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The coordinates $\varphi$ and $\psi$ are cyclic, so their canonical momenta are conserved:Because the Lagrangian has no explicit time dependence, the conserved energy isThese are the three integrals of motion requested.

If $\mathbf{r}$ is measured in the rotating frame, its inertial velocity is $\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. The masses producing the gravitational field are stationary in this frame, so their potential energy is a time-independent function $V(\mathbf{r})$. Thus kinetic minus potential energy gives the stated LagrangianPut $\mathbf{v}=\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. ThenSubstitution in the Euler-Lagrange equation yieldsThe extra terms are the Coriolis acceleration, Euler acceleration, and centrifugal acceleration of a rotating reference frame.

The canonical momentum and Hamiltonian areAccordingly, Hamilton's equations are

For a regular Lagrangian, define the canonical momentum $p_i=\partial L/\partial {\dot{q}}_i$ and invert this relation for $\dot{q}(q,p,t)$. Its Legendre transform is the HamiltonianSince $Ldt=p\cdot dq-Hdt$, the action isFor independent variations with $\delta q(t_1)=\delta q(t_2)=0$,Stationarity for arbitrary $\delta p$ and $\delta q$ gives Hamilton's equations $\dot{q}={\partial }_{p}H$ and $\dot{p}=-{\partial }_{q}H$.

For a particle of mass $m$ and charge $q$, the Hamiltonian of a charged particle in an electromagnetic field isThe canonical momentum is $\mathbf{p}=m\dot{\mathbf{x}}+q\mathbf{A}$. Hamilton's equations are equivalent to the Lorentz force law

The positive-frequency solution for $u_k=a{f}_k$ is proportional to $e^{-ick\tau }$. Since $a=-1/(H\tau )$, its scalar-field mode function has the stated form $f_k=\mathcal{N}\tau e^{-ick\tau }$. The canonical momentum in conformal time is $\pi =a^2{\phi }^{\prime }$, so the canonical commutation relation requires the Wronskian normalizationSubstitution gives $2ick{a}^2|\mathcal{N}{|}^2{\tau }^2=i$. Hence, up to an irrelevant constant phase,The choice $e^{-ick\tau }$ is the Bunch-Davies vacuum condition at early conformal time.

The equal-time canonical commutation relation for the transverse-traceless graviton and its canonical momentum is the transverse-traceless projector. At zero momentum the supplied polarization completeness relation giveswhere symmetry and tracelessness of ${\omega }_{ij}$ remove the trace term. Thus the charge generates precisely the transformation in part i:This is the soft, field-independent part of the Noether charge associated with the large diffeomorphism.

The conserved number operator isSubstitution of $\Psi =\sqrt{\rho }{e}^{-i\theta }$ into the Landau-Ginzburg theory givesThe imaginary term is a total derivative, soThus $\rho$ is the number density and is the canonical momentum conjugate to $\theta$. The canonical commutation relation is $[\theta (\mathbf{x}),\rho (\mathbf{y})]=i{\delta }^{(d)}(\mathbf{x}-\mathbf{y})$. Consequently, for ${\theta }_0=\int d^{d}x\theta$ in volume $V$,Equivalently, the averaged phase $\bar{\theta }={\theta }_0/V$ obeys $\Delta \bar{\theta }\Delta N\ge 1/2$. This number-phase conjugacy means that a state of sharp $N$ has no sharp phase, whereas a phase-selected state exhibiting spontaneous symmetry breaking must superpose different number sectors. Such sectors become effectively degenerate in the thermodynamic limit.

Put $q=\delta \theta$, $p=\delta \varphi$, so $\theta =\pi /2-q$ and $\varphi =p$. The quadratic Lagrangian is obtained by takingThe two real fluctuations form a coordinate and its canonical momentum rather than two independent modes. Their Euler-Lagrange equations combine to giveand similarly for $p$. The ferromagnetic magnon therefore has the quadratic dispersion relation

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.

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.

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.