advanced32 min studyLesson 255 in Physics

Continue beyond the electric dipole

Use source symmetry and a controlled long-wavelength expansion to identify magnetic dipole and electric quadrupole emission.

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What you’ll understand

  • Identify magnetic dipole and traceless electric quadrupole moments.
  • Separate the long-wavelength expansion from the radiation-zone limit.
  • Explain how cancellation of the electric dipole exposes weaker multipoles.

A source with zero electric dipole moment need not be silent. A circulating current can have a changing magnetic moment, while a symmetric arrangement of charges can change shape without moving its electric dipole. These mechanisms appear at the next orders of the radiation expansion. The useful question is therefore which moments vary and which symmetry cancels them. Distance selects the radiation zone; source size relative to wavelength selects how many multipoles must be retained.

Organize the expansion by source phase

Start from the retarded vector-potential integral for a localized conserved source. At a remote observer, approximate the inverse distance by inverse observer radius while retaining the source-position correction to the retarded time. Expanding that correction produces powers of source size divided by reduced wavelength. The leading integral of current is tied to the electric dipole derivative. The next current moment splits into antisymmetric circulation and symmetric shape change, giving magnetic dipole and electric quadrupole terms.

All radiative multipoles have inverse-radius far fields. A quadrupole is not identified by an inverse-radius-to-the-fourth far radiation field; that familiar power belongs to static multipole fields. Suppression of higher radiative multipoles comes from extra powers of the small source phase and the structure of the moments. Relative estimates fail when a nominally leading moment vanishes by symmetry. In that case a formally smaller multipole can supply the entire leading radiation.

∣x−x′∣≃r−n⋅x′,tr≃t−rc+n⋅x′c,kℓ≪1,kr≫1,r≫ℓ|\mathbf x-\mathbf x^{\prime}|\simeq r-\mathbf n\cdot\mathbf x^{\prime},\qquad t_r\simeq t-\frac rc+\frac{\mathbf n\cdot\mathbf x^{\prime}}{c},\qquad k\ell\ll1,\quad kr\gg1,\quad r\gg\ell
ℓ bounds source size, r is observer distance in metres, k=ω/c, and n=x/r. A controlled short-source multipole truncation and a far-zone observation are independent requirements.↔ Scroll wide equations horizontally.

Circulation produces a magnetic dipole

The magnetic moment of a localized current is one half the integral of position crossed with current density. A small planar loop gives current times oriented area. The factor one half in the volume expression is essential; using it again on the loop formula halves the answer incorrectly. A steady loop has magnetic field but no magnetic dipole radiation. An oscillating moment radiates with the same sine-squared angular pattern as an electric dipole, with electric and magnetic polarization roles exchanged.

The total magnetic dipole power depends on the squared second time derivative of magnetic moment. In SI units its coefficient contains epsilon zero and speed to the fifth power, or equivalently mu zero divided by speed cubed. Comparing magnetic and electric dipoles requires comparing magnetic moment divided by c with electric moment. Their raw numerical moments have different units. This conversion also explains why a compact ordinary current loop often radiates weakly compared with a comparable separated-charge antenna.

m=12∫x′×J d3x′,mloop=IAn^loop,PMD=∣m¨∣26πϵ0c5\mathbf m=\frac12\int\mathbf x^{\prime}\times\mathbf J\,d^3x^{\prime},\qquad \mathbf m_{\mathrm{loop}}=IA\hat{\mathbf n}_{\mathrm{loop}},\qquad P_{\mathrm{MD}}=\frac{|\ddot{\mathbf m}|^2}{6\pi\epsilon_0c^5}
m is magnetic moment in A·m², I is current in amperes, and A is loop area in m². The instantaneous magnetic dipole power is the leading term for a compact vacuum source.↔ Scroll wide equations horizontally.

Shape change produces an electric quadrupole

Choose a traceless quadrupole convention explicitly because published coefficients differ with definitions. Here the moment weights three times the product of two coordinates minus radius squared times the identity tensor. Its units are coulomb square metres. Reflection symmetry can make the electric dipole zero while leaving this even moment nonzero. To radiate as a quadrupole the shape moment must vary with a nonzero third derivative, rather than merely have a large static value.

For example, equal positive charges at opposite positions on an axis, balanced by twice the negative charge at the origin, have zero net charge and zero electric dipole at every separation. Changing separation changes their quadrupole. Their leading electric dipole radiation cancels coherently; adding the separate charge powers would miss that cancellation. A wave-addition lab can demonstrate cancellation of the leading amplitudes but does not compute the residual quadrupole term, its polarization, or its angular pattern.

Qij=∫ρ(3xi′xj′−∣x′∣2δij) d3x′,EEQ=n×[n×(Q... n)]24πϵ0c3rQ_{ij}=\int\rho(3x_i^{\prime} x_j^{\prime}-|\mathbf x^{\prime}|^2\delta_{ij})\,d^3x^{\prime},\qquad \mathbf E_{\mathrm{EQ}}=\frac{\mathbf n\times[\mathbf n\times(\dddot{Q}\,\mathbf n)]}{24\pi\epsilon_0c^3r}
Q is symmetric and traceless with units C·m²; its third derivative and all moments are evaluated at t−r/c. This is the leading quadrupole radiation field with the stated Q convention in the long-wavelength far zone.↔ Scroll wide equations horizontally.

Compare a small loop with an electric dipole

A small loop has sinusoidal peak current 2 A and area 3.0×10⁻⁴ m². An electric dipole at the same angular frequency has peak moment 1.0×10⁻¹² C·m. With c=3.0×10⁸ m/s, compare their separate cycle-averaged dipole powers.

  1. The magnetic peak moment is current times area: m₀=6.0×10⁻⁴ A·m². No additional one-half factor belongs in this loop formula.

  2. Convert to the electric-moment scale by dividing by c. This gives m₀/c=2.0×10⁻¹² C·m, twice the electric peak moment.

  3. At the same frequency each average contains the same angular integral and sinusoidal one-half factor. The magnetic-to-electric power ratio is [m₀/(cp₀)]²=4. This compares separate sources, not their coherent combined pattern.

The result

The loop emits four times the electric dipole average power in their respective leading dipole approximations, provided both sources remain compact at the specified frequency.

Explore the model

Interactive lab

Interactive lab · waves

Cancel a leading amplitude

For fresh runs Reset and set BOTH phase 0°, ratio 1; then Reset and set phase 180°, ratio 1. Predict doubled and zero summed amplitudes. These are equal-frequency same-direction sinusoids: the lab illustrates cancellation, but has no source size or quadrupole field.

Cancel a leading amplitudeTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). 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 forSymmetry can cancel leading electric dipole amplitudes while a physical source retains higher multipoles.

Check your understanding

A compact symmetric source has identically zero electric dipole but an oscillating magnetic moment. What follows?

Try it yourself

Place +q at z=+a and z=−a and −2q at the origin. With q=2 nC and a=0.10 m, compute p and Qzz in the stated convention. How does Qzz change if a doubles?

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

Give me a hint

Use Qzz=Σq(3z²−r²); the central charge contributes zero to position moments.

Compare with the solution

The two electric dipole contributions cancel, so p=0. Each outer charge contributes 2qa², giving Qzz=4qa²=8.0×10⁻¹¹ C·m². Doubling a multiplies Qzz by four to 3.2×10⁻¹⁰ C·m². Static arrangements do not radiate; a suitably time-dependent separation can produce quadrupole emission while preserving zero electric dipole.

Take it with you

  • Radiative multipoles share inverse-radius far fields; source phase controls their hierarchy.
  • A magnetic loop has moment IA, and m/c is comparable in units to an electric moment.
  • Vanishing electric dipole radiation can expose magnetic dipole or quadrupole emission.

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