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Origin shift of the source moments: Q, the dipole p - Qb, the Cartesian expansion and the traceless quadrupole

Physical scope

The exact reference-point dependence of the source moments of a static charge distribution: the total charge is reference-point independent, the dipole moment shifts by $- Q b$ , every Cartesian moment has the displayed subset expansion, and the traceless quadrupole is reference-point independent exactly when the total charge and dipole vanish. The claim is a moment identity, not a statement about fields, forces, dynamics or empirical fact.

Statement

For the density-integral branch assume Countable Choice ( ${A C}_{ω}$ ) through the Lebesgue interface (The Axiom of Countable Choice ( ${A C}_{ω}$ )); the finite-point branch of all identities below is choice-free. Let $a > 0$ and let $ρ \in L^{1} (R^{3})$ be real and vanish $λ_{3}$ -a.e. outside ${\overline{B}}_{a} (0)$ (Open ball, closed ball and sphere in a metric space), or let $(x_{k}, q_{k})_{k = 1}^{N}$ be a finite system of stationary point charges, with the source moments of Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution branches (a) and (c) respectively. For $b \in R^{3}$ define the moments about the reference point $b$ by replacing, in each integrand of Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution (a) and in each summand of (c), the vector $y$ by $y - b$ (respectively $x_{k}$ by $x_{k} - b$ ):

$Q^{(b)} : = \int_{R^{3}} ρ d λ_{3}, p^{(b)} : = \int_{R^{3}} (y - b) ρ (y) d λ_{3} (y),$

$M_{i_{1} \dots i_{l}}^{(b)} : = \int_{R^{3}} \prod_{r = 1}^{l} (y - b)_{i_{r}} ρ (y) d λ_{3} (y) (l \geq 1), M^{(b, 0)} : = Q^{(b)},$

$Q_{i j}^{(b)} : = \int_{R^{3}} (3 (y - b)_{i} (y - b)_{j} - ∣ y - b ∣^{2} δ_{i j}) ρ (y) d λ_{3} (y),$

and analogously with the finite sums $M_{i_{1} \dots i_{l}}^{(b)} = \sum_{k = 1}^{N} q_{k} \prod_{r = 1}^{l} (x_{k} - b)_{i_{r}}$ $(l \geq 1)$ and $Q_{i j}^{(b)} = \sum_{k = 1}^{N} q_{k} (3 (x_{k} - b)_{i} (x_{k} - b)_{j} - ∣ x_{k} - b ∣^{2} δ_{i j})$ in the point-charge branch; for $N = 0$ the sums are the empty sum $0$ . Every displayed integral is absolutely convergent: since $ρ = 0$ $λ_{3}$ -a.e. off ${\overline{B}}_{a} (0)$ , the estimates $∣ (y - b)_{i_{1}} \dots (y - b)_{i_{l}} ∣ \leq (∣ y ∣ + ∣ b ∣)^{l} \leq (a + ∣ b ∣)^{l}$ and $∣ 3 (y - b)_{i} (y - b)_{j} - ∣ y - b ∣^{2} δ_{i j} ∣ \leq 4 (∣ y ∣ + ∣ b ∣)^{2} \leq 4 (a + ∣ b ∣)^{2}$ hold $λ_{3}$ -a.e. on the support, so each integrand is dominated $λ_{3}$ -a.e. by a finite constant times $∣ ρ ∣ \in L^{1}$ (Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution (a), The modulus of an integral is bounded by the integral of the modulus, Monotonicity and nonnegative homogeneity of the nonnegative integral, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ). Then:

(a) $Q^{(b)} = Q$ for every $b$ ;

(b) $p^{(b)} = p - Q b$ , so the dipole moment is independent of the reference point exactly when $Q = 0$ ;

$M_{i_{1} \dots i_{l}}^{(b)} = \sum_{S \subseteq {1, \dots, l}} (- 1)^{∣ S ∣} (\prod_{r \in S} b_{i_{r}}) M_{S^{c}},$

where for a subset $S \subseteq {1, \dots, l}$ we write $S^{c} : = {1, \dots, l} ∖ S$ and $M_{S^{c}}$ for the Cartesian moment whose index list is $(i_{r})_{r \in S^{c}}$ in increasing order of $r$ , with the empty index list read as $M_{\emptyset} : = Q$ ; in particular

$M_{i j}^{(b)} = M_{i j} - b_{i} p_{j} - b_{j} p_{i} + b_{i} b_{j} Q;$

(d) $Q_{i j}^{(b)} = Q_{i j} - 3 (b_{i} p_{j} + b_{j} p_{i}) + 2 (b \cdot p) δ_{i j} + (3 b_{i} b_{j} - ∣ b ∣^{2} δ_{i j}) Q$ ; consequently $Q_{i j}^{(b)} = Q_{i j}$ for every $b$ and all $i, j$ if and only if $Q = 0$ and $p = 0$ ;

(e) [Scope] this is an exact identity between the declared source moments of Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution; it asserts nothing about fields, forces, dynamics or nature, and the reference point is a choice of coordinate origin, not an operation performed on the source.

Facts & Assumptions

Given: Assume ${A C}_{ω}$ for the density branch, let $a > 0$ , $b \in R^{3}$ , and let either a real $ρ \in L^{1} (R^{3})$ vanishing $λ_{3}$ -a.e. off ${\overline{B}}_{a} (0)$ , with source moments as in Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution (a), or a finite system $(x_{k}, q_{k})_{k = 1}^{N}$ of stationary point charges with source moments as in Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution (c), be given. All moments are read about the chart origin of the laboratory frame.

[F1]

For $ρ \in L^{1} (R^{3})$ vanishing $λ_{3}$ -a.e. outside ${\overline{B}}_{a} (0)$ , the moments $Q = \int ρ$ , $p = \int y ρ (y) d λ_{3} (y)$ , $M_{i_{1} \dots i_{l}} = \int y_{i_{1}} \dots y_{i_{l}} ρ d λ_{3}$ and $Q_{i j} = \int (3 y_{i} y_{j} - ∣ y ∣^{2} δ_{i j}) ρ d λ_{3}$ are absolutely convergent, dominated $λ_{3}$ -a.e. by $a^{l} ∣ ρ ∣$ and $4 a^{2} ∣ ρ ∣$ on the support, with $\sum_{i} Q_{i i} = 0$ (Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution).

[F2]

For a finite system of stationary point charges the moments $Q = \sum_{k} q_{k}$ , $p = \sum_{k} q_{k} x_{k}$ , $M_{i_{1} \dots i_{l}} = \sum_{k} q_{k} x_{k, i_{1}} \dots x_{k, i_{l}}$ and $Q_{i j} = \sum_{k} q_{k} (3 x_{k, i} x_{k, j} - ∣ x_{k} ∣^{2} δ_{i j})$ are finite sums with $\sum_{i} Q_{i i} = 0$ , defined without Choice, and empty sums are $0$ (Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution).

[F3]

If $f, g \in L^{1}$ and $α, β \in R$ , then $\int (α f + β g) d λ_{3} = α \int f d λ_{3} + β \int g d λ_{3}$ ; this extends to finite linear combinations (The Lebesgue integral is linear on $L^{1} (μ)$ ).

[F4]

If $f, g \in L^{1} (λ_{3})$ agree $λ_{3}$ -a.e. then $\int f d λ_{3} = \int g d λ_{3}$ (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree).

[F5]

$∣ \int f d λ_{3} ∣ \leq \int ∣ f ∣ d λ_{3}$ , and for measurable $0 \leq u \leq w$ one has $\int u d λ_{3} \leq \int w d λ_{3}$ (The modulus of an integral is bounded by the integral of the modulus, Monotonicity and nonnegative homogeneity of the nonnegative integral).

[F6]

Finite sums are distributive, may be reindexed over a bijection of their index set, and the empty sum is $0$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[F7]

With the Euclidean inner product and norm, $∣ y - b ∣ \leq ∣ y ∣ + ∣ b ∣$ and $∣ (y - b)_{i} ∣ \leq ∣ y - b ∣$ for every $y, b \in R^{3}$ ; the inner product is bilinear and the components are read in the frame basis (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, 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 $d_{2}$ ).

[F8]

A polynomial identity between the coordinate functions of $R^{3}$ that holds pointwise may be integrated and differentiated with the usual polynomial calculus; the coordinate maps of $R^{3}$ are smooth in the componentwise sense ( $C^{k}$ Euclidean maps and diffeomorphisms).

[F9]

${\overline{B}}_{a} (0) = {y \in R^{3} : ∣ y ∣ \leq a}$ (Open ball, closed ball and sphere in a metric space).

Proof

technique · direct

1.1F1F5F7F9given

Establish that every integral in the definition of the $b$ -moments is absolutely convergent. For $λ_{3}$ -a.e. $y$ outside ${\overline{B}}_{a} (0)$ the integrands vanish, and for $y \in {\overline{B}}_{a} (0)$ one has $∣ y ∣ \leq a$ by [F9], so $∣ (y - b)_{i_{1}} \dots (y - b)_{i_{l}} ∣ \leq (∣ y ∣ + ∣ b ∣)^{l} \leq (a + ∣ b ∣)^{l}$ and $∣ 3 (y - b)_{i} (y - b)_{j} - ∣ y - b ∣^{2} δ_{i j} ∣ \leq 3 ∣ y - b ∣^{2} + ∣ y - b ∣^{2} \leq 4 (a + ∣ b ∣)^{2}$ by [F7]; hence each integrand is bounded $λ_{3}$ -a.e. on the support by a finite constant times $∣ ρ ∣ \in L^{1}$ , and the triangle and monotonicity rules [F5] give absolute convergence.

1.2F6F8algebra

Prove the polynomial expansion identity: for every $l \geq 0$ , all indices $i_{1}, \dots, i_{l} \in {1, 2, 3}$ and every $y \in R^{3}$ , $\prod_{r = 1}^{l} (y - b)_{i_{r}} = \sum_{S \subseteq {1, \dots, l}} (- 1)^{∣ S ∣} (\prod_{r \in S} b_{i_{r}}) (\prod_{r \in S^{c}} y_{i_{r}})$ , where empty products are $1$ and $S^{c} = {1, \dots, l} ∖ S$ . The case $l = 0$ is $1 = 1$ , the unique subset being $S = \emptyset$ ; for $l \geq 1$ write the product as $(\prod_{r = 1}^{l - 1} (y - b)_{i_{r}}) (y_{i_{l}} - b_{i_{l}})$ , apply the induction hypothesis to the first factor and distribute: the subsets $S$ of ${1, \dots, l - 1}$ that avoid $l$ and the subsets to which $l$ is adjoined are in bijection and reindex the same finite sum, giving the displayed identity.

2.1step 1.1step 1.2F1F3F4F7

In the density branch, substitute step 1.2 into the definition of $M_{i_{1} \dots i_{l}}^{(b)}$ and integrate term by term. Each summand is a real constant $(- 1)^{∣ S ∣} \prod_{r \in S} b_{i_{r}}$ times the $L^{1}$ function $y \mapsto (\prod_{r \in S^{c}} y_{i_{r}}) ρ (y)$ , which is dominated $λ_{3}$ -a.e. by $∣ b ∣^{∣ S ∣} a^{l - ∣ S ∣} ∣ ρ ∣$ by [F1] and [F7]; finite linearity [F3] and the a.e. support convention [F4] therefore give $M_{i_{1} \dots i_{l}}^{(b)} = \sum_{S} (- 1)^{∣ S ∣} (\prod_{r \in S} b_{i_{r}}) M_{S^{c}}$ with $M_{\emptyset} : = Q$ . Specialising to $l = 0$ gives $Q^{(b)} = Q$ , which is (a). Specialising to $l = 1$ gives $p_{i}^{(b)} = p_{i} - b_{i} Q$ , which is (b): if $Q = 0$ then $p^{(b)} = p$ for every $b$ ; conversely, if $p^{(b)} = p$ for every $b \in R^{3}$ , then $Q b = 0$ for every $b$ , and taking $b = e_{1}$ gives $Q e_{1} = 0$ , whose first coordinate is $Q$ , so $Q = 0$ , and the dipole moment is independent of the reference point exactly when $Q = 0$ .

3.1step 2.1F3F7

Evaluate the $l = 2$ instance of step 2.1: $M_{i j}^{(b)} = M_{i j} - b_{i} p_{j} - b_{j} p_{i} + b_{i} b_{j} Q$ . Summing over $i = j$ and using finite linearity [F3] gives $\sum_{k = 1}^{3} M_{k k}^{(b)} = \sum_{k = 1}^{3} M_{k k} - 2 b \cdot p + ∣ b ∣^{2} Q$ , where $b \cdot p = \sum_{k} b_{k} p_{k}$ is the Euclidean inner product [F7].

4.1step 3.1F1F3algebra

Compute the traceless quadrupole about $b$ . By [F1], $Q_{i j}^{(b)} = 3 M_{i j}^{(b)} - δ_{i j} \sum_{k} M_{k k}^{(b)}$ and $Q_{i j} = 3 M_{i j} - δ_{i j} \sum_{k} M_{k k}$ , so substituting step 3.1 into the first identity gives $Q_{i j}^{(b)} = Q_{i j} - 3 (b_{i} p_{j} + b_{j} p_{i}) + 2 (b \cdot p) δ_{i j} + (3 b_{i} b_{j} - ∣ b ∣^{2} δ_{i j}) Q$ , which is the displayed identity of (d); every step is a finite algebraic rearrangement [F3].

5.1step 4.1F7algebra

If $Q = 0$ and $p = 0$ , the $b$ -dependent terms of step 4.1 vanish, so every quadrupole component is reference-point independent. Conversely, suppose $Q_{i j}^{(b)} = Q_{i j}$ for all $b, i, j$ . The $11$ components of step 4.1 at $b = e_{1}$ and $b = - e_{1}$ give $- 4 p_{1} + 2 Q = 0$ and $4 p_{1} + 2 Q = 0$ ; adding gives $Q = 0$ . With $Q = 0$ , the $i i$ component at $b = e_{i}$ is $- 4 p_{i} = 0$ for each $i$ , hence $p = 0$ . This proves the equivalence in (d).

6.1step 1.2step 5.1F2F6

Repeat the argument in the finite point-charge branch and complete the choice accounting. The same expansion identity of step 1.2 is substituted into the finite sums of [F2]; finite distributivity, reindexing and the empty-sum convention [F6] produce the identical identities (a)-(d) with sums in place of integrals, without any convergence hypothesis, and the case $N = 0$ makes both sides $0$ at every step. The identities are therefore choice-free in this branch, while the density branch used only the declared Countable Choice of the measure interface.

7.1step 6.1F4∎

Combine the branches and record the scope. Replacing $ρ$ by an equal $λ_{3}$ -a.e. density leaves every $b$ -moment unchanged by [F4], so (a)-(d) hold as stated for the declared representatives; the density branch is established in steps 2.1, 4.1 and 5.1, the finite-point branch in step 6.1, and the identities are those of the source moments of Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution about the chart origin, the reference point $b$ being a choice of coordinate origin only; no field, force or dynamical content is asserted, which is clause (e).

Origin shift of the source moments: Q, the dipole p - Qb, the Cartesian expansion and the traceless quadrupole

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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