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A dipole node does not control the relative multipole error

Physical scope

A constructed smooth compact-source example in the retarded classical Maxwell model: at one direction the electric-dipole radiation vanishes while a higher source moment gives nonzero radiation. It is a conditional model example, not a measured antenna pattern or empirical claim.

Example

There are jointly smooth charge and current densities of compact support in both space and time for which the electric dipole acceleration is parallel to $n = e_{3}$ , so the electric-dipole radiation amplitude at $n$ vanishes, while the exact retarded radiation amplitude at $n$ is nonzero. Thus an absolute source-scale error bound for the multipole expansion gives no relative error bound at a dipole node.

Example references

Finite multipole expansion with an absolute dipole-power error

Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution

Facts & Assumptions

Given: Countable Choice and the SI retarded Maxwell model with no independent homogeneous field; $a, c > 0$ with units $m$ and $m s^{- 1}$ ; $u \in R$ a fixed retarded time in seconds; independent positive amplitudes $α, β$ with units $C m^{- 2} s^{- 2}$ and $C m^{- 1} s^{- 3}$ ; a dimensionless nonnegative smooth spatial function $f$ supported in a smaller ball and equal to one on a still smaller ball; dimensionless compactly supported smooth time cutoffs; and the source formulas in the Verification below.

[A1]

Assume ${A C}_{ω}$ for the framework's Lebesgue and surface measures. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

For $0 < r_{0} < R_{0}$ there is a smooth $η : R^{3} \to [0, 1]$ equal to one on ${\overline{B}}_{r_{0}} (0)$ and supported in $B_{R_{0}} (0)$ . (A smooth bump between concentric Euclidean balls)

[F2]

For a compact set contained in an open set there is a smooth compactly supported cutoff equal to one on a neighborhood of the compact set. (Test function cutoffs and euclidean localization)

[F3]

At each time the dipole moment about the laboratory chart origin is $p (t) = \int_{B_{a}} y ρ (y, t) d y$ , with units $C m$ . Its time dependence here is that of the controlled-dipole theorem: for jointly smooth conserved sources supported in a fixed ball, $p^{''} (u) = \int_{B_{a}} J_{t} (y, u) d y$ . (Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution, Finite multipole expansion with an absolute dipole-power error)

[F4]

Mixed partial derivatives of a smooth function commute, and the divergence theorem applies to a smooth compactly supported vector field on a ball containing its support. (Clairaut--Schwarz theorem for continuous second partial derivatives, Divergence on a bounded C1 Euclidean domain)

[F5]

A Euclidean ball has positive finite Lebesgue measure; a nonnegative function equal to one on such a ball has strictly positive integral. (Euclidean balls have positive finite Lebesgue measure, Monotonicity and nonnegative homogeneity of the nonnegative integral)

[F6]

For jointly smooth conserved SI sources $ρ, J$ on $R^{3} \times R$ supported in a fixed spatial ball, in the adopted retarded Maxwell model, $E_{r a d} (n, u) = \lim_{r \to \infty} r E (r n, u + r / c) = \frac{μ_{0}}{4 π} n \times (n \times V (n, u))$ , where $V (n, u) = \int_{B_{a}} J_{t} (y, u + n \cdot y / c) d y$ and $∣ n ∣ = 1$ . The limit is uniform on every nondegenerate compact retarded-time interval and in direction. Here $μ_{0} > 0$ has units $N A^{- 2}$ , $ε_{0} = (μ_{0} c^{2})^{- 1}$ , $E : R^{3} \times R \to R^{3}$ is smooth with units $V m^{- 1}$ , and $E_{r a d}$ has units $V$ . (Transverse far field with a uniform remainder)

Verification

technique · construct one smooth electric-current pulse and one smooth divergence-free circulating pulse, then evaluate their dipole and shifted-current moments in the axial direction

1.1givenF1F2F5constructA1

Choose $0 < r_{0} < R_{0} < a$ and apply [F1] to obtain a dimensionless $f \in C^{\infty} (R^{3})$ with $0 \leq f \leq 1$ , $f = 1$ on ${\overline{B}}_{r_{0}} (0)$ and $supp f \subset B_{R_{0}} (0)$ . By [F5], $F : = \int_{R^{3}} f (y) d y \geq λ_{3} (B_{r_{0}}) > 0$ . By [F2], choose dimensionless $χ_{G}, χ_{H} \in C_{c}^{\infty} (R)$ with $χ_{G} = 1$ near $u$ and $χ_{H} = 1$ on a neighborhood of $[u - a / c, u + a / c]$ . Define $G (t) : = α χ_{G} (t) (t - u)^{2}$ and $H (t) : = β χ_{H} (t) (t - u)^{3}$ , with units $C m^{- 2}$ and $C m^{- 1}$ , respectively. Then $G^{''} (u) = 2 α$ and $H^{'''} (s) = 6 β$ for every $s \in [u - a / c, u + a / c]$ .

2.1step 1.1F4algebraA1

Let $e_{1} = (1, 0, 0)$ and $e_{3} = (0, 0, 1)$ , and define $ρ (y, t) : = - G (t) \partial_{3} f (y), J (y, t) : = G^{'} (t) f (y) e_{3} + H^{'} (t) (- \partial_{3} f (y) e_{1} + \partial_{1} f (y) e_{3}) .$ These maps $ρ : R^{3} \times R \to R$ and $J : R^{3} \times R \to R^{3}$ have units $C m^{- 3}$ and $A m^{- 2}$ : both $G^{'} f$ and $H^{'} \partial_{i} f$ have the latter units. They are jointly smooth and supported in $supp f \times (supp G \cup supp G^{'} \cup supp H^{'})$ , a compact subset of $B_{a} \times R$ . Their derivatives satisfy $\partial_{t} ρ + {div}_{y} J = - G^{'} (t) \partial_{3} f + G^{'} (t) \partial_{3} f + H^{'} (t) (- \partial_{1} \partial_{3} f + \partial_{3} \partial_{1} f) = 0$ by [F4], so the source is admissible for the controlled multipole theorem on any nondegenerate compact interval containing $u$ .

3.1step 1.1step 2.1F3F4algebraA1

The current derivative is $J_{t} = G^{''} f e_{3} + H^{''} (- \partial_{3} f e_{1} + \partial_{1} f e_{3})$ . Each integral $\int_{R^{3}} \partial_{i} f d y$ is zero by the divergence theorem applied to $f e_{i}$ on a ball containing $supp f$ , since its boundary flux vanishes. Thus [F3] gives $p^{''} (u) = \int J_{t} (y, u) d y = G^{''} (u) F e_{3} = 2 α F e_{3}$ , which is nonzero and parallel to $n = e_{3}$ . Therefore $n \times (n \times p^{''} (u)) = 0$ .

3.2step 1.1step 2.1F4F5F6algebraA1

For $n = e_{3}$ , the $e_{1}$ component of the exact shifted current moment is $V_{1} (e_{3}, u) = - \int_{B_{a}} H^{''} (u + y_{3} / c) \partial_{3} f (y) d y = \frac{1}{c} \int_{B_{a}} f (y) H^{'''} (u + y_{3} / c) d y = \frac{6 β F}{c} > 0.$ For the middle equality, apply [F4] to $f (y) H^{''} (u + y_{3} / c) e_{3}$ on a ball containing its support. Its boundary flux is zero and its divergence is $H^{''} (u + y_{3} / c) \partial_{3} f + f H^{'''} (u + y_{3} / c) / c$ . Also $∣ y_{3} ∣ \leq a$ on the support of $f$ , so $H^{'''} = 6 β$ there by step 1.1. The term $G^{''} f e_{3}$ contributes no $e_{1}$ component. Hence $V (e_{3}, u)$ has a nonzero component perpendicular to $e_{3}$ .

4.1step 3.1step 3.2F3F6algebraA1∎

By [F6], $E_{r a d} (e_{3}, u) = \frac{μ_{0}}{4 π} e_{3} \times (e_{3} \times V (e_{3}, u)) \neq 0$ , whereas the dipole coefficient $\frac{μ_{0}}{4 π} e_{3} \times (e_{3} \times p^{''} (u))$ is zero by step 3.1. Therefore the absolute multipole remainder is nonzero at a direction where the dipole reference amplitude vanishes; no finite uniform relative-error bound with that dipole amplitude as denominator can hold.

A dipole node does not control the relative multipole error

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources