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At $m_1=m_2$, the explicitly time-dependent coefficient becomes one. This restores time-translation symmetry, and the total energy becomes conserved. No further analogous restoration occurs merely from setting $m_3$ equal to them: the three arbitrary interaction functions still distinguish their arguments and particle labels.

If the question counts the extra ordinary unweighted rotations of the $x$ coordinate labels, the conventional generator count is as follows. For $m_1=m_2\ne m_3$, rotations of $(x_1,x_2)$ preserve both the kinetic and oscillator terms, so one ordinarily displayed rotation charge joins the restored energy: two displayed generators beyond the elementary list. For three equal masses, the full ordinary rotation group of $(x_1,x_2,x_3)$ has three generators; together with energy this gives four, or two more than in the preceding equal-pair case.

There is a necessary qualification to interpreting those numbers as genuinely new independent conservation laws. The weighted charges in solution deform continuously to those same ordinary rotation charges at equality. For example ${\mathcal{J}}_{12}$ is conserved for every positive $m_1,m_2$ and becomes $x_1{p}_{x_2}-x_2{p}_{x_1}$ when the masses agree. ThusThe restriction must be stated if using the first count. Equal mass does not create three previously absent independent oscillator integrals. If $\lambda =0$, energy conservation was already present; special choices of the interaction functions can add further symmetries and are excluded by the generic assumption.

A constant of motion is a function $I(\mathbf{q},\mathbf{p},t)$ whose total derivative vanishes along every physical trajectory. It may depend explicitly on time; arbitrary functions of conserved quantities are also conserved, so a count normally concerns a specified independent set or symmetry generators.

For a fixed-time infinitesimal variation $\delta q_i=ϵ{\xi }_i(\mathbf{q},t)$, suppose $\delta L=ϵdB/dt$. On a solution of the Euler-Lagrange equations,Therefore Noether theorem givesFor a transformation also changing time by $\delta t=ϵ\tau$, the action variation includes $Ld(\delta t)/dt$, and the transformed velocity varies by $ϵ({\dot{\xi }}_i-{\dot{q}}_i\dot{\tau })$. The same integration by parts gives the general point-symmetry chargeThus a time-translation symmetry gives the energy, a spatial translation gives the corresponding total momentum, and a rotation gives its angular-momentum generator. The required symmetry is a symmetry of the action, allowing a boundary term; demanding that the numerical value of the Lagrangian remain strictly unchanged is unnecessarily restrictive. Conversely, explicit time dependence of a Lagrangian generally prevents conservation of its energy even while other charges survive.^[ref]

The Poisson bracket is antisymmetric, so $\{H,H\}=0$. The Hamiltonian energy balance consequently givesFor a regular Lagrangian the Legendre transform in mechanics gives $H_t=-L_t$, evaluated at the corresponding velocity. In particular, a Lagrangian with no explicit time dependence has time-translation symmetry. Noether theorem associates this continuous symmetry of the action with the conserved quantity ${\sum }_i{\dot{q}}_i\partial L/\partial {\dot{q}}_i-L=H$. For an ordinary natural mechanical Lagrangian $L=T-V$, with $T$ quadratic in the velocities, this quantity is the total energy $T+V$.

The azimuth $\varphi$ is a cyclic coordinate, since $\partial \mathcal{L}/\partial \varphi =0$. Its Euler-Lagrange equation gives conservation of the vertical component of angular momentum:This $h$ is the actual angular momentum, not specific angular momentum. The Lagrangian also has no explicit time dependence. The conserved energy isAlong solutions, $\dot{E}=-\partial \mathcal{L}/\partial t=0$. These are the conserved quantities associated with rotational symmetry about the vertical and time-translation symmetry, respectively.

For a central potential use $L=m|\dot{\mathbf{r}}{|}^2/2-V(|\mathbf{r}|)$. An infinitesimal rotation about any constant unit vector $\mathbf{n}$ has $\varphi =0$ and $\mathbf{\psi }=\mathbf{n}\times \mathbf{r}$. Noether's theorem gives $Q=m\dot{\mathbf{r}}\cdot (\mathbf{n}\times \mathbf{r})=\mathbf{n}\cdot \mathbf{L}$, where $\mathbf{L}=\mathbf{r}\times m\dot{\mathbf{r}}$. Since $\mathbf{n}$ is arbitrary, every component of angular momentum is conserved. A time-translation symmetry has $\varphi =1$, $\mathbf{\psi }=0$, giving $Q=-E$ andas conserved quantities.

The orbiting planet produces a potential $V(r,\vartheta ,\phi -\Omega t)$. It has explicit time dependence and exerts an azimuthal torque, so generally $\dot{E}=V_t$ and ${\dot{L}}_z=-V_{\phi }$ are nonzero. Nevertheless a simultaneous time-translation symmetry $\delta t=\lambda$ and rotation $\delta \phi =\Omega \lambda$ leaves the Lagrangian and $dt$ unchanged. Noether's theorem gives $Q=\Omega p_{\phi }-E$ and henceEquivalently $V_t=-\Omega V_{\phi }$ implies $d(E-\Omega L_z)/dt=0$. This Jacobi integral assumes a rigidly rotating potential with constant $\Omega$ and negligible back-reaction from the test particle.

Invariance under a shift of the time coordinate, with other required background structures held fixed. A continuous time-translation symmetry gives an energy conservation law through Noether theorem. A generic expanding cosmological background is not stationary and need not supply a corresponding globally conserved total energy.