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External force and torque of prescribed charge data in a Lipschitz external field

Physical scope

Prescribed external force and torque for broad compact $L^{1}$ charge data and a Lipschitz external vector-field datum under an explicitly additional $ρ E$ force model, with full force/torque/origin-shift bounds. A separate finite point-force assignment needs only external values (continuity for the stated field datum); its Lipschitz bounds are retained. Exact Lorentz-postulate instantiation is restricted explicitly to its smooth-source matter branch or $C^{1}$ prescribed tensor and massive $C^{2}$ test-worldline hypotheses at an instant of rest. No self-force, trajectory existence, permanent rest, equilibrium, stability or empirical assertion is supplied.

Statement

Assume the Axiom of Countable Choice ( ${A C}_{ω}$ ), inherited from the Lebesgue interface (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $a > 0$ and let $ρ \in L^{1} (R^{3})$ be real and vanish $λ_{3}$ -a.e. outside ${\overline{B}}_{a} (0)$ ; let $E_{e x t} : R^{3} \to R^{3}$ be a prescribed external vector-field datum, with electric-field units of Framework-local SI units, the vacuum constants and the exact/measured status of their values, that is continuous on ${\overline{B}}_{a} (0)$ and Lipschitz there with constant $L \geq 0$ , i.e. $∣ E_{e x t} (x) - E_{e x t} (x^{'}) ∣ \leq L ∣ x - x^{'} ∣$ for all $x, x^{'} \in {\overline{B}}_{a} (0)$ . Consider the matter model in which the prescribed static distribution $ρ$ is carried by a body at rest in the laboratory frame and adopt additionally the prescribed external force model $f_{e x t} = ρ E_{e x t}$ . This is a declared extension to these locally integrable charge data and Lipschitz fields, motivated by the smooth continuum coupling of The Lorentz coupling: external force on a test charge and the continuum force and power densities (ii); that postulate covers smooth sources and does not itself establish this extension. The only force counted by the present model is $f_{e x t}$ ; the self-field of $ρ$ , all other fields, mechanical support and back-reaction are excluded by this model. Define the total external force and the total external torque about the origin by the componentwise Lebesgue integrals

$F : = \int_{R^{3}} ρ (y) E_{e x t} (y) d λ_{3} (y) \in R^{3}, τ : = \int_{R^{3}} ρ (y) (y \times E_{e x t} (y)) d λ_{3} (y) \in R^{3},$

with units N and N m respectively; $L$ has units V m $^{- 2}$ . The integrands are interpreted through their a.e.-equal measurable zero extensions from the closed ball: continuity of $E_{e x t}$ there gives Borel factors, and null modifications remain Lebesgue measurable by $L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ to the Borel sets. They are absolutely convergent because a.e. on the support $∣ E_{e x t} ∣ \leq ∣ E_{e x t} (0) ∣ + L a$ and $∣ y \times E_{e x t} ∣ \leq a (∣ E_{e x t} (0) ∣ + L a)$ , so each component is dominated by a constant times $∣ ρ ∣ \in L^{1}$ (Integral over a measurable subset, Integrable real and complex functions, and their integrals, The cross product in $R^{3}$ , The squared cross-product norm is the Gram determinant of two vectors, The modulus of an integral is bounded by the integral of the modulus, Monotonicity and nonnegative homogeneity of the nonnegative integral). Then:

(a) [Uniform force and torque expansions]

$F = Q E_{e x t} (0) + R_{F}, ∣ R_{F} ∣ \leq L a ∥ ρ ∥_{1}, τ = p \times E_{e x t} (0) + R_{τ}, ∣ R_{τ} ∣ \leq L a^{2} ∥ ρ ∥_{1},$

where $Q$ and $p$ are the total charge and dipole moment of Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution (a) and $∥ ρ ∥_{1} = \int ∣ ρ ∣ d λ_{3}$ ;

(b) [Neutral distribution] if $Q = 0$ then the leading force term vanishes, $F = R_{F}$ with $∣ R_{F} ∣ \leq L a ∥ ρ ∥_{1}$ , and the torque is the ideal-dipole torque $p \times E_{e x t} (0)$ up to the bound $∣ R_{τ} ∣ \leq L a^{2} ∥ ρ ∥_{1}$ ; in this sense a neutral distribution of size $a$ in a field with Lipschitz constant $L$ carries the dipole torque $p \times E_{e x t} (0)$ with an explicit remainder;

