Electric Quadrupole Moment

Carbon dioxide is as neutral as a molecule can be: zero net charge, and — by its perfect O=C=O symmetry — zero dipole moment. The first two rungs of the multipole ladder are silent. Yet CO $_2$ molecules grip each other tightly enough to shape the crystal structure of dry ice, and a radio coil can interrogate the shape of a nitrogen nucleus buried in a suitcase. When monopole and dipole both go quiet, the far field does not vanish; it drops to a $1/r^3$ whisper with an unmistakable angular pattern, and the voice is the electric quadrupole moment. This is where the ladder outgrows scalars and vectors. Total charge is one number; the dipole moment is three. The quadrupole needs five, packaged as a symmetric traceless tensor $Q_{ij}$ — our first honest tensor, and the count is no accident: five is $2\ell+1$ for $\ell=2$ , the size of the spherical-harmonic family $Y_2^m$ . We will extract $Q_{ij}$ directly from the $\ell=2$ rung of Multipole Expansion of the Potential, watch tracelessness emerge by construction rather than decree, collapse the tensor to a single scalar when the source has an axis, and run one four-charge example from definition to numbers. The payoff reaches far beyond electrostatics. The deuteron's nonzero quadrupole moment, measured by Rabi's group in 1939, forced physicists to accept that the nuclear force is not central — the tensor force was discovered inside this chapter's mathematics. And gravity's radiation ladder starts at the quadrupole: every LIGO chirp is the $\ell=2$ story retold.

Key takeaways

The $\ell=2$ term of the multipole expansion defines $Q_{ij}=\int(3x_i'x_j'-r'^2\delta_{ij})\,\rho\,\mathrm{d}\tau'$ — symmetric and traceless by construction, hence five independent components, matching the five $Y_2^m$ ; its potential is $V_{\mathrm{quad}}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{1}{2r^3}\sum Q_{ij}\hat r_i\hat r_j$ .
With an axis of symmetry one scalar $Q\equiv Q_{zz}$ suffices: $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q(3\cos^2\theta-1)}{4r^3}$ — positive lobes on the axis, a negative belt at the equator, zero on the cone $\theta=54.74°$ .
The linear array $+q,\,-2q,\,+q$ (spacing $d$ ) has $Q_{\mathrm{tot}}=0$ , $\mathbf{p}=0$ , $Q_{zz}=4qd^2$ ; on axis the quadrupole term is low by exactly $(d/r)^2$ — 1% at $r=10d$ .
Sign is shape: for positive charge, prolate (cigar) means $Q_{zz}>0$ , oblate (pancake) means $Q_{zz}<0$ . The deuteron's $Q=+0.286\ e\cdot\mathrm{fm}^2$ announced a slightly cigar-shaped nucleus — and with it the tensor force.
Convention alert: books stash the factor $\tfrac12$ differently (Griffiths inside $Q_{ij}$ , Jackson inside $V$ ); take moment and potential formula from the same source.
Key idea: the quadrupole moment is the traceless symmetric tensor $Q_{ij}=\int(3x_i'x_j'-r'^2\delta_{ij})\rho\,\mathrm{d}\tau'$ ; it produces a $1/r^3$ potential with $\ell=2$ angular shape and is the leading moment whenever both monopole and dipole vanish.

Where this lives today

Walk through a modern airport's checked-baggage screening and your suitcase may be briefly interrogated by equation (7). Nuclear quadrupole resonance detectors pulse radio waves into luggage and listen for the fingerprint frequencies at which $^{14}$ N nuclei reorient in the electric field gradients of specific crystals — near 5.2 MHz for RDX, below 1 MHz for TNT — with no applied magnet, because the explosive's own lattice supplies the gradient; the same trick authenticates pharmaceuticals through packaging. Solid-state NMR spectroscopists spin samples at exactly $54.74°$ to the field, because averaging over that cone nulls $3\cos^2\theta-1$ and erases quadrupolar broadening: the nodal cone is a dial setting on real hardware. One honest distinction: the quadrupole mass filters and lenses of mass spectrometers and accelerators borrow the same $\ell=2$ mathematics for the shape of an applied field ( $V\propto x^2-y^2$ ), not for a charge moment of the beam — same harmonic, different actor. The grandest tenant is gravity. Mass conservation forbids gravitational monopole radiation and momentum conservation forbids dipole radiation, so gravity's ladder starts at $\ell=2$ : a mass distribution radiates only insofar as its quadrupole moment accelerates. Every gravitational-wave chirp is a mass-quadrupole time series read off two black holes' final orbits — the electromagnetic version, quadrupole radiation, is deferred to the radiation chapter (E9). From baggage halls to binary mergers, the third rung is the working rung.

