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Uniform Coulomb dipole expansion of a compactly supported charge density with an explicit remainder

Physical scope

The exterior potential of a compactly supported $L^{1}$ charge density in classical electrostatics: for $∣ x ∣ \geq 2 a$ the localised Newtonian potential equals the monopole term plus the dipole term plus an explicit remainder bounded by $16 a^{2} ∥ ρ ∥_{1} / (4 π ε_{0} ∣ x ∣^{3})$ , uniformly in the exterior region. Only the potential is expanded; no derivative estimate of the remainder, no field expansion and no value on or inside the support are claimed.

Statement

Assume the Axiom of Countable Choice ( ${A C}_{ω}$ ), inherited from the cited Lebesgue and Newtonian interfaces (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $a > 0$ , let $ρ \in L^{1} (R^{3})$ be real and vanish $λ_{3}$ -a.e. outside the closed ball ${\overline{B}}_{a} (0)$ (Open ball, closed ball and sphere in a metric space, Measure-null sets and almost-everywhere statements relative to a measure), let $Q$ and $p$ be the total charge and dipole moment of $ρ$ about the chart origin (Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution (a)), let $∥ ρ ∥_{1} : = \int_{R^{3}} ∣ ρ ∣ d λ_{3}$ , and let $φ : = ε_{0}^{- 1} N ρ$ , where $N ρ$ is the Newtonian potential of Newtonian potential of compactly supported data and $ε_{0}$ is the constant of Framework-local SI units, the vacuum constants and the exact/measured status of their values. Then:

(a) [Uniform monopole-dipole expansion] for every $x$ with $r : = ∣ x ∣ \geq 2 a$ the integral $\int_{R^{3}} ρ (y) ∣ x - y ∣^{- 1} d λ_{3} (y)$ is absolutely convergent and equals $4 π N ρ (x)$ , and

$φ (x) = \frac{Q}{4 π ε_{0} r} + \frac{p \cdot x}{4 π ε_{0} r^{3}} + R (x), ∣ R (x) ∣ \leq \frac{16 a^{2} ∥ ρ ∥_{1}}{4 π ε_{0} r^{3}},$

the bound being uniform over the whole exterior region ${x : ∣ x ∣ \geq 2 a}$ ;

(b) [Leading-term form] consequently, for $r \geq 2 a$ ,

$∣ φ (x) - \frac{Q}{4 π ε_{0} r} ∣ \leq \frac{∣ p ∣}{4 π ε_{0} r^{2}} + \frac{16 a^{2} ∥ ρ ∥_{1}}{4 π ε_{0} r^{3}},$

and if $Q = 0$ and $p \neq 0$ , then for every unit direction $n \in R^{3}$ , $∣ n ∣ = 1$ , with $p \cdot n \neq 0$ and every $r \geq \max (2 a, 32 a^{2} ∥ ρ ∥_{1} / ∣ p \cdot n ∣)$ the dipole term is the leading term in the sense that $φ (r n) = \frac{p \cdot n}{4 π ε_{0} r^{2}} (1 + δ)$ with $∣ δ ∣ \leq \frac{1}{2}$ , so $∣ φ (r n) ∣ \geq \frac{∣ p \cdot n ∣}{8 π ε_{0} r^{2}}$ and the remainder bound decays one power of $r$ faster than the dipole term;

(c) [Physical reading] by The localized Newtonian potential and the ideal point-charge Coulomb field (c) the function $φ$ is finite $λ_{3}$ -a.e., locally integrable, of class $C^{\infty}$ and harmonic off ${\overline{B}}_{a} (0)$ , and satisfies $- Δ T_{φ} = T_{ρ / ε_{0}}$ in $D^{'} (R^{3})$ ; the displayed expansion is therefore the exterior expansion of the potential of the compactly supported source $ρ$ in the fixed-time reading of The static and pure electrostatic regimes and their fixed-time reading (e), and in the smooth branch (The localized Newtonian potential and the ideal point-charge Coulomb field (a)) the field $E : = - \nabla φ$ of The electric and magnetic fields as fixed-time spatial vector fields realises the pure electrostatic regime of The static and pure electrostatic regimes and their fixed-time reading (c) with source $ρ$ ;

(d) [Scope and exclusions] the hypotheses are a real compactly supported $L^{1}$ density, the region $r \geq 2 a$ , and the framework's constants; $ρ$ need not be continuous or bounded; the statement concerns $φ$ itself, and no derivative estimate for $R$ , no expansion of the field $E$ , and no assertion on $∣ x ∣ < 2 a$ or on the support of $ρ$ is claimed.

Facts & Assumptions

Given: Assume ${A C}_{ω}$ ; let $a > 0$ , let $ρ \in L^{1} (R^{3})$ be real and vanish $λ_{3}$ -a.e. outside ${\overline{B}}_{a} (0)$ , and put $φ : = ε_{0}^{- 1} N ρ$ with $ε_{0} > 0$ as in Framework-local SI units, the vacuum constants and the exact/measured status of their values.