(c) [Reference-point shift] for any reference point $o \in R^{3}$ , the torque about $o$ , $τ_{o} : = \int_{R^{3}} ρ (y) (y - o) \times E_{e x t} (y) d λ_{3} (y)$ , equals $τ - o \times F$ exactly;

(d) [Ideal pair of test charges at rest] let $q \in R$ and $d \in R^{3} ∖ {0}$ , put $a = ∣ d ∣ / 2$ , and consider the two stationary point charges $+ q$ at $d / 2$ and $- q$ at $- d / 2$ (Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution (c)), each treated as a test charge at rest in the laboratory chart in the same prescribed external field $E_{e x t}$ , with the mutual interaction and the self-field of each charge excluded by the model. Adopt for this finite prescribed-charge branch the external force assignment $F_{k} = q_{k} E_{e x t} (x_{k})$ and torque $x_{k} \times F_{k}$ . It is defined for merely continuous external field values at the two points, independently of a worldline equation. For the same Lipschitz bound as above, finite summation gives the expansions of (a),(b), with $Q = 0$ , $p = q d$ , and total variation $2 ∣ q ∣$ : $∣ R_{F} ∣ \leq 2 ∣ q ∣ L a$ , $∣ R_{τ} ∣ \leq 2 ∣ q ∣ L a^{2}$ . More generally the exact finite force/torque sum requires only the specified field values; no Lipschitz estimate is asserted without its hypothesis.

(d1) [Covariant sufficient specialization] If, in addition, an actual prescribed electromagnetic tensor $F$ of The electromagnetic two-form, its observer decomposition and the four-current is $C^{1}$ on a neighbourhood of the graph of each massive future-timelike $C^{2}$ test worldline with constant charge $q^{'}$ and supplied rest mass $m > 0$ , and the external-field coupling The Lorentz coupling: external force on a test charge and the continuum force and power densities (i) applies, then at an instant with zero coordinate velocity $U = (c, 0, 0, 0)$ its coordinate three-force $f = d p / d t$ is $q^{'} E_{e x t}$ : the four-force is $γ (f \cdot v / c, f)$ with $γ = 1$ at rest (Energy, momentum, invariant mass and four-force of relativistic particles, and necessary collision thresholds (b)), and $U_{0} = - c$ and $F^{i 0} = - E_{i} / c$ . This reproduces the prescribed point-force assignment at that instant. Similarly the smooth-source $J = 0$ matter-coupling branch of postulate (ii) reproduces the continuum force assignment when its own source/field/matter hypotheses hold. No $C^{1}$ tensor or worldline is inferred from merely continuous $E$ data, and no two-charge trajectory existence or permanent rest is claimed.

(e) [Scope] the statement is conditional on the explicitly stated matter model: only the prescribed external field acts on the distribution, the distribution is prescribed and at rest, and no self-force, back-reaction, mechanical equilibrium, stability or dynamics of the body is asserted; the external field is a datum, not solved for. Clause (d) is an instantaneous idealisation of two ideal point test charges with mutual interaction excluded; it is not a two-body dynamics, a bound-state or a stability statement. No radiation, medium response or empirical claim is made. The prescribed finite force assignment and finite moment algebra in (d) are choice-free; the volume integrals in (a)-(c) carry their explicit measure premise, while (d1) is a separate conditional instantiation of the exact physical coupling.

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 ${\overline{B}}_{a} (0)$ , let $E_{e x t} : R^{3} \to R^{3}$ be continuous on ${\overline{B}}_{a} (0)$ and Lipschitz there with constant $L \geq 0$ , and adopt the prescribed external force model $f_{e x t} = ρ E_{e x t}$ on the prescribed distribution at rest in the laboratory frame.

[F1]

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 the dipole moment $p = \int y ρ (y) d λ_{3} (y)$ about the chart origin are absolutely convergent, and in the finite system of stationary point charges $(x_{k}, q_{k})_{k = 1}^{N}$ one has $Q = \sum_{k} q_{k}$ and $p = \sum_{k} q_{k} x_{k}$ as finite sums (Total charge, dipole moment, Cartesian moments and traceless quadrupole of a static charge distribution).

[F2]

The electric field carries V m $^{- 1}$ , a force carries the newton, and $ε_{0} = (μ_{0} c^{2})^{- 1} > 0$ ; the laboratory structure is that of Laboratory geometry: a restricted inertial chart, its spatial slices and fixed-time fields (Framework-local SI units, the vacuum constants and the exact/measured status of their values).

[F3]

An electric field of class $C^{k}$ is a map $E : Ω \times I \to R^{3}$ , and at fixed $t$ the map $E (\cdot, t) : Ω \to R^{3}$ is a real spatial vector field read in the frame basis; no value at a point of singular source support is asserted (The electric and magnetic fields as fixed-time spatial vector fields).

[F4]

The postulate's external-field coupling (i) is $m d U / d τ = q Φ (U)$ with $Φ$ the raised-index operator $η (Φ (u), w) = F (w, u)$ of the field, equivalently $m d U^{μ} / d τ = q F^{μ ν} U_{ν}$ , for a constant charge, a supplied rest mass $m > 0$ and a future timelike $C^{2}$ worldline, with the particle's own field excluded; clause (ii) adopts the continuum force density $ρ E + J \times B$ for a smooth source interacting through a matter model to which the coupling applies (The Lorentz coupling: external force on a test charge and the continuum force and power densities).

[F5]

In the chart the electromagnetic two-form has $F_{0 i} = - E_{i} / c$ and $F^{i 0} = - E_{i} / c$ for the field pair $(E, B)$ (The electromagnetic two-form, its observer decomposition and the four-current).

[F6]

A massive worldline is a $C^{k}$ future timelike curve, $k \geq 1$ , with everywhere nonzero future timelike tangent; its four-velocity satisfies $η (U, U) = - c^{2}$ ; and a massive $C^{2}$ worldline with laboratory reading $t \mapsto r (t)$ satisfies $∣ r^{'} (t) ∣ < c$ with four-velocity $U = γ (c, r^{'})$ and Lorentz factor $γ = (1 - ∣ r^{'} ∣^{2} / c^{2})^{- 1 / 2}$ , so that at a parameter with $r^{'} = 0$ one has $γ = 1$ and $U = (c, 0, 0, 0)$ (Worldlines, four-velocity, chart readings and the frame/observer distinction).

[F7]

On $R^{3}$ the cross product is bilinear and alternating, and $∣ w \times z ∣^{2} = ∣ w ∣^{2} ∣ z ∣^{2} - ⟨ w, z ⟩^{2}$ , so $∣ w \times z ∣ \leq ∣ w ∣ ∣ z ∣$ (The cross product in $R^{3}$ , The cross product is bilinear, alternating, and orthogonal to both factors, The squared cross-product norm is the Gram determinant of two vectors).

[F8]

For the Euclidean inner product, $∣ ⟨ u, w ⟩ ∣ \leq ∣ u ∣ ∣ w ∣$ (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]

If $f, g \in L^{1}$ and $α, β \in R$ , then $\int (α f + β g) = α \int f + β \int g$ ; if $f$ and $g$ agree a.e. their integrals agree; and $∣ \int f ∣ \leq \int ∣ f ∣$ with monotonicity for nonnegative integrands (The Lebesgue integral is linear on $L^{1} (μ)$ , 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, Monotonicity and nonnegative homogeneity of the nonnegative integral).

[F10]

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

[F12]

Lebesgue measure is the completion of Borel Lebesgue measure under Countable Choice; any finite modification of a measurable function on a subset of a null set remains measurable ( $L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ to the Borel sets). For example, each preimage differs from its old measurable preimage only by subsets of that null set.

[F13]

For a massive constant-mass $C^{2}$ worldline, the coordinate spatial momentum is $p = m γ v$ , the coordinate three-force is $f = d p / d t$ , and the four-force $m d U / d τ$ is $γ (f \cdot v / c, f)$ (Energy, momentum, invariant mass and four-force of relativistic particles, and necessary collision thresholds (a),(b)).

[F11]

${\overline{B}}_{a} (0) = {y : ∣ y ∣ \leq a}$ and $∣ y ∣ \leq a$ holds $λ_{3}$ -a.e. on the support of $ρ$ ; the Euclidean norm obeys the reverse triangle inequality $∣ ∣ u ∣ - ∣ w ∣ ∣ \leq ∣ u - w ∣$ (Open ball, closed ball and sphere in a metric space, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , 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 $d_{2}$ ).

