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The Euler-Lagrange equations from the quadratic Lagrangian areFor a normal mode proportional to $e^{i\omega t}$, let $\lambda ={\omega }^{2}a/g$. Nonzero amplitudes requireHenceThe corresponding amplitude ratios are $\Phi /\Theta =2\lambda /(1-\lambda )$. The lower-frequency mode has $\Phi /\Theta =2(\sqrt{2}-1)$, with the angles in phase; the higher-frequency mode has $\Phi /\Theta =-2(1+\sqrt{2})$, with opposite signs. The small hoop's spin remains zero in both modes.

Direct variation of the Lagrangian givesFor a static configuration, $k\rho =qE\cos \theta$ and $qE\rho \sin \theta =0$. Since $qE>0$ and the radius is positive,The magnetic force vanishes at rest and does not shift this equilibrium. In canonical variables it has $p_{\rho }=0$ but $p_{\theta }=qB{\rho }_0^2/2$, rather than zero canonical angular momentum.

For a regular Lagrangian define $p_i=\partial L/\partial {\dot{q}}_i$ and $H={\sum }_i{p}_i{\dot{q}}_i-L$. Its differential isThe velocity differentials cancel. Using $\partial L/\partial q_i={\dot{p}}_i$ and comparing independent coefficients givesThe last equality concerns explicit partial derivatives in their respective natural variables; it does not assert that total time derivatives are always negatives.

Under the combined local transformation,The derivative of the phase cancels the added gauge potential. The conjugate derivative gains the opposite phase, so their product and $|\varphi {|}^2$ are invariant. AlsoEvery term of the interacting Lagrangian is therefore invariant.

For the supplied Lagrangian, variation with respect to ${\varphi }^{\ast }$ usesThus $\square \varphi +2ie{A}^{\mu }{\partial }_{\mu }\varphi +ie({\partial }_{\mu }{A}^{\mu })\varphi +m^2\varphi =f$. In the Lorenz gauge,For real external fields and driving, the conjugate equation has $-2ie{A}^{\mu }{\partial }_{\mu }$ acting on ${\varphi }^{\ast }$. The supplied model has no $e^2{A}^2|\varphi {|}^2$ term; one must not insert such a term from a different minimally coupled Lagrangian.

Use $\hslash =c=1$, signature $(+---)$, and define $D_{\mu }={\partial }_{\mu }+ie{A}_{\mu }$. Under a local phase transformation,the derivative of the phase cancels the change of the potential, giving $D_{\mu }^{\prime }{\varphi }^{\prime }=e^{-ie\alpha }{D}_{\mu }\varphi$. Therefore a gauge-invariant scalar-electromagnetic Lagrangian density isThe mass term is invariant, the covariant-derivative product has canceling phases, and $F_{\mu \nu }^{\prime }=F_{\mu \nu }$ because mixed partial derivatives commute. A potential depending only on ${\varphi }^{\ast }\varphi$ may be added without changing the symmetry. The Maxwell term makes the electromagnetic field dynamical; it can be omitted when $A_{\mu }$ is a prescribed external background.

Expanding the covariant-derivative term gives the interaction linear in $e$ plus $e^2{A}_{\mu }{A}^{\mu }|\varphi {|}^2$. This quadratic term is required for exact local gauge invariance. The later part (c) deliberately gives only the linear interaction, and must be solved as that different stated Lagrangian rather than silently adding the missing term.

For fields ${\phi }_A$ and a continuous infinitesimal internal symmetry $\delta {\phi }_A=ϵ\Delta {\phi }_A$, suppose the density changes by a divergence, $\delta \mathcal{L}=ϵ{\partial }_{\mu }{K}^{\mu }$. Varying the action and using the Euler-Lagrange equations gives the Noether currentThe conserved charge is $Q=\int j^0{d}^{3}x$ when the flux through the boundary vanishes; more generally $dQ/dt=-\oint \mathbf{j}\cdot d\mathbf{S}$. If the Lagrangian is exactly invariant, $K^{\mu }=0$. Treat a complex field and its conjugate as independent variables during variation, imposing conjugacy on physical solutions. Time and space translations give energy and momentum through the corresponding spacetime version of the theorem. Continuous global symmetry gives an ordinary charge; local gauge redundancy requires additional care and is not an independent conserved charge for every arbitrary function.

Treat $\varphi$ and ${\varphi }^{\ast }$ as independent fields, keep $A_{\mu }(x)$ as the prescribed background, and use the source's interaction with no added $e^2{A}^2|\varphi {|}^2$ term. The momentum derivatives areVarying ${\varphi }^{\ast }$ and $\varphi$ gives respectivelyNo Lorenz gauge condition has been assumed. The background divergence term must be retained unless such a condition is separately imposed.

The canonical momenta conjugate to the two scalar fields areso $\dot{\varphi }={\pi }^{\ast }-ie{A}_0\varphi$ and ${\dot{\varphi }}^{\ast }=\pi +ie{A}_0{\varphi }^{\ast }$. Define the physical three-vector potential by $A_{\mu }=(A_0,-\mathbf{A})$. The Legendre transform in the proper canonical variables isEquivalently the first product is $\pi {\pi }^{\ast }+ie{A}_0({\varphi }^{\ast }{\pi }^{\ast }-\varphi \pi )+e^2{A}_0^2|\varphi {|}^2$. The last quadratic term arises from replacing velocities by canonical momenta even though the original Lagrangian has no quadratic interaction. It is not permission to modify the specified Lagrangian. An explicitly time-dependent background generally makes the scalar-field energy time dependent.

Write $A=\pi a^2$ for the cross-sectional area, and use $z$ consistently for the longitudinal coordinate: the paper's later $x$ is the same coordinate. The canonical momentum is a density per reference length,The Legendre transform gives the Hamiltonian density in its proper canonical variables:The Euler-Lagrange equation isFor a twice-differentiable wave packet $\psi =f(z-ct)$, the derivatives are $\ddot{\psi }=c^2{f}^{\prime \prime }$ and ${\psi }^{\prime \prime }=f^{\prime \prime }$. Thus the nontrivial packet solves the wave equation whenThe negative sign gives the opposite propagation direction. Here $\Pi$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, $\mathcal{H}$ has units $\mathrm{J}{\mathrm{m}}^{-1}$, and $c$ has units $\mathrm{m}{\mathrm{s}}^{-1}$. The result follows from the supplied axial stiffness in the Lagrangian. For a freely contracting thin elastic rod the relevant longitudinal modulus is normally Young's modulus, rather than the bulk modulus; the question's given Lagrangian fixes which coefficient is being used.

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.

Choose Minkowski metric $(+,-,-,-)$, so $A_{\mu }=(\varphi /c,-\mathbf{A})$ and $d\tau =dt\sqrt{1-v^2/c^2}$. In laboratory time the relativistic charged-particle action has LagrangianHere $e$ denotes the signed particle charge in the action. The canonical momentum and the position derivative areThe Euler-Lagrange equation therefore givesThe last term is $(\mathbf{v}\times \mathbf{B}{)}_i$, and the first two terms form the electric field. Thus the relativistic Lorentz force equation isIts power relation and acceleration form areThe power relation follows by taking the scalar product with $\mathbf{v}$ and using the relativistic energy-momentum relation. It also shows why the magnetic field alone cannot change the particle's energy.

Use the downward $y$ direction of the PDF diagram. The bob's Cartesian coordinates areThusThe hinge and rod are massless, so only the bob contributes kinetic energy:Because gravity points along positive $y$, its gravitational potential energy is $-mgy=-mg\ell \cos \theta$. The zero-rest-length spring has elastic potential energy $k{s}^2/2$. Put $s=\ell \eta$ and divide $T-V$ by the positive constant $m{\ell }^2$. The equivalent Lagrangian isMultiplying the action by a nonzero constant leaves its stationary paths unchanged. This derivation is a direct use of kinetic energy from Cartesian position vectors; no independent kinetic term should be assigned to the massless support.

Subtracting the two linearized equations eliminates their acceleration terms:For $g,k>0$, the assumed nontrivial simple harmonic motion therefore hasThe support displacement and pendulum angle oscillate in phase. Inserting this relation into either Euler-Lagrange equation yieldsso the one positive finite angular frequency isThe negative frequency describes the same real oscillation after a phase change. For a general phase, replace the sine in both coordinates by $A\cos (\omega t)+B\sin (\omega t)$.

There is no missing second finite branch. The quadratic Lagrangian iswhose kinetic matrix is $\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$ and has rank one. Its characteristic determinant is ${\omega }_0^2{\omega }_1^2-({\omega }_0^2+{\omega }_1^2){\omega }^2$, not a quadratic polynomial in ${\omega }^2$. This singular kinetic matrix in small oscillations results from taking the support mass to be exactly zero. Adding a small support mass restores a second mode that runs to infinite frequency as that mass vanishes.

For a stiff spring,The hinge becomes fixed, recovering the simple pendulum frequency, as expected for the pendulum with a massless spring-mounted support. At finite $k$, horizontal support motion makes the frequency smaller.

Choose $A^{\mu }=(\Phi ,\mathbf{A})$, so that $F_{i0}=E_i$ and $F_{ij}={ϵ}_{ijk}{B}_k$, with ${ϵ}_{123}=1$. The definitions of the polarization-magnetization tensor imply $M^{0i}=-P_i$ and $M^{ij}={ϵ}_{ijk}{M}_k$. Separating temporal and spatial indices,The Levi-Civita symbol contraction equals $2{\delta }_{k\ell }$, givingThus $\mathbf{P}$ is the electric dipole moment per volume, or electric polarization, and $\mathbf{M}$ is the magnetic dipole moment per volume, or magnetization. The signs are those of dipole interactions in a Lagrangian; the associated fixed-dipole potential energies have the opposite signs. Consistently, the equation in part (a) has componentsso ${\rho }_b=-\nabla \cdot \mathbf{P}$ and ${\mathbf{J}}_b={\partial }_t\mathbf{P}+\nabla \times \mathbf{M}$ are the bound charge and bound current.

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.

Use the angular coordinate $\theta$ for the position on the ring. The kinetic energy is $T=m{a}^2{\dot{\theta }}^2/2$. The gravitational potential energy is constant because the ring is horizontal, and the ideal constraint force has no tangential component. Thus the Lagrangian has no dependence on $\theta$ itself: it is a cyclic coordinate.

The Euler-Lagrange equation givesEquivalently, the conserved conjugate momentum $p_{\theta }=m{a}^2\dot{\theta }$ is the vertical angular momentum, as required by rotational symmetry.

A common rotation changes neither the separation nor the spring energy. Its infinitesimal direction is $(1,1)$, exactly the zero-frequency normal mode. Thus the continuum of rotated equilibrium points requires that zero mode; a nonzero restoring frequency in that direction would contradict rotational symmetry.

The second mode has $I_1{\eta }_1+I_2{\eta }_2=0$, so it changes the relative angle while leaving the inertia-weighted common angle unchanged. It therefore carries zero perturbation in the conserved angular momentum associated with common motion. Indeed the coordinatesseparate the Lagrangian exactly:The cyclic coordinate $q$ is free, while $\delta$ has restoring or inverted curvature. This also explains why all members of either equilibrium family have the same frequencies. Reflection symmetry makes the potential even in $\delta$, and the stable oscillatory mode is consistent with time reversal in classical mechanics through its two opposite frequency signs.

Choose the current convention corresponding to $\delta \phi =i\alpha \phi$. Noether's theorem givesFor the requested terms with no $\sigma$ and $f=\sqrt{2}v$, the Lagrangian is $\frac{1}{2}(\partial \pi {)}^2$ and the current becomes $J^{\mu }=f{\partial }^{\mu }\pi$. The supplied covariant oscillator normalization cancels the invariant measure when the annihilation operator contracts with $a^{†}(p)$, soThe momentum is on the massless shell, $p^2=0$. Differentiating once more therefore givesThis Goldstone current matrix element in a phase-only scalar theory shows that a conserved current can have a nonzero matrix element proportional to a null momentum. Thus current conservation is fully consistent with the current creating a massless one-particle state from the vacuum; the nonzero coefficient $f$ is its Goldstone-current coupling. It is the divergence that vanishes, not the current matrix element. Reversing the current convention reverses the nonzero matrix element but leaves this zero divergence unchanged.

Varying the vector field gives the Proca equation,Its divergence uses antisymmetry of $F^{\mu \nu }$ to cancel the first term. Since $m_A\ne 0$,For an on-shell momentum this imposes $p_{\nu }{ϵ}^{\nu }=0$ on the four polarization components, leaving three independent solutions. In the rest frame, $p=(m_A,0)$ makes ${ϵ}^0=0$, leaving precisely the three spatial polarizations of a massive spin-one particle.

There is no fourth independent oscillator hidden in $A^0$. Its canonical momentum is zero because the Lagrangian has no ${\dot{A}}^0$, and the temporal equation constrains $A^0$ in terms of the spatial data. Equivalently the primary and secondary second-class constraints remove two variables from the eight-dimensional phase space, leaving six phase-space variables, or three physical degrees of freedom.

For this three-site Abelian Higgs chain, the potential is minimized by $|{\varphi }_1|=|{\varphi }_2|=v$. Choose their phases real in a local unitary gauge. The scalar kinetic terms then givewhere Lorentz contractions are understood andThe nonzero eigenvalues of $K$ equal those ofnamely $1+q^2-q$ and $1+q^2+q$. The third eigenvalue is zero because $R$ has rank two. Both nonzero eigenvalues are positive for $q>0$.

Mass-normalization correction: the canonical vector mass term is $\frac{1}{2}{B}^{\mathsf{T}}{\mathsf{M}}^{2}B$, so the Lagrangian actually gives ${\mathsf{M}}^2=2g^2{v}^{2}K$. Its physical masses are thereforeThe printed dimensionless eigenvalue pattern is correct, but its common scale must be $m=\sqrt{2}gv$, not $gv$, for the written kinetic terms and vacuum value. For a concrete counterexample set $q=1$ and $B=(A,0,-A)/\sqrt{2}$. This is a canonically normalized eigenmode and its mass term is $g^2{v}^2{A}_{\mu }{A}^{\mu }$, giving physical squared mass $2g^2{v}^2$ rather than the printed $g^2{v}^2$. The printed scale $gv$ would require a vacuum modulus $v/\sqrt{2}$ or a corresponding change in scalar normalization, neither of which is the specified potential minimum.

The unbroken transformations preserve both vacuum phases:Thus the local breaking pattern isTwo scalar phases supply the longitudinal polarizations of the two massive vector bosons; one massless vector remains. This is the Higgs mechanism, with its mass matrix fixed by the scalar charge matrix and the vacuum normalization.^[ref]