Momentum, work and energy balances — Examples

Fix the two-body model of Restricted interaction models: pair decompositions, antisymmetry and centrality options, potential forces and the gravitational branch clause (iv) (the general pair-potential example) with $N=2$, masses $m_1=m_2=m>0$, time window $\mathbb{R}$, configuration domain $Q={\mathbb{R}}^6$ (the potential below is smooth everywhere, so no collision exclusion is needed), and the Hooke pair potential $U(x_1,x_2):=\frac{1}{2}k\parallel x_1-x_2{\parallel }_2^2,$ with a declared spring constant $k>0$ of dimension $\mu {\tau }^{-2}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework), so that $U$ has energy dimension. The force map is $F_1(x)=-k(x_1-x_2),F_2(x)=k(x_1-x_2),$ the two negative gradients of $U$; it is linear in the positions and independent of time and velocities, with $\parallel F(x)-F(x^{\mathrm{\prime }}){\parallel }_2\le 2k\parallel x-x^{\mathrm{\prime }}{\parallel }_2$, hence continuous and state-locally Lipschitz with the uniform constant $2k$. The model therefore satisfies the standing hypotheses of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance clause (iii), it has no external forces ($F_i^{\mathrm{e}\mathrm{x}\mathrm{t}}\equiv 0$), and its pair force is central: $F_{12}=-k(x_1-x_2)$ is parallel to $x_1-x_2$. It presents its force map through the global potential $U$ in the sense of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets clause (a).

Fix an oriented orthonormal triple $(e_1,e_2,e_3)$ with $e_3=e_1\times e_2$, a constant centre-of-mass velocity $V\in {\mathbb{R}}^3$ of dimension $\mathrm{\ell }{\tau }^{-1}$, and a length $R>0$, and set $\omega :=\sqrt{2k\mathrm{/}m}$ (dimension ${\tau }^{-1}$). Define $e(t):=\cos (\omega t)e_1+\sin (\omega t)e_2,x_1(t):=Vt-\frac{R}{2}e(t),x_2(t):=Vt+\frac{R}{2}e(t)(t\in \mathbb{R})\mathrm{.}$ Then $x=(x_1,x_2)$ is the unique maximal motion of the model with the datum $x_1(0)=-\frac{R}{2}{e}_1$, $x_2(0)=\frac{R}{2}{e}_1$, ${\dot{x}}_1(0)=V-\frac{R}{2}\omega e_2$, ${\dot{x}}_2(0)=V+\frac{R}{2}\omega e_2$, and along it:

1. the total momentum is $P=2mV$, constant, and the centre of mass is $X(t)=Vt$ with $V_{\mathrm{c}\mathrm{m}}=V$;
2. the kinetic energy is $K=m\parallel V{\parallel }_2^2+\frac{1}{4}m{R}^2{\omega }^2$, and its centre-of-mass/relative splitting holds with reduced mass $\mu =m\mathrm{/}2$ and relative coordinate $r=x_1-x_2$;
3. the angular momentum about the origin is $J_0=\frac{1}{2}m{R}^2\omega e_3$, constant, while for the moving reference curve $o(t)=\mathbf{W}t$ with a constant velocity $\mathbf{W}\in {\mathbb{R}}^3$ of dimension $\mathrm{\ell }{\tau }^{-1}$ and $\mathbf{W}\perp V$ one has $J_o^{\mathrm{\prime }}=-2m(\mathbf{W}\times V)$, in agreement with the moving-origin correction; this correction is nonzero when both $\mathbf{W}$ and $V$ are nonzero;
4. the mechanical energy is $E=m\parallel V{\parallel }_2^2+k{R}^2$, constant, and the power identity $K^{\mathrm{\prime }}=\mathcal{P}$ holds with $\mathcal{P}$ equal to the negative time derivative of the potential energy $\frac{1}{2}k\parallel r{\parallel }_2^2$;
5. under the Galilean change to the centre-of-mass frame ($b=0$, $a_0=0$, $R_{\mathrm{r}\mathrm{o}\mathrm{t}}=I_3$, $v_{\mathrm{b}\mathrm{o}\mathrm{o}\mathrm{s}\mathrm{t}}=-V$) the transformed quantities are $\tilde{P}=0$, $\tilde{K}=\frac{1}{2}k{R}^2$ and $\tilde{E}=k{R}^2$, in agreement with 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 clauses 2 and 5.

• What the check exhibits. The example exercises the balances of the page on one explicit motion: the algebraic splitting, the momentum law with vanishing total external force, the angular-momentum law with a moving reference point (where the correction $-o^{\mathrm{\prime }}\times P$ is nonzero when both chosen velocities $\mathbf{W}$ and $V$ are nonzero and perpendicular, while the torque vanishes), the work-energy and mechanical-energy balances, and the transformation laws of the kinetic and mechanical energies. The centrality clause of Angular momentum balance about a fixed or moving reference point, the internal torque sum, and the centrality case applies here; the companion counterexample exhibits a model where the same conservation conclusion holds without centrality.

• Scope. The parameters $m$, $k$, $R$, $V$ and $\mathbf{W}$ are declared data of this computation; no measured value is asserted, no physical spring is claimed to obey the Hooke law in any regime, and the check is a deduction inside the model, not evidence about nature.

Work in the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with one label $N=1$, mass $m>0$, time window $I_0=\mathbb{R}$ and configuration domain $Q={\mathbb{R}}^3$. Write $q=(x,y,z)\in Q$ and $e_1=(1,0,0)$. Define $\mathbf{d}(t,q,v):=d(t,x,v)e_1$ and $\mathbf{F}(t,q,v):=F(t,x,v)e_1$ from the scalar expressions below, and set $U(t,q):=U(t,x)$, independent of $y,z$. Here the velocity argument of $d,F$ may be any $v\in {\mathbb{R}}^3$; these expressions are independent of it. Declare the potential presentation of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets clause (a) $U(t,x):=\frac{1}{2}\lambda t{x}^2,d(t,x,v):=2ma+\lambda a{t}^3,$ where $a>0$ is a constant of dimension $\mathrm{\ell }{\tau }^{-2}$ and $\lambda >0$ a constant of dimension $\mu {\tau }^{-3}$, so that $U$ has dimension $\mu {\mathrm{\ell }}^2{\tau }^{-2}$ and $d$ has the dimension of a force (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). Both maps are smooth; the force map of the model is $F(t,x,v)=-{\mathrm{\partial }}_{x}U(t,x)+d(t,x,v)=-\lambda tx+2ma+\lambda a{t}^3$, and the model satisfies $\mathbf{F}=-{\mathrm{\nabla }}_{q}U+\mathbf{d}$ on $\mathbb{R}\times {\mathbb{R}}^3\times {\mathbb{R}}^3$. On each compact time interval $\mathrm{\mid }t\mathrm{\mid }\le T$, its force has state-Lipschitz constant $\lambda T$; $\mathbf{d}$ is state-independent. Thus both satisfy the standing regularity. The one-dimensional computation below uses the rectilinear motion $q(t)=x(t)e_1$ and zero transverse coordinates and velocities. Each scalar force and residual-force expression denotes its $e_1$ component, $\mathrm{\parallel }q\mathrm{\prime }(t){\mathrm{\parallel }}_2^2=x\mathrm{\prime }(t{)}^2$, and $⟨\mathbf{d},q\mathrm{\prime }⟩=dx\mathrm{\prime }$; no constrained configuration domain or reaction force is assumed.

The curve $x(t):=a{t}^2(t\in \mathbb{R})$ is a motion of this model, and with $K$, $\mathcal{P}$ 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 $E=K+U(t,x(t))$:

For the numerical clauses below, put $s:=t\mathrm{/}(1\mathrm{s})$; displayed energy, power and work values include their SI units, and the interval $0\le s\le 1$ means $0\le t\le 1\mathrm{s}$.

1. on the interval $[0,1\mathrm{s}]$ the numerical data $m:=1\mathrm{k}\mathrm{g}$, $a:=1\mathrm{m}{\mathrm{s}}^{-2}$, $\lambda :=1\mathrm{k}\mathrm{g}{\mathrm{s}}^{-3}$ give $K(t)=2s^2\mathrm{J},U(t,x(t))=\frac{1}{2}{s}^5\mathrm{J},E(t)=(2s^2+\frac{1}{2}{s}^5)\mathrm{J},E^{\mathrm{\prime }}(t)=(4s+\frac{5}{2}{s}^4)\mathrm{W},$ and the balance of Mechanical energy balance for a potential presentation: $E^{\mathrm{\prime }}={\mathrm{\partial }}_{t}U$ plus the residual power, conservation for a static potential with no residual forces, and the integrated work form clause 1, $E^{\mathrm{\prime }}={\mathrm{\partial }}_{t}U(t,x(t))+d\dot{x}$, holds with ${\mathrm{\partial }}_{t}U(t,x(t))=\frac{1}{2}{s}^4\mathrm{W}$ and $d\dot{x}=(4s+2s^4)\mathrm{W}$;
2. the power is $\mathcal{P}(t)=4s\mathrm{W}$, the work over $[0,1\mathrm{s}]$ is $W([0,1\mathrm{s}])=2\mathrm{J}$, and the work-energy theorem $K(1\mathrm{s})-K(0)=W([0,1\mathrm{s}])$ holds; the integrated form of clause 3 of Mechanical energy balance for a potential presentation: $E^{\mathrm{\prime }}={\mathrm{\partial }}_{t}U$ plus the residual power, conservation for a static potential with no residual forces, and the integrated work form holds as $2\mathrm{J}=-\frac{1}{2}\mathrm{J}+(\frac{1}{10}+2+\frac{2}{5})\mathrm{J}$;
3. replacing $U$ by the equivalent potential $\tilde{U}(t,x)=U(t,x)+c(t)$ with $c(t):=-(1\mathrm{J})(t\mathrm{/}(1\mathrm{s}){)}^2$ (same force, different representative) gives $\tilde{E}(t)=E(t)+c(t)=(s^2+\frac{1}{2}{s}^5)\mathrm{J}$ and ${\tilde{E}}^{\mathrm{\prime }}(t)=(2s+\frac{5}{2}{s}^4)\mathrm{W}={\mathrm{\partial }}_t\tilde{U}(t,x(t))+d\dot{x}$, so the balance holds in the renormalized presentation with a different value and rate;
4. the model is driven and the energy is not conserved: $E(1\mathrm{s})-E(0)=\frac{5}{2}\mathrm{J}$ in the first normalization, while the balance attributes the change to the explicit time dependence, ${\int }_0^{1\mathrm{s}}{\mathrm{\partial }}_{t}Udt=\frac{1}{10}\mathrm{J}$, and the residual power, ${\int }_0^{1\mathrm{s}}d\dot{x}dt=\frac{12}{5}\mathrm{J}$, whose sum $\frac{1}{10}\mathrm{J}+\frac{12}{5}\mathrm{J}=\frac{5}{2}\mathrm{J}$ is the energy change $E(1\mathrm{s})-E(0)={\int }_0^{1\mathrm{s}}{E}^{\mathrm{\prime }}dt$.

• The three contributions are visible. On this data the balance reads $E^{\mathrm{\prime }}=(\frac{1}{2}{s}^4)\mathrm{W}+(4s+2s^4)\mathrm{W}$: the first term is the explicit time dependence of the potential and the remaining terms are the power of the residual force. Both are positive for $0<t\le 1\mathrm{s}$, so the model is driven; this is the behaviour clause 2 of Mechanical energy balance for a potential presentation: $E^{\mathrm{\prime }}={\mathrm{\partial }}_{t}U$ plus the residual power, conservation for a static potential with no residual forces, and the integrated work form excludes by hypothesis.

• The representative matters. The renormalized potential $\tilde{U}$ presents the same force and the same motion, yet $\tilde{E}$ differs from $E$ by $c(t)$ and its rate by $c^{\mathrm{\prime }}(t)$; a conservation statement must name the representative. The numerical data are declared model parameters, not measured values.

In the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with the external-plus-pair presentation and the antisymmetry clause (ii)1 of Restricted interaction models: pair decompositions, antisymmetry and centrality options, potential forces and the gravitational branch, if the internal torque sum ${\sum }_{i<j}(x_i-x_j)\times F_{ij}$ vanishes identically, then every pair force $F_{ij}$ is central (parallel to $x_i-x_j$).

Equivalently: centrality of the pair forces is necessary for the internal-torque conclusion used by Angular momentum balance about a fixed or moving reference point, the internal torque sum, and the centrality case.