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Four stationary point charges with vanishing total charge and dipole moment and traceless quadrupole diag(6qa^2,-6qa^2,0)

Physical scope

A finite source-moment computation for four stationary point charges in a named inertial frame: vanishing total charge and dipole moment, vanishing odd Cartesian moments, and the exact traceless quadrupole $diag (6 q a^{2}, - 6 q a^{2}, 0)$ . No field, force, self-energy, dynamics, stability or empirical claim is made.

Example

Let $q \in R$ and $a > 0$ , let $(e_{1}, e_{2}, e_{3})$ be the $δ$ -orthonormal positively oriented basis of the laboratory frame (Laboratory geometry: a restricted inertial chart, its spatial slices and fixed-time fields, The Euclidean spatial structure and cross product of an inertial frame), and consider the four stationary point charges of Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution (c): $+ q$ at $a e_{1}$ , $+ q$ at $- a e_{1}$ , $- q$ at $a e_{2}$ , $- q$ at $- a e_{2}$ . Then:

(a) $Q = 0$ and $p = 0$ ; more generally every odd Cartesian moment vanishes, $M_{i_{1} \dots i_{2 k + 1}} = 0$ for every $k \geq 0$ and all indices, because the configuration is invariant under the central reflection $x \mapsto - x$ with the charges preserved;

(b) the degree- $2$ Cartesian moment is $M_{i j} = 2 q a^{2} (δ_{i 1} δ_{j 1} - δ_{i 2} δ_{j 2})$ , i.e. $diag (2 q a^{2}, - 2 q a^{2}, 0)$ in the frame basis, and its trace vanishes;

(c) the traceless quadrupole is $Q_{i j} = 6 q a^{2} (δ_{i 1} δ_{j 1} - δ_{i 2} δ_{j 2}) = diag (6 q a^{2}, - 6 q a^{2}, 0)$ and $\sum_{i} Q_{i i} = 0$ ; since $Q = p = 0$ , it is independent of the reference point by the origin-shift theorem (Origin shift of the source moments: Q, the dipole p - Qb, the Cartesian expansion and the traceless quadrupole (d));

(d) [Scope] this is a finite source-moment computation for prescribed stationary point charges. No field, force, energy (in particular no claim that the self-energy of the point charges is finite), dynamics, stability or empirical content is asserted.

Facts & Assumptions

Given: Let $q \in R$ , $a > 0$ , and let $(e_{1}, e_{2}, e_{3})$ be the $δ$ -orthonormal positively oriented basis of the laboratory frame; the four stationary point charges are $+ q$ at $a e_{1}$ , $+ q$ at $- a e_{1}$ , $- q$ at $a e_{2}$ , $- q$ at $- a e_{2}$ , as in the finite system of stationary point charges of Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution (c).

[F1]

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

[F2]

The basis $(e_{1}, e_{2}, e_{3})$ is $δ$ -orthonormal and positively oriented, and the vector product and the inner product are those of the frame's positive spatial three-space; in particular $δ (e_{r}, e_{s}) = δ_{r s}$ (The Euclidean spatial structure and cross product of an inertial frame).

[F3]

The frame basis is expressed in the laboratory chart with Euclidean norm $∣ w ∣^{2} = δ (w, w)$ and coordinates $w = \sum_{i} w^{i} e_{i}$ (Laboratory geometry: a restricted inertial chart, its spatial slices and fixed-time fields, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F4]

$\sum_{i} Q_{i i} = 0$ for the finite system, and if $Q = 0$ and $p = 0$ then the traceless quadrupole is independent of the reference point: $Q_{i j}^{(b)} = Q_{i j}$ for every $b$ (Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution, Origin shift of the source moments: Q, the dipole p - Qb, the Cartesian expansion and the traceless quadrupole).

[F5]

Finite sums may be reindexed over a bijection of their index set and expanded by distributivity (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[F6]

The Euclidean norm of a coordinate vector $w = (w^{1}, w^{2}, w^{3})$ is $∣ w ∣ = (\sum_{i} (w^{i})^{2})^{1 / 2}$ (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

Verification

technique · direct

1.1F1F2F3F5

Compute the total charge and the dipole moment by inserting the four points into the finite sums of [F1]: $Q = q + q + (- q) + (- q) = 0$ , and $p = q (a e_{1}) + q (- a e_{1}) + (- q) (a e_{2}) + (- q) (- a e_{2}) = q a (e_{1} - e_{1}) + (- q) a (e_{2} - e_{2}) = 0$ , using linearity and the coordinate expansions of [F2], [F3] and [F5].

2.1step 1.1F1F5

Show that every odd Cartesian moment vanishes. The configuration is invariant under the central reflection $x \mapsto - x$ : the charge $+ q$ at $\pm a e_{1}$ is carried to the charge $+ q$ at $\mp a e_{1}$ , and the charge $- q$ at $\pm a e_{2}$ to the charge $- q$ at $\mp a e_{2}$ , so pairing each point with its image and reindexing the finite sum [F5] gives $M_{i_{1} \dots i_{2 k + 1}} = \sum_{j = 1}^{4} q_{j} \prod_{r = 1}^{2 k + 1} x_{j, i_{r}} = \sum_{j = 1}^{4} q_{j} (- x_{j, i_{1}}) \dots (- x_{j, i_{2 k + 1}}) = - \sum_{j = 1}^{4} q_{j} \prod_{r} x_{j, i_{r}} = - M_{i_{1} \dots i_{2 k + 1}}$ , hence $M_{i_{1} \dots i_{2 k + 1}} = 0$ ; the case $k = 0$ is the vanishing dipole $p = 0$ ; total charge $Q = 0$ was computed separately in step 1.1.

3.1step 2.1F1F2F5

Compute the degree-two Cartesian moment. By [F1] and the coordinates of [F2], $M_{i j} = q (a e_{1})_{i} (a e_{1})_{j} + q (- a e_{1})_{i} (- a e_{1})_{j} - q (a e_{2})_{i} (a e_{2})_{j} - q (- a e_{2})_{i} (- a e_{2})_{j} = q a^{2} e_{1 i} e_{1 j} + q a^{2} e_{1 i} e_{1 j} - q a^{2} e_{2 i} e_{2 j} - q a^{2} e_{2 i} e_{2 j} = 2 q a^{2} (δ_{i 1} δ_{j 1} - δ_{i 2} δ_{j 2})$ , which is $diag (2 q a^{2}, - 2 q a^{2}, 0)$ in the frame basis and has vanishing trace $2 q a^{2} (1 - 1) = 0$ .

4.1step 3.1F1F2F3F4F6

Compute the traceless quadrupole. With [F1], $Q_{i j} = \sum_{k} q_{k} (3 x_{k, i} x_{k, j} - ∣ x_{k} ∣^{2} δ_{i j}) = 3 M_{i j} - δ_{i j} \sum_{k} q_{k} ∣ x_{k} ∣^{2}$ ; each of the four points has $∣ x_{k} ∣ = a$ because $∣ \pm a e_{r} ∣^{2} = a^{2} δ (e_{r}, e_{r}) = a^{2}$ by [F2], [F3] and [F6], so $\sum_{k} q_{k} ∣ x_{k} ∣^{2} = a^{2} (q + q - q - q) = 0$ and $Q_{i j} = 3 M_{i j} = 6 q a^{2} (δ_{i 1} δ_{j 1} - δ_{i 2} δ_{j 2})$ , whose trace is $6 q a^{2} (1 - 1) = 0$ , in agreement with the general tracelessness of [F4].

5.1step 4.1F4∎

Record the reference-point behaviour and the scope. Since $Q = p = 0$ by steps 1.1 and 2.1, clause (d) of the origin-shift theorem [F4] gives $Q_{i j}^{(b)} = Q_{i j}$ for every reference point $b$ , so the value $diag (6 q a^{2}, - 6 q a^{2}, 0)$ is independent of the reference point; no field, force, energy or self-energy of the point charges is computed or claimed, and the computation is a finite source-moment identity for the prescribed stationary charges.

Four stationary point charges with vanishing total charge and dipole moment and traceless quadrupole diag(6qa^2,-6qa^2,0)

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

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