The physical picture

Stack two dipoles back to back — plus-minus against minus-plus — and every simple description cancels: no net charge, and the dipole arrows annihilate. What survives is subtler: not how much charge, not how far plus sits from minus, but how the charge is stretched or squashed along different axes. That anisotropy is what $Q_{ij}$ measures. A spherical blob has $Q_{ij}=0$ , whatever its radial profile; pull it into a cigar along $z$ and $Q_{zz}$ goes positive; flatten it into a pancake and $Q_{zz}$ goes negative. The quadrupole is the far field's first report on shape. The signature it broadcasts is a four-lobe pattern: positive potential in two cones around the axis (for $Q>0$ ), a negative belt at the equator, and exactly zero on the cone at $54.74°$ where $3\cos^2\theta-1$ changes sign — the "magic angle" at which solid-state NMR spectroscopists spin their samples, because it kills this factor.

Figure

The quadrupole signature — four lobes with a nodal cone at 54.7 degrees, and the sign-is-shape dictionary

Why does shape need a tensor while offset needed only a vector? Because a stretch has no arrowhead: flip a cigar end for end and it is the same cigar. Direction without orientation is information a vector cannot encode but a quadratic form $\sum Q_{ij}\hat r_i\hat r_j$ encodes perfectly. Carry the count into the math: nine entries, cut to six by symmetry, to five by tracelessness — exactly the $2\ell+1=5$ members of the $\ell=2$ family catalogued in Spherical Harmonics. Tensor language and harmonic language are the same five-dimensional room with different furniture; proving that is the article's structural punchline.

The third rung, in tensor language

Multipole Expansion of the Potential expanded $1/\eta$ — with $\boldsymbol{\eta}=\mathbf{r}-\mathbf{r}'$ the separation vector — in powers of $r'/r$ and delivered the ladder. We take its $\ell=2$ rung as our inheritance, exactly as that article wrote it:

V_2(\mathbf{r})=\frac{1}{4\pi\varepsilon_0}\,\frac{1}{r^3}\int (r')^2\,P_2(\cos\alpha)\,\rho(\mathbf{r}')\,\mathrm{d}\tau' \qquad (1)

where: $\alpha$ — the angle between the field direction $\hat{\mathbf r}$ and the source point $\mathbf{r}'$ ; $P_2(x)=\tfrac12(3x^2-1)$ — from Legendre Polynomials. As it stands, (1) is awkward company: $\alpha$ ties the integral to the observation direction, so it must apparently be redone for every direction. The tensor rewrite divorces the two — do the source integral once, and let the direction contract with the stored table afterward. The table is the quadrupole moment:

Q_{ij}=\int\left(3x_i'x_j'-r'^2\,\delta_{ij}\right)\rho(\mathbf{r}')\,\mathrm{d}\tau' \qquad (2)

where: $x_i'$ — the Cartesian components of $\mathbf{r}'$ ; $\delta_{ij}$ — the Kronecker delta, subtracting the isotropic part; units $\mathrm{C\,m^2}$ . For point charges, $Q_{ij}=\sum_k q_k(3x_{ki}x_{kj}-r_k^2\delta_{ij})$ . In terms of (2), the rung (1) becomes — as the first derivation proves —

V_{\mathrm{quad}}(\mathbf{r})=\frac{1}{4\pi\varepsilon_0}\,\frac{1}{2r^3}\sum_{i,j}Q_{ij}\,\hat r_i\hat r_j \qquad (3)

where: $\hat r_i$ — components of the unit vector toward the field point; the $\tfrac12$ is part of the formula in this normalization (Jackson's — see the warning). Source physics lives in the five numbers $Q_{ij}$ , geometry in the quadratic form $\hat r_i\hat r_j$ , distance in $1/r^3$ — one power faster in decay than the dipole, as the ladder promised. Two structural facts come free. Symmetry, $Q_{ij}=Q_{ji}$ , is visible in (2) at a glance. Tracelessness,

\sum_i Q_{ii}=Q_{xx}+Q_{yy}+Q_{zz}=0 \qquad (4)

where: the sum runs over the diagonal — holds identically, because the kernel's trace is $3r'^2-3r'^2=0$ . Not an imposed convention but a manufactured property, and the tensor echo of a fact we own: $r^2$ is not harmonic, so no isotropic piece may appear in a legitimate exterior solution of Laplace's equation. The counting lands where the picture predicted: nine, minus three for symmetry, minus one for the trace — five, the dimension of the $\ell=2$ harmonic family. The next rung, the octopole, needs a symmetric traceless three-index tensor with $2(3)+1=7$ components and a $1/r^4$ potential; we leave it there.

Derivation AFrom the third rung to the traceless tensor

The goal is to convert the Legendre form (1) into the tensor form (3), watching definition (2) — trace subtraction included — emerge on its own rather than being decreed. The strategy is pure componentry: write $\cos\alpha$ as a dot product, expand $P_2$ , and force both terms into a common double sum so the field direction factors out of the source integral. The one clever move is a $\delta_{ij}$ insertion that looks like bureaucracy and is actually the entire trace story; watch for it in Step 2.

Step 1. Express the angle through vectors: $r'\cos\alpha=\hat{\mathbf r}\cdot\mathbf{r}'$ , so the integrand of (1) becomes

(r')^2P_2(\cos\alpha)=\frac{3(\hat{\mathbf r}\cdot\mathbf{r}')^2-r'^2}{2}

Why: powers of $r'\cos\alpha$ turn into Cartesian components; keeping $\alpha$ would chain the source integral to the observation direction forever.

Step 2. Componentize over a shared double sum: $(\hat{\mathbf r}\cdot\mathbf{r}')^2=\sum_{i,j}\hat r_i\hat r_j\,x_i'x_j'$ , and — the key insertion, since $\hat{\mathbf r}$ is a unit vector — $r'^2=\sum_{i,j}\hat r_i\hat r_j\,\delta_{ij}\,r'^2$ . Then

(r')^2P_2(\cos\alpha)=\frac{1}{2}\sum_{i,j}\hat r_i\hat r_j\left(3x_i'x_j'-r'^2\delta_{ij}\right)

Why the insertion matters: without it the two terms live in different index worlds and nothing factors; with it, the $-r'^2\delta_{ij}$ rides inside the parenthesis — the trace subtraction is the price of a common sum.

Step 3. Integrate against the source. The $\hat r_i$ carry no primed dependence and pass through the integral, leaving $V_2=\frac{1}{4\pi\varepsilon_0}\frac{1}{2r^3}\sum\hat r_i\hat r_j\int(3x_i'x_j'-r'^2\delta_{ij})\rho\,\mathrm{d}\tau'$ . The integral is definition (2), and equation (3) stands — the $\tfrac12$ is $P_2$ 's own, never absorbed.

Step 4. Harvest tracelessness. Setting $j=i$ and summing inside the kernel: $3r'^2-3r'^2=0$ for every source point, so $\sum_iQ_{ii}=0$ identically — equation (4). Symmetry is manifest.

Step 5. Count the room. A symmetric $3\times3$ tensor has six independent entries; the trace removes one, leaving five — $2\ell+1$ for $\ell=2$ . The five quadratic forms $Q_{ij}$ can generate are precisely $xy$ , $yz$ , $zx$ , $x^2-y^2$ , $3z^2-r^2$ : the real harmonics $r^2Y_2^m$ in Cartesian dress, as catalogued in Spherical Harmonics — and $V_{\mathrm{quad}}\propto\sum Q_{ij}x_ix_j/r^5$ is harmonic because $Q_{ij}$ is traceless. As a check: contracting $Q_{ij}=\mathrm{diag}(-\tfrac12,-\tfrac12,1)Q$ with $\hat r(\theta)$ reproduces $\tfrac12(3\cos^2\theta-1)$ at $\theta=0°,30°,54.74°,90°$ to six digits, and the kernel's trace vanishes for arbitrary point-charge sets.

One scalar when there is an axis

Most quadrupoles we meet — linear molecules, spinning nuclei, symmetric arrays — have an axis of symmetry, and five components are then four too many: symmetry forces the tensor diagonal with equal transverse entries, and tracelessness fixes those from the third. A single scalar survives,

Q\equiv Q_{zz}=\int\left(3z'^2-r'^2\right)\rho\,\mathrm{d}\tau' \qquad (5)

where: $z'$ — the coordinate along the symmetry axis. This is the number tables quote (nuclear physics divides by $e$ : $Q/e$ in $\mathrm{fm}^2$ or barns — same kernel as ours). The second derivation collapses the potential accordingly:

V_{\mathrm{quad}}=\frac{1}{4\pi\varepsilon_0}\,\frac{Q\left(3\cos^2\theta-1\right)}{4r^3}=\frac{Q\,P_2(\cos\theta)}{8\pi\varepsilon_0\,r^3} \qquad (6)

where: $\theta$ — the polar angle from the symmetry axis. This is exactly the $B_2P_2(\cos\theta)/r^3$ exterior mode from Separation of Variables in Spherical Coordinates, its coefficient now tied to a property of the source. On the axis the angular factor is $+2$ ; on the equator, $-1$ : opposite signs, the axis twice as strong — the four-lobe pattern in numbers. On the cone $\cos^2\theta=\tfrac13$ , i.e. $\theta=54.74°$ , the quadrupole potential vanishes at every distance.

Go deeper · Derivation B · 1 minAxial symmetry collapses the tensor to one scalar

Derivation BAxial symmetry collapses the tensor to one scalar

The goal is equation (6): for a source symmetric about the $z$ -axis, the five-component tensor reduces to the single scalar (5) and the potential to a pure $P_2(\cos\theta)/r^3$ shape. The strategy: let symmetry kill components one reflection at a time, let the trace fix what symmetry leaves related, then contract with an arbitrary direction. The factor bookkeeping in the last step is where readers lose a factor of two, so we do it slowly.

Step 1. Kill the off-diagonals. Axial symmetry means $\rho$ is unchanged under $x'\to-x'$ ; the integrand of $Q_{xz}$ is odd under that flip, so $Q_{xz}=0$ , likewise $Q_{yz}=Q_{xy}=0$ . The symmetry axis is a principal axis. Why this generalizes: any real symmetric $Q_{ij}$ can be diagonalized; symmetry merely names the frame in advance.

Step 2. Relate the diagonals. Rotating $90°$ about $z$ swaps $x'\leftrightarrow y'$ leaving $\rho$ fixed, so $Q_{xx}=Q_{yy}$ ; tracelessness (4) then gives

Q_{ij}=\mathrm{diag}\!\left(-\tfrac{Q}{2},\,-\tfrac{Q}{2},\,Q\right),\qquad Q\equiv Q_{zz}

One measured number reconstructs the whole tensor — why a nuclear quadrupole moment fits in one table entry.

Step 3. Contract with $\hat{\mathbf r}=(\sin\theta\cos\varphi,\ \sin\theta\sin\varphi,\ \cos\theta)$ :

\sum_{i,j}Q_{ij}\hat r_i\hat r_j=-\frac{Q}{2}\sin^2\theta+Q\cos^2\theta=Q\,\frac{3\cos^2\theta-1}{2}

Why $\varphi$ disappears: $\cos^2\varphi+\sin^2\varphi=1$ — azimuth cannot survive around a symmetric source.

Step 4. Insert into (3), keeping the halves straight — one from the potential formula, one from the contraction: $V_{\mathrm{quad}}=\frac{1}{4\pi\varepsilon_0}\frac{1}{2r^3}\cdot Q\,\frac{3\cos^2\theta-1}{2}$ , which is (6). The nodal cone sits at $\theta=\arccos(1/\sqrt3)=54.7356°$ . As a check: the Step 3 contraction agrees numerically at four angles to six digits, and equation (6) is confirmed independently in the next derivation, where brute-force superposition reproduces its axis value including the factor of 4.

When the first two rungs go silent

Equation (6) matters most where it leads. The dominance logic of Multipole Expansion of the Potential says the first non-vanishing moment rules the far field; the quadrupole inherits the throne whenever $Q_{\mathrm{tot}}=0$ and $\mathbf{p}=0$ together. Nature supplies such sources in bulk: any neutral distribution with inversion symmetry qualifies, since the dipole integrand $\mathbf{r}'\rho$ is then odd. CO $_2$ and N $_2$ are the textbook molecules — their $1/r^3$ handshake governs dry-ice packing and the leading anisotropy of their van der Waals interactions — and the linear array below is the same idea as a lab exercise. A bonus rides along, stated in one line: $Q_{ij}$ is origin-independent precisely when $Q_{\mathrm{tot}}=0$ and $\mathbf{p}=0$ — the leading moment never shifts with the origin — the systematic story being told in Origin Dependence of Multipole Moments. How does the world measure a moment that whispers at $1/r^3$ ? By energies, not potential surveys. Taylor-expanding a source's energy in an external potential (the expansion that gave $-\mathbf{p}\cdot\mathbf{E}$ in Electric Dipole Moment) produces at second order

U_{\mathrm{quad}}=-\frac{1}{6}\sum_{i,j}Q_{ij}\,\frac{\partial E_j}{\partial x_i} \qquad (7)

where: the derivatives form the electric field gradient (EFG) at the source's location; the trace ambiguity is harmless since $\nabla\cdot\mathbf{E}=0$ in the external vacuum field. A quadrupole feels nothing in a uniform field and couples only to how the field changes across it — the fact powering every experiment in the history band: park a nucleus inside a molecule or crystal, let the electrons supply a ferocious gradient, and read the orientation-dependent energy as a radio-frequency spectrum.

The linear quadrupole, end to end

Theory earns trust on a concrete case: three charges on a line — $+q$ at $z=\pm d$ , $-2q$ at the origin, two back-to-back dipoles sharing their negative end.

Figure

The linear quadrupole — geometry, silent lower moments, and the on-axis comparison point

Go deeper · Derivation C · 2 minThe linear quadrupole from moments to potential

Derivation CThe linear quadrupole from moments to potential

The goal is the complete pipeline on one source: verify that monopole and dipole vanish for the $+q,-2q,+q$ array, compute its quadrupole tensor from definition (2), predict the potential from (6), then hold the prediction against the exact potential from raw superposition — including a closed form for the error. The strategy is direct computation; the thing to watch is the kernel's geometry, where a lazy $3z'^2-r'^2\to3z'^2$ substitution (forgetting that on-axis charges have $r'^2=z'^2$ ) silently corrupts the answer.

Step 1. Audit the lower rungs. Total charge: $q-2q+q=0$ . Dipole: $p_z=q(d)+(-2q)(0)+q(-d)=0$ , and $p_x=p_y=0$ since all charges sit on the axis. Both rungs silent: the quadrupole owns the far field — and is origin-independent for this source.

Step 2. Compute the tensor. For charges on the $z$ -axis, $r_k'^2=z_k'^2$ , so the kernel is $2z_k'^2$ :

Q_{zz}=\sum_k q_k\left(2z_k'^2\right)=q\left(2d^2\right)+(-2q)(0)+q\left(2d^2\right)=4qd^2

Axial symmetry and the trace fix the rest: $Q_{xx}=Q_{yy}=-2qd^2$ ; check $-2-2+4=0$ ✓. Why the near-miss $2qd^2$ haunts this calculation: it is the primitive moment $\int z'^2\rho\,\mathrm{d}\tau'$ , and also Griffiths's half-sized convention — see the warning; in our normalization the answer is $4qd^2$ , full stop.

Step 3. Exact potential on the axis by superposition, at $z=r>d$ :

V_{\mathrm{exact}}=\frac{q}{4\pi\varepsilon_0}\left[\frac{1}{r-d}+\frac{1}{r+d}-\frac{2}{r}\right]=\frac{q}{4\pi\varepsilon_0}\,\frac{2d^2}{r\left(r^2-d^2\right)}

Why it collapses so neatly: the bracket is a discrete second difference — the structure that silenced the first two moments cancels the $1/r$ and $1/r^2$ pieces.

Step 4. Quadrupole prediction on the axis. Equation (6) at $\theta=0$ gives $V_{\mathrm{quad}}=\frac{1}{4\pi\varepsilon_0}\frac{2qd^2}{r^3}$ — the leading term of Step 3 expanded in $d/r$ , confirming the factor chain of both earlier derivations by an independent route.

Step 5. Quantify the error in closed form:

\frac{V_{\mathrm{quad}}}{V_{\mathrm{exact}}}=\frac{r^2-d^2}{r^2}=1-\left(\frac{d}{r}\right)^2

so the quadrupole term is low by exactly $(d/r)^2$ on the axis — 1% at $r=10d$ . The corrections are the $\ell=4,6,\dots$ rungs: odd moments vanish by inversion symmetry, so the first correction is hexadecapole. Numerically, with $q=1$ nC and $d=1$ mm at $r=10d$ : the exact potential is $18.1567$ V, the quadrupole approximation gives $17.9751$ V, and the error is $-1.0000\%$ — matching the closed form exactly; the trace is zero to machine precision. The full table appears in Worked Example 1.

Limiting cases

case	result	meaning
spherical source	$Q_{ij}=0$	only anisotropy feeds the quadrupole
axial source	$Q_{ij}=\mathrm{diag}(-\tfrac{Q}{2},-\tfrac{Q}{2},Q)$	one scalar reconstructs the tensor; $V\propto P_2(\cos\theta)/r^3$
on the axis, $\theta=0$	$V=+\dfrac{2Q}{16\pi\varepsilon_0 r^3}$	strongest lobe; positive for prolate positive charge
on the equator, $\theta=90°$	$V=-\dfrac{Q}{16\pi\varepsilon_0 r^3}$	half the axis strength, opposite sign — the belt
nodal cone, $\theta=54.74°$	$V_{\mathrm{quad}}=0$ at every $r$	$3\cos^2\theta=1$ ; NMR's magic angle
$r\gg d$ with $Q_{\mathrm{tot}}=\mathbf{p}=0$	$V\to V_{\mathrm{quad}}$ , error $\sim(d/r)^2$	quadrupole rules, origin-independent (CO $_2$ , N $_2$ , the linear array)
uniform external field	$U_{\mathrm{quad}}=0$	quadrupoles couple only to field gradients, equation (7)
point dipole limit	$Q_{ij}\to0$	no spatial extent, no second moment

Common trapEvery book has its own quadrupole — mixing two of them halves or doubles your answer silently

The physics is unambiguous but the bookkeeping is not, and the failure mode produces clean, plausible, wrong numbers. Jackson (and this article) defines $Q_{ij}=\int(3x_i'x_j'-r'^2\delta_{ij})\rho\,\mathrm{d}\tau'$ and keeps the $\tfrac12$ in the potential, equation (3). Griffiths (Problem 3.45, 4th ed.) tucks that $\tfrac12$ inside the moment — his $Q_{ij}$ is half ours — and writes $V$ with no visible $\tfrac12$ ; the molecular-physics Buckingham convention (behind tabulated values like CO $_2$ 's $\Theta\approx-4.3\times10^{-40}\ \mathrm{C\,m^2}$ ) does the same. Nuclear tables divide by $e$ and quote $Q/e$ in $\mathrm{fm}^2$ or barns with the full kernel, matching ours. Some texts use the primitive second moments $\int x_i'x_j'\rho\,\mathrm{d}\tau'$ — no factor 3, trace intact. Take $Q$ from one source and the $V$ formula from another and you get a silent factor error: our linear array is $4qd^2$ here and $2qd^2$ in Griffiths, with identical physical potentials. The fix is mechanical: before using any tabulated moment, find the source's own $V_{\mathrm{quad}}$ formula and check the pairing; anchor — the $+q,-2q,+q$ array's on-axis potential is $2kqd^2/r^3$ in every correct pairing.

Worked examples

1 · The linear quadrupole with numbers (confidence builder)

Three charges on the $z$ -axis: $+q$ at $z=\pm d$ and $-2q$ at the origin, with $q=1.0$ nC and $d=1.0$ mm. Find $Q_{zz}$ , then compare the exact on-axis potential with the quadrupole formula at $r=3d,\ 5d,\ 10d,\ 30d$ .

Step 1. The moment, from definition (2) with the on-axis kernel $2z'^2$ :

Q_{zz}=2qd^2+0+2qd^2=4qd^2=\boxed{4.0\times10^{-15}\ \mathrm{C\,m^2}}

with $Q_{xx}=Q_{yy}=-2.0\times10^{-15}\ \mathrm{C\,m^2}$ (trace check: $-2-2+4=0$ ✓).

Step 2. The two potentials on the axis. Exact: $V_{\mathrm{exact}}=kq\cdot2d^2/[r(r^2-d^2)]$ with $k=8.988\times10^9\ \mathrm{N\,m^2/C^2}$ . Quadrupole, from (6) at $\theta=0$ : $V_{\mathrm{quad}}=kQ_{zz}/(2r^3)=2kqd^2/r^3$ ; at $r=10d=1.0$ cm this is $17.98$ V.

Step 3. The table (all values recomputed):

$r$	$V_{\mathrm{exact}}$	$V_{\mathrm{quad}}$	error
$3d$	$748.96$ V	$665.74$ V	$-11.11\%$
$5d$	$149.79$ V	$143.80$ V	$-4.00\%$
$10d$	$18.157$ V	$17.975$ V	$-1.00\%$
$30d$	$0.66649$ V	$0.66575$ V	$-0.111\%$

The error column is $-(d/r)^2$ to every displayed digit — $-1/9,\ -1/25,\ -1/100,\ -1/900$ — exactly as Derivation C's closed form predicts, the first neglected rung being $\ell=4$ .

Step 4. One off-axis check: on the equator at $r=10d$ the quadrupole formula gives $-8.988$ V (exact: $-8.921$ V) — sign flipped, magnitude half the axis value: the four-lobe pattern in volts. Ten source-sizes away, the quadrupole term is already accurate to 1%.

2 · The deuteron's quadrupole moment, and what it bought physics (exam level)

The deuteron's quadrupole moment is $Q/e=+0.286\ \mathrm{fm}^2$ . (a) Express $Q$ in SI units. (b) Inside a D $_2$ molecule the deuteron sits in the electric field gradient of the molecular electrons, $V_{zz}\approx3.25\times10^{21}\ \mathrm{V/m^2}$ . Estimate the quadrupole coupling energy and frequency. (c) What did this number teach nuclear physics?

Step 1. Units. Nuclear tables quote $Q/e$ with the kernel of equation (5), so

Q=0.286\times\left(1.602\times10^{-19}\ \mathrm{C}\right)\times\left(10^{-15}\ \mathrm{m}\right)^2=\boxed{4.58\times10^{-50}\ \mathrm{C\,m^2}}

( $0.286\ \mathrm{fm}^2=0.00286$ b.) Positive sign: the deuteron is slightly prolate — a cigar, not a pancake.

Step 2. The coupling scale. The energy (7) for an axial moment in an axial gradient is of order the spectroscopists' "quadrupole coupling constant" $eqQ\equiv Q\,V_{zz}$ (orientation factor order one; $-\tfrac14$ aligned):

Q\,V_{zz}=\left(4.58\times10^{-50}\right)\times\left(3.25\times10^{21}\right)=1.49\times10^{-28}\ \mathrm{J}=9.3\times10^{-10}\ \mathrm{eV}

As a frequency: $\nu=QV_{zz}/h=\boxed{\approx225\ \mathrm{kHz}}$ — squarely the radio-frequency splitting Kellogg, Rabi, Ramsey, and Zacharias resolved. Feel the smallness: $9.3\times10^{-10}$ eV is $\sim3\times10^{7}$ times below room-temperature thermal energy and $\sim4\times10^{-16}$ of the deuteron's 2.22 MeV binding. Only a resonance method — comparing radio frequencies rather than measuring energies outright — could see it in 1939.

Step 3. The physics purchased. A two-nucleon ground state bound by a purely central force would be pure S-wave — spherically symmetric — and (5) would integrate to zero. The measured $Q\neq0$ proved the neutron-proton force depends on the spins' orientation relative to the separation vector: the tensor force, which mixes a few percent of D-wave into the deuteron. One number of size $10^{-9}$ eV rewrote the symmetry of the strong force.

Origins · 1 minIsidor Isaac Rabi and how this was discovered

Isidor Isaac Rabi

1898–1988 · Columbia University and the molecular-beam laboratory

Rabi arrived in New York from Galicia as an infant, grew up over his father's grocery on the Lower East Side, and liked to credit his career to his mother's daily question — not "what did you learn?" but "did you ask a good question today?". The question he built a laboratory around was how to make a single molecule report its interior, and the answer — molecular-beam magnetic resonance, 1938 — let molecules flying through vacuum absorb radio waves only at frequencies matching their internal splittings. The method was meant to weigh magnetic moments; it promptly delivered a surprise nobody ordered. In 1939 Rabi, with Kellogg, Ramsey, and Zacharias, found the HD and D $_2$ resonance patterns scarred by extra structure — a few hundred kilohertz no magnetic interaction could explain, but an electric quadrupole coupling could. The deuteron was not round; the nuclear force was not central. Rabi took the 1944 Nobel Prize for the resonance method itself, whose descendants — NMR, MRI, atomic clocks, the cesium standard that defines the second — make his lab arguably the most consequential room in twentieth-century metrology. "Who ordered that?", his verdict on the muon, remains physics' best one-line review.

History

1918

— Einstein derives the quadrupole formula for gravitational radiation: gravity's ladder begins at the rung this article studies.

1935

— Schüler and Schmidt trace anomalous hyperfine structure in europium to non-spherical nuclei — the first sighting of nuclear quadrupole moments.

1939

— Kellogg, Rabi, Ramsey, and Zacharias resolve the deuteron's quadrupole coupling in HD and D2 beams; the tensor force enters nuclear physics.

1944

— Rabi receives the Nobel Prize for molecular-beam magnetic resonance.

1950

— Dehmelt and Krüger detect, and Pound analyzes, pure nuclear quadrupole resonance in crystals — no magnet required, the crystal's own field gradient does the work.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §3.4 (Multipole Expansion) · Jackson, Classical Electrodynamics, 3rd ed., §4.1–4.2 · Zangwill, Modern Electrodynamics, §4.4 · Purcell & Morin, Electricity and Magnetism, 3rd ed., §10.2 Key numerical anchors: the four-charge example gives $Q_{zz}=4qa^2$ with $\mathrm{Tr}\,Q=0$ ; the prefactors are $1/(8\pi\varepsilon_0 r^3)$ in the axial potential and $-\tfrac16$ in the gradient coupling energy. The barn ( $10^{-28}\,\mathrm{m}^2$ ) is the standard SI-derived nuclear cross-section unit. On-axis table values: $18.1567$ V (exact) vs $17.9751$ V (quadrupole) at $r=10d$ , with error exactly $-(d/r)^2$ ; nodal angle $54.7356°$ ; deuteron moment $Q=4.58\times10^{-50}\ \mathrm{C\,m^2}$ and coupling frequency $QV_{zz}/h\approx225$ kHz.

Builds on: Laplace's Equation · Separation of Variables in Spherical Coordinates · Legendre Polynomials · Poisson's Equation Leads to: Boundary Conditions for the Potential · Harmonic Functions and the Mean-Value Theorem · Method of Images

Why is the linear quadrupole 4qd² and not 2qd²? Two dipoles of moment qd separated by d suggests 2qd².

Both numbers are "right" — in different conventions, the classic trap. The two-dipole estimate computes the primitive moment $\sum q_kz_k'^2=2qd^2$ . Our kernel is $3z'^2-r'^2$ , which on the axis (where $r'^2=z'^2$ ) equals $2z'^2$ , doubling that to $4qd^2$ ; Griffiths's convention, with its $\tfrac12$ inside the definition, lands back on $2qd^2$ . The potentials all agree: the physical on-axis answer is $2kqd^2/r^3$ in every correct pairing. If your number differs from a book's by a factor of 2, audit the conventions before the arithmetic.

Does the quadrupole moment depend on where I put the origin?

In general, yes — the one-line version of a story told properly in Origin Dependence of Multipole Moments. Shifting the origin mixes lower moments upward, through terms proportional to the total charge and the dipole moment; so $Q_{ij}$ is origin-independent precisely when both vanish — exactly where it leads. The pattern is general: the first non-vanishing moment is an intrinsic, origin-free property; everything above it is bookkeeping relative to your chosen origin. For CO $_2$ or the $+q,-2q,+q$ array you may quote $Q$ without naming an origin; for an ion you may not.

What does it mean for a nucleus to have a quadrupole moment when it is a quantum object without a fixed shape?

The tabulated $Q$ is the expectation value of $\sum(3z_k^2-r_k^2)$ over the protons in the stretched state, with the nuclear spin maximally aligned ( $m=j$ ). Two consequences surprise people. A nucleus with spin 0 or ½ shows no quadrupole moment whatever its intrinsic deformation — too few orientation states to exhibit a rank-2 anisotropy (angular-momentum selection). And a measured $Q\neq0$ still reports genuine shape: the deuteron's positive sign means prolate, and its smallness — 0.286 fm² against a ~2 fm size — means a subtle cigar, a few-percent D-wave admixture.

Why does a quadrupole feel only field gradients and not the field itself?

Expand the source's energy in an external potential about its center: the zeroth-order term couples to total charge (zero here), the first-order to the dipole moment (zero here), and the surviving second-order term is $-\tfrac16\sum Q_{ij}\,\partial_iE_j$ — built from derivatives of the field. Physically, a uniform field pulls the array's positive ends one way and its doubled negative middle the other with forces that cancel exactly; only a field that differs across the source leaves a net energy. Hence quadrupole spectroscopy relies on the ferocious gradients inside molecules and crystals — of order $10^{21}\ \mathrm{V/m^2}$ — not lab electrodes.

What is the difference between a quadrupole moment and the quadrupole field in a mass spectrometer?

Same mathematics, different actor. A quadrupole moment is a property of a charge distribution — five numbers describing its shape anisotropy — producing a $1/r^3$ potential outside itself. A quadrupole field device (mass filter, ion trap, focusing magnet) is a set of external electrodes shaped so the potential in the working region is a pure $\ell=2$ harmonic like $V\propto x^2-y^2$ : an interior solution of Laplace's equation, growing as $r^2$ , with no charge in the region at all. The shared DNA is the five-member $\ell=2$ family — one in the exterior expansion ( $r^{-3}$ ), one in the interior expansion ( $r^{+2}$ ) of Separation of Variables in Spherical Coordinates.

Why does gravitational radiation start at the quadrupole?

Because gravity's "charge" is mass-energy, and its lower moments are frozen by conservation laws. The monopole is the total mass — conserved, so it cannot oscillate and radiate. The dipole is $\sum m_k\mathbf{r}_k$ , whose time derivative is the total momentum — also conserved, so no dipole radiation either (electromagnetism escapes only because charges come in two signs, letting $\mathbf{p}$ oscillate while momentum stays fixed). The first moment free to change is the mass quadrupole — Einstein's 1918 formula, and why LIGO's observables are second time-derivatives of a $Q_{ij}$ -like tensor. Two orbiting black holes are the cosmos's linear quadrupole, swung at kilohertz rates.