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A first-class constraint generates an infinitesimal canonical transformation $\delta F=\{F,{ϵ}^i{\phi }_i\}$. For the phase-space action $I=\int (p\dot{q}-{\lambda }^i{\phi }_i)dt$, the multiplier transformation $\delta {\lambda }^k={\dot{ϵ}}^k-{\lambda }^i{ϵ}^j{C}_{ij}{}^k$ cancels the bulk variation when $\{{\phi }_i,{\phi }_j\}=C_{ij}{}^k{\phi }_k$. The remaining temporal boundary term vanishes for parameters that vanish at the endpoints.

A complete integral depends on as many independent parameters as configuration coordinates and supplies a locally invertible canonical transformation to constants of motion. For a type-one generating function $G(q,Q,t)$, require a nondegenerate mixed-derivative matrix in the relevant local chart. Solving $G_t+H(q,G_q,t)=0$ makes the transformed Hamiltonian zero, so both new canonical coordinates and new canonical momenta are constant. An arbitrary parameter-deficient solution does not ensure such a transformation.

For bound Kepler orbits, the Delaunay variables obey $0<G<L$, $-G<H<G$, and three periodic angles. A canonical transformation preserves phase space volume. At fixed $L$, integrating $H$ supplies the measure $2GdG$. Since $G=L\sqrt{1-e^2}$, the fraction with orbital eccentricity less than $e$ is $L^{-2}{\int }_{L\sqrt{1-e^2}}^{L}2GdG=e^2$. Thus the probability density function is $2e$, independently of any normalizable energy-only phase-space distribution function. A Boltzmann distribution is unnecessary. Even the time-dependent density $F(E(L))[1+\eta \cos (l-n(L)t)]$, $|\eta |<1$, has this same marginal because integrating the mean anomaly removes its oscillatory part; an eccentricity marginal alone does not imply equilibrium.

For a bound noncollision Kepler orbit with gravitational parameter $\mu$, the canonical actions per unit mass are $\mathcal{L}=\sqrt{\mu a}$, $\mathcal{G}=\mathcal{L}\sqrt{1-e^2}$ and $\mathcal{H}=\mathcal{G}\cos i$. Their angles are mean anomaly $\ell$, argument of periapsis $\omega$ and longitude of ascending node $\Omega$. The Hamiltonian is $-{\mu }^2/(2{\mathcal{L}}^2)$, so only $\ell$ advances in the unperturbed problem. In the prograde spherical-action chart, $J_r=\mathcal{L}-\mathcal{G}$, $J_{\theta }=\mathcal{G}-\mathcal{H}$, $J_{\varphi }=\mathcal{H}$; the type-two canonical transformation is generated by $F_2=(\mathcal{L}-\mathcal{G})w_r+(\mathcal{G}-\mathcal{H})w_{\theta }+\mathcal{H}{w}_{\varphi }$. Retrograde spherical actions have $J_{\theta }=\mathcal{G}-|\mathcal{H}|$, so the absolute value cannot be dropped globally. Circular and coplanar limits are singular angle charts, not singular orbits.

On $K^4$, use the standard basis $\mathcal{U}=(u_1,u_2,u_3,u_4)$ and the alternating bilinear form with matrix $J$ above for $n=2$. A second symplectic basis isFor instance, $\omega (f_1,f_2)=1-1=0$, while $\omega (e_1,f_1)=\omega (e_2,f_2)=1$ and the cross-pairings vanish. The change of basis matrix, whose columns are these new basis vectors in $\mathcal{U}$, isThe symmetric matrix $C$ makes $S^{T}JS=J$, so $S$ is a symplectic matrix. Explicitly, multiplying the blocks gives a lower-right block $C^T-C=0$; the other blocks agree with $J$. For a fixed vector, change of basis means $[v{]}_{\mathcal{U}}=S[v{]}_{\mathcal{B}}$, henceTheir symplectic form gives $3\cdot 0+2\cdot 1-1\cdot 1-1\cdot 0=1$ in $\mathcal{U}$, and $1\cdot 0+1\cdot 1-1\cdot 0-1\cdot 0=1$ in $\mathcal{B}$. The separate coordinate-plane contributions change, but their sum does not.

The same matrix also represents an active symplectic linear map: keeping one basis fixed, it sends $(q,p)$ to $(q+Cp,p)$. The following diagram shows this active interpretation over the real numbers. Its preserved total pairing is precisely the test used for linear canonical transformations.

Assume positive, pairwise distinct masses and generic nonzero $\omega ,\lambda$. Let $p_{x_i}=m_i{\dot{x}}_i$, and similarly for $y,z$. The three uncoupled $x$ oscillators each conserveUniform translation of all $y_i$ conserves $P_y={\sum }_i{p}_{y_i}$. The $z_1,z_2$ interaction permits their common translation, while $z_3$ is entirely free, giving two independent momenta $P_{z,12}=p_{z_1}+p_{z_2}$ and $p_{z_3}$. The interacting $z_1,z_2$ subsystem is autonomous and has its own conserved energy. A useful basic collection is thereforeThe total energy is generally not conserved. With $\kappa =\lambda (m_1-m_2)$, its derivative is $\dot{H}=-\kappa \sin (\kappa t)U_{12}(y_1-y_2)$, which is generically nonzero.

Unequal masses do not remove every rotation charge in the oscillator sector. Put $X_i=\sqrt{m_i}{x}_i$, $P_i=p_{x_i}/\sqrt{m_i}$. This is a canonical transformation and makes that sector an isotropic harmonic oscillator. Its rotations conserveThese mass-weighted oscillator symmetries exist already at distinct masses. The energies and three rotation charges are not all functionally independent; the three-oscillator sector has at most five independent time-independent integrals on regular trajectories.

There are further simple explicitly time-dependent constants, such as $x_i\cos \omega t-({\dot{x}}_i/\omega )\sin \omega t$ and $x_i\sin \omega t+({\dot{x}}_i/\omega )\cos \omega t$, and center-of-mass boost constants. For example $MY-t{P}_y$ is constant for $M={\sum }_i{m}_i$ and $Y={\sum }_i{m}_i{y}_i/M$. These illustrate why an unrestricted raw number of conserved functions is not meaningful.

Work on $I>0$, with a local choice of the periodic angle $\theta$. The inverse formulas areIn particular, ${\theta }_q=p/(2I)$, ${\theta }_p=-q/(2I)$, $I_q=\alpha q$ and $I_p=p/\alpha$. Their Poisson bracket isThe other fundamental Poisson brackets vanish by antisymmetry. Thus this is a canonical transformation for every $\alpha >0$, on local charts away from the origin. At $I=0$ the angle is undefined.

In the new coordinates the harmonic oscillator Hamiltonian isFor $I$ to be a conserved quantity throughout phase space, Hamilton's equations require $\dot{I}=-{\partial }_{\theta }{H}_0=0$ identically. The two coefficients must coincide. For the usual $m>0$ and $\omega >0$, positivity of $\alpha$ selectsThese are harmonic oscillator action-angle variables; the action variable is $I=H_0/\omega$.

On the interior of the principal inverse-cosine branch, $0<q<\pi$, the given canonical transformation hasIntegrating $Q={\partial }_P{F}_2$ suggests $F_2=\cos q\sin P+C(q)$. Its other derivative is ${\partial }_q{F}_2=-\sin q\sin P+C^{\prime }(q)$; comparison with $p$ requires $C^{\prime }(q)=0$. Hence a suitable type-two generating function for a canonical transformation isBoth derivative equations reproduce the specified transformation exactly. Its mixed derivative is $F_{2,qP}=-\sin q\cos P$, so it defines a local canonical transformation where $0<q<\pi$ and $\cos P\ne 0$, equivalently where $|Q/\cos P|<1$. At $q=0,\pi$ or $\cos P=0$, the stated formulas fail to give a regular coordinate chart. These restrictions explain the square-root branch and the local meaning of canonicity. This is a trigonometric type-two canonical transformation.

A type-two generating function for a canonical transformation uses the mixed variables $(q,P)$ and definesFor a local invertible canonical transformation, require $F_{2,qP}\ne 0$ so that $p=F_{2,q}(q,P)$ can be solved for $P$. The differential identity $d{F}_2=pdq+QdP$ gives $d(F_2-PQ)=pdq-PdQ$, expressing preservation of the canonical symplectic form.

Use $\mu =\gamma m$ for the gravitational parameter, to distinguish it from the Delaunay action called $G$. The displayed actions are per unit reduced mass. Accordingly use specific canonical momenta; with physical relative momenta, all three actions acquire a common reduced-mass factor, which leaves the eccentricity fraction below unchanged.

The canonical transformation. In spherical coordinates, separation of the Kepler Hamilton-Jacobi equation givesA local type-two generating function, continued across turning points with the appropriate momentum branches, isThe old momenta are $\partial W/\partial (r,\theta ,\varphi )$ and the new angles are $l={\partial }_{L}W$, $g={\partial }_{G}W$, $h={\partial }_{H}W$. This defines a canonical transformation, since its generating differential is $dW=p_{r}dr+p_{\theta }d\theta +p_{\varphi }d\varphi +ldL+gdG+hdH$. With standard choices of angle origins, the resulting Delaunay variables arewith $l$ the mean anomaly, $g$ the argument of periapsis, and $h$ the longitude of ascending node. In particular, the signed $H$ includes retrograde orbits; the squared equation in the PDF must not be read as $H\ge 0$.

For bound noncollision Kepler orbits, the independent ranges areThe limiting circular, radial and coplanar cases sit on boundaries where some angle charts degenerate. They have zero measure in the generic phase-volume calculation. The ranges are constrained as above, rather than a Cartesian box with three unconstrained actions.

The eccentricity count. A canonical transformation preserves phase space volume, soIntegrating the three angles contributes $(2\pi {)}^3$, and integrating $H$ gives $2G$. At fixed $L$, the condition that the orbital eccentricity be less than $e_0$ is $G>L\sqrt{1-e_0^2}$. ThusAny energy weight cancels between these integrals. If the total population is finite, or an explicit finite action band is used,This is the Delaunay phase-volume proof of the thermal eccentricity distribution. The source's $e^2$ law is a cumulative count, not a differential eccentricity density. Also, pointwise finite $F(E)$ does not alone imply finite $N_{\mathrm{tot}}$: $F=1$ gives ${\int }_0^{\infty }{L}^{2}dL=\infty$. The finite normalization of an energy-only Kepler distribution is the additional condition needed for a global probability statement; the fraction at every fixed energy is unaffected.

Why the observation does not imply relaxation or equilibrium. No Boltzmann distribution was used: every normalizable energy-only phase-space distribution function has the same eccentricity marginal, irrespective of collisional thermalization. An even stronger counterexample is a genuinely time-dependent collisionless distributionKepler motion has $\dot{l}=\nu (L)$ and constant actions and other angles, so ${\partial }_{t}f+\nu {\partial }_{l}f=0$: this is an exact nonstationary solution of the Collisionless Boltzmann equation. Integrating $l$ removes the cosine and leaves precisely the same $e^2$ cumulative count at every time. An eccentricity marginal alone cannot establish thermal relaxation, a relaxed age, or even a stationary distribution.

Differentiating givesThereforeFor real coordinates take $q>0$ and a local branch of the angle. The transformation is locally canonical transformation because its determinant is one, but periodicity in $p$ prevents a globally one-to-one map if all real $p$ are retained.

A one-dimensional canonical transformation is a locally invertible change of phase-space variables preserving the symplectic two-form: $dQ\wedge dP=dq\wedge dp$. Equivalently it preserves Hamilton equations for the transformed Hamiltonian. Expanding the two-form yieldssoConversely this Jacobian condition makes the transformation locally invertible and symplectic in two dimensions.

Write ${\mu }_g=\gamma m$ for the gravitational parameter of the relative Kepler orbit, and use specific orbital energy and momenta per unit reduced mass. The Hamiltonian isIf physical momenta rather than these specific momenta are used, multiply the actions by the reduced mass. This only introduces a constant factor into the phase space measure and leaves the eccentricity fractions unchanged. The formulas in the question adopt the specific-action convention.

Here is an explicit local canonical transformation. A separated complete integral of the Hamilton-Jacobi equation isChoose branches along an orbit and take $p_j=\partial S/\partial q_j$, $l=\partial S/\partial L$, $g=\partial S/\partial G$, $h=\partial S/\partial H$. Thenwhose exterior derivative proves preservation of the canonical symplectic form. Thus these variables are canonical, and the absolute Jacobian determinant is one. The angular separation gives $G=|\mathbf{r}\times \mathbf{v}|$ and $H=G\cos i$, including the sign of the angular momentum along the reference axis.

For clarity, the radial action can be evaluated rather than assumed. Let $r=a(1-e\cos u)$, with $u$ the eccentric anomaly. Then $p_r=\sqrt{{\mu }_g/a}e\sin u/(1-e\cos u)$, andIn the integral, use $e^2{\sin }^{2}u/(1-e\cos u)=1+e\cos u-(1-e^2)/(1-e\cos u)$ and ${\int }_0^{2\pi }(1-e\cos u{)}^{-1}du=2\pi /\sqrt{1-e^2}$. The polar action is $J_{\theta }=G-|H|$, while the signed azimuthal action is $J_{\varphi }=H$. The separated angular coordinates have the usual Delaunay variables interpretation: $g$ is the argument of periapsis, and $h$ is the longitude of ascending node. In particular,with a suitable origin at periapsis, so $l$ is the mean anomaly. We have therefore recoverednot merely assigned names to the new coordinates.

For nondegenerate bound Kepler orbits, the ranges areThe boundaries describe circular, coplanar or radial degeneracies, where some angles cease to be defined; they have zero measure in the present calculation. The actions are canonical coordinates with nested bounds, rather than three independently unrestricted Cartesian coordinates. The square of $H$ alone would miss the distinction between prograde and retrograde inclinations.

By the canonical transformation, the population measure becomesAt fixed $L$, the condition $e^{\prime }<e$ is $L\sqrt{1-e^2}<G<L$. Integrating all three angles and then $H$ givesConsequently the Delaunay phase-volume proof of the thermal eccentricity distribution yieldsThe PDF's $e^2$ dependence must be interpreted as a cumulative distribution function, not an eccentricity probability density function or the value of the original six-dimensional phase-space distribution function. A finite value of $F(E)$ at every energy also does not suffice to make the total population finite: $F=1$ on all bound energies gives the divergent integral ${\int }_0^{\infty }{L}^{2}dL$. The fractions are well defined at each fixed energy, over a finite allowed energy range, or for an energy distribution with a finite total integral.

A Boltzmann distribution is therefore unnecessary. Any normalizable energy-only distribution has the same eccentricity marginal, although such a distribution is a collisionless steady state by Jeans theorem. The marginal alone does not even prove stationarity: for $0<|\eta |<1$,is a positive time-dependent solution of the Collisionless Boltzmann equation and has exactly the same eccentricity marginal after integration over $l$. An observed $e^2$ cumulative law establishes neither thermal relaxation nor the age of the system or Universe.

For a bound Kepler orbit, the semimajor axis is half the long diameter of the ellipse; if $r_p,r_{mathrma}$ are its apsidal distances, $a=(r_p+r_{mathrma})/2$. The orbital eccentricity is $e=(r_{mathrma}-r_p)/(r_{mathrma}+r_p)$, with $0\le e<1$ and semiminor axis $b=a\sqrt{1-e^2}$. The gravitating centre is a focus, displaced by $ae$ from the ellipse's geometric centre.

Choose a reference plane, a positive normal and a fixed reference direction in that plane. The ascending node is where the orbit crosses toward positive normal displacement. The longitude of ascending node $\Omega$ is the reference-plane angle from the fixed direction to that node. The argument of periapsis $\omega$ is the angle from the ascending node to periapsis, measured in the orbital plane in the direction of motion. The orbital inclination $i$ is the angle between the reference normal and the orbital angular-momentum normal, with $0\le i\le \pi$. The diagram separates the ellipse's shape from these orientation angles.

Let $\mu =GM$ and use actions per unit test-particle mass. In the prograde spherical-action chart underlying the stated Hamiltonian, $J_{\varphi }=L_z$ and $J_{\theta }=L-L_z$, so $J_{\theta }+J_{\varphi }=L$. The radial action then satisfies $J_r+L=\sqrt{\mu a}$, consistently with $H_K=-\mu /(2a)$. DefineIf $(w_r,w_{\theta },w_{\varphi })$ are the old angles, a type-two generating function isDifferentiation with respect to the old angles recovers $J_r,J_{\theta },J_{\varphi }$, and differentiation with respect to the new actions gives the canonical transformationIts triangular integer action matrix has determinant one, so it preserves the angle periodicities. In the new coordinates,These are the Delaunay variables:With the usual choices of angle origins, $w_1=\ell$ is mean anomaly, $w_2=\omega$ is argument of periapsis, and $w_3=\Omega$ is longitude of ascending node. The generator $\mathcal{G}=L$ rotates the orbit within its plane, whereas $\mathcal{H}=L_z$ rotates that plane about the reference normal; this explains the angular identification. The fast angle obeys Kepler's equation $\ell =\mathcal{E}-e\sin \mathcal{E}$, where $\mathcal{E}$ is eccentric anomaly.

For retrograde orbits, signed $J_{\varphi }=L_z$ gives $J_{\theta }=L-|L_z|$ and the spherical Hamiltonian uses $J_r+J_{\theta }+|J_{\varphi }|$. The stated plus-sign form is therefore a prograde chart, not a global signed-action formula. The Delaunay variables still use $\mathcal{H}=L_z$; on the retrograde branch the generating function is $(\mathcal{L}-\mathcal{G})w_r+(\mathcal{G}+\mathcal{H})w_{\theta }+\mathcal{H}{w}_{\varphi }$. At $e=0$, periapsis is undefined, and at $i=0$ or $\pi$ the node is undefined: these are angle-chart degeneracies, not physical singularities.

For the perturbed planet orbit, the radial and vertical frequencies mean small oscillations about a circular equatorial orbit of cylindrical radius $R$. Write $r=\sqrt{R^2+z^2}$ and retain the stated quadrupole potential. On the equator,At fixed axial specific angular momentum $L_z$, the radial effective potential is $\varphi (R,0)+L_z^2/(2R^2)$. Its second derivative at circular balance $L_z=R^2{\Omega }_c$ gives the radial epicyclic frequencyReflection symmetry decouples vertical motion to first order. Expanding the potential in $z$ givesso the vertical epicyclic frequency isThe circular orbit is radially stable only for $R^2>3{\alpha }^2$. For $\alpha \ne 0$ in that domain, $\nu >{\Omega }_c>\kappa$. Vertical node crossings repeat at frequency $\nu$ while azimuth advances at ${\Omega }_c$, so the line of nodes has nodal precession rate ${\Omega }_c-\nu <0$. Radial periapsis passages repeat at $\kappa$, giving apsidal precession rate ${\Omega }_c-\kappa >0$. To leading quadrupole order, with $n=\sqrt{\mu /R^3}$,This is oblate-quadrupole epicyclic precession. Here $\varpi$ denotes the nearly equatorial longitude of periapsis, not $\omega$ alone.

The precession signs above have a small-eccentricity, small-inclination domain. They are not a universal claim for every inclined Kepler ellipse. To see the distinction with the same Delaunay variables, average the perturbation over an unperturbed orbit. Since $z/r=\sin i\sin (\omega +f)$ and $dt=r^{2}df/\mathcal{G}$, the term linear in $e\cos f$ integrates to zero and the ${\sin }^2(\omega +f)$ average in the $r^{-3}$ integral is $1/2$. ThusDifferentiating at fixed other actions yields the first-order secular ratesTheir nearly equatorial sum reproduces the prograde $\dot{\varpi }$. For a polar orbit, however, the node is fixed and $\dot{\omega }<0$. This supplies a concrete counterexample to extending the requested forward periapsis precession beyond the intended epicyclic regime.

A canonical transformation preserves the symplectic form and therefore the form of Hamilton's equations for arbitrary Hamiltonians. In one degree of freedom,Thus the transformation is canonical exactly when $[Q,P]=1$. Equivalently the chain rule gives $[F,G{]}_{q,p}=[Q,P{]}_{q,p}[F,G{]}_{Q,P}$ for arbitrary transformed functions, so the bracket and Hamiltonian dynamics are preserved precisely with this factor one. The condition also makes the Jacobian nonzero and hence supplies local invertibility; a global canonical change of variables additionally requires the stipulated global bijectivity on its domain.

Let $A=Dy(x)$ be the Jacobian matrix and $K(y)=H(x(y))$. The chain rule gives ${\nabla }_{x}H=A^T{\nabla }_{y}K$. Therefore the Hamilton equations transform to $\dot{y}=AJ{A}^T{\nabla }_{y}K$, with $J=\left(\begin{array}{cc}0& -I\\ I& 0\end{array}\right)$. They have the required canonical form for every Hamiltonian exactly whenNecessity follows by choosing Hamiltonians whose gradients take arbitrary values at a point; sufficiency follows from the displayed transformed equation. This is the symplectic matrix condition for a canonical transformation, and it also forces $A$ to be invertible.

