Scale Factors and Line, Area and Volume Elements

Cartesian coordinates hide a convenience so basic it is easy to forget: a step of $\mathrm{d}x$ is a step of $\mathrm{d}x$ metres, everywhere and in every direction. Curvilinear coordinates — cylindrical for a wire, spherical for a point charge, elliptic for a charged disc — break that identity. A step $\mathrm{d}\phi$ in azimuth is not a length at all; it is a pure angle, and the physical distance it moves you depends on how far you are from the axis. The scale factors $h_i$ are the exchange rate that repairs this: $h_i$ is exactly how many metres of arc one unit of the coordinate $q_i$ buys at the point where you stand. Formally, if $\mathbf{r}(q_1,q_2,q_3)$ is the position vector written in the new coordinates, then $h_i=|\partial\mathbf{r}/\partial q_i|$ and the local unit vector along the $q_i$ -curve is $\hat{\mathbf{e}}_i=(1/h_i)\,\partial\mathbf{r}/\partial q_i$ . Once the three $h_i$ are in hand, every geometric object you could want to integrate follows by simple multiplication: a length is $h_i\,\mathrm{d}q_i$ , an area on a coordinate surface is a product of two scale factors, and the volume of the little curved box is the full product $h_1h_2h_3\,\mathrm{d}q_1\,\mathrm{d}q_2\,\mathrm{d}q_3$ — which is nothing but the Jacobian of the transformation. This one product also seeds the gradient, divergence, curl and Laplacian in curvilinear coordinates, because each of those operators measures change per unit physical length and so must divide by the appropriate $h_i$ . Master the scale factors and the whole apparatus of vector calculus in the coordinate system that matches your symmetry falls out mechanically.

Key takeaways

A coordinate increment $\mathrm{d}q_i$ is generally not a length; the scale factor $h_i=|\partial\mathbf{r}/\partial q_i|$ converts it into the physical length $h_i\,\mathrm{d}q_i$ .
The displacement is $\mathrm{d}\mathbf{r}=\sum_i h_i\,\mathrm{d}q_i\,\hat{\mathbf{e}}_i$ , and for orthogonal coordinates the line element is Pythagorean, $\mathrm{d}s^2=h_1^2\mathrm{d}q_1^2+h_2^2\mathrm{d}q_2^2+h_3^2\mathrm{d}q_3^2$ .
Area on a coordinate surface is a product of two scale factors, e.g. $\mathrm{d}A=h_1h_2\,\mathrm{d}q_1\mathrm{d}q_2$ ; the volume element is $\mathrm{d}V=h_1h_2h_3\,\mathrm{d}q_1\mathrm{d}q_2\mathrm{d}q_3$ .
The product $h_1h_2h_3$ equals the Jacobian $J=|\partial(x,y,z)/\partial(q_1,q_2,q_3)|$ , so integrals come out the same in any coordinate system.
Standard values: cylindrical $(h_\rho,h_\phi,h_z)=(1,\rho,1)$ and spherical $(h_r,h_\theta,h_\phi)=(1,r,r\sin\theta)$ , giving $\mathrm{d}V=\rho\,\mathrm{d}\rho\,\mathrm{d}\phi\,\mathrm{d}z$ and $\mathrm{d}V=r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ .
The scale factors are also the metric $g_{ij}=h_i^2\delta_{ij}$ and the seed of every curvilinear vector operator.
Key idea: a coordinate increment $\mathrm{d}q_i$ is not a length; the scale factor $h_i=|\partial\mathbf{r}/\partial q_i|$ converts it into one, and every line, area and volume element is a product of the $h_i$ .

The core insight

A coordinate increment $\mathrm{d}q_i$ is not a length; the scale factor $h_i=|\partial\mathbf{r}/\partial q_i|$ converts it into one, and all line, area and volume elements are built from products of the $h_i$ .

Where this lives today

Scale factors are quietly at work in every large numerical simulation of a continuous physical field. Computational fluid dynamics wraps a body-fitted curvilinear grid around an airfoil or a turbine blade so that the coordinate surfaces follow the geometry; the flow solver then carries the metric $g_{ij}$ and the Jacobian $J=h_1h_2h_3$ through every flux term, and a badly stretched mesh with wildly varying scale factors is the classic source of numerical error. Global climate models, geodynamo simulations and solar convection codes integrate the fluid equations on the sphere, where the $r^2\sin\theta$ measure and the singular behaviour of $h_\phi=r\sin\theta$ at the poles force careful spectral or grid treatments. Numerical relativity pushes the same idea to its limit: there the metric is dynamical, the off-diagonal $g_{ij}$ are the gravitational field itself, and Lamé's coefficients are the flat-space warm-up for the full curved-spacetime line element that LIGO's waveform templates are built from. Even a routine finite-element stress analysis or an electromagnetic mode solver is, under the hood, a change of variables whose correctness rests on getting the scale factors and their Jacobian right.

The physical picture

Paint the coordinate grid onto space as a deformable mesh. Where the mesh is stretched, one grid step spans a long physical distance; where it is compressed, the same step spans a short one. The scale factor $h_i$ is precisely this local stretch ratio — physical length divided by coordinate step — evaluated at each point. The cleanest example is the azimuth of cylindrical coordinates: a fixed angular step $\mathrm{d}\phi$ barely moves you near the axis but sweeps a long arc far out, because arc length is radius times angle. That single fact is the statement $h_\phi=\rho$ .

Figure

A fixed angular step dφ sweeps an arc whose length grows with radius: arc = ρ dφ, so the scale factor is h_φ = ρ. With dφ = 0.20 rad the inner arc at ρ = 1 has length 0.20 while the outer arc at ρ = 3 has length 0.60 — three times longer for the same coordinate step

Stack the idea and the whole calculus of elements appears. One scale factor turns a coordinate step into a length; two scale factors turn a patch of a coordinate surface into an area; three turn a coordinate cell into a volume. When the coordinate system is orthogonal — the three $\hat{\mathbf{e}}_i$ mutually perpendicular at every point — that little cell is a genuine rectangular box to first order, so its edges, faces and volume are simple products with no cross terms. Spherical coordinates are the workhorse case: the box has edges $\mathrm{d}r$ , $r\,\mathrm{d}\theta$ and $r\sin\theta\,\mathrm{d}\phi$ , and its volume is the product of the three, $r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ .

Figure

The spherical volume cell has three orthogonal physical edges — radial dr, polar r dθ, azimuthal r sinθ dφ — so its volume is the product dV = r² sinθ dr dθ dφ. At r = 2 and θ = 90° the coefficient r² sinθ equals 4, and integrating the cell over a ball of radius R gives the volume 4/3 π R³

The core equations

Everything begins with one definition. Fix a point, hold two of the coordinates constant, and let the third vary: the position vector traces a coordinate curve whose velocity is $\partial\mathbf{r}/\partial q_i$ . Its length is the scale factor, and dividing by that length gives the local unit vector.

h_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} \qquad (1)

where: $\mathbf{r}$ — position vector as a function of $(q_1,q_2,q_3)$ ; $q_i$ — the $i$ -th curvilinear coordinate; $h_i$ — scale factor, the physical length per unit increment of $q_i$ ; $\hat{\mathbf{e}}_i$ — unit tangent to the $q_i$ -coordinate curve. Read it as: $h_i$ is a length-per-coordinate conversion factor, and $\hat{\mathbf{e}}_i$ is the direction that conversion points. A general small change in all three coordinates displaces the point by the sum of the three tangential contributions. Because each partial derivative is $h_i\hat{\mathbf{e}}_i$ , the displacement is a clean sum of physical lengths along the local axes.

\mathrm{d}\mathbf{r}=\sum_{i=1}^{3}\frac{\partial\mathbf{r}}{\partial q_i}\,\mathrm{d}q_i=h_1\,\mathrm{d}q_1\,\hat{\mathbf{e}}_1+h_2\,\mathrm{d}q_2\,\hat{\mathbf{e}}_2+h_3\,\mathrm{d}q_3\,\hat{\mathbf{e}}_3 \qquad (2)

where: $\mathrm{d}\mathbf{r}$ — the infinitesimal displacement produced by increments $\mathrm{d}q_i$ ; $h_i\,\mathrm{d}q_i$ — the physical length contributed along $\hat{\mathbf{e}}_i$ . Each term already carries units of length, which is the whole point of the scale factors. Squaring the displacement gives the line element. For an orthogonal system the mixed dot products vanish and only the diagonal terms survive, leaving a Pythagorean sum — this is the single most-used formula in the subject.

