Past exam of the classical mechanics course of Part B of the physics course of the University of Oxford / 2016 / 2

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]

Assume positive, pairwise distinct masses and generic nonzero $\omega ,\lambda$. Let $p_{x_i}=m_i{\dot{x}}_i$, and similarly for $y,z$. The three uncoupled $x$ oscillators each conserveUniform translation of all $y_i$ conserves $P_y={\sum }_i{p}_{y_i}$. The $z_1,z_2$ interaction permits their common translation, while $z_3$ is entirely free, giving two independent momenta $P_{z,12}=p_{z_1}+p_{z_2}$ and $p_{z_3}$. The interacting $z_1,z_2$ subsystem is autonomous and has its own conserved energy. A useful basic collection is thereforeThe total energy is generally not conserved. With $\kappa =\lambda (m_1-m_2)$, its derivative is $\dot{H}=-\kappa \sin (\kappa t)U_{12}(y_1-y_2)$, which is generically nonzero.

Unequal masses do not remove every rotation charge in the oscillator sector. Put $X_i=\sqrt{m_i}{x}_i$, $P_i=p_{x_i}/\sqrt{m_i}$. This is a canonical transformation and makes that sector an isotropic harmonic oscillator. Its rotations conserveThese mass-weighted oscillator symmetries exist already at distinct masses. The energies and three rotation charges are not all functionally independent; the three-oscillator sector has at most five independent time-independent integrals on regular trajectories.

There are further simple explicitly time-dependent constants, such as $x_i\cos \omega t-({\dot{x}}_i/\omega )\sin \omega t$ and $x_i\sin \omega t+({\dot{x}}_i/\omega )\cos \omega t$, and center-of-mass boost constants. For example $MY-t{P}_y$ is constant for $M={\sum }_i{m}_i$ and $Y={\sum }_i{m}_i{y}_i/M$. These illustrate why an unrestricted raw number of conserved functions is not meaningful.

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.

With $m_1=m_2=m_3=m$, the explicit energy and ordinary rotation generators just discussed areThe energy is constant because the drive factor no longer depends on time. For each rotation charge, differentiating and using ${\ddot{x}}_i=-{\omega }^2{x}_i$ gives ${\dot{J}}_{ij}=m(x_i{\ddot{x}}_j-x_j{\ddot{x}}_i)=0$.

These are angular momenta in the auxiliary three-dimensional space whose coordinates are the particle labels $(x_1,x_2,x_3)$. They are not the physical spatial angular momentum ${\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$; generic $U_{12},U_{13},V$ do not have physical rotational invariance. They are precisely the equal-mass representatives of the weighted charges already present in solution. The new energy integral supplies the autonomous relative-$y$ dynamics not present in the driven case.