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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

Physical scope

The transformation laws of the model's mechanical quantities under a change of inertial frame: the centre of mass and centre-of-mass velocity transform affinely, the total momentum gains M times the boost velocity, the kinetic energy gains a cross term and the boost kinetic energy, the angular momentum and torque about corresponding C1 reference curves acquire the stated rotation and boost terms, and power and work transform according to the transferred force map of the force postulate, with the momentum-increment work form requiring a motion with the antisymmetric pair presentation; for a globally potential model whose potential is transferred by the scalar rule, the mechanical energy gains the same kinetic boost terms. The result is a conditional deduction from the adopted frame, worldline and force laws; it asserts no invariance of these quantities and no measured value.

Statement

Adopt the force model 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 let $F = (o, u, (e_{1}, e_{2}, e_{3})), F^{'} = (o^{'}, u^{'}, (e_{1}^{'}, e_{2}^{'}, e_{3}^{'}))$ be two inertial frames with coordinates $(t, x)$ and $(t^{'}, x^{'})$ related by $t^{'} = t + b, x^{'} = R x + v t + a_{0}, b \in R, a_{0}, v \in R^{3}, R \in S O (3)$ (Galilean transformation of worldlines: velocity and acceleration laws, frame-independence of zero coordinate acceleration, and the group of frame changes clause 1, The orthogonal group $O (3)$ and the rotation group $S O (3)$ ). Let $x : J \to Q$ be a curve family of the model in the first frame and let $η_{i} (t^{'}) : = R x_{i} (t^{'} - b) + v (t^{'} - b) + a_{0} (t^{'} \in J + b)$ be the description of the same worldlines in the second frame. Write $σ_{i} (t) : = F_{i} (t, x (t), w (t))$ and $Σ (t) : = \sum_{i} σ_{i} (t)$ for the evaluated total forces in the first frame, and read the second frame's quantities from the description $η$ with the transferred force map $F_{*}$ of clause (vi) of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance: ${\tilde{σ}}_{i} (t^{'}) : = F_{* i} (t^{'}, η (t^{'}), η^{'} (t^{'}))$ , $\tilde{P} : = \sum_{i} m_{i} η_{i}^{'}$ , $\tilde{X} : = M^{- 1} \sum_{i} m_{i} η_{i}$ , $\tilde{K} : = \frac{1}{2} \sum_{i} m_{i} ∥ η_{i}^{'} ∥_{2}^{2}$ , ${\tilde{J}}_{\tilde{o}} : = \sum_{i} (η_{i} - \tilde{o}) \times m_{i} η_{i}^{'}$ and ${\tilde{T}}_{\tilde{o}} : = \sum_{i} (η_{i} - \tilde{o}) \times {\tilde{σ}}_{i}$ for any $C^{1}$ reference curve $\tilde{o} : J + b \to R^{3}$ (clause 3 uses the curve corresponding to $o$ ), $\tilde{P} : = \sum_{i} ⟨ {\tilde{σ}}_{i}, η_{i}^{'} ⟩$ and $\tilde{W} ([t_{0}^{'}, t_{1}^{'}]) : = \int_{t_{0}^{'}}^{t_{1}^{'}} \tilde{P} (t^{'}) d t^{'}$ for $t_{0}^{'} \leq t_{1}^{'}$ in $J + b$ . Then, with all first-frame quantities read at the corresponding time $t^{'} - b$ :

Centre of mass and momentum. $\tilde{X} (t^{'}) = R X (t^{'} - b) + v (t^{'} - b) + a_{0}, {\tilde{X}}^{'} (t^{'}) = R V (t^{'} - b) + v, \tilde{P} (t^{'}) = R P (t^{'} - b) + M v;$
Kinetic energy. $\tilde{K} (t^{'}) = K (t^{'} - b) + ⟨ v, R P (t^{'} - b) ⟩ + \frac{1}{2} M ∥ v ∥_{2}^{2};$
Angular momentum and torque. For every $C^{1}$ reference curve $o : J \to R^{3}$ of the definition, let $\tilde{o} (t^{'}) : = R o (t^{'} - b) + v (t^{'} - b) + a_{0}$ and $X_{o} (t) : = M^{- 1} \sum_{i} m_{i} (x_{i} (t) - o (t)) = X (t) - o (t)$ . Then ${\tilde{J}}_{\tilde{o}} (t^{'}) = R J_{o} (t^{'} - b) + (R X_{o} (t^{'} - b)) \times (M v), {\tilde{T}}_{\tilde{o}} (t^{'}) = R T_{o} (t^{'} - b),$ where $T_{o} (t) : = \sum_{i} (x_{i} (t) - o (t)) \times σ_{i} (t)$ is the evaluated total torque of the first frame about $o (t)$ ;
Power and work. $\tilde{P} (t^{'}) = P (t^{'} - b) + ⟨ R Σ (t^{'} - b), v ⟩,$ and for every curve family and $t_{0} < t_{1}$ in $J$ $\tilde{W} ([t_{0} + b, t_{1} + b]) = W ([t_{0}, t_{1}]) + \int_{t_{0}}^{t_{1}} ⟨ R Σ (t), v ⟩ d t .$ If in addition $x$ is a motion and the model invokes the pair presentation of Restricted interaction models: pair decompositions, antisymmetry and centrality options, potential forces and the gravitational branch with the antisymmetry clause, then $Σ = P^{'}$ along the motion by Total momentum balance and centre-of-mass equation for a model with antisymmetric pair forces, and $\tilde{W} ([t_{0} + b, t_{1} + b]) = W ([t_{0}, t_{1}]) + ⟨ R (P (t_{1}) - P (t_{0})), v ⟩;$
Potential and mechanical energy. If the model is a global potential model of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets clause (a) with potential $U : I_{0} \times Q \to R$ of class $C^{2}$ and the second frame's model declares the scalar transfer $U_{*} (t^{'}, y) : = U (t^{'} - b, R^{T} (y_{1} - v (t^{'} - b) - a_{0}), \dots, R^{T} (y_{N} - v (t^{'} - b) - a_{0})),$ then $U_{*}$ is $C^{2}$ on its open transferred domain, $F_{* i} (t^{'}, y, u) = - \nabla_{y_{i}} U_{*} (t^{'}, y)$ at corresponding data, and the mechanical energies of clause (b) of that definition satisfy $\tilde{E} (t^{'}) : = \tilde{K} (t^{'}) + U_{*} (t^{'}, η (t^{'})) = E (t^{'} - b) + ⟨ v, R P (t^{'} - b) ⟩ + \frac{1}{2} M ∥ v ∥_{2}^{2} .$

Whether the second frame's model uses the scalar transfer of clause 5 is a modelling declaration; a model whose potential is not transferred by this rule (for instance a background potential that selects a frame) is a different model in the second frame, and no transformation law of $E$ is asserted for it. No empirical content is claimed.

Facts & Assumptions

Given: The force model and one curve family $x : J \to Q$ in the first inertial frame, a second inertial frame with the transformation parameters $(b, a_{0}, v, R)$ , and the description $η$ of the worldlines in the second frame; for clause 3, also a $C^{1}$ reference curve $o : J \to R^{3}$ .

[L1]

Kinematics (Galilean transformation of worldlines: velocity and acceleration laws, frame-independence of zero coordinate acceleration, and the group of frame changes clause 2): each $η_{i}$ is of class $C^{2}$ and $η_{i}^{'} (t^{'}) = R w_{i} (t^{'} - b) + v, η_{i}^{''} (t^{'}) = R {\ddot{x}}_{i} (t^{'} - b) (t^{'} \in J + b) .$

[L2]

Notions (Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame): $w_{i} = x_{i}^{'}$ , $p_{i} = m_{i} w_{i}$ , $P = \sum_{i} p_{i}$ , $X = M^{- 1} \sum_{i} m_{i} x_{i}$ , $V = X^{'}$ , $K = \frac{1}{2} \sum_{i} m_{i} ∥ w_{i} ∥_{2}^{2}$ , $P = \sum_{i} ⟨ σ_{i}, w_{i} ⟩$ , $J_{o} = \sum_{i} (x_{i} - o (t)) \times p_{i}$ and $T_{o} = \sum_{i} (x_{i} - o (t)) \times σ_{i}$ for a $C^{1}$ reference curve $o$ , and the work $W$ is the integral of $P$ ; the corresponding tilde-quantities of the statement are the same construction in the second frame from $η$ and ${\tilde{σ}}_{i} = F_{* i} (t^{'}, η, η^{'})$ . For a motion, $m_{i} {\ddot{x}}_{i} = σ_{i}$ and the transferred description satisfies $m_{i} η_{i}^{''} = {\tilde{σ}}_{i}$ , with ${\tilde{σ}}_{i} (t^{'}) = R σ_{i} (t^{'} - b)$ at corresponding data (clause (vi) of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance, which defines $F_{*}$ by $F_{*} (t^{'}, y, u) = R_{N} F (t^{'} - b, R_{N}^{T} (y - \overset{ˉ}{v} (t^{'} - b) - {\overset{ˉ}{a}}_{0}), R_{N}^{T} (u - \overset{ˉ}{v}))$ , where $R_{N} : = diag (R, \dots, R)$ and $\overset{ˉ}{v} : = (v, \dots, v)$ , ${\overset{ˉ}{a}}_{0} : = (a_{0}, \dots, a_{0})$ are the repeated $N$ -tuples; for a global potential model $σ_{i} = - \nabla_{x_{i}} U$ and $E = K + U (t, x (t))$ (Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets clauses (a)-(b)).

[L3]

Orthogonality consequences: for $R \in S O (3)$ one has $R^{T} R = I_{3}$ ( $S O (3)$ is a subgroup of the general linear group, with $R^{- 1} = R^{T}$ ), hence $⟨ R u, R z ⟩ = u^{T} R^{T} R z = u^{T} z = ⟨ u, z ⟩, ∥ R u ∥_{2} = ∥ u ∥_{2}$ for all $u, z \in R^{3}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, The transpose $A^{T}$ of a matrix); moreover $R (u \times z) = (R u) \times (R z),$ because $R [u]_{\times} z = R [u]_{\times} R^{T} R z = [R u]_{\times} R z$ by clause 3 and clause 1 of The skew-matrix/cross-product correspondence $[ω]_{\times} z = ω \times z$ and its conjugation law $R [ω]_{\times} R^{T} = [R ω]_{\times}$ together with $R^{T} R = I_{3}$ (The cross product in $R^{3}$ , The cross product is bilinear, alternating, and orthogonal to both factors).

[L4]

Chain rule and gradients (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder, A total derivative computes every directional derivative, and its matrix is the Jacobian, If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative, $C^{k}$ Euclidean maps and diffeomorphisms, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case): the maps of the statement are compositions of $C^{1}$ maps with the affine transfer $(t^{'}, y) \mapsto (t^{'} - b, R_{N}^{T} (y - \overset{ˉ}{v} (t^{'} - b) - {\overset{ˉ}{a}}_{0}))$ , whose Jacobian in $y$ is the orthogonal block matrix $R_{N}^{T}$ ; finite sums, scalar multiples and integrals of differentiable curves behave as in The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, Laws of finite sums and finite products and Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative.

[L5]

Balance supplier (Total momentum balance and centre-of-mass equation for a model with antisymmetric pair forces): for a motion of a model that invokes the external-plus-pair presentation with the antisymmetry clause, $\dot{P} = \sum_{i} F_{i}^{e x t}$ , and also $\sum_{i} F_{i} = \sum_{i} F_{i}^{e x t}$ because the internal pairs cancel.

Proof

technique · direct; compute each quantity of the second frame from the kinematic laws and the orthogonality of $R$, and integrate the power identity for the work law

1.1givenL1L2L4algebra

Centre of mass. By the definition of $\tilde{X}$ and the kinematic law, $\tilde{X} (t^{'}) = M^{- 1} \sum_{i} m_{i} (R x_{i} (t^{'} - b) + v (t^{'} - b) + a_{0}) = R (M^{- 1} \sum_{i} m_{i} x_{i} (t^{'} - b)) + v (t^{'} - b) + a_{0} = R X (t^{'} - b) + v (t^{'} - b) + a_{0},$ where linearity of $R$ and the finite sum are used ([L2], [L4]); its $t^{'}$ -derivative is $R V (t^{'} - b) + v$ by the chain rule applied to $t^{'} \mapsto X (t^{'} - b)$ with $V = X^{'}$ .

1.2L3

Orthogonality. For $R \in S O (3)$ and $u, z \in R^{3}$ , $⟨ R u, R z ⟩ = ⟨ u, z ⟩$ and $∥ R u ∥_{2} = ∥ u ∥_{2}$ because $⟨ R u, R z ⟩ = (R u)^{T} (R z) = u^{T} R^{T} R z = u^{T} z$ with $R^{T} R = I_{3}$ ; and $R (u \times z) = (R u) \times (R z)$ because $R (u \times z) = R [u]_{\times} z = R [u]_{\times} R^{T} R z = [R u]_{\times} (R z)$ by the skew-matrix correspondence and $R^{T} R = I_{3}$ . These two consequences are used freely below.

1.3givenL2L4algebra

Potential transfer. Let the model be a global potential model with $σ_{i} = - \nabla_{x_{i}} U$ and let $U_{*}$ be the scalar transfer of the statement. $U_{*}$ is $C^{2}$ , being the composition of the $C^{2}$ map $U$ with the affine transfer $y \mapsto R_{N}^{T} (y - \overset{ˉ}{v} (t^{'} - b) - {\overset{ˉ}{a}}_{0})$ and the time shift ([L4]). Its spatial gradient is computed by the chain rule: at corresponding data $(t^{'}, y)$ with $x = R_{N}^{T} (y - \overset{ˉ}{v} (t^{'} - b) - {\overset{ˉ}{a}}_{0})$ , $\nabla_{y_{i}} U_{*} (t^{'}, y) = R \nabla_{x_{i}} U (t, x),$ because the Jacobian of $y_{i} \mapsto R^{T} (y_{i} - v (t^{'} - b) - a_{0})$ is $R^{T}$ and $(R^{T})^{T} = R$ (The transpose $A^{T}$ of a matrix, A total derivative computes every directional derivative, and its matrix is the Jacobian). Hence $- \nabla_{y_{i}} U_{*} (t^{'}, η (t^{'})) = R (- \nabla_{x_{i}} U (t, x (t))) = R σ_{i} (t) = F_{* i} (t^{'}, η (t^{'}), η^{'} (t^{'}))$ at corresponding times, so the transferred model presents its force map through $U_{*}$ .

2.1step 1.1step 1.2L1L2algebra

Momentum and kinetic energy. By [L1], $\tilde{P} = \sum_{i} m_{i} (R w_{i} + v) = R \sum_{i} m_{i} w_{i} + M v = R P (t^{'} - b) + M v .$ For the kinetic energy, expand with bilinearity of the inner product and step 1.2: $∥ η_{i}^{'} ∥_{2}^{2} = ∥ R w_{i} + v ∥_{2}^{2} = ∥ w_{i} ∥_{2}^{2} + 2 ⟨ R w_{i}, v ⟩ + ∥ v ∥_{2}^{2},$ so, summing with the masses, $\tilde{K} = \frac{1}{2} \sum_{i} m_{i} ∥ w_{i} ∥_{2}^{2} + ⟨ R \sum_{i} m_{i} w_{i}, v ⟩ + \frac{1}{2} M ∥ v ∥_{2}^{2} = K (t^{'} - b) + ⟨ R P (t^{'} - b), v ⟩ + \frac{1}{2} M ∥ v ∥_{2}^{2},$ which with step 1.1 is clauses 1 and 2.

2.2step 1.2L1L2algebra

Angular momentum and torque. At corresponding times $t = t^{'} - b$ , the definitions give $η_{i} (t^{'}) - \tilde{o} (t^{'}) = R (x_{i} (t) - o (t))$ . By the cross-product identity of step 1.2, ${\tilde{J}}_{\tilde{o}} = \sum_{i} (R (x_{i} - o)) \times m_{i} (R w_{i} + v) = \sum_{i} m_{i} R ((x_{i} - o) \times w_{i}) + \sum_{i} (R (x_{i} - o)) \times m_{i} v .$ The first sum is $R \sum_{i} (x_{i} - o) \times m_{i} w_{i} = R J_{o} (t^{'} - b)$ ; in the second, $\sum_{i} (R (x_{i} - o)) \times m_{i} v = R [\sum_{i} m_{i} (x_{i} - o) \times R^{T} v] = R [(M X_{o}) \times R^{T} v] = (R M X_{o}) \times v = (R X_{o}) \times (M v),$ using $v = R (R^{T} v)$ , the cross-product identity, the definition of $X_{o}$ and bilinearity. Hence ${\tilde{J}}_{\tilde{o}} = R J_{o} + (R X_{o}) \times (M v)$ . For the torque, the transfer rule gives ${\tilde{σ}}_{i} (t^{'}) = R σ_{i} (t^{'} - b)$ at corresponding data, so ${\tilde{T}}_{\tilde{o}} = \sum_{i} (R (x_{i} - o)) \times R σ_{i} = \sum_{i} R ((x_{i} - o) \times σ_{i}) = R T_{o} (t^{'} - b),$ again by step 1.2; this is clause 3.

2.3step 1.2L1L2algebra

Power. By the transfer rule and [L1], $\tilde{P} = \sum_{i} ⟨ R σ_{i}, R w_{i} + v ⟩ = \sum_{i} ⟨ σ_{i}, w_{i} ⟩ + ⟨ R \sum_{i} σ_{i}, v ⟩ = P (t^{'} - b) + ⟨ R Σ (t^{'} - b), v ⟩,$ using step 1.2 in the first equality of the middle expression.

3.1step 2.3L2L4L5algebra

Work. The force map is continuous and the curve family is $C^{2}$ , so $P$ and $⟨ R Σ, v ⟩$ are continuous on $J$ and these Riemann integrals exist. Integrating the power identity of step 2.3 over $[t_{0} + b, t_{1} + b]$ and substituting $t = t^{'} - b$ gives $\tilde{W} ([t_{0} + b, t_{1} + b]) = \int_{t_{0}}^{t_{1}} P (t) d t + \int_{t_{0}}^{t_{1}} ⟨ R Σ (t), v ⟩ d t = W ([t_{0}, t_{1}]) + \int_{t_{0}}^{t_{1}} ⟨ R Σ (t), v ⟩ d t,$ the general work law in clause 4. If $x$ is a motion of a model with the antisymmetric pair presentation, then [L5] and Total momentum balance and centre-of-mass equation for a model with antisymmetric pair forces give $Σ = P^{'}$ on $J$ ; since $P$ is $C^{1}$ by that balance theorem, the Newton–Leibniz theorem applied componentwise yields $\int_{t_{0}}^{t_{1}} ⟨ R Σ (t), v ⟩ d t = ⟨ R (P (t_{1}) - P (t_{0})), v ⟩,$ which is the conditional specialization in clause 4.

4.1step 2.1step 1.3L2algebra∎

Mechanical energy. With the transfer of step 1.3, $U_{*} (t^{'}, η (t^{'})) = U (t^{'} - b, x (t^{'} - b)),$ by the definitions of $U_{*}$ and $η$ ; combining with clause 2 of step 2.1, $\tilde{E} = \tilde{K} + U_{*} (t^{'}, η (t^{'})) = K (t^{'} - b) + ⟨ R P (t^{'} - b), v ⟩ + \frac{1}{2} M ∥ v ∥_{2}^{2} + U (t^{'} - b, x (t^{'} - b)) = E (t^{'} - b) + ⟨ v, R P (t^{'} - b) ⟩ + \frac{1}{2} M ∥ v ∥_{2}^{2},$ which is clause 5.

Remarks

Conservation under a frame change has explicit conditions. If $P$ is constant, then $\tilde{P} = R P + M v$ is constant; if both $P$ and $K$ are constant, then the kinetic-energy law makes $\tilde{K}$ constant. Under the scalar potential transfer, constancy of both $E$ and $P$ makes $\tilde{E}$ constant. The angular-momentum law includes the time-dependent term $(R X_{o} (t^{'} - b)) \times (M v)$ , so constancy of $J_{o}$ alone does not generally make ${\tilde{J}}_{\tilde{o}}$ constant; it does so exactly when that added term is constant.
The transferred model is a declaration. Clause (vi) of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance fixes how a force law is read in another inertial frame; clause 5 adds the analogous scalar transfer for a potential and is asserted only when the model declares it. A background potential that does not transfer this way selects a frame, consistent with Status audit and model boundaries of the force page: open systems, frame selection, breakdown and empirical scope.

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

Statement

Facts & Assumptions

Proof

Remarks

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