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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 $H=H(p)$, 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.

The Poisson bracket is $\{f,g\}={\sum }_i(f_{q_i}{g}_{p_i}-f_{p_i}{g}_{q_i})$. Canonical variables satisfy $\{Q_i,Q_j\}=\{P_i,P_j\}=0$ and $\{Q_i,P_j\}={\delta }_{ij}$.

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 continuous symmetry is a family of transformations depending continuously on parameters for which the action is unchanged, possibly up to a spacetime boundary term. For an infinitesimal field variation $\delta \varphi =ϵ\Delta \varphi$, a Noether current $j^{\mu }$ obeyswhen the Euler-Lagrange field equation holds. Its Noether chargeis conserved provided the flux ${\int }_{S_{\infty }}{j}^i{n}_{i}dS$ vanishes.

Noether theorem gives the implication from a differentiable global variational symmetry to an on-shell conserved current. A conserved current gives a conserved charge only under suitable boundary and convergence conditions. Conversely, a conserved charge generates a continuous symmetry through Poisson brackets classically or a commutator quantum mechanically when a regular Hamiltonian formulation exists. Currents can be changed by identically conserved improvement terms, and inverse Noether statements require regularity and the exclusion of such trivial currents, so the three notions are related but not literally in one-to-one correspondence.

For spacetime translations, the canonical stress-energy tensor isForraising the second index gives the symmetric tensorTranslation invariance gives four conserved currents $T^{\mu \nu }$, one for each fixed $\nu$, and the four conserved charges are the four-momentum$P^0$ is the energy and $\mathbf{P}$ is the spatial momentum.

The Lorentz transformation variation of a scalar is generated solely by its argument. The corresponding Lorentz current isUsing ${\partial }_{\mu }{T}^{\mu \nu }=0$ and symmetry of $T^{\rho \sigma }$,In particular, conservation of the boost chargeimpliesThe quantity denoted $V^i$ in the question is therefore the conserved momentum component $P^i$, assuming the same vanishing boundary flux.

For smooth functions $f,g$ on the six-dimensional phase space with canonical coordinates $(q_i,p_i)$, the Poisson bracket is

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.

A quadratic homogeneous polynomial in the fibre variables of the cotangent bundle is a first integral of the geodesic Hamiltonian exactly when its symmetric lowered coefficients form a rank-two Killing tensor. For $v=g^{-1}p$, the Poisson bracket is $\{K,H\}={\nabla }_{(a}{K}_{bc)}{v}^a{v}^b{v}^c$, so the equivalence follows by comparing cubic coefficients at every point.

An invertible real matrix preserving the standard symplectic form. With $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$, it satisfies $A^{T}JA=J$, equivalently $AJ{A}^T=J$. The first identity describes preservation of the form; the second describes preservation of coordinate Poisson brackets.