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The Galilean boost charge of the model: exact condition, uniform centre-of-mass motion, the induced motions and the boundary of the conservation law

Physical scope

The Galilean boost charge of the force model with a declared potential presentation in one inertial frame: with dimensionless a and a velocity-valued family parameter, the time-dependent triple (tau, xi, B) = (0, t a, a . M x) satisfies the model's Noether identity exactly when the differential translation condition sum_i a . grad_i U = 0 holds; in the global potential branch the boost charge a . (t P - M X) is then constant, the component a of the centre of mass moves uniformly, and admissible boosted curves t -> x(t) + s t a are again motions precisely under the endpoint-gradient agreement in Statement clause 4; for an arbitrary presentation the exact rate of the charge is the time times the component a of the total force, while nonconservation from a nonzero translation derivative is asserted in the global potential branch. The charge has dimension mass times length (action per velocity), and its rate has momentum dimensions. The theorem is conditional on the adopted force branch, the declared potential and the absence of residual forces in the conservation clauses; it asserts no boost invariance of any particular force law and carries no empirical content.

Statement

Adopt the force model with potential presentation $(U, d)$ and mechanical Lagrangian $L = K - U$ of The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions, with masses $m_{1}, \dots, m_{N} > 0$ , open configuration domain $Q \subseteq R^{3 N}$ , time window $I_{0}$ , mass operator $M$ and $U$ of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets. For this boost family take the parameter $s$ to have units of velocity and take $a \in R^{3}$ to be dimensionless; then $s t a$ is a displacement, $ξ^{a} = t a$ with $a : = (a, \dots, a) \in R^{3 N}$ has units of time, and $B^{a}$ and $K_{a}$ have units $k g m$ , equal to action per unit velocity. Fix a nonzero vector $a \in R^{3}$ and consider the maps

$ξ^{a} (t, x) : = t a, B^{a} (t, x) : = a \cdot \sum_{i} m_{i} x_{i},$

These maps, together with $τ \equiv 0$ , form a $C^{2}$ triple on $I_{0} \times Q$ . It is a symmetry datum in the sense of Continuous symmetries of the mechanical Lagrangian of a potential presentation: symmetry data, the Noether identity, the charge and its dimensions exactly when it satisfies the Noether identity, as characterised in clause 1 below. Write $P (t) : = \sum_{i} m_{i} x_{i}^{'} (t)$ for the total momentum and $X (t)$ for the centre of mass of Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame, where the scalar total mass is $M : = \sum_{i} m_{i}$ and $M X (t) = \sum_{i} m_{i} x_{i} (t)$ ; thus along a motion $B^{a} (t, x (t)) = M a \cdot X (t)$ . Define the boost charge

$K_{a} (t) : = a \cdot (t P (t) - M X (t)) = t (a \cdot P (t)) - \sum_{i} m_{i} a \cdot x_{i} (t) .$

Then:

(the identity). The triple $(0, ξ^{a}, B^{a})$ satisfies the model's Noether identity on $O$ if and only if the translation-invariance condition $\sum_{i} a \cdot \nabla_{x_{i}} U (t, x) = 0$ for all $(t, x) \in I_{0} \times Q$ of clause 1 of Translation symmetry of the model's Lagrangian and conservation of total momentum: exact hypothesis, conservation and its sharpness holds.
(conservation). If in addition $d \equiv 0$ , then for every motion $x : J \to Q$ of the model on an interval $J$ with nonempty interior (The associated first-order system, motions, maximal motions and the Newtonian flow of a force model), the function $t \mapsto K_{a} (t)$ is constant on $J$ .
(uniform centre-of-mass motion). Under the hypotheses of clause 2, the function $t \mapsto M a \cdot X (t)$ is affine on $J$ with slope $a \cdot P$ .
(boosted motions). Under the hypotheses of clause 2, let $x : J \to Q$ be a motion and $s \in R$ satisfy $x (t) + s t a \in Q$ for every $t \in J$ . Then $\tilde{x} (t) : = x (t) + s t a$ is a motion if and only if $\nabla_{x} U (t, \tilde{x} (t)) = \nabla_{x} U (t, x (t))$ for every $t \in J$ . This equality, and hence the boosted-motion conclusion, follows if each segment between $x (t)$ and $\tilde{x} (t)$ is contained in $Q$ , or if the all-admissible-endpoint finite invariance criterion of Translation symmetry of the model's Lagrangian and conservation of total momentum: exact hypothesis, conservation and its sharpness clause 1 holds.
(the general rate and the boundary). For an arbitrary presentation $(U, d)$ and every motion $x : J \to Q$ ,

$K_{a}^{'} (t) = t a \cdot F_{t o t} (t), F_{t o t} (t) : = \sum_{i} F_{i} (t, x (t), x^{'} (t));$

consequently, when $d \equiv 0$ , the charge $K_{a}$ is constant on $J$ if and only if $t \mapsto t \sum_{i} a \cdot \nabla_{x_{i}} U (t, x (t))$ vanishes identically on $J$ , and if the translation-invariance condition fails at some state, then there are a time $t_{0} \neq 0$ and a state $x_{0}$ with $g : = \sum_{i} a \cdot \nabla_{x_{i}} U (t_{0}, x_{0}) \neq 0$ such that the maximal motion through $(t_{0}, x_{0}, 0)$ satisfies $K_{a}^{'} (t_{0}) = - t_{0} g \neq 0$ and $K_{a}$ is not constant on its interval.

The charge exists as a conserved quantity exactly under the same potential invariance that conserves the momentum component $a \cdot P$ ; the conservation clauses are conditional deductions of the global potential branch, and no empirical content is asserted.

Facts & Assumptions

Given: The model, the presentation $(U, d)$ , the vector $a$ , the triple $(0, ξ^{a}, B^{a})$ , and the motions and states occurring in the clauses.

[F1]

Model data and chart (The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions, Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets, A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations, Configuration domain of $N$ labeled particles: open admissible sets, regular holonomic constraints and the mass operator): $O = I_{0} \times Q \times R^{3 N}$ is open and $(O, L)$ is a time-dependent Lagrangian chart whenever $Q \neq \emptyset$ and $a^{'} < b^{'}$ with $[a^{'}, b^{'}] \subseteq I_{0}$ , over $[a^{'}, b^{'}] \times Q$ ; $L = K - U$ with $L_{x} = - \nabla_{x} U$ , $L_{v} = M v$ ; $F_{i} = - \nabla_{x_{i}} U + d_{i}$ ; and $U$ is $C^{2}$ so that the map $(t, x) \mapsto \sum_{i} a \cdot \nabla_{x_{i}} U (t, x)$ is continuous (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[F2]

The model's symmetry vocabulary (Continuous symmetries of the mechanical Lagrangian of a potential presentation: symmetry data, the Noether identity, the charge and its dimensions): a $C^{2}$ triple $(τ, ξ, B)$ is a symmetry datum exactly when $L_{t} τ + L_{x} \cdot ξ + L_{v} \cdot (D_{t} ξ - v D_{t} τ) + L D_{t} τ = D_{t} B$ on $O$ , with charge $(M v) \cdot ξ - E τ - B$ and $E = K + U$ ; for $τ \equiv 0$ the identity is $L_{x} \cdot ξ + L_{v} \cdot D_{t} ξ = D_{t} B$ and the charge is $(M v) \cdot ξ - B$ .

[F3]

The general Noether theorem of the model (The general Noether theorem of the model: point transformations with time reparametrisation, the charge $p \cdot ξ - E τ - B$ , its residual-force rate and its potential gauge clause 1): a symmetry datum satisfying the model identity with $d \equiv 0$ has constant charge along every motion.

[F4]

The translation statement (Translation symmetry of the model's Lagrangian and conservation of total momentum: exact hypothesis, conservation and its sharpness clauses 1 and 3): the differential condition is equivalent to potential invariance along admissible segments; invariance across excluded segments also requires agreement of the interval-component constants. Under the differential condition with $d \equiv 0$ , $a \cdot P$ is constant along every motion.

[F5]

Momentum, centre of mass and motions (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, Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance): $P = \sum_{i} m_{i} x_{i}^{'}$ , $M = \sum_{i} m_{i}$ is the scalar total mass and $M X = \sum_{i} m_{i} x_{i}$ , so $M X^{'} = P$ ; along a motion $M x^{''} = F$ and $P^{'} = F_{t o t} : = \sum_{i} F_{i}$ , and through every initial datum there is exactly one maximal motion with open maximal interval containing the initial time.