[F1]

For a measurable compactly supported $f$ , the Newtonian potential is $N f (x) = \int_{R^{n}} Φ (x - y) f (y) d y$ at each $x$ where the integral is absolutely finite (Newtonian potential of compactly supported data).

[F2]

With the positive-minus-Laplacian normalisation in dimension three, the fundamental solution is $Φ (z) = 1 / (4 π ∣ z ∣)$ away from the pole (The localized Newtonian potential and the ideal point-charge Coulomb field).

[F3]

If $ρ \in L^{1} (R^{3})$ is real and compactly supported, then $N ρ$ is finite $λ_{3}$ -a.e., belongs to $L_{l o c}^{1} (R^{3})$ , is $C^{\infty}$ and harmonic off the support of $ρ$ , and $- Δ T_{ε_{0}^{- 1} N ρ} = T_{ρ / ε_{0}}$ in $D^{'} (R^{3})$ ; if in addition $ρ$ is $C^{1}$ with compact support in ${\overline{B}}_{R}$ , then $φ = ε_{0}^{- 1} N ρ \in C^{2} (R^{3})$ satisfies $- Δ φ = ρ / ε_{0}$ , $div E = ρ / ε_{0}$ and $curl E = 0$ for $E = - \nabla φ$ (The localized Newtonian potential and the ideal point-charge Coulomb field).

[F4]

The framework's electric constant is $ε_{0} : = (μ_{0} c^{2})^{- 1} > 0$ with $μ_{0} > 0$ measured and $c > 0$ ; the potential carries the volt (Framework-local SI units, the vacuum constants and the exact/measured status of their values).

[F5]

For $ρ \in L^{1} (R^{3})$ vanishing $λ_{3}$ -a.e. outside ${\overline{B}}_{a} (0)$ , the total charge $Q = \int ρ d λ_{3}$ and dipole moment $p = \int y ρ (y) d λ_{3} (y)$ about the chart origin are absolutely convergent (Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution).

[F6]

If $ρ \in L^{1} (R^{3})$ vanishes $λ_{3}$ -a.e. outside ${y : ∣ y ∣ \leq a}$ and $r = ∣ x ∣ \geq 2 a > 0$ , then $\int_{R^{3}} ρ (y) ∣ x - y ∣^{- 1} d y = Q / r + p \cdot x / r^{3} + R_{0} (x)$ with $∣ R_{0} (x) ∣ \leq 16 a^{2} ∥ ρ ∥_{1} / r^{3}$ (Monopole-dipole expansion with an explicit remainder for compactly supported $L^{1}$ densities).

[F7]

If $f \in L^{1}$ and $α \in R$ , then $\int α f d λ_{3} = α \int f d λ_{3}$ (The Lebesgue integral is linear on $L^{1} (μ)$ ).

[F8]

$∣ ⟨ u, w ⟩ ∣ \leq ∣ u ∣ ∣ w ∣$ for the Euclidean inner product, and $∣ u + w ∣ \leq ∣ u ∣ + ∣ w ∣$ for the Euclidean norm (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

[F9]

${\overline{B}}_{a} (0) = {y : ∣ y ∣ \leq a}$ ; and if $N$ is a norm on a real vector space $V$ , then $∣ N (u) - N (w) ∣ \leq N (u - w)$ for all $u, w \in V$ (Open ball, closed ball and sphere in a metric space, The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ ).

[F10]

If $f, g \in L^{1} (λ_{3})$ agree $λ_{3}$ -a.e., their Lebesgue integrals agree, and $∣ \int f ∣ \leq \int ∣ f ∣$ (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree, The modulus of an integral is bounded by the integral of the modulus).

[F11]

In the static regime with $J = 0$ and $B = 0$ the smooth Maxwell equations reduce to $div E = ρ / ε_{0}$ , $curl E = 0$ , and the fixed-time dictionary identifies the spatial equations with the four-dimensional weak reading (The static and pure electrostatic regimes and their fixed-time reading).

Proof

technique · direct

1.1F1F2F8F9F10given

Fix $x$ with $r = ∣ x ∣ \geq 2 a$ and establish the absolute convergence and the Coulomb identification. For $λ_{3}$ -a.e. $y$ , either $ρ (y) = 0$ or $y \in {\overline{B}}_{a} (0)$ ; in the latter case $∣ y ∣ \leq a$ by [F9], so $∣ x - y ∣ \geq ∣ x ∣ - ∣ y ∣ \geq r - a \geq r / 2 > 0$ by the reverse triangle inequality for the Euclidean norm [F9] and [F8], and hence $∣ x - y ∣^{- 1} \leq 2 / r$ ; the integrand is therefore dominated $λ_{3}$ -a.e. by $2 ∣ ρ ∣ / r \in L^{1}$ , which gives absolute convergence by [F10]. Since $ρ$ vanishes a.e. off ${\overline{B}}_{a} (0)$ and the kernel $y \mapsto ∣ x - y ∣^{- 1}$ is continuous on that ball, replacing $ρ$ by its a.e.-equal representative that is zero off ${\overline{B}}_{a} (0)$ does not change the integral [F10], and the defining formula of the Newtonian potential $N ρ (x) = \int Φ (x - y) ρ (y) d y$ together with $Φ (z) = 1 / (4 π ∣ z ∣)$ for $n = 3$ gives $\int ρ (y) ∣ x - y ∣^{- 1} d λ_{3} (y) = 4 π N ρ (x)$ for every admissible $x$ [F1, F2].

2.1step 1.1F4F5F6F7

Apply the monopole-dipole remainder lemma and scale to obtain (a). By [F6] with the same $a$ and the same a.e.-support hypothesis, $\int ρ (y) ∣ x - y ∣^{- 1} d λ_{3} (y) = Q / r + p \cdot x / r^{3} + R_{0} (x)$ with $∣ R_{0} (x) ∣ \leq 16 a^{2} ∥ ρ ∥_{1} / r^{3}$ for $r \geq 2 a$ , where $Q$ and $p$ are the moments of [F5]; multiplying the identity by the constant $(4 π ε_{0})^{- 1}$ , which factors out of the integral by [F7] and is finite and positive by [F4], gives $φ (x) = Q / (4 π ε_{0} r) + p \cdot x / (4 π ε_{0} r^{3}) + R (x)$ with $R : = (4 π ε_{0})^{- 1} R_{0}$ and the displayed uniform bound.

3.1step 2.1F8F10algebra

Prove the leading-term estimate (b). By Cauchy-Schwarz and $∣ x ∣ = r$ [F8], $∣ p \cdot x ∣ \leq ∣ p ∣ r$ , so $∣ p \cdot x ∣ / (4 π ε_{0} r^{3}) \leq ∣ p ∣ / (4 π ε_{0} r^{2})$ , and the triangle inequality [F10] applied to step 2.1 gives the displayed bound. If $Q = 0$ and $p \neq 0$ , fix $n \in R^{3}$ with $∣ n ∣ = 1$ and $p \cdot n \neq 0$ and $r \geq \max (2 a, 32 a^{2} ∥ ρ ∥_{1} / ∣ p \cdot n ∣)$ ; then the dipole term equals $p \cdot n / (4 π ε_{0} r^{2})$ and the remainder obeys $∣ R ∣ \leq 16 a^{2} ∥ ρ ∥_{1} / (4 π ε_{0} r^{3}) \leq \frac{1}{2} ∣ p \cdot n ∣ / (4 π ε_{0} r^{2})$ , so writing $δ : = R / (p \cdot n / (4 π ε_{0} r^{2}))$ gives $∣ δ ∣ \leq \frac{1}{2}$ and $φ (r n) = \frac{p \cdot n}{4 π ε_{0} r^{2}} (1 + δ)$ with $∣ φ (r n) ∣ \geq \frac{∣ p \cdot n ∣}{8 π ε_{0} r^{2}}$ .

4.1step 3.1F3F10F11

Read the physical content (c). By [F3], applied to the representative that vanishes off ${\overline{B}}_{a} (0)$ , the function $N ρ$ is finite $λ_{3}$ -a.e., locally integrable, $C^{\infty}$ and harmonic off the support of that representative, hence off ${\overline{B}}_{a} (0)$ , and $- Δ T_{ε_{0}^{- 1} N ρ} = T_{ρ / ε_{0}}$ in $D^{'} (R^{3})$ ; these properties are unchanged by the a.e. choice of representative [F10], and $φ = ε_{0}^{- 1} N ρ$ is related to $ρ$ exactly as in the fixed-time dictionary of [F11]. In the smooth case of [F3], $φ$ is $C^{2}$ on $R^{3}$ with $- Δ φ = ρ / ε_{0}$ , and $E = - \nabla φ$ satisfies $div E = ρ / ε_{0}$ and $curl E = 0$ , which is the pure electrostatic regime of [F11] with source $ρ$ .

5.1step 4.1F4∎

Record the scope (d) and complete the proof. The hypotheses used were exactly the real $L^{1}$ density vanishing $λ_{3}$ -a.e. off ${\overline{B}}_{a} (0)$ , the exterior region $r \geq 2 a$ and the framework's constants [F4]; no continuity or boundedness of $ρ$ was assumed, the argument differentiates no remainder and expands no field, and no assertion inside $∣ x ∣ < 2 a$ or on the support of $ρ$ was made, so the exclusions of (d) are respected.

Uniform Coulomb dipole expansion of a compactly supported charge density with an explicit remainder

Statement

Facts & Assumptions

Proof

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