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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.

For an infinitesimal continuous symmetry $\delta q_i=\epsilon {\xi }_i(q,t)$, allow the Lagrangian to change by a total derivative, $\delta L=\epsilon dF/dt$. Direct variation givesOn a solution of the Euler-Lagrange equations, the first term vanishes. This proves Noether's theorem in the fixed-time form:Spatial translations yield total momentum, and rotations yield angular momentum. Time translations yield energy when the Lagrangian has no explicit time dependence; this follows either from the extended time-transformation form of the theorem or by differentiating $H={\sum }_i{p}_i{\dot{q}}_i-L$ to obtain $\dot{H}=-\partial L/\partial t$.

The canonical momentum is $p_i=\partial L/\partial {\dot{q}}_i$. If the velocity-momentum relation can be locally inverted, the Legendre transform defines $H(q,p,t)={\sum }_i{p}_i{\dot{q}}_i-L$, with the velocities expressed in terms of $q,p,t$. Its differential isThe Euler-Lagrange equations then yield Hamilton's equations,A singular velocity Hessian instead requires a constrained-Hamiltonian treatment.

Vary $S=\int L(q,\dot{q},t)dt$ with fixed endpoints. Integration by parts gives $\delta S=\int {\sum }_i[L_{q_i}-d{L}_{{\dot{q}}_i}/dt]\delta q_{i}dt$, hence the Euler-Lagrange equations. Define $p_i=L_{{\dot{q}}_i}$ and, assuming an invertible velocity-momentum relation, $H=p\cdot \dot{q}-L$. The equivalent action is $\int (p\cdot \dot{q}-H)dt$. Independent variations of $p$ and $q$ yieldBoundary terms vanish for fixed endpoint coordinates; constraining momenta independently at the same endpoints is unnecessary. Singular Lagrangians require constraint analysis instead of this unconstrained Legendre transform.

Use $u^{\lambda }{u}_{\lambda }=-c^2$ and a constant Minkowski metric. Vary the velocity-dependent square root before imposing the proper-time normalization. Equivalently, vary a reparametrization-invariant worldline action first and then choose proper time. Its canonical covector momentum isThe coordinate derivative at fixed velocity is $\partial \mathcal{L}/\partial x^a=q{u}^{\lambda }{\partial }_a{A}_{\lambda }$. The Euler-Lagrange equation and the chain rule from a therefore giveDefine the antisymmetric field tensor byThe particle equation becomesAntisymmetry ensures $u^a{F}_{a\lambda }{u}^{\lambda }=0$, so the force is orthogonal to the four-velocity and preserves $u^2=-c^2$. The worldline variation before proper-time normalization is essential: substituting $\sqrt{-u^2}=c$ into the Lagrangian before taking its velocity derivative would turn the kinetic term into a constant and incorrectly remove the inertial momentum. The normalization is a property of the proper-time solution, not permission to discard the variational dependence.

In Cartesian components, $\partial L/\partial v_i=m{v}_i+e{A}_i$ and $\partial L/\partial x_i=e{v}_j{\partial }_i{A}_j-e{\partial }_i\varphi$. The Euler-Lagrange equation therefore givesThus the minimally coupled particle Lagrangian yields the Lorentz force.

For an arbitrary smooth gauge function $\chi$, take $\mathbf{A}\to \mathbf{A}+\nabla \chi$ and $\varphi \to \varphi -{\partial }_t\chi$. Both fields are unchanged, whileThe action changes only by an endpoint term, so fixed-endpoint variation gives the same motion. Canonical momentum shifts by $e\nabla \chi$, whereas kinetic momentum $m\mathbf{v}$ and the Lorentz force are gauge invariant. This is the relevant gauge invariance of the classical particle dynamics.

Use metric signature $(+,-,-,-)$ and units with propagation speed one. The scalar Euler-Lagrange equation givesWith the stated spatial Fourier convention, $-{\nabla }^2$ becomes $k^2$. Each transformed mode obeysHere $F$ and $V$ are the initial transformed field and velocity. Reality requires $F(-\mathbf{k})=F(\mathbf{k}{)}^{\ast }$ and the same for $V$; the Fourier convention places no extra $(2\pi {)}^3$ in these evolution formulas.

Including the interaction gives the two coupled real Klein-Gordon fields,Adding and subtracting diagonalizes them:Both squared masses are positive for $0<g<m^2$. For arbitrary initial data $F_i={\tilde{\varphi }}_i(0)$ and $V_i={\dot{\tilde{\varphi }}}_i(0)$, defineThen the general solution is ${\tilde{\varphi }}_1=(U_{+}+U_{-})/2$, ${\tilde{\varphi }}_2=(U_{+}-U_{-})/2$, with the corresponding reality conditions on initial data.

For the specified sinusoidal initial displacement, put ${\Omega }_{\pm }=\sqrt{q^2+m^2\mp g}$. Both normal fields initially have amplitude $A\sin (\mathbf{q}\cdot \mathbf{x})$ and zero time derivative. ConsequentlyWriting $\bar{\Omega }=({\Omega }_{-}+{\Omega }_{+})/2$ and $\delta \Omega =({\Omega }_{-}-{\Omega }_{+})/2$ gives ${\varphi }_1=A\sin (\mathbf{q}\cdot \mathbf{x})\cos \bar{\Omega }t\cos \delta \Omega t$ and ${\varphi }_2=A\sin (\mathbf{q}\cdot \mathbf{x})\sin \bar{\Omega }t\sin \delta \Omega t$. This is field-amplitude beating of coupled Klein-Gordon modes. The conserved energy includes the interaction term $-g{\varphi }_1{\varphi }_2$; separate uncoupled-field energies need not be conserved, and an envelope zero is not by itself exact complete energy transfer.

Put $G=g/\ell$ and $s=\kappa /m$. Applying the Euler-Lagrange equations to the quadratic Lagrangian givesA nonzero motion satisfying the specified equal-and-opposite initial displacements can have one frequency only if $(1,-1)$ is a normal-mode vector. Its acceleration terms are proportional to $(1,0)$, while the restoring terms are proportional to $(2G-2s,2s-G)$. HenceandThe rods swing oppositely. The hinge mass moves sideways, while the final mass has no first-order displacement: the two horizontal contributions cancel and its vertical change is quadratic in angle. In the corresponding geometric ansatz it moves vertically at twice the small-oscillation frequency to leading nonzero order, but the displayed single-frequency solution is a linearized result. The trivial case $\xi =0$ imposes no parameter condition.

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.

The derivatives needed for varying $A_{\nu }$ areTherefore the electromagnetic Euler-Lagrange equation isAntisymmetry of $F$ and commutation of derivatives imply ${\partial }_{\nu }{\partial }_{\mu }{F}^{\mu \nu }=0$, hence ${\partial }_{\nu }{J}^{\nu }=0$. In ordinary derivatives, $J^{\mu }=j^{\mu }-2q^2{A}^{\mu }|\varphi {|}^2$; the interaction contribution must not be omitted.

Take positive $\theta$ toward positive $x$ from the downward vertical, as in the source diagram. The mass has position and velocityThus $T=m|\dot{\mathbf{r}}{|}^2/2$ and $V=mg(a_y-l\cos \theta )$. Expanding the squared speed givesThe prescribed support coordinates are functions of time, rather than additional dynamical variables. Terms depending only on time do not affect the Euler-Lagrange equation for $\theta$, but have been retained here to obtain the inertial-frame kinetic and potential energies directly.

The derivatives needed in the Euler-Lagrange equation areThe terms proportional to ${\dot{a}}_i\dot{\theta }$ cancel after differentiation, leavingWith $\mathbf{l}=l(\sin \theta ,-\cos \theta ,0)$ and $\mathbf{g}=-g\hat{\mathbf{y}}$,Hence the vector equation is $l^2\ddot{\theta }\hat{\mathbf{z}}=\mathbf{l}\times (\mathbf{g}-\ddot{\mathbf{a}})$. In the translating support frame, $-m\ddot{\mathbf{a}}$ is the inertial force. The rod's tension passes through the hinge and contributes no torque about it. The same torque equation therefore has an immediate mechanical interpretation.

Time-translation symmetry conserves the energy $H$, and axial rotational symmetry makes $\varphi$ a cyclic coordinate, conservingThe radial Euler-Lagrange equation before elimination isInsert $\dot{\varphi }=J/(m{r}^2)-qB/(2m)$. The cross terms linear in $JqB$ cancel, leavingEquivalently use the reduced radial potentialThe last term is constant at fixed $J$ and has no radial derivative, but is part of the full conserved energy. Conservation of $J$ is stated in the rotationally symmetric gauge; the conserved mechanical combination can be written in terms of magnetic flux in a gauge-independent way.

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.

Set $\Delta =\theta -{\theta }_B$ and $u=z-b{\theta }_B$. Applying the Euler-Lagrange equations separately to $r,\theta ,z,{\theta }_B$ givesFor example, $\partial L/\partial \theta =-kar\sin \Delta$ whereas $\partial L/\partial {\theta }_B=kar\sin \Delta +kbu$; these signs also follow from equal and opposite spring forces projected on the permitted bead motion. The angular particle equation can equivalently be written $m(r^2\ddot{\theta }+2r\dot{r}\dot{\theta })=-kar\sin \Delta$. The factor $M(a^2+b^2)$ is the bead's inertia with respect to the helical coordinate, rather than its moment of inertia about the axis alone.

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.

In a uniform-field region, $z$ is cyclic and $\gamma$ is constant. Its equation is $d(\gamma m\dot{z})/dt=0$. The radial Euler-Lagrange equation isso with $\dot{\theta }=-\Omega$ and $qB/(\gamma m)=2\Omega$,The radial equation applies on intervals with a regular signed cylindrical-coordinate description; the Cartesian solution handles axis crossings.

In a field-free region, the equations areFor this particle $p_{\theta }=0$, so $\dot{\theta }=0$ and $\ddot{\rho }=0$ outside, until a coordinate change is needed at an axis crossing.

Axial velocity is constant within each uniform region, but not across the ends. In a smooth fringe,The entrance develops an azimuthal speed $|\Omega |{\rho }_0$ while radial velocity remains zero. Conservation of total speed givesForward passage presumes the radicand is positive. The exit fringe changes axial velocity again. Neglecting these small changes is a paraxial approximation, requiring $|\Omega |{\rho }_0\ll v$, rather than an exact consequence of cylindrical symmetry.

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.

For the horizontal generalized coordinate $\eta$,The corresponding Euler-Lagrange equation isFor $\theta$, the terms proportional to $\dot{\eta }\dot{\theta }\sin \theta$ cancel between $d(\partial L/\partial \dot{\theta })/dt$ and $\partial L/\partial \theta$. Hence the other exact equation isKeeping only terms linear in small coordinates and their derivatives givesThe equilibrium is $\theta =\eta =0$. These characteristic frequencies label gravitational and spring restoring scales; they are not two independent normal mode frequencies of this massless-support system.