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Retarded potentials of a point history as the retarded convolution of the point sources, and the off-worldline Lienard-Wiechert fields

Statement

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ , $q \in R$ , $ε_{0} > 0$ , put $μ_{0} : = (c^{2} ε_{0})^{- 1}$ and $C : = q / (4 π ε_{0})$ , and let $z : R \to R^{3}$ be of class $C^{3}$ with $∣ z^{'} (s) ∣ \leq v_{*} < c$ for every $s \in R$ (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral). Write $Ω : = R^{3} \times R$ and $Γ : = {(z (s), s) : s \in R} \subseteq Ω$ . For $(x, t) \in Ω ∖ Γ$ let $s = s (x, t)$ be the unique retarded root of Global existence, uniqueness and differentiability of the retarded root with $t - s = ∣ x - z (s) ∣ / c$ , and put $R : = x - z (s)$ , $r : = ∣ R ∣ > 0$ , $n : = R / r$ , $β : = z^{'} (s) / c$ , $κ : = 1 - n \cdot β$ , $N : = n \cdot β$ , $D : = κ r$ , $\dot{β} : = z^{''} (s) / c$ and $d : = n \cdot \dot{β}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ). Then:

(a) the scalar distribution $ρ \in D^{'} (Ω)$ and the vector distribution $J \in (D^{'} (Ω))^{3}$ of Point-support spacetime densities: distributions, continuity equation and endpoint fluxes,

$⟨ ρ, ψ ⟩ = q \int_{R} ψ (z (s), s) d s, ⟨ J, ψ ⟩ = q \int_{R} z^{'} (s) ψ (z (s), s) d s,$

satisfy $\partial_{t} ρ + div J = 0$ .

(b) Let $K$ be the retarded wave kernel of The retarded wave kernel is a distribution supported on the cone, solves the point-wave equation, and convolves to the retarded solution and, for $ψ \in D (Ω)$ and $τ \in R$ , $y \in R^{3}$ , put $(K ⊛ ψ) (y, τ) : = ⟨ K, ψ (y + \cdot, τ + \cdot) ⟩$ . Define the retarded-history pairings by the explicit integrals

$⟨ Φ, ψ ⟩ : = \frac{q}{ε_{0}} \int_{R} (K ⊛ ψ) (z (s), s) d s, ⟨ Υ, ψ ⟩ : = μ_{0} q \int_{R} z^{'} (s) (K ⊛ ψ) (z (s), s) d s .$

Here the second integral is componentwise. The notation $ε_{0}^{- 1} ⟨ ρ, K ⊛ ψ ⟩$ and $μ_{0} ⟨ J, K ⊛ ψ ⟩$ , if used, means precisely these explicit history integrals, rather than ordinary distribution action on a compactly supported test. These pairings define distributions on $Ω$ , are finite for every $ψ \in D (Ω)$ and satisfy the exact retarded-convolution identities

$⟨ Φ, ψ ⟩ = \int_{Ω ∖ Γ} \frac{C ψ (x, t)}{κ (x, t) r (x, t)} d x d t, ⟨ Υ, ψ ⟩ = \int_{Ω ∖ Γ} \frac{β (x, t)}{c} \frac{C ψ (x, t)}{κ (x, t) r (x, t)} d x d t$

(for $Υ$ , componentwise against $ψ \in D (Ω)$ ; both integrands are continuous on the open set $Ω ∖ Γ$ and integrable there against $ψ$ ). Consequently the potentials are the $C^{2}$ functions

$φ : = \frac{C}{D} = \frac{C}{κ r} : Ω ∖ Γ \to R, A : = \frac{β φ}{c} = \frac{β C}{c κ r} : Ω ∖ Γ \to R^{3}$

off the worldline, and the zero extensions $\tilde{φ}, \tilde{A}$ on $Γ$ belong to $L_{l o c}^{1} (Ω)$ ; $Φ$ is the regular distribution of $\tilde{φ}$ , and $Υ$ that of $\tilde{A}$ (Regular distribution from a locally integrable function).

(c) Let $E : = - \nabla φ - \partial_{t} A$ and $B : = curl A$ on $Ω ∖ Γ$ (Scalar-vector potential pairs, gauge transformation, Lorenz divergence and the wave expression). Then, off the worldline,

$E = C [\frac{(1 - ∣ β ∣^{2}) (n - β)}{κ^{3} r^{2}} + \frac{n \times ((n - β) \times \dot{β})}{c κ^{3} r}], B = \frac{n \times E}{c} .$

The fields $E, B$ are $C^{1}$ off the worldline. More generally, if $z \in C^{k}$ , $k \geq 3$ , the retarded root is $C^{k}$ , the potentials are $C^{k - 1}$ and the fields are $C^{k - 2}$ there; if $z \in C^{\infty}$ , all these objects are smooth off the worldline. The result is a mathematical model computation for the stipulated uniformly subluminal smooth worldline and the retarded choice; no self-force law, no radiation-reaction postulate and no physical Maxwell coupling is asserted or proved.

Facts & Assumptions

Given: The constants $c, q, ε_{0}, μ_{0}, C$ , the all-time uniformly subluminal $C^{3}$ trajectory $z$ , the domain $Ω = R^{3} \times R$ with its worldline graph $Γ$ , and the retarded data $s, R, r, n, β, κ, N, D, \dot{β}, d$ of the Statement.

[F1]

For an open interval $I$ , a $C^{1}$ trajectory $z : I \to R^{3}$ whose graph map into an open $Ω$ is proper, and $ψ \in D (Ω)$ , the functionals $⟨ ρ, ψ ⟩ = q \int_{I} ψ (z (s), s) d s$ and $⟨ J, ψ ⟩ = q \int_{I} z^{'} (s) ψ (z (s), s) d s$ are distributions and satisfy $\partial_{t} ρ + div J = 0$ ; properness is automatic for $I = R$ (Point-support spacetime densities: distributions, continuity equation and endpoint fluxes).

[F2]

For $c > 0$ and $C^{1}$ $z$ with $∣ z^{'} ∣ \leq v_{*} < c$ there is exactly one retarded root $s$ of $t - s = ∣ x - z (s) ∣ / c$ , with $s \leq t$ and equality exactly on the worldline; off it $s < t$ , $r = ∣ x - z (s) ∣$ satisfies $r = c (t - s) > 0$ , $κ = 1 - n \cdot z^{'} (s) / c$ obeys $1 - v_{*} / c \leq κ \leq 1 + v_{*} / c$ , and $\partial_{t} s = 1 / κ$ , $\nabla s = - n / (c κ)$ ; a $C^{k}$ trajectory gives a $C^{k}$ root on the off-worldline domain, in particular $C^{3}$ for $C^{3}$ $z$ (Global existence, uniqueness and differentiability of the retarded root).

[F3]

For the retarded kernel $K$ one has $(K ⊛ ψ) (y, τ) = (4 π)^{- 1} \int ψ (y + w, τ + ∣ w ∣ / c) ∣ w ∣^{- 1} d w$ for $ψ \in D (Ω)$ (The retarded wave kernel is a distribution supported on the cone, solves the point-wave equation, and convolves to the retarded solution).

[F4]

For a potential pair, $E = - \nabla φ - \partial_{t} A$ , $B = curl A$ (Scalar-vector potential pairs, gauge transformation, Lorenz divergence and the wave expression).

[F5]

A locally integrable function defines the regular functional $u_{f} (φ) = \int f φ$ ; distributions are the continuous complex-linear functionals on the LF test space, whose fixed-support seminorms bound absolute pairings (Regular distribution from a locally integrable function, Distribution, Test function topology).

[F6]

Under ${A C}_{ω}$ , for nonnegative Borel $g$ on $R^{n}$ , $\int_{R^{n}} g d λ_{n} = \int_{0}^{\infty} \int_{S^{n - 1}} g (r ω) r^{n - 1} d σ (ω) d r$ ; in particular $\int_{B_{L}} ∣ w ∣^{- 1} d w = σ (S^{2}) L^{2} / 2$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F7]

Tonelli and Fubini on sigma-finite products: nonnegative measurable functions may be integrated in either order, and for $f \in L^{1} (μ \times ν)$ the three integrals agree (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Fubini's theorem for L^1 functions on a sigma-finite product).

[F8]

Euclidean Borel sets are the Borel sets of the product structure, and the Lebesgue product measure agrees with Euclidean Lebesgue measure on Borel sets (The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n}, On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}).

[F9]

Under ${A C}_{ω}$ , for a $C^{1}$ diffeomorphism $T : U \to V$ of open sets, $\int_{V} g = \int_{U} g (T) ∣ \det D T ∣$ for nonnegative Borel $g$ , and $\int_{V} f = \int_{U} f (T) ∣ \det D T ∣$ for $f \in L^{1} (V)$ (Borel change of variables from the compact-support formula and Radon uniqueness, A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions).

[F10]

The integral is linear on $L^{1}$ , satisfies $∣ \int f ∣ \leq \int ∣ f ∣$ , nonnegative integrals are monotone and scale, and a nonnegative measurable function has zero integral exactly when it vanishes almost everywhere (The Lebesgue integral is linear on $L^{1} (μ)$ , The modulus of an integral is bounded by the integral of the modulus, Monotonicity and nonnegative homogeneity of the nonnegative integral, A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[F11]

Compact subsets of $R^{n}$ are bounded and closed, and continuous real functions on compacta are bounded and attain extremes (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F12]

The gradient, the coordinate partial derivatives, and the divergence and curl on $C^{1}$ fields are those fixed by the cited definitions; vector-valued one-variable derivatives and integrals are defined componentwise (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , Divergence and curl of a $C^{1}$ vector field, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[F13]

The chain rule for total derivatives, the algebra of one-variable derivatives, equality of mixed second partials for $C^{2}$ functions, and the $C^{k}$ vocabulary hold as cited (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Clairaut--Schwarz theorem for continuous second partial derivatives, $C^{k}$ Euclidean maps and diffeomorphisms).

[F14]

The Euclidean inner product is bilinear and symmetric and the cross product is bilinear and alternating; in particular $n \times (n - β) = - n \times β$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The cross product in $R^{3}$ , The cross product is bilinear, alternating, and orthogonal to both factors).

[F15]

The divergence and curl obey $curl (f F) = \nabla f \times F + f curl F$ (Divergence and curl are linear and satisfy the scalar product rules).

[A1]

The Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF1F11

Since $z$ is $C^{3}$ , it is $C^{1}$ , and $I : = R$ is open; the graph map $s \mapsto (z (s), s)$ into $Ω = R^{3} \times R$ is proper because a compact subset of $Ω$ has bounded time coordinate, so its inverse image in $R$ is closed and bounded. Hence [F1] applies with $q, ε_{0}$ as given: the functionals $ρ, J$ of the Statement are distributions on $Ω$ and $\partial_{t} ρ + div J = 0$ , which is (a).

1.2givenF3F5F6F10F11A1

Finiteness and bounds for the history pairings. Fix $ψ \in D (Ω)$ ; by compactness of its support choose reals $M \geq 0$ and $t_{0} \leq t_{1}$ with $supp ψ \subseteq {(x, t) : ∣ x ∣ \leq M, t_{0} \leq t \leq t_{1}}$ , and put $A_{ψ} : = \sup ∣ ψ ∣ < \infty$ , $Z : = ∣ z (0) ∣$ , $S : = \max {0, (M + Z - c t_{0}) / (c - v_{*})}$ , $H : = \max {S, ∣ t_{1} ∣}$ , $L : = 1 + M + Z + v_{*} H$ , $T : = \max {0, t_{1} + S}$ and $M_{Q} : = T σ (S^{2}) L^{2} / (8 π)$ . If $g (s, w) : = (z (s) + w, s + ∣ w ∣ / c)$ lies in $supp ψ$ , then $s \leq t_{1}$ ; and for $s \leq 0$ the bound $∣ z (s) ∣ \leq Z + v_{*} ∣ s ∣$ gives $t_{0} \leq s + ∣ w ∣ / c \leq - (1 - v_{*} / c) ∣ s ∣ + (M + Z) / c$ , so $s \geq - S$ . Also $∣ w ∣ \leq M + ∣ z (s) ∣ \leq M + Z + v_{*} H < L$ . Therefore $∣ ⟨ Φ, ψ ⟩ ∣ \leq \frac{∣ q ∣}{ε_{0}} \int_{- S}^{t_{1}} \frac{1}{4 π} \int_{B_{L}} \frac{A_{ψ}}{∣ w ∣} d w d s \leq \frac{∣ q ∣ M_{Q} A_{ψ}}{ε_{0}}$ , and componentwise $∣ ⟨ Υ_{j}, ψ ⟩ ∣ \leq μ_{0} ∣ q ∣ v_{*} M_{Q} A_{ψ}$ . These bounds are fixed-support seminorms, so $Φ$ and the components of $Υ$ are distributions.

1.3givenF2F9F13F14

The retarded substitution map. Put $U : = R \times (R^{3} ∖ {0})$ , $V : = Ω ∖ Γ$ and $g (s, w) : = (z (s) + w, s + ∣ w ∣ / c)$ . Then $V$ is open: $Γ$ is closed, since a convergent sequence $(z (s_{j}), s_{j}) \to (x, t)$ has $s_{j} \to t$ , hence $x = z (t)$ by continuity. The map $g : U \to V$ is a bijection with inverse $(x, t) \mapsto (s (x, t), x - z (s (x, t)))$ : the root $s (x, t)$ of [F2] exists, satisfies $x - z (s) \neq 0$ exactly off the worldline, and $s (x, t)$ is $C^{1}$ there because $z$ is $C^{3}$ ; conversely no image point lies on $Γ$ , for $∣ w ∣ = ∣ z (t) - z (s) ∣ \leq v_{*} ∣ t - s ∣ = v_{*} ∣ w ∣ / c$ is impossible for $w \neq 0$ . The map is $C^{1}$ with derivative matrix (in the order $s, w_{1}, w_{2}, w_{3}$ ) whose last row is $(1, w / (c ∣ w ∣))$ and whose first column is $(z^{'} (s), 1)$ with the identity in the remaining block; expanding along the last row gives $\det D g (s, w) = - 1 + z^{'} (s) \cdot w / (c ∣ w ∣) = - κ (s, w)$ , so $∣ \det D g ∣ = κ$ , and $κ \geq 1 - v_{*} / c > 0$ by [F2]. If $g (s, w) = g (s^{'}, w^{'})$ then $∣ s - s^{'} ∣ = ∣ ∣ w^{'} ∣ - ∣ w ∣ ∣ / c \leq ∣ z (s^{'}) - z (s) ∣ / c \leq (v_{*} / c) ∣ s - s^{'} ∣$ , hence $s = s^{'}$ and $w = w^{'}$ . By [F9], for every nonnegative Borel $h$ on $V$ one has $\int_{V} h = \int_{U} h (g) κ d s d w$ , and for every $f \in L^{1} (V)$ , $\int_{V} f = \int_{U} f (g) κ d s d w$ .

1.4givenF2F12F13F14

Retarded derivative identities. On $V$ , with $s = s (x, t)$ the root of [F2]: $s_{t} = 1 / κ$ and $\nabla s = - n / (c κ)$ . Hence $r_{t} = n \cdot (- z^{'} (s) s_{t}) = - c N / κ$ and $\nabla r = n / κ$ ; $R_{t} = - z^{'} (s) s_{t} = - c β / κ$ and $\partial_{i} R = e_{i} + β n_{i} / κ$ ; $β_{t} = \dot{β} s_{t} = \dot{β} / κ$ and $\partial_{i} β = \dot{β} \partial_{i} s = - \dot{β} n_{i} / (c κ)$ . Since $κ = 1 - N$ and $D = κ r = r - R \cdot β$ , the product rules with the bilinearity of the inner product give $D_{t} = r_{t} - R_{t} \cdot β - R \cdot β_{t} = (c (∣ β ∣^{2} - N) - r d) / κ$ and, componentwise, $\partial_{i} D = \partial_{i} r - (\partial_{i} R) \cdot β - R \cdot \partial_{i} β = n_{i} (1 - ∣ β ∣^{2} + r d / c) / κ - β_{i}$ , that is, $\nabla D = ((1 - ∣ β ∣^{2} + r d / c) / κ) n - β$ .

1.5F14algebra

The needed triple-product identity follows directly from the coordinate definition in [F14]. For arbitrary $a, b, u \in R^{3}$ , the first component of $a \times (b \times u)$ is $a_{2} (b_{1} u_{2} - b_{2} u_{1}) - a_{3} (b_{3} u_{1} - b_{1} u_{3}) = b_{1} (a_{2} u_{2} + a_{3} u_{3}) - u_{1} (a_{2} b_{2} + a_{3} b_{3}) = b_{1} (a \cdot u) - u_{1} (a \cdot b)$ , since the terms $b_{1} a_{1} u_{1} - u_{1} a_{1} b_{1}$ cancel. Cyclic permutation gives the other two components, so $a \times (b \times u) = b (a \cdot u) - u (a \cdot b)$ . With $a = n$ , $b = n - β$ , $u = \dot{β}$ this gives $n \times ((n - β) \times \dot{β}) = d (n - β) - κ \dot{β}$ , since $n \cdot n = 1$ .

2.1step 1.2step 1.3F3F7F9F10

Scalar pairing identity. Insert the kernel identity [F3] into the definition of $Φ$ : $⟨ Φ, ψ ⟩ = \frac{q}{4 π ε_{0}} \int_{R} \int_{R^{3}} ψ (g (s, w)) ∣ w ∣^{- 1} d w d s$ . The double integral is absolutely convergent by the cylinder estimate of step 1.2, so Fubini's theorem reorders it, and the substitution identity of step 1.3 for the $L^{1}$ function $ψ / (κ r)$ on $V$ (whose absolute integral is finite by the same cylinder estimate) turns it into $⟨ Φ, ψ ⟩ = \int_{V} \frac{C ψ (x, t)}{κ (x, t) r (x, t)} d x d t$ , since $r (g (s, w)) = ∣ w ∣$ , $κ (g (s, w)) = κ (s, w)$ and the Jacobian factor $κ$ cancels $κ^{- 1}$ .

2.2step 1.2step 1.3F3F7F9F10F14

Vector pairing identity. Componentwise, $⟨ Υ_{j}, ψ ⟩ = \frac{μ_{0} q}{4 π} \int_{R} \int_{R^{3}} z_{j}^{'} (s) ψ (g (s, w)) ∣ w ∣^{- 1} d w d s$ ; the double integral is absolutely bounded by $μ_{0} ∣ q ∣ v_{*}$ times the scalar majorant of step 1.2, Fubini applies, and the substitution of step 1.3 gives $⟨ Υ_{j}, ψ ⟩ = \frac{μ_{0} q c}{4 π} \int_{V} \frac{β_{j} ψ}{κ r} d x d t$ , where $β_{j} (s) = β_{j} (g (s, w))$ and $z_{j}^{'} = c β_{j}$ . Since $\frac{μ_{0} q c}{4 π} = μ_{0} ε_{0} c \cdot \frac{q}{4 π ε_{0}} = \frac{C}{c} \cdot c^{2} μ_{0} ε_{0} = \frac{C}{c}$ , this is $⟨ Υ_{j}, ψ ⟩ = \int_{V} \frac{β_{j} (x, t)}{c} \frac{C ψ}{κ r} d x d t$ .

3.1step 1.2step 1.3step 2.1step 2.2F2F5F7F8F9F10

Potentials and regular distributions. By [F2] the root is $C^{3}$ off the worldline, so $r, n$ are $C^{3}$ , $β$ is $C^{2}$ , $κ$ is $C^{2}$ and $D = κ r$ is $C^{2}$ with $D > 0$ on $V$ ; hence $φ = C / D$ and $A = β φ / c$ are of class $C^{2}$ on $V$ , and the identities of steps 2.1 and 2.2 read $⟨ Φ, ψ ⟩ = \int_{V} φ ψ$ and $⟨ Υ_{j}, ψ ⟩ = \int_{V} A_{j} ψ$ for every test $ψ$ . The graph $Γ$ is closed, hence Borel, and null: for each time $t$ its spatial section ${z (t)}$ is a singleton, so Tonelli together with the product identification of Euclidean measure gives $λ_{4} (Γ) = \int_{R} λ_{3} ({z (t)}) d t = 0$ . Therefore $\int_{V} φ ψ = \int_{Ω} \tilde{φ} ψ$ and $\int_{V} A_{j} ψ = \int_{Ω} {\tilde{A}}_{j} ψ$ for the zero extensions. Finally $\tilde{φ}, \tilde{A}$ are locally integrable: for a compact $Q \subseteq Ω$ the substitution identity of step 1.3 and the cylinder bound of step 1.2 give $\int_{Ω} ∣ \tilde{φ} ∣ 1_{Q} \leq ∣ C ∣ \int_{U} 1_{Q} (g) ∣ w ∣^{- 1} d s d w \leq 4 π ∣ C ∣ M_{Q^{'}}$ with $M_{Q^{'}} = T^{'} σ (S^{2}) L^{' 2} / (8 π) < \infty$ for the window constants of $Q$ . Hence the regular functionals of $\tilde{φ}, \tilde{A}$ are distributions and $Φ = u_{\tilde{φ}}$ , $Υ = u_{\tilde{A}}$ .

4.1step 1.4step 1.5step 3.1F4F12F13F14

The electric field. On $V$ , $\nabla φ = - C \nabla D / D^{2}$ , $φ_{t} = - C D_{t} / D^{2}$ and $A = β φ / c$ , so $A_{t} = (\dot{β} φ / κ + β φ_{t}) / c = \dot{β} C / (c κ D) - β C D_{t} / (c D^{2})$ and $E = - \nabla φ - A_{t} = \frac{C}{D^{2}} (\nabla D + \frac{β D_{t}}{c}) - \frac{C \dot{β}}{c κ D}$ . Substituting the identities of step 1.4, the bracket equals $\frac{1}{κ} (1 - ∣ β ∣^{2} + \frac{r d}{c}) (n - β)$ , so $E = C \frac{(1 - ∣ β ∣^{2}) (n - β) + r d (n - β) / c}{κ^{3} r^{2}} - \frac{C \dot{β}}{c κ D} = \frac{C}{κ^{3} r^{2}} [(1 - ∣ β ∣^{2}) (n - β) + \frac{r}{c} (d (n - β) - κ \dot{β})]$ , using $D = κ r$ to write the second summand over the common denominator. The cross-product identity $n \times ((n - β) \times \dot{β}) = d (n - β) - κ \dot{β}$ of step 1.5 then gives the displayed electric field.

5.1step 1.4step 4.1F2F4F13F14F15∎

The magnetic field and regularity. By [F15] and the chain rule, $curl A = \nabla (φ / c) \times β + (φ / c) curl β$ with $curl β = \nabla s \times \dot{β}$ , so $curl A = - \frac{C \nabla D \times β}{c D^{2}} - \frac{C n \times \dot{β}}{c^{2} κ^{2} r}$ . Substituting $\nabla D$ from step 1.4 and using $n \times (n - β) = - n \times β$ gives $curl A = - C [\frac{1 - ∣ β ∣^{2} + r d / c}{c κ^{3} r^{2}} (n \times β) + \frac{n \times \dot{β}}{c^{2} κ^{2} r}]$ . Crossing the field of step 4.1 with $n / c$ gives the same expression, because $n \times n = 0$ , $n \times (n - β) = - n \times β$ and $n \times (d (n - β) - κ \dot{β}) = - d (n \times β) - κ (n \times \dot{β})$ ; hence $B = n \times E / c$ . Since for $z \in C^{3}$ the root is $C^{3}$ and $\nabla D, D_{t}, β, \dot{β}, κ, r, n$ are as in step 1.4 with $κ r > 0$ on $V$ , both displayed fields are $C^{1}$ on $V$ ; the same computation with a $C^{k}$ trajectory uses the $C^{k}$ root branch of [F2] and gives $C^{k - 2}$ fields, and the smooth branch gives smooth fields.

Remarks

Choice accounting. ${A C}_{ω}$ is used through the polar formula and the two change-of-variables theorems (step 1.3 and the polar bound of step 1.2); the root lemma, the current continuity and the algebraic computations are choice-free as stated in their own items.
What the retarded selection does and does not do. The pairings select the retarded history $s < t$ through the kernel $K$ and the root of [F2]. No statement is made about exactly two solutions, about arbitrary nonzero Cauchy data, about finite self-energy or about a self-force law; the fields are computed from the stipulated potential pair off the worldline only.
Signs. The conventions $E = - \nabla φ - \partial_{t} A$ , $B = curl A$ and $C = q / (4 π ε_{0})$ , $μ_{0} = (c^{2} ε_{0})^{- 1}$ are those fixed in [F4] and the Statement; the formula for $B$ is the consequence $B = n \times E / c$ , not an independent convention.

Retarded potentials of a point history as the retarded convolution of the point sources, and the off-worldline Lienard-Wiechert fields

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources