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Non-central antisymmetric pair forces with vanishing internal torque sum: the centrality clause is not necessary for angular-momentum conservation

Physical scope

A counterexample inside the force branch: a three-label model with constant antisymmetric pair forces has non-central pair forces and yet its internal torque sum vanishes identically, so its total angular momentum about a fixed origin is constant along every motion even though the centrality hypothesis of the balance theorem fails. The item refutes only the claim that centrality is necessary for the internal-torque conclusion; it asserts no measured value and no conservation for general non-central models.

Statement refuted

In the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with the external-plus-pair presentation and the antisymmetry clause (ii)1 of Restricted interaction models: pair decompositions, antisymmetry and centrality options, potential forces and the gravitational branch, if the internal torque sum $\sum_{i < j} (x_{i} - x_{j}) \times F_{i j}$ vanishes identically, then every pair force $F_{i j}$ is central (parallel to $x_{i} - x_{j}$ ).

Equivalently: centrality of the pair forces is necessary for the internal-torque conclusion used by Angular momentum balance about a fixed or moving reference point, the internal torque sum, and the centrality case.

Facts & Assumptions

Given: The force branch, one inertial frame, and the notation of Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame and of Angular momentum balance about a fixed or moving reference point, the internal torque sum, and the centrality case: evaluated pair forces $F_{i j}$ along a motion, the internal torque sum $I = \sum_{i < j} (x_{i} - x_{j}) \times F_{i j}$ , the angular momentum $J_{o} = \sum_{i} (x_{i} - o) \times p_{i}$ and the torque decomposition $T_{o} = T_{o}^{e x t} + I$ of clause 1 of that theorem.

[L1]

Model notions (The associated first-order system, motions, maximal motions and the Newtonian flow of a force model, Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame): a motion satisfies $m_{i} {\ddot{x}}_{i} = F_{i} (t, x, \dot{x})$ ; the pair presentation writes $F_{i} = F_{i}^{e x t} + \sum_{j \neq i} F_{i j}$ with $F_{j i} = - F_{i j}$ ; the dimension conventions are Dimensions and SI unit conventions of the non-relativistic classical mechanics framework and the coordinate curves are those of Coordinate velocity and acceleration of a worldline in an inertial frame, and the spatial velocity and acceleration vectors.

[L2]

Angular balance (Angular momentum balance about a fixed or moving reference point, the internal torque sum, and the centrality case clauses 1-4): $T_{o} = T_{o}^{e x t} + I$ , and $J_{o}^{'} = T_{o}^{e x t} + I - o^{'} \times P$ ; for a constant reference point with $T_{o} \equiv 0$ the angular momentum $J_{o}$ is constant; the centrality clause is sufficient for $I \equiv 0$ .

[L3]

Momentum balance (Total momentum balance and centre-of-mass equation for a model with antisymmetric pair forces clause 2): if the evaluated total external force vanishes, the total momentum is constant.

[L4]

Cross-product algebra (The cross product in $R^{3}$ , The cross product is bilinear, alternating, and orthogonal to both factors): bilinearity, $u \times u = 0$ , $u \times v = - (v \times u)$ ; for the standard basis, $e_{1} \times e_{2} = e_{3}$ , $e_{2} \times e_{3} = e_{1}$ , $e_{3} \times e_{1} = e_{2}$ . The Euclidean inner product is The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , finite sums are linear (Laws of finite sums and finite products), and $R$ is an interval with nonempty interior (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Counterexample

technique · direct; define the model and a motion, verify the model hypotheses, and exhibit the non-central pair forces with identically vanishing internal torque

1.1givenL1

The model. Take $N = 3$ , masses $m_{1} = m_{2} = m_{3} : = 1 k g$ , time window $R$ , configuration domain $Q : = R^{9}$ (no exclusion is needed), no external forces, and the constant pair force maps $F_{12} \equiv f_{0}, F_{13} \equiv - f_{0}, F_{23} \equiv f_{0}, f_{0} : = F_{0} e_{3}, F_{0} : = 1 N,$ with $F_{j i} : = - F_{i j}$ for every ordered pair. Constant maps are continuous and state-Lipschitz with constant $0$ , the antisymmetry $F_{j i} = - F_{i j}$ holds by declaration, and the presentation is $F_{i} = 0 + \sum_{j \neq i} F_{i j}$ ; hence the model satisfies the hypotheses of the force branch and of Restricted interaction models: pair decompositions, antisymmetry and centrality options, potential forces and the gravitational branch clause (i) with the antisymmetry clause (ii)1.

2.1step 1.1L1L4algebra

The motion. Set $L_{0} : = 1 m$ , $t_{*} : = 1 s$ and $s : = t / t_{*}$ . Define $x_{1} (t) : = L_{0} s e_{2}, x_{2} (t) : = L_{0} e_{1}, x_{3} (t) : = L_{0} (e_{2} + s e_{1}) (t \in R) .$ These curves are $C^{2}$ with ${\ddot{x}}_{i} = 0$ ; the net force on each label vanishes, $F_{1} = F_{12} + F_{13} = f_{0} - f_{0} = 0, F_{2} = F_{21} + F_{23} = - f_{0} + f_{0} = 0, F_{3} = F_{31} + F_{32} = f_{0} - f_{0} = 0,$ so $m_{i} {\ddot{x}}_{i} = 0 = F_{i}$ for all $t$ : the curves form a motion of the model on $R$ . Moreover the positions are pairwise distinct for every $t$ , since $x_{1} - x_{2} = L_{0} (s e_{2} - e_{1}) \neq 0$ , $x_{2} - x_{3} = L_{0} ((1 - s) e_{1} - e_{2}) \neq 0$ and $x_{1} - x_{3} = L_{0} (- s e_{1} + (s - 1) e_{2}) \neq 0$ (the first or second coordinates in each pair cannot vanish simultaneously).

2.2step 1.1L4algebra

The internal torque sum vanishes identically. Using bilinearity [L4], $I = \sum_{i < j} (x_{i} - x_{j}) \times F_{i j} = (x_{1} - x_{2}) \times f_{0} + (x_{1} - x_{3}) \times (- f_{0}) + (x_{2} - x_{3}) \times f_{0} = [(x_{1} - x_{2}) - (x_{1} - x_{3}) + (x_{2} - x_{3})] \times f_{0},$ and the bracket telescopes to $0$ . Hence $I \equiv 0$ on the whole configuration domain, without any centrality. The individual pair torques are nevertheless nonzero: at $t = 0$ , $(x_{1} - x_{2}) \times F_{12} = L_{0} F_{0} e_{2}$ , $(x_{1} - x_{3}) \times F_{13} = L_{0} F_{0} e_{1}$ and $(x_{2} - x_{3}) \times F_{23} = - L_{0} F_{0} (e_{1} + e_{2})$ , all with torque units $N m$ and with sum zero; so the vanishing is by cancellation among the pairs.

3.1step 2.1L4algebra

The pair forces are not central. The pair force $F_{12} = F_{0} e_{3}$ is parallel to $x_{1} - x_{2}$ only if $(x_{1} - x_{2}) \times e_{3} = 0$ ; but $(x_{1} - x_{2}) \times F_{12} = L_{0} F_{0} (s e_{1} + e_{2}) \neq 0$ for every $t$ , since the two components do not vanish simultaneously; this torque has units $N m$ . Hence centrality fails along the whole motion (and $F_{12}$ fails it for every configuration with $(x_{1} - x_{2}) \times e_{3} \neq 0$ ).

3.2step 2.1step 2.2L2L4algebra

Angular momentum is conserved. By clause 1 of [L2], $T_{o} = T_{o}^{e x t} + I$ ; with the origin as reference point and no external forces, $T_{o}^{e x t} = 0$ , so $T_{0} = 0$ and clause 4 of [L2] applies even though the centrality hypothesis of clause 3 is not met: $J_{0}$ is constant along the motion. Directly, $J_{0} = x_{1} \times m_{1} {\dot{x}}_{1} + x_{2} \times m_{2} {\dot{x}}_{2} + x_{3} \times m_{3} {\dot{x}}_{3} = (L_{0} s e_{2}) \times m_{1} (L_{0} / t_{*}) e_{2} + (L_{0} e_{1}) \times 0 + L_{0} (e_{2} + s e_{1}) \times m_{3} (L_{0} / t_{*}) e_{1} = - m_{3} L_{0}^{2} / t_{*} e_{3},$ constant in $t$ and nonzero, with units $k g m^{2} s^{- 1}$ .

3.3step 2.1L3algebra

Total momentum is also conserved: the total external force vanishes identically, so clause 2 of [L3] gives that $P = m_{1} {\dot{x}}_{1} + m_{2} {\dot{x}}_{2} + m_{3} {\dot{x}}_{3} = m_{1} (L_{0} / t_{*}) e_{2} + m_{3} (L_{0} / t_{*}) e_{1}$ is constant, as it is directly; this has momentum units $k g m s^{- 1}$ .

4.1step 3.1step 2.2step 3.2L2∎

Conclusion. The model satisfies the hypotheses of the balance theorems with antisymmetric, non-central pair forces; its internal torque sum vanishes identically and its angular momentum about the origin is constant along the motion. The refuted statement is therefore false: centrality is a sufficient but not necessary condition for the internal-torque conclusion of Angular momentum balance about a fixed or moving reference point, the internal torque sum, and the centrality case, the cancellation occurring among the pair torques.

Remarks

What is refuted, and what is not. Only the converse direction is refuted: cancellation does not force centrality. The sufficient direction, centrality implies $I \equiv 0$ , is clause 3 of Angular momentum balance about a fixed or moving reference point, the internal torque sum, and the centrality case and is unaffected. Nothing here claims that non-central models generally conserve angular momentum.
The cancellation is structural. The telescoping identity of step 2.2 uses only the constancy of the declared pair forces and the assignment $F_{12} \equiv w$ , $F_{13} \equiv - w$ , $F_{23} \equiv w$ ; it holds for every configuration, not only along the exhibited motion. The motion is used to exhibit nonzero individual pair torques and a nonzero conserved angular momentum.

Non-central antisymmetric pair forces with vanishing internal torque sum: the centrality clause is not necessary for angular-momentum conservation

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources