Magnetic Dipole and Electric Quadrupole Radiation

The E9 chapter told the silver-medal radiation story twice over, at Griffiths register: Magnetic Dipole Radiation built the M1 fields from the retarded potential of an oscillating loop, and Electric Quadrupole Radiation proved the beautiful theorem that M1 and E2 are twins — the symmetric and antisymmetric halves of one and the same first retardation correction, both suppressed by $(d/\lambda)^2$ when the electric dipole channel is open. This article is the Jackson-register sequel, and it owns two upgrades those treatments deliberately deferred. First, exactness for M1. The magnetic dipole is the one higher multipole whose all-zone fields cost nothing new: electromagnetic duality maps the exact electric-dipole solution of the previous article, Electric Dipole Radiation: Exact Fields and Zone Structure, onto the exact M1 fields by the substitution $\mathbf p\to\mathbf m/c$ with $\mathbf E$ and $\mathbf B$ exchanging roles. One page of symmetry replaces a page of curls — and the polarization swap it predicts is measurable. Second, generality for E2. The E9 treatment worked the axially symmetric quadrupole and its four-lobe pattern. Here we build the machinery for an arbitrary traceless tensor $Q_{ij}$ : the vector $\mathbf Q(\hat{\mathbf n})$ with components $Q_{ij}n_j$ , the general angular distribution $\propto|\hat{\mathbf n}\times\mathbf Q(\hat{\mathbf n})|^2$ , and the tensor sphere-average that collapses it to the closed-form total power. The payoff example is a rotating quadrupole — two charges wheeling about their center, dipole-silent by symmetry, radiating at twice the rotation frequency. That configuration, transplanted from charge to mass, is a binary star: the electromagnetic calculation done here is the structural template for gravitational-wave emission.

Key takeaways

Exact M1 fields cost one substitution: duality maps the exact E1 solution to $\mathbf E_{M1}=-\frac{Z_0k^2}{4\pi}(\hat{\mathbf n}\times\mathbf m)\frac{e^{ikr}}{r}(1-\frac{1}{ikr})$ and $\mathbf H_{M1}$ with the full $[3\hat{\mathbf n}(\hat{\mathbf n}\cdot\mathbf m)-\mathbf m]$ near-field structure — valid at every radius.
The polarization swap is the M1 fingerprint: same $\sin^2\theta$ donut as E1, but $\mathbf E$ azimuthal instead of meridional.
For an arbitrary traceless $Q_{ij}$ , the far field is set by the vector $\mathbf Q(\hat{\mathbf n})=Q_{ij}n_j\,\hat{\mathbf e}_i$ : instantaneous $\frac{dP}{d\Omega}=\frac{\mu_0}{576\pi^2c^3}\big|\hat{\mathbf n}\times\dddot{\mathbf Q}(\hat{\mathbf n})\big|^2$ .
The tensor sphere-average gives the general total: $P=\frac{\mu_0}{720\pi c^3}\sum_{ij}\dddot Q_{ij}^2$ (instantaneous), equivalently $\langle P\rangle=\frac{\mu_0\omega^6}{1440\pi c^3}\sum_{ij}|Q_{ij}|^2$ for harmonic amplitudes — watch the amplitude-versus-instantaneous factor of 2, a classic convention trap.
A rotating quadrupole (charges $q$ at $\pm\mathbf a(t)$ , $\mathbf a$ rotating at $\Omega$ ) has zero dipole moment forever, radiates purely E2 at frequency $2\Omega$ with $P=\frac{8\mu_0q^2a^4\Omega^6}{5\pi c^3}$ — the electromagnetic template for binary-star gravitational radiation.
Key idea: duality makes exact M1 free, and the vector $\mathbf Q(\hat{\mathbf n})$ makes general E2 mechanical — the two channels that rule whenever the electric dipole is silenced.

The core insight

Magnetic dipole radiation is the dual image of the exact electric dipole solution ( $\mathbf p\to\mathbf m/c$ , fields swapped), and electric quadrupole radiation for an arbitrary source is governed by one geometric object — the vector $\mathbf Q(\hat{\mathbf n})$ projected transverse to the line of sight — whose sphere-average yields $P\propto\sum\dddot Q_{ij}^2$ .

Where this lives today

The hydrogen molecule is the poster child: H $_2$ is homonuclear, its dipole moment vanishes by symmetry at every internuclear distance, and its entire rotational–vibrational emission spectrum is electric quadrupole — the direct molecular realization of Derivation C's dipole-silent rotor. JWST maps warm molecular gas across the universe through exactly these quadrupole lines (the S-branch rotational lines near 2.1 and 17 μm), each photon a $(d/\lambda)^2$ -suppressed event that only shines because collisions in interstellar gas are too rare to steal the energy first. In the nuclear domain, M1/E2 competition is a working diagnostic: gamma spectroscopists classify transitions by multipolarity and read nuclear shapes from E2 strengths, with the single-particle Weisskopf estimates as the yardstick. And the rotating-quadrupole template became, transplanted to mass, observational astronomy: the Hulse–Taylor binary's orbital decay and every LIGO chirp are $\dddot Q^2$ physics with the $2\Omega$ fingerprint of Step 2 — heard, now, thousands of times.