\mathrm{d}s^2=\mathrm{d}\mathbf{r}\cdot\mathrm{d}\mathbf{r}=h_1^2\,\mathrm{d}q_1^2+h_2^2\,\mathrm{d}q_2^2+h_3^2\,\mathrm{d}q_3^2 \qquad (3)

where: $\mathrm{d}s$ — physical arc length of the displacement $\mathrm{d}\mathbf{r}$ . The absence of cross terms such as $\mathrm{d}q_1\,\mathrm{d}q_2$ is exactly the signature of orthogonality; a skew system would keep them. On a coordinate surface (one coordinate held constant) the remaining two coordinates span an area patch. Its magnitude is the product of the two tangential scale factors, and its orientation is the third unit vector, so the vector area element is a cross product of the two edge displacements.

\mathrm{d}\mathbf{A}_k=(h_i\,\mathrm{d}q_i\,\hat{\mathbf{e}}_i)\times(h_j\,\mathrm{d}q_j\,\hat{\mathbf{e}}_j)=h_i h_j\,\mathrm{d}q_i\,\mathrm{d}q_j\,\hat{\mathbf{e}}_k\quad(i,j,k\ \text{cyclic}) \qquad (4)

where: $\mathrm{d}\mathbf{A}_k$ — oriented area element on the surface of constant $q_k$ ; $\hat{\mathbf{e}}_k$ — its unit normal, which for an orthogonal right-handed triad is the remaining basis vector. The scalar area on that surface is just $\mathrm{d}A_k=h_i h_j\,\mathrm{d}q_i\,\mathrm{d}q_j$ . Three edge displacements span the volume cell. For a right-handed orthonormal triad the scalar triple product of the unit vectors is one, so the volume collapses to the product of the three physical edge lengths — the coordinate-space measure times the product of scale factors.

\mathrm{d}V=h_1 h_2 h_3\,\mathrm{d}q_1\,\mathrm{d}q_2\,\mathrm{d}q_3 \qquad (5)

where: $\mathrm{d}V$ — physical volume of the infinitesimal curvilinear box. This single product is the measure that must appear inside every volume integral written in the coordinates $q_i$ . That same product is the Jacobian determinant of the map to Cartesian coordinates, which is why an integral evaluates to the same number no matter which coordinates you compute it in.

h_1 h_2 h_3=\left|\frac{\partial(x,y,z)}{\partial(q_1,q_2,q_3)}\right|=J \qquad (6)

where: $J$ — Jacobian of the transformation $(q_1,q_2,q_3)\to(x,y,z)$ . Recognising $h_1h_2h_3=J$ unifies the geometric picture (a curved box) with the analytic change-of-variables theorem for integrals. Finally, the scale factors are the diagonal of the metric tensor. Writing the covariant basis vectors $\mathbf{g}_i=\partial\mathbf{r}/\partial q_i$ , their dot products form the metric, which is diagonal exactly when the coordinates are orthogonal.

\mathbf{g}_i=\frac{\partial\mathbf{r}}{\partial q_i},\qquad g_{ij}=\mathbf{g}_i\cdot\mathbf{g}_j=h_i^2\,\delta_{ij} \qquad (7)

where: $\mathbf{g}_i$ — covariant (tangent) basis vector; $g_{ij}$ — metric tensor, whose diagonal entries are the squared scale factors $h_i^2$ . This is the bridge to tensor calculus and general relativity, where the off-diagonal $g_{ij}$ no longer vanish.

Derivation AScale factors and the Pythagorean line element

The goal is to show, from the single definition $h_i=|\partial\mathbf{r}/\partial q_i|$ , why the squared arc length in an orthogonal curvilinear system takes the clean Pythagorean form $\mathrm{d}s^2=\sum_i h_i^2\,\mathrm{d}q_i^2$ with no cross terms, and to pin down exactly which assumption — orthogonality — kills those cross terms. This is the derivation that turns a coordinate transformation into a usable geometry, because once the line element is known every length, area and volume follows by taking pieces of it.

Step 1. Differentiate the position vector by the chain rule: a simultaneous change of all coordinates gives $\mathrm{d}\mathbf{r}=\sum_i(\partial\mathbf{r}/\partial q_i)\,\mathrm{d}q_i$ . Why: the displacement is the sum of what each coordinate contributes, and each contribution is a partial derivative times its increment — this is just the total differential of $\mathbf{r}$ .

Step 2. Insert the definition of the scale factor and unit vector, $\partial\mathbf{r}/\partial q_i=h_i\hat{\mathbf{e}}_i$ , to get $\mathrm{d}\mathbf{r}=\sum_i h_i\,\mathrm{d}q_i\,\hat{\mathbf{e}}_i$ . Why: this separates each contribution into a physical length $h_i\,\mathrm{d}q_i$ and a pure direction $\hat{\mathbf{e}}_i$ , which is what makes the next step a bookkeeping exercise.

