advanced27 min studyLesson 236 in Physics

State the convention before computing a multipole

Connect normalized angular coefficients with physical charge moments and radial decay.

On this page

What you’ll understand

  • Use normalized complex spherical harmonics consistently.
  • Relate spherical coefficients to monopole and dipole moments.
  • Distinguish an exact exterior boundary expansion from a far-field truncation.

A multipole expansion separates angular shape from radial range. It becomes reliable only when its normalization and domain are explicit: a coefficient called a dipole moment in one convention may differ by a square root of four pi in another. We will use orthonormal complex spherical harmonics, recover the familiar charge and dipole terms, and solve a sphere-boundary problem without pretending that every exterior field is exactly a point dipole.

Choose an orthonormal angular basis

On the unit sphere, spherical harmonics are eigenfunctions of the angular Laplacian. Degree l is a nonnegative integer and order m runs from −l to +l. We use unit integral norm, the usual Condon–Shortley phase, and the conjugation relation between opposite m values. A real potential may use complex coefficients, provided they obey the corresponding conjugation condition. Complex notation does not turn the observable electrostatic potential into a complex physical quantity.

Separation of the charge-free spherical Laplace equation gives radial powers rˡ and r⁻ˡ⁻¹ for each harmonic. Regular interior solutions discard the inverse powers at the origin. Exterior solutions that decay at infinity discard growing powers. Neither choice applies automatically in a shell between two finite radii, where both branches may be needed. The domain and boundary conditions decide the allowed radial functions before coefficients are fitted.

∫Ylm∗Yl′m′ dΩ=δll′δmm′,Yl,−m=(−1)mYlm∗,Rl=Arl+Br−l−1\int Y_{lm}^{*}Y_{l^{\prime}m^{\prime}}\,d\Omega=\delta_{ll^{\prime}}\delta_{mm^{\prime}},\qquad Y_{l,-m}=(-1)^mY_{lm}^{*},\qquad R_l=Ar^l+Br^{-l-1}
Ylm is dimensionless and normalized over solid angle. The radial expression applies in a source-free spherical region; A and B carry the units needed for V in volts.↔ Scroll wide equations horizontally.

Source moments carry a normalization

For charges confined inside radius a, define the spherical moment by integrating charge density times rˡ and the conjugated harmonic. The exterior potential then has the displayed coefficient divided by permittivity and by 2l+1. This formula includes the factor of four pi through the normalization of the harmonics. Replacing them with unnormalized Legendre polynomials requires changing coefficients as well; one cannot swap angular functions while retaining the same numerical moments.

For l=0 the coefficient is total charge divided by the square root of four pi, recovering the ordinary Coulomb potential. For an axisymmetric dipole, the l=1,m=0 coefficient is the z-directed dipole moment times the square root of three over four pi. The leading nonzero multipole dominates far away, but dominance is an approximation unless all higher moments vanish. Changing the expansion origin changes higher moments; even the dipole depends on origin when net charge is nonzero.

qlm=∫r′lρ(r′)Ylm∗(Ω′) d3r′,V=1ϵ0∑l=0∞∑m=−llqlmYlm(Ω)(2l+1)rl+1q_{lm}=\int r^{\prime l}\rho(\mathbf r')Y_{lm}^{*}(\Omega')\,d^3r',\qquad V=\frac1{\epsilon_0}\sum_{l=0}^{\infty}\sum_{m=-l}^{l}\frac{q_{lm}Y_{lm}(\Omega)}{(2l+1)r^{l+1}}
qlm has units C·mˡ and this exterior source expansion converges for r>a containing every source. The stated conjugated-moment convention yields no further conjugation on qlm in V.↔ Scroll wide equations horizontally.

Boundary coefficients are a different object

If potential is prescribed on a sphere rather than charge density inside it, project the surface voltage onto the harmonics. Each surface coefficient then multiplies the appropriate exterior power of a/r. Those surface coefficients have volts as units; they are not the source moments, whose units depend on l. A boundary with only constant and cosine angular terms produces an exact monopole-plus-dipole exterior potential under the stated decay condition.

The spectral lab illustrates coefficient bookkeeping in a simpler basis. Reset and set N=2 and p=0; predict equal amplitudes of its two sine terms. Reset and set N=2 and p=2; predict that the second coefficient becomes one quarter of the first. Truncation changes angular or spatial detail in many expansions, but this instrument uses one-dimensional sin(nx−t), not spherical harmonics, multipole moments, or the radial powers derived here.

q00=Q4π,q10=34πpz,V(r,Ω)=∑lmflm(ar)l+1Ylm(Ω)q_{00}=\frac{Q}{\sqrt{4\pi}},\qquad q_{10}=\sqrt{\frac{3}{4\pi}}p_z,\qquad V(r,\Omega)=\sum_{lm} f_{lm}\left(\frac ar\right)^{l+1}Y_{lm}(\Omega)
Q is net charge in C, pz is dipole component in C·m, and flm=∫V(a,Ω)Y*lm dΩ is boundary coefficient in V. The last equation solves the decaying exterior Dirichlet problem.↔ Scroll wide equations horizontally.

A biased sphere with a dipolar pattern

On a sphere of radius a=0.20 m, prescribe V(a,θ)=4+6 cosθ volts. There are no exterior charges and V tends to zero at infinity. Find the exterior potential and its values at r=2a on the north axis and equator.

  1. The constant term belongs to degree zero and the cosθ term to degree one. In the normalized basis the surface coefficients are f₀₀=4√(4π) V and f₁₀=6√(4π/3) V. All other coefficients vanish.

  2. Use the decaying radial branch for each degree. On recombining the normalization factors, the exact exterior potential is 4(a/r)+6(a/r)² cosθ volts. It matches the prescribed sphere and tends to zero at infinity.

  3. At r=2a the monopole part is 2 V and the dipole amplitude is 1.5 V. The north-axis value is 3.5 V; the equatorial value is 2 V because cosθ vanishes there. These values refer to radius 0.40 m.

The result

The exterior solution is exactly 4a/r+6(a/r)²cosθ volts, giving 3.5 V on the north axis and 2 V at the equator when r=2a.

Explore the model

Interactive lab

Interactive lab · spectral

Track a coefficient convention

Reset and set N=2, p=0; predict equal mode amplitudes. Reset and set N=2, p=2; predict a second coefficient of 1/4 while the first stays 1.

Track a coefficient conventionA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe lab uses a one-dimensional finite sine series. It illustrates coefficient weighting, not spherical angular normalization or exterior radial decay.

Check your understanding

With qlm defined using the conjugated harmonic above, what are its units?

Try it yourself

An exterior sphere-boundary problem has V(a,θ)=3P₂(cosθ) volts, where P₂(u)=(3u²−1)/2. Find V at r=2a on the equator and compare its radial decay with a dipole.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

A pure degree-two boundary term acquires (a/r)³, and P₂(0)=−1/2.

Compare with the solution

At the requested radius the radial factor is 1/8, so V=3×(1/8)×(−1/2)=−3/16 V. A quadrupole potential decays as r⁻³, whereas a dipole potential decays as r⁻². Doubling radius reduces this quadrupole amplitude by eight, compared with four for a dipole.

Take it with you

  • Normalized spherical moments and surface-voltage coefficients have different units.
  • An exterior degree-l potential decays as r⁻ˡ⁻¹.
  • A far-field truncation needs sources inside the convergence sphere and a controlled scale ratio.

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