Vector operators in cylindrical and spherical coordinates

A radial field can have constant-looking components while its direction varies from point to point. This is why Cartesian derivative formulas cannot be applied to cylindrical or spherical components as if the basis were fixed. Scale factors and changing unit vectors account for the geometry. We use physical components along unit basis vectors throughout.

1 Cylindrical and spherical unit frames

In cylindrical coordinates, $x=\rho\cos\theta$, $y=\rho\sin\theta$, and $z=z$. The unit vectors are

$$e_\rho=(\cos\theta,\sin\theta,0),\quad e_\theta=(-\sin\theta,\cos\theta,0),\quad e_z=(0,0,1).$$

They form a right-handed orthonormal frame for $\rho>0$. Differentiation gives $\partial_\theta e_\rho=e_\theta$ and $\partial_\theta e_\theta=-e_\rho$. The displacement is $dx=e_\rho d\rho+\rho e_\theta d\theta+e_zdz$, so coordinate scale factors are $(1,\rho,1)$.

For spherical coordinates use $\phi$ as polar angle from the positive $z$ axis and $\theta$ as azimuth:

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

In the right-handed coordinate order $(r,\phi,\theta)$, the unit vectors are

$$e_r=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi),$$ $$e_\phi=(\cos\phi\cos\theta,\cos\phi\sin\theta,-\sin\phi),\quad e_\theta=(-\sin\theta,\cos\theta,0).$$

The scale factors are $(1,r,r\sin\phi)$ because $dx=e_rdr+r e_\phi d\phi+r\sin\phi e_\theta d\theta$. The formulas below apply on regular charts $r>0$, $0<\phi<\pi$; chart singularities on the axis are not automatically singularities of the physical field.

Physical vector components in $F=F_re_r+F_\phi e_\phi+F_\theta e_\theta$ are not the same as coefficients of the nonunit coordinate vectors $\partial x/\partial r,\partial x/\partial\phi,\partial x/\partial\theta$. Confusing them removes or duplicates scale factors.

2 Deriving the orthogonal-coordinate formulas

For right-handed orthogonal coordinates $u_1,u_2,u_3$ with positive scale factors $h_i$, write $H=h_1h_2h_3$. Since $df=\sum_i f_{u_i}du_i$ must equal $\nabla f\cdot dx$, coefficient comparison gives

$$\nabla f=\sum_i e_i\frac1{h_i}f_{u_i}.$$

A small coordinate box has volume $H\,du_1du_2du_3$. Its two faces normal to $e_i$ have area scale $H/h_i$. Subtracting opposite-face fluxes and dividing by volume gives

$$\nabla\cdot F=\frac1H\sum_i\partial_{u_i}\left(\frac H{h_i}F_i\right).$$

This local derivation is the differential limit of the divergence theorem; it includes changing face areas.

For curl, circulation around a small oriented coordinate rectangle in directions $u_j,u_k$ is the difference of the derivatives of the edge quantities $h_kF_k$ and $h_jF_j$. Dividing by its area $h_jh_kdu_jdu_k$ and applying Stokes locally gives

$$(\nabla\times F)_i=\frac1{h_jh_k}\left[\partial_{u_j}(h_kF_k)-\partial_{u_k}(h_jF_j)\right],$$

for cyclic triples $(i,j,k)$. These arguments require differentiable fields and a regular chart. They derive the scale-factor formulas rather than treating them as unrelated tables.

3 Cylindrical formulas and examples

Substituting $(1,\rho,1)$ gives

$$\nabla f=e_\rho f_\rho+e_\theta\frac{f_\theta}{\rho}+e_zf_z,$$ $$\nabla\cdot F=\frac1\rho\partial_\rho(\rho F_\rho)+\frac1\rho\partial_\theta F_\theta+\partial_zF_z,$$ $$\nabla\times F=e_\rho\left(\frac1\rho\partial_\theta F_z-\partial_zF_\theta\right) +e_\theta(\partial_zF_\rho-\partial_\rho F_z) +e_z\frac1\rho\left[\partial_\rho(\rho F_\theta)-\partial_\theta F_\rho\right].$$

Taking divergence of the gradient gives $\Delta f=\rho^{-1}\partial_\rho(\rho f_\rho)+\rho^{-2}f_{\theta\theta}+f_{zz}$.

Solid rotation. For $F=\Omega\rho e_\theta$, divergence is zero and curl is $2\Omega e_z$. In Cartesian coordinates this is $(-\Omega y,\Omega x,0)$, confirming the same answer. Differentiating only the component $F_\theta=\Omega\rho$ and ignoring the frame geometry would miss a factor in the curl.

Radial planar spreading. The field $F=Ae_\rho/\rho$ has zero divergence away from the axis, since $\rho F_\rho=A$. Its outward flux through a cylinder of radius $R$ and height $L$ is $2\pi AL$. Nonzero enclosed flux does not contradict the local zero divergence: the axis is a singularity excluded from the formula's regular region.

4 Spherical formulas and radial operators

With coordinate order $(r,\phi,\theta)$,

$$\nabla f=e_rf_r+e_\phi\frac{f_\phi}{r}+e_\theta\frac{f_\theta}{r\sin\phi},$$ $$\nabla\cdot F=\frac1{r^2}\partial_r(r^2F_r) +\frac1{r\sin\phi}\partial_\phi(\sin\phi F_\phi) +\frac1{r\sin\phi}\partial_\theta F_\theta.$$

The curl components are

$$(\nabla\times F)_r=\frac{\partial_\phi(\sin\phi F_\theta)-\partial_\theta F_\phi}{r\sin\phi},$$ $$(\nabla\times F)_\phi=\frac1r\left[\frac1{\sin\phi}\partial_\theta F_r-\partial_r(rF_\theta)\right],\qquad (\nabla\times F)_\theta=\frac1r[\partial_r(rF_\phi)-\partial_\phi F_r].$$

The scalar Laplacian is

$$\Delta f=\frac1{r^2}\partial_r(r^2f_r) +\frac1{r^2\sin\phi}\partial_\phi(\sin\phi f_\phi) +\frac{f_{\theta\theta}}{r^2\sin^2\phi}.$$

For radial $f=f(r)$, this reduces to $f''+2f'/r$. Thus $\Delta(1/r)=0$ for $r>0$. The function is singular at zero, so this is not a globally source-free statement. For $F=Ae_r/r^2$, divergence is likewise zero away from zero and sphere flux is $4\pi A$.

As a nonsingular chart check, the Cartesian field $(x,y,z)$ becomes $F=re_r$ and has divergence $r^{-2}\partial_r(r^3)=3$. The cylindrical version is $\rho e_\rho+ze_z$, also giving 3. A valid physical result agrees across regular charts.

A Cartesian chain-rule proof of the scale factors

To justify the local formulas directly, let $x=x(u)$ be a $C^2$ coordinate diffeomorphism between open subsets of $\mathbb R^3$, with mutually orthogonal nonzero coordinate vectors $x_{u_i}=h_i e_i$ and a right-handed unit frame $(e_1,e_2,e_3)$. Let $F$ be $C^1$ on the target. Write $F_i=F\cdot e_i$ and $H=h_1h_2h_3$. The reciprocal coordinate vector is $\nabla u_i=e_i/h_i$, since its dot product with $x_{u_j}$ is $\delta_{ij}$.

For $i\ne j$, equality of mixed coordinate derivatives gives $\partial_j(h_i e_i)=\partial_i(h_j e_j)$. Dotting with $e_i$ yields $\partial_jh_i=h_j e_i\cdot\partial_i e_j$, since a unit vector is orthogonal to each of its own derivatives. The Cartesian chain rule and the trace definition of divergence now give

$$\nabla\cdot F=\sum_i\frac1{h_i}e_i\cdot\partial_iF =\sum_j\left[\frac{\partial_jF_j}{h_j}+\frac{F_j}{h_j}\sum_{i\ne j}\frac{\partial_jh_i}{h_i}\right] =\frac1H\sum_j\partial_j\left(\frac H{h_j}F_j\right).$$

This accounts for every derivative of the moving frame. For a cyclic triple $(i,j,k)$, the Cartesian curl identity expressed in this right-handed orthonormal frame gives

$$(\nabla\times F)_i=\frac{e_k\cdot\partial_jF}{h_j}-\frac{e_j\cdot\partial_kF}{h_k}.$$

Differentiate $F_k=F\cdot e_k$ and $F_j=F\cdot e_j$. The remaining frame terms combine using $\partial_j(h_k e_k)=\partial_k(h_j e_j)$, or equivalently

$$-\frac{\partial_j e_k}{h_j}+\frac{\partial_k e_j}{h_k} =\frac{\partial_j h_k}{h_jh_k}e_k-\frac{\partial_k h_j}{h_jh_k}e_j.$$

Substitution gives exactly $[\partial_j(h_kF_k)-\partial_k(h_jF_j)]/(h_jh_k)$. Thus the curl formula follows by ordinary derivatives on a regular chart, without requiring an unproved theorem on an arbitrary curved coordinate box. The scalar Laplacian follows by applying the proved divergence identity to the gradient for a $C^2$ scalar field. Cylindrical and spherical substitutions retain their already specified coordinate order and positive scale-factor domains.

5 Graded practice

E1 Foundation. Express $F=(x,y,2z)$ in cylindrical components and compute its divergence.

E2 Scalar field. For $f=\rho^2+z^2$, calculate the cylindrical gradient and Laplacian. Compare with Cartesian results.

E3 Radial field. For $F=r^me_r$ on $r>0$, find its divergence and identify the exponent giving zero.

E4 Chart versus field. Use the spherical formula to calculate $\Delta(r\cos\phi)$. Explain why the coordinate denominators do not imply this field is singular on the axis.

E5 Curl. For cylindrical $F=\rho z\,e_\theta$, compute curl.

6 Full solutions

S1. $F=\rho e_\rho+2ze_z$. Divergence is $\rho^{-1}\partial_\rho(\rho^2)+\partial_z(2z)=2+2=4$, matching $1+1+2$ in Cartesian coordinates.

S2. Gradient is $2\rho e_\rho+2ze_z$, and Laplacian is $\rho^{-1}\partial_\rho(2\rho^2)+2=4+2=6$. Since $f=x^2+y^2+z^2$, these are the expected Cartesian gradient and Laplacian.

S3. Divergence is $r^{-2}\partial_r(r^{m+2})=(m+2)r^{m-1}$. It vanishes for $m=-2$, away from zero. A singular origin still requires separate flux treatment.

S4. The radial contribution is $2\cos\phi/r$. The polar contribution is $-2\cos\phi/r$, and the azimuthal contribution is zero. They cancel, giving zero. The scalar is simply Cartesian $z$, smooth everywhere; only the spherical chart degenerates on the axis.

S5. The radial component is $-\partial_z(\rho z)=-\rho$, the azimuthal component is zero, and the vertical component is $\rho^{-1}\partial_\rho(\rho^2z)=2z$. Thus curl is $-\rho e_\rho+2ze_z$.

Summary

Curvilinear vector operations require physical component conventions, coordinate orientation and scale factors. Gradients compare scalar differentials with displacement; divergence tracks changing face areas; curl tracks circulation on coordinate rectangles. Cylindrical and spherical chart singularities must be distinguished from genuine singular fields and sources.