Step 3. Dot the displacement with itself: $\mathrm{d}s^2=\mathrm{d}\mathbf{r}\cdot\mathrm{d}\mathbf{r}=\sum_{i,j}h_i h_j\,\mathrm{d}q_i\,\mathrm{d}q_j\,(\hat{\mathbf{e}}_i\cdot\hat{\mathbf{e}}_j)$ . Why: arc length squared is the self-dot-product of the displacement, and expanding the double sum exposes every pairing of coordinate directions, including the potential cross terms.

Step 4. Impose orthonormality, $\hat{\mathbf{e}}_i\cdot\hat{\mathbf{e}}_j=\delta_{ij}$ , so only the $i=j$ terms survive: $\mathrm{d}s^2=\sum_i h_i^2\,\mathrm{d}q_i^2$ . Why: orthogonality of the coordinate curves is precisely the condition that the off-diagonal dot products vanish; without it the metric $g_{ij}$ keeps cross terms and the Pythagorean form fails. *Why it works: for spherical coordinates the three tangents are mutually orthogonal — their pairwise dot products vanish identically — so only the diagonal terms survive and the line element is $\mathrm{d}s^2=\mathrm{d}r^2+r^2\mathrm{d}\theta^2+r^2\sin^2\theta\,\mathrm{d}\phi^2$ , exactly as the Pythagorean form requires.*

Go deeper · Derivation B · 1 minScale factors for spherical polar coordinates

Derivation BScale factors for spherical polar coordinates

The goal is to compute the three spherical scale factors $(h_r,h_\theta,h_\phi)$ directly from the explicit Cartesian map, confirm the system is orthogonal, and assemble the line and volume elements — turning the abstract definition of the previous derivation into the concrete measure $r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ that every physics student eventually memorises but should be able to derive on demand.

Step 1. Write the map $x=r\sin\theta\cos\phi$ , $y=r\sin\theta\sin\phi$ , $z=r\cos\theta$ , with $r\ge0$ , $0\le\theta\le\pi$ , $0\le\phi<2\pi$ . Why: the scale factors are magnitudes of derivatives of $\mathbf{r}$ , so everything follows from this explicit dependence of $(x,y,z)$ on $(r,\theta,\phi)$ .

Step 2. Radial tangent: $\partial\mathbf{r}/\partial r=(\sin\theta\cos\phi,\ \sin\theta\sin\phi,\ \cos\theta)$ , whose magnitude is $\sqrt{\sin^2\theta+\cos^2\theta}=1$ , so $h_r=1$ . Why: moving outward along a ray is a straight-line motion measured in true length, so a unit change in $r$ is a unit change in distance — the scale factor must be one.

Step 3. Polar tangent: $\partial\mathbf{r}/\partial\theta=r(\cos\theta\cos\phi,\ \cos\theta\sin\phi,\ -\sin\theta)$ , with magnitude $r\sqrt{\cos^2\theta+\sin^2\theta}=r$ , so $h_\theta=r$ . Why: a change in the polar angle carries you along a great circle of radius $r$ , and arc on that circle is $r\,\mathrm{d}\theta$ — the radius is the scale factor.

Step 4. Azimuthal tangent: $\partial\mathbf{r}/\partial\phi=r\sin\theta(-\sin\phi,\ \cos\phi,\ 0)$ , with magnitude $r\sin\theta$ , so $h_\phi=r\sin\theta$ . Why: azimuthal motion runs along a circle of latitude whose radius is the perpendicular distance to the polar axis, $r\sin\theta$ , which shrinks to zero at the poles where $\sin\theta=0$ .

Step 5. Assemble: the three tangents are mutually orthogonal, so $\mathrm{d}s^2=\mathrm{d}r^2+r^2\mathrm{d}\theta^2+r^2\sin^2\theta\,\mathrm{d}\phi^2$ and $\mathrm{d}V=h_r h_\theta h_\phi\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi=r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ . Why: with the scale factors known, the Pythagorean line element and the product volume element from the core equations are pure substitution. *Why it works: the pairwise dot products of the three tangents vanish, confirming orthogonality; their magnitudes give $(1,r,r\sin\theta)$ ; and the product $h_r h_\theta h_\phi = r^2\sin\theta$ is the standard spherical volume measure.*

Go deeper · Derivation C · 1 minThe scale-factor product is the Jacobian

Derivation CThe scale-factor product is the Jacobian

The goal is to prove the identity $h_1h_2h_3=J$ that lets the same volume integral be computed in any coordinate system with the same answer. We show that the volume of the curvilinear cell, written as a scalar triple product of its three edge vectors, is on the one hand the product of scale factors and on the other hand precisely the Jacobian determinant of the coordinate map — the geometric and analytic pictures of "volume measure" turning out to be one object.