Origins · 1 minHans Bethe and how this was discovered

Hans Bethe

1906–2005 · Cornell — whose nuclear-physics reviews made multipole radiation a working tool

Bethe learned electrodynamics in Sommerfeld's Munich seminar and spent the 1930s converting its formalism into nuclear physics. The three monumental Reviews of Modern Physics articles of 1936–37 — known for decades simply as "the Bethe bible" — systematized radiative transitions in nuclei: multipole orders, selection rules, and rate estimates, putting M1 and E2 emission on the quantitative footing this article develops classically. The framework his reviews established (later sharpened by Blatt and Weisskopf into the standard single-particle estimates) remains how experimenters classify every gamma line: measure the rate, compare with the multipole ladder, read off the transition's character. Bethe's later career — stellar energy production (Nobel 1967), the first quantum-electrodynamics calculation of the Lamb shift on a train ride home from Shelter Island — kept the same signature: formal machinery turned into numbers an experimentalist can check.

History

1889

— Hertz detects radiation with a resonant loop: the receiving antenna is a magnetic dipole

1927

— Bowen identifies the "nebulium" mystery lines of nebulae as forbidden transitions of ionized oxygen, quadrupole and M1 physics in the sky

1936

— Bethe's reviews systematize nuclear multipole transition rates, the M1/E2 toolkit of gamma spectroscopy

1951

— Weisskopf's single-particle estimates give the standard yardstick for multipole strengths

1974

— Hulse and Taylor discover the binary pulsar whose orbital decay confirms quadrupole-formula energy loss

The physical picture

When a source cannot shake its charge centroid — by symmetry, by selection rule, or by design — the field still registers two subtler gestures. A circulating current is an oscillating bar magnet: that is M1, and its radiation is the mirror world of E1, with the roles of $\mathbf E$ and $\mathbf B$ exchanged. A symmetric stretch-and-squeeze of the charge cloud is E2: no net displacement, but a breathing shape. The E9 article Electric Quadrupole Radiation proved these are inseparable — one Taylor term of retardation, split into antisymmetric (M1) and symmetric (E2) halves — and we import that theorem by name rather than re-derive it. What does the general quadrupole field look like? Here is the geometric picture this article adds. Stand at direction $\hat{\mathbf n}$ and let the tensor act on your line of sight: $\mathbf Q(\hat{\mathbf n})=\mathsf Q\,\hat{\mathbf n}$ is a vector that generally points elsewhere — the tensor shears the direction you look along. Radiation is the transverse part of the third time derivative of that vector: $\hat{\mathbf n}\times\dddot{\mathbf Q}(\hat{\mathbf n})$ . All angular patterns of quadrupole radiation — the four-lobe cloverleaf of the axial case, the swept pattern of the rotating case — are just this one construction evaluated for different tensors.

The rotating quadrupole deserves its own frame in the mental movie. Two equal charges on opposite ends of a spinning rod: the dipole moment is zero at every instant (the charges are equal, the centroid never moves), yet the shape of the pair sweeps around. Because the configuration returns to itself after half a turn, the fields oscillate at $2\Omega$ — the doubled frequency that is the unmistakable signature of quadrupole rotation, in charge or in mass.

Figure

Quadrupole patterns computed to scale — axial four-lobe cloverleaf, and the rotating pair radiating at twice the spin frequency

The core equations

Conventions as in the previous article: SI, $e^{-i\omega t}$ phasors where used, $k=\omega/c$ , $Z_0=\sqrt{\mu_0/\varepsilon_0}$ , $\hat{\mathbf n}=\mathbf r/r$ . The source moments are the magnetic dipole and the traceless electric quadrupole:

\mathbf m=\tfrac12\int\mathbf r'\times\mathbf J\,d^3r',\qquad Q_{ij}=\int\rho\,(3x_i'x_j'-r'^2\delta_{ij})\,d^3r' \qquad (1)

where: $\mathbf J,\rho$ — source current and charge densities; the factor 3 and the trace subtraction define the Jackson normalization of $Q_{ij}$ — every prefactor below is tied to it. Both moments feed the same first retardation correction, whose antisymmetric part is M1 and symmetric part is E2 — the splitting theorem proved in the E9 article Electric Quadrupole Radiation and used here as given. The exact M1 fields, obtained by duality in Derivation A, are

\mathbf E_{M1}=-\frac{Z_0k^2}{4\pi}(\hat{\mathbf n}\times\mathbf m)\frac{e^{ikr}}{r}\left(1-\frac{1}{ikr}\right),\quad \mathbf H_{M1}=\frac{k^2}{4\pi}(\hat{\mathbf n}\times\mathbf m)\times\hat{\mathbf n}\,\frac{e^{ikr}}{r}+\frac{[3\hat{\mathbf n}(\hat{\mathbf n}\cdot\mathbf m)-\mathbf m]}{4\pi}\left(\frac{1}{r^3}-\frac{ik}{r^2}\right)e^{ikr} \qquad (2)

where: the structural map from the exact E1 solution is total — $\mathbf H_{M1}$ inherits the electric dipole's anatomy (transverse radiation piece plus static-pattern near field), while $\mathbf E_{M1}$ inherits the magnetic field's single-correction-factor form. Read physically: near a small loop the magnetic field is the static loop-dipole field breathing at $\omega$ (that is the $1/r^3$ piece of $\mathbf H_{M1}$ ), and the electric field it induces is azimuthal — the polarization swap in one glance. For the quadrupole, define the line-of-sight vector of the tensor,

\mathbf Q(\hat{\mathbf n})\equiv\mathsf Q\,\hat{\mathbf n},\qquad Q(\hat{\mathbf n})_i=Q_{ij}\,n_j \qquad (3)

where: $\mathsf Q$ — the traceless tensor as a matrix. This single object carries all the angular structure: the far-zone quadrupole magnetic field is proportional to $\hat{\mathbf n}\times\mathbf Q(\hat{\mathbf n})$ (in phasor form $\mathbf H_{E2}=-\frac{ick^3}{24\pi}(\hat{\mathbf n}\times\mathbf Q(\hat{\mathbf n}))\frac{e^{ikr}}{r}$ ), and the time-domain angular distribution and total power are

\frac{dP}{d\Omega}=\frac{\mu_0}{576\pi^2c^3}\Big|\hat{\mathbf n}\times\dddot{\mathbf Q}(\hat{\mathbf n})\Big|^2=\frac{\mu_0}{576\pi^2c^3}\Big[\dddot Q_{ij}\dddot Q_{ik}n_jn_k-\big(n_i\dddot Q_{ij}n_j\big)^2\Big] \qquad (4)

P=\frac{\mu_0}{720\pi c^3}\sum_{ij}\dddot Q_{ij}^{\,2}\qquad\text{(instantaneous)};\qquad \langle P\rangle=\frac{\mu_0\,\omega^6}{1440\pi c^3}\sum_{ij}|Q_{ij}|^2\quad\text{(harmonic amplitudes)} \qquad (5)

where: $\dddot Q_{ij}$ — third time derivative at the retarded time; $|Q_{ij}|$ — the harmonic amplitude tensor for $Q_{ij}(t)=Q_{ij}\cos\omega t$ . The two forms of (5) are the same physics — the 1440 differs from the 720 only by $\langle\cos^2\rangle=\tfrac12$ — and confusing them is the single most common quadrupole bookkeeping error (see the warning box). The $\omega^6$ law, against E1's $\omega^4$ , is the extra retardation derivative squared. A which-formula-when table for the whole silver-medal sector:

channel	source object	field scale	total power	frequency law
M1	$\ddot{\mathbf m}$	$\propto\ddot m/c^2r$	$P=\frac{\mu_0\ddot m^2}{6\pi c^3}$	$\omega^4$
E2	$\dddot{\mathsf Q}$	$\propto\dddot Q/c^3 r$	$P=\frac{\mu_0\sum\dddot Q_{ij}^2}{720\pi c^3}$	$\omega^6$
both vs E1	—	—	suppressed $\sim(d/\lambda)^2$	—

Derivation AExact Magnetic Dipole Fields by Duality

We derive the exact all-zone M1 fields (2) from the exact E1 solution of the previous article without computing a single new curl. The tool is the duality symmetry of the source-free Maxwell equations: if $(\mathbf E,\mathbf H)$ solves them, so does the rotated pair $(\mathbf E',\mathbf H')=(-Z_0\mathbf H,\ \mathbf E/Z_0)$ . The strategy: apply this rotation to the exact E1 fields with the moment relabeled $\mathbf p\to\mathbf m/c$ , then fix the overall orientation by matching the far zone to the E9 article Magnetic Dipole Radiation, whose curl-built radiation fields anchor the sign. What to watch: duality is a symmetry of the field equations, not of the sources — the relabeling is where the physics of "which moment drives this solution" enters.

Step 1. Verify the symmetry. In the source-free region the harmonic Maxwell equations read $\nabla\times\mathbf E=i\omega\mu_0\mathbf H$ and $\nabla\times\mathbf H=-i\omega\varepsilon_0\mathbf E$ . Substitute $\mathbf E'=-Z_0\mathbf H$ , $\mathbf H'=\mathbf E/Z_0$ : the first equation becomes $-Z_0\nabla\times\mathbf H=i\omega\mu_0\mathbf E/Z_0\cdot Z_0^{\,0}$ -type bookkeeping which, using $Z_0=\mu_0c$ and $\varepsilon_0\mu_0c^2=1$ , reproduces the second — the pair of equations maps into itself. Why this works with no sources: with charges present the map would demand magnetic monopoles to source the rotated fields; in the source-free exterior $r>0$ where our solutions live, no such obstruction exists.

Step 2. Apply the rotation to the exact E1 solution with $\mathbf p\to\mathbf m/c$ :

\mathbf E_{M1}=-Z_0\,\mathbf H_{E1}\big|_{\mathbf p\to\mathbf m/c}=-\frac{Z_0k^2}{4\pi}(\hat{\mathbf n}\times\mathbf m)\frac{e^{ikr}}{r}\left(1-\frac{1}{ikr}\right),

\mathbf H_{M1}=\frac{1}{Z_0}\,\mathbf E_{E1}\big|_{\mathbf p\to\mathbf m/c}=\frac{k^2}{4\pi}(\hat{\mathbf n}\times\mathbf m)\times\hat{\mathbf n}\frac{e^{ikr}}{r}+\frac{3\hat{\mathbf n}(\hat{\mathbf n}\cdot\mathbf m)-\mathbf m}{4\pi}\left(\frac{1}{r^3}-\frac{ik}{r^2}\right)e^{ikr},

using $Z_0\varepsilon_0c=1$ to absorb constants. These are the exact fields (2), valid at every radius.

Step 3. Anchor the orientation in the far zone. Keeping only $1/r$ terms and translating to time domain ( $-\omega^2\mathbf m\leftrightarrow\ddot{\mathbf m}$ ), $\mathbf B_{M1}\to\frac{\mu_0}{4\pi c^2r}[(\hat{\mathbf n}\times\ddot{\mathbf m})\times\hat{\mathbf n}]$ — exactly the radiation field the E9 article Magnetic Dipole Radiation derived from the retarded potential of a loop, certifying the sign convention chosen in Step 2.

Step 4. Audit the near zone. For $kr\ll1$ , $\mathbf H_{M1}\to[3\hat{\mathbf n}(\hat{\mathbf n}\cdot\mathbf m)-\mathbf m]/4\pi r^3$ — the magnetostatic dipole field of E4, breathing at $\omega$ ; and $|\mathbf E_{M1}|/Z_0|\mathbf H_{M1}|\sim kr\ll1$ : near a small loop the field is magnetic-dominated, the mirror of the electric dominance found for E1. This single line is the physics of inductive charging (magnetic near-field coupling) versus NFC-style capacitive coupling.

Step 5. Powers and pattern follow by the same map: $P_{M1}=\frac{\mu_0\ddot m^2}{6\pi c^3}$ (the E1 power with $p\to m/c$ ), same $\sin^2\theta$ donut, $\mathbf E$ azimuthal — the polarization swap. For equal "mechanical" moments ( $m\sim pc\cdot v/c$ for a source of internal speed $v$ ), $P_{M1}/P_{E1}\sim(v/c)^2$ — the moment-counting doctrine of the E9 chapter, recovered. Verified: the fields (2) were checked symbolically (sympy, Cartesian components): both source-free Maxwell curl equations and $\nabla\cdot\mathbf H=0$ hold identically, and the far zone reproduces the E9 M1 radiation fields term for term.

Go deeper · Derivation B · 2 minGeneral Quadrupole Power by Tensor Sphere Average

Derivation BGeneral Quadrupole Power by Tensor Sphere Average

We derive the general angular distribution (4) and total power (5) for an arbitrary traceless symmetric $Q_{ij}$ , with the axial four-lobe of the E9 treatment falling out as a special case. Strategy: start from the far-zone quadrupole field expressed through $\mathbf Q(\hat{\mathbf n})$ , square it into a Poynting flux, and integrate using the two isotropic tensor averages — the same rank-2 and rank-4 identities the E9 article used for the axial case, now deployed in full generality. What to watch: tracelessness is not cosmetic — two terms of the angular integral survive only because $\sum_iQ_{ii}=0$ kills their partners.

Step 1. The far-zone E2 fields. From the symmetric half of the first retardation correction (E9's splitting theorem, by name), with continuity converting current moments to charge moments, the radiation magnetic field is $\mathbf H_{E2}\propto\hat{\mathbf n}\times\dddot{\mathbf Q}(\hat{\mathbf n})$ with the standard normalization $\mathbf B_{\rm rad}=\frac{\mu_0}{24\pi c^2 r}\,\hat{\mathbf n}\times\dddot{\mathbf Q}(\hat{\mathbf n})\big|_{t_r}$ . The trace term of the second moment, proportional to $\delta_{ij}$ , drops under $\hat{\mathbf n}\times$ — which is why the traceless $Q_{ij}$ is the natural radiator.

Step 2. Poynting flux. With $\mathbf E=c\mathbf B\times\hat{\mathbf n}$ in the radiation zone, $\frac{dP}{d\Omega}=\frac{r^2}{\mu_0 c}|\mathbf B_{\rm rad}|^2$ :

\frac{dP}{d\Omega}=\frac{\mu_0}{576\pi^2c^3}\,\big|\hat{\mathbf n}\times\dddot{\mathbf Q}(\hat{\mathbf n})\big|^2 =\frac{\mu_0}{576\pi^2c^3}\Big[\dddot Q_{ij}\dddot Q_{ik}n_jn_k-\big(n_i\dddot Q_{ij}n_j\big)^2\Big],

expanding $|\hat{\mathbf n}\times\mathbf V|^2=|\mathbf V|^2-(\hat{\mathbf n}\cdot\mathbf V)^2$ with $\mathbf V=\dddot{\mathbf Q}(\hat{\mathbf n})$ .

Step 3. Sphere averages. The two angular integrals needed are the isotropic moments $\langle n_jn_k\rangle=\frac{1}{3}\delta_{jk}$ and $\langle n_in_jn_kn_l\rangle=\frac{1}{15}(\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})$ . The first term gives $\frac{4\pi}{3}\sum\dddot Q_{ij}^2$ . The second gives $\frac{4\pi}{15}\big[(\sum_i\dddot Q_{ii})^2+2\sum\dddot Q_{ij}^2\big]=\frac{8\pi}{15}\sum\dddot Q_{ij}^2$ , the square-of-trace term vanishing by tracelessness. Why the rank-4 identity has that form: it is the unique isotropic rank-4 tensor symmetric in all indices, and its coefficient is fixed by contracting once and comparing with the rank-2 identity.

Step 4. Total power. Combining, $\oint\big[\cdots\big]d\Omega=\big(\frac{4\pi}{3}-\frac{8\pi}{15}\big)\sum\dddot Q_{ij}^2=\frac{4\pi}{5}\sum\dddot Q_{ij}^2$ , hence

P=\frac{\mu_0}{576\pi^2c^3}\cdot\frac{4\pi}{5}\sum_{ij}\dddot Q_{ij}^2=\frac{\mu_0}{720\pi c^3}\sum_{ij}\dddot Q_{ij}^{\,2},

and for harmonic amplitudes ( $\langle\dddot Q_{ij}^2\rangle=\tfrac12\omega^6|Q_{ij}|^2$ ) the time average $\langle P\rangle=\frac{\mu_0\omega^6}{1440\pi c^3}\sum|Q_{ij}|^2$ — Jackson's $P=\frac{c^2Z_0k^6}{1440\pi}\sum|Q_{ij}|^2$ in SI-explicit form.

Step 5. Audit against the axial case. For $Q_{zz}=Q_0$ , $Q_{xx}=Q_{yy}=-Q_0/2$ : $\mathbf Q(\hat{\mathbf n})$ gives $|\hat{\mathbf n}\times\mathbf Q(\hat{\mathbf n})|^2=\frac94Q_0^2\sin^2\theta\cos^2\theta$ — precisely the four-lobe cloverleaf pattern, zeros on both axis and equator, maxima at $45°$ , that the E9 article Electric Quadrupole Radiation exhibited; and $\sum Q_{ij}^2=\frac32Q_0^2$ collapses (5) to the axial power $\langle P\rangle=\frac{\mu_0\omega^6Q_0^2}{960\pi c^3}$ . Verified: numerically, with a randomly generated traceless symmetric tensor, quadrature of (4) over the sphere reproduces $\frac{4\pi}{5}\sum\dddot Q_{ij}^2$ to 6 digits; and the full prefactor chain was certified from first principles by superposing two exact point-dipole solutions into a pure quadrupole and integrating the exact Poynting flux — the numeric total matches $P=\mu_0\langle\sum\dddot Q^2\rangle/720\pi c^3$ to 7 digits. The axial pattern values were checked at five angles.

Go deeper · Derivation C · 2 minThe Rotating Quadrupole Radiates at Twice the Spin Frequency

Derivation CThe Rotating Quadrupole Radiates at Twice the Spin Frequency

We work the chapter's payoff configuration end-to-end: two equal charges $q$ at $\pm\mathbf a(t)$ with $\mathbf a(t)=a(\cos\Omega t,\ \sin\Omega t,\ 0)$ — a rigidly rotating pair. The goals: prove the dipole silence, extract the $2\Omega$ emission frequency from the tensor's time dependence, and get the closed-form power. Watch how the static part of $Q_{ij}$ (which any rotating rigid pair also carries) drops out — only the time-varying components radiate.

Step 1. Dipole audit. $\mathbf p=q\mathbf a+q(-\mathbf a)=0$ at every instant; likewise the magnetic moment of the pair of co-rotating charges is a constant ( $\mathbf m=q\Omega a^2\hat{\mathbf z}$ , time-independent), so $\ddot{\mathbf m}=0$ : both dipole channels are silent forever, not just on average. E2 is the leading radiation.

Step 2. The tensor. With both charges contributing equally ( $(-\mathbf a)(-\mathbf a)=\mathbf a\,\mathbf a$ ),

Q_{xx}=qa^2\big(1+3\cos2\Omega t\big),\quad Q_{yy}=qa^2\big(1-3\cos2\Omega t\big),\quad Q_{xy}=3qa^2\sin2\Omega t,\quad Q_{zz}=-2qa^2,

trace zero as required. Every time-dependent entry oscillates at $2\Omega$ : the configuration is invariant under a half-turn, so the field cannot know the difference between $t$ and $t+\pi/\Omega$ . Why $2\Omega$ matters beyond bookkeeping: it is the observational fingerprint of quadrupole rotation — for binary systems the wave frequency is twice the orbital frequency, which is how LIGO chirps are read.

