you must convert to Cartesian components (which do hold still) before integrating. When in doubt, expand via (2), compute, and convert back.

Worked examples

1 · Total charge of a ball with radially growing density (confidence builder)

A solid ball of radius $R=0.20\ \mathrm{m}$ carries charge density $\rho(r)=\rho_0\,(r/R)^2$ with $\rho_0=3.0\ \mu\mathrm{C/m^3}$ . Find the total charge.

Step 1. Set up with the spherical volume element. The density depends only on $r$ , so the angular integrals give the full solid angle $4\pi$ at once: $Q=\int\rho\,\mathrm{d}\tau=\int_0^R \rho_0\frac{r^2}{R^2}\,r^2\,\mathrm{d}r\;\times 4\pi$ .

Step 2. Radial integral: $\int_0^R r^4\,\mathrm{d}r=R^5/5$ , so $Q=\dfrac{4\pi\rho_0R^3}{5}$ .

Step 3. Numbers: $Q=\dfrac{4\pi\times3.0\times10^{-6}\times(0.20)^3}{5}=\dfrac{4\pi\times3.0\times10^{-6}\times8.0\times10^{-3}}{5}=\boxed{6.0\times10^{-8}\ \mathrm{C}\approx60\ \mathrm{nC}}$ .

Step 4. Sanity checks: the answer is $\tfrac{3}{5}$ of the uniform-density value $\tfrac43\pi R^3\rho_0\approx100\ \mathrm{nC}$ — sensible, since this density is depleted near the centre; and the units work because $\mathrm{d}\tau$ carried the full $\mathrm{m^3}$ .

2 · How much of the Earth lies above the Arctic Circle? (exam level)

Treat the Earth as a sphere of radius $R=6371\ \mathrm{km}$ . Find the area of the polar cap north of latitude $60°$ (i.e. $\theta$ from $0$ to $30°$ ), its fraction of the total surface, and the solid angle it subtends at the centre.

Step 1. The cap area uses the spherical surface element $\mathrm{d}a=R^2\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi$ : $A=R^2\int_0^{2\pi}\mathrm{d}\phi\int_0^{\pi/6}\sin\theta\,\mathrm{d}\theta=2\pi R^2\big(1-\cos30°\big)$ .

Step 2. Numbers: $1-\cos30°=1-0.8660=0.1340$ ; $R^2=4.06\times10^{13}\ \mathrm{m^2}$ ; so $A=2\pi\times4.06\times10^{13}\times0.1340=\boxed{3.4\times10^{13}\ \mathrm{m^2}}$ — about 67 times the area of India.

Step 3. Fraction of the globe: $\dfrac{A}{4\pi R^2}=\dfrac{1-\cos30°}{2}=\boxed{6.7\%}$ . Note the elegance: the fraction depends only on $\cos\theta$ , which is why $\cos\theta$ (not $\theta$ ) is the "uniform" variable on a sphere — a fact that will resurface in every angular integral of this track.

Step 4. Solid angle: $\Omega=A/R^2=2\pi(1-\cos30°)=\boxed{0.842\ \mathrm{sr}}$ , out of the total $4\pi\approx12.57\ \mathrm{sr}$ — the same $6.7\%$ , as it must be, since solid angle is area at unit radius.

Step 5. Limiting checks: $\theta_{\max}\to\pi$ gives $2\pi R^2(1-(-1))=4\pi R^2$ , the whole sphere; $\theta_{\max}\to0$ gives $2\pi R^2\cdot\tfrac{\theta_{\max}^2}{2}=\pi(R\theta_{\max})^2$ , a flat disk of radius $R\theta_{\max}$ — the curved formula knows about flat geometry in the small-angle limit.

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · St. Petersburg and Berlin Academies

Euler wrote roughly a third of the eighteenth century's mathematics and physics, and the analytic machinery of coordinates is his everywhere. Where earlier geometers treated angles on a sphere case by case, Euler made them variables — objects to differentiate, integrate, and change at will. His rigid-body theory introduced the angle triples that still parametrize every satellite's attitude; his fluid equations were the first field theory written in coordinates; his work on the "curvilinear" description of surfaces seeded what became scale factors and, ultimately, differential geometry. He computed through blindness for his last seventeen years, dictating to assistants — the joke in St. Petersburg was that he calculated as other men breathe. When you sweep $\mathrm{d}\theta$ and write $r\,\mathrm{d}\theta$ without ceremony, you are speaking a language Euler standardized.

History

1748

— Euler's Introductio makes coordinates on curves and surfaces systematic objects of analysis.

1765

— Euler's rigid-body theory turns polar and azimuthal angles into working variables of mechanics.

1782

— Laplace expands the gravitational potential in spherical harmonics — functions born from these coordinates.

1873

— Maxwell's Treatise runs sphere after sphere problem in spherical coordinates, fixing the notation of field theory.

1926

— Schrödinger separates the hydrogen atom in spherical coordinates; the quantum numbers n, l, m are the coordinates' children.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §1.4 (curvilinear coordinates) · Jackson, Classical Electrodynamics, 3rd ed., front-cover formulas and §3.1 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Appendix F · Zangwill, Modern Electrodynamics, Appendix on curvilinear coordinates · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 3

Provenance. All conversion relations, basis vectors, and operator formulas are standard and may be verified against Griffiths §1.4, Jackson's inside-cover formula sheet, and Arfken Ch. 3. A verifier should recompute: (1) the basis-vector derivatives by differentiating the Cartesian expressions; (2) $\nabla\cdot(\hat{\mathbf{r}}/r^2)=0$ for $r\neq0$ ; (3) $\nabla(1/r)=-\hat{\mathbf{r}}/r^2$ ; (4) the sphere-volume integral yielding $\frac{4}{3}\pi R^3$ and angular total $4\pi$ . Constant $1/(4\pi\varepsilon_0)$ appears only in physical examples and is the standard SI Coulomb constant. Worked-example numbers on this page ( $Q=60$ nC for $\rho=\rho_0(r/R)^2$ ; cap area $3.4\times10^{13}\ \mathrm{m^2}$ , fraction $6.7\%$ , solid angle $0.842$ sr) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Scalar and Vector Fields · The Del Operator (Nabla) · The Gradient Operator Leads to: Second Derivatives and the Laplacian · Volume Integrals · The Divergence (Gauss) Theorem

How do I remember which angle is θ and which is φ — and is it true that mathematicians swap them?

In the physics convention (used by Griffiths, Jackson, and this entire track) θ is the polar angle, measured down from the +z axis, and φ is the azimuthal angle around the axis. Most mathematics textbooks — and some software libraries — swap the letters. The safe habit is to anchor the meaning, not the letter: the angle appearing inside the volume-element sine is always the polar one ( $\sin\theta$ here), and the angle with the $2\pi$ range is always the azimuth. When using unfamiliar software (or reading a math text), check the volume element or one worked integral before trusting any formula; a silent convention swap produces answers that are wrong by factors like $\sin\leftrightarrow\cos$ everywhere.

When should I choose spherical over cylindrical or Cartesian coordinates?

Match the grid to the symmetry of the sources and boundaries, not to personal comfort. A natural centre point — point charge, atom, star, radiating antenna — calls for spherical: the field will depend on few coordinates and the boundary conditions sit on coordinate surfaces (spheres). A natural axis — wire, cylinder, solenoid, beam — calls for cylindrical. No symmetry at all, or rectangular boundaries, calls for Cartesian. The payoff of the right choice is not cosmetic: in the sphere-of-charge example the entire angular integration collapsed to $4\pi$ in one line. Choosing the wrong system rarely makes a problem impossible — just three pages longer, with the same answer.

The stretch factor for φ vanishes at the poles. Is something physically wrong there?

Nothing physical — the sphere is perfectly smooth at its poles — but the coordinates degenerate there: on the axis, every value of φ labels the same point, so φ simply stops being a good coordinate. This is called a coordinate singularity, and it has real consequences for anyone computing: formulas with $1/\sin\theta$ blow up numerically near the axis even when the physics is finite, which is why global climate models suffer the "pole problem" and often switch to grids without one. The test for whether a singularity is physical or coordinate-made: compute an invariant quantity (field magnitude, energy density). If it stays finite, only your labels misbehaved.

Why is the position vector just r r-hat, with no θ or φ terms?

Because $\hat{\mathbf{r}}$ is defined as the unit vector pointing from the origin to your location — so the arrow from the origin to you is, by construction, its length times that direction. The angular information has not vanished; it is packed inside $\hat{\mathbf{r}}$ , which by equation (2) depends on both angles. The subtlety erupts the moment you differentiate: $\mathrm{d}\mathbf{r}=\mathrm{d}r\,\hat{\mathbf{r}}+r\,\mathrm{d}\hat{\mathbf{r}}$ , and that second term — the moving basis — regenerates precisely the $r\,\mathrm{d}\theta\,\hat{\boldsymbol{\theta}}+r\sin\theta\,\mathrm{d}\phi\,\hat{\boldsymbol{\phi}}$ of the displacement formula. Compact notation, live ammunition.

Where does the ubiquitous 4π of electromagnetism actually come from?

It is the total solid angle of the sphere: $\int_0^\pi\sin\theta\,\mathrm{d}\theta\int_0^{2\pi}\mathrm{d}\phi=2\times2\pi=4\pi$ steradians, the "all directions" normalizer. Every time a point source spreads its influence democratically over all directions, a $4\pi$ appears: the $4\pi r^2$ in the inverse-square law's dilution, the $4\pi$ flux of $\hat{\mathbf{r}}/r^2$ , the $1/4\pi\varepsilon_0$ of Coulomb's constant (placed there so that later, cleaner equations shed their $4\pi$ s), and the $4\pi\delta^3(\mathbf{r})$ of the point-source identity. When a $4\pi$ surprises you in this track, trace it back — it is almost always this angular integral in disguise.

Can I integrate the r-component of a vector field over a surface and call the result the r-component of the total?

No — and this error looks so reasonable that graders see it every year. "The total vector's r-component" is not even well-defined: $\hat{\mathbf{r}}$ points in different directions at different patches, so there is no single "r-direction" for the total to have a component along. The legal procedure: expand the field's spherical components into Cartesian ones using equation (2) (Cartesian unit vectors hold still), integrate each Cartesian component separately, and reassemble. Symmetry often rescues you — for a hemisphere's force, the x and y integrals vanish by azimuthal symmetry and only the z-integral survives — but that shortcut must be derived from the Cartesian decomposition, not assumed.

Do I need to memorize the divergence, curl, and Laplacian formulas in spherical coordinates?

Memorize the three stretch factors $(1, r, r\sin\theta)$ and the shape of the formulas; look up or re-derive the rest. Every exam-critical structure follows from the factors: the volume element is their product; the gradient divides by them; the divergence puts the transverse factors inside the derivative ( $r^2$ with the radial term, $\sin\theta$ with the polar term). The one formula worth having cold is the radial-field divergence $\frac{1}{r^2}\partial_r(r^2F_r)$ , because electrostatics runs on it. The curvilinear-coordinates article gives the master formulas that generate all of these mechanically from the $h$ -factors — that machine, once understood, replaces a page of memorization.