Curvilinear Coordinates — Spherical and Cylindrical

⚡ Key Takeaways

Curvilinear coordinates describe points in space using systems other than Cartesian $(x,y,z)$ — most importantly cylindrical $(\rho,\varphi,z)$ and spherical $(r,\theta,\varphi)$ . They simplify problems with rotational or radial symmetry by aligning coordinate surfaces with the geometry. The price is position-dependent unit vectors and extra metric factors in gradients, divergences, Laplacians, and volume integrals. Standard tools for physics and engineering since the work of Lamé, Gauss, and Laplace in the 18th–19th centuries.

💡 The Core Insight

Choose coordinates whose level surfaces match your problem's symmetry, then pay for it with scale factors $h_i = |\partial \mathbf r/\partial q^i|$ that rescale every derivative and integral. Once you know the three $h_i$ , every differential operator — gradient, divergence, curl, Laplacian — and the volume element follow from a single universal template. Cylindrical has $h_\rho=1,\,h_\varphi=\rho,\,h_z=1$ ; spherical has $h_r=1,\,h_\theta=r,\,h_\varphi=r\sin\theta$ .

DefinitionCurvilinear coordinate system

Let $U\subseteq\mathbb{R}^3$ be open. A curvilinear coordinate system on $U$ is a smooth map $\mathbf r:V\to U$ , $\mathbf r(q^1,q^2,q^3)=(x,y,z)$ , with smooth inverse on the interior, i.e. the Jacobian is nonzero:

1

\quad J=\det\!\left[\frac{\partial(x,y,z)}{\partial(q^1,q^2,q^3)}\right]\neq 0.

Derivation hint: invertibility of $\mathbf r$ near a point is guaranteed by the Inverse Function Theorem precisely when this Jacobian determinant is nonzero. The tangent (covariant basis) vectors and scale factors are

2

\quad \mathbf e_i=\frac{\partial \mathbf r}{\partial q^i},\qquad h_i=|\mathbf e_i|=\left|\frac{\partial \mathbf r}{\partial q^i}\right|,\qquad \hat{\mathbf e}_i=\frac{1}{h_i}\frac{\partial\mathbf r}{\partial q^i}.

Derivation hint: $\mathbf e_i$ is the velocity of the curve obtained by varying only $q^i$ ; its length is how much physical distance corresponds to one unit of $q^i$ . The system is orthogonal when $\mathbf e_i\cdot\mathbf e_j=0$ for $i\neq j$ , in which case the squared line element (metric) is diagonal:

3

\quad ds^2=\sum_{i=1}^3 h_i^2\,(dq^i)^2.

Derivation hint: write $d\mathbf r=\sum_i \mathbf e_i\,dq^i$ , take $ds^2=d\mathbf r\cdot d\mathbf r$ , and drop cross terms because $\mathbf e_i\cdot\mathbf e_j=0$ .

DefinitionCylindrical coordinates

Cylindrical coordinates $(\rho,\varphi,z)$ with $\rho\ge 0$ , $\varphi\in[0,2\pi)$ , $z\in\mathbb{R}$ :

4

\quad x=\rho\cos\varphi,\quad y=\rho\sin\varphi,\quad z=z,\qquad \rho=\sqrt{x^2+y^2},\ \ \varphi=\operatorname{atan2}(y,x).

Derivation hint: this is plane polar coordinates in the $xy$ -plane with $z$ carried along unchanged. Scale factors, line element, and volume element:

5

\quad h_\rho=1,\ h_\varphi=\rho,\ h_z=1;\qquad ds^2=d\rho^2+\rho^2 d\varphi^2+dz^2;\qquad dV=\rho\,d\rho\,d\varphi\,dz.

Derivation hint: differentiate [4] to get $\mathbf e_\rho=(\cos\varphi,\sin\varphi,0)$ , $\mathbf e_\varphi=(-\rho\sin\varphi,\rho\cos\varphi,0)$ , $\mathbf e_z=(0,0,1)$ ; their magnitudes give the $h_i$ , and $dV=h_\rho h_\varphi h_z\,d\rho\,d\varphi\,dz$ .

DefinitionSpherical coordinates

Spherical coordinates $(r,\theta,\varphi)$ with $r\ge 0$ , polar angle $\theta\in[0,\pi]$ (from $+z$ axis), azimuth $\varphi\in[0,2\pi)$ — the physics ISO convention:

6

\quad x=r\sin\theta\cos\varphi,\quad y=r\sin\theta\sin\varphi,\quad z=r\cos\theta.

Derivation hint: project the radius $r$ onto the $z$ -axis ( $r\cos\theta$ ) and onto the $xy$ -plane ( $r\sin\theta$ ), then split the planar projection by azimuth. Scale factors, line element, and volume element:

7

\quad h_r=1,\ h_\theta=r,\ h_\varphi=r\sin\theta;\qquad dV=r^2\sin\theta\,dr\,d\theta\,d\varphi.

Derivation hint: differentiate [6] with respect to each coordinate; the magnitudes of the three tangent vectors give $h_r,h_\theta,h_\varphi$ , and $dV=h_rh_\theta h_\varphi\,dr\,d\theta\,d\varphi$ .

⚠️ Notation clash

The symbol $\varphi$ means the azimuthal angle in both systems, but $\theta$ is the polar angle in physics spherical coordinates while many mathematics texts swap $\theta$ and $\varphi$ (so $\theta$ is azimuth and $\varphi$ is polar). Always state your convention. Also, $r$ (spherical radius) $\neq \rho$ (cylindrical radius): $\rho=r\sin\theta$ .

TheoremUniversal differential operators in orthogonal curvilinear coordinates

For an orthogonal system with scale factors $(h_1,h_2,h_3)$ , a scalar field $f$ , and a vector field $\mathbf A=A_1\hat{\mathbf e}_1+A_2\hat{\mathbf e}_2+A_3\hat{\mathbf e}_3$ :

8

\quad \nabla f=\sum_i \frac{1}{h_i}\frac{\partial f}{\partial q^i}\,\hat{\mathbf e}_i,

Derivation hint: $df=\nabla f\cdot d\mathbf r=\sum_i (\nabla f)_i\,h_i\,dq^i$ must equal $\sum_i (\partial f/\partial q^i)\,dq^i$ ; match coefficients.

9

\quad \nabla\cdot\mathbf A=\frac{1}{h_1h_2h_3}\sum_i \frac{\partial}{\partial q^i}\!\left(\frac{h_1h_2h_3}{h_i}A_i\right),

Derivation hint: apply the divergence theorem to an infinitesimal coordinate box with faces of area $(h_jh_k\,dq^jdq^k)$ and volume $h_1h_2h_3\,dq^1dq^2dq^3$ .

10

\quad \nabla\times\mathbf A=\frac{1}{h_1h_2h_3}\begin{vmatrix} h_1\hat{\mathbf e}_1 & h_2\hat{\mathbf e}_2 & h_3\hat{\mathbf e}_3\\[4pt] \partial_{q^1} & \partial_{q^2} & \partial_{q^3}\\[4pt] h_1A_1 & h_2A_2 & h_3A_3\end{vmatrix},

Derivation hint: apply Stokes' theorem to each infinitesimal coordinate rectangle; the circulation of $\mathbf A$ around a $q^j$ – $q^k$ loop uses edge lengths $h_jdq^j$ .

11

\quad \nabla^2 f=\frac{1}{h_1h_2h_3}\sum_i \frac{\partial}{\partial q^i}\!\left(\frac{h_1h_2h_3}{h_i^2}\frac{\partial f}{\partial q^i}\right).

Derivation hint: compose [8] then [9], i.e. $\nabla^2 f=\nabla\cdot(\nabla f)$ with $A_i=(1/h_i)\partial f/\partial q^i$ .

ProofSpherical Laplacian from the universal template

Goal: specialize [11] to spherical coordinates using $h_r=1$ , $h_\theta=r$ , $h_\varphi=r\sin\theta$ . 1. Compute the product $h_rh_\theta h_\varphi = (1)(r)(r\sin\theta)=r^2\sin\theta$ . 2. Form the three quotients $\dfrac{h_1h_2h_3}{h_i^2}$ :

$i=r$ : $\dfrac{r^2\sin\theta}{1}=r^2\sin\theta$ ;
$i=\theta$ : $\dfrac{r^2\sin\theta}{r^2}=\sin\theta$ ;
$i=\varphi$ : $\dfrac{r^2\sin\theta}{r^2\sin^2\theta}=\dfrac{1}{\sin\theta}$ .

3. Insert into [11], dividing the whole sum by $h_1h_2h_3=r^2\sin\theta$ :

12

\quad \nabla^2 f=\frac{1}{r^2\sin\theta}\!\left[\frac{\partial}{\partial r}\!\Big(r^2\sin\theta\,\frac{\partial f}{\partial r}\Big)+\frac{\partial}{\partial\theta}\!\Big(\sin\theta\,\frac{\partial f}{\partial\theta}\Big)+\frac{\partial}{\partial\varphi}\!\Big(\frac{1}{\sin\theta}\frac{\partial f}{\partial\varphi}\Big)\right].

Derivation hint: factor $\sin\theta$ out of the $r$ -term since it is independent of $r$ ; this is the substitution of step 2 into the master Laplacian. 4. Simplify by pulling constant factors through each derivative:

13

\quad \nabla^2 f=\frac{1}{r^2}\frac{\partial}{\partial r}\!\Big(r^2\frac{\partial f}{\partial r}\Big)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\!\Big(\sin\theta\,\frac{\partial f}{\partial\theta}\Big)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2 f}{\partial\varphi^2}.

Derivation hint: in the first term $\sin\theta$ cancels with the prefactor; the angular part is the operator $\nabla^2_{S^2}/r^2$ on the unit sphere. $\qquad\blacksquare$

∎

ProofCylindrical Laplacian

With $h_\rho=1,h_\varphi=\rho,h_z=1$ so $h_\rho h_\varphi h_z=\rho$ :

14

\quad \nabla^2 f=\frac{1}{\rho}\frac{\partial}{\partial\rho}\!\Big(\rho\frac{\partial f}{\partial\rho}\Big)+\frac{1}{\rho^2}\frac{\partial^2 f}{\partial\varphi^2}+\frac{\partial^2 f}{\partial z^2}.

Derivation hint: quotients $h_1h_2h_3/h_i^2$ are $\rho,\,1/\rho,\,\rho$ for $\rho,\varphi,z$ ; substitute into [11] and divide by $\rho$ . $\qquad\blacksquare$

∎

ProofWhy basis vectors carry derivatives — the time derivative of position

The unit vectors depend on position, so differentiating them produces the centripetal/Coriolis terms. For cylindrical: 1. From [4], $\hat{\boldsymbol\rho}=(\cos\varphi,\sin\varphi,0)$ , $\hat{\boldsymbol\varphi}=(-\sin\varphi,\cos\varphi,0)$ . 2. Differentiate with respect to $\varphi$ : $\dfrac{\partial\hat{\boldsymbol\rho}}{\partial\varphi}=\hat{\boldsymbol\varphi}$ , $\dfrac{\partial\hat{\boldsymbol\varphi}}{\partial\varphi}=-\hat{\boldsymbol\rho}$ . 3. With $\mathbf r=\rho\hat{\boldsymbol\rho}+z\hat{\mathbf z}$ , the acceleration becomes

15

\quad \ddot{\mathbf r}=(\ddot\rho-\rho\dot\varphi^2)\hat{\boldsymbol\rho}+(\rho\ddot\varphi+2\dot\rho\dot\varphi)\hat{\boldsymbol\varphi}+\ddot z\,\hat{\mathbf z}.

Derivation hint: apply the chain rule twice to $\mathbf r=\rho\hat{\boldsymbol\rho}(\varphi)+z\hat{\mathbf z}$ , using $\dot{\hat{\boldsymbol\rho}}=\dot\varphi\hat{\boldsymbol\varphi}$ and $\dot{\hat{\boldsymbol\varphi}}=-\dot\varphi\hat{\boldsymbol\rho}$ . $\qquad\blacksquare$

∎

GL

Gabriel Lamé

1795–1870

The scale factors $h_i$ are universally called Lamé coefficients after Gabriel Lamé, who systematized orthogonal curvilinear coordinates in his 1859 Leçons sur les coordonnées curvilignes. Trained at the École Polytechnique, Lamé worked in St. Petersburg on railway engineering before returning to Paris, where he applied curvilinear coordinates to heat conduction and elasticity. He proved that Laplace's equation separates in the eleven classical orthogonal coordinate systems and introduced what are now called Lamé functions for ellipsoidal harmonics. Gauss called him the foremost French mathematician of his era.

🧠 Intuition

Think of a coordinate system as a labeling scheme stamped onto space. Cartesian uses a rigid square grid: every direction-marker $\hat{\mathbf x},\hat{\mathbf y},\hat{\mathbf z}$ is the same arrow everywhere. Curvilinear systems bend the grid to hug the geometry. On a sphere, the "outward" direction $\hat{\mathbf r}$ literally points differently in Paris than in Tokyo — that's why $\hat{\mathbf r}$ is a field of arrows, not one arrow. The scale factor $h_i$ answers: "If I bump coordinate $q^i$ by one unit, how far do I physically move?" Bumping $\varphi$ by $d\varphi$ near the $z$ -axis (small $\rho$ ) barely moves you, but far out (large $\rho$ ) it sweeps a huge arc — hence $h_\varphi=\rho$ . In spherical coordinates, near the poles ( $\theta\to 0$ ) a change in azimuth $\varphi$ moves you almost nothing, so $h_\varphi=r\sin\theta\to 0$ : this is the coordinate singularity at the poles where $\varphi$ becomes meaningless. The volume element $dV=h_1h_2h_3\,dq^1dq^2dq^3$ is just the volume of the tiny "brick" with edge lengths $h_i\,dq^i$ . The extra factors $\rho$ (cylindrical) and $r^2\sin\theta$ (spherical) you memorize for integrals are nothing but this product.

Figure

Spherical coordinates: the point P at radius r, polar angle θ from +z, azimuth φ in the xy-plane

ExampleVolume and moment of inertia of a solid sphere

Given: A uniform solid ball of radius $R=0.20\ \text{m}$ , mass $M=5.0\ \text{kg}$ . Find its volume and its moment of inertia about a diameter.

Step 1 — Volume using spherical $dV=r^2\sin\theta\,dr\,d\theta\,d\varphi$ :

V=\int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{R} r^2\sin\theta\,dr\,d\theta\,d\varphi=\Big(\tfrac{R^3}{3}\Big)(2)(2\pi)=\tfrac{4}{3}\pi R^3.

Numerically $V=\tfrac{4}{3}\pi(0.20)^3=0.0335\ \text{m}^3$ .

Step 2 — Density $\rho_m=M/V=5.0/0.0335=149.2\ \text{kg/m}^3$ .

Step 3 — Moment of inertia about the $z$ -axis. The distance from the axis is $\rho=r\sin\theta$ , so $I_z=\int \rho_m (r\sin\theta)^2\,dV$ :

I_z=\rho_m\int_0^{2\pi}\!\!d\varphi\int_0^\pi \sin^3\theta\,d\theta\int_0^R r^4\,dr=\rho_m(2\pi)\Big(\tfrac{4}{3}\Big)\Big(\tfrac{R^5}{5}\Big)=\tfrac{8\pi}{15}\rho_m R^5.

Using $\rho_m=\tfrac{3M}{4\pi R^3}$ this reduces to the textbook result $I_z=\tfrac{2}{5}MR^2$ .

Step 4 — Number: $I_z=\tfrac{2}{5}(5.0)(0.20)^2=0.080\ \text{kg}\,\cdot\,\text{m}^2$ .

\boxed{V=0.0335\ \text{m}^3,\qquad I_z=\tfrac{2}{5}MR^2=0.080\ \text{kg}\,\cdot\,\text{m}^2}

ExampleCapacitance per length of a coaxial cable via cylindrical Laplacian

Given: Inner conductor radius $a=1.0\ \text{mm}$ , outer radius $b=4.0\ \text{mm}$ , vacuum dielectric $\varepsilon_0=8.85\times10^{-12}\ \text{F/m}$ . Potential depends only on $\rho$ . Find $C/L$ .

Step 1 — Laplace in cylindrical, $V=V(\rho)$ only, so [14] gives

\frac{1}{\rho}\frac{d}{d\rho}\Big(\rho\frac{dV}{d\rho}\Big)=0\ \Rightarrow\ \rho\frac{dV}{d\rho}=\text{const}=A.

Step 2 — Integrate: $V(\rho)=A\ln\rho+B$ . Apply $V(a)=V_0$ , $V(b)=0$ :

V(\rho)=V_0\frac{\ln(b/\rho)}{\ln(b/a)}.

Step 3 — Field and surface charge. $E_\rho=-dV/d\rho=\dfrac{V_0}{\rho\ln(b/a)}$ . The inner-conductor surface charge per length: by Gauss, line charge $\lambda=2\pi\varepsilon_0\rho E_\rho=\dfrac{2\pi\varepsilon_0 V_0}{\ln(b/a)}$ .

Step 4 — Capacitance per length $C/L=\lambda/V_0$ :

\frac{C}{L}=\frac{2\pi\varepsilon_0}{\ln(b/a)}=\frac{2\pi(8.85\times10^{-12})}{\ln 4}=\frac{5.56\times10^{-11}}{1.386}.

\boxed{\dfrac{C}{L}=4.01\times10^{-11}\ \text{F/m}=40.1\ \text{pF/m}}

⚠️ Unit vectors are not constant — do not pull them out of integrals or derivatives

A frequent error: writing $\displaystyle\int \mathbf F\,dV$ with $\mathbf F=F_r\hat{\mathbf r}$ and treating $\hat{\mathbf r}$ as constant. Because $\hat{\mathbf r}$ points in different directions at different points, $\int \hat{\mathbf r}\,dV\neq \hat{\mathbf r}\int dV$ . To integrate a vector field, resolve it into Cartesian components first (which are constant), or carry the angular dependence explicitly. Likewise $\frac{d}{dt}(v_r\hat{\mathbf r})\neq \dot v_r\hat{\mathbf r}$ ; the basis derivative terms in [15] are essential.

⚠️ The Laplacian is NOT just second partials in curvilinear coordinates

$\nabla^2 f\neq \partial_{rr}f+\partial_{\theta\theta}f+\partial_{\varphi\varphi}f$ . The scale factors generate first-derivative terms (like $\frac{2}{r}\partial_r f$ from expanding [13]). Always use the full forms [13]/[14].

CounterexampleNaive "divergence" of the inverse-square field

Consider $\mathbf E=\dfrac{\hat{\mathbf r}}{r^2}$ (a point-charge field, $A_r=1/r^2$ ). A naive $\partial_r A_r=-2/r^3$ suggests nonzero divergence everywhere. The correct spherical divergence from [9] is

\nabla\cdot\mathbf E=\frac{1}{r^2}\frac{\partial}{\partial r}\Big(r^2\cdot\frac{1}{r^2}\Big)=\frac{1}{r^2}\frac{\partial}{\partial r}(1)=0\quad(r\neq 0).

The $r^2$ scale factor exactly cancels the field's $1/r^2$ decay — the field is divergence-free everywhere except the singular origin, where it produces the delta function $4\pi\delta^3(\mathbf r)$ . Ignoring scale factors gives the wrong (nonzero) answer.

connection

Orthogonal curvilinear coordinates are the bridge between Vector Calculus and Differential Geometry: the scale factors $h_i$ are the diagonal entries of the metric tensor $g_{ij}=\mathbf e_i\cdot\mathbf e_j$ , and the universal operators [8]–[11] are special cases of the tensor formulas for $\nabla f$ , $\nabla\cdot$ , and the Laplace–Beltrami operator. They are the workhorse of Separation of Variables for Partial Differential Equations: spherical coordinates yield Legendre Polynomials and Spherical Harmonics from the $\theta$ -equation, while cylindrical coordinates yield Bessel Functions from the $\rho$ -equation. The volume elements feed directly into Multiple Integrals via the change-of-variables theorem, where $h_1h_2h_3$ equals the Jacobian $|J|$ of [1].

🔬 Where this appears

⏳ History

- 1671–1676 — Newton and others use polar-type coordinates informally for orbital and curve problems, the seed of curvilinear thinking.

- 1748 — Euler systematically employs spherical and cylindrical descriptions in mechanics and fluid problems.

- 1782 — Laplace introduces what become Laplace's equation and spherical harmonics while studying gravitational potential.

- 1785 — Legendre defines the polynomials

P_\ell

arising from the spherical

\theta

-equation in potential theory.

- 1813–1828 — Gauss develops the intrinsic geometry of surfaces (the Theorema Egregium), laying groundwork for the metric viewpoint of curvilinear coordinates.

- 1822 — Fourier's heat-conduction work motivates separable solutions in symmetric geometries.

- 1824 — Bessel publishes the functions (named for him) that solve the cylindrical radial equation, in the context of planetary perturbations.

- 1841–1859 — Lamé formalizes general orthogonal curvilinear coordinates and the scale factors (Lamé coefficients), culminating in Leçons sur les coordonnées curvilignes (1859).

- 1873 — Maxwell's Treatise codifies vector operators in curvilinear systems for electromagnetism; Heaviside and Gibbs later refine vector-calculus notation.

open-problem · Coordinate singularities and the impossibility of one global chart

Spherical and cylindrical coordinates each fail to be one-to-one or smooth on a "thin" set: the $z$ -axis ( $\rho=0$ in cylindrical, $\theta=0,\pi$ in spherical) where the azimuth $\varphi$ is undefined, and the origin ( $r=0$ ). At these places the Jacobian [1] vanishes and the inverse map breaks down — yet space itself is perfectly smooth there. This is the elementary face of a deep topological fact: the sphere $S^2$ admits no single coordinate chart (it is not homeomorphic to an open subset of $\mathbb{R}^2$ ), and indeed no smooth nonvanishing vector field exists on it (the Hairy Ball Theorem), which is why $\hat{\boldsymbol\varphi}$ must blow up at the poles. Numerically these singularities cause real trouble: global atmospheric and ocean models suffer the "pole problem" (vanishing $h_\varphi$ forces tiny time steps near poles), prompting cubed-sphere, icosahedral, and yin-yang grids. Choosing how to cover a curved space with overlapping charts is exactly the program of Differential Geometry and the theory of Manifolds.

Curvilinear Coordinates — Spherical and Cylindrical

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also