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For independent generalized coordinates $q_i$, their generalized velocities ${\dot{q}}_i$ and a Lagrangian $L(q,\dot{q},t)$, the Euler-Lagrange equations areHere the partial derivatives treat $q_i,{\dot{q}}_i,t$ as independent arguments, whereas $d/dt$ follows the motion. For a conservative mechanical system, choose coordinates satisfying the mechanical constraints, express the kinetic energy $T$ and potential energy $V$ in those coordinates, set $L=T-V$, and evaluate these equations to obtain the equations of motion. If nonconservative generalized forces $Q_i$ are present, they appear on the right-hand side.

A cyclic coordinate is absent explicitly from $L$: $\partial L/\partial q_i=0$, even though $L$ may depend on ${\dot{q}}_i$. Its canonical momentum is consequently conserved,This does not require ${\dot{q}}_i=0$ or even that ${\dot{q}}_i$ be constant; the relation between canonical momentum and generalized velocity depends on the Lagrangian.

Model the cart as a particle on a fixed plane curve, remaining constrained to the track. Its vertical velocity is $\dot{y}=y^{\prime }(x)\dot{x}$, so its kinetic energy, gravitational potential energy and Lagrangian areConsequently the canonical momentum conjugate to the horizontal coordinate isIt is not simply the horizontal component $m\dot{x}$ of the mechanical momentum, because a change in $x$ also moves the cart vertically. Differentiate carefully in the Euler-Lagrange equation:Subtracting yieldsThe term involving $y^{\prime \prime }$ accounts for the changing track geometry. An independent check is conservation of energy:

Choose zero potential energy at the equilibrium position. For the simple harmonic oscillator, the kinetic energy and elastic potential energy give the LagrangianIts canonical momentum is $p=\partial L/\partial \dot{x}=m\dot{x}$. The Legendre transform defining the Hamiltonian consequently givesThe Euler-Lagrange equation is $m\ddot{x}+kx=0$. Thus along the motionThis proves conservation of energy. Equivalently, the Lagrangian has no explicit time dependence, so $dH/dt=-\partial L/\partial t=0$. The Lagrangian itself need not be constant: the kinetic energy and potential energy exchange periodically.

For the elastic pendulum, set $r=l+x>0$ and measure $\theta$ from the downward vertical. In the vertical plane, Cartesian coordinates are $X=r\sin \theta$ and $Z=-r\cos \theta$, with $Z$ upward. Differentiating gives ${\dot{X}}^2+{\dot{Z}}^2={\dot{x}}^2+r^2{\dot{\theta }}^2$. Thus the Lagrangian isFor $x$, the Euler-Lagrange equation givesFor $\theta$, the canonical momentum is $p_{\theta }=m(l+x{)}^2\dot{\theta }$. Its Euler-Lagrange equation isor, for positive length,At an equilibrium, all time derivatives vanish, so $kx=mg\cos \theta$ and $\sin \theta =0$. Modulo full rotations, the two formal possibilities areThe lower configuration always has positive length. The upper one is a physical straight-spring configuration only if $l>mg/k$. For example, $l=mg/(2k)$ would give the purported upper state a negative length $-mg/(2k)$; at $l=mg/k$ its zero length makes the angular coordinate singular. Also, the straight compressed spring is an idealization; a real unsupported compressed spring can buckle.

The potential energy is $V=k{x}^2/2-mg(l+x)\cos \theta$. At either formal stationary point its mixed second derivative vanishes, whileBoth curvatures are positive at the lower state, making it a stable equilibrium; the upper state, when it has positive length, has negative angular curvature and is unstable.

To linearize about the stable state, define the small displacement $\xi =x-mg/k$, not $x$ itself, and $r_0=l+mg/k$. Products such as ${\dot{\theta }}^2$, $\dot{\xi }\dot{\theta }$ and $\xi \ddot{\theta }$ are second order, while $\cos \theta =1+O({\theta }^2)$ and $\sin \theta =\theta +O({\theta }^3)$. The first-order equations of motion therefore decouple:There are two independent normal modes. The radial one is vertical simple harmonic motion with angular frequency ${\omega }_r=\sqrt{k/m}$. The transverse one is a simple pendulum oscillation at fixed, statically stretched length $r_0$, with angular frequency ${\omega }_{\theta }=\sqrt{g/r_0}$. Their decoupling is a first-order small-oscillation result, not an assertion about the nonlinear motion of the elastic pendulum.

Generalized coordinates incorporate holonomic constraints without solving separately for every constraint force. For conservative mechanics use $L=T-V$. The principle of stationary action makes the action $\int Ldt$ stationary for variations vanishing at the endpoints. Integration by parts givesThe canonical momentum is $p_i=\partial L/\partial {\dot{q}}_i$. A cyclic coordinate gives conserved $p_i$; absent explicit time dependence, ${\sum }_i{p}_i{\dot{q}}_i-L$ is conserved. These are symmetry–conservation relationships, not assumptions that all momenta are constant. Nonconservative generalized forces enter the right side.

For example, a pendulum has $L=m{\ell }^2{\dot{\theta }}^2/2+mg\ell \cos \theta$, giving $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0$. The method extends directly to coupled coordinates and velocity-dependent interactions when the correct Lagrangian is used; $T-V$ with a position-only potential is not the most general form.

For generalized coordinates $q_i$ with Lagrangian $L(q,\dot{q},t)$, the conjugate momentum isA coordinate absent from $L$ is cyclic, and its conjugate momentum is conserved. More generally, Noether's theorem connects continuous symmetries of the action to conserved quantities: spatial translations give linear momentum and rotations give angular momentum. If $L$ has no explicit time dependence, the conserved energy function isThe momentum conjugate to an angle is an angular momentum, not a linear velocity. Canonical momentum can also differ from mechanical momentum: for $L=\frac{1}{2}m{v}^2+q\mathbf{A}\cdot \mathbf{v}-q\varphi$, $\mathbf{p}=m\mathbf{v}+q\mathbf{A}$. Thus a conserved conjugate momentum must be identified from the actual Lagrangian and its symmetry, rather than assumed to equal $m\mathbf{v}$. Gauge-dependent canonical momenta require the corresponding gauge choice or gauge-covariant symmetry interpretation.^[ref]

Keep the electromagnetic units of the supplied Lagrangian, including its factor $c$, and regard $e$ as the signed coupling used there. Set $a_B=e{\Phi }_B/(2\pi c)$. The canonical momentum is $p_{\varphi }=I\dot{\varphi }+a_B$, so the Legendre transform givesSingle-valued periodic wavefunctions have normalized eigenstates $e^{in\varphi }/\sqrt{2\pi }$ with integer $n$. Defining dimensionless flux $f=e{\Phi }_B/(hc)$ gives $E_n={\hslash }^2(n-f{)}^2/(2I)$, and the trace is thereforeThe flux-threaded quantum rotor spectrum is periodic under $f\mapsto f+1$ after relabelling $n$, corresponding to flux period $hc/|e|$ in these units. A local gauge change can remove the magnetic vector potential from the differential operator but then imposes a twisted boundary condition with phase $e^{-2\pi if}$; it cannot remove the flux dependence from both the Hamiltonian and boundary condition simultaneously. In SI conventions the flux coupling has no $c$, so one must not mix the two unit systems.

The displayed oscillator expansion and the displayed action require two separate corrections to describe one conventional positive-energy theory. First compute what the oscillator algebra literally implies. Assume the two oscillator species commute with each other and annihilate the vacuum. Same-charge contractions vanish, soFor the two mixed contractions the printed phases give the same kernel, with $z=x-y$:Thus $⟨T\varphi (x){\varphi }^{\ast }(y){⟩}_0=⟨T{\varphi }^{\ast }(x)\varphi (y){⟩}_0=C_{\mathrm{lit}}(z)$. In $D$ dimensions replace $d^{3}p$ and $d^{4}p$ by $d^{D-1}p$ and $d^{D}p$; the source's explicit three-dimensional spatial measure is the $D=4$ specialization. The covariant integral is manifestly Lorentz invariant. Withwe have $C_{\mathrm{lit}}(z)=D_F(z{)}^{\ast }$. This reversed-frequency scalar contraction has the opposite pole prescription to the usual positive-energy Feynman propagator. For a single zero-momentum mode at positive time separation, the printed contraction is $e^{+imt}/(2m)$, while $D_F$ is $e^{-imt}/(2m)$: this is a concrete phase counterexample, rather than an immaterial change of integration variable.

There is also a normalization of a complex scalar propagator conflict independent of those phases. The displayed Lagrangian density gives the canonical momentum ${\pi }_{\varphi }={\dot{\varphi }}^{\ast }/2$. Substituting the printed modes yieldsinstead of the canonical $+i{\delta }^{(3)}$. Reversing the mode phases repairs the sign but leaves $+i{\delta }^{(3)}/2$, so it does not repair the normalization. The complex scalar field convention with unit mixed propagator normally has no $1/2$ in its quadratic action.^[ref]

To retain the actual action in the paper, use $\varphi ={\varphi }_1+i{\varphi }_2$ with canonically normalized real components, or equivalently replace its mode expansion byThe corresponding path integral with the usual Feynman i-epsilon prescription haswith the two same-charge propagators still zero. This action-consistent kernel is used in b and c. Keeping conventional unit modes instead would require removing the quadratic factors $1/2$ from the action as well as reversing the phases. Those are distinct theories unless the field and its coupling are rescaled together.

A real scalar field suffices to describe one neutral species of bosonic quasiparticle. Let $\pi (\mathbf{x})$ be its canonical momentum. In three spatial dimensions, choose the positive classical HamiltonianThe same construction works in any number of spatial dimensions by replacing $d^{3}x$ appropriately. Under a spatial Fourier transform, ${\nabla }^2\varphi$ becomes $-|\mathbf{p}{|}^2\tilde{\varphi }$, so the Hamiltonian is a collection of independent quantum harmonic oscillators with squared angular frequency ${\omega }_{\mathbf{p}}^2={\kappa }^2|\mathbf{p}{|}^4+m^2$. The usual reality condition pairs opposite momenta; it does not introduce a second species.

The canonical action is $S=\int dt[\int d^{3}x\pi \dot{\varphi }-H]$. Varying $\pi$ gives $\pi =\dot{\varphi }$; eliminating it by the Legendre transform givesThis Lagrangian density is local: only the field and finitely many derivatives at the same point occur. It contains second spatial derivatives but only first time derivatives, so no extra canonical time data are needed. A square root of a differential operator in the classical Hamiltonian is unnecessary. The fourth-order spatial term breaks Lorentz invariance while preserving spatial rotations and translations. This is a free scalar field with quartic spatial dispersion.

The canonical anticommutation relations arewith the mixed and equal-type anticommutators zero. Insert the mode expansion of a Dirac field. The phases cancel at equal time, and a givesSince $\bar{\psi }{\gamma }^0={\psi }^{†}$, the requested componentwise answer isThis is the Equal-time canonical anticommutator of a Dirac field: the canonical momentum conjugate to ${\psi }_{\alpha }$ is ${\pi }_{\alpha }=i{\psi }_{\alpha }^{†}$. It is the fermionic counterpart of the canonical field canonical commutation relation, ensuring the correct local canonical quantization. For this first-order Lagrangian density, the conjugate variables are constrained; the displayed relation is the standard result after handling that constraint. It does not assert a nonzero anticommutator between the fields at separated spatial points.

The Einstein tensor and its contractions areTo match the supplied Ricci tensor components, use the Riemann curvature tensor conventionThis convention also fixes the sign used for geodesic deviation above. In vacuum, tracing the Einstein field equations gives $-R+4\Lambda =0$. Substitution therefore yields $R_{\mu \nu }=\Lambda g_{\mu \nu }$.

Differentiate the angular field equation to obtainConsequently $R_{00}=\Lambda A=\Lambda g_{00}$ and $R_{11}=-\Lambda /A=\Lambda g_{11}$. The other angular component follows from $R_{33}={\sin }^2\theta R_{22}$. Thus the single angular equation implies all the displayed vacuum equations wherever this coordinate chart is regular.

Integrating $(rA{)}^{\prime }=1-\Lambda r^2$ gives $A=1+C/r-\Lambda r^2/3$. Match $g_{tt}\simeq 1+2\Phi$ with the Newtonian gravitational potential $\Phi =-GM/r$ in the weak-field overlap region to identify $C=-2GM$. Hence the Schwarzschild-de Sitter spacetime hasNegative $\Lambda$ instead gives the Schwarzschild--anti-de Sitter black hole case. This matching fixes the normalization locally; a spacetime with nonzero $\Lambda$ is not asymptotically flat.

For timelike geodesics in the orbital plane, take proper time as parameter and use $L=(A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2)/2$. The cyclic coordinates give $A\dot{t}=k$ and $r^2\dot{\varphi }=h$; the angular canonical momentum is $-h$ in this signature. The four-velocity normalization $2L=1$ then givesHere $k$ is conserved Killing energy per unit rest mass in the chosen time normalization, and $h$ is conserved angular momentum per unit rest mass. There is no general interpretation of $k$ as energy measured at an asymptotically flat infinity when $\Lambda \ne 0$.

