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 (Classical mechanics)

Hamilton's principle states that the physical path makes the action $S={\int }_{t_i}^{t_f}L(\mathbf{q},\dot{\mathbf{q}},t)dt$ stationary under differentiable variations $\delta q_i$ vanishing at the fixed endpoints. Its conventional “least action” name does not guarantee a minimum. Expanding to first order and integrating the velocity term by parts givesThe endpoint term vanishes. By the fundamental lemma of the calculus of variations, the coefficients of the independent interior variations must vanish, yieldingFor example, the zero path of a harmonic oscillator with endpoints zero can have both positive and negative second variations when the elapsed time exceeds a half-period; stationary action then need not be a least value. The Euler-Lagrange condition is the local statement that the variational principle actually supplies.

The Euler-Lagrange equations in cylindrical coordinates areThe last two follow because $\varphi$ and $z$ are cyclic coordinates. They give the conserved quantitiesSince the Lagrangian is autonomous, its energy function is conserved as well:The velocity-linear term cancels from this energy, but it still changes the equations of motion. A dimensionless logarithm means $\log (R/R_0)$ for some fixed reference length; replacing the printed $\log R$ by that expression changes the Lagrangian only by a total time derivative and shifts the numerical value of $k$, not the trajectory.

Write $w=a^2{\Omega }^2/C$, requiring $C\ne 0$ and $a>0$. The initial conditions fix $\ell =a^2\Omega$ and $\dot{z}=w+C\log (R/a)$. At $R=a$, the radial force is $a{\Omega }^2-Cw/a=0$. Thus the unperturbed motion is the helixTo analyse a small initial radial velocity, eliminate the cyclic velocities using the conserved quantities. The radial energy equation is ${\dot{R}}^2/2+U(R)=\mathcal{E}$, withHere $U^{\prime }(a)=0$ and $U^{\prime \prime }(a)=2{\Omega }^2+C^2/a^2>0$. The circular cylindrical radius is therefore stable. For $\eta =R-a$, the first-order equation is $\ddot{\eta }+{\omega }_r^2\eta =0$, soThe orbit acquires small radial oscillations. The other coordinates are modulated consistently with the same constants:These expressions retain the stated initial angular and axial velocities. The perturbation adds ${\delta }^2/2$ to the conserved energy. The expansion assumes $|\delta |/(a{\omega }_r)\ll 1$, which follows from the supplied smallness condition when $\Omega \ne 0$; it is not an exact finite-amplitude helix.

It can describe the motion per unit mass of a charged particle in the azimuthal magnetic field of a long straight current-carrying wire. Choose magnetic vector potentialThen $\mathbf{B}=\nabla \times \mathbf{A}=(mC/(qR))\hat{\mathbf{\varphi }}$ and $(T+q\mathbf{A}\cdot \mathbf{v})/m$ is the given Lagrangian. In terms of a straight-wire current $I$, $B_{\varphi }={\mu }_{0}I/(2\pi R)$, so $C=q{\mu }_{0}I/(2\pi m)$ with signs fixed by charge and current direction.

A static magnetic force does no work, consistent with the conserved kinetic energy. Its radial component is $-q\dot{z}{B}_{\varphi }$ and its axial component is $q\dot{R}{B}_{\varphi }$, reproducing the radial and axial couplings above. The reference length in the logarithm is a gauge convention, not a new physical wire radius in the ideal exterior-field model.

For the canonical Poisson bracket, direct substitution of $F=q_i$ and $F=p_i$ givesThereforeThis expresses the bracket as a first-order derivative acting on $F$, with coefficients determined by $G$; it is the phase-space chain-rule form of the definition.

Hamilton's equations are ${\dot{q}}_i=H_{p_i}$ and ${\dot{p}}_i=-H_{q_i}$. Using the elementary brackets from the previous part givesA constant of motion is a function that stays unchanged along each physical trajectory. For a function $C(\mathbf{q},\mathbf{p})$ with no explicit time dependence, the chain rule and Hamilton's equations giveThus $\boxed{C\text{ is conserved}⟺\{C,H\}=0}$ on the accessible phase-space domain. If a function explicitly depends on time, the criterion is instead ${\partial }_{t}C+\{C,H\}=0$; the simpler statement uses the source's time-independent $C$ assumption.

Put $\ell =p_{\varphi }$. The Hamiltonian has no explicit time or azimuthal dependence, so $\dot{H}=\{H,H\}=0$ and $\dot{\ell }=-H_{\varphi }=0$. Its equations of motion areWrite $C=p_R\ell \sin \varphi +({\ell }^2/R+A)\cos \varphi$. Differentiating along these equations givesEquivalently, the same cancellation evaluates $\{C,H\}$. Hence $\boxed{H,p_{\varphi },C\text{ are all constants of motion}.}$ The extra conserved component is associated with the Runge-Lenz symmetry of this inverse-distance central potential, rather than with conservation of radial momentum.

The Jacobi identity and the two brackets $\{C,H\}=\{p_{\varphi },H\}=0$ implyThus $\boxed{D=\{C,p_{\varphi }\}\text{ is conserved}}$ without re-solving the motion. Explicitly $D={\partial }_{\varphi }C=p_R{p}_{\varphi }\cos \varphi -(p_{\varphi }^2/R+A)\sin \varphi$. Squaring and adding the two rotated components givesConsequently the four quantities are not functionally independent: generically there are three independent constants. One may choose, locally, $H,p_{\varphi }$ and the orientation encoded by $C,D$; their magnitude is fixed by the displayed relation. This also matches the maximum of three independent trajectory-labeling constants for a nonstationary autonomous flow in four-dimensional phase space. On singular subsets the local functional rank can be lower: for example, at $A=0$ and $p_{\varphi }=0$, the differentials of $C,D$ are proportional to $d{p}_{\varphi }$, leaving at most two independent differentials together with $dH$; “three” is the generic count, not an assertion about every singular configuration.

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.