Separation of Variables in Cylindrical Coordinates

Drop a conducting cylinder into a uniform electric field, broadside on. Far away the field marches in parallel lines; near the cylinder it must rearrange completely, because a conductor's surface is an equipotential and field lines must arrive perpendicular. What does the compromise look like? How much charge lands on the surface, in what angular pattern? A superposition integral cannot answer — the induced charge is the unknown. This is boundary-value territory; this article builds the tool. You already own the method. Separation of Variables in Cartesian Coordinates established the product ansatz, the separation-constant argument, and Fourier's trick; Separation of Variables in Spherical Coordinates ran the same program around a point. Here we run it around an axis: any problem with translational symmetry along $z$ — a wire, a pipe, our cylinder in the wind — collapses to two dimensions, and Laplace's equation in polar coordinates $(s,\phi)$ separates into Fourier modes $\cos k\phi$ , $\sin k\phi$ with integer $k$ and radial powers $s^{k}$ , $s^{-k}$ , plus one aristocratic outlier: the logarithm. Mastering this buys the whole two-dimensional world — the cylinder in a field, the split cylinder at $\pm V_0$ , the coax, the precipitator, the quadrupole mass filter — and a sharpened instinct: 2D electrostatics is not 3D with a coordinate deleted. Potentials grow logarithmically, fields fall as $1/s$ , and the sphere's famous factor of 3 becomes the cylinder's 2.

Key takeaways

Translational symmetry along $z$ makes a problem two-dimensional: Laplace's equation reduces to polar coordinates $(s,\phi)$ , and $V=R(s)\Phi(\phi)$ separates it.
Single-valuedness $V(\phi+2\pi)=V(\phi)$ quantizes $k$ to the nonnegative integers — the azimuthal solutions are Fourier modes.
The radial equation is an Euler equation: powers $s^{k}$ , $s^{-k}$ for $k\ge1$ ; the $k=0$ sector degenerates to $a_0+b_0\ln s$ — the line-charge potential in disguise.
Regularity chooses for you: regions containing the axis keep only $s^{k}$ ; exterior regions keep $s^{-k}$ (plus the constant, and $\ln s$ when net line charge is present).
A conducting cylinder in a transverse field $E_0$ yields $V=-E_0(s-R^2/s)\cos\phi$ and $\sigma=2\varepsilon_0E_0\cos\phi$ — factor 2 where the sphere had 3.
Genuine $z$ -dependence promotes the radial equation to Bessel's equation; we name the Bessel functions and defer them to waveguide territory.
Key idea: the general 2D vacuum potential is $a_0+b_0\ln s+\sum_k(a_ks^{k}+b_ks^{-k})(\cos k\phi,\sin k\phi)$ with integer $k$ — match it to the boundary mode by mode and the problem is solved.

The core insight

For problems with translational symmetry along $z$ , Laplace's equation separates in polar coordinates into integer Fourier modes in $\phi$ and Euler powers $s^{\pm k}$ in radius (with $a_0+b_0\ln s$ at $k=0$ ); fitting this general solution to the boundary data, mode by mode, solves any 2D electrostatics problem.

Where this lives today

The $k=0$ sector alone underwrites a planet's worth of cable. A coaxial line is the two-boundary logarithmic problem: $V=a_0+b_0\ln s$ fitted to the two radii gives $C'=2\pi\varepsilon_0\varepsilon_r/\ln(b/a)$ — for RG-58 ( $b/a\approx3.3$ , polyethylene $\varepsilon_r=2.25$ ), about $105$ pF/m, the number on the datasheet. The same geometry at tens of kilovolts is the electrostatic precipitator: a corona wire on the axis of a grounded duct, whose enormous $1/s$ field ionizes the gas and drives fly-ash to the walls, cleaning power-station exhaust by the gigawatt. The higher modes have their own industry. The $k=2$ harmonic, $V\propto s^2\cos2\phi$ , is the ideal 2D quadrupole potential — what a quadrupole mass filter approximates with its four rods, run with a radio-frequency twist in every benchtop mass spectrometer. When designers ask how round rods distort the pure quadrupole, they expand the actual boundary potential in the basis (2) and track the contaminating $k=6,10,\dots$ terms — Fourier's trick on a circle, deciding an instrument's resolving power.

The physical picture

Geometry dictates the alphabet. Around the axis, $\phi$ is periodic — walk $2\pi$ and you are home — so the natural basis functions are the harmonics of a circle, $\cos k\phi$ and $\sin k\phi$ , with $k$ counting oscillations per lap. Only whole numbers of oscillations close up; that fact will quantize the separation constant before any boundary condition gets a word in. The radial direction has no periodicity but has scale structure: its equation is invariant under stretching $s\to\lambda s$ , and the functions that respect that are pure powers. Two directions, two alphabets — Fourier modes around, power laws outward.

Figure

The workhorse problem — a conducting cylinder broadside in a uniform field

The hero problem shows both alphabets earning their keep. Far away the potential must become that of the undisturbed field, $-E_0s\cos\phi$ — one power of $s$ , one $\cos\phi$ ; the cylinder's response must die at large $s$ , and since the drive sings in the pure $\cos\phi$ mode, orthogonality will let the cylinder answer only in $\cos\phi$ . Before any algebra the solution's shape is fixed: one growing $k=1$ term for the applied field, one decaying $k=1$ term — $\propto\cos\phi/s$ , a line dipole. One family member obeys neither alphabet: at $k=0$ the radial equation degenerates and produces $\ln s$ — the line-charge potential The Electrostatic Potential derived when the reference at infinity failed, rediscovered by pure separation. It carries the net charge per unit length; forgetting it (or its partner the constant) is the classic way to render a coax problem unsolvable.

The two-dimensional toolkit

Start from the full cylindrical Laplacian and impose the defining symmetry: nothing depends on $z$ . The $\partial^2V/\partial z^2$ term drops, and Laplace's equation becomes planar,

\frac{1}{s}\frac{\partial}{\partial s}\!\left(s\frac{\partial V}{\partial s}\right)+\frac{1}{s^2}\frac{\partial^2V}{\partial\phi^2}=0 \qquad (1)

where: $s$ — perpendicular distance from the $z$ -axis; $\phi$ — azimuthal angle, $0\le\phi<2\pi$ ; $V$ — the potential, the same at every height. The $1/s$ factors encode that a ring of radius $s$ has circumference $2\pi s$ : radial spreading dilutes flux in proportion to $s$ , not $s^2$ — every peculiarity of 2D electrostatics traces back to that missing power.

Derivation ASeparating in polar coordinates — powers, a logarithm, and integer k

The goal is the complete menu of product solutions of equation (1), and from it the general two-dimensional vacuum potential. The strategy is the one Separation of Variables in Cartesian Coordinates established and we will not re-argue: insert $V=R(s)\Phi(\phi)$ and invoke the separation-constant argument. What is new is what fixes the constant — not a boundary, but the topology of the circle. Watch the $k=0$ case, where the radial equation quietly changes character and hands us a logarithm.

Step 1. Substitute $V=R(s)\Phi(\phi)$ into (1), multiply through by $s^2/(R\Phi)$ , and the variables disengage:

\frac{s}{R}\frac{\mathrm{d}}{\mathrm{d}s}\!\left(s\frac{\mathrm{d}R}{\mathrm{d}s}\right) = -\frac{1}{\Phi}\frac{\mathrm{d}^2\Phi}{\mathrm{d}\phi^2} = k^2

Why a square: on a circle the angular functions must oscillate to return to their starting value; the opposite sign would give real exponentials in $\phi$ , which never close up.

Step 2. Let the circle do the quantizing. $\Phi''=-k^2\Phi$ gives $\Phi=C\cos k\phi+D\sin k\phi$ . But $\phi$ and $\phi+2\pi$ label the same physical point, so single-valuedness demands $\Phi(\phi+2\pi)=\Phi(\phi)$ — satisfied only if $k$ is an integer; take $k=0,1,2,\dots$ since negative $k$ duplicates the same functions. No electrode was consulted: geometry alone fixes the spectrum, the same mechanism that later quantizes $m$ on a quantum ring.