For the given transformation, $A=\left(\begin{array}{cc}1& 2\\ -1/4& 1/2\end{array}\right)$ has determinant one. For any $2\times 2$ matrix, $AJ{A}^T=(\det A)J$, so the proposed transformation is canonical.

A canonical transformation is a differentiable invertible change of phase-space coordinates preserving the symplectic form, equivalently the canonical Poisson brackets. At each fixed time, let $D$ be its Jacobian in the ordering $(q,p)\mapsto (Q,P)$ and put $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$. The condition iswhich is equivalent to $D^{T}JD=J$. Time dependence changes the Hamiltonian by a generating-function time derivative, but not this fixed-time symplectic condition.

The Hamilton-Jacobi equation is $S_t+\frac{1}{2}[(S_q{)}^2+{\omega }^2{q}^2]=0$. Separating $S=W(q,E)-Et$ gives $(W_q{)}^2=2E-{\omega }^2{q}^2$. On the positive-momentum branch, therefore,Choose the integration constant so that the given angle relation is obtained. For $E>0$, $\beta =S_E$ yields $q=\sqrt{2E}\sin [\omega (t+\beta )]/\omega$, while $p=S_q=\sqrt{2E-{\omega }^2{q}^2}$ on this branch. Continuing the angle across the turning points gives the full orbitThe square-root expression is locally the positive branch; the cosine formula correctly changes sign on the rest of the orbit.

The generating function has differential $dS=pdq+\beta d\alpha -Hdt$. At fixed time, taking an exterior derivative gives $dq\wedge dp=d\beta \wedge d\alpha$. Thus the canonical new coordinate is $Q=\beta$ and its conjugate momentum is $P=\alpha$. Directly, $\alpha =(p^2+{\omega }^2{q}^2)/2$ and $\beta =atan2(\omega q,p)/\omega -t$ locally implyThe angle is a local chart or a periodic variable, and is undefined at the zero-energy point. The source's phrase “variables $(\alpha ,\beta )$” lists the new momentum before its coordinate: if interpreted instead as the conventional coordinate-then-momentum ordering, that ordered pair would have $\{\alpha ,\beta \}=-1$ and be anti-canonical.

The Poisson bracket in the original coordinates isFor $z=(q,p)$, setThen ${\omega }_0(u,v)=u^{T}Jv$ and $\{F,G\}=\nabla F^{T}J\nabla G$. The pullback definition of a canonical transformation gives the pointwise matrix conditionSince $A$ is invertible, this is equivalent to $AJ{A}^T=J$. Indeed, invert $A^{T}JA=J$ and use $J^{-1}=-J$ to obtain $A^{-1}J{A}^{-T}=J$, then multiply by $A$ and $A^T$. Applying the same calculation to the inverse implication gives equivalence. These are the two equivalent symplectic matrix identities that connect the symplectic form with Poisson brackets.

The chain rule yieldsIf $f$ is a canonical transformation, the symplectic matrix identity $AJ{A}^T=J$ makes this expression $\{F,G{\}}_{Q,P}\circ f$. Thus preservation of the symplectic form implies preservation of every Poisson bracket, for all $C^1$ functions.

Conversely, assume this Poisson bracket identity for every $F,G$. Take $F(Z)=u\cdot Z$ and $G(Z)=v\cdot Z$, for arbitrary constant vectors $u,v\in {\mathbb{R}}^{2n}$. It gives $u^{T}AJ{A}^{T}v=u^{T}Jv$. Since this holds for every $u,v$, $AJ{A}^T=J$, hence $A^{T}JA=J$. Therefore preservation of all Poisson brackets is equivalent to canonicity. Only first derivatives of $f,F,G$ enter this argument.

Apply preservation of Poisson brackets to the coordinate functions. It givesConversely, the entries of $AJ{A}^T$ are exactly the Poisson brackets between all transformed coordinate functions. The displayed relations therefore say $AJ{A}^T=J$, which implies the canonical transformation condition and, by the chain rule, preservation of every Poisson bracket. Thus all three conditions are equivalent:In particular the original coordinate bracket $\{q_i,p_j\}$ is ${\delta }_{ij}$ with the chosen Poisson bracket convention.