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Angular momentum balance about a fixed or moving reference point, the internal torque sum, and the centrality case

Physical scope

The angular momentum balance of a model that declares the external-plus-pair presentation with antisymmetric pair forces: the internal torque sum is independent of the reference point and the balance about a possibly moving reference point carries the correction minus odot cross P; when the evaluated pair forces are central the internal torque sum vanishes; about a fixed point with central evaluated pair forces and vanishing external torque the angular momentum is constant. The result is conditional on the declared pair structure and centrality where invoked; it asserts no necessity of centrality and no measured value.

Statement

Adopt the force model and one inertial frame 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 suppose that the model invokes the external-plus-pair presentation of clause (i) of Restricted interaction models: pair decompositions, antisymmetry and centrality options, potential forces and the gravitational branch with the antisymmetry clause (ii)1 (as in the momentum balance of this page). Let $x : J \to Q$ be a motion on an interval $J$ with nonempty interior and let $o : J \to R^{3}$ be a $C^{1}$ reference curve, with $J_{o}$ , $T_{o}$ , $T_{o}^{e x t}$ , $I$ and $P$ as in the definition. Then:

Internal torque identity. For every $t \in J$ , $\sum_{i} (x_{i} (t) - o (t)) \times \sum_{j \neq i} F_{i j} (t, x_{i} (t), x_{j} (t), w_{i} (t), w_{j} (t)) = I (t), so T_{o} (t) = T_{o}^{e x t} (t) + I (t);$
Balance. $J_{o}$ is differentiable on $J$ and $J_{o}^{'} (t) = T_{o}^{e x t} (t) + I (t) - o^{'} (t) \times P (t) = T_{o} (t) - o^{'} (t) \times P (t) (t \in J);$
Centrality. Suppose that for every $i < j$ and every $t \in J$ the evaluated pair force is parallel to $x_{i} (t) - x_{j} (t)$ , that is, there is a real $λ_{i j} (t)$ with $F_{i j} (t, x_{i} (t), x_{j} (t), w_{i} (t), w_{j} (t)) = λ_{i j} (t) (x_{i} (t) - x_{j} (t)) .$ Then $I \equiv 0$ on $J$ , so $J_{o}^{'} (t) = T_{o}^{e x t} (t) - o^{'} (t) \times P (t) (t \in J);$ if moreover $o$ is constant, then $J_{o}^{'} = T_{o}^{e x t}$ ;
Conservation. If $o$ is constant and $T_{o} \equiv 0$ on $J$ — equivalently $T_{o}^{e x t} \equiv - I$ there, by clause 1 — then $J_{o}$ is constant on $J$ . In particular, if the evaluated pair forces are central (clause 3) and $T_{o}^{e x t} \equiv 0$ on $J$ — the case of a model with no external forces — then $T_{o} \equiv 0$ and $J_{o}$ is constant on $J$ .

Clause 3 allows cancellation: the vanishing of $I$ is a hypothesis on the evaluated pair forces and does not require centrality in general; the companion page exhibits antisymmetric non-central pair forces whose internal torque sum vanishes identically. No empirical content is claimed.

Facts & Assumptions

Given: A force model of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance that invokes the external-plus-pair presentation of Restricted interaction models: pair decompositions, antisymmetry and centrality options, potential forces and the gravitational branch with the antisymmetry clause, one inertial frame, a motion $x : J \to Q$ and a $C^{1}$ reference curve $o : J \to R^{3}$ .

[L1]

Notation and notions (Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame, The associated first-order system, motions, maximal motions and the Newtonian flow of a force model): $p_{i} = m_{i} w_{i}$ , $P = \sum_{i} p_{i}$ , $J_{o} = \sum_{i} (x_{i} - o) \times p_{i}$ , $T_{o}^{e x t} = \sum_{i} (x_{i} - o) \times F_{i}^{e x t}$ , $I = \sum_{i < j} (x_{i} - x_{j}) \times F_{i j}$ (all evaluated along the motion), $T_{o} = \sum_{i} (x_{i} - o) \times F_{i}$ ; the equation of motion is $m_{i} {\ddot{x}}_{i} = F_{i} (t, x, w)$ ; the presentation writes $F_{i} = F_{i}^{e x t} + \sum_{j \neq i} F_{i j}$ and its pair maps satisfy $F_{j i} = - F_{i j}$ .

[L2]

Product rule (Differentiation of sums, scalar multiples, inner products, cross products and squared norms of differentiable curves in $R^{3}$ clause 3): if $u, v : I \to R^{3}$ are differentiable at $t_{0}$ , then $u \times v$ is differentiable at $t_{0}$ with derivative $u^{'} (t_{0}) \times v (t_{0}) + u (t_{0}) \times v^{'} (t_{0})$ ; sums and scalar multiples of differentiable curves are differentiable with the expected derivatives (clause 1 of the same lemma, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[L3]

Cross-product algebra (The cross product in $R^{3}$ , The cross product is bilinear, alternating, and orthogonal to both factors): the cross product is bilinear, so $a \times (u + v) = a \times u + a \times v$ and $a \times (λ u) = λ (a \times u)$ ; it is alternating, so $u \times u = 0$ and $u \times v = - (v \times u)$ . These identities distribute the force sums in step 1.1 and cancel each term in step 3.1.

