Chapter 4 — Electric Multipoles

Quote

“The electrical phenomena outside the sphere are identical with those arising from an imaginary series of singular points all at the center of the sphere.”

— James Clerk Maxwell (1891)

Motivation

Sub-atomic particles, atoms, molecules, and cells produce potentials that are often sampled far from the source — nuclei from electrons, molecules from detectors, cells from electrodes. This motivates us to develop machinery to systematically approximate the Coulomb integral:

$$\phi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\int {\mathrm{d}}^3{r}^{\prime }\frac{\rho ({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

Localization means $\rho$ is confined to a volume of linear dimension $R$, and the observation point satisfies $r\gg R$. The ratio $R/r$ is weighted by multipole moments encoding the geometry of the source.

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The Primitive Cartesian Multipole Expansion

For $\rho$ confined to a sphere of radius $R$ and $r^{\prime }<R\ll r$, Taylor-expand the Green function about the origin:

$$\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}=\frac{1}{r}-r_i^{\prime }{\nabla }_i\frac{1}{r}+\frac{1}{2}{r}_i^{\prime }{r}_j^{\prime }{\nabla }_i{\nabla }_j\frac{1}{r}-\cdots$$

Substituting into the Coulomb integral:

$$\phi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\left[\frac{Q}{r}+\frac{p_i{r}_i}{r^3}+Q_{ij}\frac{3r_i{r}_j-r^2{\delta }_{ij}}{r^5}+\cdots \right]$$

where the primitive Cartesian multipole moments are:

Moment	Definition	Point charges
Monopole $Q$	$\int {\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime })$	${\sum }_{\alpha }{q}_{\alpha }$
Dipole $\mathbf{p}$	$\int {\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime }){\mathbf{r}}^{\prime }$	${\sum }_{\alpha }{q}_{\alpha }{\mathbf{r}}_{\alpha }$
Quadrupole $Q_{ij}$	$\frac{1}{2}\int {\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime })r_i^{\prime }{r}_j^{\prime }$	$\frac{1}{2}{\sum }_{\alpha }{q}_{\alpha }{r}_{\alpha i}{r}_{\alpha j}$

Convergence and Dominance

Since $p_i\sim QR$ and $Q_{ij}\sim Q{R}^2$, each successive term is smaller by a factor $R/r\ll 1$. The asymptotic behavior is governed by the first non-vanishing term. For $Q\ne 0$, the object looks like a point charge at large distances.

The Complete Primitive Expansion

The full series can be written compactly as:

$$\phi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\sum _{\ell =0}^{\infty }{C}_{ij\cdots m}^{(\ell )}{N}_{ij\cdots m}^{(\ell )}(\mathbf{r})$$

where the $\ell$th-order primitive moment is

$$C_{ij\cdots m}^{(\ell )}=\frac{1}{\ell !}\int {\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime })r_i^{\prime }{r}_j^{\prime }\cdots r_m^{\prime }(\ell \text{ indices})$$

and

$$N_{ij\cdots m}^{(\ell )}(\mathbf{r})=(-1{)}^{\ell }{\nabla }_i{\nabla }_j\cdots {\nabla }_m\frac{1}{r}\text{.}$$

This works because the Taylor expansion of $1/|\mathbf{r}-{\mathbf{r}}^{\prime }|$ is exactly a power series in ${\mathbf{r}}^{\prime }$ with coefficients built from derivatives of $1/r$.

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Electric Dipole

The Dipole Potential and Field

The second term in the expansion dominates for most electrically neutral ($Q=0$) objects — neutrons, atoms, molecules, plasmas, ordinary matter:

$$\phi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\frac{\mathbf{p}\cdot \mathbf{r}}{r^3},r\gg R\text{.}$$

The same $\cos \theta /r^2$ structure governs the far field of a magnetic dipole — see Chapter 10 — with the replacement $\mathbf{p}/{ϵ}_0\to {\mu }_0\mathbf{m}$.

When Is $\mathbf{p}$ Well-Defined?

The dipole moment depends on the choice of origin when $Q\ne 0$: shifting the origin by $\mathbf{d}$ gives ${\mathbf{p}}^{\prime }=\mathbf{p}-Q\mathbf{d}$. For a neutral system ($Q=0$), $\mathbf{p}$ is origin-independent.

A system has a permanent dipole moment when $\mathbf{p}\ne 0$ in its ground state (e.g., water). An induced dipole moment arises from an external field (conductors, dielectrics).

The electric field follows from $\mathbf{E}=-\nabla \phi$. Choosing $\mathbf{p}=p\hat{\mathbf{z}}$ and using $\hat{\mathbf{z}}=\hat{\mathbf{r}}\cos \theta -\hat{\mathbf{\theta }}\sin \theta$:

$$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\frac{3\hat{\mathbf{r}}(\hat{\mathbf{r}}\cdot \mathbf{p})-\mathbf{p}}{r^3}=\frac{p}{4\pi {ϵ}_0{r}^3}\left(2\cos \theta \hat{\mathbf{r}}+\sin \theta \hat{\mathbf{\theta }}\right)$$

Field Lines and Circular Orbits

The field line equation $\mathrm{d}r/E_r=r\mathrm{d}\theta /E_{\theta }$ integrates to $r=k{\sin }^2\theta$.

The $1/r^3$ dependence and angle-dependence of the dipole field lead to non-central forces and non-trivial dynamics. To see this, decompose $\mathbf{E}$ into cylindrical components (with the dipole along $\hat{\mathbf{z}}$, using $\hat{\mathbf{r}}=\sin \theta \hat{\mathbf{\rho }}+\cos \theta \hat{\mathbf{z}}$ and $\hat{\mathbf{\theta }}=\cos \theta \hat{\mathbf{\rho }}-\sin \theta \hat{\mathbf{z}}$):

$$E_{\rho }=\frac{3p\sin \theta \cos \theta }{4\pi {ϵ}_0{r}^3},E_z=\frac{p(3{\cos }^2\theta -1)}{4\pi {ϵ}_0{r}^3}$$

A circular orbit around the dipole axis (in a plane perpendicular to $\hat{\mathbf{z}}$) requires the force to be purely centripetal — pointing only toward the axis, with no component that would push the particle up or down out of the orbital plane. This means we need $E_z=0$, which occurs at:

$$3{\cos }^2{\theta }_0=1⟹{\theta }_0\approx 54.7°$$

At this cone of angles, the field points purely in the $\pm \hat{\mathbf{\rho }}$ direction. For the right sign of charge (so the radial force is inward), this provides exactly the centripetal force needed for circular motion. An entire family of such orbits exists at different radii — all at the same polar angle ${\theta }_0$.

However, these orbits are only marginally stable: the total energy $U=\frac{1}{2}m{v}^2+q\phi =0$ for all of them (see Problem 4.5), so any perturbation causes the particle to drift to a different radius. Despite this marginal character, the quantum-mechanical version of this problem does yield true bound states when the dipole moment exceeds a critical value $\sim 1.625\text{D}$ — rationalizing the experimental fact that neutral polar molecules can capture electrons to form stable negative ions.

Example — Average Field Inside a Sphere

Let $\mathbf{E}(\mathbf{r})$ be the electric field produced by a charge density $\rho ({\mathbf{r}}^{\prime })$, and let $V$ be a sphere of radius $R$ centered at the origin. Then:

$$\frac{1}{V}{\int }_V{\mathrm{d}}^{3}r\mathbf{E}(\mathbf{r})=-\frac{{\mathbf{p}}_{\text{in}}}{3{ϵ}_{0}V}+{\mathbf{E}}_{\text{out}}(0)$$

where ${\mathbf{p}}_{\text{in}}$ is the dipole moment of the charge inside $V$ and ${\mathbf{E}}_{\text{out}}(0)$ is the field at the center due to charge outside $V$.

Proof: Start from:

$$\frac{1}{V}{\int }_V{\mathrm{d}}^{3}r\mathbf{E}(\mathbf{r})=\frac{1}{V}{\int }_V{\mathrm{d}}^{3}r\frac{1}{4\pi {ϵ}_0}\int {\mathrm{d}}^{3}s\rho (\mathbf{s})\frac{\mathbf{r}-\mathbf{s}}{|\mathbf{r}-\mathbf{s}{|}^3}$$

Reverse the order of integration and extract a minus sign:

$$=-\frac{1}{V}\int {\mathrm{d}}^{3}s\left[\frac{1}{4\pi {ϵ}_0}{\int }_V{\mathrm{d}}^{3}r\rho (\mathbf{s})\frac{\mathbf{s}-\mathbf{r}}{|\mathbf{s}-\mathbf{r}{|}^3}\right]$$

The quantity in brackets is the electric field at $\mathbf{s}$ produced by a uniformly charged sphere of density $\rho (\mathbf{s})$ filling $V$ (since $\rho (\mathbf{s})$ is constant with respect to the $\mathbf{r}$-integration). By Gauss’s law:

$${\mathbf{E}}_{\text{sphere}}(\mathbf{s})=\{\begin{array}{ll}\frac{\rho (\mathbf{s})}{3{ϵ}_0}\mathbf{s}& s<R\\ \frac{V\rho (\mathbf{s})}{4\pi {ϵ}_0}\frac{\mathbf{s}}{s^3}& s>R\end{array}$$

Substituting back and splitting the $\mathbf{s}$-integration into inside and outside $V$:

$$\frac{1}{V}{\int }_V{\mathrm{d}}^{3}r\mathbf{E}(\mathbf{r})=-\frac{1}{3{ϵ}_{0}V}\underset{{\mathbf{p}}_{\text{in}}}{\underbrace{{\int }_{s<R}{\mathrm{d}}^{3}s\rho (\mathbf{s})\mathbf{s}}}-\frac{1}{4\pi {ϵ}_0}{\int }_{s>R}{\mathrm{d}}^{3}s\rho (\mathbf{s})\frac{\mathbf{s}}{s^3}$$

The second integral is $-{\mathbf{E}}_{\text{out}}(0)$ (the field at the origin from exterior charges, with an extra minus sign from the one we extracted). Collecting signs:

$$\frac{1}{V}{\int }_V{\mathrm{d}}^{3}r\mathbf{E}(\mathbf{r})=-\frac{{\mathbf{p}}_{\text{in}}}{3{ϵ}_{0}V}+{\mathbf{E}}_{\text{out}}(0)$$

Two important special cases:

Condition	Result	Name
All charge inside $V$ (${\mathbf{E}}_{\text{out}}=0$)	$\mathbf{p}=-3{ϵ}_0{\int }_V{\mathrm{d}}^{3}r\mathbf{E}(\mathbf{r})$	Dipole moment from field average
No charge inside $V$ (${\mathbf{p}}_{\text{in}}=0$)	$\mathbf{E}(0)=\frac{1}{V}{\int }_V{\mathrm{d}}^{3}r\mathbf{E}(\mathbf{r})$	Mean value theorem for harmonic functions

The second case states that in a charge-free region, the field at any point equals its average over any surrounding charge-free sphere — a direct consequence of ${\nabla }^2\phi =0$ (each Cartesian component of $\mathbf{E}$ is itself harmonic). This is intimately connected to Earnshaw’s theorem and the result of Problem 4.21.

The Point Dipole

Maxwell’s construction: place $\pm q/s$ at the ends of a vector $s\mathbf{b}$ and let $s\to 0$ with $\mathbf{p}=q\mathbf{b}$ fixed. The resulting potential is exact at every point except ${\mathbf{r}}_0$:

$$\phi (\mathbf{r})=-\frac{1}{4\pi {ϵ}_0}\mathbf{p}\cdot \nabla \frac{1}{|\mathbf{r}-{\mathbf{r}}_0|}$$

The associated singular charge density follows from Poisson’s equation and ${\nabla }^2(1/|\mathbf{r}-{\mathbf{r}}_0|)=-4\pi \delta (\mathbf{r}-{\mathbf{r}}_0)$:

$${\rho }_D(\mathbf{r})=-\mathbf{p}\cdot \nabla \delta (\mathbf{r}-{\mathbf{r}}_0)\text{.}$$

Deriving ${\rho }_D$ by the Limiting Process

Start from $\rho =(q/s)[\delta (\mathbf{r}-{\mathbf{r}}_0-s\mathbf{b})-\delta (\mathbf{r}-{\mathbf{r}}_0)]$. Taylor-expand the first delta function: all terms beyond first order vanish as $s\to 0$, leaving ${\rho }_D=-\mathbf{p}\cdot \nabla \delta (\mathbf{r}-{\mathbf{r}}_0)$.

Verification: Insert ${\rho }_D$ into the Coulomb integral and integrate by parts. The gradient ${\nabla }^{\prime }$ acts on $1/|\mathbf{r}-{\mathbf{r}}^{\prime }|$; changing ${\nabla }^{\prime }\to -\nabla$ (since the function depends on $\mathbf{r}-{\mathbf{r}}^{\prime }$), we recover the exact point dipole potential.

The Complete Point Dipole Field (Including the Delta Function)

The field from the $1/r^3$ formula alone integrates to zero over any origin-centered sphere (by the angular symmetry of Figure 4.3). But Example 4.1 demands ${\int }_V{\mathrm{d}}^{3}r\mathbf{E}=-\mathbf{p}/(3{ϵ}_0)$. Consistency requires a contact term:

$$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\left[\frac{3\hat{\mathbf{n}}(\hat{\mathbf{n}}\cdot \mathbf{p})-\mathbf{p}}{|\mathbf{r}-{\mathbf{r}}_0{|}^3}-\frac{4\pi }{3}\mathbf{p}\delta (\mathbf{r}-{\mathbf{r}}_0)\right]$$

where $\hat{\mathbf{n}}=(\mathbf{r}-{\mathbf{r}}_0)/|\mathbf{r}-{\mathbf{r}}_0|$.

One can verify this directly: the delta function contributes $-\mathbf{p}/(3{ϵ}_0)$ to the volume integral, matching the required result.

Dipole Force, Torque, and Energy

Consider a neutral distribution $\rho ({\mathbf{r}}^{\prime })$ in an external field $\mathbf{E}({\mathbf{r}}^{\prime })$. For a point dipole, the Taylor expansion of $\mathbf{E}$ about $\mathbf{r}$ is exact (the source is truly at a point, so all higher terms vanish in Maxwell’s limit). For a spatially extended distribution, we require $\mathbf{E}$ to vary slowly over its extent.

Dipole Force

Expanding $\mathbf{E}({\mathbf{r}}^{\prime })$ to first order about $\mathbf{r}$ and using $Q=0$:

$$\mathbf{F}=\int {\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime })\mathbf{E}({\mathbf{r}}^{\prime })=(\mathbf{p}\cdot \nabla )\mathbf{E}(\mathbf{r})\text{.}$$

For constant $\mathbf{p}$ (permanent dipole), using $\nabla \times \mathbf{E}=0$:

$$\mathbf{F}=\nabla (\mathbf{p}\cdot \mathbf{E})\text{.}$$

Note: This form is not valid for induced moments where $\mathbf{p}\sim \mathbf{E}$ varies in space.

An electric field gradient is needed to generate a force on a dipole.

Dipole Torque

Using the same approach with ${\rho }_D$ in $\mathbf{N}=\int {\mathrm{d}}^3{r}^{\prime }{\mathbf{r}}^{\prime }\times \mathbf{E}({\mathbf{r}}^{\prime }){\rho }_D({\mathbf{r}}^{\prime })$:

$$\mathbf{N}=\mathbf{p}\times \mathbf{E}+\mathbf{r}\times \mathbf{F}\text{.}$$

The first term rotates $\mathbf{p}$ toward $\mathbf{E}$; the second rotates the center of mass about the origin.

Dipole Potential Energy

$$V_E(\mathbf{r})=\int {\mathrm{d}}^3{r}^{\prime }\phi ({\mathbf{r}}^{\prime }){\rho }_D({\mathbf{r}}^{\prime })=-\mathbf{p}\cdot \mathbf{E}(\mathbf{r})\text{.}$$

Checks: $\mathbf{F}=-\nabla V_E=\nabla (\mathbf{p}\cdot \mathbf{E})$ ✓. For torque: a rigid rotation $\delta \mathbf{p}=\delta \mathbf{\alpha }\times \mathbf{p}$ gives $\delta V_E=-(\mathbf{p}\times \mathbf{E})\cdot \delta \mathbf{\alpha }$, confirming $\mathbf{N}=\mathbf{p}\times \mathbf{E}$ ✓.

