ED formulas (2)

Symbol Convention

Symbol Convention%

In accordance with ISO 80000-2, the following font conventions are employed:
- Scalars and components for vectors or tensors are represented by lightface italic type ( $a, ϕ, a_{i}, T_{i j}$ ).
- Vectors are represented by boldface italic type ( $a, ω$ ).
- Second-order tensors are represented by boldface sans-serif type ( $T, I$ ).
- Operators & Constants: Roman (upright) type is used for fixed mathematical constants (e.g., Pi $π$ , the imaginary unit $i$ ) and differential operators (e.g., the differential $d$ in $d x$ ).
- Calculus Notation: For integrals, a thin space (\,) is used to separate the integrand from the differential operator, e.g., $\int f (x) d x$ .
Minkowski Metric: The Minkowski metric tensor $η_{μ ν}$ is defined using the mostly-plus signature convention:

η_{μ ν} = diag (- 1, + 1, + 1, + 1)

Consequently, the invariant spacetime interval is given by $d s^{2} = - c^{2} d t^{2} + d x^{2} + d y^{2} + d z^{2} = - d τ^{2}$ .

Lorentz transformation

K^{'} \to K : x^{α} = Λ_{β}^{α} x^{' β}

For $x$ -axis boost (where $K^{'}$ moves with velocity $v {\hat{e}}_{x}$ relative to $K$ ), the transformation matrix $Λ$ is:

Λ_{β}^{α} = [\begin{matrix} γ & γ β \\ γ β & γ \\ 1 \\ 1 \end{matrix}]

1 Electrostatic Field

1.1 Multipole expansion

Consider electrostatic potential expansion at an external observation point $x$ , with localized charge source $ρ (x^{'})$ confined within a volume $V^{'}$ , and $R = x - x^{'}, r = | x | (f i e l d), r^{'} = | x^{'} | (s o u r c e)$ , then

\begin{aligned} \frac{1}{R} = \frac{1}{| x - x^{'} |} = e^{- x^{'} \cdot \nabla} \frac{1}{r} & = [1 - x^{'} \cdot \nabla + \frac{1}{2!} (x^{'} x^{'} : \nabla \nabla) + \dots] \frac{1}{r} \\ I : \nabla \nabla \frac{1}{r} = 0 ⟹ & = \frac{1}{r} - x^{'} \cdot \nabla \frac{1}{r} + \frac{1}{6} (3 x^{'} x^{'} - r^{' 2} I) : \nabla \nabla \frac{1}{r} + \dots \end{aligned}

⇓

\begin{aligned} φ (x) = \frac{1}{4 π ϵ_{0}} \int \frac{d q}{R} & = \frac{1}{4 π ϵ_{0}} [Q - p \cdot \nabla + \frac{1}{6} D : \nabla \nabla + \dots] \frac{1}{r} \\ = \frac{Q}{4 π ϵ_{0} r} + \frac{p \cdot \hat{r}}{4 π ϵ_{0} r^{2}} + \frac{D : \hat{r} \hat{r}}{8 π ϵ_{0} r^{3}} + \dots \end{aligned}

Where $Q = \int d q = \int ρ (x^{'}) d V^{'}$ is total charge, $p = \int x^{'} d q$ is electric dipole moment, $D = \int (3 x^{'} x^{'} - r^{' 2} I) d q$ is electric quadrupole moment.

Electrostatic intensity expansion

E = - \nabla φ = \frac{Q \hat{r}}{4 π ϵ_{0} r^{2}} + \frac{3 (p \cdot \hat{r}) \hat{r} - p}{4 π ϵ_{0} r^{3}} + \frac{5 (D : \hat{r} \hat{r}) \hat{r} - 2 D \cdot \hat{r}}{8 π ϵ_{0} r^{4}} + \dots

1.2 Small charged body in electric field

Assuming the charge of charged bodies $x$ is so small compared with the source $x^{'}$ which generates the electric field that $\nabla^{2} φ (x) = 0$ . Let $ξ$ be the relative coordinate of a charge element $d q$ from the body's center $x$ . The potential energy $U = \int φ (x + ξ) d q$ of the small body is,

U \approx Q φ |_{x} + p \cdot \nabla φ |_{x} + \frac{1}{6} D : \nabla \nabla φ |_{x} = (Q φ) |_{x} - (p \cdot E) |_{x} - \frac{1}{6} (D : \nabla E) |_{x}

Note that $Q = \int d q = \int ρ (ξ) d V$ , $p = \int ξ d q$ , $D = \int (3 ξ ξ - ξ^{2} I) d q$ are intrinsic properties of the small charged body itself, not the external source.