Proof

technique · direct

1.1F7F8F9F11F12given

Establish that $F$ and $τ$ are well defined. Their integrands on the closed ball are measurable products of $ρ$ with continuous coordinate functions; the zero extensions are measurable, and the displayed full-space integrands differ from them only on the null exceptional set where $ρ$ fails to vanish outside the ball, so [F12] supplies measurability of those integrands as well. For $λ_{3}$ -a.e. $y$ either $ρ (y) = 0$ or $y \in {\overline{B}}_{a} (0)$ , and then $∣ y ∣ \leq a$ by [F11]; the Lipschitz property gives $∣ E_{e x t} (y) ∣ \leq ∣ E_{e x t} (0) ∣ + L ∣ y ∣ \leq ∣ E_{e x t} (0) ∣ + L a$ and hence $∣ y \times E_{e x t} (y) ∣ \leq ∣ y ∣ ∣ E_{e x t} (y) ∣ \leq a (∣ E_{e x t} (0) ∣ + L a)$ by [F7] and [F8]. Each component of the integrands is therefore dominated $λ_{3}$ -a.e. by a finite constant times $∣ ρ ∣ \in L^{1}$ , so the componentwise integrals converge absolutely by the triangle and monotonicity rules, and no value of $E_{e x t}$ on a null set matters [F9].

2.1step 1.1F1F8F9F11

Expand the force and bound its remainder. Write $E_{e x t} (y) = E_{e x t} (0) + (E_{e x t} (y) - E_{e x t} (0))$ and integrate: by finite linearity of the componentwise integral [F9], $F = (\int ρ d λ_{3}) E_{e x t} (0) + \int ρ (y) (E_{e x t} (y) - E_{e x t} (0)) d λ_{3} (y) = Q E_{e x t} (0) + R_{F}$ with $Q$ as in [F1]. If $R_{F} \neq 0$ , test against the unit vector $u : = R_{F} / ∣ R_{F} ∣$ : $∣ R_{F} ∣ = u \cdot R_{F} = \int ρ (y) u \cdot (E_{e x t} (y) - E_{e x t} (0)) d λ_{3} (y) \leq \int ∣ ρ (y) ∣ ∣ E_{e x t} (y) - E_{e x t} (0) ∣ d λ_{3} (y) \leq L \int ∣ y ∣ ∣ ρ (y) ∣ d λ_{3} (y) \leq L a ∥ ρ ∥_{1}$ , using [F8] for $u \cdot v \leq ∣ v ∣$ , the Lipschitz hypothesis, the monotonicity rule, and $∣ y ∣ \leq a$ $λ_{3}$ -a.e. on the support [F11]; the case $R_{F} = 0$ is trivial. This is the force statement of (a).

3.1step 2.1F1F7F9F11

Expand the torque and bound its remainder. Since $y \times E_{e x t} (y) = y \times E_{e x t} (0) + y \times (E_{e x t} (y) - E_{e x t} (0))$ , finite linearity of the componentwise integral and the bilinearity of the cross product [F7] give $\int ρ (y) y \times E_{e x t} (0) d λ_{3} (y) = (\int y ρ (y) d λ_{3} (y)) \times E_{e x t} (0) = p \times E_{e x t} (0)$ with $p$ as in [F1], so $τ = p \times E_{e x t} (0) + R_{τ}$ with $R_{τ} : = \int ρ (y) y \times (E_{e x t} (y) - E_{e x t} (0)) d λ_{3} (y)$ . For the bound, apply the unit-vector pairing argument of step 2.1 to $R_{τ}$ (with the zero case immediate): $∣ R_{τ} ∣ \leq \int ∣ ρ (y) ∣ ∣ y ∣ ∣ E_{e x t} (y) - E_{e x t} (0) ∣ d λ_{3} (y) \leq L \int ∣ y ∣^{2} ∣ ρ ∣ d λ_{3} \leq L a^{2} ∥ ρ ∥_{1}$ by [F7], [F9] and [F11]. This is the torque statement of (a).

4.1step 3.1

Specialise to the neutral case (b): if $Q = 0$ , the leading force term $Q E_{e x t} (0)$ vanishes and step 2.1 leaves $F = R_{F}$ with $∣ R_{F} ∣ \leq L a ∥ ρ ∥_{1}$ , while step 3.1 gives the torque as the ideal-dipole torque plus the stated bound.

5.1step 4.1F7F9

Prove the reference-point shift (c). For any $o \in R^{3}$ , $(y - o) \times E_{e x t} (y) = y \times E_{e x t} (y) - o \times E_{e x t} (y)$ by bilinearity [F7]; integrating, $o$ is a constant vector and $\int ρ (y) E_{e x t} (y) d λ_{3} (y) = F$ , so finite linearity [F9] gives $τ_{o} = τ - o \times F$ exactly.

6.1step 5.1F1F7F10

Treat the ideal pair of test charges (d). Under the explicitly adopted finite force assignment $F_{k} = q_{k} E_{e x t} (x_{k})$ , the total force is $F = q E_{e x t} (d / 2) - q E_{e x t} (- d / 2) = q (E_{e x t} (d / 2) - E_{e x t} (- d / 2))$ , and inserting $\pm E_{e x t} (0)$ gives $F = R_{F}$ with $R_{F} = q [(E_{e x t} (d / 2) - E_{e x t} (0)) - (E_{e x t} (- d / 2) - E_{e x t} (0))]$ and $∣ R_{F} ∣ \leq ∣ q ∣ (L ∣ d / 2 ∣ + L ∣ d / 2 ∣) = 2 ∣ q ∣ L a$ ; the total torque is $τ = (d / 2) \times q E_{e x t} (d / 2) + (- d / 2) \times (- q) E_{e x t} (- d / 2) = q (d / 2) \times (E_{e x t} (d / 2) + E_{e x t} (- d / 2)) = q d \times E_{e x t} (0) + R_{τ}$ with $R_{τ} = q (d / 2) \times [(E_{e x t} (d / 2) - E_{e x t} (0)) + (E_{e x t} (- d / 2) - E_{e x t} (0))]$ and $∣ R_{τ} ∣ \leq ∣ q ∣ a \cdot 2 L a = 2 ∣ q ∣ L a^{2}$ , using [F7], the Lipschitz hypothesis and [F10]; the finite-branch moments are $Q = 0$ and $p = q d$ by [F1], and no convergence or Lipschitz hypothesis is needed for the bare finite sums themselves. All these sums are finite and choice-free.

7.1step 6.1F4F5F6F13

Instantiate the exact coupling at an instant of rest, proving (d1). Under the additional hypotheses of (d1) the postulate gives $m d U^{i} / d τ = q^{'} Φ (U)^{i} = q^{'} F^{i ν} U_{ν}$ [F4]; at an instant with $r^{'} = 0$ the chart reading of the four-velocity is $U = (c, 0, 0, 0)$ by [F6], so the spatial components $U_{j}$ vanish, $U_{0} = η_{00} U^{0} = - c$ , and only the term $q^{'} F^{i 0} U_{0}$ survives; with $F^{i 0} = - E_{i} / c$ from [F5] this equals $q^{'} (- E_{i} / c) (- c) = q^{'} E_{i}$ . By [F13], $m d U^{i} / d τ = γ f_{i}$ ; since $γ = 1$ at rest, the computed spatial four-force is exactly the coordinate three-force. Hence the covariant equation reproduces the prescribed assignment $F_{k} = q_{k} E_{e x t} (x_{k})$ at that instant and nothing more: no $C^{1}$ tensor or massive worldline is inferred from the continuous datum $E_{e x t}$ , and no trajectory or permanent-rest claim is made. In the same way the smooth source branch of the postulate reproduces the continuum density $ρ E$ only when its own stated smoothness and matter hypotheses hold [F4].

8.1step 7.1F2F3given

Record the scope (e) and close the argument. The prescribed data carry the field units and laboratory conventions of [F2]; only the prescribed external field acts in the model, the self-field, other fields, mechanical support and back-reaction are excluded by hypothesis, the external field is a datum rather than a solution of any field equation, and clauses (a)-(c) concern the declared source moments of a prescribed distribution without asserting force dynamics, equilibrium or stability; clause (d) is an instantaneous idealisation with mutual interaction and self-fields excluded, and clause (d1) is a conditional instantiation of the exact coupling, so no empirical, radiative or medium-response content is introduced. The volume integrals used the declared Countable Choice through the Lebesgue interface, while the finite-point algebra of (d) is choice-free.

9.1step 8.1∎

Combine the clauses: step 2.1 and step 3.1 give (a) with the stated bounds, step 4.1 gives (b), step 5.1 gives the exact reference-point shift (c), step 6.1 gives the finite two-charge expansions and their choice-free status (d), and step 7.1 gives the conditional covariant specialization (d1), all under the model assumptions of the statement and with the exclusions of step 8.1, which is (e).

External force and torque of prescribed charge data in a Lipschitz external field

Statement

Facts & Assumptions

Proof

Depends on

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