Origin Dependence of Multipole Moments

Three articles in a row have now made the same small confession. Monopole Moment proved in one line that $Q$ never depends on the origin. Electric Dipole Moment checked that $\mathbf{p}$ survives a shift only when $Q=0$ , and deferred "the systematic story" here. Electric Quadrupole Moment quoted the pattern — the first non-vanishing moment is origin-free — and owed a proof. This article pays all three debts at once. The issue is not pedantry. Every multipole moment is an integral of $\rho$ weighted by powers of the source position $\mathbf{r}'$ — and $\mathbf{r}'$ is measured from somewhere. Move that somewhere, and the numbers move: the "dipole moment" of a single point charge can be made anything you like, including zero, by parking the origin cleverly. Yet the far field is a physical fact that no bookkeeping choice can touch. Something in the ladder must be invariant. What, exactly, and when? The answer has the clean structure of a triangular matrix. Shifting the origin by $\mathbf{a}$ feeds each moment corrections built only from the moments below it — monopole into dipole, monopole and dipole into quadrupole, never upward. So the first rung that isn't zero receives no corrections at all: it is a genuine property of the charge distribution, quotable without naming an origin. Everything above it is convention. Along the way we earn a practical tool — the center of charge $\mathbf{p}/Q$ , the unique origin that kills a charged system's dipole term, exact analogue of the center of mass — and we see in worked numbers why a well-chosen origin makes a truncated multipole series dramatically more accurate for free.

Key takeaways

Multipole moments are integrals of $\rho$ weighted by powers of $\mathbf{r}'$ , so their values are tied to the origin the source coordinates are measured from.
Under a shift of origin by $\mathbf{a}$ : $Q' = Q$ always; $\mathbf{p}' = \mathbf{p} - Q\mathbf{a}$ ; the traceless quadrupole picks up corrections proportional to $\mathbf{p}$ and $Q$ .
The mixing is strictly one-way — lower moments contaminate higher ones, never the reverse — so the first non-vanishing moment is origin-independent, and everything above it is bookkeeping.
For $Q \neq 0$ the origin $\mathbf{a}_c = \mathbf{p}/Q$ (the center of charge) makes $\mathbf{p}' = 0$ — the electrostatic twin of the center of mass, and the standard reference point for quoting an ion's higher moments.
Practically: expanding about a good origin kills whole rungs of the ladder; our worked numbers show the monopole-only estimate improving from $-2.0\%$ error to $+0.03\%$ just by moving the origin.
KEY IDEA: an origin shift feeds each multipole moment corrections built only from lower moments, so the first non-vanishing moment is an origin-independent invariant — and all higher moments are meaningful only once a reference point is declared.

The core insight

Under a translation of the origin by $\mathbf{a}$ , the multipole moment of order $\ell$ changes by terms built from moments of order $< \ell$ ; hence the first nonvanishing moment is an origin-independent invariant and all higher ones are convention-dependent.

Where this lives today

Origin bookkeeping is a daily, load-bearing concern wherever multipoles are tabulated. Quantum-chemistry packages print a molecule's dipole and quadrupole with the origin declared in the output header (center of mass, center of nuclear charge, or a user-set point) precisely because for ions and polar species the numbers are otherwise undefined; comparing a computed quadrupole of water against experiment means first shifting both to a common origin with exactly the laws of this article. Standard tabulations of molecular quadrupoles — the tradition runs through Buckingham's mid-century formalism — quote CO $_2$ 's famous negative $Q_{zz}$ with no origin footnote only because its $Q$ and $\mathbf{p}$ vanish; the moment the molecule is protonated, the footnote becomes mandatory. The theorem's reach extends well past electrostatics, because it is really a statement about translations acting on moments of any density. In gravity every mass distribution has $M > 0$ , so the "dipole" can always be transformed away — the center of charge is the center of mass there, and expanding Earth's field about it makes the leading correction the quadrupole $J_2$ , the oblateness term that satellite orbits and GRACE gravimetry measure. The same triangle underlies gravitational-wave physics: the invariant mass monopole and the killable dipole cannot radiate, so the first moment free to oscillate — the quadrupole — carries every LIGO chirp. One lower-triangular matrix, three fields of physics.

