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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)

Physical scope

The algebraic identities of a curve family of a force model in one inertial frame: the total momentum is the total mass times the centre-of-mass velocity, the mass-weighted relative positions sum to zero together with their derivatives, and the total kinetic energy equals the centre-of-mass kinetic energy plus the mass-weighted relative kinetic energy, with the two-body reduced-mass form. The result is a conditional identity for the model's masses and curves and asserts no conservation law, no force structure and no measured value.

Statement

Adopt the force model, the 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: masses $m_{1}, \dots, m_{N} > 0$ , a curve family $x : J \to Q$ of class $C^{2}$ on an interval $J$ with nonempty interior, coordinate velocities $w_{i} = x_{i}^{'}$ , and the quantities $M = \sum_{i} m_{i}, P = \sum_{i} m_{i} w_{i}, X = M^{- 1} \sum_{i} m_{i} x_{i}, ρ_{i} = x_{i} - X, V = X^{'}, K = \frac{1}{2} \sum_{i} m_{i} ∥ w_{i} ∥_{2}^{2} .$ Then:

$P = M V$ on $J$ , that is $X^{'} = M^{- 1} P$ , and the identities $\sum_{i} m_{i} ρ_{i} = 0, \sum_{i} m_{i} ρ_{i}^{'} = 0$ hold on $J$ ;
for every $t \in J$ , $K (t) = \frac{1}{2} M ∥ V (t) ∥_{2}^{2} + \frac{1}{2} \sum_{i} m_{i} ∥ ρ_{i}^{'} (t) ∥_{2}^{2};$
if $N = 2$ , then with $r : = x_{1} - x_{2}$ and $μ : = m_{1} m_{2} / M$ one has, for every $t \in J$ , $K (t) = \frac{1}{2} M ∥ V (t) ∥_{2}^{2} + \frac{1}{2} μ ∥ r^{'} (t) ∥_{2}^{2} .$

The statement is an identity for the stated curves and masses in the fixed frame; it uses no equation of motion, no property of the force map and no conservation claim. The values are frame-dependent; the transformation laws are the frame-transformation item of this page.

Facts & Assumptions

Given: A force model of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance, one inertial frame, a $C^{2}$ curve family $x : J \to Q$ , and the masses $m_{i} > 0$ .

[L1]

Model 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): $M$ , $P$ , $X$ , $ρ_{i}$ , $V$ and $K$ are defined by the displayed finite sums from $x$ and the masses; $x_{i}$ is $C^{2}$ ; $w_{i} = x_{i}^{'}$ ; dimensions are read by Dimensions and SI unit conventions of the non-relativistic classical mechanics framework.

[L2]

Mathematical supplier (Centre of mass, the identity $\sum_{i} m_{i} ρ_{i} = 0$ , and the splitting of the kinetic sum $\sum_{i} m_{i} ∥ x_{i}^{'} ∥_{2}^{2}$ into centre-of-mass and relative parts (with the two-body reduced-mass case)): for $N \geq 1$ , an interval $I$ with at least two points, masses $m_{i} > 0$ and curves $x_{i} : I \to R^{3}$ differentiable at every point, the centre $X = M^{- 1} \sum_{i} m_{i} x_{i}$ and the relative positions $ρ_{i} = x_{i} - X$ are differentiable with $X^{'} = M^{- 1} \sum_{i} m_{i} x_{i}^{'}$ and $ρ_{i}^{'} = x_{i}^{'} - X^{'}$ ; the identities $\sum_{i} m_{i} ρ_{i} = 0$ and $\sum_{i} m_{i} ρ_{i}^{'} = 0$ hold; and for every point $\sum_{i} m_{i} ∥ x_{i}^{'} ∥_{2}^{2} = M ∥ X^{'} ∥_{2}^{2} + \sum_{i} m_{i} ∥ ρ_{i}^{'} ∥_{2}^{2},$ with the two-body case $\sum_{i} m_{i} ∥ ρ_{i}^{'} ∥_{2}^{2} = μ ∥ r^{'} ∥_{2}^{2}$ for $r = x_{1} - x_{2}$ , $μ = m_{1} m_{2} / M$ .

[L3]

The Euclidean norm satisfies $∥ u ∥_{2}^{2} = ⟨ u, u ⟩$ and scalar factors pull out of finite sums (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Laws of finite sums and finite products).

Proof

technique · direct; apply the mathematical supplier to the model's curves and masses and translate its notation through the definition

1.1givenL2

The interval $J$ has nonempty interior, hence at least two points, and each $x_{i}$ is differentiable on $J$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral); so the hypotheses of [L2] are met with $I : = J$ . Its first clause gives differentiability of $X$ and each $ρ_{i}$ , the formula $X^{'} = M^{- 1} \sum_{i} m_{i} x_{i}^{'}$ , the formula $ρ_{i}^{'} = x_{i}^{'} - X^{'}$ , and the identities $\sum_{i} m_{i} ρ_{i} = 0$ and $\sum_{i} m_{i} ρ_{i}^{'} = 0$ on $J$ .

2.1step 1.1L1algebra

Writing $x_{i}^{'} = w_{i}$ by [L1], the formula of step 1.1 reads $X^{'} = M^{- 1} \sum_{i} m_{i} w_{i} = M^{- 1} P$ , so $P = M V$ and $X^{'} = M^{- 1} P$ ; together with step 1.1 this is clause 1.

3.1step 2.1L1L2L3algebra

Clause 2: multiply the splitting identity of [L2] by $\frac{1}{2}$ and substitute $x_{i}^{'} = w_{i}$ and $X^{'} = V$ ; the left side is $K$ by [L1], and the two right-hand terms are $\frac{1}{2} M ∥ V ∥_{2}^{2}$ and $\frac{1}{2} \sum_{i} m_{i} ∥ ρ_{i}^{'} ∥_{2}^{2}$ .

4.1step 3.1L2algebra∎

Clause 3: for $N = 2$ , the two-body clause of [L2] gives $\sum_{i} m_{i} ∥ ρ_{i}^{'} ∥_{2}^{2} = μ ∥ r^{'} ∥_{2}^{2}$ with $r = x_{1} - x_{2}$ and $μ = m_{1} m_{2} / M$ ; substituting into step 3.1 gives the displayed two-body splitting.

Remarks

The splitting is instantaneous. Clause 2 is an identity at each time between the kinetic sums, not a decomposition of the motion; the two terms are not separately conserved in general, and no force hypothesis enters.
Frame dependence. The identity holds in every inertial frame with that frame's $X$ and $ρ_{i}$ . Under a Galilean change, $V$ gains the boost velocity while each relative velocity $ρ_{i}^{'}$ is rotated without a boost; thus the centre-of-mass kinetic term generally changes and the relative kinetic term is unchanged. The total kinetic-energy law is Transformation of the state quantities under a change of inertial frame: centre of mass, total momentum, kinetic energy, angular momentum, torque, power, work and mechanical energy clause 2.

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)

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources