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Gauss multiplier and smooth periodic Coulomb solver

Statement

Assume Countable Choice and the conventions of Charged gauge data and SI-natural conventions. Let $I \subseteq R$ be a nonempty open time interval, with $t$ measured in seconds and $x^{0} = c t$ , and let $T_{L}^{3} = (R / L Z)^{3}$ be the flat spatial torus with period $L > 0$ metres and periodic coordinates $x$ . Take jointly smooth periodic maps $A : I \times T_{L}^{3} \to R^{3}, ϕ : I \times T_{L}^{3} \to R, ρ : I \times T_{L}^{3} \to R, J : I \times T_{L}^{3} \to R^{3} .$ Write $\dot{A} : = \partial_{t} A$ for the time partial derivative at fixed $x$ , and $B : = \nabla \times A$ . All spatial differential operators below are the periodic coordinate derivatives, and each displayed spatial integral and Hamiltonian is evaluated at a fixed $t \in I$ ; the time argument is suppressed in the formulas. In SI units, $A$ has unit $T m$ , $ϕ$ has unit $V$ , $ρ$ has unit $C m^{- 3}$ , and $J$ has unit $A m^{- 2}$ . Thus $\dot{A}$ and $\nabla ϕ$ both have unit $V m^{- 1} = T m s^{- 1}$ , and every term of the formal classical Lagrangian density has unit $J m^{- 3}$ . Let $c > 0$ have unit $m s^{- 1}$ , let $μ_{0} > 0$ be the symbolic magnetic constant with unit $N A^{- 2}$ , and put $ϵ_{0} = (μ_{0} c^{2})^{- 1}$ with unit $F m^{- 1}$ ; no numerical value for these constants is adopted. Define the formal classical density $L = \frac{ϵ_{0}}{2} ∣ \dot{A} + \nabla ϕ ∣^{2} - \frac{1}{2 μ_{0}} ∣ \nabla \times A ∣^{2} - ρ ϕ + J \cdot A .$ The canonical momenta are $π_{A} = ϵ_{0} (\dot{A} + \nabla ϕ)$ , with unit $C m^{- 2}$ , and $π_{ϕ} = 0$ . Periodic integration by parts gives the fixed-time Hamiltonian $H (t) = \int_{T_{L}^{3}} (\frac{∣ π_{A} ∣^{2}}{2 ϵ_{0}} + \frac{∣ B ∣^{2}}{2 μ_{0}} + ϕ (\nabla \cdot π_{A} + ρ) - J \cdot A) d^{3} x .$ Variation of the multiplier $ϕ$ imposes $\nabla \cdot π_{A} + ρ = 0$ . In Coulomb gauge $\nabla \cdot A = 0$ , this becomes $- Δ ϕ = ρ / ϵ_{0}$ . At each fixed $t$ , a smooth periodic solution can exist only if $\int_{T_{L}^{3}} ρ (x, t) d^{3} x = 0$ . If $\int_{T_{L}^{3}} ρ (x, t) d^{3} x = 0$ for every $t \in I$ , there is a unique jointly smooth periodic solution with zero spatial mean, whose Fourier coefficients are ${\hat{ϕ}}_{0} (t) = 0, {\hat{ϕ}}_{k} (t) = \frac{{\hat{ρ}}_{k} (t)}{ϵ_{0} ∣ 2 π k / L ∣^{2}} (k \neq 0), {\hat{ρ}}_{k} (t) = L^{- 3} \int_{[0, L]^{3}} ρ (x, t) e^{- 2 π i k \cdot x / L} d^{3} x .$ For every $t \in I$ it satisfies $\frac{ϵ_{0}}{2} \int_{T_{L}^{3}} ∣ \nabla ϕ ∣^{2} d^{3} x = \frac{1}{2} \int_{T_{L}^{3}} ρ ϕ d^{3} x = \frac{L^{3}}{2 ϵ_{0}} \sum_{k \neq 0} \frac{∣ {\hat{ρ}}_{k} (t) ∣^{2}}{∣ 2 π k / L ∣^{2}} \geq 0.$ This is a classical conditional solver for smooth periodic neutral data; it makes no claim about quantum density products or nonperiodic boundary conditions.

Facts & Assumptions

Given: Countable Choice; the positive symbolic SI constants and units stated in the theorem; jointly smooth periodic maps on $I \times T_{L}^{3}$ ; the fixed-time convention for $\dot{A}$ ; and the Fourier convention stated in the theorem.

[A1]

The convention sets $ϵ_{0} = (μ_{0} c^{2})^{- 1}$ with positive symbolic SI constants. Charged gauge data and SI-natural conventions

[F1]

On the normalized torus, the Fourier characters form a complete orthonormal basis and Parseval holds; the theorem's torus measure is normalized to total mass one. The Fourier basis and Parseval's identity on the finite torus

[A2]

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

Proof

Given: [A1], [F1], [A2], the smooth periodic data, and periodic boundary conditions.