[L4]

Reindexing (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule clause 2): the ordered pairs $(i, j)$ with $i \neq j$ are the disjoint union of the pairs ${(i, j), (j, i)}$ over $i < j$ , so $\sum_{i} \sum_{j \neq i} a_{i j} = \sum_{i < j} (a_{i j} + a_{j i})$ for vectors $a_{i j}$ ; finite sums are linear (Laws of finite sums and finite products, [L3]).

[L5]

A continuous real-valued function on an interval with zero derivative at every interior point is constant (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant); the interval $J$ has nonempty interior (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Proof

technique · direct; reindex the internal pair sums using antisymmetry, differentiate the angular momentum by the product rule, and apply the centrality and vanishing-torque hypotheses

1.1givenL1L3L4

Internal torque identity. Using [L4] and the antisymmetry of the pair maps, $\sum_{i} (x_{i} - o) \times \sum_{j \neq i} F_{i j} = \sum_{i < j} [(x_{i} - o) \times F_{i j} + (x_{j} - o) \times F_{j i}] = \sum_{i < j} (x_{i} - x_{j}) \times F_{i j} = I,$ because $F_{j i} = - F_{i j}$ and the cross product is bilinear by [L3]; the first equality also shows $I$ does not depend on $o$ . Adding $T_{o}^{e x t} = \sum_{i} (x_{i} - o) \times F_{i}^{e x t}$ and substituting the presentation $F_{i} = F_{i}^{e x t} + \sum_{j \neq i} F_{i j}$ gives $T_{o} = T_{o}^{e x t} + I$ , completing clause 1.

2.1step 1.1givenL1L2L3algebra

Balance. Each summand $(x_{i} - o) \times p_{i}$ of $J_{o}$ is a cross product of curves differentiable on $J$ (since $x_{i}$ is $C^{2}$ and $o$ is $C^{1}$ ), so by [L2] $J_{o}$ is differentiable on $J$ with $J_{o}^{'} = \sum_{i} (w_{i} - o^{'}) \times p_{i} + \sum_{i} (x_{i} - o) \times p_{i}^{'} .$ In the first sum, $\sum_{i} w_{i} \times p_{i} = \sum_{i} m_{i} w_{i} \times w_{i} = 0$ by alternation [L3], so it equals $- o^{'} \times P$ ; in the second sum, substitute $p_{i}^{'} = m_{i} {\ddot{x}}_{i} = F_{i} (t, x, w)$ from the equation of motion, which gives $\sum_{i} (x_{i} - o) \times F_{i} = T_{o} = T_{o}^{e x t} + I$ by clause 1. Therefore $J_{o}^{'} = T_{o}^{e x t} + I - o^{'} \times P$ , which is clause 2.

3.1step 2.1givenL3algebra

Centrality. Under the hypothesis of clause 3, for each $i < j$ $(x_{i} - x_{j}) \times F_{i j} = λ_{i j} (x_{i} - x_{j}) \times (x_{i} - x_{j}) = 0$ by alternation and bilinearity [L3]; hence $I = 0$ and step 2.1 gives $J_{o}^{'} = T_{o}^{e x t} - o^{'} \times P$ . If $o$ is constant then $o^{'} = 0$ and $J_{o}^{'} = T_{o}^{e x t}$ .

4.1step 2.1step 3.1L1L5∎

Conservation. Let $o$ be constant and $T_{o} \equiv 0$ on $J$ . Then $o^{'} = 0$ and step 2.1 gives $J_{o}^{'} = T_{o}^{e x t} + I = T_{o} = 0$ at every interior point of $J$ . $J_{o}$ is continuous on $J$ , being a finite sum of products of continuous curves; [L5] gives that $J_{o}$ is constant on $J$ . In the particular case, step 3.1 gives $T_{o} = T_{o}^{e x t} + I = T_{o}^{e x t} \equiv 0$ when the evaluated pair forces are central and the external torque vanishes; this includes a model with no external forces.

Remarks

The moving-origin correction. For a fixed origin the balance reads $J_{o}^{'} = T_{o}$ ; for a moving reference curve the extra term $- o^{'} \times P$ appears. The exact conservation condition is $T_{o}^{e x t} + I = o^{'} \times P$ along the motion. When the pair torque vanishes and the external torque is also zero, this reduces to $o^{'} \times P = 0$ ; the companion page checks the correction on explicit data.
Centrality is sufficient, not necessary. Clause 3 is one sufficient condition for $I \equiv 0$ ; clause 4 records that the conservation conclusion holds whenever the total torque about the fixed reference point vanishes, which can happen without centrality by cancellation among the pair torques. The companion counterexample makes this explicit.
No claim about nature. The pair structure, antisymmetry and centrality are properties a model may or may not declare; this item is a conditional deduction from them, and the framework's Restricted interaction models: pair decompositions, antisymmetry and centrality options, potential forces and the gravitational branch clause (i) records that an arbitrary force map need not admit such a presentation.

Angular momentum balance about a fixed or moving reference point, the internal torque sum, and the centrality case

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources