I. Mathematical Foundations

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 $d s^{2} = - c^{2} d t^{2} + d x^{2} + d y^{2} + d z^{2} = - c^{2} d τ^{2}$ .

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 Preparatory Math.

1.1 Vector and tensor analysis (Euclidean geometry)

1.1.1 Basics

Kronecker & Levi-Civita symbols

δ_{i j} = {\hat{e}}_{i} \cdot {\hat{e}}_{j}, ϵ_{i j k} = ({\hat{e}}_{i} \times {\hat{e}}_{j}) \cdot {\hat{e}}_{k} \Leftrightarrow {\hat{e}}_{i} \times {\hat{e}}_{j} = ϵ_{i j k} {\hat{e}}_{k}

Determinant

ϵ_{l m n} det A_{3 \times 3} = ϵ_{i j k} A_{i l} A_{j m} A_{k n} = ϵ_{i j k} A_{l i} A_{m j} A_{n k}

⟹ det A_{3 \times 3} = \frac{1}{6} ϵ_{i j k} ϵ_{l m n} A_{i l} A_{j m} A_{k n} or det A_{3 \times 3} = ϵ_{i j k} A_{i 1} A_{j 2} A_{k 3} = ϵ_{i j k} A_{1 i} A_{2 j} A_{3 k}

Contraction identities

ϵ_{i j k} ϵ_{i m n} = δ_{j m} δ_{k n} - δ_{j n} δ_{k m}, ϵ_{i j k} ϵ_{i j n} = 2 δ_{k n}, ϵ_{i j k} ϵ_{i j k} = 3! = 6

More generally*

ϵ_{i j k} ϵ_{l m n} = det (\begin{matrix} δ_{i l} & δ_{i m} & δ_{i n} \\ δ_{j l} & δ_{j m} & δ_{j n} \\ δ_{k l} & δ_{k m} & δ_{k n} \end{matrix})

and for high-dimensional case, we have generalized Kronecker delta

ϵ_{i_{1} i_{2} \dots i_{n}} ϵ_{j_{1} j_{2} \dots j_{n}} = det (\begin{matrix} δ_{i_{1} j_{1}} & δ_{i_{1} j_{2}} & \dots & δ_{i_{1} j_{n}} \\ δ_{i_{2} j_{1}} & δ_{i_{2} j_{2}} & \dots & δ_{i_{2} j_{n}} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ δ_{i_{n} j_{1}} & δ_{i_{n} j_{2}} & \dots & δ_{i_{n} j_{n}} \end{matrix}) ≜ δ_{i_{1} i_{2} \dots i_{n}}^{j_{1} j_{2} \dots j_{n}}

Axial Vector / Pseudovector

A_{i}^{'} = det (R) R_{i j} A_{j}, R \in O (3)

Symmetric-Antisymmetric decomposition

T_{i j} = T_{(i j)} + T_{[i j]} = \frac{T_{i j} + T_{j i}}{2} + \frac{T_{i j} - T_{j i}}{2}

Double Contraction (adopt to the proximity rule)

A : B ≜ A_{i j} B_{j i} = tr (A B)

Convention: throughout these notes, : follows the proximity contraction rule and therefore differs from the common Frobenius convention $A : B ≜ A_{i j} B_{i j} = tr (A^{T} B)$ .

1.1.2 Cross & dot product

Scalar Triple product

a \cdot (b \times c) = b \cdot (c \times a) = c \cdot (a \times b)

Vector Triple product / BAC-CAB formula

a \times (b \times c) = b a \cdot c - c a \cdot b or (b \times c) \times a = c a \cdot b - b a \cdot c

Lagrange's identity

(a \times b) \cdot (c \times d) = (a \cdot c) (b \cdot d) - (a \cdot d) (b \cdot c)

Associative Law (the tensor remains centered in the contraction)

a \cdot T \cdot b = (a \cdot T) \cdot b = a \cdot (T \cdot b)

a \cdot T \times b = (a \cdot T) \times b = a \cdot (T \times b)

a \times T \times b = (a \times T) \times b = a \times (T \times b)

Cross $\to$ Dot

ω \times I = I \times ω = (\begin{matrix} 0 & - ω_{3} & ω_{2} \\ ω_{3} & 0 & - ω_{1} \\ - ω_{2} & ω_{1} & 0 \end{matrix}) ≜ Ω \Leftrightarrow Ω_{i j} = ϵ_{i k j} ω_{k} ⟹ ω \times v = Ω \cdot v, \forall v \in R^{3}

1.1.3 $\nabla$

Differential operations

Coordinate Component Expansion of Differential Operators

grad Φ ≜ {\hat{e}}_{i} \partial_{i} Φ

div v ≜ ({\hat{e}}_{i} \partial_{i}) \cdot v = \partial_{i} v_{i}

rot v (or curl) ≜ ({\hat{e}}_{i} \partial_{i}) \times v = ϵ_{i j k} (\partial_{j} v_{k}) {\hat{e}}_{i} = | \begin{matrix} {\hat{e}}_{1} & {\hat{e}}_{2} & {\hat{e}}_{3} \\ \partial_{1} & \partial_{2} & \partial_{3} \\ v_{1} & v_{2} & v_{3} \end{matrix} |

\nabla^{2} Φ = \nabla \cdot \nabla Φ = \sum_{i} \partial_{i}^{2} Φ

Two critical identities

\nabla \times \nabla φ = 0, \nabla \cdot (\nabla \times a) = 0

Leibniz Rule

\begin{aligned} \nabla (φ ψ) & = (\nabla φ) ψ + φ (\nabla ψ) \\ \nabla (a \cdot b) & = a \cdot \nabla b + b \cdot \nabla a + a \times (\nabla \times b) + b \times (\nabla \times a) \\ ⟹ a \times (\nabla \times a) = \frac{1}{2} \nabla (a \cdot a) - a \cdot \nabla a \end{aligned}

\begin{aligned} \nabla (φ a) & = (\nabla φ) a + φ (\nabla a) \\ \nabla \cdot (φ a) & = (\nabla φ) \cdot a + φ (\nabla \cdot a) \\ \nabla \times (φ a) & = (\nabla φ) \times a + φ (\nabla \times a) \end{aligned}

\begin{aligned} \nabla (a \times b) & = (\nabla a) \times b - (\nabla b) \times a \\ \nabla \cdot (a \times b) & = (\nabla \times a) \cdot b - (\nabla \times b) \cdot a \\ \nabla \times (a \times b) & = (\nabla \cdot b + b \cdot \nabla) a - (\nabla \cdot a + a \cdot \nabla) b \end{aligned}

\begin{aligned} \nabla \cdot (φ T) & = (\nabla φ) \cdot T + φ (\nabla \cdot T) \\ \nabla \cdot (a \cdot T) & = (\nabla a) : T + a \cdot (\nabla \cdot T) \end{aligned}

\begin{aligned} \nabla \cdot (a b) & = (\nabla \cdot a) b + a \cdot \nabla b \\ \nabla \cdot (a b c) & = (\nabla \cdot a) b c + (a \cdot \nabla b) c + b (a \cdot \nabla c) \end{aligned}

\nabla \times (a \times b) = \nabla \cdot (b a - a b) = (\nabla \cdot b + b \cdot \nabla) a - (\nabla \cdot a + a \cdot \nabla) b

Taylor Series in Operator Form

φ (x + ϵ) = e^{ϵ \cdot \nabla} φ (x) = [1 + (ϵ \cdot \nabla) + \frac{1}{2!} (ϵ \cdot \nabla)^{2} + \frac{1}{3!} (ϵ \cdot \nabla)^{3} + \dots] φ (x)

Here, the terms in the parentheses are

\frac{1}{2} (ϵ \cdot \nabla) (ϵ \cdot \nabla) = \frac{1}{2} ϵ_{i} ϵ_{j} \frac{\partial^{2}}{\partial x_{i} \partial x_{j}}, \frac{1}{6} (ϵ \cdot \nabla)^{3} = \frac{1}{6} ϵ_{i} ϵ_{j} ϵ_{k} \frac{\partial^{3}}{\partial x_{i} \partial x_{j} \partial x_{k}}, \dots

$T (ϵ) = e^{ϵ \cdot \nabla}$ is so-called "Translation Operator" in quantum mechanics or lie group ( $\hat{T} (ϵ) = e^{\frac{i}{ℏ} ϵ \cdot \hat{p}}$ ).

Integral operations

Coordinate-independent definition

d φ (r) = \nabla φ \cdot d l

\nabla \cdot F ≜ lim_{V \to 0} [\frac{1}{V} \oint_{\partial V} F \cdot d S]

\hat{n} \cdot (\nabla \times F) ≜ lim_{S \to 0} [\frac{1}{S} \oint_{\partial S} F \cdot d l]

Fundamental theorem of gradients

\int_{P}^{Q} d l \cdot \nabla = |_{P}^{Q}, = φ, a, T, \dots

⟹ \oint_{C} d l \cdot \nabla = 0 (conservative field)

Generalized Gauss's Theorem

∭_{V} d V \nabla = \oint_{\partial V} d σ, = φ, \times a, \cdot a, \times T \dots, with d σ = n d S

⟹ \oint_{\partial V} d σ = 0

also ⟹ \nabla = lim_{V \to 0} [\frac{1}{V} \oint_{\partial V} d σ] ⟹ {\begin{aligned} \nabla φ & = lim_{V \to 0} [\frac{1}{V} \oint_{\partial V} d σ φ] \\ \nabla \cdot F & = lim_{V \to 0} [\frac{1}{V} \oint_{\partial V} d σ \cdot F] (Classic Gauss) \\ \nabla \times F & = lim_{V \to 0} [\frac{1}{V} \oint_{\partial V} d σ \times F] \end{aligned}

In Gauss's Divergence Theorem $∭_{V} d V \nabla \cdot F = \oint_{\partial V} d σ \cdot F$ , if $F = φ \nabla ψ$ , one finds the Green's first identity:

\int_{V} d V [φ \nabla^{2} ψ + \nabla φ \cdot \nabla ψ] = \oint_{\partial V} d σ \cdot φ \nabla ψ

and substituting $F = φ \nabla ψ - ψ \nabla φ$ yields the Green's second identity:

\int_{V} d V [φ \nabla^{2} ψ - ψ \nabla^{2} φ] = \oint_{\partial V} d σ \cdot [φ \nabla ψ - ψ \nabla φ]

Generalized Stokes' Theorem

\oint_{S} (d σ \times \nabla) = \oint_{\partial S} d l, = φ, \times a, \cdot a, \times T \dots

⟹ \oint_{\partial S} d l \cdot F = \oint_{S} (d σ \times \nabla) \cdot F = \oint_{S} d σ \cdot (\nabla \times F) (Classic Stokes)

Helmholtz decomposition

For any continuous differentiable vector field $F$ , if $lim_{r \to \infty} r F (x) \to 0$ ,

\begin{aligned} F & = \underset{longitudinal/irrotational part}{\underset{⏟}{- \nabla φ}} + \underset{transverse/solenoidal part}{\underset{⏟}{\nabla \times A}} \\ = - \frac{1}{4 π} \nabla \int \frac{\nabla^{'} \cdot F (x^{'})}{| x - x^{'} |} d V^{'} + \frac{1}{4 π} \nabla \times \int \frac{\nabla^{'} \times F (x^{'})}{| x - x^{'} |} d V^{'} \end{aligned}

for static magnetic field, Biot-Savart Law:

⟹ B (x) = \frac{1}{4 π} \nabla \times \int \frac{μ_{0} J (x^{'})}{| x - x^{'} |} d V^{'} = \frac{μ_{0}}{4 π} \int \frac{J (x^{'}) \times (x - x^{'})}{| x - x^{'} |^{3}} d V^{'}, x : field; x^{'} : source

for static electric field, Coulomb's Law

⟹ E (x) = - \frac{1}{4 π} \nabla \int \frac{ρ (x^{'}) / ϵ_{0}}{| x - x^{'} |} d V^{'} = \frac{1}{4 π ϵ_{0}} \int \frac{ρ (x^{'}) (x - x^{'})}{| x - x^{'} |^{3}} d V^{'}

1.1.4 Cases

Determinant

det (a b) = 0, det (I + a b) = 1 + a \cdot b

For $r = | r |, r = (x, y, z) = r_{i} {\hat{e}}_{i} = r \hat{r}$ & $\nabla$

\nabla r = \hat{r}, \nabla r = I, \nabla \cdot r = 3, \nabla \times r = 0

\nabla \times \hat{r} = 0, \nabla \cdot \hat{r} = \frac{2}{r}, \nabla \hat{r} = \frac{I - \hat{r} \hat{r}}{r}, \nabla \times [f (r) \hat{r}] = 0

\nabla^{2} (- \frac{1}{r}) = \nabla \cdot (\frac{\hat{r}}{r^{2}}) = 4 π δ^{3} (r), \nabla \frac{1}{r} = - \frac{\hat{r}}{r^{2}}, \nabla \frac{\hat{r}}{r^{2}} = \frac{I - 3 \hat{r} \hat{r}}{r^{3}} + \frac{4 π δ^{3} (x)}{3} I

Double Contraction (adopt to the proximity rule)

I : a b = tr (a b) = a \cdot b ⟹ I : \nabla a = \nabla \cdot a

T : I = tr (T) ⟹ I : I = 3

$\nabla$

\nabla \cdot φ I = \nabla φ

\nabla \cdot (T \times r) = - r \times (\nabla \cdot T), with T_{i j} = T_{j i}

1.1.5 $\nabla$ in the orthogonal curvilinear coordinates

Definition

In this section, we adopt $e_{i}$ rather than ${\hat{e}}_{i}$ to indicate that the basis vectors are orthogonal but not normalized: $e_{i} = h_{i} {\hat{e}}_{i} (no summation)$ . Replace the Cartesian coordinate values $(x_{1}, x_{2}, x_{3})$ with the Curvilinear coordinate values $(u_{1}, u_{2}, u_{3})$ .

Lamé Parameters

h_{i} = | \frac{\partial r}{\partial u_{i}} |, H = \prod_{i = 1}^{3} h_{i}, {\hat{e}}_{i} = \frac{\partial r / \partial u_{i}}{| \partial r / \partial u_{i} |} = \frac{1}{h_{i}} \frac{\partial r}{\partial u_{i}} = \frac{1}{h_{i}} e_{i} (no summation)

for Cylindrical coordinates, $r = (ρ \cos ϕ, ρ \sin ϕ, z)$ :

h_{ρ}, h_{ϕ}, h_{z} = 1, ρ, 1 and {\begin{cases} {\hat{e}}_{ρ} = \frac{1}{h_{ρ}} \frac{\partial r}{\partial ρ} = (\cos ϕ, \sin ϕ, 0) \\ {\hat{e}}_{ϕ} = \frac{1}{h_{ϕ}} \frac{\partial r}{\partial ϕ} = (- \sin ϕ, \cos ϕ, 0) \\ {\hat{e}}_{z} = \frac{1}{h_{z}} \frac{\partial r}{\partial z} = (0, 0, 1) \end{cases}

for Spherical coordinates, $r = (r \sin θ \cos ϕ, r \sin θ \sin ϕ, r \cos θ)$ :

h_{r}, h_{θ}, h_{ϕ} = 1, r, r \sin θ and {\begin{cases} {\hat{e}}_{r} = \frac{1}{h_{r}} \frac{\partial r}{\partial r} = (\sin θ \cos ϕ, \sin θ \sin ϕ, \cos θ) \\ {\hat{e}}_{θ} = \frac{1}{h_{θ}} \frac{\partial r}{\partial θ} = (\cos θ \cos ϕ, \cos θ \sin ϕ, - \sin θ) \\ {\hat{e}}_{ϕ} = \frac{1}{h_{ϕ}} \frac{\partial r}{\partial ϕ} = (- \sin ϕ, \cos ϕ, 0) \end{cases}

$\nabla$

Coordinate Component Expansion of Differential Operators

grad Φ = \sum_{i} \frac{1}{h_{i}} e_{i} \partial_{i} Φ = {\hat{e}}_{i} \partial_{i} Φ

div v = \sum_{i} \frac{1}{H} \partial_{i} (\frac{H v_{i}}{h_{i}}) = \frac{1}{h_{1} h_{2} h_{3}} [\frac{\partial}{\partial u_{1}} (h_{2} h_{3} v_{1}) + \frac{\partial}{\partial u_{2}} (h_{3} h_{1} v_{2}) + \frac{\partial}{\partial u_{3}} (h_{1} h_{2} v_{3})]

curl v = \frac{1}{h_{1} h_{2} h_{3}} | \begin{matrix} h_{1} {\hat{e}}_{1} & h_{2} {\hat{e}}_{2} & h_{3} {\hat{e}}_{3} \\ \frac{\partial}{\partial u_{1}} & \frac{\partial}{\partial u_{2}} & \frac{\partial}{\partial u_{3}} \\ h_{1} v_{1} & h_{2} v_{2} & h_{3} v_{3} \end{matrix} |

\nabla^{2} v = \nabla \cdot \nabla v = \sum_{k} \frac{1}{H} \partial_{k} (\frac{H}{h_{k}} (\nabla v)_{k}) = \sum_{k} \frac{1}{H} \partial_{k} (\frac{H}{h_{k}} \partial_{k}^{2} v)

The differential identity of the basis vector

\nabla u_{i} = {\hat{e}}_{i} = \frac{e_{i}}{h_{i}}, \nabla \times \frac{e_{i}}{h_{i}} = \nabla \times {\hat{e}}_{i} = 0, \nabla \cdot \frac{h_{i} e_{i}}{H} = 0 (no summation)

For Cylindrical coordinates, $r = (r \cos ϕ, r \sin ϕ, z)$

\begin{aligned} \nabla v = e_{r} \frac{\partial v}{\partial r} + e_{θ} \frac{1}{r} \frac{\partial v}{\partial θ} + e_{z} \frac{\partial v}{\partial z} \\ \nabla \cdot v = \frac{1}{r} \frac{\partial}{\partial r} (r v_{r}) + \frac{1}{r} \frac{\partial v_{θ}}{\partial θ} + \frac{\partial v_{z}}{\partial z} \\ \nabla \times v = \frac{1}{r} | \begin{array}{c} e_{r} & r e_{θ} & e_{z} \\ \frac{\partial}{\partial r} & \frac{\partial}{\partial θ} & \frac{\partial}{\partial z} \\ v_{r} & r v_{θ} & v_{z} \end{array} | \\ \nabla^{2} v = \frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial v}{\partial r}) + \frac{1}{r^{2}} \frac{\partial^{2} v}{\partial θ^{2}} + \frac{\partial^{2} v}{\partial z^{2}} \end{aligned}

and there are more formulas which may be useful in your further study, *

\nabla^{2} v = (\nabla^{2} v_{r} - \frac{v_{r}}{r^{2}} - \frac{2}{r^{2}} \frac{\partial v_{θ}}{\partial θ}) {\hat{e}}_{r} + (\nabla^{2} v_{θ} - \frac{v_{θ}}{r^{2}} + \frac{2}{r^{2}} \frac{\partial v_{r}}{\partial θ}) {\hat{e}}_{θ} + (\nabla^{2} v_{z}) {\hat{e}}_{z}

v \cdot \nabla b = (v \cdot \nabla b_{r} - \frac{v_{θ} b_{θ}}{r}) {\hat{e}}_{r} + (v \cdot \nabla b_{θ} + \frac{v_{θ} b_{r}}{r}) {\hat{e}}_{θ} + (v \cdot \nabla b_{z} + v_{z} \frac{\partial b_{z}}{\partial z}) {\hat{e}}_{z}

For Spherical coordinates, $r = (r \sin θ \cos φ, r \sin θ \sin φ, r \cos θ)$

\begin{aligned} \nabla v = e_{r} \frac{\partial v}{\partial r} + e_{θ} \frac{1}{r} \frac{\partial v}{\partial θ} + e_{φ} \frac{1}{r \sin θ} \frac{\partial v}{\partial φ} \\ \nabla \cdot v = \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} v_{r}) + \frac{1}{r \sin θ} \frac{\partial}{\partial θ} (\sin θ v_{θ}) + \frac{1}{r \sin θ} \frac{\partial v_{φ}}{\partial φ} \\ \nabla \times v = \frac{1}{r^{2} \sin θ} | \begin{array}{c} e_{r} & r e_{θ} & r \sin θ e_{φ} \\ \frac{\partial}{\partial r} & \frac{\partial}{\partial θ} & \frac{\partial}{\partial φ} \\ v_{r} & r v_{θ} & r \sin θ v_{φ} \end{array} | \\ \nabla^{2} v = \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial v}{\partial r}) + \frac{1}{r^{2} \sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial v}{\partial θ}) + \frac{1}{r^{2} \sin^{2} θ} \frac{\partial^{2} v}{\partial φ^{2}} \end{aligned}

and there are more formulas as well, *

\begin{aligned} \nabla^{2} v = & [\nabla^{2} v_{r} - \frac{2}{r^{2}} (v_{r} + \frac{1}{\sin θ} \frac{\partial \sin θ v_{θ}}{\partial θ} + \frac{1}{\sin θ} \frac{\partial v_{φ}}{\partial φ})] {\hat{e}}_{r} \\ + & [\nabla^{2} v_{θ} + \frac{1}{r^{2}} (2 \frac{\partial v_{r}}{\partial θ} - \frac{v_{θ}}{\sin^{2} θ} - \frac{2 \cos θ}{\sin^{2} θ} \frac{\partial v_{φ}}{\partial φ})] {\hat{e}}_{θ} \\ + & [\nabla^{2} v_{φ} + \frac{1}{r^{2}} (\frac{2}{\sin θ} \frac{\partial v_{r}}{\partial φ} + \frac{2 \cos θ}{\sin^{2} θ} \frac{\partial v_{θ}}{\partial φ} - \frac{v_{φ}}{\sin^{2} θ})] {\hat{e}}_{φ} \end{aligned}

\begin{aligned} v \cdot \nabla b = & (v \cdot \nabla b_{r} - \frac{v_{θ} b_{θ} + v_{φ} b_{φ}}{r}) {\hat{e}}_{r} \\ + & (v \cdot \nabla b_{θ} + \frac{v_{θ} b_{r} - v_{φ} b_{φ} \cot θ}{r}) {\hat{e}}_{θ} \\ + & (v \cdot \nabla b_{φ} + \frac{v_{φ} b_{r} + v_{φ} b_{θ} \cot θ}{r}) {\hat{e}}_{φ} \end{aligned}

1.2 Dirac $δ$ function

1.2.1 Definition

1D definition

\int_{R} δ (x - a) φ (x) d x = φ (a) ⟹ \int_{R} δ^{(n)} (x - a) φ (x) d x = (- 1)^{n} φ^{(n)} (a)

⟹ \int_{R} δ^{(n)} (x - a) d x = {\begin{cases} 1, n = 1 \\ 0, n > 1 \end{cases}

δ (x) = \frac{d}{d x} Θ (x), Θ (x) = {\begin{cases} 0, & x < 0 \\ 1, & x > 0 \end{cases}

3D definition

δ^{3} (x - a) = δ (x_{1} - a_{1}) δ (x_{2} - a_{2}) δ (x_{3} - a_{3})

⟹ \int_{R^{3}} δ^{3} (x - a) φ (x) d^{3} x = φ (a)

\int_{R^{3}} \partial^{α} δ^{3} (x - a) φ (x) d^{3} x = (- 1)^{α} \partial^{α} φ (a), with \partial^{α} = {(\frac{\partial}{\partial x_{1}})}^{α_{1}} {(\frac{\partial}{\partial x_{2}})}^{α_{2}} {(\frac{\partial}{\partial x_{3}})}^{α_{3}}, α = α_{1} + α_{2} + α_{3}

⟹ \int_{R^{3}} [\nabla δ^{3} (x - a)] φ (x) d^{3} x = - \nabla φ (a)

⟹ \int_{R^{3}} [\nabla δ^{3} (x - a)] \cdot A (x) d^{3} x = - (\nabla \cdot A) |_{x = a}

⟹ \int_{R^{3}} [\nabla^{2} δ^{3} (x - a)] φ (x) d^{3} x = \nabla^{2} φ (a)

1.2.2 Fundamental characteristics of $δ$

x δ (x) = 0, x^{n} δ^{(n)} (x) = (- 1)^{n} n! δ (x), x_{i} δ^{3} (x) = 0, x_{i} \partial_{j} δ^{3} (x) = - δ_{i j} δ^{3} (x)

for $f (x) : R \to R$ , one finds

δ (f (x)) = \sum_{n} \frac{δ (x - x_{i})}{| f^{'} (x_{i}) |}, with f (x_{i}) = 0, f^{'} (x_{i}) \neq 0

for $f (x) : R^{n} \to R$ *

\int_{R^{n}} g (x) δ (f (x)) d^{n} x = \int_{f (x) = 0} \frac{g (x)}{| \nabla f (x) |} d S, with \nabla f (x_{i}) \neq 0

for $F (x) : R^{n} \to R^{n}$

δ (F (x)) = \sum_{i} δ (x - x_{i}) / {| \frac{\partial (F_{1}, F_{2}, \dots, F_{n})}{\partial (x_{1}, x_{2}, \dots, x_{n})} |}_{x = x_{i}}, with F (x_{i}) = 0, det (\frac{\partial F_{i}}{\partial x_{j}}) \neq 0

$⟹$ for the orthogonal curvilinear coordinates

δ^{3} (x - x^{'}) = \frac{1}{H} δ^{3} (u - u^{'})

1.2.3 Cases

the point dipole. Let the point dipole be at the origin, and its charge density distribution is:

ρ (x) = - p \cdot \nabla δ^{3} (x)

Because the total charge is zero,

Q = \int_{R^{3}} ρ (x) d^{3} x = \int_{R^{3}} [- p \cdot \nabla δ^{3} (x)] d^{3} x = - \nabla (- p \cdot \int_{R^{3}} δ^{3} (x) d^{3} x) = 0

and the first-order moment (dipole moment) is $p$ ,

\begin{aligned} d & = \int x ρ (x) d^{3} x = \int_{R^{3}} x [- p \cdot \nabla δ^{3} (x)] d^{3} x \\ = \int_{R^{3}} x [- p_{i} \frac{\partial δ^{3}}{\partial x_{i}}] d^{3} x = - p_{i} \int_{R^{3}} x \frac{\partial δ^{3}}{\partial x_{i}} d^{3} x \\ = p_{i} {\frac{\partial x}{\partial x_{i}} |}_{x = 0} = p_{i} {\hat{e}}_{i} = p \end{aligned}

2 Fundamentals of Electromagnetism

2.1 Maxwell's equation

In the cosmos

{\begin{aligned} \nabla \cdot E & = \frac{ρ}{ϵ_{0}} & (Gauss’s Law) \\ \nabla \cdot B & = 0 & (Gauss’s Law for Magnetism) \\ \nabla \times E & = - \frac{\partial B}{\partial t} & (Faraday Law) \\ \nabla \times B & = μ_{0} J + \frac{1}{c^{2}} \frac{\partial E}{\partial t} & (Ampere-Maxwell Law) \end{aligned}

with $c = \frac{1}{\sqrt{μ_{0} ϵ_{0}}}$ .

In the medium

{\begin{aligned} \nabla \cdot D & = ρ_{f} & (Gauss’s Law) \\ \nabla \cdot B & = 0 & (Gauss’s Law for Magnetism) \\ \nabla \times E & = - \frac{\partial B}{\partial t} & (Faraday’s Law) \\ \nabla \times H & = J_{f} + \frac{\partial D}{\partial t} & (Ampere-Maxwell Law) \end{aligned}

\nabla \cdot E = \frac{ρ_{f} + ρ_{p}}{ϵ_{0}}, ρ_{p} = - \nabla \cdot P

\nabla \times B = μ_{0} (J_{f} + J_{M} + J_{P}) + \frac{1}{c^{2}} \frac{\partial E}{\partial t} = μ_{0} (J_{f} + \nabla \times M + \frac{\partial P}{\partial t}) + \frac{1}{c^{2}} \frac{\partial E}{\partial t}

Lorentz force

\frac{d p}{d t} = q E + q v \times B

Ohm law

j = σ (\underset{constitutive term}{\underset{⏟}{E}} + \underset{convective term}{\underset{⏟}{u \times B}} - \underset{hall term}{\underset{⏟}{\frac{1}{n e} j \times B}} + \underset{electron pressure term}{\underset{⏟}{\frac{\nabla p_{e}}{n e}}} - \underset{inertial term}{\underset{⏟}{\frac{m_{e}}{n e^{2}} \frac{\partial j}{\partial t}}} + \underset{more and more}{\underset{⏟}{\dots}})

In most cases of electrodynamics, it suffices to retain the first term $j = σ E$ . Don't be curious about exploring how many more terms are left behind.

2.2 Polarization and magnetization

Polarization and magnetization intensity

D = ϵ_{0} E + P, H = \frac{1}{μ_{0}} B - M

for linear isotropic media,

D = ϵ E, H = \frac{1}{μ} B

Charge / Current

{\begin{aligned} J_{P} & = \frac{\partial P}{\partial t} & (Polarizing Current) \\ J_{M} & = \nabla \times M, K_{M} = \hat{n} \times (M_{1} - M_{2}) & (Magnetizing Current) \\ ρ_{p} & = - \nabla \cdot P, σ_{p} = \hat{n} \cdot (P_{1} - P_{2}) & (Polarizing Charge) \end{aligned}

2.3 Boundary conditions

In cosmos ( $\hat{n} = {\hat{n}}_{1 \to 2}$ )

\hat{n} \cdot ϵ_{0} (E_{2} - E_{1}) = σ, \hat{n} \cdot (B_{2} - B_{1}) = 0, \hat{n} \times (E_{2} - E_{1}) = 0, \hat{n} \times \frac{1}{μ_{0}} (B_{2} - B_{1}) = K

or simply

ϵ_{0} (E_{2} - E_{1}) = σ \hat{n}, \frac{1}{μ_{0}} (B_{2} - B_{1}) = K \times \hat{n}

In media

\hat{n} \cdot (D_{2} - D_{1}) = σ_{f}, \hat{n} \cdot (B_{2} - B_{1}) = 0, \hat{n} \times (E_{2} - E_{1}) = 0, \hat{n} \times (H_{2} - H_{1}) = K_{f}

\hat{n} \cdot ϵ_{0} (E_{2} - E_{1}) = σ_{f} + σ_{p}, \hat{n} \cdot (B_{2} - B_{1}) = 0, \hat{n} \times (E_{2} - E_{1}) = 0, \frac{1}{μ_{0}} \hat{n} \times (B_{2} - B_{1}) = K_{f} + K_{M}

with $J_{P} = \frac{\partial P}{\partial t} = 0$ .

2.4 Electromagnetic potential

Definition

E = - \nabla φ - \frac{\partial A}{\partial t}, B = \nabla \times A

Coulomb gauge

C ≜ \nabla \cdot A = 0

Lorentz gauge

L ≜ \nabla \cdot A + \frac{1}{c^{2}} \frac{\partial φ}{\partial t} = 0

Gauge invariance

φ^{'} = φ - \frac{\partial ψ}{\partial t}, A^{'} = A + \nabla ψ

Electromagnetic potential equation

\begin{aligned} ◻ φ + \partial_{t} L & = - \frac{ρ}{ϵ_{0}} \\ ◻ A - \nabla L & = - μ_{0} J \end{aligned}

with $◻ ≜ \nabla^{2} - \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}} (d’Alembert operator)$ .

2.5 EM Wave

2.5.1 Wave equation

Wave equation for time-varying field

\begin{aligned} ◻ E = (\nabla^{2} - \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}}) E & = \frac{1}{ϵ_{0}} \nabla ρ + μ_{0} \frac{\partial J}{\partial t} \\ ◻ B = (\nabla^{2} - \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}}) B & = - μ_{0} (\nabla \times J) \end{aligned}

Definition: phase of monochromatic wave

ϕ = k \cdot x - ω t

Dispersion relation

ω = k c

⟹ v_{p} = \frac{ω}{k} = c, v_{g} = \frac{d ω}{d k} = c

2.5.2 Polarization

Real description

For the monochromatic planar electromagnetic wave in a vacuum,

E_{1} = A_{1} \cos ϕ, E_{2} = A_{2} \cos (ϕ + δ)

⟹ {(\frac{E_{1}}{A_{1}})}^{2} + {(\frac{E_{2}}{A_{2}})}^{2} - 2 \frac{E_{1} E_{2}}{A_{1} A_{2}} \cos δ = \sin^{2} δ

for certain $k \cdot x$ ,

E_{1} = A_{1} \cos ω t, E_{2} = A_{2} \cos (ω t - δ)

⟹ {\begin{cases} \sin δ > 0, Right-hand \\ \sin δ < 0, Left-hand \end{cases}

Complex description

{\tilde{E}}_{1} = A_{1} e^{i ϕ}, {\tilde{E}}_{2} = A_{2} e^{i (ϕ + δ)}

for certain $k \cdot x$ ,

{\tilde{E}}_{1} = A_{1} e^{i (- ω t)}, {\tilde{E}}_{2} = A_{2} e^{i (- ω t + δ)}

define the degree of polarization

\tilde{R} = \frac{{\tilde{E}}_{2}}{{\tilde{E}}_{1}} = \frac{A_{2}}{A_{1}} e^{i δ}

⟹ {\begin{cases} Im \tilde{R} > 0, Right-hand \\ Im \tilde{R} < 0, Left-hand \end{cases}

Note: $k \to - k$ indicates that the wave vector undergoes inversion. According to the principle that the thumb points in the direction of the wave vector, the polarization also experiences inversion at this time. If $k \to - k$ and $ω t \to - ω t$ , the polarization handedness remains unchanged.

Circularly polarized basis vectors*

{\begin{cases} {\hat{e}}_{+} ≜ \frac{{\hat{e}}_{1} + i {\hat{e}}_{2}}{\sqrt{2}}, Right-hand \\ {\hat{e}}_{-} ≜ \frac{{\hat{e}}_{1} - i {\hat{e}}_{2}}{\sqrt{2}}, Left-hand \end{cases}

satisfy

{\hat{e}}_{-} = {\hat{e}}_{+}^{*}, {\hat{e}}_{\pm}^{*} \cdot {\hat{e}}_{\pm} = 1, {\hat{e}}_{\pm} \cdot {\hat{e}}_{\pm} = 0

Monochromatic waves can be transformed from 2D Cartesian coordinates to circularly polarized coordinates,

E = E_{0} e^{i ϕ} = (E_{+} {\hat{e}}_{+} + E_{-} {\hat{e}}_{-}) e^{i ϕ}, E_{\pm} = {\hat{e}}_{\pm}^{*} \cdot E_{0}

2.5.3 Complex description

for ${\tilde{E}}_{1} = A_{1} e^{i ϕ}, {\tilde{E}}_{2} = A_{2} e^{i (ϕ + δ)}$ , if $A_{1, 2}$ is independent with $(t, x)$ ,

\nabla ⟶ i k, \nabla \cdot ⟶ i k \cdot, \nabla \times ⟶ i k \times, \frac{\partial}{\partial t} ⟶ - i ω

Long - term average: for $\tilde{f} = {\tilde{f}}_{0} e^{- i ω t}, \tilde{g} = {\tilde{g}}_{0} e^{- i ω t}$ ,

⟨ ℜ [\tilde{f}] \cdot ℜ [\tilde{g}] ⟩ = \frac{1}{2} ℜ [{\tilde{f}}_{0} {\tilde{g}}_{0}^{*}]

In the vector scenario,

⟨ ℜ [\tilde{f}] \cdot ℜ [\tilde{g}] ⟩ = \frac{1}{2} ℜ [\tilde{f} \cdot {\tilde{g}}^{*}], ⟨ ℜ [\tilde{f}] \times ℜ [\tilde{g}] ⟩ = \frac{1}{2} ℜ [\tilde{f} \times {\tilde{g}}^{*}]

2.6 Conservation / Continuity equation

Charge conservation

\nabla \cdot j + \frac{\partial ρ}{\partial t} = 0

Energy conservation

- \frac{\partial w}{\partial t} = \nabla \cdot S + f \cdot v

with EMF (electromagnetic field) energy, Poynting vector and Lorentz force

w = \frac{1}{2} ϵ_{0} E^{2} + \frac{B^{2}}{2 μ_{0}}, S = \frac{1}{μ_{0}} E \times B, f = ρ_{e} E + J \times B

Momentum conservation & Maxwell stress tensor

- \frac{\partial g}{\partial t} = \nabla \cdot T + f

with EMF momentum and Maxwell stress tensor $T_{M}$ :

g = ϵ_{0} E \times B = S / c^{2}, T_{M} = - T = - w I + (ϵ_{0} E E + \frac{B B}{μ_{0}})

In a steady state where the electromagnetic momentum is constant over time, the total force acting on the particles within volume $V$ can be expressed as

F = ∭_{V} f d V = - \oint_{\partial V} d σ \cdot T = \oint_{\partial V} d σ \cdot T_{M}

Angular momentum conservation

- \frac{\partial}{\partial t} (r \times g) = \underset{= r \times (\nabla \cdot T)}{\underset{⏟}{\nabla \cdot (r \times T)}} + r \times f

Conservation / Continuity equation in the linear homogeneous media

Charge conservation

\nabla \cdot j_{f} + \frac{\partial ρ_{f}}{\partial t} = 0, \nabla \cdot (j_{P} + j_{M}) + \frac{\partial ρ_{p}}{\partial t} = 0

Energy conservation (Poynting's Theorem in Media)

\frac{\partial w}{\partial t} = \nabla \cdot S + j_{f} \cdot E

with EMF energy and Poynting vector and Lorentz force

w = \frac{1}{2} (E \cdot D + B \cdot H), S = E \times H

Momentum conservation

\frac{\partial g}{\partial t} = \nabla \cdot T + f_{f}

with EMF momentum (Minkowski form), Maxwell stress tensor and Lorentz force acting on free particles

g = D \times B, T_{M} = - T = - w I + (E D + H B), f_{f} = ρ_{f} E + J_{f} \times B

In a steady state where the electromagnetic momentum is constant over time, the total force on free particles within volume V is

F_{f} = ∭_{V} f_{f} d V = - \oint_{\partial V} d σ \cdot T

Angular momentum conservation

\frac{\partial}{\partial t} (r \times g) = \nabla \cdot (r \times T) + r \times f_{f}

For inhomogeneous media, $f_{f}$ above becomes,

f_{t o t a l} = f_{f} - \frac{1}{2} E^{2} \nabla ϵ - \frac{1}{2} H^{2} \nabla μ

3 Special Relativity & Tensor analysis (Minkowski spacetime)

3.1 Fundamental definition

(g)_{α β} = (g)^{α β} = d i a g (- 1, + 1, + 1, + 1)_{α β}

x^{α} = (c t, x)

Normalization of metric

g_{α β} g^{β γ} = δ_{α}^{γ}

Metric invariance condition

g_{α β} Λ_{μ}^{α} Λ_{ν}^{β} = g_{μ ν}

⟹ g^{μ ν} Λ_{μ}^{α} Λ_{ν}^{β} = g^{α β}, g_{α β} Λ_{μ}^{α} Λ_{ν}^{β} = g_{μ ν}, g^{μ ν} Λ_{μ}^{α} Λ_{ν}^{β} = g^{α β}

3.2 Lorentz transformation

Invariant interval

d s^{2} = g_{α β} d x^{α} d x^{β} = d x_{α} d x^{β} = - c^{2} d t^{2} + | d x |^{2} = - c^{2} d τ^{2} {\begin{cases} < 0, & timelike \\ = 0, & lightlike \\ < 0, & spacelike \end{cases}

Christoffel symbols (the first kind)*

Γ_{ρ μ ν} = \frac{1}{2} (\frac{\partial g_{ρ μ}}{\partial x^{ν}} + \frac{\partial g_{ρ ν}}{\partial x^{μ}} - \frac{\partial g_{μ ν}}{\partial x^{ρ}}) = g_{α β} \frac{\partial^{2} x^{α}}{\partial x^{' ν} \partial x^{' μ}} \frac{\partial x^{β}}{\partial x^{' ρ}} = 0

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

Where $K$ is the laboratory frame and $K^{'}$ is the moving frame with speed $v = v {\hat{e}}_{x}$ .

The inverse of the matrix $Λ_{β}^{α}$ is $Λ_{β}^{α}$ :

(Λ^{- 1})_{α}^{μ} = g_{α β} Λ_{ν}^{β} g^{ν μ} = Λ_{α}^{μ}

Λ_{β}^{α} Λ_{α}^{γ} = δ_{β}^{γ} or Λ_{γ}^{α} Λ_{β}^{γ} = δ_{β}^{α}

for $x$ -axis boost:

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

for general $β$ boost:

Λ_{β}^{α} = (\begin{matrix} γ & γ β \\ γ β & I + (γ - 1) \hat{β} \hat{β} \end{matrix}), β = \frac{v}{c}, \hat{β} = \frac{v}{| v |}

3.2.1 Velocity transformation

Suppose frame $K^{'}$ moves with velocity $v = v {\hat{e}}_{x}$ relative to frame $K$ along the positive x-axis. An object has velocity $u = (u_{x}, u_{y}, u_{z})$ in frame $K$ and $u^{'} = (u_{x}^{'}, u_{y}^{'}, u_{z}^{'})$ in frame $K^{'}$ , then the velocity $u$ transforms as:

u_{x} = \frac{u_{x}^{'} + v}{1 + \frac{v u_{x}^{'}}{c^{2}}}, u_{y} = \frac{u_{y}^{'}}{γ (1 + \frac{v u_{x}^{'}}{c^{2}})}, u_{z} = \frac{u_{z}^{'}}{γ (1 + \frac{v u_{x}^{'}}{c^{2}})}

If the boost is in an arbitrary direction $β = v / c$ *

u = \frac{1}{1 + \frac{v \cdot u^{'}}{c^{2}}} [u^{'} + \frac{1}{γ} u_{⊥} + v]

Where $u_{⊥}$ represents the component of $u^{'}$ perpendicular to the boost direction. Assuming $θ = ⟨ u, v ⟩, θ^{'} = ⟨ u^{'}, v ⟩$ , *

\tan θ = \frac{u^{'} \sin θ^{'}}{γ_{v} (u^{'} \cos θ^{'} + v)} \overset{if u^{'} = c}{⟹} \tan θ = \frac{\sin θ^{'}}{γ_{v} (\cos θ^{'} + β_{v})} (Aberration of Light Formula)

3.2.2 The Proper Orthochronous Lorentz Group*

The Full Lorentz Group $O (1, 3)$ consists of all transformations that preserve the Minkowski metric. From the property $Λ^{T} η Λ = η$ above, it follows that $| det (Λ) | = 1, | Λ_{0}^{0} | \geq 1$ . $O (1, 3)$ is a Lie group that possesses four disjoint, connected components. It can be expressed as the union of the Proper Orthochronous Lorentz Group $S O (1, 3)^{↑}$ (the identity component) and its cosets:

P = d i a g (1, - 1, - 1, - 1), T = d i a g (- 1, 1, 1, 1)

O (1, 3) = {S O (1, 3)^{↑}, P S O (1, 3)^{↑}, T S O (1, 3)^{↑}, P T S O (1, 3)^{↑}}

Infinitesimal Lorentz transformation

Λ_{ν}^{μ} = δ_{ν}^{μ} + ω_{ν}^{μ}, with ω_{ν}^{μ} = (\begin{matrix} 0 & β_{x} & β_{y} & β_{z} \\ β_{x} & 0 & θ_{z} & - θ_{y} \\ β_{y} & - θ_{z} & 0 & θ_{x} \\ β_{z} & θ_{y} & - θ_{x} & 0 \end{matrix})

K^{'} \to K : x^{α} = Λ_{β}^{α} x^{' β} = [\begin{matrix} c t^{'} + β \cdot x^{'} \\ x^{'} + β c t^{'} - θ \times x^{'} \end{matrix}]

For infinite transformation $Ω = lim_{n \to \infty} n ω$ ,

Λ = l i m_{n \to \infty} {(1 + \frac{Ω}{n})}^{n} = e^{Ω} = \sum_{k = 0}^{\infty} \frac{Ω^{k}}{k!} = {\begin{cases} [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & \cos θ & \sin θ & 0 \\ 0 & - \sin θ & \cos θ & 0 \\ 0 & 0 & 0 & 1 \end{array}], & z-axis rotation \\ [\begin{array}{c} ch ϕ & sh ϕ \\ sh ϕ & ch ϕ \\ 1 \\ 1 \end{array}], & x-axis boost, with \tanh ϕ = β \end{cases}

3.3 Tensor analysis (Minkowski spacetime)

3.3.1 4-description of vector / tensor

Primary 4-vector / tensor

4-velocity / accelerated velocity

Let $a_{i n}$ to be the accelerated speed in the instantaneous co-moving frame and $a$ in the lab frame, one finds

u^{α} = \frac{d x^{α}}{d τ} = γ (c, v), a^{α} = \frac{d u^{α}}{d τ} = (γ^{4} (β \cdot a), γ^{2} a + γ^{4} (β \cdot a) β)

u^{α} u_{α} = - c^{2}, a^{μ} u_{μ} = 0

a^{μ} a_{μ} = | a_{i n} |^{2} = γ^{4} [| a |^{2} + γ^{2} (β \cdot a)^{2}] ⟹ | a_{i n} | = {\begin{cases} γ^{2} a, β ⊥ a \\ γ^{3} a, β ∥ a \end{cases}

4-momentum

p^{α} = m u^{α} = (γ m c, γ m v) = (\frac{E}{c}, p)

p^{α} p_{α} = - m^{2} c^{2} ⟹ E^{2} = (| p | c)^{2} + (m c^{2})^{2}

4-force

F^{α} = \frac{d p^{α}}{d τ} = m a^{α} = (γ^{4} m (β \cdot a), γ^{2} m a + γ^{4} m (β \cdot a) β)

Since $F = \frac{d p}{d t}, F \cdot β = \frac{d E / c}{d t} ⟶ F^{α} = γ (F \cdot β, F)$ ,

⟹ F = γ m a + γ^{3} m (β \cdot a) β ⟹ F = {\begin{cases} γ m a, β ⊥ a \\ γ^{3} m a, β ∥ a \end{cases}

4-operator

\partial_{μ} = \frac{\partial}{\partial x^{μ}} = (\frac{1}{c} \frac{\partial}{\partial t}, \nabla), \partial_{μ} \partial^{μ} = ◻ = - \frac{1}{c^{2}} \frac{\partial}{\partial t^{2}} + \nabla^{2}

Three fundamental invariant tensor in the Minkowski space

g^{α β}, δ_{β}^{α}, ϵ^{α β μ ν}

The recursion formula of decomposition

T^{(α_{1} α_{2} \dots α_{s})} = \frac{1}{s} \sum_{k = 1}^{s} T^{α_{k} (α_{1} α_{2} \dots α_{k - 1} α_{k + 1} \dots α_{s})}

T^{[α_{1} α_{2} \dots α_{s}]} = \frac{1}{s} \sum_{k = 1}^{s} (- 1)^{k - 1} T^{α_{k} [α_{1} α_{2} \dots α_{k - 1} α_{k + 1} \dots, α_{s}]}

with base case is $T^{(α)} = T^{[α]} = T^{α}$ . for ex.,

T^{(α β)} = \frac{T^{α β} + T^{β α}}{2}, T^{[α β]} = \frac{T^{α β} - T^{β α}}{2}

T^{[α β]} ≜ (\begin{matrix} 0 & p \\ - p & a \end{matrix}) ≜ {p, a}, with a_{i j} = ϵ_{i j k} a_{k} (or a = - I \times a)

\begin{aligned} T^{(α β γ)} & = \frac{1}{3} [T^{α (β γ)} + T^{β (α γ)} + T^{γ (α β)}] \\ = \frac{1}{6} [T^{α β γ} + T^{α γ β} + T^{β α γ} + T^{β γ α} + T^{γ α β} + T^{γ β α}] \end{aligned}

\begin{aligned} T^{[α β γ]} & = \frac{1}{3} [(- 1)^{0} T^{α [β γ]} + (- 1)^{1} T^{β [α γ]} + (- 1)^{2} T^{γ [α β]}] \\ = \frac{1}{6} [T^{α β γ} - T^{α γ β} - T^{β α γ} + T^{β γ α} + T^{γ α β} - T^{γ β α}] \end{aligned}

Dual tensor of asymmetric matrix

Dual tensor of asymmetric matrix $A^{α β} ≜ (\begin{matrix} 0 & p \\ - p & a \end{matrix}) ≜ {p, a}$ is

A^{α β} ≜ \frac{1}{2} ϵ^{α β μ ν} A_{μ ν}, with ϵ^{α β μ ν} = {\begin{cases} + 1, & (α β μ ν) is an even permutation of 0123 \\ - 1, & (α β μ ν) is an odd permutation of 0123 \\ 0, & (α β μ ν) is in other case \end{cases}

A^{α β} ⟹ A^{α β} = (\begin{matrix} 0 & a \\ - a & - p \end{matrix}) ≜ {a, - p}

Eigenvalue equation of second-order tensors

A^{α β} X_{β} = λ X^{α} ⟹ (A^{α β} - λ g^{α β}) X_{β} = 0

the Principal Invariants

⟹ \sum_{α} λ_{α}, \sum_{α < β} λ_{α} λ_{β}, \sum_{α < β < μ} λ_{α} λ_{β} λ_{μ}, \sum_{α < β < μ < ν} λ_{α} λ_{β} λ_{μ} λ_{ν}

for an asymmetric matrix $A^{α β} ≜ (\begin{matrix} 0 & p \\ - p & a \end{matrix}) ≜ {p, a}$ ,

I_{1} = a^{2} - p^{2} = - \frac{1}{2} A^{α β} A_{α β}, I_{2} = p \cdot a = - \frac{1}{4} A^{α β} A_{α β}

4 Lagrangian Formulation of the EM Field

4.1 Covariant EM equation

4.1.1 Maxwell's equation

Electromagnetic Field Tensor

A^{μ} = (\frac{φ}{c}, A) ⟹ F^{μ ν} = \partial^{μ} A^{ν} - \partial^{ν} A^{μ} = [\begin{matrix} 0 & E_{x} / c & E_{y} / c & E_{z} / c \\ - E_{x} / c & 0 & B_{z} & - B_{y} \\ - E_{y} / c & - B_{z} & 0 & B_{x} \\ - E_{z} / c & B_{y} & - B_{x} & 0 \end{matrix}]

or simply

F^{μ ν} = (\begin{matrix} 0 & \frac{E}{c} \\ - \frac{E}{c} & B \end{matrix}) = {\frac{E}{c}, B}, B_{i j} = ϵ_{i j k} B_{k}

and two invariants

I_{1} = \frac{E^{2}}{c^{2}} - B^{2} = - \frac{1}{2} F^{α β} F_{α β}, I_{2} = \frac{E \cdot B}{c} = - \frac{1}{4} F^{α β} F_{α β}

4-form Maxwell's equation

{\begin{aligned} \begin{aligned} \nabla \cdot E & = \frac{ρ}{ϵ_{0}} \\ \nabla \times B - \frac{1}{c^{2}} \frac{\partial E}{\partial t} & = μ_{0} J \end{aligned}} & ⟺ & \partial_{α} F^{α β} = - μ_{0} J^{β} \\ \begin{aligned} \nabla \cdot B & = 0 \\ \nabla \times E + \frac{\partial B}{\partial t} & = 0 \end{aligned}} & ⟺ & \partial_{[μ} F_{ν ρ]} = 0 ⟺ \partial_{α} F^{α β} = 0 \end{aligned}

Differential 2-form of electromagnetic field tensor $F$ *

F = \frac{1}{2} F_{μ ν} d x^{μ} \land d x^{ν} ⟹ d F = 0, d ⋆ F = J

Gauge invariance

\exists A^{α} = \partial^{α} ψ, if curl (A) = \partial^{α} A^{β} - \partial^{β} A^{α} = 0

Thus, the field strength tensor $F^{μ ν} = \partial^{μ} A^{ν} - \partial^{ν} A^{μ}$ remains invariant under the gauge transformation $A^{' α} = A^{α} + \partial^{α} ψ$

A^{' α} = A^{α} + \partial^{α} ψ

4-form Maxwell's equation under the Lorenz gauge $\partial_{α} A^{α} = 0$ ,

\partial_{α} \partial^{α} A^{β} = - μ_{0} J^{β} (covariant elctromagnetic potential equation)

⟹ ◻ φ = - \frac{ρ}{ϵ_{0}}, ◻ A = - μ_{0} J

4.1.2 Lorentz transformation of EM field

from Lorentz transformation of $F^{α β}$ :

K^{'} \to K : F^{α β} = Λ_{μ}^{α} Λ_{ν}^{β} F^{' μ ν}, with Λ_{β}^{α} = [\begin{matrix} γ & γ β \\ γ β & γ \\ 1 \\ 1 \end{matrix}]

one finds

{\begin{cases} E_{∥}^{'} = E_{∥}, & E_{⊥}^{'} = γ (E_{⊥} + β \times c B) \\ B_{∥}^{'} = B_{∥}, & c B_{⊥}^{'} = γ (c B_{⊥} - β \times E) \end{cases}

or more neatly (define $B_{⊥} ≜ \hat{β} \times c B, B_{∥} ≜ c B_{∥}$ ),

{\begin{cases} E_{∥}^{'} = E_{∥}, & E_{⊥}^{'} = γ (E_{⊥} + β B_{⊥}) \\ B_{∥}^{'} = B_{∥}, & B_{⊥}^{'} = γ (B_{⊥} + β E_{⊥}) \end{cases}

Electric-like / Magnetic-like / Light-like field

E^{2} - c^{2} B^{2} is ivariant ⟹ {\begin{cases} E > c B, & (Electric-like) \\ E = c B, & (Light-like) \\ E < c B, & (Magnetic-like) \end{cases}

In the lab frame $K$ , if

$E \cdot B = 0$ and $E > c B$ , $⟹$ $B^{'} = 0$ in the inertial frame with speed $β = \frac{E \times c B}{E^{2}}$ . The cycloidal motion will transform into a hyperbolic motion.
$E \cdot B = 0$ and $E < c B$ , $⟹$ $E^{'} = 0$ in the inertial frame with speed $β = \frac{E \times c B}{(c B)^{2}}$ , which is precisely the electric field drift velocity.
$E \cdot B = 0$ and $E = c B$ , $⟹$ $E^{'} = \infty \cdot 0, B^{'} = \infty \cdot 0$ in the photon frame with speed $β = 1$ (well, at present, this cannot be achieved).
$E \cdot B \neq 0$ , $⟹$ $E^{'} \cdot B^{'} = 0$ in the inertial frame with speed satisfy $\frac{β}{1 + β^{2}} = \frac{E \times c B}{E^{2} + c^{2} B^{2}}$ .

4.1.3 4-wave vector

4-form Wave equation for the field strength tensor can be derived from 4-form Maxwell's equation,

\begin{array}{ll} \partial_{α} F^{α β} = - μ_{0} J^{β} \\ \partial_{[μ} F_{ν ρ]} = 0 \to \partial^{α} (\partial_{[α} F_{μ ν]}) = 0 \end{array}} ⟹ \partial_{α} \partial^{α} F^{μ ν} = - μ_{0} (\partial^{μ} J^{ν} - \partial^{ν} J^{μ})

Thus, the monochromatic plane wave satisfies $F^{μ ν} = F_{0}^{μ ν} e^{i (k \cdot x - ω t)} ≜ F_{0}^{μ ν} e^{i k^{α} x_{α}}$ with the definition of 4-wave vector,

k^{α} ≜ (\frac{ω}{c}, k)

k^{α} k_{α} = \frac{ω^{2}}{c^{2}} - | k |^{2} is invariant ⟹ k c \equiv ω in any inertial frame

Lorentz transformation of $k^{α}$ . assuming $θ = ⟨ k, β ⟩, θ^{'} = ⟨ k^{'}, β ⟩$ ,

K \to K^{'} : k^{' α} = Λ_{β}^{α} k^{β} ⟺ \begin{array}{l} {\begin{cases} \frac{ω^{'}}{c} = γ (\frac{ω}{c} - β \cdot k) & ⟺ & ω^{'} = γ ω (1 - β \cos θ) & (doppler frequency shift) \\ \begin{array}{ll} k_{∥}^{'} = γ (k_{∥} - β \frac{ω}{c}) \\ k_{⊥}^{'} = k_{⊥} \end{array}} & ⟺ & \tan θ^{'} = \frac{\sin θ}{γ (\cos θ - β)} & (Aberration of Light Formula) \end{cases} \end{array}

The relativistic Doppler formula demonstrates that whether the observer approaches the source or the source approaches the observer, as long as their relative velocity $β$ and the rest frequency $ω_{0}$ of the source are identical, the observed frequency remains the same. This differs from the classical case; the observed frequency is always,

ω_{o b s} = ω_{0} \sqrt{\frac{1 + β}{1 - β}}

4.1.4 Conservation / Continuity equation

4-current & charge conservation

Define the 4-current,

J^{α} = ρ_{e 0} u^{α} = (ρ_{e} c, j), with ρ_{e} = γ ρ_{e 0}, j = ρ_{e} v = γ ρ_{e 0} v

Note: Since the total charge $Q$ is a Lorentz invariant but the volume $V$ undergoes length contraction ( $V = V_{0} / γ$ ), the charge density must increase by the same factor, $ρ_{e} = Q / V = γ ρ_{e 0}$ .

Furthermore, while the time interval dilates ( $d t = γ d τ$ ) and the spatial volume element contracts ( $d^{3} x = d^{3} x_{0} / γ$ ), these two $γ$ factors exactly cancel each other out. This ensures that the four-dimensional spacetime volume $d^{4} x = c d t d^{3} x$ is a Lorentz invariant (scalar).

The current's 4-divergence is zero,

\partial_{α} J_{e}^{α} = \nabla \cdot j_{e} + \frac{\partial ρ_{e}}{\partial t} = 0

A 4-vector $J^{α}$ is referred to as a conserved current if $\partial_{α} J^{α} = 0$ . The corresponding conserved charge, defined as $Q = \frac{1}{c} ∭_{R^{3}} J^{0} d^{3} x$ , is a Lorentz invariant.

Mass conservation

J_{m}^{α} = ρ_{m 0} u^{α} = (ρ_{m} c, j_{m}), with ρ_{m} = γ ρ_{m 0}, j_{m} = ρ_{m} v = γ ρ_{m 0} v

\partial_{α} j_{m}^{α} = \nabla \cdot j_{m} + \frac{\partial ρ_{m}}{\partial t} = 0

Energy-momentum tensor conservation

4-Lorentz force

f^{α} = F^{α β} J_{β} = (f^{0}, f) = (\frac{1}{c} E \cdot J, ρ_{e} E + J \times B)

for single particle with charge $+ e$ , $J_{β} = e u_{β} = γ e (c, v), ρ_{e} = γ e$ . thus,

f^{α} = e F^{α β} u_{β} = γ (F \cdot β, F) = γ (e E \cdot β, e E + e v \times B)

Energy-momentum tensor of EM field

T_{f}^{μ ν} ≜ - \frac{1}{4 μ_{0}} g^{μ ν} F^{α β} F_{α β} - \frac{1}{μ_{0}} F^{μ α} F_{α}^{ν} = (\begin{matrix} w & \frac{S}{c} \\ \frac{S}{c} & T \end{matrix})

Energy-momentum tensor of particles

T_{p}^{μ ν} ≜ ρ_{m 0} u^{μ} u^{ν} = γ ρ_{m} (\begin{matrix} c^{2} & c v \\ c v & v v \end{matrix}) ⟹ f^{ν} = ρ_{m} \frac{d u^{ν}}{d τ} = \partial_{μ} T_{p}^{μ ν}

Energy-momentum conservation

\partial_{μ} T_{f}^{μ ν} = - f^{ν} = - \partial_{μ} T_{p}^{μ ν}

\partial_{μ} T^{μ ν} = 0 with T^{μ ν} = T_{f}^{μ ν} + T_{p}^{μ ν}

relevant conserved charges are

P^{ν} = \frac{1}{c} ∭_{R^{3}} T^{0 ν} d^{3} x = (\frac{E}{c}, P) = (∭ \frac{w + γ ρ_{m} c^{2}}{c}, ∭ c g + γ ρ_{m} v)

Angular energy-momentum conservation*

4-angular momentum conservation

M^{μ α β} ≜ x^{α} T^{μ β} - x^{β} T^{μ α} ⟹ \partial_{μ} M^{μ α β} = 0

4-moment of force

τ^{α β} ≜ x^{α} f^{β} - x^{β} f^{α} ⟹ \partial_{μ} M_{f}^{μ α β} = - τ^{α β} = - \partial_{μ} M_{p}^{μ α β}

relevant conserved charges are

L^{i j} = \frac{1}{c} ∭_{R^{3}} M^{0 i j} d^{3} x ⟹ L = ∭ [r \times (g + γ ρ_{m} v)] d^{3} x

K^{i} = L^{0 i} = \frac{1}{c} ∭_{R^{3}} M^{00 i} d^{3} x = c t ∭ \underset{total momentum}{\underset{⏟}{(g + γ ρ_{m} v)}} d^{3} x - \frac{1}{c} ∭ \underset{total energy}{\underset{⏟}{[w + γ ρ_{m} c^{2}]}} x d^{3} x

The second term in the right side of $K^{i}$ is the velocity of total energy, which is proportional to $t$ , namely energy center moving in a straight line.

4.1.5 Maxwell's equation in the media*

#todo

4.2 Lagrangian Formulation

4.2.1 Lagrange equation

Lagrange equation for Particles

For the action $S = \int_{t_{1}}^{t_{2}} L (q_{α} (t), {\dot{q}}_{α} (t), t) d t$ , provided that the boundaries $t_{1}$ , $t_{2}$ are fixed and $δ q_{α} = 0$ at the boundaries, then the principle of least action $δ S = 0$ yields the Lagrange equation of particles as follows:

\frac{δ L}{δ q_{α}} ≜ \frac{\partial L}{\partial q_{α}} - \frac{d}{d t} \frac{\partial L}{\partial {\dot{q}}_{α}} = 0

Gauge transformation

Δ L = \tilde{L} - L = \frac{d f (q_{α}, t)}{d t}

Covariant Lagrange equation for Relativistic Particle

For the action $S = \int_{τ_{1}}^{τ_{2}} L (x^{μ} (τ), {\dot{x}}^{μ} (τ)) d τ$ , provided that the endpoints $τ_{1}, τ_{2}$ are fixed and $δ x^{μ} = 0$ at the boundaries, then

\frac{δ L}{δ x^{μ}} ≜ \frac{\partial L}{\partial x^{μ}} - \frac{d}{d τ} \frac{\partial L}{\partial {\dot{x}}^{μ}} = 0

Gauge transformation is $Δ L = \frac{d f (x^{μ}, τ)}{d τ}$ .

Lagrange equation for Scalar Field

For the action $S = \int_{Ω} L (φ (x^{μ}), \partial_{μ} φ (x^{μ}), x^{μ}) d^{4} x$ , provided that the boundary of the spacetime region $Ω$ is fixed and $δ φ = 0$ on the boundary, then

\frac{δ L}{δ φ} ≜ \frac{\partial L}{\partial φ} - \partial_{μ} (\frac{\partial L}{\partial (\partial_{μ} φ)}) = 0

Gauge transformation is $Δ L = \partial_{α} C^{α} (φ, x^{μ})$ .

Lagrange equation for Vector Field

For the action $S = \int_{Ω} L (A^{α} (x^{μ}), \partial_{μ} A^{α} (x^{μ}), x^{μ}) d^{4} x$ , provided that the boundary of the spacetime region $Ω$ is fixed and $δ A^{α} = 0$ on the boundary, then

\frac{δ L}{δ A^{α}} ≜ \frac{\partial L}{\partial A^{α}} - \partial_{μ} (\frac{\partial L}{\partial (\partial_{μ} A^{α})}) = 0

Gauge transformation is $Δ L = \partial_{α} C^{α} (A^{μ}, x^{μ})$ .

Generalized Lagrange Equation for Vector Field

For the action $S = \int_{Ω} L (A^{α}, \partial_{μ_{1}} A^{α}, \partial_{μ_{1} μ_{2}} A^{α}, \dots, \partial_{μ_{1} μ_{2} \dots μ_{s}} A^{α}; x^{μ}) d^{4} x$ , provided that the boundary of the spacetime region $Ω$ is fixed, and the field variation along with its derivatives up to $s - 1$ order vanish on the boundary (i.e., $δ (\partial_{μ_{1}} \dots \partial_{μ_{m}} A^{α}) = 0$ on $\partial Ω$ for all $0 \leq m \leq s - 1$ ), then the generalized Euler-Lagrange equation reads,

\frac{δ L}{δ A^{α}} ≜ \frac{\partial L}{\partial A^{α}} + \sum_{k = 1}^{s} (- 1)^{k} \partial_{μ_{1}} \partial_{μ_{2}} \dots \partial_{μ_{k}} (\frac{\partial L}{\partial (\partial_{μ_{1}} \partial_{μ_{2}} \dots \partial_{μ_{k}} A^{α})}) = 0

Gauge transformation is $Δ L = \partial_{μ} C^{μ} (A^{α}, \partial_{μ_{1}} A^{α}, \dots, \partial_{μ_{1} \dots μ_{s - 1}} A^{α}; x^{μ})$ .

4.2.2 Particles' motion

L (x, v) = - m c^{2} \sqrt{1 - \frac{v^{2}}{c^{2}}} - e φ + e v \cdot A, p = \frac{\partial L}{\partial v} = γ m v + e A

L (x^{μ}, u^{μ}) = - m c \sqrt{- u^{α} u_{α}} + e A^{α} u_{α}, p^{α} = \frac{\partial L}{\partial u_{α}} = m u^{α} + e A^{α}

Substituting into the Lagrange equation yields $\frac{d p}{d t} = e E + e v \times B$ and $\frac{d p^{α}}{d τ} = e F^{α β} u_{β}$ , respectively.

4.2.3 Field equation

Klein-Gordon field *

L = - \frac{1}{2} (\partial^{μ} φ \partial_{μ} φ + κ^{2} φ^{2}) + ρ (x^{μ}) φ (x^{μ})

Substituting into the Lagrange equation yields Klein-Gordon equation $(\partial^{μ} \partial_{μ} - κ^{2}) φ (x^{μ}) = - ρ (x^{μ})$ , which has a monochromatic plane solution $φ (x^{μ}) = φ_{0} e^{i k_{α} x^{α}} = φ_{0} e^{i (k \cdot x - ω t)}$ without the mass source $ρ$ , and

(i k^{μ}) (i k_{μ}) - κ^{2} = 0 ⟹ ω = c \sqrt{κ^{2} + k^{2}} ⟹ {\begin{cases} v_{p} = \frac{ω}{k} = c \sqrt{\frac{κ^{2}}{k^{2}} + 1} > c \\ v_{g} = \frac{d ω}{d k} = c / \sqrt{\frac{κ^{2}}{k^{2}} + 1} < c \end{cases}

For a static, spherically symmetric source, the field equation reduces to $\frac{1}{r} (\frac{\partial^{2}}{\partial r^{2}} - κ^{2}) φ (r) = 0$ . The physically meaningful solution (vanishing at infinity) is the Yukawa potential,

φ (r) \propto \frac{1}{r} e^{- κ r}

EM field

L = L_{f} + L_{p} = - \frac{1}{4 μ_{0}} F_{α β} F^{α β} + A^{α} J_{α} \equiv \frac{1}{2} (ϵ_{0} E^{2} - \frac{1}{μ_{0}} B^{2}) - (ρ φ - J \cdot A)

with F^{α β} = \partial^{α} A^{β} - \partial^{β} A^{α}

Substituting into the Lagrange equation yields the covariant Maxwell's equations with a source term,

\partial_{α} F^{α β} = - μ_{0} J^{β}

4.3 Covariant single particle motion equation

4.3.1 Single particle in uniform electric field

Assuming $E \equiv E {\hat{e}}_{x}$ and the particle is at rest and $x = 0$ at time $t = 0$ (at every subsequent moment $t > 0$ , $a$ , $v$ and $x$ are parallel to $E$ ) . One finds

\frac{d p}{d t} = e E, p = p {\hat{e}}_{x} = γ m v {\hat{e}}_{x}

thus $e E d t = d (γ m v)$ ,

e E x = \frac{m v}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} ⟹ v (t) = \frac{ω_{E} t \cdot c}{\sqrt{1 + {(ω_{E} t)}^{2}}}, ω_{E} = \frac{e E}{m c}

⟹ x (t) = \frac{c}{ω_{E}} [\sqrt{1 + (ω_{E} t)^{2}} - 1] ⟹ {(\frac{x}{c / ω_{E}} + 1)}^{2} - ω_{E}^{2} t^{2} = 1

This is a hyperbola, which can also be expressed in the form with the parameter $τ$ . First above,

γ (t) = \frac{1}{\sqrt{1 + \frac{v^{2}}{c^{2}}}} ⟹ γ (t) = \sqrt{1 + ω_{E}^{2} t^{2}}

Since $\frac{d t}{d τ} = γ$ ,

⟹ c t (τ) = \frac{c}{ω_{E}} \sinh (ω_{E} τ)

and substituting into $x (t) = \frac{c}{ω_{E}} [\sqrt{1 + (ω_{E} t)^{2}} - 1]$ yields,

x (τ) = \frac{c}{ω_{E}} [\cosh (ω_{E} τ) - 1]

4.3.2 Single particle in uniform magnetic field

Assuming $B \equiv B {\hat{e}}_{z}$ and the particle satisfies $x = 0$ , $p_{x} = 0$ , $p_{y} = p_{⊥} = γ m v_{⊥}$ and $p_{z} = p_{∥} = γ m v_{∥}$ at time $t = 0$ . One finds

\frac{d E}{d t} = 0 ⟹ γ = const, \frac{d p_{x}}{d t} = e v_{y} B, \frac{d p_{y}}{d t} = - e v_{x} B, \frac{d p_{z}}{d t} = 0

⟹ γ m \frac{d v_{x}}{d t} = e B v_{y}, γ m \frac{d v_{y}}{d t} = - e B v_{x}

⟹ v_{x} (t) = v_{⊥} \sin (ω_{c} t), v_{y} (t) = v_{⊥} \cos (ω_{c} t), v_{z} (t) \equiv v_{∥}

with relativistic cyclotron frequency $ω_{c} = \frac{e B}{γ m}$ , then integral

⟹ x (t) = \frac{v_{⊥}}{ω_{c}} [1 - \cos (ω_{c} t)], y (t) = \frac{v_{⊥}}{ω_{c}} \sin (ω_{c} t), z (t) = v_{∥} t

This depicts a spiral line moving along the $z$ axis. Since $γ \equiv \frac{t}{τ}$ ,

x (τ) = R [1 - \cos (ω_{B} τ)], y (τ) = R \sin (ω_{B} τ), z (τ) = γ v_{∥} τ,

with cyclotron radius $R = \frac{v_{⊥}}{ω_{c}}$ and classic gyro-frequency $ω_{B} = γ ω_{c} = \frac{e B}{m}$ .

4.3.3 Single particle in uniform electromagnetic field *

Perpendicular field

Assuming $E ⊥ B$ with $B \equiv B {\hat{e}}_{z}$ , $E \equiv E {\hat{e}}_{x}$ . Refer to 4-Lorentz force $f^{α} = e F^{α β} u_{β} = γ (e E \cdot β, e E + e v \times B)$ and Lorentz equation $\frac{d p^{α}}{d τ} = f^{α}$ , One finds

m c \frac{d u^{0}}{d τ} = e E u^{1}, m \frac{d u^{1}}{d τ} = \frac{e E}{c} u^{0} + e u^{2} B, m \frac{d u^{2}}{d τ} = - e u^{1} B, m \frac{d u^{3}}{d τ} = 0

Take the derivative of the second expression above with respect to the proper time $τ$ ,

\frac{d^{2} u^{1}}{d τ^{2}} = ω_{E} \frac{d u^{0}}{d τ} + ω_{B} \frac{d u^{2}}{d τ}

Substitute the 1-st and 3-rd equation into the above equation:

\frac{d^{2} u^{1}}{d τ^{2}} = (ω_{E}^{2} - ω_{B}^{2}) u^{1} = \frac{e^{2} I}{m^{2}} u^{1}

$I = \frac{E^{2}}{c^{2}} - B^{2}$ is a Lorentz invariant that we are all familiar with. Define $Ω = | ω_{E}^{2} - ω_{B}^{2} |^{\frac{1}{2}}$ , then the solutions fall into three categories:

Magnetic dominance

If $ω_{B} > ω_{E}$ , namely $c B > E$ , the equation becomes the harmonic oscillator equation ${\ddot{u}}^{1} = - Ω^{2} u^{1}$ . However, it's complex to write out the parametric equations for the particle's motion in full. We adopt to a boost with $v_{B} = \frac{E \times c B}{(c B)^{2}} c$ since $E^{'} = 0$ in the new inertial frame, with $B^{'} = \sqrt{- I} = \frac{m}{e} Ω$ . 4-velocity component equation is simplified to

\frac{d u^{' 0}}{d τ} = 0, \frac{d u^{' 1}}{d τ} = Ω u^{' 2}, \frac{d u^{' 2}}{d τ} = - Ω u^{' 1}, \frac{d u^{' 3}}{d τ} = 0

Surely it's a cyclotron motion like previous section #4.3.2 Single particle in uniform magnetic field. Using $γ_{B} = \frac{1}{\sqrt{1 - (v_{B} / c)^{2}}} = \frac{c B}{\sqrt{c^{2} B^{2} - E^{2}}}$ , we can reversely transform to the lab frame,

\begin{aligned} u^{0} & = γ_{B} (u^{' 0} - β_{B} u^{' 2}) \\ u^{1} & = u^{' 1} \\ u^{2} & = γ_{B} (u^{' 2} - β_{B} u^{' 0}) \\ u^{3} & = u^{' 3} \end{aligned}

Electric dominance

If $ω_{B} < ω_{E}$ , namely $c B < E$ , the equation becomes the hyperbolic equation ${\ddot{u}}^{1} = Ω^{2} u^{1}$ . Adopt to a boost with $v_{E} = \frac{E \times c B}{E^{2}} c$ since $B^{'} = 0$ in the new inertial frame, with $E^{'} = \sqrt{I} = \frac{m}{e} Ω$ . 4-velocity component equation is simplified to

\frac{d u^{' 0}}{d τ} = \frac{Ω}{c} u^{' 1}, \frac{d u^{' 1}}{d τ} = \frac{Ω}{c} u^{' 0}, u^{' 2}, u^{' 3} = const

Surely it's is a hyperbola like previous section #4.3.1 Single particle in uniform electric field.

Light case

If $ω = ω_{B} = ω_{E}$ , namely $c B = E$ , the equation becomes the light equation ${\ddot{u}}^{1} = 0$ . 4-velocity component equation is simplified to

\frac{d u^{0}}{d τ} = ω u^{1}, \frac{d u^{1}}{d τ} = ω (u^{0} + u^{2}), \frac{d u^{2}}{d τ} = - ω u^{1}, \frac{d u^{3}}{d τ} = 0

Thus

u^{1} (τ) = α τ + u^{1} (0), with α = {\dot{u}}^{1} (0) = ω (u^{0} (0) + u^{2} (0))

u^{0} (τ) = \frac{1}{2} ω α τ^{2} + ω u^{1} (0) τ + u^{0} (0), u^{2} (τ) = - \frac{1}{2} ω α τ^{2} - ω u^{1} (0) τ + u^{2} (0)

Then integral

\begin{aligned} x (τ) & = \int u^{1} d τ = \frac{1}{2} α τ^{2} + u^{1} (0) τ \\ y (τ) & = \int u^{2} d τ = - \frac{1}{6} ω α τ^{3} - \frac{1}{2} ω u^{1} (0) τ^{2} + u^{2} (0) τ \\ c t (τ) & = \int u^{0} d τ = \frac{1}{6} ω α τ^{3} + \frac{1}{2} ω u^{1} (0) τ^{2} + u^{0} (0) τ \end{aligned}

⟹ y = - c t + \frac{α}{ω} τ

Since

α \propto (u^{0} + u^{2}) = lim_{v_{y} \to - c} \frac{c + v_{y}}{\sqrt{1 - \frac{v_{y}^{2}}{c^{2}}}} = c \cdot lim_{v_{y} \to - c} \sqrt{\frac{c + v_{y}}{c - v_{y}}} = 0 ⟹ y = - c t

namely it's a particle propagating in the $E \times B$ -direction at the speed of light as $u^{1} (0) = 0$ .

Non-Perpendicular field

If $E \cdot B \neq 0$ , one can find a inertial frame with the speed satisfying $\frac{β}{1 + β^{2}} = \frac{E \times c B}{E^{2} + c^{2} B^{2}}$ where $E^{'} \cdot B^{'} = 0$ . Therefore, Assuming $E ∥ B$ with $B \equiv B {\hat{e}}_{z}$ , $E \equiv E {\hat{e}}_{z}$ , the solution is a spiral motion that undergoes hyperbolic acceleration in the $z$ direction.

4.4 Noether's theorem of field *

In this section, $φ_{I}$ ( $I = 1, \dots, N$ ) denotes the components of a multicomponent field, but $φ_{I}$ will be abbreviated as $φ$ . When $φ$ appears twice, it implies summation over $I$ .

4.4.1 Variation of field

Formal variation

δ φ ≜ φ^{'} (x^{μ}) - φ (x^{μ})

Total variation

\bar{δ} φ ≜ φ^{'} (x^{' μ}) - φ (x^{μ}), with x^{' μ} = x^{μ} + \bar{δ} x^{μ}

Relation of the two variation

\bar{δ} φ = δ φ + (\partial_{μ} φ) \bar{δ} x^{μ}

4.4.2 Formal variation of Lagrangian

δ L (φ (x^{μ}), \partial^{α} (x^{μ}), x^{μ}) = \partial^{α} (Π_{α} δ φ) + \underset{Lagrange equation =0}{\underset{⏟}{(\frac{\partial L}{\partial φ} - \partial^{α} Π_{α})}} δ φ, with Π_{α} = \frac{\partial L}{\partial (\partial^{α} φ)}

Since the field equation wouldn't change under the gauge $δ L = \partial^{α} C_{α}$ , one find a conserved current / Noether current,

J^{α} (x^{μ}) = Π^{α} δ φ - C^{α}, with \partial_{α} J^{α} = 0

4.4.3 Total variation of Lagrangian

\begin{aligned} \bar{δ} L (φ (x^{μ}), \partial^{α} φ (x^{μ}), x^{μ}) & = δ L + (\partial_{α} L) \bar{δ} x^{α} \\ for infinitesimal Poincare transformation: \partial_{α} \bar{δ} x^{α} = 0 \to & = \partial_{α} [L \bar{δ} x^{α} + Π^{α} δ φ] \\ = \partial_{α} J^{α} \end{aligned}

where the conserved current density is

J^{α} = Π^{α} \bar{δ} φ + \underset{T_{β}^{α}}{\underset{⏟}{(Π^{α} \partial_{β} φ - δ_{β}^{α} L)}} \bar{δ} x^{β}

J^{α} = T^{α β} \bar{δ} x_{β} + Π^{α} \bar{δ} φ, with T^{α β} = - Π^{α} \partial^{β} φ + g^{α β} L

Warning: the regular energy-momentum tensor $T^{α β}$ is not symmetric generally not symmetric for fields with spin (such as the electromagnetic field with spin-1). An additional divergence term $\frac{1}{μ_{0}} \partial_{λ} (F^{μ λ} A^{ν})$ needs to be introduced to ensure symmetry (refer to Electrodynamics: 3. Symmetric correction).

Addition: Under the $(-, +, +, +)$ metric, conventionally $T^{α β} = - Π^{α} \partial^{β} φ + g^{α β} L$ , whereas for the $(+, -, -, -)$ , it's $T^{α β} = Π^{α} \partial^{β} φ - g^{α β} L$ . This convention ensures that the energy density always satisfies the standard relation $T^{00} \equiv \frac{\partial L}{\partial (\partial^{0} φ)} \partial^{0} φ - L$ . Furthermore, the distinction between $T^{α β}$ and its transposed counterpart $T^{' α β} = - Π^{β} \partial^{α} φ + g^{α β} L = T^{β α}$ amounts to a total divergence term (i.e., $T^{α β} - T^{' α β} = \partial_{μ} Ψ^{μ α β}$ ). Consequently, $T^{' α β}$ serves as an equally valid alternative for describing the conserved currents.

4.4.4 Noether's theorem of field

Noether's theorem: If $\bar{δ} L = \partial_{α} C^{α}$ under the transformation $x^{α} \to x^{' α} + \bar{δ} x^{α}$ , then $J^{α}$ is the conserved current satisfying $\partial_{α} J^{α} = 0$

Space-time Translation

Consider a global infinitesimal translation $x^{' μ} = x^{μ} + ϵ^{μ}$ . The field transforms as a scalar, hence $\bar{δ} φ = 0$ . The Noether current is:

J^{α} = - T^{α β} ϵ_{β} ⟹ \partial_{α} T^{α β} = 0

The conserved 4-momentum $P^{β}$ is,

P^{β} = \frac{1}{c} \int T^{0 β} d^{3} x = (\frac{E}{c}, P) is conserved with E = ∭ (\frac{\partial L}{\partial (\partial^{t} φ)} \partial^{t} φ - L) d^{3} x

$T^{α β}$ needs to be corrected before we can obtain the physically meaningful expression of $P^{i}$ as well.

Rotations and Boosts

Consider an infinitesimal Lorentz transformation: $x^{' μ} = Λ_{ν}^{μ} x^{ν} \approx (δ_{ν}^{μ} + ω_{ν}^{μ}) x^{ν}$ , hence $\bar{δ} x^{μ} = ω_{ν}^{μ} x^{ν}$ and $\bar{δ} A^{μ} = ω_{ν}^{μ} A^{ν}$ ,

J^{α} = T^{α β} \bar{δ} x_{β} + Π^{α β} \bar{δ} A_{β}, with T^{α β} = - Π^{α μ} \partial^{β} A_{μ} + g^{α β} L, Π_{α β} = \frac{\partial L}{\partial (\partial_{α} A_{β})}

\begin{aligned} ⟹ J^{α} & = T^{α β} ω_{β γ} x^{γ} + Π^{α β} ω_{β γ} A^{γ} \\ ω_{β γ} = - ω_{γ β} \to & = \frac{1}{2} ω_{β γ} \underset{M^{α β γ}}{\underset{⏟}{[(x^{γ} T^{α β} - x^{β} T^{α γ}) + (A^{γ} Π^{α β} - A^{β} Π^{α γ})]}} \end{aligned}

Thus, $\partial_{α} M^{α β γ} = 0$ , the conserved charges are

L^{α β} = \frac{1}{c} ∭ M^{0 α β} d^{3} x

The latter section is available at II.