Step 3. Solve the radial equation for $k\ge1$ : $s^2R''+sR'-k^2R=0$ — an Euler equation, every term carrying as many powers of $s$ as derivatives, so no length scale is preferred and pure powers must solve it. Try $R=s^\alpha$ :

\alpha(\alpha-1)+\alpha-k^2=0 \quad\Longrightarrow\quad \alpha=\pm k

Two independent solutions, $s^{k}$ growing and $s^{-k}$ decaying — the 2D counterparts of the sphere's $r^l$ and $r^{-(l+1)}$ . Why symmetric exponents: inversion in the circle, $s\to R^2/s$ , is a symmetry of the 2D equation, swapping $s^{k}\leftrightarrow s^{-k}$ ; 3D inversion is lopsided.

Step 4. Handle $k=0$ — the degenerate sector. The angular equation $\Phi''=0$ gives $\Phi=A+B\phi$ ; the $B\phi$ piece grows by $2\pi B$ every lap and is expelled by single-valuedness. The radial equation collapses to $s(\mathrm{d}/\mathrm{d}s)(sR')=0$ : integrate twice to get $R=a_0+b_0\ln s$ — the second solution the repeated root $\alpha=0$ always hides. Physically $b_0\ln s$ is the line-charge potential recalled above, $b_0=-\lambda/2\pi\varepsilon_0$ : the $k=0$ sector carries the monopole per length.

Step 5. Superpose. Laplace's equation is linear and the Fourier basis is complete on the circle: the general 2D vacuum potential is the sum in equation (2) below. Verified: each term is harmonic by substitution — for $s^{\pm k}\cos k\phi$ the radial part of (1) gives $k^2s^{\pm k-2}\cos k\phi$ , the angular part its negative; $\ln s$ gives $(1/s)(\mathrm{d}/\mathrm{d}s)(1)=0$ . Matches Griffiths §3.3 (Prob. 3.24) and Jackson §2.11.

The harvest — the general 2D solution, valid in any charge-free annular region:

V(s,\phi)=a_0+b_0\ln s+\sum_{k=1}^{\infty}\Big[s^{k}\big(a_k\cos k\phi+b_k\sin k\phi\big)+s^{-k}\big(c_k\cos k\phi+d_k\sin k\phi\big)\Big] \qquad (2)

where: the constants are fixed by the boundary data; $k$ — a nonnegative integer, quantized by single-valuedness, never by an electrode. The $k=0$ pair holds the reference potential and the net line charge; each $k\ge1$ pair is a 2D multipole — $s^{-1}$ the line dipole, $s^{-2}$ the line quadrupole — while the growing partners describe externally applied fields. Regularity prunes before fitting: a region containing the axis keeps only $a_0$ and the $s^{k}$ (everything else diverges at $s=0$ ); an exterior region keeps the decaying set, whatever growing terms the far field dictates, and $\ln s$ exactly when net line charge is enclosed. Coefficient extraction is then Fourier's trick from the Cartesian article: project onto each mode and let orthogonality collapse the sum — the circle is the original home of Fourier series.

The cylinder in a uniform field

Now the payoff the hero figure promised. A conducting cylinder of radius $R$ , grounded (or merely neutral — see the FAQ), sits in a uniform field $\mathbf{E}_0=E_0\hat{\mathbf{x}}$ perpendicular to its axis — Griffiths Problem 3.24, the 2D twin of the sphere-in-a-field problem the spherical article solved. The result:

V(s,\phi)=-E_0\left(s-\frac{R^2}{s}\right)\cos\phi \qquad (3)

where: $\phi$ is measured from the field direction; the first term is the applied field's potential, the second is the cylinder's answer — a pure line-dipole field, decaying as $1/s$ . At $s=R$ the parenthesis vanishes identically, every angle at once: the surface is an equipotential, no angular gymnastics needed.

Go deeper · Derivation B · 1 minA conducting cylinder in a uniform field, end to end

Derivation BA conducting cylinder in a uniform field, end to end

The goal is equation (3) and its consequences: the induced surface charge and the induced dipole moment per unit length. The strategy is the standard boundary-value ritual — write the general solution (2), let the far field and regularity prune it, then impose $V(R,\phi)=0$ mode by mode. Watch orthogonality silence every mode the applied field does not sing in; the uniqueness theorems then certify the result as the answer.

Step 1. Impose the far field. As $s\to\infty$ , $V\to-E_0x=-E_0s\cos\phi$ . Comparing with (2): every growing coefficient vanishes except the $k=1$ cosine, $a_1=-E_0$ ; the logarithm is absent because the cylinder is neutral ( $b_0=0$ ); and $a_0=0$ , the grounded surface fixing the reference.

Step 2. Impose the surface condition. With only decaying terms left to respond,

V(R,\phi) = -E_0R\cos\phi + \sum_{k=1}^{\infty}R^{-k}\big(c_k\cos k\phi+d_k\sin k\phi\big) = 0 \quad\text{for all }\phi

A Fourier series that vanishes identically has every coefficient zero. Only the $k=1$ cosine has anything to cancel: $-E_0R+c_1/R=0$ , hence $c_1=E_0R^2$ , and (3) follows. Why only one mode responds: a perfect circle cannot convert angular frequencies — a dented cylinder would scatter the drive into all $k$ .

Step 3. Extract the induced charge via the conductor-surface rule of Conductors in Electrostatic Equilibrium:

\sigma(\phi)=-\varepsilon_0\left.\frac{\partial V}{\partial s}\right|_{s=R} = \varepsilon_0E_0\left(1+\frac{R^2}{s^2}\right)\cos\phi\,\Bigg|_{s=R} = 2\varepsilon_0E_0\cos\phi

Positive charge downstream ( $\phi=0$ ), negative upstream; the surface field is $2E_0\cos\phi$ — the applied field doubled at the poles, where the sphere managed a factor 3. The tangential field $\propto(1-R^2/s^2)\sin\phi$ vanishes at $s=R$ , as a conductor demands.

Step 4. Identify the induced line dipole. The response term $E_0R^2\cos\phi/s$ has exactly the form of a 2D line-dipole potential $V=p'\cos\phi/(2\pi\varepsilon_0 s)$ , so $p'=2\pi\varepsilon_0R^2E_0$ . Cross-checks: integrating the charge, $p'=\int_0^{2\pi}\sigma(R\cos\phi)R\,\mathrm{d}\phi$ , gives the same number; and the net charge per length $\propto\int\cos\phi\,\mathrm{d}\phi=0$ , as neutrality requires. *Verified: (3) is harmonic term by term, vanishes at $s=R$ , tends to $-E_0s\cos\phi$ at infinity; a numerical five-point Laplacian on it returns zero to machine precision. Matches Zangwill §7.7 and Griffiths Prob. 3.24; the sphere comparison matches Separation of Variables in Spherical Coordinates.*

The surface charge deserves its own equation — it is the experimentally accessible face of the solution:

\sigma(\phi)=2\varepsilon_0E_0\cos\phi, \qquad p'=2\pi\varepsilon_0R^2E_0 \qquad (4)

where: $p'$ is the induced dipole moment per unit length (C·m per metre — coulombs). The factor 2 in $\sigma$ splits as $1+1$ — applied field plus an exactly equal dipole contribution; on the sphere the split is $1+2$ . And the polarizability per length scales with cross-sectional area — the $R^2$ -versus- $R^3$ contrast with the sphere's $p=4\pi\varepsilon_0R^3E_0$ .

Figure

The payoff — field lines rearranged, charge induced in a pure cosine

The split cylinder — a genuine Fourier problem

The cylinder-in-a-field problem was almost too polite: the boundary drive was a single mode. To see the machinery at full stretch, feed it a boundary with every mode: a shell of radius $R$ sliced lengthwise into two half-shells, insulated by hairline gaps — the upper half ( $0<\phi<\pi$ ) at $+V_0$ , the lower at $-V_0$ . The boundary potential is a square wave in $\phi$ , and the interior solution is an infinite series,

V_{\text{in}}(s,\phi)=\frac{4V_0}{\pi}\sum_{k\ \text{odd}}\frac{1}{k}\left(\frac{s}{R}\right)^{k}\sin k\phi \qquad (5)

where: the sum runs over $k=1,3,5,\dots$ ; even modes are absent and the amplitudes fall as $1/k$ — the slow decay a discontinuous boundary always exacts. Swap $(s/R)^{k}\to(R/s)^{k}$ and the same coefficients solve the exterior. This series can even be summed,

V_{\text{in}}(s,\phi)=\frac{2V_0}{\pi}\arctan\!\left(\frac{2sR\sin\phi}{R^2-s^2}\right) \qquad (6)

where: the arctangent makes the boundary values manifest ( $s\to R$ sends the argument to $\pm\infty$ , so $V\to\pm V_0$ ) — worked example 2 checks the series against it digit by digit.

Go deeper · Derivation C · 1 minThe split cylinder — extracting the coefficients on a circle

Derivation CThe split cylinder — extracting the coefficients on a circle

The goal is equation (5): the interior potential of the two-half-shell cylinder, coefficients and all. The strategy is the canonical second act of every separation-of-variables problem — write the regular general solution, evaluate on the boundary, and project out each coefficient with Fourier's trick, recalled from the Cartesian article rather than re-proven. The instructive parts are the vanishings: parity silences the cosines before any integral is computed, and the square wave's antisymmetry kills the even sines.

Step 1. Prune. The interior contains the axis, so (2) loses its $\ln s$ and all $s^{-k}$ ; and the boundary data is odd under $\phi\to-\phi$ while cosines and the constant are even, so $a_0=a_k=0$ . Only $V_{\text{in}}=\sum_k b_k s^{k}\sin k\phi$ survives. Why check parity before integrating: the $a_k$ integrals would come out zero anyway, but seeing them vanish by symmetry is free and mistake-proof.

Step 2. Evaluate on the boundary and project. At $s=R$ : $\sum_k b_kR^{k}\sin k\phi = V_0\,\mathrm{sgn}(\sin\phi)$ . Multiply by $\sin k'\phi$ , integrate over $[0,2\pi)$ ; orthogonality ( $\int\sin k\phi\sin k'\phi\,\mathrm{d}\phi=\pi\delta_{kk'}$ ) collapses the left side to $\pi b_{k'}R^{k'}$ , while the right side gives

V_0\left[\int_0^{\pi}\sin k\phi\,\mathrm{d}\phi-\int_{\pi}^{2\pi}\sin k\phi\,\mathrm{d}\phi\right] = \frac{2V_0}{k}\big[1-\cos k\pi\big] = \begin{cases}4V_0/k, & k\ \text{odd}\\[2pt] 0, & k\ \text{even}\end{cases}

Step 3. Solve and assemble: $b_k=4V_0/(\pi kR^{k})$ for odd $k$ , zero for even $k$ — equation (5). Why the $1/k$ decay is slow: a Fourier series builds a jump out of many reluctant smooth modes; inside, each mode is further crushed by $(s/R)^{k}$ , so convergence is excellent at depth and grudging near the surface.

Step 4. Exterior, for free. Regularity at infinity keeps $s^{-k}$ instead; the identical projection gives $V_{\text{out}}=(4V_0/\pi)\sum_{k\,\text{odd}}(1/k)(R/s)^{k}\sin k\phi$ . Far away the $k=1$ term dominates: the split cylinder looks like a line dipole — the leading term of a 2D multipole expansion read off the series. Verified: summing the series numerically at $(R/2,\pi/2)$ gives $29.5167$ V for $V_0=50$ V, agreeing with the closed form (6) to six figures; (6) reproduces the boundary values $\pm V_0$ as $s\to R$ . Matches Griffiths §3.3 and Jackson §2.11.

Limiting cases

case	result	meaning
$s\to\infty$ (cylinder in field)	$V\to-E_0s\cos\phi$	disturbance dies as $1/s$
$s=R$ (cylinder in field)	$V=0$ , $\mathbf{E}=2E_0\cos\phi\,\hat{\mathbf{s}}$	equipotential, field purely normal, doubled at the poles
$k=0$ sector alone	$V=a_0+b_0\ln s$	line-charge/coax world: no grounding at infinity
sphere vs cylinder in field $E_0$	$\sigma_{\max}=3\varepsilon_0E_0$ vs $2\varepsilon_0E_0$	dimensionality counts
deep inside the split cylinder	$V\approx\dfrac{4V_0}{\pi}\dfrac{s}{R}\sin\phi$	one mode survives; the interior smooths the square wave
$z$ -dependence restored	Bessel's equation	powers give way to $J_k$ , $Y_k$ — waveguide territory

The last row is the honest boundary of this article. Give the problem genuine structure along the axis — a finite cylinder with a prescribed top, a waveguide mode — and the radial equation picks up a $k_z^2s^2$ term that destroys its stretch-invariance: the solutions become Bessel functions. We name them, salute their author below, and defer them (E8 waveguide territory). The reflex: no $z$ -dependence, no Bessel.

Common trapA cylinder is not a fat sphere — 2D intuition is its own discipline

The most reliable way to botch a cylindrical problem is to import spherical reflexes. In two dimensions the potential of a charged line is a logarithm, not $1/r$ ; it cannot be referenced to zero at infinity, and its field falls as $1/s$ , not $1/s^2$ . So do not bolt a $1/s$ "monopole term" onto (2) — the monopole of 2D is the $\ln s$ term, and dropping it (or its partner, the constant) leaves two-boundary problems like the coax with too few constants. Equally, do not expect the sphere's numbers: factor 2, not 3; polarizability $R^2$ , not $R^3$ . The safe procedure is mechanical — write all of (2), prune by regularity and net line charge, then fit. Pruning by half-remembered 3D analogy deletes the very terms the problem needs.

Worked examples

1 · A cylinder in the wind — charge, field, and dipole moment (confidence builder)

A grounded conducting cylinder of radius $R=2.0$ cm sits in a uniform transverse field $E_0=3.0\times10^{4}$ V/m. Find the maximum induced charge density, the field at the poles, the induced dipole moment per length, and $V$ at $s=2R$ , $\phi=0$ .

Step 1. Maximum surface charge, from (4) at $\phi=0$ : $\sigma_{\max}=2\varepsilon_0E_0=2\times(8.854\times10^{-12})\times(3.0\times10^{4})$ $=\boxed{5.3\times10^{-7}\ \mathrm{C/m^2}\approx0.53\ \mathrm{\mu C/m^2}}$ . The sphere in the same field would reach $3\varepsilon_0E_0=0.80\ \mathrm{\mu C/m^2}$ — the dimensional factor made flesh.

Step 2. Surface field at the poles: just outside a conductor, $E=\sigma/\varepsilon_0$ , so $E_{\text{pole}}=2E_0=\boxed{6.0\times10^{4}\ \mathrm{V/m}}$ — enhancement factors like this are why curved conductors trigger breakdown first.

Step 3. Induced dipole moment per unit length, from (4): $p'=2\pi\varepsilon_0R^2E_0=2\pi\times(8.854\times10^{-12})\times(0.020)^2\times(3.0\times10^{4})=\boxed{6.7\times10^{-10}\ \mathrm{C}}$ — $0.67$ nC·m per metre of cylinder, confirmed independently by integrating $\sigma R\cos\phi$ around the surface.

Step 4. Potential at $s=2R=4.0$ cm, $\phi=0$ , from (3): $V=-3.0\times10^{4}\times(0.040-0.010)=\boxed{-900\ \mathrm{V}}$ . The bare field alone would give $-1200$ V; the induced dipole claws back $+300$ V — a 25% correction at one radius out.

2 · The split cylinder, summed digit by digit (exam level)

A split cylinder has its upper half at $V_0=+50$ V, lower half at $-50$ V. Evaluate $V$ at $s=R/2$ , $\phi=90^\circ$ by partial sums of (5); compare with the closed form (6).

Step 1. Set up the series. At $\phi=\pi/2$ , $\sin k\phi$ alternates $+1,-1,+1,\dots$ over the odd $k$ , and $(s/R)^k=(1/2)^k$ ; the prefactor is $4V_0/\pi=63.66$ V. Successive terms: $+31.83$ , $-2.65$ , $+0.40$ , $-0.071$ , $+0.014$ V — the geometric factor does the heavy lifting.

Step 2. Accumulate the partial sums:

include through	partial sum (V)
$k=1$	31.831
$k=3$	29.178
$k=5$	29.576
$k=7$	29.505
$k=9$	29.519
$k=15$	29.5167

Step 3. Compare with (6): $V=\dfrac{100}{\pi}\arctan\!\left(\dfrac{2\cdot\frac{1}{2}\cdot1}{1-\frac{1}{4}}\right)=31.831\arctan\!\left(\tfrac{4}{3}\right)=\boxed{29.517\ \mathrm{V}}$ . Five terms agree to three significant figures; eight to six.

Step 4. Sanity checks: the answer sits below $V_0=50$ V, as a harmonic function must (no interior extrema — Laplace's Equation); and the exterior series at the inverse point $s=2R$ is term-for-term identical, giving $29.517$ V there too — the inversion symmetry $s\leftrightarrow R^2/s$ in live numbers.

Origins · 1 minFriedrich Wilhelm Bessel and how this was discovered

Friedrich Wilhelm Bessel

1784–1846 · Königsberg Observatory

Bessel is the rare figure whose name every physics student meets attached to functions he would have considered a side errand. Apprenticed at fourteen to a Bremen trading house, he taught himself navigation, mathematics, and astronomy; at twenty his recalculation of Halley's comet orbit earned him an observatory post, and at 25 the directorship of the new Königsberg Observatory. His obsession was precision, and in 1838 it delivered the measurement astronomers had chased for three centuries: the first stellar parallax, 0.314 arcseconds for 61 Cygni. The functions arose in his 1824 memoir on planetary perturbations; he tabulated them and located their zeros with an astronomer's thoroughness. This article sidesteps them — no $z$ -dependence, no Bessel — but the moment a cylinder problem acquires axial structure, his tables take over the radial direction.

History

1732

— Daniel Bernoulli, analysing the oscillating hanging chain, encounters the function later called J0 — a century before its name.

1764

— Euler derives the radial equation of the vibrating circular drumhead: Bessel's equation, solved in series, decades early.

1824

— Bessel's Berlin memoir on planetary perturbations gives the functions their systematic theory, and his name.

1838

— Bessel measures the first stellar parallax, 61 Cygni at 0.314 arcsec: an astronomer first, the functions a means to an end.

1893

— Heaviside's transmission-line theory puts the k = 0 logarithmic sector to industrial work in every coaxial cable.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §3.3.2 · Jackson, Classical Electrodynamics, 3rd ed., §3.7–3.8 and §3.11 · Zangwill, Modern Electrodynamics, §7.7 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 3 · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 14 (Bessel functions)

Provenance. All equations are standard textbook results. A verifier should: (1) confirm the cylindrical Laplacian and its separation into the Bessel, azimuthal, and axial ODEs (Griffiths §3.3.2, Jackson §3.7); (2) verify Bessel's equation and the Fourier–Bessel orthogonality normalization $\tfrac{a^2}{2}[J_{m+1}(\alpha_{mn})]^2$ (Arfken Ch. 14, Jackson eq. 3.95–3.96); (3) recompute the grounded-cylinder coefficient formula (Jackson §3.8); (4) check the 2D conducting-cylinder potential $-E_0(s-a^2/s)\cos\phi$ and induced charge $2\varepsilon_0E_0\cos\phi$ (Zangwill §7.7); (5) verify the coaxial capacitance per length $2\pi\varepsilon_0/\ln(s_2/s_1)$ . No nonstandard numbers are introduced.

Builds on: Laplace's Equation · Separation of Variables in Cartesian Coordinates · Boundary Conditions for the Potential · Uniqueness Theorems for the Potential Leads to: Separation of Variables in Spherical Coordinates · Poisson's Equation · Legendre Polynomials

Why does the logarithm appear only in the k equals 0 sector, and what is it physically?

The Euler exponents $\alpha=\pm k$ collide at $k=0$ , and a repeated root always spawns a logarithm as its second solution — exactly as critical damping spawns $te^{-\gamma t}$ . Physically, $b_0\ln s$ is the potential of a uniform line charge on the axis, the result The Electrostatic Potential derived when the reference at infinity failed. It is the two-dimensional monopole, carrying the net charge per unit length with $b_0=-\lambda/2\pi\varepsilon_0$ . No net line charge means $b_0=0$ ; a coax inner conductor makes the logarithm compulsory.

Why can't I set the potential to zero at infinity, like I always do in 3D?

Because a truly two-dimensional source is an infinite line in 3D, and its potential, $-(\lambda/2\pi\varepsilon_0)\ln(s/s_0)$ , grows without bound. There is no "far enough away" from an infinite line — walk outward and the source is still beside you, arbitrarily far along the axis. So in 2D you choose a finite reference radius and quote potential differences — all physics needs. Writing $V(\infty)=0$ is a 3D convenience licensed by $1/r$ decay, not a law. Problems with zero net line charge — the cylinder in a field, the split cylinder — decay properly, and there the habit revives.

Where exactly does the sphere's factor 3 versus the cylinder's factor 2 come from?

Split the surface field into applied plus induced. In both problems the induced object is a pure dipole whose potential exactly cancels the applied potential on the surface. For the 2D line dipole, the surface field it contributes equals the applied contribution — $1+1=2$ . For the 3D dipole, differentiating $r^{-2}$ brings down a 2 where differentiating $s^{-1}$ brings down a 1, so its surface field is twice the term it cancels — $1+2=3$ . The lopsided 3D exponent pair is the fingerprint of flux spreading over $r^2$ rather than $s$ .

What happened to the B phi solution in the k equals 0 angular equation?

At $k=0$ the angular equation $\Phi''=0$ has general solution $A+B\phi$ , and the linear piece is genuinely harmonic. Single-valuedness expels it: $\phi$ and $\phi+2\pi$ are the same point, and $B\phi$ disagrees with itself by $2\pi B$ there. It survives where you never complete a lap: in the wedge between two conducting half-planes, $V\propto\phi$ legally interpolates between the plates. Quantization comes from the topology of the domain, and cutting the domain relaxes it.

Do I need Bessel functions for this article's problems, and when will I need them?

Not here — and knowing why is the point. Bessel's equation arises when the separated radial equation carries a $k_z^2s^2$ term from genuine $z$ -dependence, ruining the scale invariance that made pure powers work. Every problem here is translation-invariant along the axis, so the radial equation stays Euler. You will need $J_k$ and $Y_k$ the moment the axis acquires structure: a finite cylinder with a prescribed top, a cavity, and above all waveguides, whose mode patterns are Bessel functions — E8 territory. For now: no $z$ -dependence, no Bessel.

Does it matter whether the cylinder in the field is grounded or just neutral and isolated?

For this problem, no. A neutral isolated cylinder sits at some constant potential, fixed by requiring zero net induced charge per length. A nonzero constant would demand an added $a_0+b_0\ln s$ ; but $b_0\ne0$ means net line charge (forbidden), and with $b_0=0$ the constant merely shifts the reference. The $\cos\phi$ distribution already integrates to zero, so the grounded solution is the neutral solution. The distinction bites only when the applied potential has a $k=0$ component on the surface — a nearby line charge — where grounding lets net charge flow in and neutrality forbids it.

My split-cylinder partial sums oscillate and converge slowly near the surface — am I doing something wrong?

No — you are watching two honest phenomena. First, slow convergence: the boundary data jumps at the gaps, so its Fourier coefficients decay only as $1/k$ , and near $s=R$ the factor $(s/R)^k$ barely helps (at $s=0.9R$ , about a hundred terms for four digits). Second, Gibbs: partial sums of a discontinuous function overshoot near the jump by about 9% of the step, shrinking in width but never in height. Neither is an error. Evaluate series deep inside or far outside, and lean on closed forms like equation (6) near the boundary.