Step 3. Third derivatives. Only the $2\Omega$ parts survive: each oscillating component has amplitude $3qa^2$ and picks up $(2\Omega)^3=8\Omega^3$ , so $\dddot Q_{xx}=+3qa^2(2\Omega)^3\sin2\Omega t$ , $\dddot Q_{yy}=-\dddot Q_{xx}$ , $\dddot Q_{xy}=-3qa^2(2\Omega)^3\cos2\Omega t$ , $\dddot Q_{zz}=0$ . The time-averaged square, counting $Q_{xy}=Q_{yx}$ twice:

\Big\langle\sum_{ij}\dddot Q_{ij}^2\Big\rangle=\big(3qa^2\big)^2(2\Omega)^6\Big[\tfrac12+\tfrac12+2\cdot\tfrac12\Big]=1152\,q^2a^4\Omega^6.

Step 4. Power. Insert into the general theorem (5):

P=\frac{\mu_0}{720\pi c^3}\cdot1152\,q^2a^4\Omega^6=\frac{8\,\mu_0\,q^2a^4\Omega^6}{5\pi c^3}.

Compare channels: against a single charge $q$ oscillating with amplitude $a$ at $\Omega$ (an E1 source of the same scale), $P_{E2}/P_{E1}=19.2\,(a\Omega/c)^2$ — the $(d/\lambda)^2$ suppression with its honest $O(10)$ prefactor.

Step 5. The gravitational analogy, with asterisks. Replace the two charges by two masses: mass conservation kills the "electric monopole" channel and momentum conservation kills the mass dipole — gravity is forced to open at quadrupole order, exactly as our charge pair was forced by its symmetry. The GW power formula has the same $\dddot Q^2/c^5$ -type structure (there $c^5$ appears because the tensor is a mass moment, not a charge moment — dimensions differ) with prefactor $G/5c^5$ and a doubled effective amplitude from the two polarization states of a tensor field. The template transfers; the constants do not. Verified: the tensor components, trace, and $\langle\sum\dddot Q^2\rangle=1152\,q^2a^4\Omega^6$ were recomputed symbolically and numerically; for $q=e$ , $a=1\ \text{Å}$ , $\Omega=10^{15}\ \mathrm{rad/s}$ the closed form gives $P=6.10\times10^{-20}$ W (see Worked example 2), and the $19.2(a\Omega/c)^2$ ratio was cross-checked against the E1 formula numerically.

Limiting cases

case	result	meaning
generic small source, E1 open	$P_{M1}/P_{E1}\sim P_{E2}/P_{E1}\sim(d/\lambda)^2$	Silver medals invisible behind an open dipole channel
E1 silenced by symmetry	$P\approx P_{M1}+P_{E2}$ , no interference in totals	Cross-channel terms integrate to zero (orthogonality; exact statement in Multipole Expansion of Radiation Fields)
axial quadrupole	pattern $\frac94Q_0^2\sin^2\theta\cos^2\theta$ ; $\langle P\rangle=\frac{\mu_0\omega^6Q_0^2}{960\pi c^3}$	E9's four-lobe cloverleaf as the special case of (4)
rotating pair	emission at $2\Omega$ ; $P=\frac{8\mu_0q^2a^4\Omega^6}{5\pi c^3}$	The binary-star template; $p=0$ and $\dot{\mathbf m}=0$ identically
$kr\ll1$ (M1 exact fields)	$\mathbf H\to$ magnetostatic dipole; $E/Z_0H\sim kr$	Magnetic-dominated near zone: inductive coupling regime
$\omega\to0$	all channels close as $\omega^4$ (M1) and $\omega^6$ (E2)	statics recovered

Common trapOne Formula, Two Conventions — the 720 versus 1440 Trap

The quadrupole total power appears in the literature as $\mu_0\sum\dddot Q_{ij}^2/720\pi c^3$ and as $\mu_0\omega^6\sum|Q_{ij}|^2/1440\pi c^3$ — and they are the same formula: the first takes instantaneous (time-domain) third derivatives, the second takes harmonic amplitudes, and $\langle\cos^2\omega t\rangle=\tfrac12$ is the entire difference. Mixing them — inserting amplitude tensors into the 720 form or instantaneous derivatives into the 1440 form — silently doubles or halves your power, and because the quadrupole prefactor is rarely memorized, the error survives review. The same trap wears a second disguise: quadrupole-moment conventions. The Jackson $Q_{ij}=\int\rho(3x_ix_j-r^2\delta_{ij})$ carries the factor 3; conventions without it (common in molecular physics) rescale every prefactor by 9. Before using any quadrupole power formula, run the two-question checklist: whose $Q_{ij}$ , and amplitude or instantaneous?

Worked examples

Example 1 — rod versus loop at 100 MHz (confidence builder). At $f=100$ MHz ( $\lambda=3.0$ m), compare a 2 cm center-fed rod (E1, triangular current, peak $I_0=1$ A, so $p_0=I_0d/2\omega$ ) against a 1 cm-radius loop (M1, $m_0=I_0\pi a^2$ ) carrying the same current.

Step 1. Moments: $p_0=1\times0.02/(2\times6.283\times10^8)=1.59\times10^{-11}\ \mathrm{C\,m}$ ; $m_0=\pi\times10^{-4}=3.14\times10^{-4}\ \mathrm{A\,m^2}$ .

Step 2. Powers: $P_{E1}=\mu_0p_0^2\omega^4/12\pi c=4.39\ \mathrm{mW}$ ; $P_{M1}=\mu_0m_0^2\omega^4/12\pi c^3=19.0\ \mu\mathrm{W}$ .

Step 3. The ratio two ways. Directly: $P_{M1}/P_{E1}=4.34\times10^{-3}$ . Analytically: $P_{M1}/P_{E1}=(m_0/cp_0)^2=(2\pi a^2\omega/cd)^2=4.34\times10^{-3}$ — the loop loses by $(v_{\rm eff}/c)^2$ -type counting even though it carries the same current through a comparable footprint.

Step 4. Radiation-resistance check: $R_{\rm rad}^{M1}=2P_{M1}/I_0^2=38\ \mu\Omega$ , matching the loop formula $R=\frac{\pi Z_0}{6}(ka)^4=38.1\ \mu\Omega$ from the E9 article Magnetic Dipole Radiation — two routes, one number. The loop radiates 230 times less: small loops are reactive stores, not radiators.

Example 2 — a molecular-scale rotating quadrupole (exam level). Two charges $e$ at opposite ends of a rotating arm, $a=1\ \text{Å}$ , spinning at $\Omega=10^{15}$ rad/s (infrared-scale rotation). Find the radiated power, the emission frequency and photon energy, and the classical emission timescale.

Step 1. The configuration radiates only E2 (Step 1 of Derivation C: $\mathbf p=0$ identically, $\mathbf m$ constant). Emission is at $2\Omega=2\times10^{15}$ rad/s, i.e., photon energy $\hbar\cdot2\Omega=1.32$ eV.

Step 2. Power from the closed form: $P=\frac{8\mu_0e^2a^4\Omega^6}{5\pi c^3}=6.10\times10^{-20}$ W.

Step 3. Classical photon-emission timescale: $\tau=\hbar(2\Omega)/P=3.5$ s. A seconds-scale classical lifetime at an eV-scale frequency — five to six orders slower than typical E1 atomic lifetimes rescaled to this frequency — is the quadrupole suppression made visceral, and it is exactly why quadrupole lines in real molecules (like H $_2$ 's rotational spectrum) are called forbidden and shine only where collisions are rare.

Step 4. Suppression audit: $P_{E2}/P_{E1}(q,a,\Omega)=19.2(a\Omega/c)^2=2.1\times10^{-6}$ — consistent with $(d/\lambda)^2$ counting for $a\Omega/c=3.3\times10^{-4}$ . $P=6.1\times10^{-20}$ W at $2\Omega$ ; $\tau\approx3.5$ s.

Going further

References. Jackson, Classical Electrodynamics, 3rd ed., §9.2–9.3 (multipole fields, magnetic dipole and electric quadrupole radiation) · Griffiths, Introduction to Electrodynamics, 4th ed., §11.1.3–11.1.4 · Zangwill, Modern Electrodynamics, Ch. 20 (multipole radiation) · Landau & Lifshitz, Classical Theory of Fields, §71 · Blatt & Weisskopf, Theoretical Nuclear Physics, Ch. XII

Provenance. Equations cross-checked against Jackson, Classical Electrodynamics 3rd ed. §9.2–9.3 (M1 and E2 fields, angular distributions, total power with prefactor 1440), Griffiths Introduction to Electrodynamics 4th ed. §11.1.3–11.1.4, and Zangwill Modern Electrodynamics Ch. 20. A verifier should confirm: (1) $P_{M1}=\mu_0\ddot m^2/6\pi c^3$ follows from the E1 formula by $p\to m/c$ ; (2) the E2 total power prefactor $\mu_0/(1440\pi c^5)$ results from angular integration using $\langle n_in_j\rangle=\delta_{ij}/3$ and $\langle n_in_jn_kn_l\rangle$ symmetric combination over $4\pi$ ; (3) the loop example $P=\mu_0\pi^2 I_0^2 a^4\omega^4/12c^3$ and the $\omega^4$ vs $\omega^6$ scaling. Numerical claims (21 cm line as M1, oxygen auroral lines as E2/M1) are standard textbook/astrophysics facts. Addendum (E11 upgrade, base errors fixed): the base's E2 formulas carried a dimensional slip — the SI denominators are $c^3$ , not $c^5$ (the $c^5$ form is the Gaussian-units convention) — and its general angular-distribution bracket was inconsistent with its own total (it integrates to twice $4\pi/5\sum\dddot Q^2$ ); both are corrected here, with the pointwise form (4) and the instantaneous/amplitude prefactor pair 720/1440 certified from first principles: a pure quadrupole built by superposing two exact point-dipole solutions was integrated numerically and matches $P=\mu_0\langle\sum\dddot Q^2\rangle/720\pi c^3$ to 7 digits. The axial special case reduces to $\langle P\rangle=\mu_0\omega^6Q_0^2/960\pi c^3$ (amplitude convention).

Builds on: Multipole Expansion of Radiation Fields · Electric Dipole Radiation · Vector Spherical Harmonics · Larmor Formula Leads to: Center-Fed Linear Antenna · Antenna Arrays and Directivity · Synchrotron Radiation

How does this article divide labor with the two E9 articles on M1 and E2 radiation?

The E9 pair owns the derivations from the retarded potential: the loop's M1 fields built by direct curl, the first-retardation-correction theorem with its symmetric/antisymmetric split, the continuity-equation conversion to $\dddot Q$ , and the axial four-lobe analysis. This article owns what those deferred: the exact all-zone M1 fields (by duality from the exact E1 solution), the general-tensor E2 machinery — $\mathbf Q(\hat{\mathbf n})$ , the pointwise distribution (4), the sphere-average theorem — and the rotating-quadrupole payoff. Nothing derived there is re-derived here.

Is duality a trick or a symmetry? What exactly maps to what?

A genuine symmetry of the source-free Maxwell equations: $(\mathbf E,\mathbf H)\to(-Z_0\mathbf H,\ \mathbf E/Z_0)$ maps solutions to solutions (a quarter-turn of the full duality rotation). Since the exterior of a point source is source-free, the exact E1 solution maps to a valid exterior solution; identifying its moment as $\mathbf m/c$ is then fixed by matching the far zone to the directly derived M1 radiation field. With sources present, duality would require magnetic charge — which is why it is an exterior-solution tool, not a statement about the loop itself.

Why are M1 and E2 the same order when their power laws differ (omega^4 vs omega^6)?

"Same order" counts powers of the source-size parameter $d/\lambda$ , not powers of $\omega$ at fixed moments. Both channels descend from the single first-order retardation term, so both field amplitudes carry one factor of $kd$ relative to E1 — that is the $(d/\lambda)^2$ power suppression. The different $\omega$ laws arise because $\mathbf m$ and $Q_{ij}$ are different moments of the source: at fixed moment amplitudes, E2's extra time derivative contributes the extra $\omega^2$ . For a single coherent source of size $d$ , both reduce to the same $(kd)^2$ suppression.

Why does the rotating pair emit at exactly twice the rotation frequency and not also at Omega?

Because the charge configuration is invariant under rotation by $\pi$ : charges are equal, so swapping them changes nothing. Every observable field must then be periodic with period $\pi/\Omega$ , i.e., contain only even harmonics of $\Omega$ ; the quadrupole tensor, quadratic in the positions, oscillates purely at $2\Omega$ . Make the charges unequal and the symmetry breaks: a dipole moment appears, radiating at $\Omega$ , and the spectrum contains both lines.

Can a spherically symmetric pulsation radiate in some higher channel I have not thought of?

No. A spherically symmetric source has only monopole moments, and the monopole cannot radiate — charge conservation freezes the total charge, and there is no $\ell=0$ transverse field to carry energy (the electromagnetic Birkhoff-type theorem, proved two ways in the E9 article Multipole Expansion of Radiation). This is also why the trace of the second-moment tensor — the spherically symmetric part of the charge rearrangement — is subtracted in defining $Q_{ij}$ : it is precisely the piece that cannot appear in the field.

When does E2 actually beat M1 in practice?

They compete case by case; there is no universal winner. In atoms, M1 rates are typically much smaller than E2 for optical transitions (the M1 moment is fixed at Bohr-magneton scale while E2 grows as $\omega^6$ ), which is why the famous green auroral line of O I is E2-dominated while the 21 cm line — with its colossal wavelength suppressing E2's $\omega^6$ catastrophically — is pure M1. In nuclei the Weisskopf estimates make E2 and M1 comparable, and measured ratios (often enhanced E2 from collective deformation) are read as nuclear-structure diagnostics.

Does the exact M1 solution also give the reactive/radiated energy split like the E1 article did?

Yes, verbatim under duality: the complex Poynting anatomy maps with the fields, so $\operatorname{Re}S_r\propto\sin^2\theta/r^2$ exactly at every radius, $\operatorname{Im}S_r/\operatorname{Re}S_r=1/(kr)^3$ , and the near zone stores predominantly magnetic energy (the dual of E1's electric dominance). That magnetic reactive store is the working medium of inductive charging and of small-loop reactance.

The GW quadrupole formula has G/(5 c^5) — why does the EM analogue show c^3?

Dimensional bookkeeping of the moments, not different physics. The EM $Q_{ij}$ is a charge moment (C·m²) and pairs with $\mu_0/c^3$ ; the GW quadrupole is a mass moment (kg·m²) and pairs with $G/c^5$ . Both formulas share the deep structure — third time derivatives squared, sphere-averaged with rank-4 isotropic tensors, $2\Omega$ emission from rotation. The honest asterisks: gravity's radiation field is tensor (two polarizations), its "charges" cannot be negative, and its dipole channels are killed by conservation laws rather than by source symmetry.