Furthermore, The total electrostatic force $F$ acting on the small charged body is,

F = - \nabla U = Q E_{ext} |_{x} + (p \cdot \nabla) E_{ext} |_{x} + \frac{1}{6} (D : \nabla \nabla) E_{ext} |_{x} + \dots

The total force moment $N$ acting on the small charged body about its center $x$ is,

N = \int ξ \times E_{ext} (x + ξ) d q = p \times E_{ext} + \frac{1}{3} \nabla \cdot (D \times E_{ext}) + \dots

1.3 Spherical harmonic expansion

1.3.1 Orthonormal Complete Set of Functions

Definition

Orthonormality

\int_{a}^{b} ψ_{n}^{*} (x) ψ_{m} (x) d x = δ_{n m}, with Integration Limits [a, b]

Completeness: Any square-integrable function $f (x)$ can be expanded as

f (x) = \sum_{n} C_{n} ψ_{n} (x), where C_{n} = \int_{a}^{b} ψ_{n}^{*} (x) f (x) d x

which implies

\sum_{n} ψ_{n}^{*} (x^{'}) ψ_{n} (x) = δ (x - x^{'})

Trigonometric & Complex Exponential Functions

ψ_{n} (x) = \frac{1}{\sqrt{a}} \sin \frac{n π x}{a}, \frac{1}{\sqrt{a}} \cos \frac{n π x}{a}, \frac{1}{\sqrt{2 a}} \exp (i \frac{n π x}{a}), with Integration Limits [- a, a]

Note: $n \in N_{+}$ for sine set; $n \in N$ for cosine set; $n \in Z$ for complex set.

Legendre Polynomials

ψ_{l} (x) = \sqrt{\frac{2 l + 1}{2}} P_{l} (x), l \in N, with Integration Limits [- 1, 1]

P_{l} (x) = \frac{1}{2^{l} l!} \frac{d^{l}}{d x^{l}} (x^{2} - 1)^{l} = {\begin{cases} 1, l = 0 \\ x, l = 1 \\ \frac{1}{2} (3 x^{2} - 1), l = 2 \\ \frac{1}{2} (5 x^{3} - 3 x), l = 3 \\ \dots \end{cases}

Spherical Harmonics

ψ_{l m} (θ, φ) = Y_{l}^{m} (Ω) = \sqrt{\frac{1}{2 π} \frac{2 l + 1}{2} \frac{(l - m)!}{(l + m)!}} P_{l}^{m} (\cos θ) e^{i m φ},

l \in N, m \in Z & | m | \leq l, with θ \in [0, π], φ \in [0, 2 π]

P_{l}^{m} (x) = (- 1)^{m} (1 - x^{2})^{\frac{m}{2}} \frac{d^{m}}{d x^{m}} P_{l} (x), for m \geq 0

For negative $m$ , the relations are given by:

P_{l}^{- m} (x) = (- 1)^{m} \frac{(l - m)!}{(l + m)!} P_{l}^{m} (x) or Y_{l}^{- m} (θ, φ) = (- 1)^{m} Y_{l}^{m *} (θ, φ)

The first few associated Legendre functions are,

\begin{aligned} P_{0}^{0} (x) & = 1, \\ P_{1}^{0} (x) & = x, P_{1}^{1} (x) = - \sqrt{1 - x^{2}}, \\ P_{2}^{0} (x) & = \frac{1}{2} (3 x^{2} - 1), P_{2}^{1} (x) = - 3 x \sqrt{1 - x^{2}}, P_{2}^{2} (x) = 3 (1 - x^{2}), \\ P_{3}^{0} (x) & = \frac{1}{2} (5 x^{3} - 3 x), P_{3}^{1} (x) = - \frac{3}{2} (5 x^{2} - 1) \sqrt{1 - x^{2}}, P_{3}^{2} (x) = 15 x (1 - x^{2}), P_{3}^{3} (x) = - 15 (1 - x^{2})^{3 / 2} \\ \dots \end{aligned}

And the first few spherical harmonics are,

\begin{aligned} Y_{0}^{0} (θ, φ) & = \sqrt{\frac{1}{4 π}}, \\ Y_{1}^{0} (θ, φ) & = \sqrt{\frac{3}{4 π}} \cos θ, Y_{1}^{1} (θ, φ) = - \sqrt{\frac{3}{8 π}} \sin θ e^{i φ}, \\ Y_{2}^{0} (θ, φ) & = \sqrt{\frac{5}{16 π}} (3 \cos^{2} θ - 1), Y_{2}^{1} (θ, φ) = - \sqrt{\frac{15}{8 π}} \sin θ \cos θ e^{i φ}, Y_{2}^{2} (θ, φ) = \sqrt{\frac{15}{32 π}} \sin^{2} θ e^{2 i φ}, \\ Y_{3}^{0} (θ, φ) & = \sqrt{\frac{7}{16 π}} (5 \cos^{3} θ - 3 \cos θ), Y_{3}^{1} (θ, φ) = - \sqrt{\frac{21}{64 π}} (5 \cos^{2} θ - 1) \sin θ e^{i φ}, \\ Y_{3}^{2} (θ, φ) & = \sqrt{\frac{105}{32 π}} \cos θ \sin^{2} θ e^{2 i φ}, Y_{3}^{3} (θ, φ) = - \sqrt{\frac{35}{64 π}} \sin^{3} θ e^{3 i φ} \\ \dots \end{aligned}

Orthonormality

\oint Y_{l^{'}}^{m^{'} *} (Ω) Y_{l}^{m} (Ω) d Ω = δ_{l l^{'}} δ_{m m^{'}}

Completeness

\sum_{l = 0}^{+ \infty} \sum_{m = - l}^{l} Y_{l}^{m} (Ω^{'}) Y_{l}^{m} (Ω) = δ (Ω - Ω^{'}) = δ (\cos θ - \cos θ^{'}) δ (φ - φ^{'})

1.3.2 Multipole expansion (Legendre function) *

By using the Legendre generating function, one finds

\frac{1}{R} = \frac{1}{| x - x^{'} |} = \sum_{l = 0}^{\infty} \frac{r_{<}^{l}}{r_{>}^{l + 1}} P_{l} (\cos γ), γ = \arccos \frac{x \cdot x^{'}}{| x \cdot x^{'} |}, r_{<} = min (x, x^{'}), r_{>} = max (x, x^{'})

P_{l} (\cos γ) = 2 π \cdot \frac{2}{2 l + 1} \cdot \sum_{m = - l}^{l} Y_{l}^{m *} (θ^{'}, φ^{'}) Y_{l}^{m} (θ, φ)

Consider a localized charge distribution $ρ (x^{'})$ confined within a volume $V^{'}$ . The exact electrostatic potential $Φ (x)$ at an external observation point $x$ (where $r = | x | > r^{'} = | x^{'} |$ , hence $r_{<} = r^{'}$ and $r_{>} = r$ ) is given by,

Φ (x) = \frac{1}{4 π ε_{0}} \int_{V^{'}} \frac{ρ (x^{'})}{| x - x^{'} |} d^{3} x^{'} = \frac{1}{4 π ε_{0}} \sum_{l = 0}^{\infty} \frac{1}{r^{l + 1}} \int_{V^{'}} ρ (x^{'}) (r^{'})^{l} P_{l} (\cos γ) d^{3} x^{'}

Monopole Term (Point Charge). For $l = 0$ , since $P_{0} (\cos γ) = 1$ , the first term simplifies to:

Φ^{(0)} (x) = \frac{1}{4 π ε_{0} r} \int_{V^{'}} ρ (x^{'}) d^{3} x^{'} = \frac{1}{4 π ε_{0}} \frac{q}{r}, with q = \int_{V^{'}} ρ (x^{'}) d^{3} x^{'}

Dipole Term. For $l = 1$ , since $P_{1} (\cos γ) = \cos γ = \frac{x \cdot x^{'}}{r r^{'}}$ , the second term becomes:

Φ^{(1)} (x) = \frac{1}{4 π ε_{0} r^{2}} \int_{V^{'}} ρ (x^{'}) r^{'} (\frac{x \cdot x^{'}}{r r^{'}}) d^{3} x^{'} = \frac{1}{4 π ε_{0}} \frac{p \cdot \hat{r}}{r^{2}}

with p = \int_{V^{'}} ρ (x^{'}) x^{'} d^{3} x^{'}

Quadrupole Term. For $l = 2$ , since $P_{2} (\cos γ) = \frac{1}{2} (3 \cos^{2} γ - 1)$ , substituting $\cos γ$ into the expression yields:

\begin{aligned} Φ^{(2)} (x) & = \frac{1}{4 π ε_{0} r^{3}} \int_{V^{'}} ρ (x^{'}) (r^{'})^{2} \cdot \frac{1}{2} [3 \frac{(x \cdot x^{'})^{2}}{r^{2} (r^{'})^{2}} - 1] d^{3} x^{'} \\ = \frac{1}{8 π ε_{0} r^{5}} x_{i} x_{j} \int_{V^{'}} ρ (x^{'}) [3 x_{i}^{'} x_{j}^{'} - δ_{i j} (r^{'})^{2}] d^{3} x^{'} \\ = \frac{1}{4 π ε_{0}} \frac{1}{2 r^{5}} \sum_{i, j} D_{i j} x_{i} x_{j} \\ = \frac{D : \hat{r} \hat{r}}{8 π ϵ_{0} r^{3}} \end{aligned}

with D_{i j} = \int_{V^{'}} ρ (x^{'}) [3 x_{i}^{'} x_{j}^{'} - δ_{i j} (r^{'})^{2}] d^{3} x^{'}

And extending to $l = 3, 4, \dots$ yields the octupole ( $l = 3$ ), hexadecapole ( $l = 4$ ), and more high $2^{l}$ -pole moments $\dots$

1.3.3 Rotation of harmonics & Wigner $D$ -Matrices *

See Appendix for ED formulas#2 Rotation of harmonics & Wigner $D$ -Matrices *.

1.4 Method: separation of variables

For this section, everyone is advised to review more example problems.

1.4.1 Cartesian coordinate system

The fundamental solution of Laplace's equation $\nabla^{2} φ = 0$

\frac{\nabla^{2} φ}{φ} = \frac{1}{X} \frac{d^{2} X}{d x^{2}} + \frac{1}{Y} \frac{d^{2} Y}{d y^{2}} + \frac{1}{Z} \frac{d^{2} Z}{d z^{2}} = 0

Let:

\frac{1}{X} \frac{d^{2} X}{d x^{2}} = - k_{x}^{2}, \frac{1}{Y} \frac{d^{2} Y}{d y^{2}} = - k_{y}^{2}, \frac{1}{Z} \frac{d^{2} Z}{d z^{2}} = k_{z}^{2}, k_{z}^{2} = k_{x}^{2} + k_{y}^{2}

Note: The sign of the separation constant is entirely determined by the boundary conditions: directions bounded in space require a negative sign to yield trigonometric functions that can zero out at both boundaries, while directions extending to infinity require a positive sign to yield exponential functions (such as the wave propagating in the z-direction).

In directions where the separation constant is negative (e.g., $x, y$ ),

X (x) = A \cos (k_{x} x) + B \sin (k_{x} x)

In the direction where the separation constant is positive (e.g., $z$ ),

Z (z) = C \cosh (k_{z} z) + D \sinh (k_{z} z)

See the thought from ppt int Appendix for ED formulas#3 Example 2D Laplace Equation in a Rectangular Pipe.

1.4.2 Cylindrical coordinate system

The fundamental solution of Laplace's equation $\nabla^{2} φ = 0$

\frac{1}{s R} \frac{d}{d s} (s \frac{d R}{d s}) + \frac{1}{s^{2} Φ} \frac{d^{2} Φ}{d ϕ^{2}} + \frac{1}{Z} \frac{d^{2} Z}{d z^{2}} = 0

$z$ -direction equation: $\frac{1}{Z} \frac{d^{2} Z}{d z^{2}} = k^{2} ⟹ Z (z) = e^{\pm k z}$ (or monotonic hyperbolic functions)
$ϕ$ -direction equation: $\frac{1}{Φ} \frac{d^{2} Φ}{d ϕ^{2}} = - ν^{2} ⟹ Φ (ϕ) = A \cos (ν ϕ) + B \sin (ν ϕ)$

Note: Since $ϕ$ possesses a $2 π$ periodicity in physical space (i.e., $Φ (ϕ) = Φ (ϕ + 2 π)$ ), the separation constant $ν$ must be an integer.

$s$ -direction equation (Bessel Equation): $\frac{d^{2} R}{d s^{2}} + \frac{1}{s} \frac{d R}{d s} + (k^{2} - \frac{ν^{2}}{s^{2}}) R = 0$

The solutions to this equation are the Bessel functions of the first kind $J_{ν} (k s)$ and the Bessel functions of the second kind (Neumann functions) $N_{ν} (k s)$ . If the physical domain includes the origin ( $s = 0$ ), the coefficient of $N_{ν} (k s)$ must be 0 because $N_{ν} (k s) \to - \infty$ as $s \to 0$ .

Axisymmetric & $z$ -Independent Case ( $ν = 0, k = 0$ )

The Laplace equation collapses to a 1D ODE:

\frac{d}{d s} (s \frac{d φ}{d s}) = 0 ⟹ \frac{d^{2} φ}{d s^{2}} + \frac{1}{s} \frac{d φ}{d s} = 0

$z$ -Independent Case (General Solution for $k = 0$ ):

The problem reduces to a 2D Laplace equation in polar coordinates, where the radial equation transitions from a Bessel equation to an Euler-Cauchy equation:

s^{2} \frac{d^{2} R}{d s^{2}} + s \frac{d R}{d s} - ν^{2} R = 0

the general solution for $k = 0$ is obtained by superimposing all possible harmonic components:

φ (s, ϕ) = \underset{as ν = 0}{\underset{⏟}{A_{0} + B_{0} \ln s}} + \sum_{ν = 1}^{\infty} (A_{ν} s^{ν} + B_{ν} s^{- ν}) \cos (ν ϕ) + \sum_{ν = 1}^{\infty} (C_{ν} s^{ν} + D_{ν} s^{- ν}) \sin (ν ϕ)

1.4.3 Spherical coordinate system

In the spherical coordinate system $(r, θ, ϕ)$ , Laplace's equation is given by:

\nabla^{2} φ = \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial φ}{\partial r}) + \frac{1}{r^{2} \sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial φ}{\partial θ}) + \frac{1}{r^{2} \sin^{2} θ} \frac{\partial^{2} φ}{\partial ϕ^{2}} = 0

Let the solution be $φ (r, θ, ϕ) = R (r) Θ (θ) Φ (ϕ)$ . Usually, the angular parts are combined and expressed in terms of spherical harmonics (See 3-D Laplace Equation And Spherical Harmonic Function).

Radial Equation

\frac{d}{d r} (r^{2} \frac{d R}{d r}) = l (l + 1) R ⟹ R (r) = A r^{l} + B r^{- (l + 1)}

where $l$ is a non-negative integer.

Angular Equation
- Azimuthal angle $ϕ$ direction: Similar to the cylindrical system, the solution is $e^{\pm i m ϕ}$ (where $m$ is an integer, and $| m | \leq l$ ).
- Polar angle $θ$ direction: Corresponds to the associated Legendre equation, whose solutions are the associated Legendre polynomials $P_{l}^{m} (\cos θ)$ .
Axisymmetric Systems ( $\partial / \partial ϕ = 0 ⟹ m = 0$ )

φ (r, θ) = \sum_{l = 0}^{\infty} (A_{l} r^{l} + \frac{B_{l}}{r^{l + 1}}) P_{l} (\cos θ)

General Case Solution

φ (r, θ, ϕ) = \sum_{l = 0}^{\infty} \sum_{m = - l}^{l} (A_{l m} r^{l} + \frac{B_{l m}}{r^{l + 1}}) Y_{l}^{m} (θ, ϕ)

[\frac{1}{\sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial}{\partial θ}) + \frac{1}{\sin^{2} θ} \frac{\partial^{2}}{\partial ϕ^{2}}] Y_{l}^{m} = - l (l + 1) Y_{l}^{m}, \frac{\partial^{2}}{\partial ϕ^{2}} Y_{l}^{m} = - m^{2} Y_{l}^{m}

1.4.4 Green function *

Main equation

\nabla^{2} φ (r) = - \frac{ρ (r)}{ϵ_{0}}

\nabla^{2} G (r, r^{'}) = - \frac{δ (r - r^{'})}{ϵ_{0}}

Fundamental solution in infinite space,

G_{0} (r, r^{'}) = \frac{1}{4 π ϵ_{0} | r - r^{'} |}

φ (r) = \int ρ (r^{'}) G (r, r^{'}) d^{3} r^{'}

With boundaries (Dirichlet condition):

When $φ |_{S} = φ_{0} (r)$ , and the Green's function satisfies

\nabla^{2} G (r, r^{'}) = - \frac{δ (r - r^{'})}{ϵ_{0}}, G (r, r^{'}) |_{r \in S} = 0

then the solution for any volume charge density $ρ$ is,

φ (r) = \int_{V} ρ (r^{'}) G (r, r^{'}) d^{3} r^{'} - ϵ_{0} \oint_{S} φ_{0} (r^{'}) \frac{\partial G (r, r^{'})}{\partial n^{'}} d S^{'}

With boundaries (Neumann condition):

When $\frac{\partial φ}{\partial n} |_{S} = g (r)$ , the Neumann Green's function $G_{N}$ satisfies

\nabla^{2} G_{N} (r, r^{'}) = - \frac{δ (r - r^{'})}{ϵ_{0}}, {\frac{\partial G_{N}}{\partial n} |}_{r \in S} = - \frac{1}{ϵ_{0} S}

the potential is,

φ (r) = \int_{V} ρ (r^{'}) G_{N} (r, r^{'}) d^{3} r^{'} + ϵ_{0} \oint_{S} g (r^{'}) G_{N} (r, r^{'}) d S^{'} + ⟨ φ ⟩_{S}

where $⟨ φ ⟩_{S}$ is the average potential on the boundary.

Refer to Green Function for several basic examples in infinite space.

1.5 Cases

If all charges are contained within the sphere $V$ (see Appendix for ED formulas#1 $ iiint_V boldsymbol{E}( boldsymbol{x}) , mathrm{d} 3x = - frac{ boldsymbol{p}}{3 epsilon_0}$ or T 2.3 in the textbook)

∭_{V} E (x) d^{3} x = - \frac{p}{3 ϵ_{0}}

An ellipsoid uniformly charged with Q (in the principal inertia axis frame)

D_{11} = \frac{Q}{5} (2 a_{1}^{2} - a_{2}^{2} - a_{3}^{2}), D_{22} = \dots, D_{i \neq j} = 0

2 Magnetostatic Field

2.1 magnetic energy

#todo

2.2 Multipole expansion (scalar potential)

2.2.1 magnetic scalar potential *

Without free/conductive current

\nabla \times H = J_{f} ⟹ \nabla \times H = 0, H = - \nabla ψ

then for linear media $B = μ H$ ,

\nabla \cdot H = 0 ⟹ \nabla^{2} ψ = 0

Consequently, the multipole expansion can be applied to magnetic scalar potential $ψ$ to derive $H$ as well.

For a current-carrying coil,

ψ (x) = - \frac{I}{4 π} Ω, Ω = - \iint_{S} \frac{\hat{R} \cdot d σ^{'}}{R^{2}}

where $S = \frac{1}{2} \oint_{C} x^{'} \times d l^{'}$ is the arbitrary geometric vector area enclosed by the current loop, and $Ω$ is the solid angle of the surface $S$ with respect to field point $x$ .

2.2.2 Multipole expansion

Consider magnetostatic potential expansion at an external observation point $x$ , with localized current coil/source $J (x^{'})$ confined within a volume $V^{'}$ , and $R = x - x^{'}, r = | x | (f i e l d), r^{'} = | x^{'} | (s o u r c e)$ , then

\begin{aligned} \frac{1}{R} = \frac{1}{| x - x^{'} |} = e^{- x^{'} \cdot \nabla} \frac{1}{r} & = [1 - x^{'} \cdot \nabla + \dots] \frac{1}{r} = \frac{1}{r} - x^{'} \cdot \nabla \frac{1}{r} + \dots \end{aligned}

⇓

\begin{aligned} ψ (x) = \frac{I}{4 π} \iint_{S} \frac{\hat{R} \cdot d σ^{'}}{R^{2}} & = - \frac{I}{4 π} \iint_{S} d σ^{'} \cdot \nabla (\frac{1}{R}) \\ = - \frac{I}{4 π} [(\iint_{S} d σ^{'}) \cdot \nabla - (\iint_{S} d σ^{'} x^{'}) : \nabla \nabla + \dots] \frac{1}{r} \\ = \frac{m \cdot \hat{r}}{4 π r^{2}} + \frac{D_{m} : \hat{r} \hat{r}}{8 π r^{3}} + \dots \end{aligned}

Where there is no magnetic monopole term ( $1 / r$ ). $m = I \iint_{S} d σ^{'}$ is the magnetic dipole moment of the shell (area vector), and $D_{m} = I \iint_{S} [3 (d σ^{'} x^{'} + x^{'} d σ^{'}) - 2 (x^{'} \cdot d σ^{'}) I]$ is the trace-free magnetic quadrupole moment tensor.

Magnetostatic intensity expansion ( $H = - \nabla ψ$ )

H = \frac{3 (m \cdot \hat{r}) \hat{r} - m}{4 π r^{3}} + \frac{5 (D_{m} : \hat{r} \hat{r}) \hat{r} - 2 D_{m} \cdot \hat{r}}{8 π r^{4}} + \dots

2.3 Multipole expansion (vector potential)

The quadrupole moment of the magnetic field is relatively difficult to calculate and can be skipped. However, the dipole moment needs to be understood.

2.3.1 Moment equation for localized current *

Using $\nabla \cdot (J r) = (- \partial_{t} ρ) r + J$ ,

∭_{V} J d V = \dot{p}, p = ∭_{V} r d q

Using $\nabla \cdot (J r r) = (- \partial_{t} ρ) r r + J r + r J$ ,

∭_{V} J r = m \times I + \frac{1}{6} \dot{D} + \frac{1}{6} \dot{g} I

where $m = \frac{1}{2} ∭_{V} (r \times J) d V$ is magnetic dipole moment, $D = \int (3 r r - r^{2} I) d q$ is electric quadrupole moment, and $g = ∭ r^{2} d q$ is the trace of the second spatial moment of the charge distribution.

for steady current, using $\partial_{t} ρ = 0$ ,

∭_{V} J d V = 0

∭_{V} J r d V = m \times I ⟹ ∭_{V} J_{i} r_{j} d V = - ϵ_{i j k} m_{k}

2.3.2 Multipole expansion *

Consider magnetostatic potential expansion at an external observation point $x$ , with localized current source $J (x^{'})$ confined within a volume $V^{'}$ , and $R = x - x^{'}, r = | x | (f i e l d), r^{'} = | x^{'} | (s o u r c e)$ , then

\begin{aligned} \frac{1}{R} = \frac{1}{| x - x^{'} |} = e^{- x^{'} \cdot \nabla} \frac{1}{r} & = [1 - x^{'} \cdot \nabla + \frac{1}{2!} (x^{'} x^{'} : \nabla \nabla) + \dots] \frac{1}{r} \\ I : \nabla \nabla \frac{1}{r} = 0 ⟹ & = \frac{1}{r} - x^{'} \cdot \nabla \frac{1}{r} + \frac{1}{6} (3 x^{'} x^{'} - r^{' 2} I) : \nabla \nabla \frac{1}{r} + \dots \end{aligned}

⇓

\begin{aligned} A (x) & = \frac{μ_{0}}{4 π} \int_{V^{'}} \frac{J (x^{'})}{R} d V^{'} \\ = \frac{μ_{0}}{4 π} {\frac{1}{r} \int_{V^{'}} J (x^{'}) d V^{'} + \frac{1}{r^{3}} \int_{V^{'}} J (x^{'}) x^{'} d V^{'} \cdot x + \frac{1}{6} \int_{V^{'}} J (x^{'}) [(3 x^{'} x^{'} - r^{' 2} I) : \nabla \nabla \frac{1}{r}] d V^{'} + \dots} \\ using Coulomb Gauge \to & = \frac{μ_{0}}{4 π r^{3}} (m \times I) \cdot x + \frac{μ_{0}}{24 π} \nabla \times \frac{D_{m} \cdot x}{r^{3}} + \dots \\ = \frac{μ_{0}}{4 π r^{2}} (m \times \hat{r}) - \frac{μ_{0}}{8 π r^{3}} \hat{r} \times (D_{m} \cdot \hat{r}) + \dots \end{aligned}

Where $m = \frac{1}{2} \int_{V^{'}} (x^{'} \times J (x^{'})) d V^{'}$ is the magnetic dipole moment, and $D_{m} = \int_{V^{'}} [(x^{'} \times J) x^{'} + x^{'} (x^{'} \times J)] d V^{'}$ is the trace-free magnetic quadrupole moment tensor.

Magnetostatic intensity expansion

B (x) = \nabla \times A = \frac{μ_{0}}{4 π r^{3}} [3 (m \cdot \hat{r}) \hat{r} - m] + \frac{μ_{0}}{8 π r^{4}} [5 (D_{m} : \hat{r} \hat{r}) \hat{r} - 2 D_{m} \cdot \hat{r}] + \dots

2.3.3 Cases

Ring current

m = I S

where $S = \frac{1}{2} \oint_{C} r \times d l$ is the geometric vector area enclosed by the current loop.

Moving point charge, using $J (r) = q v δ (r - r_{e})$ ,

m = \frac{q}{2 m_{p}} L

where $L = m_{p} (r \times v)$ is the orbital angular momentum.

Rotating charged body, using $J = ρ_{e} (ω \times r)$ , and for any rigid body with $\frac{ρ_{e}}{ρ_{m}} \equiv \frac{Q}{M}$ ,

m = \frac{Q}{2 M} L, L = \int ρ_{m} (r \times v) d V = I_{m} ω

where $I_{m}$ is the moment of inertia tensor. For a sphere with constant $ρ_{e} = \frac{Q}{4 π a^{3} / 3}$ , $I_{m} = \frac{2}{5} M a^{2}$ , thus

m = \frac{1}{5} Q a^{2} ω

for a spherical shell with constant $σ_{e} = \frac{Q}{4 π a^{2}}$ , $I_{m} = \frac{2}{3} M a^{2}$ , thus

m = \frac{1}{2} \oint (r \times K) d σ = \frac{1}{3} Q a^{2} ω

If all currents are contained within the sphere $V$ (see Appendix for ED formulas #1 $ iiint_V boldsymbol{E}( boldsymbol{x}) , mathrm{d} 3x = - frac{ boldsymbol{p}}{3 epsilon_0}$)

∭_{V} B (x) d^{3} x = \frac{2 μ_{0}}{3} m

2.4 Small current-carrying conductor in magnetic field

Assuming the spatial size of the current-carrying body $x$ is so small compared with the distance to the source $x^{'}$ which generates the magnetic field that $\nabla \times B_{ext} = 0$ and $\nabla \cdot B_{ext} = 0$ within the body. Let $ξ$ be the relative coordinate of a current element $J (ξ) d V$ from the body's center $x$ . The interaction magnetic energy $U = \int J (ξ) \cdot A (x + ξ) d V$ of the small body is,

U \approx m \cdot B_{ext} |_{x} + \frac{1}{6} D_{m} : \nabla B_{ext} |_{x} + \dots

Note that $m = \frac{1}{2} \int (ξ \times J (ξ)) d V$ and $D_{m}$ are the magnetic dipole and quadrupole moment tensor, which are intrinsic properties of the small current-carrying body itself, not the external source.

Furthermore, for a system with constant currents, the total magnetostatic force $F$ acting on the small current-carrying body is given by the generalized force relation $F = + \nabla U$ (why use "+"? see Appendix for ED formulas#4 Physical Interpretation of the Sign and Energy Definitions *),

F = + \nabla U = (m \cdot \nabla) B_{ext} |_{x} + \frac{1}{6} (D_{m} : \nabla \nabla) B_{ext} |_{x} + \dots

For a current loop/coil $m = I S$ ,

m \cdot \nabla B = I \nabla (B \cdot S) = I \nabla Φ

The total force moment $N$ acting on the small current-carrying body about its center $x$ is,

N = \int ξ \times (J (ξ) \times B_{ext} (x + ξ)) d V = m \times B_{ext} + \dots

3 EM Wave

3.1 retarded potential

4 Electromagnetic Radiation