Origins · 1 minJames Joseph Sylvester and how this was discovered

James Joseph Sylvester

1814–1897 · London, Baltimore, Oxford

Sylvester was the great algebraist of invariants — coiner of the words "matrix," "discriminant," and "invariant" itself — and the multipole ladder's origin structure is invariant theory in its natural habitat: quantities that transform triangularly under a group (here, translations), with the leading survivor an honest invariant. His direct contribution to this article's subject is the elegant 1876 note proving what is now called the Sylvester–Maxwell theorem: every harmonic polynomial of degree $\ell$ — every multipole — corresponds to a unique set of $\ell$ directions, "poles" on a sphere, tying his canonical forms for binary and ternary quantics to Maxwell's Treatise picture of multipoles as successive derivatives of $1/r$ along chosen axes. Sylvester's career was itself displaced repeatedly from its origin: barred from a Cambridge degree as a Jew, he detoured through actuarial work and the law (where he met his lifelong collaborator Cayley), founded the American Journal of Mathematics at Johns Hopkins at sixty-one, and took Oxford's Savilian chair at sixty-nine — proof by example that what matters is invariant under translation.

History

1812

— Poisson's memoirs on electrified conductors organize far fields by successive powers of distance, the ladder in embryo

1873

— Maxwell's Treatise constructs multipoles as repeated directional derivatives of the point-charge potential — poles on a sphere

1876

— Sylvester's note on spherical harmonics proves the canonical pole decomposition, gluing invariant theory to electrostatics

1912

— Debye's polar-molecule theory makes the neutral-system dipole a tabulated, origin-free material constant

1959

— Buckingham's molecular multipole formalism standardizes origin conventions for quoting quadrupoles of polar species

The physical picture

Stand far from a charge cloud and squint. The first feature you resolve is its net charge; if that vanishes, its lopsidedness (the dipole); if that too vanishes, its shape (the quadrupole). Whatever leads is a fact about the cloud — you never told the cloud where your coordinate origin was, so the leading signal cannot possibly depend on it. That is the invariance half of the story, and it is forced by physics before any algebra is done. The origin-dependence half has an old friend from mechanics. A body's mass is origin-free, but its moment of inertia — a second moment of the mass distribution — depends on the axis, through the parallel-axis theorem $I' = I_{\rm cm} + Ma^2$ . Note the anatomy of that correction: it is built from the lower moment $M$ , times powers of the shift. Electrostatics repeats the pattern rung by rung: $\mathbf{p}' = \mathbf{p} - Q\mathbf{a}$ is a parallel-axis theorem for the dipole, with the monopole $Q$ playing the mass; the quadrupole's shift law is the same idea one floor up, fed by both $\mathbf{p}$ and $Q$ .

Figure

One cloud, two origins — what moves and what cannot

The one-way direction of the mixing is worth internalizing before the proof, because it is the whole theorem. Replace $\mathbf{r}'$ by $\mathbf{r}' - \mathbf{a}$ inside an $\ell$ -th moment and expand: every cross term has fewer factors of $\mathbf{r}'$ than you started with — the shift can only demote position factors to constants, never mint new ones. So an $\ell$ -pole integrand spills downward into $(\ell-1), (\ell-2), \dots$ pole integrals, and information flows strictly from low moments to the corrections of high ones. Run the logic backward and the invariance theorem falls out: if every moment below $\ell$ vanishes, there is nothing to spill, and the $\ell$ -th moment stands untouched by any shift whatsoever.

Figure

The triangular mixing rule — corrections flow upward from lower rungs only

There is also a constructive payoff. If $Q \neq 0$ , the dipole shift law $\mathbf{p}' = \mathbf{p} - Q\mathbf{a}$ is not a nuisance but a knob: dial $\mathbf{a} = \mathbf{p}/Q$ and the dipole term is gone. This center of charge is the charge-weighted mean position, the exact electrostatic analogue of the center of mass, and expanding about it hands you a series that jumps straight from monopole to quadrupole — one whole rung of accuracy for free, as worked example 1 demonstrates with numbers.

The transformation laws

Everything begins from the ladder built in Multipole Expansion of the Potential — recalled here, not re-derived. For a localized $\rho$ and field points outside it,

V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\left[\frac{Q}{r} + \frac{\mathbf{p}\cdot\hat{\mathbf{r}}}{r^2} + \frac{1}{2}\,\frac{\hat{r}_i\hat{r}_j\,Q_{ij}}{r^3} + \cdots\right] \qquad (1)

where: $Q = \int\rho\,\mathrm{d}\tau'$ — the monopole moment (total charge); $\mathbf{p} = \int\mathbf{r}'\rho\,\mathrm{d}\tau'$ — the dipole moment; $Q_{ij} = \int(3r'_ir'_j - r'^2\delta_{ij})\rho\,\mathrm{d}\tau'$ — the traceless Cartesian quadrupole (Griffiths-family convention; summation over repeated indices). Each coefficient weighs $\rho$ by a homogeneous polynomial of degree $\ell$ in $\mathbf{r}'$ — degree $0$ , $1$ , $2$ respectively — and that degree is what decides how the moment responds to moving the origin. Now move the origin to $O'$ , displaced by $\mathbf{a}$ from $O$ . The physics — the cloud, the field, the potential at every point of space — is untouched; the only thing that changes is the label each source point wears: $\mathbf{r}'_{\rm new} = \mathbf{r}' - \mathbf{a}$ . The three laws that result:

Q' = Q, \qquad \mathbf{p}' = \mathbf{p} - Q\,\mathbf{a}, \qquad Q'_{ij} = Q_{ij} - 3(a_ip_j + a_jp_i) + 2(\mathbf{a}\cdot\mathbf{p})\,\delta_{ij} + Q\,(3a_ia_j - a^2\delta_{ij}) \qquad (2)

where: primes denote moments computed about $O'$ . Read the structure, not the clutter: the correction to each moment is a polynomial in $\mathbf{a}$ whose coefficients are lower moments only — $Q$ feeds $\mathbf{p}'$ ; $\mathbf{p}$ and $Q$ feed $Q'_{ij}$ ; nothing feeds $Q'$ . Note also that each correction has the right tensor character and the quadrupole's correction is itself traceless (contract $i=j$ : $-6\,\mathbf{a}\cdot\mathbf{p} + 6\,\mathbf{a}\cdot\mathbf{p} + Q(3a^2 - 3a^2) = 0$ ), as it must be for the shifted moment to remain a legal traceless quadrupole.

Derivation AThe shift laws for monopole, dipole, and quadrupole

We derive all three laws of (2) by the same two-step move — substitute the relabeled coordinate, expand, and recognize the surviving integrals as lower moments — and then close with a consistency check the formulas must pass: performing two shifts in a row has to equal performing their sum, or the "laws" are not laws. Watch where each correction's moment-order comes from: every factor of $\mathbf{a}$ pulled out of the integrand costs one factor of $\mathbf{r}'$ , which is precisely the demotion mechanism behind the triangular structure.

Step 1. The monopole. About the new origin,

Q' = \int \rho(\mathbf{r}')\,\mathrm{d}\tau' = Q.

There is nothing to substitute into: the integrand carries no position factor, so relabeling positions cannot touch it. (Relabeling also leaves $\rho$ and $\mathrm{d}\tau'$ alone — we are renaming points, not moving charge.) The total charge is origin-independent for every distribution, with no hypothesis needed.

Step 2. The dipole. Substitute the new coordinate into the degree-one weight and split:

\mathbf{p}' = \int(\mathbf{r}' - \mathbf{a})\,\rho\,\mathrm{d}\tau' = \int\mathbf{r}'\rho\,\mathrm{d}\tau' - \mathbf{a}\int\rho\,\mathrm{d}\tau' = \mathbf{p} - Q\,\mathbf{a}.

Why the correction is a monopole term: pulling the constant $\mathbf{a}$ outside leaves $\int\rho$ — the shift traded the one position factor for a constant, demoting a dipole integrand to a monopole one. If $Q = 0$ the correction dies for every $\mathbf{a}$ : a neutral system's dipole moment is an intrinsic property.

Step 3. The quadrupole. Substitute into the degree-two weight, expanding both pieces:

3(r'_i - a_i)(r'_j - a_j) = 3r'_ir'_j - 3(a_ir'_j + a_jr'_i) + 3a_ia_j, \qquad |\mathbf{r}' - \mathbf{a}|^2 = r'^2 - 2\,\mathbf{a}\cdot\mathbf{r}' + a^2.

Assemble the integrand $3(r'_i - a_i)(r'_j - a_j) - |\mathbf{r}'-\mathbf{a}|^2\delta_{ij}$ , integrate term by term, and recognize $\int r'_i\rho = p_i$ , $\int\rho = Q$ :

Q'_{ij} = Q_{ij} - 3(a_ip_j + a_jp_i) + 2(\mathbf{a}\cdot\mathbf{p})\,\delta_{ij} + Q\,(3a_ia_j - a^2\delta_{ij}).

Why two flavors of correction: the terms linear in $\mathbf{a}$ kept one factor of $\mathbf{r}'$ and so integrate to dipole moments; the terms quadratic in $\mathbf{a}$ kept none and integrate to $Q$ . Degree counting alone dictates which lower moments can appear — the coefficients just fill in the tensor bookkeeping.

Step 4. The composition check. Shift by $\mathbf{a}$ , then by $\mathbf{b}$ : the dipole law applied twice gives $\mathbf{p}'' = \mathbf{p}' - Q'\mathbf{b} = \mathbf{p} - Q(\mathbf{a}+\mathbf{b})$ , exactly the single-shift law — because $Q' = Q$ . The same bookkeeping closes for $Q''_{ij}$ , the cross terms assembling precisely because the second shift is fed by the already-shifted $\mathbf{p}'$ . Translations compose; the laws respect it. Three consistency checks close the derivation: worked example 2 confirms the quadrupole law by direct recomputation from shifted coordinates, giving $1.200\times10^{-14}\,\mathrm{C\,m^2}$ both ways; contracting the correction with $\delta_{ij}$ gives zero, confirming the shifted moment remains traceless; and setting $Q=0$ , $\mathbf{p}=0$ kills every correction term, so the quadrupole is invariant whenever both lower moments vanish.

The pattern begs to be proved once and for all, at every order. The clean way is to work with the raw (pre-traceless) moments — the trace subtractions are linear combinations of the same integrals and inherit the result automatically.

Go deeper · Derivation B · 2 minThe invariance theorem and the center of charge

Derivation BThe invariance theorem and the center of charge

The goal is the general theorem: if every moment of order $k < \ell$ vanishes, the $\ell$ -th moment is origin-independent — for all $\ell$ , in one argument — followed by its constructive corollary, the center of charge. The strategy is the multinomial expansion of the shifted weight: we show every correction term is proportional to some lower-order moment, so the hypothesis wipes them all out at a stroke. The only subtlety worth watching is that the theorem needs the lower moments to vanish entirely (all components), not merely the ones along $\mathbf{a}$ .

Step 1. Define the raw $\ell$ -th moments, the full symmetric array of degree- $\ell$ position weights:

M^{(\ell)}_{i_1\cdots i_\ell} = \int r'_{i_1}\,r'_{i_2}\cdots r'_{i_\ell}\;\rho(\mathbf{r}')\,\mathrm{d}\tau'.

$M^{(0)} = Q$ , $M^{(1)}_i = p_i$ , and the traceless $Q_{ij} = 3M^{(2)}_{ij} - \delta_{ij}M^{(2)}_{kk}$ — every physical moment is a fixed linear combination of raw ones, so whatever we prove for the $M$ 's transfers.

Step 2. Shift and expand. Each factor in the weight becomes $(r'_{i_n} - a_{i_n})$ ; multiply out the product of $\ell$ binomials:

\prod_{n=1}^{\ell}\left(r'_{i_n} - a_{i_n}\right) = r'_{i_1}\cdots r'_{i_\ell} \;+\; \sum_{k=0}^{\ell-1}\;(\pm)\,\big(\ell-k\text{ factors of } \mathbf{a}\big)\times\big(k\text{ factors of } \mathbf{r}'\big).

The leading term reproduces the old integrand. Every other term swapped at least one $\mathbf{r}'$ for an $\mathbf{a}$ — its degree in the source position is some $k$ strictly less than $\ell$ .

Step 3. Integrate term by term. Constants come out; what remains under the integral in a degree- $k$ term is exactly a raw moment of order $k$ :

M'^{(\ell)} = M^{(\ell)} + \sum_{k=0}^{\ell-1} \big(\text{polynomial in } \mathbf{a}\big)\times M^{(k)}.

This is the triangular structure in full generality: the shifted $\ell$ -th moment equals the old one plus corrections fed only by orders $k < \ell$ . No expansion step can increase the number of position factors, so no higher moment ever appears — the matrix of the transformation, ordered by $\ell$ , is lower-triangular with ones on the diagonal.

Step 4. Apply the hypothesis. If $M^{(k)} = 0$ for every $k < \ell$ — all components, which is what "the lower moments vanish" must mean — then every correction term dies identically, for every shift $\mathbf{a}$ :

M'^{(\ell)} = M^{(\ell)}.

The first non-vanishing moment is an invariant of the distribution. And the converse bites too: if some lower moment survives, one can always find a shift that changes $M^{(\ell)}$ — higher moments of a charged or polar system are conventions until an origin is declared.

Step 5. The corollary. For $Q \neq 0$ , ask which origin kills the dipole: $\mathbf{p}' = \mathbf{p} - Q\mathbf{a} = 0$ has the unique solution

\mathbf{a}_c = \frac{\mathbf{p}}{Q} = \frac{1}{Q}\int\mathbf{r}'\rho\,\mathrm{d}\tau',

the charge-weighted mean position — the center of charge. About it the expansion runs monopole, skip, quadrupole: the leading correction to the point-charge picture starts at $1/r^3$ . For a single point charge the recipe returns the origin to the charge itself ( $\mathbf{a}_c = \mathbf{r}_0$ ), as it must. Note what the corollary does not promise: no shift can also kill the quadrupole — one vector $\mathbf{a}$ cannot zero a five-component tensor — so "the origin that simplifies everything" stops at the dipole rung. As a check, the general theorem reproduces all three special cases: $Q' = Q$ for $\ell = 0$ (the hypothesis is empty), $\mathbf{p}' = \mathbf{p}$ when $Q = 0$ , and $Q'_{ij} = Q_{ij}$ when $Q = 0$ and $\mathbf{p} = 0$ ; worked example 1 confirms $\mathbf{a}_c$ numerically, with the dipole about the center of charge equal to zero to machine precision.

For the record, the same theorem holds in the spherical language of Spherical Harmonics: the moments $q_{\ell m}$ shift by mixing only $q_{km'}$ with $k < \ell$ , because $r'^\ell Y^*_{\ell m}$ is a degree- $\ell$ harmonic polynomial and shifting re-expands it into lower degrees. Same triangle, fancier indices — the result is a fact about translations, not about Cartesian bookkeeping.

Limiting cases

Case	Result	Meaning
Any distribution	$Q' = Q$	total charge is origin-free, always — the $\ell=0$ hypothesis is empty
$Q \neq 0$	$\mathbf{p}' = \mathbf{p} - Q\mathbf{a}$	the dipole is pure convention; $\mathbf{a} = \mathbf{p}/Q$ (center of charge) zeroes it
$Q = 0$	$\mathbf{p}' = \mathbf{p}$	a neutral system's dipole is intrinsic — why water's $1.85\,\mathrm{D}$ needs no footnote
$Q = 0,\ \mathbf{p} = 0$	$Q'_{ij} = Q_{ij}$	the quadrupole leads and is invariant — CO $_2$ , N $_2$ , the linear array
$Q = 0,\ \mathbf{p} = 0,\ Q_{ij}=0$	octupole invariant	the triangle continues rung by rung, same proof
single point charge $q$ at $\mathbf{r}_0$	$\mathbf{p} = q\mathbf{r}_0$ about $O$ , $= 0$ about the charge	the extreme case: every higher moment is pure origin bookkeeping
shift then shift ( $\mathbf{a}$ , then $\mathbf{b}$ )	same as one shift by $\mathbf{a}+\mathbf{b}$	the laws compose — translations form a group and the algebra respects it

Common trapA quoted dipole moment is meaningless until someone names the origin — unless the system is neutral

The reflex "p is a property of the distribution" is only true when $Q = 0$ . For an ion, $\mathbf{p}' = \mathbf{p} - Q\mathbf{a}$ can be given any value whatsoever — including zero — by choosing the origin, so a paper reporting "the dipole moment of OH $^-$ " without a reference point has reported nothing. (Conventions differ, too: quantum chemists usually quote ion moments about the center of mass, which is generally not the center of charge — the two agree only when charge tracks mass.) The same trap climbs the ladder: a polar neutral molecule's quadrupole is origin-dependent because $\mathbf{p} \neq 0$ feeds it. Before comparing any two tabulated higher moments, check three things: the net charge, the lower moments, and the declared origin.

Worked examples

1 · An ion's dipole is a knob — turn it to zero (confidence builder)

Two point charges on the $x$ -axis: $q_1 = +3.0\,\mathrm{nC}$ at $x = +1.0\,\mathrm{cm}$ and $q_2 = -1.0\,\mathrm{nC}$ at $x = -1.0\,\mathrm{cm}$ . Find $Q$ , $\mathbf{p}$ about the origin, the center of charge, and the dipole there; then test what the shift buys at a field point $x = 1.0\,\mathrm{m}$ .

Step 1. The moments about $O$ . $Q = 3.0 - 1.0 = +2.0\,\mathrm{nC}$ . Dipole: $p_x = (3.0\,\mathrm{nC})(0.01\,\mathrm{m}) + (-1.0\,\mathrm{nC})(-0.01\,\mathrm{m}) = 4.0\times10^{-11}\,\mathrm{C\,m}$ .

Step 2. Center of charge: $a_c = p_x/Q = (4.0\times10^{-11})/(2.0\times10^{-9}) = 0.020\,\mathrm{m}$ — two centimeters down the axis, outside the segment between the charges (nothing wrong with that; it is a weighted mean with a negative weight in the mix). Dipole about $\mathbf{a}_c$ : $p'_x = p_x - Qa_c = 4.0\times10^{-11} - (2.0\times10^{-9})(0.020) = 0$ ✓.

Step 3. The payoff at $x = 1.0\,\mathrm{m}$ . Exact potential: $V = k\left[\frac{3.0\,\mathrm{nC}}{0.99\,\mathrm{m}} + \frac{-1.0\,\mathrm{nC}}{1.01\,\mathrm{m}}\right] = 18.336\,\mathrm{V}$ . Monopole-only estimate about $O$ : $kQ/r = 17.975\,\mathrm{V}$ — off by $-2.0\%$ . Monopole-only about the center of charge (distance $0.98\,\mathrm{m}$ ): $kQ/0.98 = 18.342\,\mathrm{V}$ — off by $+0.03\%$ .

The same one-term formula, sixty-odd times more accurate, purely because the origin was chosen to silence the dipole rung. (About $\mathbf{a}_c$ the next correction is the quadrupole, $Q'_{zz}$ -type terms of order $(d/r)^2$ ; adding it lands within $0.002\%$ of exact. About $O$ you would have needed the dipole term just to reach $-0.01\%$ .)

2 · Shift a quadrupole's origin — formula versus honest recount (exam level)

Three charges on the $z$ -axis: $+q$ at $z = +d$ , $+q$ at $z = -d$ , $-q$ at the origin, with $q = 1.0\,\mathrm{nC}$ , $d = 1.0\,\mathrm{mm}$ . (Net charge $+q$ : this is a charged system, so the ladder above the monopole is up for grabs.) Compute $\mathbf{p}$ and $Q_{zz}$ about $O$ ; then about $O'$ at $z = +2.0\,\mathrm{mm}$ , both by the transformation laws and by direct recomputation.

Step 1. About $O$ . $Q = +1.0\,\mathrm{nC}$ . By symmetry $p_z = q(d) + q(-d) - q(0) = 0$ . Quadrupole (axial charges have $r'^2 = z'^2$ , so $3z'^2 - r'^2 = 2z'^2$ ): $Q_{zz} = 2q d^2 + 2q d^2 - 0 = 4qd^2 = 4.0\times10^{-15}\,\mathrm{C\,m^2}$ .

