Cylindrical Coordinates

Half of electromagnetism happens near a line. Currents flow in straight wires, charge rides on cables, solenoids and coaxial lines and optical fibers are all tubes around an axis — and for every one of them, Cartesian coordinates are the wrong language and spherical coordinates are the wrong shape. The natural grid is cylindrical: distance from the axis $s$ , angle around it $\phi$ , and height along it $z$ . It is plane-polar coordinates with a third dimension bolted on, and after the two-angle gymnastics of the sphere it will feel like a rest day — only one stretch factor to track instead of two. But that single factor does real work. Sweep the angle $\phi$ by $\mathrm{d}\phi$ and you travel an arc $s\,\mathrm{d}\phi$ — farther out, farther traveled — so the volume element is $s\,\mathrm{d}s\,\mathrm{d}\phi\,\mathrm{d}z$ , the gradient divides its $\phi$ -derivative by $s$ , and the divergence wraps its radial term into $\frac{1}{s}\partial_s(sF_s)$ . That last structure is the axial twin of the sphere's $\frac{1}{r^2}\partial_r(r^2F_r)$ , and it explains at a glance why the signature fields of this geometry — the line charge's $\mathbf{E}\propto\hat{\mathbf{s}}/s$ and the straight wire's $\mathbf{B}\propto\hat{\boldsymbol{\phi}}/s$ — are divergence-free and curl-free everywhere off the axis, with all their sources hiding in the line itself. One warning before the fun: notation. This track follows Griffiths in writing the axial distance as $s$ — because $r$ collides with the spherical radius and $\rho$ collides with charge density, and both collisions cause real accidents. Learn the triad $(s,\phi,z)$ , and translation to any other book is painless.

Key takeaways

Cylindrical coordinates locate a point by $(s,\phi,z)$ : perpendicular distance from the $z$ -axis, azimuth around it, and height; $x=s\cos\phi$ , $y=s\sin\phi$ , $z=z$ .
The stretch factors are $(1,\,s,\,1)$ : only the angular sweep is stretched, giving $\mathrm{d}\boldsymbol{\ell}=\mathrm{d}s\,\hat{\mathbf{s}}+s\,\mathrm{d}\phi\,\hat{\boldsymbol{\phi}}+\mathrm{d}z\,\hat{\mathbf{z}}$ and the volume element $\mathrm{d}\tau=s\,\mathrm{d}s\,\mathrm{d}\phi\,\mathrm{d}z$ .
The unit vectors $\hat{\mathbf{s}}$ and $\hat{\boldsymbol{\phi}}$ rotate with the point ( $\partial_\phi\hat{\mathbf{s}}=\hat{\boldsymbol{\phi}}$ , $\partial_\phi\hat{\boldsymbol{\phi}}=-\hat{\mathbf{s}}$ ); only $\hat{\mathbf{z}}$ is a genuine constant.
The divergence packs the stretch factor into the radial term, $\nabla\cdot\mathbf{F}=\frac{1}{s}\partial_s(sF_s)+\frac{1}{s}\partial_\phi F_\phi+\partial_zF_z$ , which is why $\hat{\mathbf{s}}/s$ fields are source-free off the axis.
This is the coordinate system of wires, cables, solenoids, waveguides, and beams — anything with an axis.
Key idea: cylindrical coordinates are plane polar plus height, with one stretch factor s converting angle sweeps into arc lengths; the lone factor generates the volume element s ds dφ dz and every operator formula of the system.

The core insight

Cylindrical coordinates $(s,\phi,z)$ are plane-polar coordinates with height: the single stretch factor $s$ converts angular sweeps into arc lengths, and it alone generates the volume element $s\,\mathrm{d}s\,\mathrm{d}\phi\,\mathrm{d}z$ and all the differential operators of the system.

Where this lives today

The world's data moves through cylindrical coordinates. An optical fiber is a glass cylinder a few microns across, and the light modes it carries are solutions of Maxwell's equations in exactly the $(s,\phi,z)$ system of this article — radial Bessel profiles, azimuthal phase windings $e^{im\phi}$ , and a propagation factor along $z$ . When engineers design a fiber to carry more channels, they are choosing which of those cylindrical harmonics may propagate. The same mathematics, scaled up, runs the coaxial cables of every lab and the waveguides of every radar. At the big-machine frontier, cylindrical coordinates are the native language of confinement and acceleration. Particle-accelerator beam dynamics is written in $(s,\phi,z)$ around the beam axis; the superconducting solenoids of MRI scanners and collider detectors are engineered around the on-axis and off-axis field expansions in these coordinates; and tokamak plasma physics starts from cylindrical coordinates before bending the cylinder into a torus. Wherever there is an axis — and technology loves axes — this grid is already drawn.

The physical picture

Stand a wire vertically and ask where a point is relative to it. Three answers locate it completely: how far from the wire ( $s$ ), how far around ( $\phi$ ), how far up ( $z$ ). The coordinate surfaces are the natural furniture of axial problems — cylinders of constant $s$ wrap the wire like insulation, half-planes of constant $\phi$ fan out from the axis, and planes of constant $z$ slice across it. A field with the wire's symmetry depends on $s$ alone, and calculus in these coordinates collapses to one variable.

Figure

The cylindrical coordinate system — distance from the axis, angle around it, height along it

Lengths behave exactly as the sphere taught you, only simpler. Radial steps and vertical steps are honest lengths already: $\mathrm{d}s$ and $\mathrm{d}z$ . The angular step alone is stretched — sweeping $\mathrm{d}\phi$ at distance $s$ from the axis moves you along an arc $s\,\mathrm{d}\phi$ . One stretch factor, in the one place symmetry demands it: near the axis, a degree of azimuth is a hair's breadth; at the rim of a tokamak it is meters.

Figure

The volume element — a curved brick with edges ds, s dφ, and dz

The dictionary and the elements

The transformation to Cartesian is plane polar in the $xy$ -plane with $z$ carried along unchanged:

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

where: $s=\sqrt{x^2+y^2}\ge0$ is the distance to the axis (not to the origin — that is the spherical $r$ ; the two agree only in the $z=0$ plane), and $\phi\in[0,2\pi)$ is the same azimuth spherical coordinates use. The unit vectors follow by differentiating and normalizing:

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

where: the triad is right-handed, $\hat{\mathbf{s}}\times\hat{\boldsymbol{\phi}}=\hat{\mathbf{z}}$ . Note who moves and who does not: $\hat{\mathbf{s}}$ and $\hat{\boldsymbol{\phi}}$ swing around with the azimuth — differentiating (2) gives the workhorse identities $\partial_\phi\hat{\mathbf{s}}=\hat{\boldsymbol{\phi}}$ and $\partial_\phi\hat{\boldsymbol{\phi}}=-\hat{\mathbf{s}}$ — while $\hat{\mathbf{z}}$ is the one honest constant in the system. The position vector is $\mathbf{r}=s\,\hat{\mathbf{s}}+z\,\hat{\mathbf{z}}$ , with the azimuthal information hidden inside $\hat{\mathbf{s}}$ , exactly as the sphere hid its angles inside $\hat{\mathbf{r}}$ . The general displacement collects the stretch factors $(1,s,1)$ :

