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For $\mathcal{L}={\dot{\varphi }}^2/2-{\mathcal{V}}_2+{\mathcal{L}}_3(\varphi ,\dot{\varphi })$, where ${\mathcal{L}}_3$ is cubic in perturbation amplitude, the canonical momentum has $\pi =\dot{\varphi }+O({\varphi }^2)$. Put $\dot{\varphi }=\pi +\Delta$ with $\Delta$ second order. Then $\pi \dot{\varphi }-{\dot{\varphi }}^2/2={\pi }^2/2-{\Delta }^2/2$, so through cubic order the Legendre transform in mechanics gives ${\mathcal{H}}_3=-{\mathcal{L}}_3$ evaluated with the free momentum. At fourth order there are extra terms, so derivative interactions do not obey this simple rule at all orders.

A Hamiltonian density is a local integrand for the field-theory Hamiltonian. For canonical fields with invertible velocity-momentum relation it is the Legendre transform in mechanics of the Lagrangian density, summing over all conjugate field components. Gauge theories require constraints and multiplier terms as well.

Using the lapse function $N$, shift vector $N^i$ and spatial metric $h_{ij}$, the Legendre transform in mechanics of the Einstein-Hilbert action gives, in units $16\pi G=1$,Here $\pi =h_{ij}{\pi }^{ij}$ and ${\pi }^{ij}$ is the canonical momentum of the spatial metric. Varying lapse and shift imposes the Hamiltonian constraint and momentum constraint. An asymptotic time translation requires an Arnowitt-Deser-Misner energy boundary term. The bulk constraints generate gauge evolution; the nonzero asymptotic energy comes from the boundary.

A Lagrangian is hyperregular when its fiber derivative, sending a velocity to its conjugate momentum, is a global diffeomorphism from the tangent bundle to the cotangent bundle. This gives a globally defined Legendre transform in mechanics and a global equivalence between the Euler-Lagrange equations and Hamilton's equations. With time dependence this requirement is imposed at each time, with a smooth inverse depending on time.

Let $X={\partial }^{-2}\zeta$, with an inverse spatial Laplacian defined by a boundary or zero-mode prescription. It depends on $\zeta$, not on its time derivative. Differentiation gives the canonical momentumCount $\zeta ,\pi ,X$ as first order. Inverting perturbatively givesFor the Legendre transform in mechanics, write $v=\dot{\zeta }=\pi +\Delta$ with $\Delta$ second order. The quadratic kinetic contribution is $\pi v-v^2/2={\pi }^2/2-{\Delta }^2/2$, whose correction is fourth order; in the cubic terms one may consequently replace $v$ by $\pi$. The Hamiltonian density is thereforeThe cubic interaction Hamiltonian density isIn the free interaction picture, ${\pi }_I={\dot{\zeta }}_I$. Thus here ${\mathcal{H}}_{\mathrm{int}}=-{\mathcal{L}}_3$ with the time derivatives interpreted as free fields. This equality follows from the cubic Legendre transform for scalar derivative interactions; at higher orders additional terms can appear.

The passage to Hamiltonian mechanics uses the Legendre transform in mechanics. Assume a regular Lagrangian: the velocity Hessian matrix $({\partial }^{2}L/\partial v^i\partial v^j)$ is invertible. Thendefines a locally invertible map from the tangent bundle to the cotangent bundle. Write its local inverse as $v=v(t,q,p)$ and define the HamiltonianDifferentiating this expression, all terms containing $dv$ cancel because $p_i=L_{v^i}$. HenceThe Euler-Lagrange equations become Hamilton's equations:Conversely, a solution of Hamilton's equations satisfies $\dot{q}=v(t,q,p)$, hence $p=L_v(t,q,\dot{q})$, and $\dot{p}=L_q$ recovers the Euler-Lagrange equations. This proves local equivalence of the two descriptions.

On the cotangent bundle, choose the canonical symplectic form ${\omega }_{\mathrm{can}}={\sum }_{i}d{q}^i\wedge d{p}_i=-d\lambda$ and the convention ${\iota }_{X_H}{\omega }_{\mathrm{can}}=dH$. Then $X_H={\sum }_i(H_{p_i}{\partial }_{q^i}-H_{q^i}{\partial }_{p_i})$, so its integral curves are exactly the phase-space equations above. This sign convention is used throughout these solutions. For a hyperregular Lagrangian, the Legendre transform in mechanics is globally invertible and gives global equivalence; regularity alone only gives local equivalence. A singular velocity Hessian matrix may instead produce constraints, so the ordinary unconstrained argument does not apply to every Lagrangian.

Use natural units and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. A real scalar field assigns a real variable $\varphi (t,\mathbf{x})$ to each spatial point. Its Lagrangian density can be taken to beThe principle of stationary action gives the Euler-Lagrange equation $\square \varphi +V^{\prime }(\varphi )=0$. With $V(\varphi )=m^2{\varphi }^2/2$, this is the Klein-Gordon equation. An additional nonlinear part of $V$ describes interactions.

The canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$. The Legendre transform in mechanics gives the canonical Hamiltonian density of a real scalar fieldThe Hamiltonian equations $\dot{\varphi }=\pi$ and $\dot{\pi }={\nabla }^2\varphi -V^{\prime }(\varphi )$ recover the same field equation. In canonical quantization, the fields become operators satisfying the equal-time canonical commutation relationsA spatial lattice makes the analogy with many coupled quantum-mechanical coordinates precise. Each lattice field value is a coordinate, with its own conjugate momentum. The path integral is another representation of the same quantum evolution.

To see its origin, first consider one coordinate with $H=p^2/(2M)+V(q)$. Split a time interval into $N$ steps of length $\epsilon$ and insert position and momentum resolutions of the identity. The short-time kernel isMultiplying the kernels and integrating over intermediate positions gives the phase-space path integralThe endpoints of $q$ are fixed. The momentum integrals are Gaussian integrals; completing the square produces the configuration-space path integralAt finite slicing its normalization contains $(M/(2\pi i\epsilon ){)}^{N/2}{\prod }_{j=1}^{N-1}d{q}_j$. This fixes the composition law and the initial delta-function kernel. One sums over all paths, not merely solutions of the classical equation. Restoring $\hslash$ replaces the weight by $e^{iS/\hslash }$; stationary phase explains the emergence of classical trajectories.

For the field, use scalar field configuration eigenstates $|\phi ⟩$, satisfying $\hat{\varphi }(\mathbf{x})|\phi ⟩=\phi (\mathbf{x})|\phi ⟩$. Insert their completeness relations on every time slice. This givesThe endpoint field configurations are fixed. Integrating the Gaussian momentum variables leaves the scalar field path integral $\mathcal{N}\int \mathcal{D}\varphi e^{iS[\varphi ]}$. The functional measure means a regulated product over the field variables. A spacetime lattice or another ultraviolet cutoff makes this product finite before the continuum limit; interacting continuum calculations may require renormalization. The oscillatory Minkowski weight is an amplitude, not a positive probability density.

For vacuum expectation values, the boundaries must select the vacuum rather than arbitrary field configurations. Long imaginary-time evolution suppresses excited states: $e^{-TH}|n⟩=e^{-T{E}_n}|n⟩$, so after normalization only the lowest-energy component remains as $T\to \infty$. This is vacuum projection by imaginary time. The corresponding Feynman i-epsilon prescription in the real-time integral specifies the vacuum boundary conditions and the poles of the propagator. With $t=-i\tau$, the Euclidean path integral has the weight $e^{-S_E}$, whereIt is often a useful regulated starting point; analytic continuation returns the vacuum time-ordered quantities.

Introduce a classical source $J(x)$ and define the normalized vacuum generating functionalwith the same vacuum prescription in numerator and denominator. A functional derivative brings down $i\varphi (x)$. The order of the time slices makes the operator insertion time-ordered. Thus source differentiation inserts time-ordered field operators:The denominator removes vacuum diagrams and gives normalized expectation values. It is essential that these are time-ordered products; differentiating this vacuum functional does not directly give every possible operator ordering.

The free theory illustrates the method. Its quadratic kernel is $K=-\square -m^2$ with the vacuum pole prescription, and completing the square gives the Gaussian evaluation of a free scalar generating functionalTwo source derivatives give ${\Delta }_F=⟨\Omega |T\{\hat{\varphi }(x)\hat{\varphi }(y)\}|\Omega ⟩$. Higher derivatives give all pairings, the content of Wick theorem. For an interaction $S_{\mathrm{int}}$, one may use path-integral perturbation by source derivatives:Expanding this expression generates Feynman diagrams and their Wick contractions. The connected generating functional $W[J]=-i\log Z[J]$ retains connected contributions; in particular $\delta W/\delta J=⟨\varphi {⟩}_J$. These functionals turn the computation of field-operator expectations into source differentiation of an ordinary regulated integral.

Use signature $(+---)$ and take the covariant spatial components $A_i$ as canonical coordinates. The canonical quantization of the electromagnetic field begins with the canonical momentaThus the primary momentum constraint of the electromagnetic potential is ${\Pi }^0=0$: $A_0$ has no independent velocity. The Hamiltonian obtained by the Legendre transform in mechanics, up to a boundary term, isPreserving the primary constraint requires the Gauss law constraint in gauge theory, ${\partial }_i{\Pi }^i=0$. It also follows by varying $A_0$. These are two first-class constraints; they generate the gauge freedom and remove two canonical pairs from the four potential components. The reduced phase space has four dimensions per spatial mode, hence two propagating photon degrees of freedom.

Impose Coulomb gauge, ${\partial }_i{A}_i=0$. With no charges, Gauss's law then gives ${\nabla }^2{A}_0=0$; vanishing boundary conditions set $A_0=0$. This is radiation gauge. The remaining components are transverse and obey the massless wave equation. Let ${ϵ}_i^{(r)}(\mathbf{k})$, $r=1,2$, be orthonormal transverse polarization vectors. Their photon polarization completeness relation isThe canonical transverse photon field is the Hermitian operatorwhereThe field and its conjugate momentum have the transverse equal-time commutatorThis is the quantized reduced bracket, or equivalently the Dirac bracket after imposing the constraints and gauge conditions. The normal-ordered Hamiltonian is ${\sum }_r\int d^{3}k{\omega }_{\mathbf{k}}{a}_r^{†}{a}_r/(2\pi {)}^3$. Its excitations are photons; circular combinations of the two transverse polarizations have helicity $+1$ and $-1$. The scalar and longitudinal potential components do not create additional physical photons.

The Feynman propagator is the vacuum expectation of a time-ordered product. The mode expansion directly gives the radiation-gauge photon propagator, with $z=x-y$:The first expression comes from the creation-annihilation commutator; the second is its contour-integral representation. The positive-energy pole lies below the real axis and the negative-energy pole above it. In this reduced free-field description, the temporal operator is zero. A photon propagator must specify its gauge; the spatial transverse propagator is not the same tensor as the covariant four-potential propagator.

For the commonly used covariant form, add the gauge fixing term $-({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$. The resulting Fourier-space kinetic operator isThe inversion of the gauge-fixed Maxwell kinetic operator gives $K^{\mu \nu }{D}_{\nu \rho }^F=i{\delta }^{\mu }{}_{\rho }$. In Feynman gauge, $\xi =1$, the photon propagator isEquivalently, $\square D_{\mu \nu }^F=i{\eta }_{\mu \nu }{\delta }^4(x-y)$ with the vacuum pole prescription. For general $\xi$, the momentum-space numerator is ${\eta }_{\mu \nu }-(1-\xi )k_{\mu }{k}_{\nu }/(k^2+i0)$.

A covariant canonical realization uses four polarization oscillators with $[a_r,a_s^{†}]=-(2\pi {)}^3{\eta }_{rs}{\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$. The resulting indefinite inner product is auxiliary. In Gupta-Bleuler quantization, impose $({\partial }_{\mu }{A}^{\mu }{)}^{(+)}|\mathrm{phys}⟩=0$ and take the Gupta-Bleuler null-state quotient. The scalar-longitudinal combination is thereby removed from the physical state space, leaving the same two transverse photon states. Thus the four-component Feynman-gauge numerator does not imply four physical polarization states.

The canonical momentum is $p=L_{\dot{q}}$, and a regular Legendre transform in mechanics defines $H(q,p,t)=p\dot{q}-L(q,\dot{q},t)$ after solving for $\dot{q}$. Its differential isbecause the $d\dot{q}$ terms cancel. With the Euler-Lagrange equation $\dot{p}=L_q$, this gives Hamilton's equationsIn particular $H_q=-L_q$ at corresponding velocities, so a cyclic coordinate for the Lagrangian is also cyclic for the Hamiltonian.

The canonical momenta are $p_x=m\dot{x}$, $p_y=m\dot{y}+qBx$, $p_z=m\dot{z}$. The Legendre transform in mechanics givesHamilton's equations areThus $\ddot{x}=\Omega \dot{y}$, $\ddot{y}=-\Omega \dot{x}-qE/m$, and $\ddot{z}=0$, where $\Omega =qB/m$. The signs also follow from $\mathbf{E}=-E{\mathbf{e}}_y$ and $\mathbf{B}=B{\mathbf{e}}_z$.

For $Bq\ne 0$, put $v_d=-E/B$, $a=v_{x0}-v_d$, $b=v_{y0}$. The motion isDifferentiation verifies both equations and all initial conditions. The centre of the circular motion has the electric drift velocity $-E{\mathbf{e}}_x/B$. For $B=0$, take $x=x_0+v_{x0}t$, $y=y_0+v_{y0}t-qE{t}^2/(2m)$ and the same $z$. For $q=0$, the particle moves freely.

Perform the Legendre transform in mechanics, using ${\dot{h}}_{ij}=2N{K}_{ij}+D_i{N}_j+D_j{N}_i$. The trace contraction ${\pi }^{ij}{K}_{ij}=\sqrt{h}(K_{ij}{K}^{ij}-K^2)$ givesTo avoid differentiating a tensor density as though it were an ordinary tensor, set $p^{ij}={\pi }^{ij}/\sqrt{h}$. Integration by parts with the Levi-Civita connection of $h$ givesafter dropping the surface term. AlsoTherefore the requested Hamiltonian isVarying the lapse function and shift vector imposes the Hamiltonian constraint $\mathcal{H}=0$ and momentum constraint ${\mathcal{H}}_i=0$. The total canonical generator also contains multipliers for the vanishing lapse and shift canonical momenta. With an asymptotic boundary, surface terms must be restored to define the Arnowitt-Deser-Misner energy.

For a regular Lagrangian define the canonical momentum components $p_j=\partial L/\partial {\dot{q}}_j$. Require the velocity Hessian to be invertible locally so that the velocities can be expressed in terms of $(q,p,t)$. The Legendre transform in mechanics definesThe explicit time argument cannot generally be omitted when the given Lagrangian depends on time. Its differential is $dH={\sum }_j{\dot{q}}_{j}d{p}_j-{\sum }_j(\partial L/\partial q_j)d{q}_j-(\partial L/\partial t)dt$, since the $d{\dot{q}}_j$ terms cancel. The Euler-Lagrange equations give $\partial L/\partial q_j={\dot{p}}_j$, and comparison yields Hamilton's equations:For the time-independent stellar gravitational potential below, $H$ is conserved. Its sign convention differs from the binding energy in question 2: here $E=H=v^2/2-\psi$ is the ordinary specific orbital energy.

The prescribed nondegenerate prolate spheroidal coordinates require $0<\alpha <\beta$; equality collapses the coordinate family. Their physical velocity components are $v_j=h_j{\dot{q}}_j$, and $h_{\varphi }=R$. Per unit stellar mass the Lagrangian is $L=\frac{1}{2}{\sum }_j{h}_j^2{\dot{q}}_j^2+\psi$, so $p_j=h_j^2{\dot{q}}_j=h_j{v}_j$. Its Hamiltonian isThis uses acceleration $+\nabla \psi$, as specified, rather than importing the opposite gravitational potential convention.

For the Stäckel potential, use the Hamilton-Jacobi equation with principal function $S=W_{\lambda }(\lambda )+W_{\mu }(\mu )+l\varphi -Et$. Axisymmetry makes $p_{\varphi }=l=R{v}_{\varphi }$ constant. Write $d=\lambda -\mu$ and $A(\tau )=(\tau +\alpha )(\tau +\beta )$. Multiplying $H=E$ by $d$ and usingseparates it as $F(\lambda )-E\lambda =F(\mu )-E\mu$, where $F(\tau )=2A(\tau )(W_{\tau }^{\prime }{)}^2-(\alpha -\beta )l^2/[2(\tau +\alpha )]-G(\tau )$. Each side depends on its own coordinate only and therefore equals a constant, chosen as $-I$. Here $G(\tau )$ is the generating function of the gravitational potential, not the gravitational constant. HenceThis proves the conserved axisymmetric Stäckel third integral. Eliminating $E$ between the two expressions and using $p_j=h_j{v}_j$ givesAt regular generic phase-space points the three separation constants $E,l,I$ are independent; special circular or degenerate orbits can reduce their local independence. Thus the model is an integrable Hamiltonian system.

For the intended regular, phase-mixed steady model, Jeans theorem permits a galactic distribution function $f=F(E,l,I)$. At any fixed point, $E$ and $I$ depend on $v_{\lambda }$ and $v_{\mu }$ only through their squares, and $l$ depends only on $v_{\varphi }$. Thus $f$ is even separately in $v_{\lambda }$ and $v_{\mu }$, whether or not it is even in $v_{\varphi }$. Odd integration proves velocity-ellipsoid alignment in a Stäckel potential:The argument allows azimuthal streaming: it does not require zero mean $v_{\varphi }$. The means in the two meridional directions vanish, so the off-diagonal velocity covariance also vanishes. Second moments must exist for this conclusion.

