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For a rightgoing wave packet, put $\xi =z-ct$ and assume $f^{\prime }$ decays sufficiently rapidly. ThenThe shift charge becomesIt is time independent. It vanishes for a localized displacement packet that returns to the same displacement at both ends, but need not vanish for a localized strain pulse joining different asymptotic displacements.

For the dilation charge, substitute $z=\xi +ct$:The change of variables removes all time dependence. This charge also need not vanish; choosing an origin at the packet's energy centroid can make it zero. Conservation assumes the integral exists and there is no boundary flux, rather than merely assuming that the energy is finite.

Position-dependent density. The density ${\mathcal{L}}_1$ remains independent of $\psi$ itself, so the shift Noether charge survives with its altered canonical momentum:A prescribed generic function $\rho (z)$ supplies a fixed spatial scale and breaks scale invariance. For the dilation-current candidate with the corresponding ${\mathcal{H}}_1$ and ${\Pi }_1$, the scale-breaking sources in derivative field theories giveThus $Q_2$ is generally not conserved. Constant density recovers the earlier case; specially scale-invariant profiles and suitably invariant domains would require separate treatment.

Quartic strain. The extra term $-\alpha ({\psi }^{\prime }{)}^4$ in ${\mathcal{L}}_2$ again preserves a uniform field shift. Its time momentum is unchanged, while its spatial current becomes $-AK{\psi }^{\prime }-4\alpha ({\psi }^{\prime }{)}^3$. ThereforeThe new Hamiltonian density is ${\mathcal{H}}_2={\mathcal{H}}_0+\alpha ({\psi }^{\prime }{)}^4$. Under waveform dilation, the quartic term scales with $b^4$ rather than $b^2$, andHence there is no corresponding conserved $Q_2$ for nonzero generic $\alpha$, even if the energy in its candidate expression is updated.

Field-dependent quadratic stiffness. In ${\mathcal{L}}_3$, the term $-\beta {\psi }^2({\psi }^{\prime }{)}^2$ breaks the uniform field shift:when spatial flux vanishes. Thus $Q_1$ is not generally conserved. However all terms still contain exactly two derivatives in total, with a field of zero dilation weight. The factor ${\psi }^2$ gains no extra power of $b$, so waveform dilation remains an action symmetry. Its charge isHere $\Pi =A\rho \dot{\psi }$ as before. Relative to the original charge, the energy term changes; the symmetry survives despite the nonlinear field dependence. These statements concern symmetry conservation, separately from any restrictions on parameters needed for positive elastic energy.

Write $\rho$ for cylindrical radius in this question, not the rod's mass density. A charged relativistic particle in a time-independent magnetic field has the charged-particle electromagnetic Lagrangianwith zero electrostatic potential. The cylindrical speed is ${\mathbf{v}}^2={\dot{\rho }}^2+{\rho }^2{\dot{\theta }}^2+{\dot{z}}^2$. Inside a uniform axial field, choose the symmetric magnetic vector potential $A_{\theta }=B\rho /2$, $A_{\rho }=A_z=0$. Its cylindrical curl gives $B_z={\rho }^{-1}{\partial }_{\rho }(\rho A_{\theta })=B$. HenceIn a field-free region far from either end one can use $A_{\theta }=0$, recovering the requested outside expression. The question's $V=-qB{\rho }^2\dot{\theta }/2$ is a velocity-dependent generalized potential, not electrostatic potential energy.

There is an important end-field qualification. A finite axial field that jumps to zero with no radial component would violate Gauss's law for magnetism. The consistent near-axis idealisation is the solenoid fringe field from an axisymmetric vector potential, $A_{\theta }=\rho B(z)/2$, giving $B_{\rho }=-\rho B^{\prime }(z)/2$ and $B_z=B(z)$. The piecewise Lagrangian is its hard-edge limit and retains fringe impulses through the jump in $A_{\theta }$. Those impulses are essential to the focusing calculation.

The azimuth is a cyclic coordinate, so its canonical momentum is conserved:Initially the velocity is parallel to the axis in a field-free region, so $p_{\theta }=0$. Conservation of canonical angular momentum across a solenoid fringe therefore gives, inside the uniform field,This holds away from the coordinate singularity $\rho =0$; the Cartesian orbit will be smooth there.

Time independence also conserves the energy Hamiltonian,The velocity-linear magnetic term cancels in this Legendre transform. Thus $\gamma ={\gamma }_v=(1-v^2/c^2{)}^{-1/2}$ is constant, as expected because a static magnetic Lorentz force does no work. Mechanical angular momentum can change even though $p_{\theta }$ and the total speed are conserved.

Let $v=\sqrt{-\ln \lambda /\lambda }>0$ and write $\varphi =v+\chi$. ThenSince $V^{\prime }(v^2)=0$ and $V^{\prime \prime }(v^2)=\lambda$, the quadratic potential isThereforeThe constant can be dropped from the fluctuation equations. For canonical real fields, use $\chi =(h+i\pi )/\sqrt{2}$:These are the radial and angular masses of a phase-invariant scalar potential. The radial Higgs mode changes the field magnitude and costs energy to second order. The angular Goldstone boson changes the phase along the vacuum circle without a quadratic restoring force. Its linear dispersion is ${\omega }^2=|\mathbf{k}{|}^2$, whereas the radial mode has ${\omega }^2=|\mathbf{k}{|}^2+m_h^2$. The zero angular mass follows the continuous broken global symmetry, not from an accidental cancellation in a particular coordinate choice.

The sign can be checked independently by multiplying the equation by $\dot{x}$. With $E=({\dot{x}}^2+x^2)/2$,For periodic motion the total-derivative near-field term averages to zero, so positive $\epsilon$ gives a radiative energy loss, consistent with the slow poles' damping. Negative $\epsilon$ acts as energy gain in this model.

For the comparison equation as printed, $\ddot{x}+x-\epsilon \dot{x}=f$, the response denominator instead is $1-{\omega }^2+i\epsilon \omega$, with exact polesThus ordinary positive damping in that sign convention requires $\boxed{\epsilon <0}$, the opposite sign from slow radiation damping in the third-derivative equation. The word “damped” does not override the explicitly printed minus sign. With the usual equation $\ddot{x}+x+\gamma \dot{x}=f$, positive $\gamma$ gives the familiar lower-half-plane poles.

With the PDF's inverse Fourier transform convention, the derivatives act asso its forced differential operator becomes ${\omega }^2-k^2+i\gamma \omega k^2$. ThusFor the physical impulse response choose the retarded fundamental solution, vanishing before $t=0$. For $k\ne 0$ and $\gamma >0$, its poles lie in the lower half-plane. The retarded Fourier pole prescription $\omega \mapsto \omega +i0$ specifies the zero mode and the zero-damping limit as well; a Green function without a temporal or boundary prescription would otherwise permit homogeneous additions.

This is the Kelvin-Voigt rod Green function. Multiplying the homogeneous real-space equation by $-1$ givesA spatial Fourier mode therefore satisfies ${\ddot{\varphi }}_k+\gamma k^2{\dot{\varphi }}_k+k^2{\varphi }_k=0$: positive $\gamma$ really damps the motion. For sufficiently decaying fields its energy satisfiesThis checks the sign of the imaginary term in $\hat{G}$ independently of the transform calculation.

Constraint and example. A holonomic constraint can be written $f_{\alpha }(q_1,\dots ,q_n,t)=0$, with no irreducible dependence on velocities. A velocity relation is holonomic if it integrates to such an equation. For example, a particle constrained to a stationary circle satisfies $x^2+y^2={\ell }^2$, reducing its two Cartesian degrees of freedom to one angle. A nonholonomic constraint cannot be integrated in this way.

Kinematics and kinetic energy. Set $d=R-r>0$. In the plane of the original diagram, choose the orbital angle $\theta$ positive counterclockwise from vertically downward. The marked radius of the inner cylinder then has absolute orientation $\alpha =\theta -\varphi$ measured from the same downward direction. Its signed spin angular velocity is $\Omega =\dot{\theta }-\dot{\varphi }$, whileThe moment of inertia of a uniform solid cylinder is $I=m{r}^2/2$. Applying the kinetic energy decomposition about the center of mass gives

No slip. With tangential unit vector ${\mathbf{e}}_{\theta }=(\cos \theta ,\sin \theta )$, the inner contact point has velocity $(d\dot{\theta }+r\Omega ){\mathbf{e}}_{\theta }$. The fixed wall requires this to vanish. HenceBecause $r,R$ are constant, integration giveswhere the integration constant is zero with the marked-dot reference in the diagram and continuously unwrapped angles. This is a holonomic constraint, even though it was first obtained from a condition on velocities.

Oscillations. The gravitational potential energy above the lowest position is $V=mgd(1-\cos \theta )$. Eliminating $\varphi$ gives the reduced LagrangianThe Euler-Lagrange equation becomesThus the bottom is a stable equilibrium, and for $|\theta |\ll 1$ there is simple harmonic motion, withThe extra rotational kinetic energy reduces the frequency compared with a particle sliding without friction along the same circular path. The small oscillations inside a circular cylinder formula applies with inertia ratio $\gamma =1/2$.

Friction and energy. Let $F_t$ be the signed tangential contact force in the ${\mathbf{e}}_{\theta }$ direction. Applying Newton's second law to translation and the torque equation to rotation givesUsing the rolling without slipping condition yieldsExcept at special instants, static friction is needed to produce the spin acceleration. There is no sliding at the contact point, whose instantaneous velocity is zero; therefore this force does no total work. Its translational and rotational powers cancel: $F_{t}d\dot{\theta }+F_{t}r\Omega =0$. The stationary normal reaction likewise does no work. With a rigid wall, adequate static friction, and no rolling resistance, conservation of energy gives

Angular momentum. The cylinder's spin angular momentum $I\Omega$ is generally not conserved because the contact friction exerts a torque about its center of mass. About the fixed outer-cylinder center $O$, the angular momentum decomposition about the centre of mass givesGravity and friction exert the external torqueThus the rolling cylinder does not generally conserve angular momentum, although it conserves mechanical energy. There is a special geometric cancellation at $R=3r/2$: then $L_O$ and its torque vanish identically for this planar motion. Angular momentum of a larger isolated system including the wall and Earth is a separate conservation statement.

Scope of the holonomic reduction. The integrated constraint is exact at every angle for planar rolling without slipping with sustained contact; it is not restricted to small oscillations. Completing a full turn requires tracking angle winding rather than reducing each angle separately modulo $2\pi$. This global bookkeeping does not make the local velocity relation nonintegrable.

However, arbitrary initial conditions need not sustain this ideal rolling motion. The inward normal reaction and the required friction areContact requires $N\ge 0$, and a finite coefficient ${\mu }_s$ of static friction requires $|F_t|\le {\mu }_{s}N$. If either fails, the cylinder detaches or slips and the one-coordinate rolling Lagrangian no longer applies. Slipping contact allows $\theta$ and $\varphi$ to be independent; detachment removes the fixed center-distance constraint. These changes alone do not establish a nonholonomic constraint. Allowing general three-dimensional reorientation can introduce nonintegrable rolling constraints, but that motion is outside the diagram's two-angle planar model. Thus the stated planar rolling constraint is holonomic throughout its regime of validity; it cannot be imposed on every physically possible motion.

The two continuous, time-independent symmetries are time translation and a common rotation of both angles. Their Noether charges are respectively the total energy and vertical angular momentum:For the second symmetry, $\theta \mapsto \theta +ϵ$ and $\phi \mapsto \phi +ϵ$, so the Noether theorem gives the sum of their conjugate momenta. Independent rotations are not symmetries of the spring coupling.

There are also discrete symmetries: simultaneous reflection $(\theta ,\phi )\mapsto (-\theta ,-\phi )$, and time reversal in classical mechanics, which reverses both angular velocities. These leave the dynamics invariant but provide no additional infinitesimal Noether charges. Combined with common rotations, the reflections give the full planar orthogonal symmetry of the angular geometry. If $I_1=I_2$, interchange of the two angular coordinates is an additional discrete symmetry of the reduced dynamics.

There is a further useful time-dependent symmetry, up to a total derivative, on a continuous lift of the angles: adding the same constant angular velocity, $(\theta ,\phi )\mapsto (\theta +vt,\phi +vt)$. It changes $L$ byIts conserved quantity is $\boxed{I_1\theta +I_2\phi -tJ}$. It is the integration constant for free common rotation, rather than a new globally single-valued angle charge on the configuration torus. The globally defined continuous charges are $E$ and $J$ above.

Above the threshold, $c^2{k}_j^2>1/(4{\tau }^2)$, the square root in part (d) is real. Both fields oscillate, but the first has envelope $\boxed{e^{-t/(2\tau )}}$ and the second $\boxed{e^{+t/(2\tau )}}$. Thus the first is damped and the second amplified.

Below the threshold, set ${\sigma }_j=\sqrt{1/(4{\tau }^2)-c^2{k}_j^2}$. The temporal factors areThey are nonoscillatory. For nonzero $k_j$, $0<{\sigma }_j<1/(2\tau )$, so both first-field branches decay and both second-field branches grow. At $k_j=0$, one branch is constant; the other is $e^{-t/\tau }$ for ${\varphi }_1$ or $e^{t/\tau }$ for ${\varphi }_2$.

At equality, the repeated temporal root gives $\boxed{(A+Bt)e^{-t/(2\tau )}}$ for the first field and $\boxed{(C+Dt)e^{t/(2\tau )}}$ for the second. The full two-field system is generically unstable on both sides because of the amplified sector. This is compatible with its autonomous Lagrangian: the total conserved energy is an indefinite quadratic form, rather than a positive energy forcing every component to remain bounded.

Square the operator solutions, retaining the symmetrized cross term rather than treating $x^{\prime }(0)$ and $p^{\prime }(0)$ as commuting variables. The initial symmetrized expectation vanishes, soThe means remain zero. These are also the variances. Unless $\alpha =m\omega /\hslash$, the Gaussian state breathes: its position and momentum widths exchange periodically while the mean energy remains constant.

For $b>0$, the exponential variational trial for a square-root half-line well satisfies the wall boundary condition and has norm ${\int }_0^{\infty }{x}^2{e}^{-2bx}dx=1/(4b^3)$. integration by parts has no boundary term and givesThe supplied gamma integral givesSetting $E^{\prime }(b)=0$ yields the unique minimum,At the minimum $⟨V⟩=4⟨T⟩$, so the variational ground-state estimate isThe energy diverges as $b\to 0$ or $b\to \infty$, confirming that the stationary point minimizes the trial energy.

Write $E_{\ast }=({\hslash }^2{a}^4/(2m){)}^{1/5}$. Then $E_0^{\mathrm{WKB}}=1.811694E_{\ast }$ and $E_{\mathrm{var}}=1.876493E_{\ast }$. The variational method guarantees $E_0\le E_{\mathrm{var}}$, since the trial satisfies the wall condition and has finite energy. WKB gives no comparable upper-bound guarantee and is least controlled for the lowest state. The two estimates differ by about $3.6\%$ of the WKB value; their closeness is encouraging but does not prove that the exact energy lies between them.

Write $\kappa =\sqrt{2m{V}_0}/\hslash$ and $q=\kappa a$. At low energy, the $S$ wave normally dominates. Under the additional stated near-resonance assumption $ka\ll |{\delta }_0|$, neglect $ka$ inside the cotangent of the matching condition and set $k^{\prime }\simeq \kappa$:Substitution in ${\sigma }_0=4\pi {\sin }^2{\delta }_0/k^2$ givesThis is a near-resonance square-well approximation, not a uniform formula for every well radius. A phase shift is defined modulo $\pi$; its representative must be chosen consistently when applying the comparison with $ka$.

For small fixed positive $E$, the stated approximate curve has maxima atIts formal minima are zero at $\kappa a=n\pi$. The peak height is the $S$-wave unitarity limit, and the peaks narrow as $E/V_0$ decreases.

To identify the bound-state threshold, let the bound-state energy be $-\epsilon$ with $\epsilon >0$. The reduced radial wave is proportional to $\sin (Kr)$ inside and $e^{-\chi r}$ outside, with $K=\sqrt{2m(V_0-\epsilon )}/\hslash$ and $\chi =\sqrt{2m\epsilon }/\hslash$. Matching gives $K\cot (Ka)=-\chi$. As a new bound state emerges at $\epsilon \downarrow 0$,Thus the limiting peak locations coincide with the appearance of successive $S$-wave bound states. At the exact threshold the zero-energy radial solution is not square-integrable; a normalizable bound state appears just beyond it. Finite-energy maxima shift slightly from the zero-energy positions.

The formal minima require care because the assumption used in part d fails there. Retaining $ka$ in the exact matching condition gives the spherical square-well scattering lengthaway from a threshold. At $\kappa a=n\pi$, $a_s=a$, so the true low-energy S-wave scattering contribution is $4\pi a^2$, not zero. Genuine leading-order zeros of that contribution instead satisfy $\tan q=q$, for example $q\simeq 4.493409$ and $7.725252$. The original sketch compares the exact finite-energy $S$ wave with the requested approximation, making this nonuniformity visible. Dominance by the $S$ wave must be checked separately near a higher-partial-wave threshold. The scattering-length divergence and its bound-state interpretation are also derived in institutional lecture notes.^[ref]

Put $T=t-t^{\prime }$. momentum eigenstates evolve with energy $p^2/(2m)$, so the retarded momentum kernel is $\theta (T)\delta (p-p^{\prime })e^{-i{p}^{2}T/(2m\hslash )}$. Its Fourier transform reduces toCompleting the square and defining the oscillatory Gaussian integral by a convergent limiting prescription givesFor $T>0$ the square root has quantum phase $e^{-i\pi /4}$. As $T\downarrow 0$, this kernel tends to $\delta (x-x^{\prime })$ in the distributional sense. Differentiating the Heaviside step function consequently supplies the required $i\hslash \delta (T)\delta (x-x^{\prime })$ source; an extra $-i/\hslash$ factor would correspond to a different Green-function convention.

A constant electrostatic energy offset is removed locally by the quantum phase substitution $\psi =e^{-i{V}_{0}t/\hslash }{\psi }_0$, since $(i\hslash {\partial }_t-V_0)\psi =e^{-i{V}_{0}t/\hslash }i\hslash {\partial }_t{\psi }_0$. Therefore the two frequency branches becomeThe normalization uses the kinetic frequency ${\omega }_k$, while the canonical energy eigenvalues are $E=V_0\pm \hslash {\omega }_k$. The opposite-frequency branch is not obtained by replacing ${\omega }_k$ by the same shifted positive frequency in both terms. A globally constant offset produces no electric field, whereas different offsets on opposite sides of a step cannot be removed by a single smooth global phase.

The negative-kinetic-energy branch is interpreted as an antiparticle channel. In the overlap $m{c}^2<E<V_0-m{c}^2$, particles can escape to the left and oppositely charged antiparticles to the right. This interval exists only when $V_0>2m{c}^2$. The external electrostatic source supplies the energy for pair creation; the reflected particle flux can exceed the incident one because the outgoing state need not contain only the incident particle.

The conserved quantity is signed charge current, not a positive one-particle probability in both regions. Negative $\mathcal{T}$ denotes rightward transport of opposite charge, so it is consistent with $\mathcal{R}>1$ and $\mathcal{R}+\mathcal{T}=1$. A quantized charged field with particle and antiparticle creation operators gives the appropriate interpretation of this Klein paradox. This bosonic result must not be replaced by a fermionic reflection law or interpreted as a failure of charge conservation.^[ref]

Set ${\vartheta }_{\alpha }(t)={\hslash }^{-1}{\int }^t{E}_{\alpha }(s)ds$. Differentiating the expansion produces ${\dot{a}}_{\beta }|{\phi }_{\beta }⟩+a_{\beta }|{\dot{\phi }}_{\beta }⟩-i{E}_{\beta }{a}_{\beta }|{\phi }_{\beta }⟩/\hslash$ for each term. The energy term cancels $H|\Psi ⟩$ in the Schrödinger equation. Projecting the remainder with $⟨{\phi }_{\alpha }|$ and using orthonormality yieldsThe diagonal term is purely imaginary and supplies the geometric phase. Off-diagonal instantaneous eigenstate couplings transfer population; differentiating the eigenvalue equation gives $⟨m|\dot{n}⟩=⟨m|\dot{H}|n⟩/(E_n-E_m)$ when the levels are distinct.

The quantum propagator is the position-space kernel of the time-evolution operator: $\Psi (x,t)=\int K(x,t|x^{\prime },t^{\prime })\Psi (x^{\prime },t^{\prime })d{x}^{\prime }$. For a time-independent Quantum Hamiltonian, $U=e^{-iH(t-t^{\prime })/\hslash }$. Inserting a complete orthonormal energy eigenbasis givesA continuous spectrum requires the corresponding integral. At equal times, completeness makes this kernel a Dirac delta function.