Multipole Expansion of the Potential

Derivation

Multipole Expansion of the Potential

D-047 Home PU-102 Threads fields · symmetry Depends on Scalar Potential, Poisson and Laplace Equations, legendre-polynomial-generating-function

Statement

For a static charge distribution \(\rho(\vec r')\) of bounded (compact) support, the electrostatic potential at a field point \(\vec r\) lying outside the source expands as an absolutely convergent series in inverse powers of \(r=|\vec r|\), \[ V(\vec r)=\frac{1}{4\pi\epsilon_0}\sum_{n=0}^{\infty}\frac{1}{r^{\,n+1}}\int (r')^{n}\,P_n(\cos\theta')\,\rho(\vec r')\,d^3r', \] whose first three terms are the monopole (\(\propto r^{-1}\)), dipole (\(\propto r^{-2}\)) and quadrupole (\(\propto r^{-3}\)) contributions, with \(\theta'\) the angle between \(\vec r\) and the source point \(\vec r'\), and \(P_n\) the Legendre polynomials.

Why it matters

Almost every real source — a molecule, a nucleus, an antenna's charge, a lump of dielectric — is neutral or nearly neutral, so its far field is not a simple Coulomb tail. The multipole expansion tells you exactly what survives: the leading non-vanishing moment dominates at large distance and sets the entire character of the far field (\(1/r^2\) for a dipole, \(1/r^3\) for a quadrupole).

It converts an intractable integral over an unknown shape into a handful of numbers — total charge, dipole moment, quadrupole tensor — that carry all the far-field information. These moments are the natural language of atomic, molecular and nuclear physics, of dielectrics, and of the systematic expansion that becomes radiation multipoles in the time-dependent theory.

Assumptions

The distribution has compact support (bounded in space).If \(\rho\) extends to infinity — an infinite line or plane — there is no largest \(r'\), the moments \(\int (r')^n\rho\,d^3r'\) can diverge, and the ordering into monopole, dipole, … collapses.

The field point lies outside the source: \(r>r'_{\max}\) for every point where \(\rho\neq0\).The generating-function series converges only for \(r'/r<1\). Inside the distribution the expansion in powers of \(r'/r\) diverges and one must instead expand in powers of \(r/r'\) (interior multipoles) or solve directly.

The sources are static (electrostatics).Time variation introduces retardation and radiation; the potential no longer reduces to the instantaneous Coulomb integral and the static multipole series must be replaced by radiation multipoles.

Vacuum or a linear, homogeneous medium, so the free-space Green's function \(1/(4\pi\epsilon_0|\vec r-\vec r'|)\) applies.In an inhomogeneous or nonlinear medium the Green's function changes and the simple Legendre expansion no longer represents the true potential.

The origin is fixed once and lies at or near the distribution.The dipole and quadrupole moments depend on the choice of origin whenever a lower moment is non-zero; only the first non-vanishing moment is origin-independent. A poor origin choice merely slows convergence, but does not invalidate the series.

Derivation

\[ V(\vec r)=\frac{1}{4\pi\epsilon_0}\int \frac{\rho(\vec r')}{|\vec r-\vec r'|}\,d^3r' \]

Starting point: the solution of Poisson's equation for a localized source with the boundary condition \(V\to0\) at infinity (prior result electrostatic-potential-and-poisson). A

\[ |\vec r-\vec r'|=\sqrt{r^{2}+r'^{2}-2\,\vec r\cdot\vec r'}=r\sqrt{1-2\frac{r'}{r}\cos\theta'+\left(\frac{r'}{r}\right)^{2}} \]

Law of cosines, with \(\theta'\) the angle between \(\vec r\) and \(\vec r'\) so that \(\vec r\cdot\vec r'=r\,r'\cos\theta'\); then factor out \(r\). A

\[ \frac{1}{|\vec r-\vec r'|}=\frac{1}{r}\,\big(1-2\xi u+\xi^{2}\big)^{-1/2},\qquad \xi\equiv\frac{r'}{r},\quad u\equiv\cos\theta' \]

Pure relabelling to expose the generating-function form. Because the source is bounded and the field point is outside, \(\xi<1\). A

\[ \big(1-2\xi u+\xi^{2}\big)^{-1/2}=\sum_{n=0}^{\infty}P_n(u)\,\xi^{\,n}\qquad(|\xi|<1) \]

Generating function of the Legendre polynomials (prior result legendre-polynomial-generating-function). The series converges absolutely and uniformly on \(|u|\le1\) for any fixed \(\xi<1\), which is guaranteed by \(r>r'_{\max}\). B

\[ V(\vec r)=\frac{1}{4\pi\epsilon_0}\sum_{n=0}^{\infty}\frac{1}{r^{\,n+1}}\int (r')^{n}P_n(\cos\theta')\,\rho(\vec r')\,d^3r' \]

Insert the expansion, factor \(\xi^n=(r'/r)^n\), and interchange sum and integral. The interchange is legal because the convergence is uniform over the compact support of \(\rho\) (dominated convergence). B

\[ V_{0}(\vec r)=\frac{1}{4\pi\epsilon_0}\frac{1}{r}\int \rho(\vec r')\,d^3r'=\frac{1}{4\pi\epsilon_0}\frac{Q}{r},\qquad Q\equiv\int\rho\,d^3r' \]

The \(n=0\) term: \(P_0=1\). To leading order the distribution looks like a point charge \(Q\) at the origin — the monopole. A

\[ V_{1}(\vec r)=\frac{1}{4\pi\epsilon_0}\frac{1}{r^{2}}\int r'\cos\theta'\,\rho\,d^3r' =\frac{1}{4\pi\epsilon_0}\frac{\hat r\cdot\vec p}{r^{2}},\qquad \vec p\equiv\int \vec r'\,\rho(\vec r')\,d^3r' \]

The \(n=1\) term: \(P_1(\cos\theta')=\cos\theta'\) and \(r'\cos\theta'=\hat r\cdot\vec r'\). This defines the dipole moment \(\vec p\). B

\[ (r')^{2}P_2(\cos\theta')=\tfrac12\big(3(\hat r\cdot\vec r')^{2}-r'^{2}\big)=\tfrac12\sum_{i,j}\hat r_i\hat r_j\big(3x_i'x_j'-r'^{2}\delta_{ij}\big) \] \[ V_{2}(\vec r)=\frac{1}{4\pi\epsilon_0}\frac{1}{2r^{3}}\sum_{i,j}\hat r_i\hat r_j\,Q_{ij},\qquad Q_{ij}\equiv\int\big(3x_i'x_j'-r'^{2}\delta_{ij}\big)\rho\,d^3r' \]

The \(n=2\) term: \(P_2(x)=\tfrac12(3x^2-1)\). Writing \((\hat r\cdot\vec r')^2=\sum_{ij}\hat r_i\hat r_j\,x_i'x_j'\) and \(r'^{2}=\sum_{ij}\hat r_i\hat r_j\,\delta_{ij}\,r'^2\) (using \(|\hat r|=1\)) exposes the symmetric traceless quadrupole tensor \(Q_{ij}\). C

\[ V(\vec r)=\frac{1}{4\pi\epsilon_0}\left[\frac{Q}{r}+\frac{\hat r\cdot\vec p}{r^{2}}+\frac{1}{2r^{3}}\sum_{i,j}\hat r_i\hat r_j\,Q_{ij}+\cdots\right] \]

Collect the first three orders. Each successive term falls off one power of \(r\) faster and carries the geometry of the source at the corresponding angular order. B

Result

\[ V(\vec r)=\frac{1}{4\pi\epsilon_0}\left[\frac{Q}{r}+\frac{\hat r\cdot\vec p}{r^{2}}+\frac{1}{2r^{3}}\sum_{i,j}\hat r_i\hat r_j\,Q_{ij}+\cdots\right] \]

Reading. Far from any bounded static source the potential is an ordered sum: a Coulomb tail set by the net charge \(Q\); if \(Q=0\), a dipole field set by \(\vec p=\int\vec r'\rho\,d^3r'\); if \(\vec p=0\) too, a quadrupole field set by the tensor \(Q_{ij}\); and so on. The leading non-vanishing moment controls the far field and, uniquely, is independent of where the origin is placed.

Units check. In SI, \([\epsilon_0]=\mathrm{C^2\,N^{-1}\,m^{-2}}\). Monopole: \(\dfrac{Q}{4\pi\epsilon_0 r}\to\dfrac{\mathrm{C}}{(\mathrm{C^2\,N^{-1}\,m^{-2}})\,\mathrm{m}}=\mathrm{N\,m\,C^{-1}}=\mathrm{J\,C^{-1}}=\mathrm{V}\). Dipole: \([\vec p]=\mathrm{C\,m}\), so \(\dfrac{p}{4\pi\epsilon_0 r^2}\to\dfrac{\mathrm{C\,m}}{(\mathrm{C^2N^{-1}m^{-2}})\,\mathrm{m^2}}=\mathrm{V}\). Quadrupole: \([Q_{ij}]=\mathrm{C\,m^2}\), so \(\dfrac{Q_{ij}}{4\pi\epsilon_0 r^3}\to\mathrm{V}\). Every term is a potential, as required.

Limiting cases

Point charge at the origin: only \(Q\neq0\); all higher moments vanish and \(V=Q/4\pi\epsilon_0 r\) exactly, for all \(r>0\).
Neutral source, \(Q=0\): the monopole term dies and the dipole \(1/r^2\) term leads; the potential is now origin-independent to leading order.
Ideal (point) dipole: take \(d\to0\), \(q\to\infty\) with \(\vec p=q\vec d\) fixed; all terms beyond the dipole vanish and \(V=\hat r\cdot\vec p/4\pi\epsilon_0 r^2\) everywhere.
Symmetric linear quadrupole (\(Q=0,\ \vec p=0\)): the \(1/r^3\) term leads and the potential is set entirely by \(Q_{ij}\).
Very large \(r\) (\(r\gg r'_{\max}\)): successive terms are suppressed by \((r'_{\max}/r)^n\); a single term usually suffices.

Breaks when

The field point lies inside the charge distribution (\(r<r'_{\max}\)). Then \(r'/r>1\) for part of the source, the generating-function series diverges, and the whole expansion is invalid — one must use the interior expansion in powers of \(r/r'\) or solve the boundary-value problem directly.
The source is unbounded (infinite line, plane, or a slowly decaying tail). There is no largest \(r'\), the moment integrals \(\int (r')^n\rho\,d^3r'\) can diverge, and the multipole ordering fails; such sources need their own methods (e.g. Gauss's law by symmetry).
The charges are time-dependent. Retardation makes the instantaneous Coulomb integral wrong; the correct object is the retarded potential and a radiation-multipole expansion, not this static series.
Too few terms in the near far-field. When \(r\) is only modestly larger than \(r'_{\max}\), truncating at dipole or quadrupole leaves an error of order \((r'_{\max}/r)^{n+1}\) that may be unacceptably large — the series converges, but slowly.

Failure modes

Origin-dependence confusion: computing \(\vec p\) for a source with \(Q\neq0\) and expecting it to be origin-independent. It is not — shifting the origin by \(\vec a\) changes \(\vec p\to\vec p-Q\vec a\). Only the first non-zero moment is origin-independent.
Misreading \(\theta'\): treating \(\theta'\) as the fixed polar angle of the field point. It is the angle between \(\vec r\) and the running source point \(\vec r'\), and it varies over the integration.
Dropping the trace term: writing \(Q_{ij}=\int 3x_i'x_j'\rho\,d^3r'\) without the \(-r'^2\delta_{ij}\). The tensor must be traceless; omitting it keeps an isotropic piece already counted in the monopole average.
Losing the factor \(\tfrac12\) in front of the quadrupole term, or absorbing it inconsistently into \(Q_{ij}\) (conventions differ — state yours and keep it).
Inverting the ratio: expanding in \(r/r'\) instead of \(r'/r\) for an exterior point (or vice versa), giving a series that diverges everywhere it is applied.
Pulling the angular factor out: using \(P_1=\cos\theta'\) but then treating \(\cos\theta'\) as the constant field-point angle and taking it outside the integral.

Discussion

The expansion is really a statement about symmetry. Each term is the \(\ell\)-th angular harmonic of the source seen from far away: the monopole is the spherically symmetric average, the dipole is the first-order "lopsidedness", the quadrupole is the second-order "stretch/squash", and so on. Because the Legendre polynomials form a complete orthogonal set on the sphere, no far-field information is lost — the moments \(\{Q,\vec p,Q_{ij},\dots\}\) are simply the coordinates of the source in this angular basis.

A key structural fact is the origin-independence of the leading non-vanishing moment. If \(Q\neq0\) the monopole field dominates and swamps everything at large \(r\), so the precise value of \(\vec p\) barely matters; but for a neutral molecule (\(Q=0\)) the dipole moment is a genuine, coordinate-free physical property — this is why molecular dipole moments are tabulated quantities. The same logic elevates \(Q_{ij}\) to physical status only when both \(Q\) and \(\vec p\) vanish, as in the deuteron or in centrosymmetric molecules.

Written in spherical harmonics the expansion is exact and manifestly complete: using the addition theorem, \(\dfrac{1}{|\vec r-\vec r'|}=\sum_{\ell,m}\dfrac{4\pi}{2\ell+1}\dfrac{r'^{\ell}}{r^{\ell+1}}Y_{\ell m}^{*}(\hat r')\,Y_{\ell m}(\hat r)\), so the potential becomes \(V=\dfrac{1}{\epsilon_0}\sum_{\ell,m}\dfrac{1}{2\ell+1}\dfrac{q_{\ell m}}{r^{\ell+1}}Y_{\ell m}(\hat r)\) with spherical multipole moments \(q_{\ell m}=\int r'^{\ell}Y_{\ell m}^{*}(\hat r')\rho\,d^3r'\). The Cartesian tensor \(Q_{ij}\) is symmetric and traceless — five independent components — matching exactly the \(2\ell+1=5\) values of \(q_{2m}\). The series is asymptotic in the sense that it is an exact convergent expansion for \(r>r'_{\max}\), while truncations are approximations whose error is controlled by \((r'_{\max}/r)^{\ell+1}\).

Common misconceptions. "A neutral object has no far field" — false; it has a dipole (or quadrupole) field that merely falls off faster than Coulomb. "Higher multipoles are always negligible" — only when \(r\gg r'_{\max}\); close to the source many terms matter and the series may converge slowly. "The dipole moment of any object is well defined" — only if the object is neutral; otherwise it depends on the origin.

Worked examples

Physical dipole. Charges \(+q\) at \(z=+d/2\) and \(-q\) at \(z=-d/2\). Find the leading far-field potential and evaluate it for \(q=1\ \mathrm{nC}\), \(d=1\ \mathrm{mm}\) at \(r=10\ \mathrm{cm}\) on the axis (\(\theta=0\)). A

\[ Q=q+(-q)=0,\qquad \vec p=\sum_k q_k\vec r'_k=q\left(\tfrac{d}{2}\hat z\right)+(-q)\left(-\tfrac{d}{2}\hat z\right)=q\,d\,\hat z \]

Net charge vanishes, so the dipole term leads; \(\vec p\) is origin-independent here. A

\[ V(\vec r)\simeq\frac{1}{4\pi\epsilon_0}\frac{\hat r\cdot\vec p}{r^{2}}=\frac{1}{4\pi\epsilon_0}\frac{q\,d\,\cos\theta}{r^{2}} \]

Insert \(\vec p=qd\,\hat z\) and \(\hat r\cdot\hat z=\cos\theta\). A

\[ V=\big(8.99\times10^{9}\big)\frac{(10^{-9})(10^{-3})(1)}{(0.10)^{2}}\ \mathrm{V}=\big(8.99\times10^{9}\big)\frac{10^{-12}}{10^{-2}}\ \mathrm{V} \]

Symbols first, then numbers; \(\cos0=1\), \(p=qd=10^{-12}\ \mathrm{C\,m}\), \(r^2=10^{-2}\ \mathrm{m^2}\), and \(1/4\pi\epsilon_0=8.99\times10^{9}\ \mathrm{N\,m^2\,C^{-2}}\). A

\[ V\approx 0.90\ \mathrm{V}\quad(\text{on axis},\ \theta=0) \]

Reading. The neutral pair produces a \(1/r^2\) potential; on the axis toward \(+q\) it is positive, it reverses sign at \(\theta=\pi\), and it scales as \(\cos\theta\) off-axis.

Linear quadrupole. Charges \(+q\) at \(z=+a\), \(-2q\) at the origin, \(+q\) at \(z=-a\). Show \(Q=0\) and \(\vec p=0\), find the leading potential on the axis, and evaluate for \(q=1\ \mathrm{nC}\), \(a=2\ \mathrm{cm}\), \(r=20\ \mathrm{cm}\). B

\[ Q=q-2q+q=0,\qquad \vec p=q(a\hat z)+(-2q)(0)+q(-a\hat z)=0 \]

Both lower moments vanish by the symmetric placement, so the quadrupole term leads and is origin-independent. A

\[ Q_{zz}=\sum_k q_k\big(3z_k^{2}-r_k^{2}\big)=\sum_k q_k\,2z_k^{2}=q(2a^{2})+0+q(2a^{2})=4qa^{2} \]

On the axis \(r_k^2=z_k^2\) for each charge, so \(3z_k^2-r_k^2=2z_k^2\). Axial symmetry gives \(Q_{xx}=Q_{yy}=-\tfrac12 Q_{zz}\). B

\[ V(\vec r)\simeq\frac{1}{4\pi\epsilon_0}\frac{1}{2r^{3}}\,\hat r_z\hat r_z\,Q_{zz}\Big|_{\hat r=\hat z}=\frac{1}{4\pi\epsilon_0}\frac{Q_{zz}}{2r^{3}}=\frac{1}{4\pi\epsilon_0}\frac{2qa^{2}}{r^{3}} \]

On axis \(\hat r=\hat z\), so only \(Q_{zz}\) contributes; substitute \(Q_{zz}=4qa^2\). Direct summation of the three Coulomb potentials confirms the leading \(2qa^2/r^3\) behaviour. B

\[ V=\big(8.99\times10^{9}\big)\frac{2(10^{-9})(0.02)^{2}}{(0.20)^{3}}\ \mathrm{V}=\big(8.99\times10^{9}\big)\frac{8\times10^{-13}}{8\times10^{-3}}\ \mathrm{V} \]

Numbers last: \(2qa^2=2(10^{-9})(4\times10^{-4})=8\times10^{-13}\ \mathrm{C\,m^2}\), \(r^3=8\times10^{-3}\ \mathrm{m^3}\). A

\[ V\approx 0.90\ \mathrm{V}\quad(\text{on axis},\ r=20\ \mathrm{cm}) \]

Reading. A neutral, dipole-free cluster still has a far field, but it now falls as \(1/r^3\); doubling \(r\) cuts \(V\) by a factor of eight rather than four.

Problems

Monopole dominance. A body carries net charge \(Q=5\ \mathrm{nC}\) together with an internal dipole \(p=2\times10^{-11}\ \mathrm{C\,m}\). At what on-axis distance are the monopole and dipole potentials equal in magnitude?
Solution
Set \(\dfrac{Q}{r}=\dfrac{p}{r^2}\Rightarrow r=\dfrac{p}{Q}=\dfrac{2\times10^{-11}}{5\times10^{-9}}=4\times10^{-3}\ \mathrm{m}=4\ \mathrm{mm}\). For \(r\gg4\ \mathrm{mm}\) the monopole dominates; the dipole is a small correction. (The common \(1/4\pi\epsilon_0\) cancels.)
Origin dependence. Two charges: \(+3q\) at \(x=0\) and \(-q\) at \(x=L\). Compute \(\vec p\) about the origin, then about the point \(x=L\), and reconcile the two.
Solution
\(Q=2q\neq0\). About the origin: \(\vec p=3q(0)+(-q)(L\hat x)=-qL\,\hat x\). About \(x=L\): positions become \(-L\hat x\) and \(0\), so \(\vec p'=3q(-L\hat x)+(-q)(0)=-3qL\,\hat x\). They differ because \(Q\neq0\): the shift rule \(\vec p'=\vec p-Q\vec a\) with \(\vec a=L\hat x\) gives \(\vec p'=-qL\hat x-2q(L\hat x)=-3qL\hat x\). Consistent. The dipole moment is not a physical property here.
Quadrupole of a ring. A total charge \(Q_{\rm tot}\) is spread uniformly on a ring of radius \(b\) in the \(xy\)-plane, centred at the origin. Find \(Q_{zz}\).
Solution
Every element has \(z'=0\) and \(r'=b\), so \(3z'^2-r'^2=-b^2\). Then \(Q_{zz}=\int(3z'^2-r'^2)\,dq=-b^2\!\int dq=-Q_{\rm tot}\,b^2\). By symmetry \(Q_{xx}=Q_{yy}=+\tfrac12 Q_{\rm tot}b^2\), and the trace \(Q_{xx}+Q_{yy}+Q_{zz}=0\) as required. If \(Q_{\rm tot}\neq0\) the leading far field is still the monopole; the quadrupole is the first anisotropic correction.
Truncation error. For the linear quadrupole of Worked Example 2 (\(a=2\ \mathrm{cm}\)), estimate the fractional size of the next non-zero term relative to the quadrupole at \(r=20\ \mathrm{cm}\).
Solution
Successive terms scale as \((a/r)^{n}\). For this reflection-symmetric arrangement the odd multipoles vanish, so the next non-zero term is the hexadecapole (\(n=4\)); its size relative to the quadrupole (\(n=2\)) is of order \((a/r)^{2}=(0.02/0.20)^{2}=(0.1)^2=10^{-2}\), i.e. about \(1\%\). So the quadrupole alone is good to roughly a percent at \(20\ \mathrm{cm}\).
Interior versus exterior. For a field point at \(r=1\ \mathrm{cm}\) inside a spherical shell of radius \(b=5\ \mathrm{cm}\) carrying surface charge, explain why the exterior multipole expansion cannot be used and state the correct expansion parameter.
Solution
The source sits at \(r'=b=5\ \mathrm{cm}>r=1\ \mathrm{cm}\), so \(r'/r=5>1\) and the exterior series, built on \(r'/r<1\), diverges. Inside the source region one expands \(\dfrac{1}{|\vec r-\vec r'|}=\dfrac1{r'}\sum_n P_n(\cos\theta')\left(\dfrac{r}{r'}\right)^{n}\), i.e. in powers of \(r/r'<1\). For a uniform spherical shell this interior series collapses to a constant potential (zero field) inside. The correct small parameter is \(r/r'\), not \(r'/r\).