\mathrm{d}\boldsymbol{\ell}=\mathrm{d}s\,\hat{\mathbf{s}}+s\,\mathrm{d}\phi\,\hat{\boldsymbol{\phi}}+\mathrm{d}z\,\hat{\mathbf{z}} \qquad (3)

where: each term is (stretch factor) × (coordinate change) × (direction). Multiplying perpendicular edges gives every element you will need: the volume $\mathrm{d}\tau=s\,\mathrm{d}s\,\mathrm{d}\phi\,\mathrm{d}z$ ; the outward element on a cylinder of radius $s$ , $\mathrm{d}\mathbf{a}=s\,\mathrm{d}\phi\,\mathrm{d}z\,\hat{\mathbf{s}}$ (total side area $2\pi s L$ — the circumference-times-length you learned in school, now derived); and the cap element $s\,\mathrm{d}s\,\mathrm{d}\phi\,\hat{\mathbf{z}}$ . As a first audit, the cylinder's volume: $\int_0^R s\,\mathrm{d}s\int_0^{2\pi}\mathrm{d}\phi\int_0^L\mathrm{d}z=\tfrac{R^2}{2}\cdot2\pi\cdot L=\pi R^2L$ . Correct — the machinery is calibrated.

The differential operators

The del operator divides each derivative by its stretch factor — only the azimuthal one is nontrivial:

\nabla T=\frac{\partial T}{\partial s}\,\hat{\mathbf{s}}+\frac{1}{s}\frac{\partial T}{\partial\phi}\,\hat{\boldsymbol{\phi}}+\frac{\partial T}{\partial z}\,\hat{\mathbf{z}} \qquad (4)

where: the $1/s$ converts "per radian" into "per metre of arc". Try it on $T=\ln(s)$ : $\nabla T=\hat{\mathbf{s}}/s$ — which is, up to constants, the field of a line charge, and the reason logarithmic potentials rule two-dimensional electrostatics. The divergence, as always, is flux-per-volume, and the derivation below shows the stretch factor forcing its way inside the radial derivative:

\nabla\cdot\mathbf{F}=\frac{1}{s}\frac{\partial}{\partial s}\big(s\,F_s\big)+\frac{1}{s}\frac{\partial F_\phi}{\partial\phi}+\frac{\partial F_z}{\partial z} \qquad (5)

where: the $s$ inside $\partial_s(sF_s)$ is the area factor of the curved faces. Feed it the line-charge shape $\mathbf{F}=\hat{\mathbf{s}}/s$ : $sF_s=1$ , and the divergence vanishes for $s\neq0$ — all sources compressed into the axis, the cylindrical echo of the Coulomb-field story. The curl and the Laplacian follow the same scale-factor grammar (catalogued in the cheat sheet; generated mechanically in the curvilinear-coordinates article). One curl fact deserves its own sentence: for the wire field $\mathbf{F}=\hat{\boldsymbol{\phi}}/s$ , the $z$ -component of the curl is $\frac{1}{s}\partial_s(s\cdot\frac{1}{s})=0$ off the axis — the magnetic field of a straight wire is curl-free everywhere except in the wire, where a delta function of current lives.

Derivation AThe cylindrical divergence from the flux out of a curved brick

The goal is equation (5), and specifically the structure $\frac{1}{s}\partial_s(sF_s)$ , derived from the coordinate-free meaning of divergence — net flux out per unit volume — applied to the brick of the second figure. The strategy repeats the spherical-cell argument with one stretch factor instead of two, which makes this the cleanest place to see the mechanism: the inner and outer curved faces of the brick have different areas, so even a constant radial field leaks net flux. Watch the product rule carry the geometry.

Step 1. Take the cell spanning $[s,s+\mathrm{d}s]\times[\phi,\phi+\mathrm{d}\phi]\times[z,z+\mathrm{d}z]$ . Its inner and outer curved faces have areas $s\,\mathrm{d}\phi\,\mathrm{d}z$ and $(s+\mathrm{d}s)\,\mathrm{d}\phi\,\mathrm{d}z$ — the outer face is longer in arc. Net radial outflow:

F_s(s+\mathrm{d}s)\,(s+\mathrm{d}s)\,\mathrm{d}\phi\,\mathrm{d}z - F_s(s)\,s\,\mathrm{d}\phi\,\mathrm{d}z = \frac{\partial}{\partial s}\big(sF_s\big)\,\mathrm{d}s\,\mathrm{d}\phi\,\mathrm{d}z

Why the product rule is physics: even with $F_s$ constant, more area faces outward than inward, so flux escapes — that geometric leak is the $F_s\,\partial_s(s)$ piece, and it is exactly what a $1/s$ field must fight to be source-free.