There is a hypothesis behind the printed universal wording. On a generic nonresonant regular invariant torus, a continuous steady distribution is constant along a dense orbit and hence depends only on the torus integrals; this justifies the phase-mixed form. Bare stationarity alone is not sufficient in a degenerate resonant gravitational potential with additional integrals. For example $H=\frac{1}{2}(v^2+{\omega }^2{r}^2)$ is a Stäckel Hamiltonian: choose $G(\tau )=-\frac{1}{2}{\omega }^2[{\tau }^2+(\alpha +\beta )\tau ]$, up to a constant vacuum shift. The additional conserved quantity $K_{xy}=v_x{v}_y+{\omega }^{2}xy$ gives a positive normalizable steady tracer $f\propto \exp (-aH+ϵK_{xy})$ for $a>|ϵ|>0$. Its local $⟨v_x{v}_y⟩=ϵ/(a^2-{ϵ}^2)\ne 0$. At azimuth zero and a regular point where the $\lambda$ direction has a radial component, this gives nonzero $⟨v_{\lambda }{v}_{\varphi }⟩$. This tracer need not itself source the harmonic gravitational potential. It exhibits why the stronger, ordinary phase-mixed three-integral assumption is necessary instead of claiming that every imaginable steady distribution has the asserted alignment.

The geometry follows by multiplying the coordinate-defining equation and evaluating at $\tau =-\alpha ,-\beta$:Constant $\lambda$ surfaces are prolate spheroids, with equatorial semiaxis $\sqrt{\lambda +\alpha }$ and polar semiaxis $\sqrt{\lambda +\beta }$. Constant interior $\mu$ surfaces are two-sheeted hyperboloids. Their common foci lie at $z=\pm \sqrt{\beta -\alpha }$. The orbit volume in an axisymmetric Stäckel potential is determined byIts allowed coordinate intervals have turning boundaries or coordinate endpoints. A typical bound orbit with $l\ne 0$ oscillates between inner and outer spheroids, oscillates vertically between hyperboloidal limits, and circulates about the symmetry axis with $\dot{\varphi }=l/R^2$. The equatorial fold $\mu =-\beta$ joins the positive and negative $z$ branches; it is not necessarily a physical turning point in $z$.

For three incommensurate frequencies, the regular trajectory densely samples its projected allowed volume, with nonuniform occupation determined by its speeds. It does not follow one coordinate surface, and it need not close. Resonant orbits may be closed or lie in a smaller projected set; $l=0$ confines azimuthal motion to a meridional plane (with the usual axis coordinate qualification), and circular or equatorial limits are also lower-dimensional. Smooth Stäckel separation removes the generic chaotic behavior of nonintegrable potentials; it does not assert that every orbit fills a three-dimensional volume.

Apply the Legendre transform in mechanics pointwise to the field Lagrangian density. Using ${\pi }_1={\dot{\varphi }}_1$, the Hamiltonian density isThe free Hamiltonian operator is its spatial integral:A constant vacuum-energy subtraction, if made, does not affect the commutators or the ensuing field equation.

In the paper's convention, ${\dot{h}}_{ij}=2N{K}_{ij}+D_i{N}_j+D_j{N}_i$. The Legendre transform in mechanics therefore givesIndeed ${\pi }^{ij}{K}_{ij}=\sqrt{h}(K_{ij}{K}^{ij}-K^2)$, and inversion of the canonical momentum of the spatial metric givesTo integrate the shift term without ambiguity about tensor densities, put $p^{ij}={\pi }^{ij}/\sqrt{h}$. Metric compatibility of the spatial covariant derivative gives, up to a boundary term,Thus the Hamiltonian formulation of general relativity hasVariation of the lapse function imposes the Hamiltonian constraint $\mathcal{H}=0$, while variation of the shift vector imposes the momentum constraint ${\mathcal{H}}_i=0$.

The action contains neither $\dot{N}$ nor ${\dot{N}}^i$, so their conjugate momentum vanishes. They impose constraints rather than independent evolution equations: after the Legendre transform in mechanics, the lapse function and shift vector are Lagrange multipliers in the Hamiltonian formulation of general relativity.

Use precisely the paper's sign convention for extrinsic curvature, $K_{ij}=({\dot{h}}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$. At fixed $h_{ij},N,N^i$,Consequently the canonical momentum of the spatial metric isThis is a weight-one tensor density, with indices raised and lowered using $h_{ij}$. Its trace and inverse relation areThe sign of $K_{ij}$ used here is opposite to the convention in extrinsic curvature of a spatial hypersurface; mixing the two conventions would reverse the momentum formula.