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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.

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.

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.

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.

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.

At fixed time, the coordinate field and the canonical momentum are independent variables. Their functional derivatives satisfyPutting $F=Q$, $G={\varphi }_a(\mathbf{x})$ in the Poisson bracket definition therefore givesThe superscript on the species label $a$ in the printed denominator has no spacetime-metric significance; with the ordinary species Kronecker metric it is the same expression.

An observer with future-directed unit velocity $u^{\mu }$ measures particle energy $E=-u_{\mu }{p}^{\mu }$ and spatial physical momentum given by the orthogonal projection of its four-momentum in the displayed formula. With signature $(-+++)$ and $p_{\mu }{p}^{\mu }=-m^2$, its squared magnitude is $p_{\perp \mu }{p}_{\perp }^{\mu }=E^2-m^2$. Components in the observer's orthonormal spatial frame agree with $\gamma m\mathbf{v}$. They differ in general from coordinate canonical momentum components. For a comoving observer in flat FRW, a conserved conventional spatial comoving covector has locally measured magnitude divided by the scale factor.

The Maxwell Lagrangian density contains no time derivative of $A_0$, so its canonical momentum vanishes. The condition is a first-class constraint. Its time preservation yields the Gauss law constraint in gauge theory; $A_0$ acts as the corresponding multiplier in the Hamiltonian.

The conjugate momentum to a periodic internal soliton angle generates shifts of that angle. Single-valued wavefunctions, or Bohr-Sommerfeld quantization with zero Maslov index, quantize this canonical momentum. A theta term can shift the relation between canonical momentum and mechanical charge without changing integer canonical labels.

A scalar-field shift symmetry acts by $\varphi (x)\mapsto \varphi (x)+c$ for constant $c$. Its Noether charge is the spatial integral of the canonical momentum, up to the normalization of $c$, and interactions invariant under the symmetry contain derivatives of the scalar field.

A collection of Hamiltonian constraints is second class when its Poisson bracket matrix $C_{ab}=\{{\chi }_a,{\chi }_b\}$ is invertible on the constraint surface. These conditions eliminate canonical variables rather than generate a gauge symmetry. Their consistent reduced bracket is the Dirac bracket, $\{F,G{\}}_D=\{F,G\}-\{F,{\chi }_a\}(C^{-1}{)}^{ab}\{{\chi }_b,G\}$. The zero temporal canonical momentum of a massive Proca field and its massive Gauss law form such a pair; imposing an independent temporal canonical commutation relation would contradict the constraint.

For positive constants $m,c$ and nonzero velocity $\dot{\mathbf{x}}$, consider the Lagrangian $L=mc|\dot{\mathbf{x}}|-V(\mathbf{x})$. Its canonical momentum and canonical energy areThe canonical momentum records direction but loses speed: every positive rescaling of $\dot{\mathbf{x}}$ gives the same $\mathbf{p}$. Its velocity Hessian matrix is $mc|\dot{\mathbf{x}}{|}^{-1}(I-\hat{\mathbf{v}}{\hat{\mathbf{v}}}^{\mathsf{T}})$, where $\hat{\mathbf{v}}=\dot{\mathbf{x}}/|\dot{\mathbf{x}}|$. It has the velocity direction in its kernel, so the Lagrangian is not a regular Lagrangian and has no ordinary inverse Legendre transform in mechanics.

The expression $H=V$ is an energy value on the constraint surface ${\mathbf{p}}^2=m^2{c}^2$, not an unconstrained Hamiltonian describing all positions and momenta. Using it alone in ordinary Hamilton's equations would wrongly set the velocity to zero. A constrained Hamiltonian treatment must enforce the momentum relation and its consistency with the chosen $V$. For constant $V$, the Euler-Lagrange equations keep the direction fixed while leaving the positive speed along the path undetermined; that freedom is exactly the information the momentum map lost.

Locally this density is a total derivative, so it changes neither the bulk field equations nor the energy. Its contribution to the angular canonical momentum is a spatial boundary derivative. With ${ϵ}^{01}=+1$, it produces the theta-angle shift of a sigma-model kink charge.

The wavevector times wave action gives the canonical momentum associated with translations of a wave's phase. It describes momentum exchange with a slowly varying mean fluid flow, rather than simply the spatial average of oscillating fluid momentum. For horizontal wavenumber $k$, its horizontal component is $kE/\hat{\omega }$. A convention using displacement pseudomomentum of an internal gravity wave instead has the opposite sign.

For a free massless pseudoclassical spinning particle, let $p_m$ be the canonical momentum and ${\psi }^m$ odd coordinates in a Grassmann algebra. Choose the Dirac bracket conventionwith all $p_m$ commuting with the ${\psi }^n$ and with each other. The odd supercharge $\mathcal{Q}$ then obeysThe first equality follows immediately by contracting the elementary Dirac bracket with $p_m{p}_n$. It expresses the closure of two odd symmetry transformations into an even worldline translation.

In units $\hslash =1$, canonical quantization replaces the graded Dirac bracket by the corresponding graded commutator multiplied by $i$. Thus $\{{\hat{\psi }}^m,{\hat{\psi }}^n\}={\eta }^{mn}$ andConsequently the physical-state constraint $\hat{\mathcal{Q}}|\Psi ⟩=0$ implies $\hat{\mathcal{H}}|\Psi ⟩=0$, the massless mass-shell condition. Specifying these bracket and unit conventions fixes the normalization of the worldline supersymmetry algebra.