Step 2. The two flat $\phi$ -faces are congruent rectangles $\mathrm{d}s\times\mathrm{d}z$ ; only the field difference contributes: $\big[F_\phi(\phi+\mathrm{d}\phi)-F_\phi(\phi)\big]\mathrm{d}s\,\mathrm{d}z=\partial_\phi F_\phi\,\mathrm{d}\phi\,\mathrm{d}s\,\mathrm{d}z$ . Likewise the caps give $\partial_zF_z\,\mathrm{d}z\cdot s\,\mathrm{d}\phi\,\mathrm{d}s$ .

Step 3. Divide the total outflow by the cell volume $\mathrm{d}\tau=s\,\mathrm{d}s\,\mathrm{d}\phi\,\mathrm{d}z$ :

\nabla\cdot\mathbf{F}=\frac{1}{s}\frac{\partial}{\partial s}(sF_s)+\frac{1}{s}\frac{\partial F_\phi}{\partial\phi}+\frac{\partial F_z}{\partial z}

Why $F_z$ enters undressed: the top and bottom caps are congruent — vertical geometry is Cartesian, and only the field difference drives flux.

Step 4. Payoff check: $\mathbf{F}=\hat{\mathbf{s}}/s$ gives $sF_s=1$ , so $\nabla\cdot\mathbf{F}=0$ for $s\neq0$ , yet the flux through any cylinder around the axis is $\frac{1}{s}\cdot2\pi sL=2\pi L\neq0$ . The sources hide in the axis — a line of delta functions, the two-dimensional sibling of $\nabla\cdot(\hat{\mathbf{r}}/r^2)=4\pi\delta^3(\mathbf{r})$ . Quick check: setting $F_s=s$ gives $\nabla\cdot\mathbf{F}=\frac{1}{s}\partial_s(s\cdot s)=2$ , which agrees with the Cartesian divergence of the field $(x,y,0)$ — a reassuring calibration of the formula.

Go deeper · Derivation B · 1 minGauss's law around a line charge — the coordinate system earning its keep

Derivation BGauss's law around a line charge — the coordinate system earning its keep

The goal is the electric field of an infinite line charge with linear density $\lambda$ , the signature calculation of cylindrical coordinates, using the integral form of Gauss's law with a coaxial cylindrical surface. The strategy: let symmetry dictate the field's shape before any integral is written, then choose the one surface on which the flux integral is trivial. Watch how each piece of the Gaussian surface either catches the field head-on or dodges it entirely — that clean split is what "matching the coordinate system to the symmetry" means in practice.

Step 1. Symmetry first. Translation along $z$ and rotation in $\phi$ leave the line invariant, so $E$ can depend only on $s$ . Reflection through any plane containing the axis kills $E_\phi$ ; reflection through any plane perpendicular to the axis kills $E_z$ . Hence $\mathbf{E}=E(s)\,\hat{\mathbf{s}}$ — pure radial fan. Why this step comes first: it converts a vector unknown into a single scalar function of one variable, before any calculus is spent.

Step 2. Choose the Gaussian surface to match: a coaxial cylinder of radius $s$ and length $L$ . On the curved side, $\mathbf{E}\parallel\mathrm{d}\mathbf{a}=\hat{\mathbf{s}}\,s\,\mathrm{d}\phi\,\mathrm{d}z$ and $E(s)$ is constant over the face; on the two caps, $\mathbf{E}\perp\mathrm{d}\mathbf{a}$ and the flux is zero.

Step 3. Evaluate the flux: $\oint\mathbf{E}\cdot\mathrm{d}\mathbf{a}=E(s)\times(\text{side area})=E(s)\,2\pi sL$ . Enclosed charge: $Q_{\mathrm{enc}}=\lambda L$ . Gauss's law gives

E(s)\,2\pi sL=\frac{\lambda L}{\varepsilon_0} \qquad\Longrightarrow\qquad \mathbf{E}=\frac{\lambda}{2\pi\varepsilon_0 s}\,\hat{\mathbf{s}}

Why $1/s$ and not $1/s^2$ : the field spreads over cylinders whose area grows like $s^1$ , not spheres growing like $r^2$ — dimensionality of the source sets the fall-off. The arbitrary length $L$ cancels, as it must for an infinite line.

Step 4. Consistency: this is exactly the $\hat{\mathbf{s}}/s$ shape shown divergence-free off the axis in the previous derivation — Gauss's law and the differential machinery agree everywhere they should. Consistency check: applying equation (4) to $V=-\frac{\lambda}{2\pi\varepsilon_0}\ln(s/s_0)$ gives $-\nabla V=\frac{\lambda}{2\pi\varepsilon_0 s}\hat{\mathbf{s}}$ , confirming that the logarithmic potential and the radial field are mutually consistent.

Which formula answers which question

you want	use	notes
convert points or fields	equations (1) and (2)	$s$ measures to the axis
any length, area, volume element	equation (3)	multiply the swept edges
side of a cylinder	$\mathrm{d}\mathbf{a}=s\,\mathrm{d}\phi\,\mathrm{d}z\,\hat{\mathbf{s}}$ , total $2\pi sL$	the Gauss-law workhorse
slope of a scalar field	equation (4)	only the $\phi$ term divides by $s$
sources of a vector field	equation (5)	$s$ goes inside the radial derivative
line-symmetric fields	$E=\lambda/2\pi\varepsilon_0s$ , $B=\mu_0I/2\pi s$	the two signature results

Limiting cases

Case	Result	Meaning
$z$ ignored (2D problems)	plane-polar coordinates $(s,\phi)$	electrostatics of long parallel systems is two-dimensional
on the axis ( $s=0$ )	$h_\phi=0$ , $\phi$ undefined	coordinate singularity; $1/s$ formulas need care numerically
far from a finite line ( $s\gg L$ )	field $\to$ point-like $1/r^2$	cylindrical symmetry is an approximation with a range
radial field $f(s)\,\hat{\mathbf{s}}$	$\nabla\cdot\mathbf{F}=\frac{1}{s}\partial_s(sf)$	source-free requires $f\propto1/s$
azimuthal field $g(s)\,\hat{\boldsymbol{\phi}}$	$(\nabla\times\mathbf{F})_z=\frac{1}{s}\partial_s(sg)$	curl-free requires $g\propto1/s$ — the wire's $B$
$s\to$ spherical equator	$(s,\phi,z)\to(r,\phi,z{=}0)$	the two systems agree in the midplane

Common trapWhose r is it anyway — and the φ-hat that will not sit still

Two accidents account for most cylindrical-coordinate casualties. First, notation collision: many books call the axial distance $r$ (colliding with the spherical radius — fatal when a problem contains both a wire and a sphere) or $\rho$ (colliding with charge density — fatal in the middle of Gauss's law, where $\rho$ means two things in one equation). This track writes $s$ , following Griffiths; whenever you open another text, identify its convention before using its formulas. Second, the rotating basis: $\hat{\mathbf{s}}$ and $\hat{\boldsymbol{\phi}}$ turn with the azimuth — $\partial_\phi\hat{\mathbf{s}}=\hat{\boldsymbol{\phi}}$ , $\partial_\phi\hat{\boldsymbol{\phi}}=-\hat{\mathbf{s}}$ — so components at different $\phi$ live in different bases. The field $\hat{\boldsymbol{\phi}}$ at $\phi=0$ points along $+\hat{\mathbf{y}}$ ; at $\phi=\pi$ it points along $-\hat{\mathbf{y}}$ . Add or integrate vector components across different points only after converting to Cartesian, where the basis holds still. Only $\hat{\mathbf{z}}$ may be treated as a constant.

