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Total momentum balance and centre-of-mass equation for a model with antisymmetric pair forces

Physical scope

The momentum balance of a model that declares the external-plus-pair force presentation with antisymmetric pair forces: along every motion the total momentum changes at the rate of the total external force, it is constant on an interval exactly when the total external force vanishes there, and the centre of mass obeys the single-particle equation with the total mass and the total external force. The result is a conditional deduction from the adopted force and interaction posts; it asserts no conservation for models without the antisymmetry presentation 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 in addition 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, $F_{i} (t, z) = F_{i}^{e x t} (t, x_{i}, v_{i}) + \sum_{j \neq i} F_{i j} (t, x_{i}, x_{j}, v_{i}, v_{j}),$ with the antisymmetry clause (ii)1 of that postulate, $F_{i j} (t, x_{i}, x_{j}, v_{i}, v_{j}) = - F_{j i} (t, x_{j}, x_{i}, v_{j}, v_{i})$ on the common domain. Let $x : J \to Q$ be a motion of the model on an interval $J$ with nonempty interior, with quantities $M$ , $P$ , $X$ , $V$ as in the definition, and write $σ (t) : = \sum_{i} F_{i}^{e x t} (t, x_{i} (t), w_{i} (t)) (t \in J)$ for the total external force along the motion. Then:

$P$ is of class $C^{1}$ on $J$ and $\dot{P} (t) = σ (t) (t \in J) :$ the total momentum changes at the rate of the total external force, the internal pair forces cancelling in the sum;
if $σ \equiv 0$ on $J$ , then $P$ is constant on $J$ and $V = M^{- 1} P$ is constant on $J$ ;
$X$ is of class $C^{2}$ on $J$ and $M X^{''} (t) = σ (t) (t \in J) :$ the centre of mass obeys the equation of a single particle of mass $M$ under the total external force.

The statement assumes the pair presentation and its antisymmetry; for a model whose force map is not presented through such pairs it asserts nothing, and the pair forces of the presentation need not be central (centrality is used only in the angular-momentum balance of this page). 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, and a motion $x : J \to Q$ .

[L1]

Notions and notation (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): $w_{i} = x_{i}^{'}$ , $p_{i} = m_{i} w_{i}$ , $P = \sum_{i} p_{i}$ , $M = \sum_{i} m_{i}$ , $X = M^{- 1} \sum_{i} m_{i} x_{i}$ , $V = X^{'}$ ; a motion satisfies $m_{i} {\ddot{x}}_{i} (t) = F_{i} (t, x (t), w (t))$ for all $i$ and $t \in J$ ; the external-plus-pair presentation is an identity for $F_{i}$ on its declared domain, and its pair maps satisfy the antisymmetry of clause (ii)1.

[L2]

Mathematical supplier (Total momentum and total angular momentum balances for a finite system with antisymmetric pair forces; pairwise central forces make the internal torque vanish clause 1): if $x_{i} : I \to R^{3}$ are twice differentiable, $p_{i} = m_{i} x_{i}^{'}$ and there are maps $f_{i}, f_{i j} : I \to R^{3}$ with $f_{i j} = - f_{j i}$ and $p_{i}^{'} = f_{i} + \sum_{j \neq i} f_{i j}$ , then $P = \sum_{i} p_{i}$ is differentiable on $I$ with $P^{'} = \sum_{i} f_{i}$ .

[L3]

A continuous real function on an interval whose derivative vanishes 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). Differentiability, sums and scalar multiples of curves behave as in The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral; finite sums are linear (Laws of finite sums and finite products).

Proof

technique · direct; evaluate the model's pair presentation along the motion and apply the mathematical balance identity

1.1givenL1

Define, for $t \in J$ , $f_{i} (t) : = F_{i}^{e x t} (t, x_{i} (t), w_{i} (t)), f_{i j} (t) : = F_{i j} (t, x_{i} (t), x_{j} (t), w_{i} (t), w_{j} (t)) (i \neq j) .$ These are well-defined functions $J \to R^{3}$ : the evaluated arguments lie in the stated domains of the pair presentation and of the model's state domain, and the presentation is an identity for $F_{i}$ there. By the antisymmetry clause, $f_{i j} (t) = - f_{j i} (t)$ for all $i \neq j$ and $t \in J$ .

2.1step 1.1L1L2algebra

Each $p_{i} = m_{i} w_{i}$ is differentiable on $J$ with $p_{i}^{'} = m_{i} {\ddot{x}}_{i}$ ; by the equation of motion and the presentation this equals $F_{i} (t, x (t), w (t)) = f_{i} (t) + \sum_{j \neq i} f_{i j} (t) (t \in J) .$ Thus the data of [L2] are met on the interval $J$ with the maps $f_{i}$ , $f_{i j}$ of step 1.1, and [L2] yields that $P$ is differentiable on $J$ with $P^{'} (t) = \sum_{i} f_{i} (t) = σ (t);$ $P^{'}$ is continuous because the $f_{i}$ are continuous, so $P$ is $C^{1}$ on $J$ , and clause 1 is proved.

3.1step 2.1L3algebra

Clause 2: if $σ \equiv 0$ on $J$ , then $P^{'} = 0$ at every interior point of $J$ by step 2.1, and $P$ is continuous on $J$ ; [L3] gives that $P$ is constant on $J$ ; then $V = M^{- 1} P$ is constant on $J$ as a constant multiple.

4.1step 2.1L1algebra∎

Clause 3: by clause 1 of Total momentum equals total mass times centre-of-mass velocity, and the kinetic energy splits into centre-of-mass and relative parts (two-body reduced mass) applied to the motion (which is a $C^{2}$ curve family), $X$ is differentiable on $J$ with $X^{'} = M^{- 1} P$ ; since $P$ is $C^{1}$ by step 2.1, $X^{'}$ is $C^{1}$ , so $X$ is $C^{2}$ with $X^{''} = M^{- 1} P^{'} = M^{- 1} σ,$ that is $M X^{''} = σ$ , which is clause 3.

Remarks

The hypothesis is the presentation, not the physics. The cancellation $\sum_{i} \sum_{j \neq i} f_{i j} = 0$ is a consequence of the declared antisymmetry of the model's pair maps; a force map for which no such presentation is declared has no momentum balance of this form. In particular, the framework does not claim that every interaction admits an antisymmetric pair presentation (Restricted interaction models: pair decompositions, antisymmetry and centrality options, potential forces and the gravitational branch clause (i)).
Open systems. For a proper subsystem, forces to labels outside the declared set appear among the $F_{i}^{e x t}$ and need satisfy no antisymmetry; the subsystem's momentum may be exchanged with the environment. Clause 1 is a balance, and clause 2 gives conservation when the evaluated total external force vanishes. This is the open-system boundary recorded in Status audit and model boundaries of the force page: open systems, frame selection, breakdown and empirical scope.
No centrality is used. The balance of this item holds for velocity- and position-dependent antisymmetric pair forces alike; centrality is required only for the internal-torque statement of Angular momentum balance about a fixed or moving reference point, the internal torque sum, and the centrality case, and is sufficient but not necessary there.

Total momentum balance and centre-of-mass equation for a model with antisymmetric pair forces

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources