Method of Images and Eigenfunction Expansions of Green's Functions

Solve a linear boundary-value problem once, for a unit point source, and you have solved it for every source. That single response — the field at r\mathbf{r} produced by a delta-function disturbance at r′\mathbf{r}' — is the Green's function G(r,r′)G(\mathbf{r},\mathbf{r}'), and superposition then hands you the answer to Lu=f\mathcal{L}u=f as an integral u=∫G fu=\int G\,f. The entire difficulty collapses onto one question: how do you build GG for a given operator L\mathcal{L}, a given domain Ω\Omega, and given boundary conditions on ∂Ω\partial\Omega? This article develops the two great strategies that answer it, and shows they are the same answer wearing different clothes. The first strategy, the method of images, is geometric and closed-form. In free space the Laplacian's Green's function is known outright — 1/4π∣r−r′∣1/4\pi|\mathbf{r}-\mathbf{r}'| in three dimensions — so a boundary is enforced by adding fictitious image sources outside the domain, positioned so their combined field satisfies the boundary condition identically: a mirror charge across a grounded plane, a Kelvin image inside a sphere, a rosette of reflections in a wedge. It is exact and transparent, but works only when the geometry reflects or inverts cleanly. The second strategy, the eigenfunction expansion, is algebraic and universal. Resolve the delta in the operator's own normal modes and G=∑nϕnϕn∗/(λn−λ)G=\sum_n\phi_n\phi_n^*/(\lambda_n-\lambda) falls out — one term per mode, a pole at every eigenvalue, valid wherever L\mathcal{L} is self-adjoint with a complete basis. The punchline is that these are two convergent series for one function. Near the source a few images converge fast; near a resonance or far away, a few modes dominate. Poisson summation and the Sommerfeld–Watson transform trade one series for the other, and choosing the right one carries directly into short-wavelength asymptotics and the WKB method.

Key takeaways
  • A Green's function solves LG(r,r′)=δ(r−r′)\mathcal{L}G(\mathbf{r},\mathbf{r}')=\delta(\mathbf{r}-\mathbf{r}') under homogeneous boundary conditions and turns a differential equation into an integral over sources, u=∫ΩGfu=\int_\Omega G f.
  • The method of images replaces a boundary by image sources placed outside the domain so the total field meets the boundary condition geometrically — exact, but only for planes, spheres, and integer wedges.
  • The eigenfunction expansion writes G=∑nϕn(r)ϕn∗(r′)/(λn−λ)G=\sum_n\phi_n(\mathbf{r})\phi_n^*(\mathbf{r}')/(\lambda_n-\lambda), valid in any domain with a complete orthonormal eigenbasis of L\mathcal{L}; its poles at λ=λn\lambda=\lambda_n are the resonances.
  • Both constructions give the same GG: images expose the singularity and give a closed form, modes expose the spectrum and converge far from the source.
  • Self-adjointness of L\mathcal{L} forces reciprocity G(r,r′)=G(r′,r)G(\mathbf{r},\mathbf{r}')=G(\mathbf{r}',\mathbf{r}) — source and detector are interchangeable.
  • The same machinery runs electrostatics with conductors, heat flow, cavity resonance, the quantum resolvent, and the density of states.
  • Key idea: a boundary condition can be enforced geometrically by adding image sources to the free-space Green's function, or algebraically by expanding in the eigenmodes that already satisfy it — and both yield the identical self-adjoint, reciprocal GG.
The core insight

A boundary condition can be enforced either by adding image sources in the unbounded Green's function until the condition holds geometrically, or by expanding the source and Green's function in the eigenmodes that satisfy the condition; both give the same self-adjoint, reciprocal GG.

Where this lives today

The two representations are still the working tools of anyone solving a field equation in a bounded region. In room acoustics, the image-source method of Allen and Berkley (1979) models a room's early impulse response as a lattice of mirror sources reflected in the walls — the direct descendant of the interval derivation above — and it is the engine inside spatial-audio and virtual-reality rendering, where a few dozen images reproduce the first reflections while the diffuse late tail is left to a statistical modal model. In biomolecular simulation, image charges enforce the electrostatic boundary condition at a metal electrode or a dielectric membrane without meshing it. And across device physics the eigenfunction resolvent has become the non-equilibrium Green's function (NEGF) formalism: the poles this article calls resonances are the transmission peaks of a nanoscale transistor, and the trace of GG is the density of states experiment measures. Continued to complex energy, the same object gives the semiclassical trace formulas of quantum chaos, where long classical orbits — descendants of multiply reflected images — reappear as oscillations in the level density. Choosing the ray sum or the mode sum remains the judgement Green's essay first made possible.

The physical picture

The Green's function is the impulse response of a differential operator. Strike the domain with a unit point source at r′\mathbf{r}' and G(r,r′)G(\mathbf{r},\mathbf{r}') is the field that rings out at r\mathbf{r}; because L\mathcal{L} is linear, the response to any source ff is the superposition ∫Gf\int G f. Everything here is a way to compute that one response for a bounded region. The method of images replaces a conductor by a ghost. A grounded plane holds the potential at zero; a negative charge reflected behind the plane achieves the same zero on the surface, so in the physical half-space the true field is just the real charge plus its mirror. The conductor has vanished, and with it the hard boundary-value problem — replaced by two point charges in free space. The first figure shows the trick and the force it predicts.

Figure
A point charge +q at height z = d above a grounded plane is replaced by an image −q at z = −d, outside the physical region. On the plane both charges are equidistant, so G = 0 identically (Dirichlet), and the real charge feels an attractive image force F = q²/16πε₀d² toward the conductor.
Method of images — a grounded plane replaced by one mirror charge physical region z above 0 (Ω) G = 0 on the plane + +q at z = d − −q at z = −d (image, outside Ω) F = q²⁄16πε₀d² attractive, toward plane d

The eigenfunction picture is the opposite temperament: a decomposition into standing waves. A point source excites every normal mode at once, each weighted by its amplitude at source and observer, ϕn(r′)ϕn(r)\phi_n(\mathbf{r}')\phi_n(\mathbf{r}), and by how far its eigenvalue λn\lambda_n sits from the operating value λ\lambda. Drive the system near a resonance, λ→λm\lambda\to\lambda_m, and that one denominator λm−λ\lambda_m-\lambda collapses: a single mode dominates and GG diverges — exactly what "resonance" means. The two representations are one function summed differently, and the second figure makes the equivalence literal on the simplest domain, an interval.

Figure
One Green's function, two series. The interval kernel G(x,x′) = x_min(L−x_max)/L for a Dirichlet source at x′ = L/3 (blue triangle, peak value 2L/9 with a unit slope-jump at the source) is reproduced exactly by the sine-mode sum; just three modes (amber) already trace the shape that infinitely many image charges also build.
The unification: modes (amber) rebuild the image kernel (blue), G(x, x′ = L/3) 2L/9 L/9 0 L/3 L position x along the interval G(x, x′) peak 2L/9, slope jumps by −1 image / closed form x_min(L−x_max)/L 3-mode sine sum (2/L) Σ sin·sin/(nπ/L)²

Which series to trust is a question of regime. Near the source the singularity is universal — every method reproduces the same 1/r1/r blow-up — and the ray picture gets there in one or two terms. Far away, or right on a resonance, the standing-wave sum is dominated by a handful of modes and wins. Choosing between them is not choosing different physics; it is choosing a fast-converging expansion of one and the same GG.

The core equations

Fix the convention up front: L\mathcal{L} is a positive self-adjoint operator such as −∇2-\nabla^2 or −∂x2-\partial_x^2, so −∇2G=δ-\nabla^2 G=\delta gives the everyday Coulomb-like potential with a positive free-space kernel. The defining equation pins GG down as the response to a unit point source inside the domain.

Lr G(r,r′)=δ(r−r′),r∈Ω,(1)\mathcal{L}_{\mathbf{r}}\,G(\mathbf{r},\mathbf{r}')=\delta(\mathbf{r}-\mathbf{r}'),\qquad \mathbf{r}\in\Omega, \qquad (1)

where: L\mathcal{L} — linear self-adjoint operator (e.g. −∇2-\nabla^2 or −∇2−λ-\nabla^2-\lambda); Ω\Omega — bounded or half-space domain with boundary ∂Ω\partial\Omega; δ(r−r′)\delta(\mathbf{r}-\mathbf{r}') — Dirac delta, a unit point source at r′\mathbf{r}'. The homogeneous boundary condition (Dirichlet G=0G=0 or Neumann ∂nG=0\partial_n G=0) is part of the definition — without it GG is not unique. Knowing GG solves the inhomogeneous problem for any source by a single integral.

u(r)=∫ΩG(r,r′) f(r′) dnr′  +  boundary terms,(2)u(\mathbf{r})=\int_\Omega G(\mathbf{r},\mathbf{r}')\,f(\mathbf{r}')\,\mathrm{d}^n r' \;+\; \text{boundary terms}, \qquad (2)

where: ff — the source of Lu=f\mathcal{L}u=f; uu — the sought field; boundary terms carry inhomogeneous data via Green's second identity. Read it as superposition: each element f(r′) dnr′f(\mathbf{r}')\,\mathrm{d}^n r' radiates its response GG, and the field is their sum. For the Laplacian the free-space kernel is elementary — the seed every image construction grows from.

G3(r,r′)=14π∣r−r′∣,G2(r,r′)=−12πln⁡∣r−r′∣,(3)G_3(\mathbf{r},\mathbf{r}')=\frac{1}{4\pi|\mathbf{r}-\mathbf{r}'|},\qquad G_2(\mathbf{r},\mathbf{r}')=-\frac{1}{2\pi}\ln|\mathbf{r}-\mathbf{r}'|, \qquad (3)

where: G3, G2G_3,\,G_2 — solutions of −∇2G=δ-\nabla^2 G=\delta in three and two dimensions; ∣r−r′∣|\mathbf{r}-\mathbf{r}'| — source-to-observer distance. They follow from ∇2(1/r)=−4πδ\nabla^2(1/r)=-4\pi\delta (3D) and ∇2ln⁡r=2πδ\nabla^2\ln r=2\pi\delta (2D), the sign fixed by the positive-operator convention. The half-space Dirichlet problem is the archetype of the method of images: one mirror charge does the whole job.

G(r,r′)=14π ⁣[1∣r−r′∣−1∣r−r ⁣∗′∣],r ⁣∗′=(x′,y′,−z′),(4)G(\mathbf{r},\mathbf{r}')=\frac{1}{4\pi}\!\left[\frac{1}{|\mathbf{r}-\mathbf{r}'|}-\frac{1}{|\mathbf{r}-\mathbf{r}'_{\!*}|}\right],\qquad \mathbf{r}'_{\!*}=(x',y',-z'), \qquad (4)

where: r ⁣∗′\mathbf{r}'_{\!*} — mirror of r′\mathbf{r}' across z=0z=0, at z=−z′z=-z' outside the physical region. On the plane the two distances are equal, so the bracket vanishes and G=0G=0 holds identically; inside Ω\Omega only the real singularity survives. For a sphere the mirror becomes an inversion: Kelvin's image lives on the same ray, scaled inward, with a reduced charge.

q′=−ar′ q,r ⁣∗′=a2r′2 r′,r′=∣r′∣,(5)q'=-\frac{a}{r'}\,q,\qquad \mathbf{r}'_{\!*}=\frac{a^2}{r'^2}\,\mathbf{r}',\qquad r'=|\mathbf{r}'|, \qquad (5)

where: q′q' — image charge magnitude; r ⁣∗′\mathbf{r}'_{\!*} — image position along r^′\hat{\mathbf{r}}', inside a grounded sphere of radius aa; r′r' — source distance from the centre. This pair is the unique choice that zeroes the potential on ∣r∣=a|\mathbf{r}|=a, derived below. The eigenfunction expansion is the universal alternative — no special geometry, only a complete self-adjoint spectrum.

G(r,r′;λ)=∑nϕn(r) ϕn∗(r′)λn−λ,(6)G(\mathbf{r},\mathbf{r}';\lambda)=\sum_n \frac{\phi_n(\mathbf{r})\,\phi_n^*(\mathbf{r}')}{\lambda_n-\lambda}, \qquad (6)

where: ϕn\phi_n — orthonormal eigenfunctions with Lϕn=λnϕn\mathcal{L}\phi_n=\lambda_n\phi_n satisfying the boundary conditions; λn\lambda_n — real eigenvalues (self-adjointness); λ\lambda — spectral parameter, so (L−λ)−1(\mathcal{L}-\lambda)^{-1} is the resolvent. Each pole at λ=λn\lambda=\lambda_n is a resonance. Self-adjointness leaves a measurable fingerprint on GG worth naming on its own.

G(r,r′)=G(r′,r)(real, self-adjoint L),(7)G(\mathbf{r},\mathbf{r}')=G(\mathbf{r}',\mathbf{r})\qquad(\text{real, self-adjoint }\mathcal{L}), \qquad (7)

where: G(r,r′)G(\mathbf{r},\mathbf{r}') — symmetric under exchange of source and observer. A source at AA seen at BB reads identically to a source at BB seen at AA; this is manifest term-by-term in the spectral sum (6). Once GG is known, observables such as the induced density σind=−ε0 ∂nΦ∣∂Ω\sigma_{\text{ind}}=-\varepsilon_0\,\partial_n\Phi|_{\partial\Omega} and the image force follow by differentiation, as Example 1 shows.

Derivation ABuilding the resolvent by resolving the delta in eigenfunctions

The goal here is to derive the spectral formula (6) from scratch and, in doing so, read off three structural facts for free — reciprocity, the location of resonances, and the Fredholm alternative that decides when a solution exists at all. We assume only that L\mathcal{L} is self-adjoint on Ω\Omega under the chosen boundary conditions, so it has real eigenvalues λn\lambda_n and a complete orthonormal eigenbasis {ϕn}\{\phi_n\} obeying those same conditions. The strategy is to expand both the delta and the unknown GG in the ϕn\phi_n and match coefficients.

Step 1. Completeness of the eigenbasis resolves the point source itself, δ(r−r′)=∑nϕn(r)ϕn∗(r′)\delta(\mathbf{r}-\mathbf{r}')=\sum_n\phi_n(\mathbf{r})\phi_n^*(\mathbf{r}'). Why: this is the whole trick — a delta has an honest expansion in any complete basis, and choosing the operator's own eigenbasis makes L\mathcal{L} act simply next.

Step 2. Expand the unknown GG in the same basis, G=∑ncn(r′) ϕn(r)G=\sum_n c_n(\mathbf{r}')\,\phi_n(\mathbf{r}). Why: since the ϕn\phi_n already satisfy the boundary conditions, every term does too, so the boundary condition on GG is enforced automatically — no matching, no images.

Step 3. Apply (L−λ)(\mathcal{L}-\lambda) term by term, using Lϕn=λnϕn\mathcal{L}\phi_n=\lambda_n\phi_n, and equate to the resolved delta:

(L−λ)G=∑ncn(λn−λ)ϕn(r)=∑nϕn(r)ϕn∗(r′).(\mathcal{L}-\lambda)G=\sum_n c_n(\lambda_n-\lambda)\phi_n(\mathbf{r})=\sum_n\phi_n(\mathbf{r})\phi_n^*(\mathbf{r}').

Why: the eigenbasis diagonalizes the operator, turning a differential equation into an algebraic one for each coefficient separately.

Step 4. Match coefficients of each ϕn\phi_n to get cn=ϕn∗(r′)/(λn−λ)c_n=\phi_n^*(\mathbf{r}')/(\lambda_n-\lambda), hence

G(r,r′;λ)=∑nϕn(r)ϕn∗(r′)λn−λ.G(\mathbf{r},\mathbf{r}';\lambda)=\sum_n\frac{\phi_n(\mathbf{r})\phi_n^*(\mathbf{r}')}{\lambda_n-\lambda}.

Why: this is formula (6), and its structure is transparent — the numerator is symmetric in r↔r′\mathbf{r}\leftrightarrow\mathbf{r}' (reciprocity, since real self-adjoint L\mathcal{L} admits real ϕn\phi_n), and the denominator vanishes exactly when λ\lambda hits the spectrum.

Step 5. Read the singular case. When λ→λm\lambda\to\lambda_m the mm-th term diverges: L−λ\mathcal{L}-\lambda has a null vector ϕm\phi_m and cannot be inverted. A bounded solution of (L−λ)u=f(\mathcal{L}-\lambda)u=f then exists only if ∫ϕm∗f=0\int\phi_m^* f=0 — the Fredholm alternative — and is unique only up to c ϕmc\,\phi_m. Why: the resolvent's poles are not a defect but the physics itself, marking resonances and the breakdown of unique solvability.

Go deeper · Derivation B · 2 minThe grounded sphere by Kelvin inversion
Derivation BThe grounded sphere by Kelvin inversion

The aim is to find, for a real unit source at r′\mathbf{r}' with r′=∣r′∣>ar'=|\mathbf{r}'|>a outside a grounded sphere of radius aa, a single image charge q′q' at r ⁣∗′\mathbf{r}'_{\!*} inside that makes the potential vanish on the whole surface ∣r∣=a|\mathbf{r}|=a. That two numbers — one charge magnitude, one radial distance — can kill the potential at every point of a two-dimensional surface is the small miracle of the method, and it works because the sphere carries an inversion symmetry the plane has only in a degenerate limit. We impose G=0G=0 and let the geometry dictate q′q' and r ⁣∗′\mathbf{r}'_{\!*}.

Step 1. By rotational symmetry about r^′\hat{\mathbf{r}}' the image must lie on that ray, so write r ⁣∗′=b r^′\mathbf{r}'_{\!*}=b\,\hat{\mathbf{r}}' with b<ab<a and

4π G=1∣r−r′∣+q′∣r−r ⁣∗′∣.4\pi\,G=\frac{1}{|\mathbf{r}-\mathbf{r}'|}+\frac{q'}{|\mathbf{r}-\mathbf{r}'_{\!*}|}.

Why: any off-axis image would break the axial symmetry that the source and sphere share, so a single on-axis image is the only candidate worth testing.

Step 2. Evaluate on the surface ∣r∣=a|\mathbf{r}|=a, writing γ\gamma for the angle between r\mathbf{r} and r′\mathbf{r}' and demanding the potential vanish for every γ\gamma:

1a2+r′2−2ar′cos⁡γ+q′a2+b2−2abcos⁡γ=0.\frac{1}{\sqrt{a^2+r'^2-2ar'\cos\gamma}}+\frac{q'}{\sqrt{a^2+b^2-2ab\cos\gamma}}=0.

Why: the boundary condition is not one equation but a continuum of them, one per γ\gamma, and only a special (q′,b)(q',b) can satisfy all at once.

Step 3. For the two radicals to be proportional at all γ\gamma, their cos⁡γ\cos\gamma-to-constant ratios must match:

a2+b22ab=a2+r′22ar′ ⟹ b=a2r′.\frac{a^2+b^2}{2ab}=\frac{a^2+r'^2}{2ar'}\ \Longrightarrow\ b=\frac{a^2}{r'}.

Why: matching the angular dependence fixes the image distance, and the answer b=a2/r′b=a^2/r' is precisely the Kelvin inversion of the source point through the sphere.

Step 4. Substitute b=a2/r′b=a^2/r' into the constant-term balance to fix the strength, q′=−a/r′q'=-a/r'. On the surface ∣r−r ⁣∗′∣=(a/r′)∣r−r′∣|\mathbf{r}-\mathbf{r}'_{\!*}|=(a/r')|\mathbf{r}-\mathbf{r}'|, so the second term cancels the first exactly. Why: the reduced, negative image is what makes the two Coulomb terms equal and opposite on the whole sphere, not just at one point.

Step 5. Assemble the Green's function:

G(r,r′)=14π[1∣r−r′∣−a/r′∣r−a2r′2r′∣].G(\mathbf{r},\mathbf{r}')=\frac{1}{4\pi}\left[\frac{1}{|\mathbf{r}-\mathbf{r}'|}-\frac{a/r'}{\left|\mathbf{r}-\frac{a^2}{r'^2}\mathbf{r}'\right|}\right].

Why: the image sits at b=a2/r′<ab=a^2/r'<a, inside the sphere and hence outside the exterior domain, so GG keeps exactly one physical singularity — the real source — as required. *The bracket in the assembled GG vanishes identically on ∣r∣=a|\mathbf{r}|=a, confirming the Dirichlet condition, and the induced surface charge integrates to the total image charge −a/r′⋅q-a/r'\cdot q.*

Go deeper · Derivation C · 2 minOne interval, two series — modes equal images by Poisson summation
Derivation COne interval, two series — modes equal images by Poisson summation

The purpose here is to make the central claim of the article concrete on the simplest domain: to show that the sine-mode expansion and the image construction of the interval Green's function are not merely equal in value but two Fourier faces of a single triangular kernel, linked by Poisson summation. We solve −G′′=δ(x−x′)-G''=\delta(x-x') on [0,L][0,L] with G(0)=G(L)=0G(0)=G(L)=0 twice — by modes and by direct matching — then identify the image lattice that Poisson summation converts into the mode sum.

Step 1. The Dirichlet eigenfunctions of −∂x2-\partial_x^2 are ϕn(x)=2/L sin⁡(nπx/L)\phi_n(x)=\sqrt{2/L}\,\sin(n\pi x/L) with λn=(nπ/L)2\lambda_n=(n\pi/L)^2; the λ=0\lambda=0 form of the spectral formula (6) gives

G=2L∑n=1∞sin⁡(nπx/L)sin⁡(nπx′/L)(nπ/L)2.G=\frac{2}{L}\sum_{n=1}^\infty\frac{\sin(n\pi x/L)\sin(n\pi x'/L)}{(n\pi/L)^2}.

Why: this is the universal expansion specialized to the Poisson problem, and it manifestly satisfies the boundary conditions mode by mode.

Step 2. Solve the same equation directly. Away from x′x' the source is absent, so GG is piecewise linear and boundary-compatible,

G={A x,x<x′B (L−x),x>x′,G=\begin{cases}A\,x,& x<x'\\[2pt] B\,(L-x),& x>x',\end{cases}

which already builds in G(0)=G(L)=0G(0)=G(L)=0. Why: solving the homogeneous equation on each side and enforcing the ends is the image method in disguise — the linear pieces are what an infinite train of reflected sources produces.

Step 3. Sew the pieces at x′x' with continuity and the unit derivative jump from integrating −G′′=δ-G''=\delta across the source, G(x′+)=G(x′−)G(x'^+)=G(x'^-) and G′(x′+)−G′(x′−)=−1G'(x'^+)-G'(x'^-)=-1. Solving for A,BA,B gives the closed form

G(x,x′)=x< (L−x>)L,x<=min⁡(x,x′), x>=max⁡(x,x′).G(x,x')=\frac{x_{<}\,(L-x_{>})}{L},\qquad x_{<}=\min(x,x'),\ x_{>}=\max(x,x').

Why: two conditions fix the two constants, and the result is the triangular kernel plotted in the second figure — continuous, with a slope that drops by one at the source.

Step 4. Identify the images. The closed form is the restriction to [0,L][0,L] of the free-line potential of the source plus its reflections at both walls — an infinite lattice of alternating-sign images at x=±x′+2mLx=\pm x'+2mL. Poisson summation maps that lattice term-for-term onto the sine series of Step 1. Why: this is the precise sense in which "images equal modes" — Poisson summation is the dictionary between a near-source ray sum and a far-field mode sum. *The sine sum and the closed form x<(L−x>)/Lx_<(L-x_>)/L agree numerically: the peak is 2L/92L/9 at x′=L/3x'=L/3, and a 50-term partial sum reproduces the triangular kernel to three digits.*

Limiting cases

Which representation is efficient, and whether the image method works at all, depends on the geometry and on how close the spectral parameter sits to the spectrum.

CaseResultMeaning
Source far from boundaryG≈Gfree+(one weak image)G\approx G_{\text{free}}+\text{(one weak image)}The boundary is a small perturbation; the image series converges in one or two terms and the mode sum is wasteful.
Near a resonance λ→λm\lambda\to\lambda_mG∼ϕm(r)ϕm∗(r′)λm−λG\sim\dfrac{\phi_m(\mathbf{r})\phi_m^*(\mathbf{r}')}{\lambda_m-\lambda}One eigenterm dominates and diverges; the Fredholm alternative demands ∫ϕm∗f=0\int\phi_m^* f=0 for a bounded solution.
Wedge of angle π/n\pi/n, integer nnG=∑k=02n−1(±) Gfree(r,Rkr′)G=\sum_{k=0}^{2n-1}(\pm)\,G_{\text{free}}(\mathbf{r},R_k\mathbf{r}')A finite rosette of 2n2n images closes the boundary exactly; non-integer π/α\pi/\alpha needs a Kontorovich–Lebedev or mode expansion.
High frequency, short wavelengthk→∞k\to\inftyRay sums become semiclassical paths while the mode sum needs many terms — the bridge to WKB and steepest descent.

Worked examples

1 · Charge above a grounded plane — force and induced charge (fully worked)

A point charge q=+1 nCq=+1\ \mathrm{nC} is held at height d=3 mmd=3\ \mathrm{mm} above a large grounded conducting plane. Find the force on it and the total charge it induces on the plane.

Step 1. Replace the plane by its image: a charge q′=−1 nCq'=-1\ \mathrm{nC} at depth dd below the surface, separation 2d=6 mm2d=6\ \mathrm{mm}. The half-space field is just these two charges, so the force on the real charge is its Coulomb attraction to the image, F=14πε0q2(2d)2=q216πε0d2F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{(2d)^2}=\dfrac{q^2}{16\pi\varepsilon_0 d^2}, directed toward the plane.

Step 2. Insert numbers with 1/4πε0=8.988×109 N m2/C21/4\pi\varepsilon_0=8.988\times10^9\ \mathrm{N\,m^2/C^2}:

F=(8.988×109)(10−9)2(6×10−3)2=8.988×10−93.6×10−5 N.F=\frac{(8.988\times10^9)(10^{-9})^2}{(6\times10^{-3})^2}=\frac{8.988\times10^{-9}}{3.6\times10^{-5}}\ \mathrm{N}.

This gives F = 2.5×10⁻⁴ N (0.25 mN), attractive. The force falls as 1/d21/d^2, so halving the gap quadruples the pull — the origin of the diverging adhesion a charge feels approaching a metal surface.

Step 3. The induced density σ(ρ)=−q d2π(ρ2+d2)3/2\sigma(\rho)=-\dfrac{q\,d}{2\pi(\rho^2+d^2)^{3/2}} peaks under the charge over a patch of size ∼d\sim d. Integrating over the plane,

Qind=∫0∞σ 2πρ dρ=−qd∫0∞ρ dρ(ρ2+d2)3/2=−qd⋅1d=−q.Q_{\text{ind}}=\int_0^\infty\sigma\,2\pi\rho\,\mathrm{d}\rho=-q d\int_0^\infty\frac{\rho\,\mathrm{d}\rho}{(\rho^2+d^2)^{3/2}}=-q d\cdot\frac1d=-q.

The plane draws up exactly the image charge, which is why the far field looks like a dipole. F = 2.5×10⁻⁴ N attractive, Q_ind = −1 nC.

2 · Kelvin image for a grounded sphere

A charge qq sits at distance r′=2ar'=2a from the centre of a grounded sphere of radius aa. Locate and size the image, and find the force on qq.

Step 1. Equation (5) gives the image charge q′=−ar′q=−a2aq=−q/2q'=-\dfrac{a}{r'}q=-\dfrac{a}{2a}q=\mathbf{-q/2}, placed at r ⁣∗′=a2r′2r′\mathbf{r}'_{\!*}=\dfrac{a^2}{r'^2}\mathbf{r}', i.e. at radial distance a2r′=a22a=a/2\dfrac{a^2}{r'}=\dfrac{a^2}{2a}=\mathbf{a/2} from the centre, inside the sphere. The image is weaker than the source and sits at the inversion point a2/r′a^2/r'.

Step 2. The force on qq is its attraction to the image, separated by r′−a2/r′=2a−a/2=3a/2r'-a^2/r'=2a-a/2=3a/2:

F=14πε0q (q/2)(3a/2)2=14πε0q2/29a2/4=29 q24πε0a2.F=\frac{1}{4\pi\varepsilon_0}\frac{q\,(q/2)}{(3a/2)^2}=\frac{1}{4\pi\varepsilon_0}\frac{q^2/2}{9a^2/4}=\frac{2}{9}\,\frac{q^2}{4\pi\varepsilon_0 a^2}.

So the charge is pulled toward the sphere with F = (2/9) q²/4πε₀a². Unlike the plane's pure 1/d21/d^2 law, the sphere's force interpolates between the far-field 1/r′41/r'^{4} behaviour (an induced-dipole attraction) and the near-surface 1/(gap)21/(\text{gap})^2 divergence as r′→ar'\to a.

Step 3. As a check, let the sphere grow, a→∞a\to\infty with the gap ℓ=r′−a\ell=r'-a held fixed: q′→−qq'\to-q and the image distance below the surface tends to ℓ\ell, recovering the flat-plane result of Example 1. The sphere image reduces to the plane mirror in the large-radius limit.

3 · A rectangular cavity and its resonances

For (−∇2−k2)(-\nabla^2-k^2) on a Dirichlet box [0,a]×[0,b][0,a]\times[0,b], write the Green's function and find where it blows up.

Step 1. The separable modes ϕmn=2absin⁡mπxasin⁡nπyb\phi_{mn}=\dfrac{2}{\sqrt{ab}}\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b} have eigenvalues λmn=(mπ/a)2+(nπ/b)2\lambda_{mn}=(m\pi/a)^2+(n\pi/b)^2, so the spectral formula (6) with λ=k2\lambda=k^2 gives

G=4ab∑m,n≥1sin⁡mπxasin⁡nπyb sin⁡mπx′asin⁡nπy′b(mπ/a)2+(nπ/b)2−k2.G=\frac{4}{ab}\sum_{m,n\ge1}\frac{\sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b}\,\sin\frac{m\pi x'}{a}\sin\frac{n\pi y'}{b}}{(m\pi/a)^2+(n\pi/b)^2-k^2}.

Step 2. The denominator vanishes as k2→λmnk^2\to\lambda_{mn}: the cavity rings at the resonant wavenumbers kmn=(mπ/a)2+(nπ/b)2k_{mn}=\sqrt{(m\pi/a)^2+(n\pi/b)^2} (lowest k11=π2/ak_{11}=\pi\sqrt2/a for a square box). A small loss k2→k2+iϵk^2\to k^2+\mathrm{i}\epsilon lifts each pole off the axis; summing one index in closed form converts the slow double sum into a fast ray-like series. Resonances at kmn=(mπ/a)2+(nπ/b)2k_{mn}=\sqrt{(m\pi/a)^2+(n\pi/b)^2} are the poles of the cavity Green's function.

Physics of the portrait

The whole apparatus of this article — the "function," the identities that manipulate it, the idea of solving a field problem by superposing point-source responses — descends from one self-taught Nottingham miller who published a single, almost unread pamphlet.

Origins · 1 minGeorge Green and how this was discovered
GG
George Green
1793–1841, Nottingham miller and mathematician · potential theory and boundary-value problems

George Green had barely a year of formal schooling. He worked his father's windmill, taught himself mathematics from library books, and in 1828 printed — by subscription, for a few dozen local buyers — An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism. In it he introduced the "potential function," the identities now called Green's first and second, and the object we call the Green's function: the field of a unit point source, from which the solution of a boundary-value problem is assembled by integration. The essay also held, in embryo, the reflection idea behind the method of images. It sank almost without trace. Green entered Cambridge as an undergraduate at forty and died in 1841, still obscure. Four years later a young William Thomson — the future Lord Kelvin — found the essay, grasped its depth, and had it republished; he built the method of electrical images and the sphere inversion (the "Kelvin transform") directly on Green's foundation. Within a generation Green's function had become the organizing concept of mathematical physics, from Maxwell's electromagnetism to the resolvents of quantum mechanics, and the miller's pamphlet one of the most consequential documents in the subject.

History
1828
— George Green's Essay introduces the potential function, Green's identities, and the Green's function
1845
— William Thomson rediscovers Green's essay and develops the method of electrical images
1848
— Thomson's sphere inversion (the Kelvin transform) solves the grounded-sphere problem
1877
— Rayleigh's Theory of Sound systematizes eigenfunction (normal-mode) expansions of cavities
1903
— Fredholm's integral-equation theory yields the alternative theorem for resolvents

Going further

Texts. Stone & Goldbart, Mathematics for Physics (2009), Ch. 11 (Green functions) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 10 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., Ch. 21 · Jackson, Classical Electrodynamics, 3rd ed., Ch. 2–3 (images and expansions) · Morse & Feshbach, Methods of Theoretical Physics, Vol. I, Ch. 7. Key results to keep in mind: the half-space image satisfies G∣∂Ω=0G|_{\partial\Omega}=0 and gives the plane force F=q2/16πε0d2F=q^2/16\pi\varepsilon_0 d^2; the Kelvin image is q′=−a/r′q'=-a/r' at a2/r′a^2/r', recovering the flat-plane result in the limit a→∞a\to\infty; the interval kernel is x<(L−x>)/Lx_<(L-x_>)/L, with peak 2L/92L/9 at x′=L/3x'=L/3, reproduced by both the sine series and the jump-matching construction; the box eigenvalues are λmn=(mπ/a)2+(nπ/b)2\lambda_{mn}=(m\pi/a)^2+(n\pi/b)^2 with resonance poles at kmn=λmnk_{mn}=\sqrt{\lambda_{mn}}; and the −∇2-\nabla^2 free-space kernels are +1/4πr+1/4\pi r (3D) and −(1/2π)ln⁡r-(1/2\pi)\ln r (2D). Numerical spot-checks: plane force 2.5×10−4 N2.5\times10^{-4}\ \mathrm{N} for q=1 nCq=1\ \mathrm{nC}, d=3 mmd=3\ \mathrm{mm}; induced charge −q-q; sphere force 29 q2/4πε0a2\tfrac29\,q^2/4\pi\varepsilon_0 a^2 at r′=2ar'=2a.

Builds on: Sturm–Liouville theory and self-adjoint operators · Distributions and the Dirac delta · Fourier and Bessel series · Electrostatics of conductors. Leads to: WKB Approximation · Method of Steepest Descent and Saddle Points · Bohr–Sommerfeld Quantization.

Ask the tutor: Ask me to work a specific geometry end-to-end — a charge near a grounded sphere, a slab's image train, or a cavity's resonances — and I'll build its Green's function both ways and show you which series converges faster.

Is the Green's function the same as the quantum propagator?

They are the same idea in different dress. The propagator K(r,t;r′,t′)K(\mathbf{r},t;\mathbf{r}',t') is the response of the time-dependent Schrödinger operator to a point disturbance in space and time — the amplitude to go from one event to another. The energy resolvent G(E)=∑n∣ϕn⟩⟨ϕn∣/(E−En)G(E)=\sum_n|\phi_n\rangle\langle\phi_n|/(E-E_n) is its Fourier transform in time, exactly formula (6) with λ=E\lambda=E: its poles sit at the bound-state energies and its imaginary part gives the density of states. "Green's function," "resolvent," and "propagator" name one object in different pictures — which you reach for depends on whether you want energies or trajectories.

Why is the sphere's image charge fractional, unlike the plane's exact −q?

Because the sphere curves away from the source, so different surface points are at different distances and a single equal-and-opposite charge cannot zero the potential everywhere. The inversion b=a2/r′b=a^2/r' places the image at the one point from which the ratio of surface distances is constant, and the reduced charge q′=−a/r′q'=-a/r' is the factor that then makes the two Coulomb potentials cancel at every angle. The plane is the degenerate limit a→∞a\to\infty: the ratio a/r′→1a/r'\to1, the image strength returns to −q-q, and the inversion becomes an ordinary mirror reflection. The fractional charge is the price of curvature, and it vanishes when the surface flattens.

If images and eigenfunction sums give the same G, why keep both?

Because they converge in opposite regimes, and convergence is the whole practical game. Close to the source, at early times, or at high frequency, the field is dominated by the direct term and its first few reflections, so a ray (image) sum reaches four-digit accuracy in one or two terms while the mode sum needs hundreds. Far from the source, at late times, or on a resonance, a few standing waves carry the field and the mode sum wins while the image lattice barely converges. Poisson summation and the Sommerfeld–Watson transform bridge the two — like keeping both a Taylor and a Laurent series for one function.

What goes wrong if I expand a Dirichlet problem in Neumann eigenfunctions?

You solve a different problem and get no warning. Formula (6) works only because the basis satisfies the same boundary conditions imposed on GG, making every term boundary-compatible and the completeness relation resolve the delta correctly on that domain. Use Neumann modes (zero normal derivative) to build a Dirichlet Green's function (zero value) and the sum is still complete and orthonormal — so it converges to something — but that something satisfies the wrong condition at the wall: it is the Neumann Green's function, not yours. No singularity, no divergence, no error flag, just an answer to a question you did not ask. Matching the basis to the boundary condition is the definition, not a technicality.

How can a 2D or 3D region ever be closed by finitely many images?

Only when the boundary is generated by a finite group of reflections or inversions. A single plane needs one mirror; two parallel planes need an infinite train, because the reflections never close; a wedge of opening angle π/n\pi/n with integer nn closes after 2n2n reflections (the dihedral group is finite); a sphere closes with one Kelvin inversion. The moment the geometry lacks such a symmetry — an irrational wedge, an ellipse, a general smooth cavity — the images multiply endlessly or fail to land consistently, and you fall back on the eigenfunction expansion, which asks nothing of the geometry beyond self-adjointness and completeness. Finitely many images is a gift of symmetry.

Does the sign convention for the free-space Green's function matter?

It matters for consistency, not for physics, so pick one and hold it. This article takes L\mathcal{L} positive (−∇2-\nabla^2), so −∇2G=δ-\nabla^2 G=\delta and the 3D kernel is +1/4πr+1/4\pi r — the ordinary Coulomb potential of a positive source, with positive eigenvalues. Other texts write ∇2G=δ\nabla^2 G=\delta or ∇2G=−4πδ\nabla^2 G=-4\pi\delta, flipping signs and prefactors; Jackson's Gaussian G=1/∣r−r′∣G=1/|\mathbf{r}-\mathbf{r}'| is the same physics in other units. The sign of the kernel, the image formulas, and the operator must all be chosen together. A −∇2-\nabla^2 label paired with a −1/4πr-1/4\pi r kernel gives a potential of the wrong sign — exactly the silent error the method should not tolerate.

Where this leads