Worked examples

1 · Charge on a cylinder with density growing off the axis (confidence builder)

A solid cylinder of radius $R=2.0\ \mathrm{cm}$ and length $L=10\ \mathrm{cm}$ carries charge density $\rho(s)=\rho_0\,(s/R)$ with $\rho_0=4.0\ \mu\mathrm{C/m^3}$ . Find the total charge.

Step 1. Set up with the cylindrical volume element — the density depends only on $s$ , so $\phi$ and $z$ integrate immediately: $Q=\int\rho\,\mathrm{d}\tau=\int_0^R\rho_0\frac{s}{R}\,s\,\mathrm{d}s\times2\pi\times L$ . Note the two powers of $s$ in the integrand: one from the density, one from the volume element — omitting the element's $s$ is the classic slip.

Step 2. Radial integral: $\int_0^R s^2\,\mathrm{d}s=R^3/3$ , so $Q=\dfrac{2\pi\rho_0LR^2}{3}$ .

Step 3. Numbers: $Q=\dfrac{2\pi\times4.0\times10^{-6}\times0.10\times(0.020)^2}{3}=\boxed{3.4\times10^{-10}\ \mathrm{C}\approx0.34\ \mathrm{nC}}$ .

Step 4. Check against the uniform case: a uniform $\rho_0$ would give $\pi R^2L\rho_0=5.0\times10^{-10}$ C; our density averages $\tfrac23\rho_0$ over the cross-section (weighted by the $s$ element), and indeed $\tfrac23\times5.0=3.4\times10^{-10}$ C. The volume element's $s$ -weighting is doing exactly what the geometry demands.

2 · The field of a charged cable (exam level)

A long straight cable carries line charge density $\lambda=30\ \mathrm{nC/m}$ . Find $\mathbf{E}$ at $s=5.0\ \mathrm{cm}$ , verify the field is divergence-free there, and find the potential difference between $s_1=5.0$ cm and $s_2=15$ cm.

Step 1. From the Gauss-law derivation above, $\mathbf{E}=\dfrac{\lambda}{2\pi\varepsilon_0 s}\,\hat{\mathbf{s}}$ . Numbers: $E=\dfrac{30\times10^{-9}}{2\pi\times8.854\times10^{-12}\times0.050}=\boxed{1.1\times10^{4}\ \mathrm{V/m}}$ , pointing radially away from the cable.

Step 2. Divergence check with equation (5): $sE_s=\dfrac{\lambda}{2\pi\varepsilon_0}$ is constant, so $\nabla\cdot\mathbf{E}=\frac{1}{s}\partial_s(sE_s)=0$ at $s=5$ cm — no sources off the axis, exactly as Gauss's differential law demands in vacuum.

Step 3. Potential difference along a radial path, using $\mathrm{d}\boldsymbol{\ell}=\mathrm{d}s\,\hat{\mathbf{s}}$ from equation (3): $V(s_1)-V(s_2)=\int_{s_1}^{s_2}E\,\mathrm{d}s=\dfrac{\lambda}{2\pi\varepsilon_0}\ln\dfrac{s_2}{s_1}$ .

Step 4. Numbers: $\dfrac{30\times10^{-9}}{2\pi\times8.854\times10^{-12}}=539\ \mathrm{V}$ , and $\ln3=1.099$ , so $V(s_1)-V(s_2)=\boxed{5.9\times10^{2}\ \mathrm{V}}$ . The logarithm is the cable's signature: the potential of a line charge has no finite reference at infinity, which is why coaxial systems always quote potential differences — and why the capacitance per length of a coax cable carries a $\ln(b/a)$ .

Origins · 1 minFriedrich Wilhelm Bessel and how this was discovered

Friedrich Wilhelm Bessel

1784–1846 · Königsberg Observatory

Bessel left school at fourteen, apprenticed to an import-export firm, and taught himself astronomy from navigation manuals — by twenty-five he was director of Prussia's new Königsberg Observatory. The functions bearing his name are the natural harmonics of this coordinate system: solve Laplace's equation inside a cylinder — a drumhead, a waveguide, an optical fiber — and the radial profiles that emerge are Bessel functions, oscillations whose spacing the $\frac{1}{s}\partial_s(s\,\partial_s\cdot)$ radial operator slowly stretches. Bessel met them studying planetary perturbations and, characteristically, tabulated them with a precision instrument-makers could use. His masterpiece came in 1838: the first measured distance to a star, the parallax of 61 Cygni — six hundred thousand times the Earth-Sun distance, announced from careful cylindrical-mount instruments he had recalibrated himself. Every fiber-optic mode profile computed today is drawn from his function tables' descendants.

History

1822

— Fourier's treatise on heat solves conduction in a cylinder; the radial solutions are the functions later named for Bessel.

1824

— Bessel systematically develops the functions J_n while analyzing planetary perturbations, publishing the tables that made them tools.

1838

— Bessel measures the first stellar parallax, 61 Cygni at 10.3 light-years — the yardstick of the universe, from an observatory of cylindrical mounts.

1880

— Heaviside analyzes signal propagation in coaxial cable, the geometry that made cylindrical coordinates industrial.

1936

— Southworth at Bell Labs demonstrates microwave waveguides; the mode patterns are Bessel functions in the flesh.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §1.4 (Curvilinear Coordinates) and Appendix A · Jackson, Classical Electrodynamics, 3rd ed., front-cover formulas and §3.7 (Cylindrical Coordinates, Bessel Functions) · Purcell & Morin, Electricity and Magnetism, 3rd ed., Appendix F · Zangwill, Modern Electrodynamics, §1.2 and Appendix on curvilinear coordinates · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 3 The divergence formula rests on $\partial_\phi\hat{\mathbf{s}}=\hat{\boldsymbol{\phi}}$ and $\partial_\phi\hat{\boldsymbol{\phi}}=-\hat{\mathbf{s}}$ , obtainable by differentiating the explicit basis-vector expressions. The line-charge field $E=\lambda/(2\pi\varepsilon_0 s)$ , the coaxial capacitance per length $2\pi\varepsilon_0/\ln(b/a)$ , and the wire field $\mu_0 I/(2\pi s)$ are standard results. The cylinder volume integral evaluates to $\pi R^2 L$ , and $\nabla\cdot\mathbf{E}=0$ holds for the $1/s$ radial field off the axis. Worked-example numbers: $Q=0.34$ nC for $\rho=\rho_0 s/R$ ; $E=1.1\times10^4$ V/m and $\Delta V=5.9\times10^2$ V for $\lambda=30$ nC/m.

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

Why does this track write the radial coordinate as s instead of r or ρ?

Pure accident-prevention, following Griffiths. Writing $r$ invites collision with the spherical radius — deadly in problems containing both a wire and a sphere, or when comparing the line charge's $1/s$ with the point charge's $1/r^2$ . Writing $\rho$ collides with volume charge density, so that Gauss's law would contain the same symbol meaning two different things in one line. The letter $s$ is unclaimed by anything nearby. When reading other books (Jackson uses $\rho$ , many engineering texts use $r$ ), translate at the door: identify which symbol measures distance-to-axis, and rewrite formulas in your own convention before computing.

How do I decide between cylindrical and spherical coordinates for a given problem?

Ask what the sources and boundaries look like. A distinguished point — charge, atom, star — wants spherical; a distinguished axis — wire, cable, solenoid, beam — wants cylindrical; rectangular boxes want Cartesian. The tell is which coordinates the answer will not depend on: line symmetry makes fields functions of $s$ alone (two coordinates ignorable), point symmetry makes them functions of $r$ alone. Mixed problems (a finite rod, a charged disk) genuinely need judgment: near a finite rod's middle it looks cylindrical, far away it looks spherical, and the honest answer interpolates — a theme the multipole expansion will make quantitative in E2.

What happens to the coordinate system on the axis itself?

The azimuth degenerates: at $s=0$ every value of $\phi$ labels the same point, so $\phi$ (and with it $\hat{\mathbf{s}}$ and $\hat{\boldsymbol{\phi}}$ ) is undefined there. This is a coordinate singularity — the geometry of space is perfectly smooth on the axis — but it has practical teeth: formulas carrying $1/s$ misbehave numerically near the axis, and any physical field that is smooth at $s=0$ must have special structure (its $s$ - and $\phi$ -components must vanish on the axis; only $F_z$ may survive). Simulation codes handle the axis with special stencils or by excising a thin core, precisely because the coordinates, not the physics, break there.

Where do the identities for the derivatives of s-hat and φ-hat come from, and why do they matter?

Differentiate the explicit forms in equation (2): $\partial_\phi\hat{\mathbf{s}}=(-\sin\phi,\cos\phi,0)=\hat{\boldsymbol{\phi}}$ and $\partial_\phi\hat{\boldsymbol{\phi}}=(-\cos\phi,-\sin\phi,0)=-\hat{\mathbf{s}}$ ; all other derivatives vanish. They matter because every dynamical statement in cylindrical coordinates threads through them: the velocity of a particle is $\dot{\mathbf{r}}=\dot s\,\hat{\mathbf{s}}+s\dot\phi\,\hat{\boldsymbol{\phi}}+\dot z\,\hat{\mathbf{z}}$ — the $s\dot\phi$ term exists only because $\hat{\mathbf{s}}$ turns — and the operator formulas (4) and (5) encode the same derivatives. Forget that the basis moves and you lose the centripetal terms of mechanics and the $1/s$ structures of field theory in a single stroke.

Why do potentials in cylindrical problems always come with logarithms?

Because the field of a line source falls as $1/s$ , and the integral of $1/s$ is $\ln s$ . This has a consequence that surprises everyone once: the potential of an infinite line charge cannot be referenced to infinity — $\ln s$ diverges there — so one must pick a finite reference radius $s_0$ and quote $V(s)=-\frac{\lambda}{2\pi\varepsilon_0}\ln(s/s_0)$ . Nothing physical depends on the choice, since only potential differences matter, and the $\ln(b/a)$ in every coaxial-cable capacitance formula is this same logarithm with both radii physical. The deeper reason: two-dimensional space is one dimension short of diluting a source's influence fast enough for infinity to be a clean zero.

What are Bessel functions and when will I actually need them?

They are to cylinders what sines and cosines are to boxes and spherical harmonics are to spheres: the natural vibration profiles of the radial coordinate. Solve Laplace's equation (or the wave equation) in cylindrical coordinates by separation of variables and the $s$ -dependence obeys Bessel's equation, whose solutions $J_m(ks)$ oscillate like slowly-stretching cosines. In this course they surface in E2 (boundary-value problems with cylindrical boundaries) and again for waveguides and fibers in E8. Nothing about them needs memorizing now — but when you meet a drumhead's vibration pattern or a fiber mode's ringed cross-section, recognize them as this coordinate system's home orchestra.

Can I mix cylindrical and Cartesian components in one calculation?

Only with explicit conversion at the seams. The safe workflow: choose one system for the calculation; if inputs arrive in the other, convert them first using equation (2) (for vectors, convert components, not just coordinates: $F_s=F_x\cos\phi+F_y\sin\phi$ , $F_\phi=-F_x\sin\phi+F_y\cos\phi$ ). The one shortcut that is always legal: $\hat{\mathbf{z}}$ is shared by both systems, so purely axial quantities pass freely. The illegal move is adding $F_s$ values from points at different azimuths as if they were numbers on a common axis — they are components along different directions, and only a Cartesian intermediate step can add them honestly.