Step 2. About $O'$ , by the laws (2) with $\mathbf{a} = a\hat{\mathbf{z}}$ , $a = 2.0\,\mathrm{mm}$ . Dipole: $p'_z = p_z - Qa = 0 - (1.0\times10^{-9})(2.0\times10^{-3}) = -2.0\times10^{-12}\,\mathrm{C\,m}$ — a dipole moment appears from nowhere, courtesy of the monopole. Quadrupole: with $\mathbf{p} = 0$ the dipole-fed terms vanish, leaving $Q'_{zz} = Q_{zz} + Q(3a^2 - a^2) = 4.0\times10^{-15} + (1.0\times10^{-9})(2)(4.0\times10^{-6}) = 1.2\times10^{-14}\,\mathrm{C\,m^2}$ — tripled by the shift.

Step 3. The honest recount. Relative to $O'$ the charges sit at $z = -1.0$ , $-3.0$ , $-2.0\,\mathrm{mm}$ (charges $+q, +q, -q$ ). Dipole: $q(-0.001) + q(-0.003) - q(-0.002) = -2.0\times10^{-12}\,\mathrm{C\,m}$ ✓. Quadrupole: $2q\left[(0.001)^2 + (0.003)^2 - (0.002)^2\right] = 2q\,(6.0\times10^{-6}) = 1.2\times10^{-14}\,\mathrm{C\,m^2}$ ✓. The laws are just pre-packaged recounting.

Step 4. Same physics, different bookkeeping. On the axis at $z = 5.0\,\mathrm{cm}$ , the exact potential is $179.895\,\mathrm{V}$ . The three-term series about $O$ : $k[Q/r + 0 + Q_{zz}/2r^3] = 179.751 + 0.144 = 179.895\,\mathrm{V}$ — agreement to $3\times10^{-5}\,\%$ . About $O'$ (field distance $48\,\mathrm{mm}$ ): $187.24 - 7.80 + 0.49 = 179.93\,\mathrm{V}$ — the same answer to the order retained ( $0.018\%$ off, the truncation error of a series whose moments the shift inflated). Both origins tell the truth; the symmetric one just tells it in fewer words.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §3.4 (Multipole Expansion) · Jackson, Classical Electrodynamics, 3rd ed., §4.1–4.2 · Purcell & Morin, Electricity and Magnetism, 3rd ed., §10.3 · Zangwill, Modern Electrodynamics, §4.2–4.4. The water dipole moment $1.85\,\mathrm{D} = 6.2\times10^{-30}\,\mathrm{C\,m}$ is a standard tabulated constant. The spherical moments $q_{\ell m}$ are defined following the Jackson convention. Key numerical results: example 1 — $\mathbf{a}_c = 0.020\,\mathrm{m}$ , $V_{\rm exact} = 18.336\,\mathrm{V}$ , monopole about $O$ : $17.975\,\mathrm{V}$ ( $-1.97\%$ ), monopole about $\mathbf{a}_c$ : $18.342\,\mathrm{V}$ ( $+0.03\%$ ), monopole+quadrupole about $\mathbf{a}_c$ : $18.336\,\mathrm{V}$ ( $-0.0012\%$ ); example 2 — $p'_z = -2.0\times10^{-12}\,\mathrm{C\cdot m}$ and $Q'_{zz} = 1.2\times10^{-14}\,\mathrm{C\cdot m^2}$ by both routes, on-axis $V(5\,\mathrm{cm})$ : exact $179.895\,\mathrm{V}$ , series about $O$ : $179.895\,\mathrm{V}$ , about $O'$ : $179.927\,\mathrm{V}$ .

Builds on: Multipole Expansion of the Potential · Monopole Moment · Electric Dipole Moment · Electric Quadrupole Moment · Legendre Polynomials

Leads to: Dielectric Materials and Insulators (the E3 chapter, where the dipole becomes matter's voice) · Torque and Force on a Dipole in a Field · Boundary Conditions for the Potential · Uniqueness Theorems for the Potential

If moments above the leading one are "just bookkeeping," why does anyone bother computing them?

Because bookkeeping with a declared convention is still information. The full set of moments about a stated origin reconstructs the exterior potential exactly — the higher moments are indispensable the moment you want more accuracy than the leading term provides. What the theorem forbids is treating a higher moment as an intrinsic property: the number is a property of the pair (distribution, origin). State the origin — center of mass is the common contract — and higher moments become perfectly good, comparable, publishable quantities.

Is the center of charge the same as the center of mass?

Only by coincidence. The center of charge weights positions by charge, the center of mass by mass, and the two densities need not track each other — in an ion the electrons carry most of the imbalance of charge and almost none of the mass. This matters practically: quantum-chemistry conventions usually quote ion multipoles about the center of mass (it is what rotational spectroscopy physically references), which does not kill the dipole term. An ion can therefore have a perfectly well-defined, nonzero, tabulated "dipole moment about the center of mass" — a convention-laden but useful number.

Can I choose an origin that kills both the dipole and the quadrupole?

Generally no, and counting tells you why. The shift vector $\mathbf{a}$ has three components; killing the dipole spends all three ( $\mathbf{a} = \mathbf{p}/Q$ , one exact solution). The traceless quadrupole has five independent components, and after the dipole is dead you have no knobs left — the quadrupole about the center of charge is whatever it is. (For special symmetric distributions it may vanish on its own, but that is the distribution's doing, not the origin's.) The mechanical analogy holds: you can move the origin to the center of mass, but the inertia tensor there is a fixed property you then diagonalize, not adjust.

Why does the negative weight in example 1 push the center of charge outside the charges?

A weighted mean with mixed-sign weights is not confined to the "hull" of the points — the $-1\,\mathrm{nC}$ charge subtracts its position, slinging the mean past the positive charge. Nothing pathological follows: the center of charge is an algebraic reference point, not a place where anything physical sits. (For all-positive or all-negative distributions the weights are one-signed and $\mathbf{a}_c$ does lie inside, just like a center of mass.) If the net charge is small, $\mathbf{a}_c = \mathbf{p}/Q$ can even fly arbitrarily far away — a warning that for nearly neutral systems the center of charge is numerically ill-conditioned even though the formalism is exact.

Does the invariance theorem survive for time-dependent sources and radiation?

The algebraic statement survives verbatim at each instant — moments are spatial integrals, and translations mix them triangularly whether or not $\rho$ depends on time. Its most famous consequence lives in radiation theory: total charge is conserved (monopole cannot oscillate) and for an isolated system momentum conservation freezes the relevant dipole bookkeeping, so electric-monopole radiation is forbidden and gravity — where the "charge" is mass — additionally forbids dipole radiation. The first moment free to oscillate radiates first: electric dipole in electromagnetism, mass quadrupole in gravity. The triangle of this article is the statics half of that story.

The quadrupole correction in the shift law has terms with delta_ij — where do they come from?

From keeping the shifted moment traceless. The raw second moment $\int r'_ir'_j\rho$ shifts by clean terms $-a_ip_j - a_jp_i + a_ia_jQ$ ; but $Q_{ij}$ subtracts the trace, and the shift of the trace $\int r'^2\rho$ contributes the $2(\mathbf{a}\cdot\mathbf{p})\delta_{ij}$ and $-a^2Q\,\delta_{ij}$ pieces. Contract the full correction with $\delta_{ij}$ and everything cancels — the shifted tensor is exactly as traceless as the original, so the five-component structure of the quadrupole is preserved by translations, as it must be for the convention to be self-consistent.

How do I decide, in practice, which origin to expand about?

Ask what you are optimizing. For fastest convergence of a truncated series, put the origin where it silences the biggest killable term: the center of charge for $Q \neq 0$ (kills the dipole), or the natural symmetry center when one exists (often kills several rungs at once, as in example 2's symmetric origin). For comparability with tabulated data, use the tabulated convention — center of mass in molecular work — even though it converges slower. And for a source near a boundary-value problem's geometry (a sphere, a plane), the geometry wins: put the origin at the center the boundary machinery of the previous articles wants, and let the moments fall where they may.