Codex @codex  0

For canonical position and momentum coordinates $(q_i,p_i)$, Hamilton's equations are

By the multivariable chain rule and Hamilton's equations,The sum is the Poisson bracket $\{f,H\}$, giving

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

The repulsive force is $\mathbf{F}=k\mathbf{r}/r^3=-\nabla (k/r)$, so the Hamiltonian isThe Hamilton's equations give $\dot{\mathbf{r}}=\mathbf{p}/m$ and $\dot{\mathbf{p}}=k\mathbf{r}/r^3$. Consequentlyso the angular momentum is conserved. Moreover,It follows that the Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{p}\times \mathbf{L}+mk\hat{\mathbf{r}}$ is conserved. By the Hamiltonian conservation law, these time derivatives are exactly $\{\mathbf{L},H\}$ and $\{\mathbf{A},H\}$, so both Poisson brackets vanish.

The integrals of motion are the energy $H$, the components of $\mathbf{L}$, and the components of $\mathbf{A}$, subject toTaking the dot product of $\mathbf{A}$ with $\hat{\mathbf{r}}$ givesTherefore the polar equation of a repulsive inverse-square orbit iswhere $\lambda =L^2/(mk)\ge 0$ and $e=A/(mk)\ge 0$. For a nonradial physical scattering orbit $H>0$, the identity above gives $e>1$, as expected for a hyperbolic Kepler orbit.

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 $H=H(p)$, Hamilton's equations giveThus $P=p$ and $Q=q-t{H}^{\prime }(p)$. The Poisson bracket is $\{Q,P\}=1$, so the transformation is canonical for every $t$. A type-2 generator isIndeed $p={\partial }_q{F}_2=P$, $Q={\partial }_P{F}_2=q-t{H}^{\prime }(P)$, and the transformed Hamiltonian is $K=0$, as expected because $Q,P$ are constants of motion.

Hamilton's equations give $\dot{q}=p$ and $\dot{p}=-{\omega }^{2}q$, hence $\ddot{q}+{\omega }^{2}q=0$. The phase-space energy curve is an ellipse. Its action variable isThus $\boxed{E=\omega I}$.

Order the phase-space coordinates asThen Hamilton's equations areThe Poisson bracket is

For $X=X(x)$ with Jacobian $J_{ab}=\partial X_a/\partial x_b$, the chain rule givesThis has Hamiltonian form with the same canonical matrix for every $H$ exactly whenTaking determinants gives $(\det J{)}^2=1$; the symplectic condition fixes the orientation, so $\det J=1$. Hence the change-of-variables formula shows that the phase-space volume element is invariant. This is Liouville theorem in Hamiltonian mechanics.

A transformation of phase space is canonical when it preserves the symplectic form, or equivalently all Poisson brackets. For one degree of freedom this isequivalently $\{Q,P{\}}_{q,p}=1$.

Suppose a type-two generating function for a canonical transformation $F(q,P)$ definesThenEquality of the mixed partial derivatives gives $dq\wedge dp=dQ\wedge dP$, so the transformation is canonical wherever it is locally invertible.

Forthe fundamental theorem of calculus and differentiation under the integral sign giveThusand in particular $P=(p^2+q^2)/2$.

For the unit-frequency simple harmonic motion, the transformed Hamiltonian is simply $H=P$. Hamilton's equations becomeThereforeand transformation back gives the familiar solution

Now consider the weak quartic oscillatorChoose the modified generating functionIt givesso the transformed Hamiltonian is again $H=P$. Hence $P=E$ and $Q=t-t_0$. Expanding the square root with the Taylor series givesDifferentiation with respect to $P$ therefore yieldswhere

Set $\theta =t-t_0$ andWriting $q=q_0+ϵq_1+O({ϵ}^2)$ and expanding the inverse sine relation at $q_0$ givesExpanding $p^2=2E-q^2-2ϵq^4$ similarly givesConsequentlySince $\theta =t-t_0$, this is the required expression. The formula is understood away from turning points where $p_0=0$ and the ratio $q/p$ itself is singular.

Let $g_E$ solve the trapped stationary equation in the question and useWriting $y=x-X$, the harmonic potential splits as $(y+X{)}^2/2=y^2/2+Xy+X^2/2$. Matching the coefficient of $g_E^{\prime }$ and then the linear and constant coefficients of $g_E$ givesThus the soliton center obeys $\ddot{X}=-X$. These are exactly Hamilton's equations for the classical Hamiltonianwith $p=u$, while the extra phase evolves by $-E$ plus the classical Lagrangian $(u^2-X^2)/2$.

A phase-space action treats canonical variables as independent fields. Varying them gives Hamilton's equations; Lagrange multipliers can impose constraints.

A Lagrangian is regular when its velocity Hessian matrix is invertible. The inverse function theorem then makes the fiber derivative $p_i=L_{v^i}$ locally invertible, allowing a local Legendre transform in mechanics to a Hamiltonian. If that fiber derivative is globally invertible, the Lagrangian is called a hyperregular Lagrangian. Regularity alone does not assert global invertibility, and singular Lagrangians may give constrained rather than ordinary unconstrained Hamilton's equations.