Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S7 / 3

Let $p_i=\partial L/\partial {\dot{q}}_i$ and consider the infinitesimal fixed-time transformation $\delta q_i=ϵK_i(\mathbf{q})$. It induces $\delta {\dot{q}}_i=ϵd{K}_i/dt$. Expanding the Lagrangian givesOn a solution of the Euler-Lagrange equations, $\partial L/\partial q_i={\dot{p}}_i$, so this equals $d({\sum }_i{p}_i{K}_i)/dt$. Exact invariance makes the left side zero, proving the fixed-time form of Noether's theorem:The symmetry direction need not coincide with one original coordinate axis; $\mathbf{K}$ can depend on the generalized coordinates. The conclusion is evaluated on physical trajectories, while the assumed invariance is a property of the Lagrangian under the transformation.

The canonical momentum for the stated symmetric-gauge Lagrangian isSince $\mathbf{B}$ is uniform and time independent, its Euler-Lagrange equation becomesThis is the magnetic Lorentz force, which is unchanged by constant spatial translations. The Lagrangian itself changes byA total time derivative changes the action only by $ϵ[F_{\mathbf{n}}(t_f)-F_{\mathbf{n}}(t_i)]$. For variations with fixed endpoints this boundary contribution has zero variation, so the equations of motion are unchanged. The translation is therefore a symmetry of the dynamics even though this particular gauge makes the Lagrangian only quasi-invariant.

Repeat the expansion in the strict-invariance proof, but now use $\delta L/ϵ=dF/dt$. Along Euler-Lagrange trajectories,It follows that the conserved Noether quantity isThe minus sign matters. It subtracts the boundary term created by the symmetry transformation. More explicitly, before using the equations of motion one hasThe final sum vanishes on shell, proving the result for arbitrary coordinate-dependent $\mathbf{K}(\mathbf{q})$ under the source's fixed-time assumptions. It is this boundary-term form, rather than the uncorrected canonical momentum alone, that applies to translations in the magnetic example.

For translation in a fixed direction $\hat{\mathbf{n}}$, take $\mathbf{K}=\hat{\mathbf{n}}$ and the preceding $F_{\mathbf{n}}$. The triple-product identity gives $F_{\mathbf{n}}=-(q/2)\hat{\mathbf{n}}\cdot (\mathbf{B}\times \mathbf{x})$. ThusChoosing the three Cartesian directions gives the three constantsThese are magnetic pseudomomentum components, not the three components of kinetic momentum $m\dot{\mathbf{x}}$. A direct check is $d{\mathbf{K}}_{\mathrm{mag}}/dt=q\dot{\mathbf{x}}\times \mathbf{B}+q\mathbf{B}\times \dot{\mathbf{x}}=0$. In the stated gauge, ${\mathbf{K}}_{\mathrm{mag}}=\mathbf{p}+(q/2)\mathbf{B}\times \mathbf{x}$, consistent with the Noether boundary correction.