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At thermal equilibrium, microscopic reversibility equates the probabilities of a path and its reversed path when both include their Boltzmann distribution initial weights. Denote their endpoint states by $z_1,z_2$, including velocity if needed. Since the Hamiltonian is even under time reversal in classical mechanics,This detailed balance condition and the energy balance for an autonomous Lagrangian yieldThis is the fluctuation-dissipation relation for a Langevin particle: the strength of Gaussian white noise is fixed by the damping and temperature, with $k_B$ the Boltzmann constant.

For an equilibrium coarse-grained variable, the unresolved microscopic states contribute entropy; their statistical weight is encoded in the Helmholtz free energy, rather than in a single microscopic energy. Relative to the same reference measure, ${\rho }_{\mathrm{eq}}(x)\propto e^{-\beta F(x)}$, so microscopic reversibility becomesThis extension assumes an equilibrium coarse-grained description with reversible path statistics; externally driven dynamics need not obey this relation.

The equilibrium equations give $y=0$ and $x-x^3=0$, so the fixed points areThe Jacobian isAt the origin its determinant is $-1$, so the origin is a saddle equilibrium. At either of the other points the determinant is two and the trace is $\epsilon (1-\alpha )$. For small positive $\epsilon$, they are unstable foci when $\alpha <1$ and stable foci when $\alpha >1$; at $\alpha =1$ the linearization is nonhyperbolic. When $\epsilon =0$, each is a center equilibrium.

For $\epsilon =0$, the system is Hamiltonian withbecause $\dot{x}=H_y$ and $\dot{y}=-H_x$. Its conservative planar phase portrait has two wells centered at $(\pm 1,0)$, a saddle at the origin, periodic level curves around each center for $-1/4<H<0$, two homoclinic loops on $H=0$, and outer periodic curves surrounding both wells for $H>0$.

For positive small $\epsilon$,On the unperturbed level $H=H_0$,The upper and lower halves contribute equally because $dt=dx/y$. Hence the energy balance for the weakly perturbed double-well oscillator isso one may take

For the right homoclinic loop, $H_0=0$, $x_1=0$, and $x_2=\sqrt{2}$. Its persistence requiresThe substitution $u=1-x^2/2$ givesThus the homoclinic balance for the weakly perturbed double-well oscillator occurs atThe left loop gives the same condition by symmetry.

For an orbit surrounding one center, the corresponding balance ratio tends to $\alpha =1$ as the orbit shrinks to the center and to $\alpha =5/4$ as it approaches the homoclinic loop. The single-well periodic orbit range for the weakly perturbed double-well oscillator is therefore

Define the canonical momentaWhen the velocity Hessian is nonsingular, these equations can be inverted for ${\dot{q}}_i(q,p,t)$, and the Hamiltonian is the Legendre transformIts differential isbecause the $p_{i}d{\dot{q}}_i$ terms cancel. The Euler-Lagrange equation identifies $\partial \mathcal{L}/\partial q_i={\dot{p}}_i$, so coefficient comparison gives Hamilton's equations

For the symmetric top, the momenta areThusThe Legendre transform gives

Both $\varphi$ and $\psi$ are cyclic coordinates, so $p_{\varphi }$ and $p_{\psi }$ are conserved. The Hamiltonian has no explicit time dependence and is also conserved. Since $H$ is independent of $\varphi ,\psi$,under the Poisson bracket. Hence three independent first integrals in involution are

Letbe the Hamiltonian. The Hamiltonian Monte Carlo proposal is accepted with the Metropolis–Hastings acceptance probabilityThe leapfrog map is volume preserving and reversible after momentum reversal, so no proposal-density Jacobian appears.

Write the Hamiltonian as the sum of separated one-dimensional energiesEach $(q_i,p_i)$ system is a simple harmonic oscillator and is integrable away from $(q_i,p_i)=(0,0)$ by part (b). The separated energies satisfyand their differentials are independent whenever every oscillator pair is nonzero. The Poisson-commuting separated energies therefore make the $n$-dimensional oscillator integrable. The exceptional coordinate strata have measure zero, so this proves Liouville integrability in its usual almost-everywhere sense; the origin is the only point where $dH$ itself vanishes.

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 $\dot{p}=-{\partial }_{q}H$.

A Bravais lattice is the integer span of three linearly independent primitive vectors. Its reciprocal lattice consists of vectors $G$ satisfying $G\cdot r\in 2\pi \mathbb{Z}$ for every lattice vector $r$.

Translations by lattice vectors commute with the Hamiltonian. Their simultaneous eigenvalues have the form $e^{ik\cdot r}$, so an energy eigenfunction may be chosen withThus ${\psi }_k=e^{ik\cdot x}{u}_k$ with lattice-periodic $u_k$, which is Bloch theorem. Wavevectors differing by a reciprocal-lattice vector describe the same translation character, and the Brillouin zone is a fundamental Wigner-Seitz cell of the reciprocal lattice.

The radial action isWith the variable $x$ from part b this becomesApplying the supplied integral twice, with $c=0$ and $c=2$, and using the root sum and products from parts b and c givesThis is the radial action of the spherical isochrone model. Rearranging,and squaring yields the Hamiltonian in action-angle variables:As a check, differentiating this Hamiltonian gives ${\Omega }_r=(-2E{)}^{3/2}/(GM)$ and the frequency ratio found in part c.

For the unperturbed Hamiltonian system, choose a conserved energy $H(x,y)$. Along the perturbed system,The energy balance method integrates this expression around each unperturbed periodic orbit. A zero of the resulting $O(\epsilon )$ energy change is a necessary leading-order condition for a nearby periodic orbit. A simple zero at which the drift changes from positive inside to negative outside predicts a stable orbit; the opposite change predicts an unstable one.

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.

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.

For a general Lagrangian,Along an Euler-Lagrange trajectory,so energy is conserved when $L$ has no explicit time dependence.

For the given Lagrangian, write $v=|\dot{x}|$. At points where $v\ne 0$,The Euler-Lagrange equation is thereforeIn expanded form its left-hand side isThe momentum has fixed magnitude,so this is constant without needing to solve the equation of motion. The Hamiltonian is

The singular Legendre transform of a degree-one velocity Lagrangian applies here: $p$ determines only the direction of $\dot{x}$, not its magnitude, and the velocity Hessian is not invertible. Thus the usual inverse Legendre transform cannot reconstruct the Lagrangian from this Hamiltonian.

Now $g=1$, so $c=⟨\psi |\varphi ⟩$ is real andThe perturbation is diagonal in this basis:Hence the instantaneous eigenvalues are $1+2V(t)$ and $-1$ with time-independent eigenvectors, so Hamiltonians at different times commute. In units $\hslash =1$, putExpanding the initial state and evolving its two components givesTaking its overlap with $|\varphi ⟩$ and simplifying gives the exact transition probabilityWith explicit $\hslash$, each phase in the final sine is divided by $\hslash$.

Using one even and one odd site per unit cell, Bloch theorem reduces the Hamiltonian toIts eigenvalues are

With $H=p_j{p}_j/(2m)$, the stated brackets giveandThus $m\ddot{\mathbf{q}}=e\dot{\mathbf{q}}\times \mathbf{B}$, the Lorentz-force equation with zero electric field. The magnetic interaction has been moved from the Hamiltonian into the noncanonical symplectic structure.

Within each degenerate unperturbed eigenspace of fixed $n_2$, the perturbation has matrixIts eigenvectors are $|+{⟩}_1$ and $|-{⟩}_1$, with first-order shifts $+g$ and $-g$. Thus degenerate perturbation theory gives exactly the states and energies found in part (i); again, all higher corrections vanish because the perturbation commutes with the second-oscillator Hamiltonian and is exactly diagonalized within each degenerate block.

For a one-dimensional closed energy orbit, defineAfter locally inverting this relation to write $E=H(I)$, use the Hamilton--Jacobi generating functionand define the angle $\varphi =\partial W/\partial I$ modulo $2\pi$. The transformation is canonical, the Hamiltonian depends only on $I$, and

The adiabatic-invariance principle says that if an external parameter changes on a time scale much longer than the orbital period, and no separatrix is crossed, the action $I$ remains approximately constant. Its small within-cycle changes do not accumulate at leading adiabatic order.