1.1A1givenalgebra

Since the displayed density contains no $\dot{ϕ}$ , its canonical momentum is $π_{ϕ} = \partial L / \partial \dot{ϕ} = 0$ . Differentiating with respect to $\dot{A}$ gives $π_{A} = ϵ_{0} (\dot{A} + \nabla ϕ)$ . The Legendre expression $π_{A} \cdot \dot{A} - L$ equals $\frac{∣ π_{A} ∣^{2}}{2 ϵ_{0}} - π_{A} \cdot \nabla ϕ + \frac{∣ \nabla \times A ∣^{2}}{2 μ_{0}} + ρ ϕ - J \cdot A .$ The periodic boundary has no surface term, and integrating $- π_{A} \cdot \nabla ϕ$ gives $ϕ \nabla \cdot π_{A}$ . This yields the displayed $H$ ; varying the scalar multiplier gives the Gauss constraint. If $\nabla \cdot A = 0$ , then $\nabla \cdot π_{A} = ϵ_{0} Δ ϕ$ , so the constraint is $- Δ ϕ = ρ / ϵ_{0}$ .

1.2givenalgebra

Integrating this equation over the periodic box at any fixed $t$ gives $0 = L^{- 3} \int (- Δ ϕ) d^{3} x = L^{- 3} ϵ_{0}^{- 1} \int ρ d^{3} x$ , since each periodic first derivative has equal boundary values on opposite faces. Thus neutrality is necessary at each time. For neutral data, fix a compact interval $K \subset I$ and integers $r, M \geq 0$ . Differentiating the Fourier coefficient under the integral gives $\partial_{t}^{r} {\hat{ρ}}_{k} (t) = L^{- 3} \int_{[0, L]^{3}} \partial_{t}^{r} ρ (x, t) e^{- 2 π i k \cdot x / L} d^{3} x$ . For $k \neq 0$ , repeated periodic integration by parts $M$ times in a coordinate $j$ with $∣ k_{j} ∣ \geq ∣ k ∣ / \sqrt{3}$ gives $∣ \partial_{t}^{r} {\hat{ρ}}_{k} (t) ∣ \leq C_{r, M, K} (1 + ∣ k ∣)^{- M}$ uniformly for $t \in K$ ; boundedness of the finitely many low modes is absorbed in the constant. Since ${\hat{ϕ}}_{k} = {\hat{ρ}}_{k} / (ϵ_{0} ∣ 2 π k / L ∣^{2})$ for $k \neq 0$ , a mixed derivative $\partial_{t}^{r} \partial_{x}^{α}$ of its Fourier series is bounded termwise on $K \times T_{L}^{3}$ by a constant times $∣ k ∣^{∣ α ∣ - M - 2}$ . Taking $M > ∣ α ∣ + 1$ makes this series absolutely and uniformly summable over $Z^{3} ∖ {0}$ . Thus every mixed derivative series converges uniformly on compact time intervals, and the Fourier construction is jointly smooth.

2.1A1A2F1step 1.2given

Under Countable Choice [A2], apply [F1] after rescaling $x = L y$ : the characters $e^{2 π i k \cdot x / L}$ form a complete basis for $L^{- 3} d^{3} x$ . The coefficient formula then gives $∣ 2 π k / L ∣^{2} {\hat{ϕ}}_{k} (t) = {\hat{ρ}}_{k} (t) / ϵ_{0}$ for $k \neq 0$ and the zero mode is zero. Step 1.2 permits termwise differentiation in space and time to every order, so the resulting real jointly smooth periodic $ϕ$ solves $- Δ ϕ = ρ / ϵ_{0}$ at every $t$ . Reality follows from ${\hat{ρ}}_{- k} (t) = \overline{{\hat{ρ}}_{k} (t)}$ .

3.1A1F1step 2.1given∎

At any fixed $t$ , if two zero-mean solutions existed, their difference $u$ would be smooth, periodic, satisfy $Δ u = 0$ , and have ${\hat{u}}_{0} = 0$ . Periodic integration by parts in each Fourier coefficient gives ${\hat{Δ u}}_{k} = - ∣ 2 π k / L ∣^{2} {\hat{u}}_{k}$ ; since $Δ u = 0$ , every nonzero-mode coefficient vanishes, and the zero-mode coefficient vanishes by the zero-mean condition. Completeness of the Fourier basis [F1] therefore gives $u = 0$ in $L^{2}$ , hence everywhere by continuity. Parseval applied to $\nabla ϕ$ and to $ϕ, ρ$ yields $\int ∣ \nabla ϕ ∣^{2} = L^{3} \sum_{k \neq 0} ∣ 2 π k / L ∣^{2} ∣ {\hat{ϕ}}_{k} ∣^{2}, \int ρ ϕ = L^{3} \sum_{k \neq 0} \overline{{\hat{ρ}}_{k}} {\hat{ϕ}}_{k} .$ Substituting the coefficients from step 2.1 proves the displayed positive-energy identity.

Gauss multiplier and smooth periodic Coulomb solver

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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