Symmetry, Noether and conditional conservation — Examples

Work in the one-label uniform-force model of The uniform-force model: explicit global motion of a label under a constant declared field, with uniqueness, dimensions and frame covariance: the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with $N=1$, mass $m>0$, time window $I_0=\mathbb{R}$, configuration domain $Q={\mathbb{R}}^3$ and force map $F(t,x,v)=m\mathbf{g}$ for a declared constant parameter $\mathbf{g}\in {\mathbb{R}}^3$ of dimension $\mathrm{\ell }{\tau }^{-2}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). Adopt the potential presentation

$U(t,x):=-m\mathbf{g}\cdot x((t,x)\in \mathbb{R}\times {\mathbb{R}}^3),$

of class $C^{\mathrm{\infty }}$ and dimension $\mu {\mathrm{\ell }}^2{\tau }^{-2}$, so that $-{\mathrm{\nabla }}_{x}U=m\mathbf{g}=F$ and the model is a global potential model $d\equiv 0$ 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; declare a nonzero dimensionless $a\in {\mathbb{R}}^3$ and consider the motion. For the boost family, its parameter $s$ is measured in velocity units, so $K_a$ has units $\mathrm{k}\mathrm{g}\mathrm{m}$ and $K_a^{\mathrm{\prime }}$ has units $\mathrm{k}\mathrm{g}\mathrm{m}{\mathrm{s}}^{-1}$. For the time-translation datum use a parameter measured in seconds, so its generator value $\tau \equiv 1$ is dimensionless and its charge $-E$ is measured in joules. Consider the motion

$x(t):=x_0+v_{0}t+\frac{1}{2}\mathbf{g}{t}^2(t\in \mathbb{R})$

with $x_0,v_0\in {\mathbb{R}}^3$, which is the unique maximal motion through $(0,x_0,v_0)$ by The uniform-force model: explicit global motion of a label under a constant declared field, with uniqueness, dimensions and frame covariance. Verify:

1. the directional derivative of the potential is $a\cdot {\mathrm{\nabla }}_{x}U(t,x)=-ma\cdot \mathbf{g}$, so the translation condition of Translation symmetry of the model's Lagrangian and conservation of total momentum: exact hypothesis, conservation and its sharpness clause 1 holds exactly when $a\cdot \mathbf{g}=0$;
2. the momentum component is $a\cdot P(t)=ma\cdot v_0+mt(a\cdot \mathbf{g})$, hence constant on $\mathbb{R}$ exactly when $a\cdot \mathbf{g}=0$, with rate $ma\cdot \mathbf{g}$ otherwise;
3. the boost charge is $K_a(t)=t(a\cdot P(t))-ma\cdot x(t)=\frac{1}{2}m{t}^2(a\cdot \mathbf{g})-ma\cdot x_0$, with rate $\frac{d}{dt}{K}_a(t)=tm(a\cdot \mathbf{g})$;
4. the mechanical energy is $E(t)=\frac{1}{2}m\mid x^{\mathrm{\prime }}(t){\mid }^2+U(t,x(t))=\frac{1}{2}m\mid v_0{\mid }^2-m\mathbf{g}\cdot x_0$, constant on $\mathbb{R}$, and the potential has no explicit time dependence, so the time-translation identity of Time translation and the mechanical energy of the model: conservation in the autonomous global potential branch and the sharp failure for a time-dependent potential clause 1 holds;
5. for the numerical instance $m=1\mathrm{k}\mathrm{g}$, $\mathbf{g}=(0,0,-1)\mathrm{m}{\mathrm{s}}^{-2}$, $x_0=0$, $v_0=(1,0,0)\mathrm{m}{\mathrm{s}}^{-1}$ and $a=(0,0,1)$: at $t=2\mathrm{s}$ one has $P(2)=(1,0,-2)$ in $\mathrm{k}\mathrm{g}\mathrm{m}{\mathrm{s}}^{-1}$, $a\cdot P(2)=-2\mathrm{k}\mathrm{g}\mathrm{m}{\mathrm{s}}^{-1}$, $K_a(2)=-2\mathrm{k}\mathrm{g}\mathrm{m}$ and $K_a^{\mathrm{\prime }}(2)=-2\mathrm{k}\mathrm{g}\mathrm{m}{\mathrm{s}}^{-1}$, while the component along $(1,0,0)$ is constant at $1\mathrm{k}\mathrm{g}\mathrm{m}{\mathrm{s}}^{-1}$.

All values are model values of the declared idealization: the constant field is an adopted approximation of an external influence and no measurement is reported.

• Reading of the failure. For this balance comparison, declare the external-plus-pair presentation of Restricted interaction models: pair decompositions, antisymmetry and centrality options, potential forces and the gravitational branch clause (i) with $F_1^{\mathrm{e}\mathrm{x}\mathrm{t}}:\mathbb{R}\times {\mathbb{R}}^3\times {\mathbb{R}}^3\to {\mathbb{R}}^3$, $F_1^{\mathrm{e}\mathrm{x}\mathrm{t}}(t,x,v):=m\mathbf{g}$. This map is constant, so it satisfies the standing regularity. There are no pairs for $N=1$: the pair sum is zero and the antisymmetry condition of clause (ii)1 is vacuous. Thus $F_1=F_1^{\mathrm{e}\mathrm{x}\mathrm{t}}$ on the whole state domain, and the total external force along the motion is $m\mathbf{g}$. The rate $m(a\cdot \mathbf{g})$ of the momentum component is the component of the external force along $a$, in agreement with the balance identity of Total momentum balance and centre-of-mass equation for a model with antisymmetric pair forces read on this single label; the Noether sharpness clause and the balance describe the same failure from different starting points. Directions perpendicular to $\mathbf{g}$ retain a conserved momentum component, so the model retains a two-dimensional family of translation charges when $\mathbf{g}\mathrm{\ne }0$, while failing the third. When $\mathbf{g}=0$, all three components are conserved.

• Energy versus momentum. The same model conserves its mechanical energy: the potential is autonomous, and the time-translation datum is available. The example therefore exhibits the independence of the charges of this page: some symmetry data of the framework's model classes hold here and others do not, and the conservation statements must be applied one datum at a time.

Declare a two-label model of the force branch Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with $N=2$, masses $m_1,m_2>0$, total mass $M:=m_1+m_2$ and reduced mass $\mu :=m_1{m}_2\mathrm{/}M$, time window $I_0=\mathbb{R}$, configuration domain $Q=\{(x_1,x_2)\in ({\mathbb{R}}^3{)}^2:x_1\mathrm{\ne }{x}_2\}$ (open), and let $\psi :(0,\mathrm{\infty })\to \mathbb{R}$ be of class $C^2$; its argument is the squared separation of dimension ${\mathrm{\ell }}^2$, and its values have energy dimension. Adopt the potential

$U(x_1,x_2):=\psi (s(x)),s(x):=\mid x_1-x_2{\mid }^2,$

of dimension $\mu {\mathrm{\ell }}^2{\tau }^{-2}$, so that the model is a global potential model $d\equiv 0$ of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets. Verify:

For the translation family use a dimensionless direction and a length-valued parameter; for rotations use a dimensionless generator and a parameter in radians; for time translations use a parameter in seconds; for boosts use a dimensionless direction and a velocity-valued parameter. In the explicit potential below, $c_0$ has energy units $\mathrm{k}\mathrm{g}{\mathrm{m}}^2{\mathrm{s}}^{-2}$ and $\kappa$ has units $\mathrm{k}\mathrm{g}{\mathrm{s}}^{-2}$ (energy per squared length).

1. gradients and pair reading. For every $x\in Q$,

${\mathrm{\nabla }}_{x_1}U(x)=2{\psi }^{\mathrm{\prime }}(s(x))(x_1-x_2),{\mathrm{\nabla }}_{x_2}U(x)=-2{\psi }^{\mathrm{\prime }}(s(x))(x_1-x_2);$

so the model invokes the external-plus-pair presentation of Restricted interaction models: pair decompositions, antisymmetry and centrality options, potential forces and the gravitational branch with $F_i^{\mathrm{e}\mathrm{x}\mathrm{t}}=0$ and a single pair force $F_{12}=-F_{21}=-2{\psi }^{\mathrm{\prime }}(s)(x_1-x_2)$ that is antisymmetric and central. 2. translation. ${\sum }_i{\mathrm{\nabla }}_{x_i}U=0$ identically, so the differential condition of Translation symmetry of the model's Lagrangian and conservation of total momentum: exact hypothesis, conservation and its sharpness holds for every $a\in {\mathbb{R}}^3$; consequently $P$ and the boost charge $a\cdot (tP-MX(t))$ of The Galilean boost charge of the model: exact condition, uniform centre-of-mass motion, the induced motions and the boundary of the conservation law are constant along every motion and the centre of mass moves uniformly. 3. rotation. For every $a\in {\mathbb{R}}^3$, ${\sum }_i⟨a\times x_i,{\mathrm{\nabla }}_{x_i}U⟩=0$ identically, so the condition of Rotation symmetry of the model's Lagrangian and conservation of angular momentum about the origin holds for every axis; consequently the total angular momentum $J={\sum }_i{x}_i\times m_i{x}_i^{\mathrm{\prime }}$ is constant along every motion. 4. energy. $U$ has no explicit time dependence, so the mechanical energy $E=K+U$ is constant along every motion by Time translation and the mechanical energy of the model: conservation in the autonomous global potential branch and the sharp failure for a time-dependent potential. 5. explicit rigid-rotation motion (Hooke-type pair). Specialise to $\psi (s)=c_0+\kappa s$ with constants $c_0\in \mathbb{R}$ and $\kappa >0$, and put $\mathrm{\Omega }:=\sqrt{2\kappa \mathrm{/}\mu }$, $X_0\in {\mathbb{R}}^3$, $e(t):=(\cos \mathrm{\Omega }t,\sin \mathrm{\Omega }t,0)$, $r_1:=(\mu \mathrm{/}{m}_1)\rho$ and $r_2:=-(\mu \mathrm{/}{m}_2)\rho$ for a constant $\rho >0$; $X_0$ and $\rho$ have length dimension and $\mathrm{\Omega }$ has dimension ${\tau }^{-1}$. Then $x_i(t):=X_0+r_{i}e(t)$ defines a motion of the model on $\mathbb{R}$, and along it $P\equiv 0$, $J=\mathrm{\Omega }\mu {\rho }^2(0,0,1)$ and $E=c_0+2\kappa {\rho }^2$ are constant. 6. numerical instance. For $m_1=m_2=1\mathrm{k}\mathrm{g}$, $\kappa =1\mathrm{k}\mathrm{g}{\mathrm{s}}^{-2}$, $\rho =1\mathrm{m}$, $c_0=0\mathrm{J}$, $X_0=0$: $\mu =\frac{1}{2}$, $\mathrm{\Omega }=2{\mathrm{s}}^{-1}$, $r_1=\frac{1}{2}\mathrm{m}$, $r_2=-\frac{1}{2}\mathrm{m}$ and at $t=0$ one has $x_1=(\frac{1}{2},0,0)$, $x_2=(-\frac{1}{2},0,0)$, $x_1^{\mathrm{\prime }}=(0,1,0)$, $x_2^{\mathrm{\prime }}=(0,-1,0)$ in the stated units, $P=0$, $J=(0,0,1)\mathrm{k}\mathrm{g}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ and $E=2\mathrm{J}$; also ${\mathrm{\nabla }}_{x_1}U(0)=2\kappa (x_1-x_2)=(2,0,0)$ in newtons and ${\mathrm{\nabla }}_{x_2}U(0)=(-2,0,0)$.

All values are model values of the declared idealization; no measurement is reported and no property of a physical pair interaction is asserted.

• Both routes agree here. For this model the pair presentation is antisymmetric and central, so the balance theorems Total momentum balance and centre-of-mass equation for a model with antisymmetric pair forces and Angular momentum balance about a fixed or moving reference point, the internal torque sum, and the centrality case also apply and yield the same conservation statements under their own hypotheses. The Noether route verified here uses the potential presentation; the two are independent deductions, and the example checks the hypotheses of both at once.

• Representative of the potential. The additive constant $c_0$ enters the energy statement but not the forces; the numerical instance fixes $c_0=0$ as a declared normalization, in the sense of clause (c) of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets. No physical meaning is attached to the value $E=0$.

Work in the one-label force model of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with $N=1$, mass $m>0$, time window $I_0=\mathbb{R}$, configuration domain $Q={\mathbb{R}}^3$, and adopt the potential

$U(x):=\frac{k}{2}\mid x{\mid }^2,k>0\text{ with }[k]=\mu {\tau }^{-2},$

so that $F(t,x,v)=-\mathrm{\nabla }U(x)=-kx$ and the model is a global potential model $d\equiv 0$ of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets whose motions satisfy $m{x}^{\mathrm{\prime }\mathrm{\prime }}=-kx$ (The associated first-order system, motions, maximal motions and the Newtonian flow of a force model). Fix a rotation rate $\omega \mathrm{\ne }0$ of dimension ${\tau }^{-1}$ and let $\mathit{\omega }:=\omega e_3$. Define

$R(t):=\left(\begin{array}{ccc}\cos (\omega t)& -\sin (\omega t)& 0\\ \sin (\omega t)& \cos (\omega t)& 0\\ 0& 0& 1\end{array}\right)\mathrm{.}$

This matrix is the rotation about $e_3$ by angle $\omega t$; the identity ${\sin }^2(\omega t)+{\cos }^2(\omega t)=1$ gives $R(t)\in SO(3)$, and direct differentiation gives $R^{\mathrm{\prime }}(t)z=\mathit{\omega }\times R(t)z$ (The orthogonal group $O(3)$ and the rotation group $SO(3)$, $SO(3)$ is a subgroup of the general linear group, with $R^{-1}=R^{\mathsf{T}}$, The cross product in ${\mathbb{R}}^3$). Consider the time-dependent chart

$X(t,q):=R(t)q,\mathrm{\Omega }:=\mathbb{R}\times {\mathbb{R}}^3,$

which is of class $C^{\mathrm{\infty }}$ with $X_q(t,q)=R(t)$ invertible, and the substituted Lagrangian of Time-dependent coordinate maps transform the potential-part equation into the substituted Euler-Lagrange equation:

$\tilde{L}(t,q,v)=\frac{m}{2}\mid v+\mathit{\omega }\times q{\mid }^2-\frac{k}{2}\mid q{\mid }^2\mathrm{.}$

Verify:

1. Legendre data and chart Hamiltonian. $\tilde{L}$ is of mechanical type with $G=m{I}_3$, $A(q)=m\mathit{\omega }\times q$ and $V(q)=\frac{k}{2}\mid q{\mid }^2-\frac{m}{2}\mid \mathit{\omega }\times q{\mid }^2$, hence hyperregular with chart momentum $p=m(v+\mathit{\omega }\times q)$ and Hamiltonian

$H(t,q,p)=\frac{\mid p{\mid }^2}{2m}-p\cdot (\mathit{\omega }\times q)+\frac{k}{2}\mid q{\mid }^2\mathrm{.}$

2. Difference from the inertial mechanical energy. Along every $C^2$ curve $q:J\to {\mathbb{R}}^3$ on an interval $J\subseteq \mathbb{R}$ with nonempty interior, the inertial mechanical energy $E_{\mathrm{i}\mathrm{n}}=\frac{m}{2}\mid x^{\mathrm{\prime }}{\mid }^2+\frac{k}{2}\mid x{\mid }^2$ with $x=X(t,q)$ and the chart energy function $e=q^{\mathrm{\prime }}\cdot {\tilde{L}}_v-\tilde{L}$ satisfy the difference identity of The energy function of a moving chart: exact difference from the inertial mechanical energy and conservation along the chart solutions with $⟨X_t,\mathsf{M}{x}^{\mathrm{\prime }}⟩=p\cdot (\mathit{\omega }\times q)$, i.e. $E_{\mathrm{i}\mathrm{n}}=H+p\cdot (\mathit{\omega }\times q)$ along solutions.
3. Explicit circular motion. If $k=m{\omega }^2$, then $x(t):=rR(t)e_1$ with $r>0$ of dimension $\mathrm{\ell }$ is a motion of the model on $\mathbb{R}$, its chart description is the constant curve $q(t)=r{e}_1$ with $v\equiv 0$ and $p\equiv m\mathit{\omega }\times (r{e}_1)$, and along it $H\equiv 0$ while $E_{\mathrm{i}\mathrm{n}}\equiv m{\omega }^2{r}^2$; the substituted Lagrangian has ${\tilde{L}}_t\equiv 0$, so $H$ is a conserved quantity of the chart equations that is not the inertial mechanical energy.
4. Numerical instance. For $m=1\mathrm{k}\mathrm{g}$, $\omega =2{\mathrm{s}}^{-1}$, $r=1\mathrm{m}$ and $k=4\mathrm{k}\mathrm{g}{\mathrm{s}}^{-2}$ one has at $t=0$ the data $x=(1,0,0)\mathrm{m}$, $x^{\mathrm{\prime }}=(0,2,0)\mathrm{m}{\mathrm{s}}^{-1}$, $q=(1,0,0)$, $v=0$ and $p=(0,2,0)\mathrm{k}\mathrm{g}\mathrm{m}{\mathrm{s}}^{-1}$, with $H=0\mathrm{J}$ and $E_{\mathrm{i}\mathrm{n}}=4\mathrm{J}$.

All values are model values of the declared idealization; no measurement is reported.