Step 1. The cell is spanned by three edge displacements $\partial\mathbf{r}/\partial q_i\,\mathrm{d}q_i=\mathbf{g}_i\,\mathrm{d}q_i$ , and its volume is the scalar triple product $\mathrm{d}V=|\mathbf{g}_1\cdot(\mathbf{g}_2\times\mathbf{g}_3)|\,\mathrm{d}q_1\mathrm{d}q_2\mathrm{d}q_3$ . Why: the volume of a parallelepiped is the absolute triple product of its edge vectors, and to first order the curved cell is such a parallelepiped.

Step 2. For an orthogonal right-handed system $\mathbf{g}_i=h_i\hat{\mathbf{e}}_i$ with $\hat{\mathbf{e}}_1\cdot(\hat{\mathbf{e}}_2\times\hat{\mathbf{e}}_3)=1$ , so the triple product reduces to $h_1h_2h_3$ and $\mathrm{d}V=h_1h_2h_3\,\mathrm{d}q_1\mathrm{d}q_2\mathrm{d}q_3$ . Why: the orthonormal triad has unit triple product, so all the geometry collapses onto the three magnitudes — the scale factors.

Step 3. Now write the triple product as the determinant whose rows are the components of $\mathbf{g}_i=(\partial x/\partial q_i,\ \partial y/\partial q_i,\ \partial z/\partial q_i)$ ; that determinant is by definition the Jacobian $J=|\partial(x,y,z)/\partial(q_1,q_2,q_3)|$ . Why: the scalar triple product of three vectors equals the determinant of the matrix of their components, and that matrix is exactly the Jacobian matrix of the coordinate transformation.

Step 4. Equate the two evaluations of the same $\mathrm{d}V$ to conclude $h_1h_2h_3=J$ . Why: a single geometric quantity cannot have two values, so the scale-factor product and the Jacobian are identical — which is why $\mathrm{d}V=J\,\mathrm{d}q_1\mathrm{d}q_2\mathrm{d}q_3$ is the change-of-variables factor in multiple integrals. *Why it works: for spherical coordinates $h_r h_\theta h_\phi=1\cdot r\cdot r\sin\theta=r^2\sin\theta$ , which equals the directly computed Jacobian $\partial(x,y,z)/\partial(r,\theta,\phi)=r^2\sin\theta$ ; for cylindrical coordinates $1\cdot\rho\cdot1=\rho$ likewise matches its Jacobian — confirming the identity $h_1h_2h_3=J$ in both cases.*

Limiting cases

Specialising the general orthogonal formulas to the standard systems recovers every measure a physicist routinely uses.

Case	Result	Meaning
Cartesian	$(h_x,h_y,h_z)=(1,1,1)$	All scale factors unity; $\mathrm{d}s^2=\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2$ and $\mathrm{d}V=\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$ .
Cylindrical	$(h_\rho,h_\phi,h_z)=(1,\rho,1)$	$\mathrm{d}V=\rho\,\mathrm{d}\rho\,\mathrm{d}\phi\,\mathrm{d}z$ and $\mathrm{d}s^2=\mathrm{d}\rho^2+\rho^2\mathrm{d}\phi^2+\mathrm{d}z^2$ .
Spherical	$(h_r,h_\theta,h_\phi)=(1,r,r\sin\theta)$	$\mathrm{d}V=r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ and $\mathrm{d}s^2=\mathrm{d}r^2+r^2\mathrm{d}\theta^2+r^2\sin^2\theta\,\mathrm{d}\phi^2$ .
Plane polar	$(h_r,h_\phi)=(1,r)$	Two-dimensional reduction; area element $\mathrm{d}A=r\,\mathrm{d}r\,\mathrm{d}\phi$ , the familiar polar measure.

Common trapA coordinate increment is already a length

The single most common error in curvilinear calculus is treating $\mathrm{d}\theta$ or $\mathrm{d}\phi$ as if it were a distance and writing a volume integral as $\int\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ . It is not — an angle is dimensionless, and the missing scale factors carry the length. Compute a sphere's volume that way and you get $\int_0^{2\pi}\!\int_0^{\pi}\!\int_0^{R}\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi=R\cdot\pi\cdot2\pi=2\pi^2 R$ , which has units of length, not volume, and is numerically meaningless. The cure is mechanical: never guess the measure. Always write $\mathrm{d}V=h_1h_2h_3\,\mathrm{d}q_1\,\mathrm{d}q_2\,\mathrm{d}q_3$ , substitute the scale factors of your system, and only then integrate. The same discipline supplies the $\rho$ in cylindrical area and volume elements and the $r^2\sin\theta$ in spherical ones.

Worked examples

1 · Volume of a ball by integrating the spherical cell (fully worked)

Find the volume enclosed by a sphere of radius $R=3\ \mathrm{m}$ by integrating the spherical volume element, and confirm the general result $\tfrac{4}{3}\pi R^3$ .

Step 1. Write the measure. With $(h_r,h_\theta,h_\phi)=(1,r,r\sin\theta)$ the volume element is $\mathrm{d}V=r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ . The full ball is covered by $0\le r\le R$ , $0\le\theta\le\pi$ , $0\le\phi<2\pi$ .

Step 2. The integrand factorises because the limits are constants and the measure is a product, so $V=\left(\int_0^{R}r^2\,\mathrm{d}r\right)\left(\int_0^{\pi}\sin\theta\,\mathrm{d}\theta\right)\left(\int_0^{2\pi}\mathrm{d}\phi\right)$ .

Step 3. Evaluate each factor: $\int_0^{R}r^2\,\mathrm{d}r=R^3/3$ , then $\int_0^{\pi}\sin\theta\,\mathrm{d}\theta=[-\cos\theta]_0^{\pi}=2$ , and $\int_0^{2\pi}\mathrm{d}\phi=2\pi$ . The middle factor of $2$ is entirely the $\sin\theta$ from the scale factors — drop it and the answer is wrong.

Step 4. Combine: $V=\dfrac{R^3}{3}\cdot2\cdot2\pi=\dfrac{4}{3}\pi R^3$ . Substituting $R=3\ \mathrm{m}$ gives $V=\tfrac{4}{3}\pi(27)=36\pi\approx\mathbf{113.10\ m^3}$ . The general result is $V=\tfrac{4}{3}\pi R^3$ , and for $R=3\ \mathrm{m}$ it is $36\pi\approx113.1\ \mathrm{m}^3$ .

2 · Lateral surface area of a cylinder from the area element

A cylindrical tank has radius $a=0.5\ \mathrm{m}$ and height $L=2\ \mathrm{m}$ . Find the area of its curved side.

Step 1. The side is the coordinate surface $\rho=a$ (constant). The two tangential coordinates there are $\phi$ and $z$ , with scale factors $h_\phi=\rho=a$ and $h_z=1$ , so the area element is $\mathrm{d}A=h_\phi h_z\,\mathrm{d}\phi\,\mathrm{d}z=a\,\mathrm{d}\phi\,\mathrm{d}z$ .

Step 2. Integrate over the full surface, $0\le\phi<2\pi$ and $0\le z\le L$ : $A=\int_0^{L}\!\int_0^{2\pi}a\,\mathrm{d}\phi\,\mathrm{d}z=a\,(2\pi)(L)=2\pi a L$ .

Step 3. Substitute the numbers: $A=2\pi(0.5)(2)=2\pi\approx\mathbf{6.28\ m^2}$ . The factor $a$ came entirely from the scale factor $h_\phi$ ; forgetting it would have left a dimensionless "area" of $2\pi L$ . The lateral area is $2\pi a L=2\pi\approx6.28\ \mathrm{m}^2$ .

3 · Scale factors of a non-standard system: parabolic cylindrical coordinates

The coordinates $(u,v,z)$ map to Cartesian by $x=\tfrac12(u^2-v^2)$ , $y=uv$ , $z=z$ . Find the scale factors and the volume element, demonstrating that the method works beyond the textbook table.

Step 1. Compute the tangents. $\partial\mathbf{r}/\partial u=(u,\ v,\ 0)$ and $\partial\mathbf{r}/\partial v=(-v,\ u,\ 0)$ and $\partial\mathbf{r}/\partial z=(0,0,1)$ .

Step 2. Take magnitudes: $h_u=\sqrt{u^2+v^2}$ , $h_v=\sqrt{u^2+v^2}$ , $h_z=1$ . The first two are equal, a hallmark of these conformal coordinates.

Step 3. Check orthogonality: $(\partial\mathbf{r}/\partial u)\cdot(\partial\mathbf{r}/\partial v)=(u)(-v)+(v)(u)+0=0$ , and both are perpendicular to $\hat z$ . The system is orthogonal, so the Pythagorean line element applies.

Step 4. Assemble the volume element: $\mathrm{d}V=h_u h_v h_z\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}z=(u^2+v^2)\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}z$ . As a spot check, at $u=v=1$ the in-plane scale factors are $h_u=h_v=\sqrt2\approx1.414$ and the local area magnification is $h_u h_v=2$ . The scale factors are $(h_u,h_v,h_z)=(\sqrt{u^2+v^2},\ \sqrt{u^2+v^2},\ 1)$ and $\mathrm{d}V=(u^2+v^2)\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}z$ .

Physics of the portrait

Scale factors are one of those tools whose name records their inventor. They are the Lamé coefficients, after the French mathematical physicist who first treated general curvilinear coordinates as objects worth a theory of their own.

Origins · 1 minGabriel Lamé and how this was discovered

Gabriel Lamé

1795–1870, French mathematician and engineer · curvilinear coordinates, elasticity, and the coefficients h_i that bear his name

Gabriel Lamé trained as a mining engineer at the École Polytechnique and the École des Mines, spent a decade building roads and bridges in Russia at the new Institute of Engineers of Ways of Communication in St Petersburg, and returned to Paris to become one of the most versatile physicists of his generation. His deepest and most lasting idea was to stop treating cylindrical and spherical coordinates as special tricks and instead ask what any smooth system of curvilinear coordinates has in common. The answer was the set of coefficients $h_i$ — the scale factors — that convert coordinate increments into physical lengths, and with them Lamé expressed the Laplacian, the heat equation and the equations of elasticity in whatever coordinate system a problem's symmetry demanded. He gathered this programme in the 1859 Leçons sur les coordonnées curvilignes et leurs diverses applications, the first systematic textbook of the subject. His name survives across physics: Lamé coefficients in vector calculus, the Lamé parameters of linear elasticity, Lamé functions and the Lamé differential equation for the potential of an ellipsoid, and Lamé's stress theory. Gauss had already written the line element of a surface in his 1828 study of curvature, and after Lamé the idea would be generalised by Ricci and Levi-Civita into the metric tensor $g_{ij}$ of tensor calculus — but the practical machinery every physics student learns first is Lamé's.

History

1828

— Gauss's first fundamental form expresses the line element ds² of a curved surface

1833

— Lamé introduces general curvilinear coordinates and the coefficients h_i

1859

— Lamé's Leçons sur les coordonnées curvilignes systematises the whole subject

1892

— Ricci begins the absolute differential calculus and the metric tensor g_ij

1900

— Ricci and Levi-Civita codify tensor calculus, generalising h_i² to g_ij

Going further

Texts. Arfken and Weber, Mathematical Methods for Physicists, 7th ed., Ch. 2 (Curvilinear Coordinates) · Hassani, Mathematical Physics, Ch. on vector analysis and curvilinear coordinates · Stone and Goldbart, Mathematics for Physics, curvilinear-coordinates chapter · Byron and Fuller, Mathematics of Classical and Quantum Physics, Ch. 1 · Nakahara, Geometry, Topology and Physics, Ch. 5 for the metric-tensor generalisation · Riley, Hobson and Bence, Mathematical Methods for Physics and Engineering, Ch. 10 · Boas, Mathematical Methods in the Physical Sciences, Ch. 5–6. Every scale factor is computed from the explicit Cartesian map as $h_i=|\partial\mathbf{r}/\partial q_i|$ : spherical $(1,r,r\sin\theta)$ , cylindrical $(1,\rho,1)$ , plane polar $(1,r)$ , and parabolic cylindrical $(\sqrt{u^2+v^2},\sqrt{u^2+v^2},1)$ — the last confirmed orthogonal by direct dot-product check. The worked results $V=\tfrac{4}{3}\pi R^3$ (for $R=3\ \mathrm{m}$ , $36\pi\approx113.10\ \mathrm{m}^3$ ), the cylinder lateral area $2\pi a L=2\pi\approx6.28\ \mathrm{m}^2$ , and the sphere surface $4\pi R^2$ all follow from integrating the elements. The identity $h_1h_2h_3=J$ is confirmed against the directly evaluated Jacobians for both spherical and cylindrical cases.

Prerequisites. Scalar and Vector Fields · Line Integrals. Leads to. The Gradient Operator · Divergence of a Vector Field · The Laplacian Operator · Volume Integrals.

Ask the tutor: Give me any coordinate system — elliptic, toroidal, oblate spheroidal — and I will walk you through computing its scale factors and building the line, area and volume elements from scratch.

Why is a scale factor a magnitude of a derivative rather than the derivative itself?

Because a length is a magnitude. The derivative $\partial\mathbf{r}/\partial q_i$ is a vector: it carries both how fast the point moves as $q_i$ changes and in which direction. The direction is what becomes the unit vector $\hat{\mathbf{e}}_i$ ; the speed — how many metres of arc per unit of $q_i$ — is its length, and that is the scale factor $h_i=|\partial\mathbf{r}/\partial q_i|$ . Splitting the derivative into magnitude times direction is exactly what lets you write displacements as physical length times unit direction, which is the whole bookkeeping of curvilinear calculus.

What changes when the coordinate system is not orthogonal?

The line element keeps its cross terms. In general $\mathrm{d}s^2=\sum_{i,j}g_{ij}\,\mathrm{d}q_i\,\mathrm{d}q_j$ with a full symmetric metric $g_{ij}=\mathbf{g}_i\cdot\mathbf{g}_j$ , and the off-diagonal entries no longer vanish. You can still define scale factors as $h_i=\sqrt{g_{ii}}$ , but they no longer tell the whole story, and the volume element becomes $\sqrt{\det g}\,\,\mathrm{d}q_1\mathrm{d}q_2\mathrm{d}q_3$ rather than the simple product $h_1h_2h_3$ . Orthogonality is exactly the condition $g_{ij}=h_i^2\delta_{ij}$ that makes the metric diagonal and restores the clean product formulas; the skew case is the gateway to full tensor calculus.

Is the scale-factor product really the same as the Jacobian?

Yes, for an orthogonal right-handed system, and the reason is geometric. The Jacobian is the local volume-magnification factor of the coordinate map, and the volume of the curvilinear cell is the scalar triple product of its edge vectors $\mathbf{g}_i=h_i\hat{\mathbf{e}}_i$ . For an orthonormal triad that triple product is just $h_1h_2h_3$ , and written out in components the same triple product is the determinant of the Jacobian matrix. So $h_1h_2h_3=J$ is not a coincidence but two names for one number — the factor that appears in the change-of-variables theorem, which is why $\mathrm{d}V=J\,\mathrm{d}q_1\mathrm{d}q_2\mathrm{d}q_3$ .

Why do the vector operators divide by scale factors?

Because gradient, divergence, curl and Laplacian all measure change per unit physical length, and a coordinate derivative measures change per unit coordinate. To convert, you divide by the length that one coordinate step spans, namely $h_i$ . That is why the gradient reads $(\nabla f)_i=(1/h_i)\,\partial f/\partial q_i$ , and why the divergence carries the more elaborate factor $\nabla\cdot\mathbf{A}=\frac{1}{h_1h_2h_3}[\partial_{q_1}(h_2h_3A_1)+\cdots]$ — the outer $1/h_1h_2h_3$ turns a coordinate volume derivative into a physical one, and the inner products $h_2h_3$ are the face areas the field flows through. Knowing the $h_i$ is the only prerequisite for writing any of these operators explicitly.

Why does h_φ vanish at the poles in spherical coordinates?

Because the azimuthal circle collapses there. The scale factor $h_\phi=r\sin\theta$ is the radius of the circle of constant latitude, which is the perpendicular distance from the point to the polar axis. At the poles $\theta=0$ or $\theta=\pi$ , that distance is zero, so a change in $\phi$ moves you nowhere and the scale factor is zero. This is a coordinate singularity, not a physical one: space is perfectly smooth at the poles, but the $(\theta,\phi)$ grid degenerates because infinitely many values of $\phi$ label the same axial point. Numerical codes on the sphere handle this with pole-free grids or spectral methods.

How do I find the scale factors for a coordinate system I have never seen?

Follow the same three-line recipe every time. First write the explicit map giving Cartesian $(x,y,z)$ as functions of your coordinates $(q_1,q_2,q_3)$ . Second, differentiate to get the tangent vectors $\partial\mathbf{r}/\partial q_i$ and take their magnitudes — those are the scale factors $h_i$ . Third, check orthogonality by dotting the tangents pairwise; if the dot products vanish, the Pythagorean line element and $\mathrm{d}V=h_1h_2h_3\,\mathrm{d}q_1\mathrm{d}q_2\mathrm{d}q_3$ apply directly. Worked example 3 does exactly this for parabolic cylindrical coordinates. The method never changes; only the algebra does.