[F6]

Calculus (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, 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, $C^{k}$ Euclidean maps and diffeomorphisms, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ): sums, scalar multiples and pairings differentiate by the usual rules; a $C^{1}$ function on an interval is constant exactly when its derivative vanishes. For $C^{2}$ $U$ , differentiating the vanishing directional derivative gives $\nabla_{x} (a \cdot \nabla_{x} U) = 0$ , so the spatial gradient is constant along every translation segment contained in $Q$ . Global finite invariance likewise gives equality of the endpoint gradients by differentiating its identity on an open neighbourhood of the two admissible endpoints.

[F7]

Parameter units (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework, Continuous symmetries of the mechanical Lagrangian of a potential presentation: symmetry data, the Noether identity, the charge and its dimensions): $[s] = ℓ τ^{- 1}$ and $[a] = 1$ , so $[K_{a}] = μ ℓ$ and $[K_{a}^{'}] = μ ℓ τ^{- 1}$ , as also follows from $K_{a}^{'} = t a \cdot F_{t o t}$ .

Proof

technique · direct; evaluate the model identity at the boost triple and reduce its validity to the translation condition by continuity in time, transfer conservation through the model's general Noether theorem, verify the boosted curves, and compute the charge rate from the equation of motion

1.1F1F2F4F6algebra

Clause 1, evaluation of the identity: for the triple $(0, ξ^{a}, B^{a})$ one has $D_{t} ξ^{a} = (a, \dots, a)$ , $L_{x} \cdot ξ^{a} = - t \sum_{i} a \cdot \nabla_{x_{i}} U (t, x)$ and $L_{v} \cdot D_{t} ξ^{a} = \sum_{i} m_{i} a \cdot v_{i} = D_{t} B^{a}$ , since $D_{t} B^{a} = 0 + \sum_{i} m_{i} a \cdot v_{i}$ ; hence the two sides of the strict identity are $t g (t, x)$ apart, with $g (t, x) : = \sum_{i} a \cdot \nabla_{x_{i}} U (t, x)$ , and the identity holds on $O$ if and only if $t g (t, x) = 0$ for all $(t, x) \in I_{0} \times Q$ . Since $g$ is continuous and $I_{0}$ contains points $t \neq 0$ , this holds if and only if $g (t, x) = 0$ for every $t \neq 0$ in $I_{0}$ and every $x \in Q$ ; and if $0 \in I_{0}$ then continuity of $g$ at $(0, x)$ forces $g (0, x) = 0$ for every $x \in Q$ . Hence the identity is equivalent to $g \equiv 0$ on $I_{0} \times Q$ , which is the translation-invariance condition of [F4].

1.2F1F5F6algebra

Clause 5, the rate: along a motion, $t \mapsto K_{a} (t) = t (a \cdot P (t)) - M a \cdot X (t)$ is $C^{1}$ by [F5] and [F6], and differentiating gives $K_{a}^{'} = a \cdot P + t a \cdot P^{'} - M a \cdot X^{'} = a \cdot P + t a \cdot F_{t o t} - a \cdot P = t a \cdot F_{t o t},$ because $M X^{'} = P$ and $P^{'} = F_{t o t} = \sum_{i} F_{i} (t, x, x^{'})$ along the motion. In the case $d \equiv 0$ one has $F_{i} (t, x, x^{'}) = - \nabla_{x_{i}} U (t, x)$ , so $K_{a}^{'} = - t g (t, x (t))$ and, $K_{a}$ being $C^{1}$ on the interval $J$ , it is constant exactly when $t g (t, x (t))$ vanishes identically on $J$ ; if $g$ is not identically zero on $I_{0} \times Q$ , continuity of $g$ supplies a point $(t_{0}, x_{0})$ with $t_{0} \neq 0$ and $g (t_{0}, x_{0}) \neq 0$ (a point with $g \neq 0$ at time $0$ has a neighbourhood on which $g \neq 0$ , and $I_{0}$ contains times $\neq 0$ ), and the maximal motion through $(t_{0}, x_{0}, 0)$ has $K_{a}^{'} (t_{0}) = - t_{0} g (t_{0}, x_{0}) \neq 0$ , so $K_{a}$ is not constant on its interval.

1.3F1F4F5F6algebra

Clause 4, boosted motions: with $d \equiv 0$ , the curve $\tilde{x} = x + s t a$ is $C^{2}$ , takes values in $Q$ by hypothesis and satisfies ${\tilde{x}}^{''} = x^{''}$ . Its motion equation $M {\tilde{x}}^{''} = - \nabla_{x} U (t, \tilde{x})$ is therefore equivalent to $\nabla_{x} U (t, \tilde{x}) = \nabla_{x} U (t, x)$ , since $x$ is a motion. If every joining segment lies in $Q$ , the derivative along that segment of $\nabla_{x} U$ is $\nabla_{x} (a \cdot \nabla_{x} U) = 0$ by the differential hypothesis and [F6], giving the equality. Alternatively, under global finite invariance, for fixed $t$ and shift $s t a$ both endpoints have an open neighbourhood on which $U (t, y + s t a) = U (t, y)$ ; differentiation with respect to $y$ gives the same gradient equality. Thus both stated sufficient routes produce a boosted motion, and the exact criterion holds without either extra route.

2.1F2F3step 1.1

Clause 2, conservation: by step 1.1 and the hypothesis, the datum $(0, ξ^{a}, B^{a})$ satisfies the model identity on $O$ , so with $d \equiv 0$ clause 1 of [F3] gives that its charge is constant along every motion; by [F2] that charge is $(M x^{'} (t)) \cdot t a - (K (x^{'} (t)) + U (t, x (t))) \cdot 0 - B^{a} (t, x (t)) = t (a \cdot P (t)) - M a \cdot X (t) = K_{a} (t)$ .

2.2F6step 1.2

Clause 5, the non-constancy statement: by step 1.2, in the case $d \equiv 0$ a point $(t_{0}, x_{0})$ with $t_{0} \neq 0$ and $g (t_{0}, x_{0}) \neq 0$ yields a maximal motion with $K_{a}^{'} (t_{0}) = - t_{0} g (t_{0}, x_{0}) \neq 0$ , and by [F6] a function constant on the interval $J_{\max}$ would have vanishing derivative at $t_{0}$ .

3.1F4step 2.1algebra

Clause 3: by step 2.1 the function $t \mapsto K_{a} (t) = t (a \cdot P (t)) - M a \cdot X (t)$ is constant on $J$ , and by [F4] the slope $a \cdot P$ is constant on $J$ ; hence $M a \cdot X (t) = t (a \cdot P) - K_{a}$ is affine in $t$ with slope $a \cdot P$ on $J$ .

4.1step 1.1step 1.2step 1.3step 2.1step 2.2step 3.1F7∎

Clause 1 is step 1.1, clause 2 is step 2.1, clause 3 is step 3.1, clause 4 is step 1.3, and the two statements of clause 5 are step 1.2 and step 2.2; the charge is conserved exactly under the potential invariance that conserves the momentum component, and no boost invariance of any particular force law and no empirical claim is asserted. By [F7], $[K_{a}] = μ ℓ$ and $[K_{a}'] = μ ℓ τ^{- 1}$ , the charge and rate dimensions for a velocity-valued boost parameter.

Remarks

Kinematics versus model symmetry. Every model of the framework is described in every inertial frame, and the Galilean transformation of the description is available whatever the force law (Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance clause (vi)); the boost charge, by contrast, is conserved in the global potential branch when the differential translation condition of Translation symmetry of the model's Lagrangian and conservation of total momentum: exact hypothesis, conservation and its sharpness holds for its generator. For the one-label confining oscillator $U = \frac{k}{2} ∣ x ∣^{2}$ , $k > 0$ with units $μ τ^{- 2}$ , the global translation condition fails for every nonzero direction $a$ (evaluate it at a nonzero position parallel to $a$ ). At positions with $a \cdot x \neq 0$ the directional derivative is nonzero, and in the global potential branch the centre of mass oscillates rather than moves uniformly. This is the precise boundary of the boost statement: it is a property of the model's Lagrangian, not of Galilean kinematics.
Relation to the momentum clause. Clauses 1 and 2 show that for the global potential branch the boost charge adds no independent conservation hypothesis beyond the differential translation condition already required by Translation symmetry of the model's Lagrangian and conservation of total momentum: exact hypothesis, conservation and its sharpness: the two charges $a \cdot P$ and $K_{a}$ stand or fall together, and clause 3 says that the content of the boost charge is the uniform motion of the component $a$ of the centre of mass. With residual forces the rate formula of clause 5 replaces conservation, and the charge fails to be conserved along every motion on which the component $a$ of the total force does not vanish identically.
Empirical status. All statements are deductions inside the model under the declared hypotheses; the phrase "boost" names the transformation $x_{i} \mapsto x_{i} + s t a$ of the description, not any measurement, and no claim is made that a physical force law is boost invariant.

The Galilean boost charge of the model: exact condition, uniform centre-of-mass motion, the induced motions and the boundary of the conservation law

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

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