Multipole Expansion
Multipole moments, spherical harmonics, quadrupole tensors, and interaction energies at the Jackson level.
2.1 The Multipole Expansion
For a localized charge distribution, the potential at large distances can be expanded in powers of $1/r$. Using the generating function for Legendre polynomials:
where $\gamma$ is the angle between $\mathbf{r}$ and $\mathbf{r}'$, and $r > r'$. Substituting into the Coulomb potential integral:
Using the addition theorem for spherical harmonics, $P_l(\cos\gamma) = \frac{4\pi}{2l+1}\sum_{m=-l}^{l}Y_{lm}^*(\theta',\phi')Y_{lm}(\theta,\phi)$, this becomes:
where the multipole moments are $q_{lm} = \int r'^l Y_{lm}^*(\theta',\phi')\rho(\mathbf{r}')\,d^3r'$.
Derivation 1: Multipole Expansion from Taylor Series
We derive the multipole expansion by systematically Taylor-expanding $1/|\mathbf{r} - \mathbf{r}'|$ for $r' \ll r$.
Step 1: Write the exact expression
$$\frac{1}{|\mathbf{r} - \mathbf{r}'|} = \frac{1}{\sqrt{r^2 - 2\mathbf{r}\cdot\mathbf{r}' + r'^2}} = \frac{1}{r}\frac{1}{\sqrt{1 - 2(r'/r)\cos\gamma + (r'/r)^2}}$$
Step 2: Expand using the generating function
The generating function for Legendre polynomials is $(1 - 2xt + t^2)^{-1/2} = \sum_{l=0}^{\infty}t^l P_l(x)$ for $|t| < 1$. With $t = r'/r$ and $x = \cos\gamma$:
$$\frac{1}{|\mathbf{r} - \mathbf{r}'|} = \frac{1}{r}\sum_{l=0}^{\infty}\left(\frac{r'}{r}\right)^l P_l(\cos\gamma)$$
Step 3: Identify the first three terms
The $l = 0$ term (monopole): $\Phi_0 = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}$ where $Q = \int\rho\,d^3r'$.
The $l = 1$ term (dipole): $\Phi_1 = \frac{1}{4\pi\epsilon_0}\frac{\mathbf{p}\cdot\hat{r}}{r^2}$ where $\mathbf{p} = \int\mathbf{r}'\rho\,d^3r'$.
The $l = 2$ term (quadrupole): $\Phi_2 = \frac{1}{4\pi\epsilon_0}\frac{1}{2r^3}\sum_{ij}Q_{ij}\hat{r}_i\hat{r}_j$.
Step 4: Define the quadrupole tensor
The traceless quadrupole moment tensor is:
$$Q_{ij} = \int(3r'_i r'_j - r'^2\delta_{ij})\rho(\mathbf{r}')\,d^3r'$$
It has 5 independent components (symmetric, traceless $3\times 3$ tensor), matching the $2l+1 = 5$ spherical harmonics for $l = 2$.
Step 5: Convergence and far-field dominance
Each successive term falls off as $1/r^{l+1}$. At large distances, the lowest nonvanishing multipole dominates: monopole ($\sim 1/r$), dipole ($\sim 1/r^2$), quadrupole ($\sim 1/r^3$), etc. The expansion converges for $r > r'_{\max}$.
2.2 Spherical Harmonics
The spherical harmonics $Y_{lm}(\theta,\phi)$ are eigenfunctions of the angular part of the Laplacian. They form a complete orthonormal set on the unit sphere:
Orthonormality: $\int Y_{lm}^*(\theta,\phi)Y_{l'm'}(\theta,\phi)\,d\Omega = \delta_{ll'}\delta_{mm'}$. Completeness: $\sum_{l=0}^{\infty}\sum_{m=-l}^{l}Y_{lm}^*(\theta',\phi')Y_{lm}(\theta,\phi) = \delta(\cos\theta - \cos\theta')\delta(\phi - \phi')$.
Derivation 2: Addition Theorem for Spherical Harmonics
The addition theorem relates $P_l(\cos\gamma)$ to products of spherical harmonics evaluated at the two directions.
Step 1: Express $P_l(\cos\gamma)$ in terms of coordinates
Let $\hat{r}$ point in direction $(\theta,\phi)$ and $\hat{r}'$ in direction $(\theta',\phi')$. The angle $\gamma$ between them satisfies $\cos\gamma = \hat{r}\cdot\hat{r}'$.
Step 2: Expand $P_l(\cos\gamma)$ in harmonics of $(\theta,\phi)$
Fix $(\theta',\phi')$. Then $P_l(\cos\gamma)$ is a function of $(\theta,\phi)$ that satisfies the angular Laplace equation with eigenvalue $l(l+1)$. Therefore it can be expanded:
$$P_l(\cos\gamma) = \sum_{m=-l}^{l}A_{lm}(\theta',\phi')Y_{lm}(\theta,\phi)$$
Step 3: Determine the coefficients
Using orthonormality, $A_{lm} = \int P_l(\cos\gamma)Y_{lm}^*(\theta,\phi)\,d\Omega$. Evaluating by choosing $\hat{r}' = \hat{z}$ (so $\gamma = \theta$) gives $A_{lm} = \frac{4\pi}{2l+1}Y_{lm}^*(\theta',\phi')$.
Step 4: Verify by rotation invariance
The result must be rotationally invariant. Since $P_l(\cos\gamma)$ depends only on the angle between $\hat{r}$ and $\hat{r}'$, and the combination $\sum_m Y_{lm}^*(\theta',\phi')Y_{lm}(\theta,\phi)$ is a rotational scalar, the form is confirmed.
Step 5: The addition theorem
$$\boxed{P_l(\cos\gamma) = \frac{4\pi}{2l+1}\sum_{m=-l}^{l}Y_{lm}^*(\theta',\phi')Y_{lm}(\theta,\phi)}$$
This is one of the most important identities in mathematical physics. It converts a single angular function into a separable sum of products.
2.3 Electric Dipole
The dipole moment $\mathbf{p} = \int\mathbf{r}'\rho(\mathbf{r}')\,d^3r'$ is the leading-order correction when $Q = 0$. The dipole potential and field are:
In spherical coordinates with $\mathbf{p} = p\hat{z}$: $E_r = \frac{2p\cos\theta}{4\pi\epsilon_0 r^3}$, $E_\theta = \frac{p\sin\theta}{4\pi\epsilon_0 r^3}$. The field falls as $1/r^3$, one power faster than the monopole field.
Derivation 3: Electric Field of a Dipole
Step 1: Start from the dipole potential
$$\Phi = \frac{p\cos\theta}{4\pi\epsilon_0 r^2}$$
Step 2: Compute $E_r = -\partial\Phi/\partial r$
$$E_r = -\frac{\partial}{\partial r}\left(\frac{p\cos\theta}{4\pi\epsilon_0 r^2}\right) = \frac{2p\cos\theta}{4\pi\epsilon_0 r^3}$$
Step 3: Compute $E_\theta = -\frac{1}{r}\partial\Phi/\partial\theta$
$$E_\theta = -\frac{1}{r}\frac{\partial}{\partial\theta}\left(\frac{p\cos\theta}{4\pi\epsilon_0 r^2}\right) = \frac{p\sin\theta}{4\pi\epsilon_0 r^3}$$
Step 4: Write in coordinate-free form
Using $\mathbf{p}\cdot\hat{r} = p\cos\theta$, reconstruct $\mathbf{E} = E_r\hat{r} + E_\theta\hat{\theta}$ as:
$$\mathbf{E} = \frac{1}{4\pi\epsilon_0 r^3}[3(\mathbf{p}\cdot\hat{r})\hat{r} - \mathbf{p}]$$
Step 5: Include the contact term
The complete dipole field, valid at all points including $r = 0$, includes a delta function term:
$$\mathbf{E}_{\text{dip}} = \frac{1}{4\pi\epsilon_0}\left[\frac{3(\mathbf{p}\cdot\hat{r})\hat{r} - \mathbf{p}}{r^3} - \frac{4\pi}{3}\mathbf{p}\,\delta^3(\mathbf{r})\right]$$
The delta function term is essential for the hyperfine splitting of hydrogen (the 21 cm line).
2.4 Quadrupole Moment
The quadrupole tensor $Q_{ij} = \int(3r'_ir'_j - r'^2\delta_{ij})\rho\,d^3r'$ is traceless and symmetric. Its potential is:
For an axially symmetric charge distribution, only $Q_{33} \equiv Q$ survives (the others being related by tracelessness). The potential simplifies to:
Derivation 4: Multipole Interaction Energy
The energy of a charge distribution in an external potential $\Phi_{\text{ext}}$ can be expanded in multipole contributions.
Step 1: Taylor expand the external potential about the origin
$$\Phi_{\text{ext}}(\mathbf{r}') = \Phi_0 + \sum_i \left(\frac{\partial\Phi}{\partial x_i}\right)_0 x'_i + \frac{1}{2}\sum_{ij}\left(\frac{\partial^2\Phi}{\partial x_i\partial x_j}\right)_0 x'_ix'_j + \cdots$$
Step 2: Compute the interaction energy
$$W = \int\rho(\mathbf{r}')\Phi_{\text{ext}}(\mathbf{r}')\,d^3r'$$
Step 3: Identify monopole, dipole, quadrupole contributions
$$W = Q\Phi_0 - \mathbf{p}\cdot\mathbf{E}_0 - \frac{1}{6}\sum_{ij}Q_{ij}\frac{\partial E_i}{\partial x_j}\bigg|_0 + \cdots$$
Using $\mathbf{E}_0 = -\nabla\Phi_0$ and the tracelessness of $Q_{ij}$.
Step 4: Dipole-dipole interaction
For two dipoles $\mathbf{p}_1$ and $\mathbf{p}_2$ separated by $\mathbf{r}$:
$$W_{12} = \frac{1}{4\pi\epsilon_0 r^3}\left[\mathbf{p}_1\cdot\mathbf{p}_2 - 3(\mathbf{p}_1\cdot\hat{r})(\mathbf{p}_2\cdot\hat{r})\right]$$
Step 5: Torque on a dipole
The torque on a dipole in an external field is $\boldsymbol{\tau} = \mathbf{p}\times\mathbf{E}$, and the force is $\mathbf{F} = \nabla(\mathbf{p}\cdot\mathbf{E})$. For a quadrupole, the torque and force involve gradients of $\mathbf{E}$.
2.5 Origin Dependence of Multipole Moments
A crucial subtlety: multipole moments depend on the choice of origin. If we shift the origin by $\mathbf{a}$, the new coordinates are $\mathbf{r}'' = \mathbf{r}' - \mathbf{a}$.
The monopole moment $Q$ (total charge) is origin-independent. The dipole moment transforms as $\mathbf{p}' = \mathbf{p} - Q\mathbf{a}$. If $Q = 0$, the dipole moment is origin-independent. Similarly, the quadrupole moment is origin-independent if both $Q = 0$ and $\mathbf{p} = 0$.
Derivation 5: Exterior Multipole Expansion in Spherical Harmonics
We derive the full exterior expansion and connect it to the interior expansion for matching at boundaries.
Step 1: General solution to Laplace's equation in spherical coordinates
By separation of variables, $\nabla^2\Phi = 0$ has solutions:
$$\Phi(r,\theta,\phi) = \sum_{l=0}^{\infty}\sum_{m=-l}^{l}\left(A_{lm}r^l + B_{lm}r^{-(l+1)}\right)Y_{lm}(\theta,\phi)$$
Step 2: Exterior solution ($r > R$)
For the region outside a sphere of radius $R$ containing all charges, regularity at infinity requires $A_{lm} = 0$:
$$\Phi_{\text{ext}}(r,\theta,\phi) = \frac{1}{4\pi\epsilon_0}\sum_{l,m}\frac{4\pi}{2l+1}\frac{q_{lm}}{r^{l+1}}Y_{lm}(\theta,\phi)$$
Step 3: Interior solution ($r < R$)
Inside a charge-free region, regularity at the origin requires $B_{lm} = 0$:
$$\Phi_{\text{int}}(r,\theta,\phi) = \sum_{l,m}A_{lm}r^l Y_{lm}(\theta,\phi)$$
Step 4: Matching at the boundary
At a spherical boundary with surface charge $\sigma(\theta,\phi)$, continuity of the tangential $\mathbf{E}$ and the discontinuity condition $\epsilon_0(E_r^{\text{out}} - E_r^{\text{in}}) = \sigma$ determine all coefficients in terms of $\sigma_{lm}$.
Step 5: Connection to the free-space Green's function
The expansion of $1/|\mathbf{r} - \mathbf{r}'|$ in spherical harmonics is the free-space Green's function:
$$\frac{1}{|\mathbf{r} - \mathbf{r}'|} = 4\pi\sum_{l=0}^{\infty}\sum_{m=-l}^{l}\frac{1}{2l+1}\frac{r_<^l}{r_>^{l+1}}Y_{lm}^*(\theta',\phi')Y_{lm}(\theta,\phi)$$
where $r_< = \min(r,r')$ and $r_> = \max(r,r')$.
Historical Context
The multipole expansion has roots in celestial mechanics, where Laplace and Legendre developed the mathematical tools for gravitational potential theory in the late 18th century. The spherical harmonics were systematically studied by Laplace (1782) and Legendre (1784). Thomson (Lord Kelvin) and Maxwell applied these techniques to electrostatics in the 1860s-70s. The modern tensor formulation of the quadrupole moment was developed by Debye and others in the early 20th century for studying molecular charge distributions. Today, multipole expansions are essential in computational chemistry (molecular force fields), astrophysics (gravitational wave detection), and antenna theory.
Applications
Molecular Interactions
Van der Waals forces arise from fluctuating dipole-dipole interactions. The $1/r^6$ attraction comes from second-order perturbation theory with dipole matrix elements.
NMR & MRI
Nuclear quadrupole moments interact with electric field gradients, causing line splitting in NMR spectra. This provides structural information about molecular environments.
Gravitational Waves
Gravitational radiation is quadrupolar (no monopole or dipole radiation for gravity). The quadrupole formula is the leading approximation for gravitational wave emission.
Fast Multipole Method
The FMM algorithm uses multipole expansions to compute N-body interactions in O(N) time instead of O(N^2), revolutionizing computational physics.
Simulation: Multipole Field Patterns
This simulation visualizes the electric field lines and patterns for monopole, dipole, linear quadrupole, and square quadrupole charge configurations. Notice how the field structure becomes increasingly complex and falls off more rapidly with distance for higher-order multipoles.
Multipole Expansion: Monopole, Dipole & Quadrupole Fields
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Practice Problems
Problem 1:Find the dipole moment of the charge distribution: $+q$ at $(0, 0, d)$, $-2q$ at the origin, and $+q$ at $(0, 0, -d)$.
Solution:
1. The dipole moment is $\mathbf{p} = \sum_i q_i \mathbf{r}_i$.
2. $\mathbf{p} = (+q)(d\hat{z}) + (-2q)(0) + (+q)(-d\hat{z}) = qd\hat{z} - qd\hat{z} = 0$.
3. The dipole moment vanishes! This distribution is a linear quadrupole, not a dipole.
4. The total charge (monopole moment) is also zero: $Q = q - 2q + q = 0$.
5. The leading multipole is the quadrupole. The quadrupole moment tensor has $Q_{zz} = \sum_i q_i(3z_i^2 - r_i^2) = q(3d^2 - d^2) + 0 + q(3d^2 - d^2) = 4qd^2$.
6. The potential falls off as $\Phi \propto 1/r^3$ (quadrupole) rather than $1/r^2$ (dipole) at large distances. The field falls off as $1/r^4$.
Problem 2:Write the potential of a linear quadrupole (charges $+q, -2q, +q$ separated by $d$ along the $z$-axis) at a point $(r, \theta)$ for $r \gg d$.
Solution:
1. Since $Q = 0$ and $\mathbf{p} = 0$, the leading term is the quadrupole: $\Phi = \frac{1}{4\pi\epsilon_0}\frac{Q_{ij}\hat{r}_i\hat{r}_j}{2r^3}$.
2. For the linear quadrupole along $z$: $Q_{zz} = 4qd^2$, $Q_{xx} = Q_{yy} = -2qd^2$ (traceless), all off-diagonal elements zero.
3. The contraction $Q_{ij}\hat{r}_i\hat{r}_j = Q_{zz}\cos^2\theta + Q_{xx}\sin^2\theta\cos^2\phi + Q_{yy}\sin^2\theta\sin^2\phi$.
4. Simplifying: $Q_{ij}\hat{r}_i\hat{r}_j = 4qd^2\cos^2\theta - 2qd^2\sin^2\theta = 2qd^2(3\cos^2\theta - 1)$.
5. The quadrupole potential: $\Phi(r, \theta) = \frac{qd^2}{4\pi\epsilon_0}\frac{3\cos^2\theta - 1}{r^3} = \frac{qd^2}{4\pi\epsilon_0}\frac{2P_2(\cos\theta)}{r^3}$.
6. This is proportional to the Legendre polynomial $P_2(\cos\theta) = \frac{1}{2}(3\cos^2\theta - 1)$. The potential vanishes on the "magic angle" cone $\theta = \arccos(1/\sqrt{3}) \approx 54.7°$.
Problem 3:For a charge distribution with total charge $Q$, dipole moment $p$, and quadrupole moment $\mathcal{Q}$, determine how the electric field scales with distance for each multipole contribution at large $r$.
Solution:
1. The general multipole expansion: $\Phi = \frac{1}{4\pi\epsilon_0}\left(\frac{Q}{r} + \frac{\mathbf{p}\cdot\hat{r}}{r^2} + \frac{Q_{ij}\hat{r}_i\hat{r}_j}{2r^3} + \cdots\right)$.
2. Since $\mathbf{E} = -\nabla\Phi$, each $1/r^{n+1}$ term in $\Phi$ gives a $1/r^{n+2}$ term in $\mathbf{E}$.
3. Monopole ($\ell = 0$): $\Phi \propto 1/r$, $\mathbf{E} \propto 1/r^2$ (Coulomb law).
4. Dipole ($\ell = 1$): $\Phi \propto 1/r^2$, $\mathbf{E} \propto 1/r^3$. Explicitly: $\mathbf{E} = \frac{1}{4\pi\epsilon_0}\frac{3(\mathbf{p}\cdot\hat{r})\hat{r} - \mathbf{p}}{r^3}$.
5. Quadrupole ($\ell = 2$): $\Phi \propto 1/r^3$, $\mathbf{E} \propto 1/r^4$.
6. General pattern: the $2^\ell$-pole has $\Phi \propto 1/r^{\ell+1}$ and $\mathbf{E} \propto 1/r^{\ell+2}$. Higher multipoles die off faster, so only the lowest nonzero multipole dominates at large distance.
Problem 4:A dipole with moment $\mathbf{p} = p\hat{z}$ is placed in a non-uniform electric field $\mathbf{E} = E_0(1 + \alpha z)\hat{z}$. Find the torque and the force on the dipole.
Solution:
1. Torque on a dipole: $\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}$.
2. Since $\mathbf{p} = p\hat{z}$ and $\mathbf{E}$ is along $\hat{z}$: $\boldsymbol{\tau} = p\hat{z} \times E_0(1 + \alpha z)\hat{z} = 0$.
3. The torque vanishes because $\mathbf{p}$ is parallel to $\mathbf{E}$. The dipole is in a stable orientation.
4. Force on a dipole in a non-uniform field: $\mathbf{F} = \nabla(\mathbf{p} \cdot \mathbf{E})$.
5. $\mathbf{p} \cdot \mathbf{E} = pE_0(1 + \alpha z)$. So $\mathbf{F} = \nabla[pE_0(1 + \alpha z)] = pE_0\alpha\hat{z}$.
6. The dipole is attracted toward the region of stronger field ($\alpha > 0$ means field increases with $z$). This is the principle behind dielectrophoresis and explains why neutral atoms are attracted to charged objects.
Problem 5:A conducting sphere of radius $R$ is placed in a uniform electric field $\mathbf{E}_0 = E_0\hat{z}$. Using the multipole expansion, find the induced surface charge distribution and the dipole moment of the sphere.
Solution:
1. The potential outside must satisfy $\nabla^2 \Phi = 0$ with boundary conditions: $\Phi = 0$ on $r = R$ (grounded) and $\Phi \to -E_0 r\cos\theta$ as $r \to \infty$.
2. General solution: $\Phi = -E_0 r\cos\theta + \sum_\ell \frac{A_\ell}{r^{\ell+1}}P_\ell(\cos\theta)$. Only $\ell = 1$ matches the angular dependence.
3. $\Phi = -E_0 r\cos\theta + \frac{A}{r^2}\cos\theta$. At $r = R$: $0 = (-E_0 R + A/R^2)\cos\theta$, so $A = E_0 R^3$.
4. $\Phi = -E_0\cos\theta\left(r - \frac{R^3}{r^2}\right)$. The second term is a dipole field with moment $\mathbf{p} = 4\pi\epsilon_0 R^3 E_0\hat{z}$.
5. Surface charge: $\sigma = -\epsilon_0\frac{\partial\Phi}{\partial r}\big|_{r=R} = 3\epsilon_0 E_0\cos\theta$.
6. The induced dipole moment is $p = 4\pi\epsilon_0 R^3 E_0$, corresponding to a polarizability $\alpha = 4\pi\epsilon_0 R^3 = 3\epsilon_0 V$ where $V$ is the sphere volume.