Dipole Interaction Energy

For $N$ point dipoles:

$$U_E=\frac{1}{4\pi {ϵ}_0}\frac{1}{2}\sum _{i\ne j}\left[\frac{{\mathbf{p}}_i\cdot {\mathbf{p}}_j}{|{\mathbf{r}}_i-{\mathbf{r}}_j{|}^3}-\frac{3({\mathbf{p}}_i\cdot {\hat{\mathbf{r}}}_{ij})({\mathbf{p}}_j\cdot {\hat{\mathbf{r}}}_{ij})}{|{\mathbf{r}}_i-{\mathbf{r}}_j{|}^3}\right]=-\frac{1}{2}\sum _i{\mathbf{p}}_i\cdot \mathbf{E}({\mathbf{r}}_i)$$

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Electric Dipole Layers

A dipole layer is a charge-neutral macroscopic surface $S$ endowed with a dipole moment per unit area:

$$\mathbf{\tau }=\frac{\mathrm{d}\mathbf{p}}{\mathrm{d}S}$$

Physical Occurrence

Dipole layers arise wherever mobile charges accumulate near surfaces or boundary layers. The canonical example is the surface of a metal, where electron wavefunctions spill out beyond the last layer of positive ions, creating a thin layer of separated charge.

The treatment is fully general: $S$ need not be flat, and $\mathbf{\tau }$ need not be perpendicular to the surface. The potential is a superposition of infinitesimal contributions $\mathrm{d}\mathbf{p}$:

$$\phi (\mathbf{r})=-\frac{1}{4\pi {ϵ}_0}{\int }_S\mathrm{d}S\mathbf{\tau }({\mathbf{r}}_S)\cdot {\nabla }^{\prime }\frac{1}{|\mathbf{r}-{\mathbf{r}}_S|}$$

Converting to a volume integral via a delta function (for a layer at $z=0$) and integrating by parts reveals the equivalent volume charge density:

$${\rho }_L(\mathbf{r})=-\nabla \cdot [\mathbf{\tau }({\mathbf{r}}_S)\delta (z)]=-{\tau }_z({\mathbf{r}}_S){\delta }^{\prime }(z)+\sigma ({\mathbf{r}}_S)\delta (z)$$

where the effective surface charge density arises from in-plane variation of the layer:

$$\sigma ({\mathbf{r}}_S)=-{\nabla }_S\cdot \mathbf{\tau }({\mathbf{r}}_S)\text{.}$$

Origin of $\sigma ({\mathbf{r}}_S)$

The negative end of each in-plane dipole cancels the positive end of its neighbor — unless there is a variation in magnitude or direction along the surface. The resulting incomplete cancellation is the source of $\sigma$.

Potential Discontinuity at a Dipole Layer

Key Matching Condition

The perpendicular component ${\tau }_z$ produces a jump in the potential across the layer:

$$\phi ({\mathbf{r}}_S,0^{+})-\phi ({\mathbf{r}}_S,0^{-})=\frac{{\tau }_z({\mathbf{r}}_S)}{{ϵ}_0}$$

Proof: Integrate ${ϵ}_0{\phi }^{\prime \prime }={\tau }_z{\delta }^{\prime }(z)$ once to get ${ϵ}_0{\phi }^{\prime }={\tau }_z\delta (z)$, then integrate from $0^{-}$ to $0^{+}$.

The parallel components produce a discontinuity in the tangential electric field:

$${\hat{\mathbf{n}}}_2\times [{\mathbf{E}}_1-{\mathbf{E}}_2]=\nabla \times \{{\hat{\mathbf{n}}}_2({\hat{\mathbf{n}}}_2\cdot \mathbf{\tau })\}/{ϵ}_0$$

Under certain conditions, this generates corrections to the Fresnel formulae.

Application — Monolayer Electric Dipole Drops

Rectangular drops of polar molecules (width $w$, length $L\gg w$) on a surface. Each row of dipoles (pointing out of the surface) is modeled as carrying dipole moment per unit length $\tau a$.

Chemical perimeter energy: $U_C\approx 2L\lambda =2\lambda A/w$.

Electrostatic energy: Sum pairwise row–row interactions. For two rows separated by $y$, $U(y)\approx {\tau }^2{a}^{2}L/(2\pi {ϵ}_0{y}^2)$. Summing over all $\sim w/a$ rows:

$$U_E\approx \frac{{\tau }^{2}L}{2\pi {ϵ}_0}\left(\frac{w}{a}-\ln \frac{w}{a}\right)$$

Minimizing $U_T=U_C+U_E$ at fixed area $A=Lw$:

$$w_0\approx a\exp \left(\frac{4\pi {ϵ}_0\lambda }{{\tau }^2}\right)$$

The width is independent of $L$ and depends exponentially on material parameters — characteristic of 2D dipole systems. Adding more molecules creates new drops of width $w_0$ rather than widening existing ones.

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Electric Quadrupole

The quadrupole term dominates when $Q=\mathbf{p}=0$. This applies to all atomic nuclei (parity invariance forces $\mathbf{p}=0$) and all homonuclear diatomic molecules (like N${}_2$, where inversion symmetry forbids both $Q\ne 0$ and $\mathbf{p}\ne 0$).

Why Do Molecules Have Dipole Moments Despite Parity?

In the body-fixed frame, the Hamiltonian is NOT parity-invariant (e.g., the two H atoms sit on one side of the O atom in water). The electron wavefunction does not have definite parity in this frame, so $\mathbf{p}\ne 0$. In the lab frame, rotational averaging over all orientations would restore parity — but at finite temperature, the thermal ensemble retains a net $\mathbf{p}$ in an external field.

Primitive Quadrupole Tensor

$$Q_{ij}=\frac{1}{2}\int {\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime })r_i^{\prime }{r}_j^{\prime }$$

This is origin-independent when $Q=\mathbf{p}=0$ (by the same shift argument as for $\mathbf{p}$). Despite the Cartesian indices, the integral need not be evaluated in Cartesian coordinates.

Symmetry and Principal Axes

$Q_{ij}=Q_{ji}$ (symmetric), so only 6 of 9 components are independent. One can always diagonalize $Q$ by rotating to the principal axis system $(x^{\prime },y^{\prime },z^{\prime })$: ${\mathbf{Q}}^{\prime }={\mathbf{U}}^{-1}\mathbf{QU}$ where $\mathbf{U}$ is unitary (built from eigenvectors of $\mathbf{Q}$, which exist since $\mathbf{Q}$ is real symmetric). In principal axes, only 3 numbers are needed — analogous to the moment of inertia tensor.

Maxwell's Construction of the Quadrupole and Intuition

Place dipole $+\mathbf{p}/s$ at ${\mathbf{r}}_0+s\mathbf{c}$ and $-\mathbf{p}/s$ at ${\mathbf{r}}_0$; take $s\to 0$:

$$\phi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}(\mathbf{c}\cdot \nabla )(\mathbf{p}\cdot \nabla )\frac{1}{r}$$

This gives $Q_{ij}=\frac{1}{2}(c_i{p}_j+c_j{p}_i)$ and ${\rho }_Q=Q_{ij}{\nabla }_i{\nabla }_j\delta (\mathbf{r})$.

Intuition: The simplest quadrupole to visualize is two anti-aligned dipoles: $+--+$ in a line. $Q_{kk}$ measures how stretched/squeezed the charge is along axis $k$; off-diagonal $Q_{ij}$ measures tilting/twisting.

Traceless Quadrupole Tensor

Using the identity $r_i^{\prime }{r}_j^{\prime }=\frac{1}{3}(3r_i^{\prime }{r}_j^{\prime }-r^{\prime 2}{\delta }_{ij})+\frac{1}{3}{r}^{\prime 2}{\delta }_{ij}$, the quadrupole potential can be rewritten using the symmetric, traceless tensor:

$${\Theta }_{ij}=\frac{1}{2}\int {\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime })(3r_i^{\prime }{r}_j^{\prime }-r^{\prime 2}{\delta }_{ij})=3Q_{ij}-Q_{kk}{\delta }_{ij}$$

The quadrupole potential simplifies to:

$$\phi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\frac{{\Theta }_{ij}{r}_i{r}_j}{r^5}$$

Why Traceless?

The constraint ${\Theta }_{xx}+{\Theta }_{yy}+{\Theta }_{zz}=0$ reduces the independent components from 6 to 5 — matching the number of spherical harmonics $Y_2^m$. In Maxwell’s construction, this corresponds to the freedom to choose $|\mathbf{c}|=|\mathbf{p}|$, reducing 6 parameters (two 3-vectors) to 5 (two unit vectors plus one shared magnitude, minus one).

Quadrupole Force, Torque, and Energy

$$\mathbf{F}=Q_{ij}{\nabla }_i{\nabla }_j\mathbf{E},\mathbf{N}=2(\mathsf{Q}\cdot \nabla )\times \mathbf{E}+\mathbf{r}\times \mathbf{F}$$ $$V_E(\mathbf{r})=-Q_{ij}{\nabla }_i{E}_j(\mathbf{r})=-\frac{1}{3}{\Theta }_{ij}{\nabla }_i{E}_j(\mathbf{r})$$

Example 4.2 — Traceless Quadrupole of a Uniform Ellipsoid

For an ellipsoid with volume $V$, uniform charge density $\rho$, and semi-axes $a_x,a_y,a_z$:

Trick: Scale variables $x^{\prime }=a_{x}u$, $y^{\prime }=a_{y}v$, $z^{\prime }=a_{z}w$ to convert the ellipsoid integral to a sphere integral. By the symmetry of the sphere, off-diagonal components vanish, and:

$${\Theta }_{zz}=\frac{Q}{5}(2a_z^2-a_x^2-a_y^2),\text{etc. (cyclic)}$$

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Spherical Multipole Expansion

The Cartesian expansion is poorly suited to spherical symmetry. Instead, use the binomial expansion of the Green function for $r^{\prime }<r$:

$$\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}=\frac{1}{r}{\left(1-2(\hat{\mathbf{r}}\cdot {\hat{\mathbf{r}}}^{\prime })\frac{r^{\prime }}{r}+\frac{r^{\prime 2}}{r^2}\right)}^{-1/2}$$

This is recognized as the generating function for Legendre polynomials:

$$\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}=\frac{1}{r}\sum _{\ell =0}^{\infty }{\left(\frac{r^{\prime }}{r}\right)}^{\ell }{P}_{\ell }(\hat{\mathbf{r}}\cdot {\hat{\mathbf{r}}}^{\prime }),r^{\prime }<r$$

Key properties of $P_{\ell }$:

Property	Formula
Orthogonality	${\int }_{-1}^1\mathrm{d}x{P}_{\ell }(x)P_{{\ell }^{\prime }}(x)=\frac{2}{2\ell +1}{\delta }_{\ell {\ell }^{\prime }}$
Completeness	${\sum }_{\ell =0}^{\infty }\frac{2\ell +1}{2}{P}_{\ell }(x)P_{\ell }(x^{\prime })=\delta (x-x^{\prime })$
First few	$P_0=1,P_1=x,P_2=\frac{1}{2}(3x^2-1)$

Example — Force Between Non-Overlapping Spherical Distributions

Two non-overlapping spherical charge distributions ${\rho }_1(\mathbf{r})$ (centered at origin) and ${\rho }_2({\mathbf{r}}^{\prime })$ (centered at $\mathbf{R}$), with ${\rho }_2$ exterior to ${\rho }_1$. By Gauss’s law, ${\rho }_2$ produces a point-charge potential outside itself.

$$V_E=\frac{Q_2}{4\pi {ϵ}_0}\int {\mathrm{d}}^{3}r\frac{{\rho }_1(\mathbf{r})}{|\mathbf{r}-\mathbf{R}|}$$

Expanding with $P_0(\cos \theta )=1$ and using orthogonality: only $\ell =0$ survives, giving $V_E=Q_1{Q}_2/(4\pi {ϵ}_{0}R)$. The force is Coulomb’s law — as if each distribution were a point charge at its center.

Spherical Harmonic Expansion

To separate the angular dependence of $\hat{\mathbf{r}}$ and ${\hat{\mathbf{r}}}^{\prime }$, apply the addition theorem:

$$P_{\ell }(\hat{\mathbf{r}}\cdot {\hat{\mathbf{r}}}^{\prime })=\frac{4\pi }{2\ell +1}\sum _{m=-\ell }^{\ell }{Y}_{\ell m}^{\ast }({\theta }^{\prime },{\varphi }^{\prime })Y_{\ell m}(\theta ,\varphi )$$

This yields the full angular expansion:

$$\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}=\sum _{\ell =0}^{\infty }\frac{4\pi }{2\ell +1}\frac{r_{<}^{\ell }}{r_{>}^{\ell +1}}\sum _{m=-\ell }^{\ell }{Y}_{\ell m}^{\ast }({\Omega }_{<})Y_{\ell m}({\Omega }_{>})$$

where $r_{<}=\min (r,r^{\prime })$, $r_{>}=\max (r,r^{\prime })$.

Key spherical harmonic properties:

Property	Formula
Orthonormality	$\int \mathrm{d}\Omega Y_{\ell m}^{\ast }{Y}_{{\ell }^{\prime }{m}^{\prime }}={\delta }_{\ell {\ell }^{\prime }}{\delta }_{m{m}^{\prime }}$
Completeness	${\sum }_{\ell ,m}{Y}_{\ell m}(\Omega )Y_{\ell m}^{\ast }({\Omega }^{\prime })=\frac{1}{\sin \theta }\delta (\theta -{\theta }^{\prime })\delta (\varphi -{\varphi }^{\prime })$
Conjugation	$Y_{\ell }^{-m}=(-1{)}^m{Y}_{\ell m}^{\ast }$
First few	$Y_0^0=\frac{1}{\sqrt{4\pi }},Y_1^0=\sqrt{\frac{3}{4\pi }}\cos \theta ,Y_1^{\pm 1}=\mp \sqrt{\frac{3}{8\pi }}\sin \theta e^{\pm i\varphi }$

Exterior and Interior Multipole Moments

Substituting into the Coulomb integral:

Exterior ($r>R$):

$$\phi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\sum _{\ell =0}^{\infty }\sum _{m=-\ell }^{\ell }\frac{A_{\ell m}{Y}_{\ell m}(\theta ,\varphi )}{r^{\ell +1}},A_{\ell m}=\frac{4\pi }{2\ell +1}\int {\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime })r^{\prime \ell }{Y}_{\ell m}^{\ast }({\theta }^{\prime },{\varphi }^{\prime })$$

Interior ($r<R$):

$$\phi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\sum _{\ell =0}^{\infty }\sum _{m=-\ell }^{\ell }{B}_{\ell m}{r}^{\ell }{Y}_{\ell m}^{\ast }(\theta ,\varphi ),B_{\ell m}=\frac{4\pi }{2\ell +1}\int {\mathrm{d}}^3{r}^{\prime }\frac{\rho ({\mathbf{r}}^{\prime })}{r^{\prime \ell +1}}{Y}_{\ell m}({\theta }^{\prime },{\varphi }^{\prime })$$

Relationship to Cartesian Moments

The information in $A_{1,0}$, $A_{1,\pm 1}$ is identical to that in $\mathbf{p}$; the five $A_{2m}$ carry the same information as ${\Theta }_{ij}$. Each $A_{\ell m}$ is a linear combination of the $2\ell +1$ traceless Cartesian components $T_{ij\cdots }^{(\ell )}$ and vice versa.

Azimuthal Symmetry

When $\rho (r^{\prime },{\theta }^{\prime })=\rho (r^{\prime },{\theta }^{\prime })$ (no ${\varphi }^{\prime }$-dependence), only $m=0$ survives:

$$\phi (r,\theta )=\frac{1}{4\pi {ϵ}_0}\sum _{\ell =0}^{\infty }\frac{A_{\ell }}{r^{\ell +1}}{P}_{\ell }(\cos \theta ),A_{\ell }=\int {\mathrm{d}}^3{r}^{\prime }{r}^{\prime \ell }\rho (r^{\prime },{\theta }^{\prime })P_{\ell }(\cos {\theta }^{\prime })$$ $$\phi (r,\theta )=\frac{1}{4\pi {ϵ}_0}\sum _{\ell =0}^{\infty }{B}_{\ell }{r}^{\ell }{P}_{\ell }(\cos \theta ),B_{\ell }=\int {\mathrm{d}}^3{r}^{\prime }\frac{\rho (r^{\prime },{\theta }^{\prime })}{r^{\prime \ell +1}}{P}_{\ell }(\cos {\theta }^{\prime })$$

Matching at a Spherical Surface

These are linear combinations of elementary solutions of Laplace’s equation. The coefficients $A_{\ell }$ and $B_{\ell }$ are chosen so the interior and exterior potentials satisfy the matching conditions at $r=R$.

Application — The Potential Produced by $\sigma (\theta )={\sigma }_0\cos \theta$

A spherical shell of radius $R$ with surface charge density $\sigma (\theta )={\sigma }_0\cos \theta$.

Using $P_1(x)=x=\cos \theta$ and orthogonality, only $\ell =1$ survives:

$$A_{\ell }=\frac{4\pi }{3}{R}^3{\sigma }_0{\delta }_{\ell ,1},B_{\ell }=\frac{4\pi }{3}{\sigma }_0{\delta }_{\ell ,1}$$ $$\phi (r,\theta )=\frac{{\sigma }_0}{3{ϵ}_0}\{\begin{array}{ll}r\cos \theta & r<R\\ R^3\cos \theta /r^2& r>R\end{array}$$

Interior: constant field $\mathbf{E}=-({\sigma }_0/3{ϵ}_0)\hat{\mathbf{z}}$. Exterior: exact dipole field. Both expressions agree at $r=R$ (continuity of $\phi$ through a charge layer).

Application — Liquid Drop Model of Nuclear Fission

Bohr and Wheeler (1939): a uniform-charge sphere undergoes a small shape distortion as a precursor to fission.

Claim A: The most general azimuthally symmetric quadrupole distortion preserving volume and center of mass:

$$R(\theta )=R_0\left(1-\frac{{\alpha }^2}{5}+\alpha P_2(\cos \theta )\right),\alpha \ll 1$$

Proof of Claim A $R(\theta )=R_0[1+{\alpha }_0{P}_0+{\alpha }_1{P}_1+{\alpha }_2{P}_2]$. The ${\alpha }_1$ term is a rigid translation (since $P_1=\cos \theta =z$ on the unit sphere, so $r=1+{\alpha }_{1}z$ gives a shifted sphere). Volume preservation via $\frac{4\pi }{3}{R}_0^3=\frac{2\pi }{3}{R}_0^3{\int }_{-1}^1\mathrm{d}x[1+{\alpha }_0+{\alpha }_2{P}_2(x){]}^3$ and Legendre orthogonality forces ${\alpha }_0=-{\alpha }_2^2/5$ to second order.

Start with

Claim B: The change in Coulomb energy is:

$$\Delta U_E=-\frac{1}{5}{U}_0{\alpha }^2,U_0=\frac{3}{5}\frac{Q^2}{4\pi {ϵ}_0{R}_0}$$

Proof of Claim B $\rho ={\rho }_0+{\rho }_1+{\rho }_2$ (expanding the step function $\Theta (R(\theta )-r)$) and $\phi ={\phi }_0+{\phi }_1+{\phi }_2$.

Write

${\phi }_0$ is the standard uniform-sphere potential. ${\rho }_1=\alpha \varrho R_0{P}_2(\cos \theta )\delta (R_0-r)$ acts as an effective surface charge, producing ${\phi }_1$ via interior/exterior $\ell =2$ Legendre expansions matched at $r=R_0$.

Green’s reciprocity eliminates cross-terms: $\int {\rho }_0{\phi }_1=\int {\rho }_1{\phi }_0$, etc. The three surviving energy integrals evaluate (using Legendre orthogonality) to:

$${\mathcal{U}}_0=0,{\mathcal{U}}_1=\frac{2\pi }{25{ϵ}_0}{\varrho }^2{R}_0^5{\alpha }^2,{\mathcal{U}}_2=-\frac{2\pi }{15{ϵ}_0}{\varrho }^2{R}_0^5{\alpha }^2$$

Summing and using $\varrho =Q/(4\pi R_0^3/3)$ gives $\Delta U_E=-U_0{\alpha }^2/5$.

Physical upshot: Surface tension opposes fission ($\Delta U_{\text{surface}}>0$); Coulomb energy favors it ($\Delta U_E<0$). Fission occurs when the nucleus is charged enough that the Coulomb term wins.

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Traceless Cartesian vs. Spherical: Counting Components

The primitive Cartesian moment $C^{(\ell )}$ is symmetric in its $\ell$ indices. The number of independent components is equal to the number in which the indices can values, regardless of order:

$$M=\frac{1}{2}(\ell +2)(\ell +1)$$

(equivalent to $\left(\genfrac{}{}{0ex}{}{\ell +3-1}{3-1}\right)$ by a bars and stars argument as we are choosing which of three possibilities will each of $\ell$ indices correspond to, i.e., placing $l$ indistinguishable objects into three distinguishable boxes).

TODO: The book argument seems wrong in saying that vanishing traces are independent.

The traceless tensor $T^{(\ell )}$ equivalent vanishing trace conditions ${\delta }_{ij}{T}_{ij\dots m}^{(\ell )}=0$ etc., which give the same constrain for any choice of indices, and taking any one of them, we get vanishing symmetric tensor of rank $\ell -2$, which results in $\frac{1}{2}(\ell -2+2)(\ell -2+1)=\frac{1}{2}\ell (\ell -1)$ constraints so:

$$\overline{M}=M-\frac{1}{2}\ell (\ell -1)=2\ell +1$$

Thus, beginning with the quadrupole term, the traceless moments represent the potential with increasing efficiency.

Irreducibility

The spherical expansion has $2\ell +1$ moments $A_{\ell m}$ per order. The traceless Cartesian expansion also has $2\ell +1$ components per order. Both representations are irreducible (maximally efficient). The primitive Cartesian representation is reducible — it carries redundant information for $\ell \ge 2$.

Maxwell’s generalization: the point $2^{\ell }$-pole potential is

$$\phi (\mathbf{r})=\frac{(-1{)}^{\ell }}{4\pi {ϵ}_0}(\mathbf{a}\cdot \nabla )(\mathbf{b}\cdot \nabla )\cdots (\mathbf{p}\cdot \nabla )\frac{1}{r}(\ell \text{ vectors with }|\mathbf{a}|=|\mathbf{b}|=\cdots =|\mathbf{p}|)$$

This has $2\ell +1$ independent parameters: 2 angles per vector (due to fixed magnitude) plus 1 shared magnitude.

The Polarization $\mathbf{P}(\mathbf{r})$ from Multipole Expansion

A collection of $N$ point charges has $\rho (\mathbf{r})={\sum }_{\alpha }{q}_{\alpha }\delta (\mathbf{r}-{\mathbf{r}}_{\alpha })$. Taylor-expanding each delta function about a common origin:

$$\rho (\mathbf{r})=Q\delta (\mathbf{r})-\nabla \cdot \mathbf{P}(\mathbf{r})$$

where $P_i(\mathbf{r})={\sum }_{\ell =1}^{\infty }(-1{)}^{\ell -1}{C}_{ij\cdots m}^{(\ell )}{\nabla }_j\cdots {\nabla }_m\delta (\mathbf{r})$. This supports the general decomposition $\rho ={\rho }_c-\nabla \cdot \mathbf{P}$ used in Section 2.4.1 and developed further in Ch. 6.

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Problems

Foundations

4.2 — Smoluchowski’s Model of a Metal Surface

A semi-infinite metal ($z=0$ as macroscopic surface) with positive charge density $n_{+}(z)=\bar{n}\Theta (-z)$ and negative charge density $n_{-}(z)=\bar{n}[\frac{1}{2}{e}^{-\kappa |z|}+\Theta (-z)(1-\frac{1}{2}{e}^{\kappa z})]$.

Solution Sketch

(a) $n_{+}$ models the sharp termination of positive ions at $z=0$. $n_{-}$ models electron wavefunctions spilling into vacuum beyond the last ion layer, with decay length $1/\kappa$.

(b) The net charge density is $\rho (z)=-\frac{\bar{n}}{2}\mathrm{sgn}(z)e^{-\kappa |z|}$. The dipole moment per unit area:

$$p_z={\int }_{-\infty }^{\infty }\mathrm{d}zz\rho (z)=-\frac{\bar{n}}{{\kappa }^2}$$

(c) By Gauss’s law ($\mathrm{d}E/\mathrm{d}z=\rho /{ϵ}_0$) and using $\mathrm{sgn}(z)=\mathrm{d}|z|/\mathrm{d}z$:

$$E(z)=-\frac{\bar{n}}{2{ϵ}_0\kappa }{e}^{-\kappa |z|},\phi (z)=-\frac{\bar{n}}{2{ϵ}_0{\kappa }^2}\mathrm{sgn}(z)(e^{-\kappa |z|}-1)$$

(d) $\phi (\infty )-\phi (-\infty )=-\bar{n}/({ϵ}_0{\kappa }^2)=p_z/{ϵ}_0$: the potential change across the double layer, spread over the full system rather than as a sudden jump.

(e) Total energy per unit area: $U_E=\frac{1}{2}\int \mathrm{d}z\rho \phi ={\bar{n}}^2/(8{ϵ}_0{\kappa }^3)$.

4.3 — Charge Density of a Point Dipole

The charge density of a point electric dipole with moment $\mathbf{p}$ at ${\mathbf{r}}_0$ is ${\rho }_D(\mathbf{r})=-\mathbf{p}\cdot \nabla \delta (\mathbf{r}-{\mathbf{r}}_0)$. (a) Derive this using a limiting process applied to two point charges $\pm q/s$ separated by $s\mathbf{b}$ as $s\to 0$. (b) Confirm by inserting ${\rho }_D$ into the Coulomb integral and recovering the dipole potential.

Solution Sketch

(a) From $\rho =(q/s)[\delta (\mathbf{r}-{\mathbf{r}}_0-s\mathbf{b})-\delta (\mathbf{r}-{\mathbf{r}}_0)]$, Taylor-expand to first order; terms $\propto s,s^2,\dots$ vanish as $s\to 0$:

$${\rho }_D(\mathbf{r})=-\mathbf{p}\cdot \nabla \delta (\mathbf{r}-{\mathbf{r}}_0)$$

(b) Insert into the Coulomb integral, integrate by parts:

$$\phi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\int {\mathrm{d}}^3{r}^{\prime }\frac{-\mathbf{p}\cdot {\nabla }^{\prime }\delta ({\mathbf{r}}^{\prime }-{\mathbf{r}}_0)}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}=\frac{1}{4\pi {ϵ}_0}\mathbf{p}\cdot {\nabla }^{\prime }\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}{|}_{{\mathbf{r}}^{\prime }={\mathbf{r}}_0}$$

Change ${\nabla }^{\prime }\to -\nabla$ (since $|\mathbf{r}-{\mathbf{r}}^{\prime }|$ depends on the difference), recovering the exact dipole potential.

4.4 — No Self-Force (Stress Tensor Proof)

Use the electric stress tensor formalism to prove that no isolated charge distribution $\rho (\mathbf{r})$ can exert a net force on itself. Distinguish the cases when $\rho$ has a net charge and when it does not.

Solution Sketch

Let $S$ enclose $\rho (\mathbf{r})$ entirely. The net force is $\mathbf{F}={ϵ}_0{\oint }_S\mathrm{d}\mathbf{S}\cdot [(\hat{\mathbf{n}}\cdot \mathbf{E})\mathbf{E}-\frac{1}{2}\hat{\mathbf{n}}{E}^2]$. Expand $S$ to infinity (permissible since no other charges exist). If $\rho$ has net charge, $E\sim 1/r^2$ so $\mathrm{d}S{E}^2\sim 1/r^2\to 0$. If $\rho$ is neutral, $E$ falls off even faster. Either way, $\mathbf{F}=0$.

4.5 — Point Charge Motion in a Dipole Field

Place a point electric dipole $\mathbf{p}=p\hat{\mathbf{z}}$ at the origin and release a point charge $q$ (initially at rest) from the point $(x_0,y_0,0)$ in the $x$-$y$ plane. Show that the particle moves periodically in a semicircular arc.

Solution Sketch

Only $L_z$ is conserved (torque $\mathbf{N}\propto \hat{\mathbf{r}}\times \hat{\mathbf{z}}$ vanishes only on $z$-axis). The problem specifies $L_z=0$ (charge starts at rest in the $x$-$y$ plane).

With $\dot{r}=0$ and $L_z=0$, the equation of motion in $\theta$ gives:

$${\dot{\theta }}^2=-\frac{2pq\cos \theta }{4\pi {ϵ}_{0}m{r}_0^4}$$

For $pq>0$: motion requires $\cos \theta <0$, so $\theta \in [\pi /2,3\pi /2]$ — a semicircular arc with turning points at the equatorial crossings.

Why exactly circular? Energy conservation $\frac{1}{2}m{v}^2=-qp\cos \theta /(4\pi {ϵ}_0{r}^2)$ and the centripetal condition $m{v}^2/r=2qp\cos \theta /(4\pi {ϵ}_0{r}^3)$ are identical, so $r=r_0$ is automatically maintained.

The total energy $U=0$ on this orbit, making it marginally unstable — any energy nudge causes drift to a different radius. For $L_z\ne 0$, circular orbits exist at ${\theta }_0=125.3°$ ($3{\cos }^2{\theta }_0=1$), but with $L_z$ independent of radius, these are also unstable.

4.6 — Work to Assemble a Point Dipole

Show that $\mathrm{d}W=-\mathbf{E}(\mathbf{r})\cdot \mathrm{d}\mathbf{p}$ is the work increment required to assemble a point electric dipole with moment $\mathrm{d}\mathbf{p}$ at $\mathbf{r}$, beginning with charge dispersed at infinity.

Solution Sketch

The work to bring $\mathrm{d}q$ to $\mathbf{r}$ and $-\mathrm{d}q$ to $\mathbf{r}+\mathbf{\delta }$:

$$\mathrm{d}W=\phi (\mathbf{r})\mathrm{d}q-\phi (\mathbf{r}+\mathbf{\delta })\mathrm{d}q=-\mathbf{\delta }\mathrm{d}q\cdot \nabla \phi (\mathbf{r})$$

With $\mathrm{d}\mathbf{p}=-\mathbf{\delta }\mathrm{d}q$ (in the limit $\mathrm{d}q\to \infty$, $\delta \to 0$, product finite):

$$\mathrm{d}W=-\mathbf{E}(\mathbf{r})\cdot \mathrm{d}\mathbf{p}$$

Source: A.M. Portis, Electromagnetic Fields (Wiley, 1978).

4.7 — Dipoles at Vertices of Platonic Solids

Identical point electric dipoles are placed at the vertices of each of the five regular polyhedra. All dipoles are parallel (but the common direction is arbitrary). Show that the electric field at the center of each polyhedron is zero.

Solution Sketch

All five Platonic solids have cubic or icosahedral point group symmetry. The only vector invariant under all rotations of such a group is $0$. Since the dipole field at the center is a vector that must be invariant under all symmetry operations of the polyhedron (which permute identical dipoles), it must vanish.

Alternatively (constructive): For the tetrahedron, verify by direct computation. The cube is the union of two interlocking tetrahedra, so ${\mathbf{E}}_{\text{cube}}={\mathbf{E}}_{\text{tet1}}+{\mathbf{E}}_{\text{tet2}}=0$. The octahedron, dodecahedron, and icosahedron follow from their symmetry groups containing the tetrahedral group.

4.9 — Potential of a Double Layer

(a) Show that the potential due to a double-layer surface $S$ with dipole density $\mathbf{\tau }({\mathbf{r}}_S)=\tau ({\mathbf{r}}_S)\hat{\mathbf{n}}$ is $\phi (\mathbf{r})=-\frac{1}{4\pi {ϵ}_0}{\int }_S\tau ({\mathbf{r}}_S)\mathrm{d}\Omega$, where $\mathrm{d}\Omega$ is the solid angle element subtended by $\mathrm{d}S$ as seen from $\mathbf{r}$. (b) Use this to derive the matching condition at a double-layer surface.

Solution Sketch

(a) Starting from $\phi (\mathbf{r})=-\frac{1}{4\pi {ϵ}_0}{\int }_S\mathrm{d}S\tau ({\mathbf{r}}_S)\hat{\mathbf{n}}\cdot {\nabla }^{\prime }\frac{1}{|\mathbf{r}-{\mathbf{r}}_S|}$, recognize $-\hat{\mathbf{n}}\cdot {\nabla }^{\prime }(1/|\mathbf{r}-{\mathbf{r}}_S|)\mathrm{d}S=\mathrm{d}\Omega$:

$$\phi (\mathbf{r})=-\frac{1}{4\pi {ϵ}_0}{\int }_S\tau ({\mathbf{r}}_S)\mathrm{d}\Omega$$

(b) For uniform $\tau$: approaching from one side, the surface subtends $+2\pi$ solid angle; from the other, $-2\pi$. The jump is:

$$\Delta \phi =\frac{\tau }{{ϵ}_0}$$

Multipole Moments and Symmetry

4.12 — Two Neutral Disks

Two identical, charge-neutral, origin-centered disks with radially symmetric charge density. One lies in the $x$-$z$ plane; the other is tipped away from the first by angle $\alpha$ about the $z$-axis. Find $\alpha$ such that the asymptotic potential in the $x$-$y$ plane has the form $\phi (x,y)=A/s^3$ where $s=\sqrt{x^2+y^2}$.

