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Far-field dipole expansion of a compact conserved current with a uniform explicit remainder

Physical scope

A far-field dipole estimate for the magnetostatic field of a compact smooth conserved current, uniform for $r \geq 2 a$ , with an explicit remainder and absolute constants. The leading term is not an exact field; nothing is asserted inside $r < 2 a$ or at the support, and no experiment, apparatus or uncertainty is involved.

Statement

Assume Countable Choice ( ${A C}_{ω}$ ), with the exact meaning of The Axiom of Countable Choice ( ${A C}_{ω}$ ). Let $J \in C_{c}^{\infty} (Ω; R^{3})$ with $div J = 0$ and $supp J \subset B_{a} (0)$ for some $a > 0$ , let $A$ be the vector potential of Coulomb-gauge Newtonian vector potential for a compact smooth conserved current and $B = curl A$ its magnetostatic field, and let $m = \frac{1}{2} \int_{R^{3}} y \times J (y) d y$ be the magnetic moment of The magnetic dipole moment of a compact conserved current and of the declared ideal thin loop(a). Then for $r = ∣ x ∣ \geq 2 a$ , uniformly in the unit vector $n = x / r$ , $A (x) = \frac{μ_{0}}{4 π} \frac{m \times x}{r^{3}} + O ⁣ (\frac{μ_{0} a^{2} ∥ J ∥_{1}}{r^{3}}), B (x) = \frac{μ_{0}}{4 π} \frac{3 n (m \cdot n) - m}{r^{3}} + O ⁣ (\frac{μ_{0} a^{2} ∥ J ∥_{1}}{r^{4}}),$ where $∥ J ∥_{1} = \int_{R^{3}} ∣ J ∣ d λ_{3}$ and the constants implicit in the $O$ terms are absolute: one constant works for all $r \geq 2 a$ and all unit $n$ . This is a controlled two-term far-field estimate: nothing is asserted for $r < 2 a$ or at the support, no higher multipole is defined here, and the leading dipole term is not claimed to be exact.

Facts & Assumptions

Given: Assume ${A C}_{ω}$ ; let $J \in C_{c}^{\infty} (R^{3}; R^{3})$ with $div J = 0$ and $supp J \subset B_{a} (0)$ ; let $A$ , $B = curl A$ be as in Coulomb-gauge Newtonian vector potential for a compact smooth conserved current and $m$ be as in The magnetic dipole moment of a compact conserved current and of the declared ideal thin loop(a).

[F1]

The vector potential satisfies $A_{i} = μ_{0} N (J_{i}) = (μ_{0} / 4 π) \int_{R^{3}} J_{i} (y) ∣ x - y ∣^{- 1} d y$ componentwise and $B = curl A$ is its magnetostatic field (Coulomb-gauge Newtonian vector potential for a compact smooth conserved current, clauses (a)-(c)).

[F2]

The magnetic moment of the compact smooth current is the source moment $m = \frac{1}{2} \int_{R^{3}} y \times J (y) d y$ of The magnetic dipole moment of a compact conserved current and of the declared ideal thin loop(a), computed componentwise as a Riemann integral over a nondegenerate closed bounding rectangle $Q \supset supp J$ of the continuous compactly supported integrand, with the cross product of The cross product in $R^{3}$ ; clause (c) records that $m$ is a model quantity and that its far-field interpretation is the content of this theorem under its hypotheses.

[F3]

Under the stated hypotheses, with $m = \frac{1}{2} \int_{R^{3}} y \times J (y) d y$ (Lebesgue), $r = ∣ x ∣ \geq 2 a$ and $n = x / r$ , one has uniformly in $n$ : $A = \frac{μ_{0}}{4 π} \frac{m \times x}{r^{3}} + O (μ_{0} a^{2} ∥ J ∥_{1} r^{- 3})$ and $B = \frac{μ_{0}}{4 π} \frac{3 n (m \cdot n) - m}{r^{3}} + O (μ_{0} a^{2} ∥ J ∥_{1} r^{- 4})$ , with absolute constants and $∥ J ∥_{1} = \int ∣ J ∣$ (Magnetic dipole expansion with an explicit uniform remainder).

[F4]

If $f$ is a bounded Borel real function on a nondegenerate closed rectangle $Q$ and $f$ is Riemann integrable there, then $f$ is Lebesgue integrable with the same integral; equivalently, for smooth compactly supported integrands the Riemann integral over a bounding rectangle equals the Lebesgue integral of their zero extension (Riemann–Lebesgue comparison for distribution test integrands).

[F5]

${A C}_{ω}$ is the countable-choice principle assumed here and inherited by the cited comparison and measure interfaces (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct transport of the explicit uniform-remainder lemma to the framework's objects, with a Riemann-Lebesgue comparison for the two displayed moments

1.1givenF1F2F3F4

The hypotheses of [F3] are exactly the hypotheses stated here for $J$ (smooth, compactly supported, divergence-free, contained in $B_{a} (0)$ ), and the objects of [F3] are the objects of this theorem: $A = \frac{μ_{0}}{4 π} \int J (y) ∣ x - y ∣^{- 1} d y$ componentwise and $B = curl A$ by [F1]. The moment of [F2] agrees with the Lebesgue moment of [F3]: each component $y \mapsto (y \times J (y))_{i}$ is smooth and compactly supported, hence bounded Borel and Riemann integrable on any nondegenerate closed rectangle $Q$ containing $supp J$ ; by [F4] its Riemann integral over $Q$ equals its Lebesgue integral over $Q$ , and since the integrand vanishes off $supp J \subset Q$ this equals the Lebesgue integral over $R^{3}$ . Hence $\frac{1}{2} \int_{R^{3}} y \times J (y) d y$ is the same vector in both items.

2.1step 1.1F3

Substituting the identifications of step 1.1 into the conclusion of [F3] gives, for $r = ∣ x ∣ \geq 2 a$ and uniformly in the unit vector $n = x / r$ , the displayed estimates $A (x) = \frac{μ_{0}}{4 π} \frac{m \times x}{r^{3}} + O (μ_{0} a^{2} ∥ J ∥_{1} r^{- 3})$ and $B (x) = \frac{μ_{0}}{4 π} \frac{3 n (m \cdot n) - m}{r^{3}} + O (μ_{0} a^{2} ∥ J ∥_{1} r^{- 4})$ for the framework's $A$ and $B$ , with the lemma's uniformity and absolute constants and with $∥ J ∥_{1} = \int_{R^{3}} ∣ J ∣ d λ_{3}$ as in the Statement.

3.1step 1.1step 2.1F3F4F5∎

The estimates of step 2.1 are the content of the Statement; no assertion is made for $r < 2 a$ or at the support, no higher multipole is defined, and the leading dipole term is not claimed to be exact. Countable Choice is used only through the Riemann-Lebesgue comparison [F4] and the measure and Newtonian interfaces [F5] of the cited suppliers; no stronger choice principle occurs.

Source notes

Griffiths, Section 5.4.3, printed pp. 242-247, expands the vector potential and gives the dipole field (5.85)-(5.87) with the coordinate-free form and its delta term (5.90); Tong, Sections 3.3.1-3.3.2, printed pp. 54-57, gives the loop far field and the general moment (3.24); Landau-Lifshitz, Section 44, printed pp. 103-105, gives the first-order average field in Gaussian units, re-expressed here in SI. The uniform remainder is carried by the explicit mathematical supplier Magnetic dipole expansion with an explicit uniform remainder, not by the textbook expansions, and the Riemann moment of the definition is reconciled with the supplier's Lebesgue moment in step 1.1.

Far-field dipole expansion of a compact conserved current with a uniform explicit remainder

Statement

Facts & Assumptions

Proof

Source notes

Depends on

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Sources