For radial motion $h=0$, differentiation gives $2\dot{r}\ddot{r}=-A^{\prime }\dot{r}$. The radial Euler-Lagrange equation extends the resulting expression through a turning point where division by $\dot{r}$ would fail. ThusThis is the proper-time second derivative of the areal radius, not a nonzero four-acceleration of the freely falling particle. Positive $\Lambda$ contributes an outward term; negative $\Lambda$ reinforces inward attraction. Restoring units in the slow-motion, weak-field Newtonian limit givesThe extra term describes the gravitational effect of a cosmological constant, not drag or a contact force. The approximation requires $GM/(r{c}^2)\ll 1$, $|\Lambda |r^2\ll 1$ and speed much less than $c$.

Define the canonical momentum $p_i=\partial L/\partial {\dot{q}}_i$ and the energy function $E={\sum }_i{p}_i{\dot{q}}_i-L$. Along an Euler-Lagrange equation solution,Thus an autonomous Lagrangian has conserved $E$. For a simple harmonic oscillator, $L=m{\dot{x}}^2/2-k{x}^2/2$ and $\boxed{E=m{\dot{x}}^2/2+k{x}^2/2}$ is constant. For ordinary time-independent mechanical coordinates this is the total mechanical energy.

Suppress primes on the angular velocity while retaining rotating components. The canonical momentum is $\partial L^{\prime }/\partial {\dot{\mathbf{r}}}^{\prime }=m({\dot{\mathbf{r}}}^{\prime }+\mathbf{\Omega }\times {\mathbf{r}}^{\prime })$. Varying $L^{\prime }$ with respect to position and applying the Euler-Lagrange equations givesThe terms are the physical force, the velocity-dependent Coriolis force, the outward centrifugal force, and the Euler force caused by changing angular velocity. The common origin is inertial; an accelerating origin would introduce another translational apparent force.

To match the printed order $df/dt=\{H,f\}$, define the Poisson bracket with the sign conventionHamilton's equations imply $\dot{f}={\sum }_i(f_{q_i}{H}_{p_i}-f_{p_i}{H}_{q_i})=\{H,f{\}}_{\mathrm{paper}}$. With the other common Poisson bracket convention the correct statement is instead $\dot{f}=\{f,H{\}}_{\mathrm{standard}}$. For example $H=p^2/(2m)$, $f=q$ gives $\{H,q{\}}_{\mathrm{standard}}=-p/m$, contradicting the printed order unless its sign convention is adopted. Since $H_t=0$ and $\{H,H\}=0$, the Hamiltonian is conserved. Also ${\dot{p}}_{\varphi }=-H_{\varphi }=0$: this canonical momentum is the axial angular momentum.

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.

The canonical momentum is $p=\gamma m\dot{r}$, where $\gamma =(1-{\dot{r}}^2/c^2{)}^{-1/2}$. Eliminating velocity givesIts Hamilton equations are $\dot{r}=c^{2}p/H$ and $\dot{p}=0$. In the limit $|p|\ll mc$, the Taylor expansion is $H=m{c}^2+p^2/(2m)+O(p^4/(m^3{c}^2))$, and $\dot{r}\simeq p/m$. The additive rest energy does not change the classical free-particle equations.

Minimal electromagnetic coupling changes the canonical momentum to $p=\gamma m\dot{r}+qA$. The mechanical momentum is $\pi =p-qA$. The Legendre transform yieldsThe cancellation of $q\dot{r}\cdot A$ in the Legendre transform does not remove magnetic effects: $A$ remains in the relation between canonical and mechanical momenta. The expression is the relativistic kinetic-plus-rest energy with an electrostatic potential-energy term, expressed in canonical variables.

Velocity differentiation gives $\boxed{p_r=m\dot{r},p_{\theta }=m{r}^2\dot{\theta },p_z=m\dot{z}+qA(r)}$. The coordinates $\theta$ and $z$ are cyclic, so $p_{\theta },p_z$ are conserved. The radial momentum generally changes. In particular the conserved axial canonical momentum differs from the axial mechanical momentum by $qA$.

With zero scalar potential, the nonrelativistic charged-particle Lagrangian in SI units is kinetic energy plus $q\mathbf{A}\cdot \dot{\mathbf{r}}$. ThusThe velocity-dependent magnetic term changes canonical momentum but does not contribute to the mechanical energy.

Differentiate with respect to the generalized velocities:The Lagrangian has no explicit $x$, so ${\dot{p}}_x=\partial L/\partial x=0$. This canonical momentum is also the system's total horizontal mechanical momentum. Horizontal translation symmetry conserves $p_x$ even though the support and bob separately accelerate.