Solution Sketch

$Q=0$ for each disk. $\mathbf{p}=0$ because the charge density is radially symmetric (any integral $\int \mathrm{d}S\sigma (s){\hat{\mathbf{r}}}_S$ vanishes by azimuthal averaging). The first nonzero contribution is quadrupole.

The $1/s^3$ dependence in the $x$-$y$ plane is the quadrupole signature. For $\phi$ to be independent of angle in the $x$-$y$ plane (i.e., $\phi =A/s^3$ with no angular modulation), the combined quadrupole must be isotropic in that plane.

Let $\theta$ be the polar angle for the horizontal disk and ${\theta }^{\prime }$ for the tipped disk. The total quadrupole potential goes as $3({\cos }^2\theta +{\cos }^2{\theta }^{\prime })-2$. This is angle-independent when ${\cos }^2\theta +{\cos }^2{\theta }^{\prime }=\text{const}$, which requires ${\theta }^{\prime }=\theta +\pi /2$. Hence $\alpha =\pi /2$.

Intuition: Two perpendicular quadrupoles cancel each other’s angular variation, just as two perpendicular dipoles would create a circularly symmetric field pattern.

Source: J.A. Greenwood, British Journal of Applied Physics 17, 1621 (1966).

4.13 — Interaction Energy of Adsorbed Molecules

Molecules adsorbed on a crystal surface at low temperature occupy the centers of an $a\times a$ checkerboard. The orientation of each molecule is set by the lowest-order electrostatic interaction with its neighbors. (a) CO has a small dipole moment $p$. For the arrangement parameterized by angle $\alpha$ (all dipoles in-plane, tilted by $\alpha$ from one lattice direction), find $\alpha$ that minimizes the total energy considering each dipole’s 8 nearest neighbors, and show the energy/dipole is

$$U=\frac{p^2}{8\pi {ϵ}_0{a}^3}(1/\sqrt{2}-6)$$

(b) N${}_2$ has a small quadrupole moment (from electron buildup in the bond region). Sketch the expected orientational order.

Solution Sketch

(a) Evaluating the dipole–dipole interaction $U_{12}=\frac{1}{4\pi {ϵ}_0}\frac{{\mathbf{p}}_1\cdot {\mathbf{p}}_2-3({\mathbf{p}}_1\cdot \hat{\mathbf{n}})({\mathbf{p}}_2\cdot \hat{\mathbf{n}})}{|{\mathbf{r}}_1-{\mathbf{r}}_2{|}^3}$ for 4 nearest neighbors (distance $a$) and 4 next-nearest (distance $\sqrt{2}a$):

$$U_{nn}=-6,U_{nnn}=+2$$

Total energy per dipole (with $1/2$ for double counting):

$$U_t=\frac{1}{8\pi {ϵ}_0}\frac{p^2}{a^3}\left(\frac{1}{\sqrt{2}}-6\right)$$

The energy is independent of $\alpha$ — the angle drops out entirely. This is analogous to how a ferromagnet’s exchange energy depends on spin alignment but not on the global magnetization direction.

(b) For N${}_2$ (quadrupole: $+-+$ charge pattern), maximizing nearest-neighbor Coulomb attraction suggests a herringbone pattern with alternating molecular orientations at $90°$ to each other.

Source: V.M. Rozenbaum and V.M. Ogenko, Soviet Physics Solid State 26, 877 (1984).

4.15 — Quadrupole Moments: Cartesian Evaluation Only

Show explicitly that the primitive quadrupole moment $Q_{ij}=\frac{1}{2}\int {\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime })r_i^{\prime }{r}_j^{\prime }$ is defined using Cartesian components $r_i^{\prime }$, and that the integrals must be evaluated in Cartesian coordinates even if the final result is desired in another coordinate system.

Solution Sketch

The definition $Q_{ij}=\frac{1}{2}\int {\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime })r_i^{\prime }{r}_j^{\prime }$ uses Cartesian components $r_i^{\prime }$, so the integrals must be evaluated in Cartesian coordinates (where $r_x^{\prime }=x^{\prime }$, etc.). To express results in spherical coordinates, simply substitute $\hat{\mathbf{x}}\hat{\mathbf{y}}\to$ their spherical representations in $Q_{ij}$, after computing the integrals in Cartesian form.

4.16 — Properties of a Point Electric Quadrupole

The singular charge density of a point quadrupole at ${\mathbf{r}}_0$ is ${\rho }_Q(\mathbf{r})=Q_{ij}{\nabla }_i{\nabla }_j\delta (\mathbf{r}-{\mathbf{r}}_0)$. (a) Verify this produces the correct quadrupole potential via the Coulomb integral. (b) Find the force $\mathbf{F}$ on ${\rho }_Q$ in an external field $\mathbf{E}$. (c) Find the torque $\mathbf{N}$. (d) Find the interaction energy $V_E$.

Solution Sketch

(a) Insert ${\rho }_Q$ into the Coulomb integral and integrate by parts twice. Each ${\nabla }^{\prime }$ passes through the delta function onto $1/|\mathbf{r}-{\mathbf{r}}^{\prime }|$, recovering the quadrupole potential.

(b)

$$\mathbf{F}=\int {\mathrm{d}}^{3}r{\rho }_Q\mathbf{E}=Q_{ij}{\nabla }_i{\nabla }_j\mathbf{E}({\mathbf{r}}_0)$$

(c) $N_k={ϵ}_{ijk}{Q}_{m\ell }\int {\mathrm{d}}^{3}r\delta (\mathbf{r}-{\mathbf{r}}_0){\partial }_m{\partial }_{\ell }(r_j{E}_k)$. Expanding the derivatives and using

Q_{mj} = Q_{jm}$: $\mathbf{N} = 2(\mathsf{Q}\cdot\nabla)\times\mathbf{E} + \mathbf{r}\times\mathbf{F}

(d) $V_E=\int {\mathrm{d}}^{3}r{\rho }_Q\phi =-Q_{ij}{\nabla }_i{E}_j({\mathbf{r}}_0)$.

4.17 — Interaction Energy of Nitrogen Molecules

How does the leading contribution to the electrostatic interaction energy between two N${}_2$ molecules depend on the distance $R$ between them?

Solution Sketch

Each N${}_2$ is a quadrupole ($Q=0$, $\mathbf{p}=0$): $\phi \sim 1/R^3$, $\mathbf{E}\sim 1/R^4$, $\nabla E\sim 1/R^5$.

$V_E=-Q_{ij}{\nabla }_i{E}_j\sim 1/R^5$. The interaction energy falls off as $1/R^5$.

Alternatively: The potential of molecule A is ${\phi }_A\propto Q_{ij}^A{\nabla }_i{\nabla }_j(1/|\mathbf{r}-{\mathbf{r}}_A|)$. The charge density of B is ${\rho }_B=Q_{km}^B{\nabla }_k{\nabla }_m\delta (\mathbf{r}-{\mathbf{r}}_B)$. The interaction $\int {\rho }_B{\phi }_A$ involves 4 derivatives on $1/|{\mathbf{r}}_A-{\mathbf{r}}_B|$, giving $1/R^5$.

4.18 — A Black Box of Charge

A charge distribution inside a black box has all exterior multipole moments for $\ell =1,2,\dots$ equal to zero (in coordinates centered on the box). This does not imply spherical symmetry. Prove by constructing a counterexample.

Solution Sketch

Counterexample: Place a point charge $q$ at the center (gives only $\ell =0$). Surround it by a concentric spherical shell of total charge $-q$ uniformly distributed — this “point-plus-shell” (PPS) object produces zero field outside itself. Now place any number of PPS objects at arbitrary positions inside the box. The resulting charge distribution is non-spherically symmetric, yet all exterior multipole moments for $\ell \ge 1$ vanish.

Alternative: Two concentric shells with $\sigma (\theta )={\sigma }_{0,i}\cos \theta$ and radii $R_1\ne R_2$, with amplitudes chosen so ${\sigma }_{0,1}{R}_1^3+{\sigma }_{0,2}{R}_2^3=0$. Each shell only produces $\ell =1$; canceling these gives a non-spherically symmetric distribution with all exterior moments vanishing — yet the internal charge arrangement is manifestly not spherically symmetric.

4.20 — Practice with Spherical Multipoles

A spherical shell of radius $R$ carries surface charge density $\sigma (\theta ,\varphi )={\sigma }_0\sin \theta \cos \varphi$. (a) Evaluate the exterior spherical multipole moments. (b) Write $\phi (r>R)$ in the form $\phi (x,y,z,r)$. (c)–(d) Repeat for interior moments and potential. (e) Check the matching conditions at $r=R$. (f) Extract the dipole moment $\mathbf{p}$.

Solution Sketch

Key step: Rewrite the charge density in terms of spherical harmonics. Since $\sin \theta \cos \varphi =\frac{1}{2}\sin \theta (e^{i\varphi }+e^{-i\varphi })$, and $Y_1^{\pm 1}=\mp \sqrt{3/(8\pi )}\sin \theta e^{\pm i\varphi }$:

$$\sigma ={\sigma }_0\sqrt{\frac{2\pi }{3}}(-Y_1^1+Y_1^{-1})$$

Only $\ell =1$, $m=\pm 1$ contribute.

(a) Exterior moments: $A_{1,\pm 1}=\frac{4\pi }{3}{R}^3\int \mathrm{d}\Omega \sigma Y_{1,\pm 1}^{\ast }=\mp \frac{4\pi }{3}{\sigma }_0{R}^3\sqrt{\frac{2\pi }{3}}$.

(b) Exterior potential: $\phi (r>R)=\frac{{\sigma }_0{R}^3}{3{ϵ}_0}\frac{\sin \theta \cos \varphi }{r^2}=\frac{{\sigma }_0{R}^3}{3{ϵ}_0}\frac{x}{r^3}$.

(c)–(d) Interior moments: same procedure with $r^{\ell }$ instead of $r^{-(\ell +1)}$, giving $\phi (r<R)=\frac{{\sigma }_0}{3{ϵ}_0}r\sin \theta \cos \varphi =\frac{{\sigma }_0}{3{ϵ}_0}x$.

(e) Both agree at $r=R$ (continuity ✓). The normal derivative jump gives $-{ϵ}_0({\partial }_r{\phi }_{\text{out}}-{\partial }_r{\phi }_{\text{in}}){|}_R={\sigma }_0\sin \theta \cos \varphi$ ✓.

(f) The exterior potential $\propto x/r^3=\mathbf{p}\cdot \hat{\mathbf{r}}/(4\pi {ϵ}_0{r}^2)$ identifies $\mathbf{p}=\frac{4\pi }{3}{\sigma }_0{R}^3\hat{\mathbf{x}}$.

4.21 — Mean Value Theorem via Interior Multipole Expansion

Let $V$ be a charge-free volume of space. Use an interior spherical multipole expansion to show that the average of $\phi (\mathbf{r})$ over the surface of any spherical sub-volume inside $V$ equals the potential at the center of that sub-volume.

Solution Sketch

Choose the origin at the center of a charge-free spherical sub-volume of radius $R$. All source charge is outside, so an interior expansion applies:

$$\phi (r,\theta ,\varphi )=\frac{1}{4\pi {ϵ}_0}\sum _{\ell ,m}{B}_{\ell m}{r}^{\ell }{Y}_{\ell m}^{\ast }(\theta ,\varphi )$$

Average over the sphere’s surface using $Y_0^0=1/\sqrt{4\pi }$ and orthonormality:

$$⟨\phi {⟩}_S=\frac{1}{4\pi }\int \mathrm{d}\Omega \phi (R,\Omega )=\frac{1}{4\pi {ϵ}_0}\sum _{\ell ,m}{B}_{\ell m}{R}^{\ell }\int \frac{\mathrm{d}\Omega }{4\pi }{Y}_{\ell m}^{\ast }=\frac{1}{4\pi {ϵ}_0}\frac{B_{00}}{\sqrt{4\pi }}$$

Only $\ell =0$, $m=0$ survives. But $\frac{B_{00}}{\sqrt{4\pi }}\cdot \frac{1}{4\pi {ϵ}_0}=\phi (0)$, so $⟨\phi {⟩}_S=\phi (0)$: the average over any charge-free sphere equals the value at its center.

4.23 — Exterior Multipoles for Specified Potential on a Sphere

(a) Let $\phi (R,\theta ,\varphi )$ be specified on a sphere of radius $R$. Show that the exterior potential is

$$\phi (\mathbf{r})=\sum _{\ell ,m}(R/r{)}^{\ell +1}{Y}_{\ell m}(\Omega )\int \mathrm{d}{\Omega }^{\prime }\phi (R,{\Omega }^{\prime })Y_{\ell m}^{\ast }({\Omega }^{\prime })$$

(b) The eight octants of a spherical shell are held at alternating potentials $\pm V$. Find the asymptotic form of the exterior potential.

Solution Sketch

(a) The general exterior expansion is $\phi (r,\Omega )={\sum }_{\ell ,m}{A}_{\ell m}{Y}_{\ell m}(\Omega )/r^{\ell +1}$. At $r=R$:

$$\phi (R,\Omega )=\sum _{\ell ,m}\frac{A_{\ell m}}{R^{\ell +1}}{Y}_{\ell m}(\Omega )$$

Orthonormality gives $A_{\ell m}=R^{\ell +1}\int \mathrm{d}{\Omega }^{\prime }\phi (R,{\Omega }^{\prime })Y_{\ell m}^{\ast }({\Omega }^{\prime })$, so:

$$\phi (\mathbf{r})=\sum _{\ell =0}^{\infty }\sum _{m=-\ell }^{\ell }{\left(\frac{R}{r}\right)}^{\ell +1}{Y}_{\ell m}(\Omega )\int \mathrm{d}{\Omega }^{\prime }\phi (R,{\Omega }^{\prime })Y_{\ell m}^{\ast }({\Omega }^{\prime })$$

(b) Eight octants at alternating $\pm V$. By inspection of the symmetries:

• Potential flips sign under $\varphi \to \varphi +\pi /2$ → $m=\pm 2,\pm 6,\dots$
• Potential flips sign under $\theta \to \pi -\theta$ → $\ell$ must be odd
• Minimum $\ell$ with $|m|=2$: $\ell =3$

Check: $Y_3^{\pm 2}\propto {\sin }^2\theta \cos \theta e^{\pm 2i\varphi }$ flips sign under both $\theta \to \pi -\theta$ and $\varphi \to \varphi +\pi /2$ ✓.

The asymptotic potential is:

$$\phi (\mathbf{r})\sim V{\left(\frac{R}{r}\right)}^4{\sin }^2\theta \cos \theta \cos 2\varphi$$

Key insight: Exact numerical coefficients are tedious, but the symmetry analysis alone determines which harmonics appear. The leading term falls off as $1/r^4$.

4.25 — Analyze This Potential

An asymptotic potential has the form $\phi (r,\theta ,\varphi )=A/r+B{x}^2/r^5+\text{higher-order terms}$. (a) Use the traceless Cartesian expansion to show no localized source can produce this. (b) Repeat using the primitive Cartesian expansion. (c) Show that a suitable $\rho (\mathbf{r})$ can be found if we drop the requirement that the source is localized.

Solution Sketch

(a) Traceless Cartesian: The quadrupole term is ${\Theta }_{ij}{r}_i{r}_j/r^5$. Having only $B{x}^2/r^5$ requires ${\Theta }_{xx}\ne 0$ with ${\Theta }_{yy}={\Theta }_{zz}=0$ and all off-diagonals zero. But the traceless condition ${\Theta }_{xx}+{\Theta }_{yy}+{\Theta }_{zz}=0$ then forces ${\Theta }_{xx}=0$ — contradiction. No localized source exists.

(b) Primitive Cartesian: The quadrupole term is $Q_{ij}(3r_i{r}_j-r^2{\delta }_{ij})/r^5$. Setting off-diagonals to zero and demanding only $x^2/r^5$ appear:

$$\text{coefficient of }{x}^2:2Q_{xx}-Q_{yy}-Q_{zz},y^2:2Q_{yy}-Q_{xx}-Q_{zz},z^2:2Q_{zz}-Q_{xx}-Q_{yy}$$

Requiring the $y^2$ and $z^2$ coefficients to vanish gives two equations whose only solution is $Q_{xx}=Q_{yy}=Q_{zz}$, which then makes the $x^2$ coefficient zero too. Same conclusion: impossible.

(c) If $\rho$ extends to infinity, we can always find a source via Poisson’s equation: $\rho (\mathbf{r})=-{ϵ}_0{\nabla }^2\phi (\mathbf{r})$. The localization requirement is what makes it impossible — the traceless constraint is a direct consequence of ${\nabla }^2(1/r)=0$ for $r\ne 0$, which underpins all multipole expansions of localized sources.