Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford / 2019 / S7 / 1

Define the canonical momentum $p_i=\partial L/\partial {\dot{q}}_i$ and the energy function $E={\sum }_i{p}_i{\dot{q}}_i-L$. Along an Euler-Lagrange equation solution,Thus an autonomous Lagrangian has conserved $E$. For a simple harmonic oscillator, $L=m{\dot{x}}^2/2-k{x}^2/2$ and $\boxed{E=m{\dot{x}}^2/2+k{x}^2/2}$ is constant. For ordinary time-independent mechanical coordinates this is the total mechanical energy.

The appropriate invariance is invariance of the action element $Ldt$, including the change of the time measure. Under $\delta t=\lambda \varphi$ and $\delta q_i=\lambda {\psi }_i$, differentiation of $q_i^{\prime }(t^{\prime })$ gives $\delta {\dot{q}}_i=\lambda (D_t{\psi }_i-{\dot{q}}_i{D}_t\varphi )$. ThereforeUsing the Euler-Lagrange equations and $D_{t}E=-L_t$, this becomesAn action symmetry therefore yields the Noether chargeFor $\varphi =0$ this reduces to the usual spatial form of Noether's theorem. The same on-shell calculation permits velocity-dependent generators when they define a legitimate variational symmetry. Merely preserving the numerical value of $L$ is insufficient for transformations of time: the Jacobian matrix and determinant $d{t}^{\prime }/dt$ matters. If the action changes by a boundary term $\lambda D_{t}B$, the conserved quantity is $Q-B$.

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.