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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.

For $X$ in the dual cone $C$, copositivity of $Q-\lambda I$ gives $⟨X,Q-\lambda I{⟩}_F\ge 0$. An upper-bounding Lagrangian for the maximization is thereforeIts supremum over unrestricted $\lambda$ is finite precisely when $\mathrm{tr}X=1$. Minimizing that upper bound gives the conic programThis is completely positive optimization, not simply semidefinite programming: $C$ is the completely positive cone.

There is also a direct equality certificate. Every feasible $X={\sum }_j{u}_j{u}_j^T$, $u_j\ge 0$, has ${\sum }_j\Vert u_j{\Vert }_2^2=1$. Its objective is a weighted average of the nonnegative-sphere Rayleigh quotients, hence is at least $\alpha$. For a minimizing unit vector $x_{\ast }$ from part (g), $X_{\ast }=x_{\ast }{x}_{\ast }^T$ is feasible and has objective $\alpha$. Thus $\boxed{X_{\ast }=x_{\ast }{x}_{\ast }^T}$ attains the dual and both values agree, without relying on unverified regularity assumptions.

Put $h(r)=1-\mu /r^2$. For the equatorial geodesic, the geodesic Lagrangian is $\mathcal{L}=\frac{1}{2}[-h{\dot{t}}^2+h^{-1}{\dot{r}}^2+r^2{\dot{\varphi }}^2]$, where dots denote derivatives with respect to the affine parameter. The cyclic coordinates $t,\varphi$ give constants $p_t=-h\dot{t}=-E$ and $p_{\varphi }=r^2\dot{\varphi }=L$. A third constant is the tangent norm $g_{ab}{\dot{x}}^a{\dot{x}}^b=-\kappa$, preserved by the geodesic equation and metric compatibility. For a timelike geodesic rescale the affine parameter to proper time so $\kappa =1$; for a null geodesic $\kappa =0$.

Substitute $\dot{t}=E/h$, $\dot{\varphi }=L/r^2$ into the norm identity and multiply by $h$. It gives the effective potential equationOutside the horizon $r=\sqrt{\mu }$, its derivative is $V^{\prime }=-(L^2-\kappa \mu )/r^3+2\mu L^2/r^5$. A nontrivial circular null geodesic requires $L\ne 0$, givingFor a circular timelike geodesic,At it $V^{\prime \prime }=-4\mu L^2/r^6<0$, and its radius exceeds $\sqrt{2\mu }$. If $L^2\le \mu$ no finite circular timelike radius exists. Thus all finite circular null and timelike orbits in this five-dimensional exterior are radially unstable. The energy of a circular orbit is fixed by $E^2=2V(r)$, not independently arbitrary.

For comparison, with four-dimensional Schwarzschild horizon radius $r_s$, the photon circle is at $3r_s/2$ and is unstable. Circular timelike orbits exist for $r>3r_s/2$; they are unstable below $3r_s$, marginal at $3r_s$ and stable above $3r_s$. The five-dimensional potential has no corresponding stable outer circular timelike branch.

Assume the holonomic constraints have linearly independent gradients, so their common level set is locally a smooth constraint manifold parametrized by $q_1,\dots ,q_n$. The multiplier form of the Euler-Lagrange equations isAt fixed time ${\sum }_A{\partial }_A{f}_{\alpha }{\partial }_i{x}_A=0$. The chain rule for the reduced Lagrangian givesConversely, if these reduced equations hold, the Cartesian Euler-Lagrange residual annihilates every tangent direction. Linear independence of the constraint gradients implies that it is a linear combination of those gradients, giving the required Lagrange multipliers. This proves the local equivalence, including time-dependent constraints; singular constraints would need separate analysis.

For the bead, $\dot{y}=f^{\prime }(x)\dot{x}$, soThe autonomous conservation of energy gives $E=\frac{1}{2}(1+f^{\prime 2}){\dot{x}}^2$. If $E>0$, the motion has a fixed sign $s=\mathrm{sgn}\dot{x}$ andThe integral is the signed arc length divided by the constant speed, and is locally invertible. The special solution $E=0$ is a stationary bead, for which writing time as a function of its unchanging position is not meaningful.

For variations vanishing at both endpoints, differentiating the action and using integration by parts givesThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationsFor the specified Lagrangian, the gauge transformation changes it byFor arbitrary $F$, it is a total derivative precisely when $\boxed{\alpha =-1}$. The corresponding change in action is then an endpoint term whose variation is zero. In general time-dependent notation, the equations areThe electric field $-\nabla \varphi -{\partial }_t\mathbf{A}$ and magnetic field $\nabla \times \mathbf{A}$ are unchanged by the gauge transformation with $\alpha =-1$. Although the original potentials have no time dependence, their transformed versions generally do; allowing this is essential to the stated invariance.

The Euler-Lagrange equations for the kinetic Lagrangian areExpanding and multiplying by the inverse metric gives the geodesic equationThe kinetic variational principle uses an affine parameter; the energy $g_{ab}{\dot{x}}^a{\dot{x}}^b/2$ is constant along its solutions. Constant curves are also geodesics.

For the Poincare half-plane model, translation in $x$ gives the conserved quantity $p=\dot{x}/y^2$, and the energy is $E=({\dot{x}}^2+{\dot{y}}^2)/(2y^2)$. If $p=0$, the nonconstant geodesics are vertical lines, with $y=y_0{e}^{\pm \sqrt{2E}t}$. If $p\ne 0$,Writing $R^2=2E/p^2$ and integrating $dx/dy=\pm y/\sqrt{R^2-y^2}$ yields $(x-x_0{)}^2+y^2=R^2$. Thus nonconstant geodesic images are vertical lines or upper semicircles meeting the real axis orthogonally.

For the two specified points, the circle centre is $(6,0)$ and the radius is five:For example, an affine unit-speed parametrization is $x=6+5\tanh t$, $y=5\mathrm{sech}t$. The endpoints on the real axis are $(1,0)$ and $(11,0)$, outside the hyperbolic surface.

Write $E(t)=\delta A/\delta x(t)$ for the functional derivative of the action. Under time reversal in classical mechanics, $\dot{x}$ changes sign whereas $E$ is even by assumption. Thus the forward and reversed Gaussian white noise histories associated with the same geometric path areThe Onsager--Machlup path probability therefore givesHere the path probabilities are densities conditional on their respective initial states; they are not separately normalized bridges conditioned on both endpoints. The noise-to-path Jacobian determinant must be treated with a consistent discretization: it cancels when it is invariant under time reversal in classical mechanics. This is the usual additive-noise Langevin dynamics convention, including a constant-mass Underdamped Langevin dynamics or the time-reversal invariance of a path Jacobian in midpoint Overdamped Langevin dynamics. A completely arbitrary velocity-dependent Lagrangian would require specifying that measure rather than deducing its cancellation solely from the parity of $E$.

For an autonomous Lagrangian, the Euler-Lagrange equation usesApply the chain rule to the Hamiltonian along an arbitrary smooth path, without assuming the unforced Euler-Lagrange equation:Consequently the energy balance for an autonomous Lagrangian isWith explicit time dependence, an additional $-{\partial }_{t}L$ enters $dH/dt$. For ideal Gaussian white noise, the identity is understood through smooth-noise regularization or the Stratonovich chain rule. The original PDF correctly differentiates $L$ with respect to $\dot{x}$ in $H$; the supplied TeX's derivative with respect to $x$, its endpoint $t^2$, and its ordinary derivative of $A$ are transcription errors.

Let a one-parameter infinitesimal transformation be $\delta q_i=ϵX_i(q,t)$ and suppose its change in the Lagrangian is a total derivative,With canonical momentum $p_i=\partial \mathcal{L}/\partial {\dot{q}}_i$, the chain rule and the Euler-Lagrange equations give, along a motion,Comparison with the assumed total derivative proves Noether theorem:

The given Lagrangian has no explicit time dependence, so its energy function is conserved:It is also invariant under the dilation $(x,y)\mapsto (e^{ϵ}x,e^{ϵ}y)$: the velocities and $y$ all scale by the same factor, while $x/y$ is unchanged. The generator is $(X_x,X_y)=(x,y)$ and $F=0$. SinceNoether's theorem supplies the second first integralThe integrals $E$ and $D$ are independent on an open dense subset of phase space.

For unit mass, the Lagrangian in cylindrical coordinates isIntroduce shearing sheet coordinates by $r=r_0+x$ and $\phi ={\Omega }_{0}t+y/r_0$. Evaluate all derivatives of $\Phi$ at $(r_0,0)$. The reference circular orbit obeys ${\Phi }_r=r_0{\Omega }_0^2$, while midplane symmetry gives ${\Phi }_z={\Phi }_{rz}=0$. Its Taylor expansion isExpanding the kinetic energy to the same order givesThe terms linear in $x$ cancel by circular-orbit balance. Discard the constant and the term $r_0{\Omega }_0\dot{y}$ by total-time-derivative invariance of a Lagrangian; these do not change the Euler-Lagrange equations. The particle Lagrangian in a shearing sheet is thereforewith shearing-sheet tidal potentialThe second form follows from ${\Phi }_r=r{\Omega }^2$, the orbital shear parameter $q_0=-d\log \Omega /d\log r$ and the vertical epicyclic frequency ${\Omega }_z^2={\Phi }_{zz}$.

Keep $y\ge 0$ as the domain restriction, introduce nonnegative Lagrange multipliers $s_i$ for $y_i\le 1$, and introduce a free multiplier $t\in \mathbb{R}$ for ${\sum }_i{y}_i=k$. The maximizing Lagrangian isIts supremum over $y\ge 0$ is finite exactly when $s+t1\ge x$, and then equals $kt+{\sum }_i{s}_i$. By linear programming duality, since the primal is feasible and bounded,Minimizing over $s$ for fixed $t$ gives $s_i=(x_i-t{)}_{+}$, the positive part of a real-valued function. Thus the threshold formula for the sum of the largest components isFor a direct attainment check, order the coordinates $x_{[1]}\ge \cdots \ge x_{[n]}$. Any $t\in [x_{[k+1]},x_{[k]}]$ is optimal for $k<n$, while any $t\le x_{[n]}$ is optimal for $k=n$.

Replace coordinate inequalities by the Loewner order and coordinate sums by matrix trace. The maximizing semidefinite program isIts feasible set is the fantope. In an orthonormal eigenbasis of $X$, its objective is ${\sum }_i{\lambda }_i(X)Y_{ii}$, and the diagonal entries of $Y$ belong to the capped simplex. Choosing $Y$ as the orthogonal projection matrix onto the largest $k$ eigenvectors attains the sum of the largest eigenvalues, as in the Ky Fan maximum principle.

The minimizing semidefinite program isFor example, its Lagrangian arises by assigning $S⪰0$ to $I-Y⪰0$, keeping $Y⪰0$ in the domain, and using free $t$ for $\mathrm{tr}Y=k$. Finiteness of the supremum over $Y$ requires $S+tI-X⪰0$.

Equality follows directly by choosing $S$ to have eigenvalues $({\lambda }_i(X)-t{)}_{+}$ in the same orthonormal eigenbasis, with $t$ between the $k$th and $(k+1)$st eigenvalues, or below the smallest one for $k=n$. This is the threshold semidefinite program for the largest eigenvalues, and it also verifies the endpoint case $k=n$.

Under the gauge transformation,because the curl of a gradient vanishes, andThe transformed Lagrangian isA total time derivative changes the action only by endpoint terms, whose variation vanishes for fixed endpoints, so the Euler-Lagrange equations are unchanged.

The new momentum isUse the type-two generating function for a canonical transformationIt gives $\mathbf{p}=\partial F_2/\partial \mathbf{r}=\tilde{\mathbf{p}}-q\nabla f$, $\tilde{\mathbf{r}}=\partial F_2/\partial \tilde{\mathbf{p}}=\mathbf{r}$, andThis equals $|\tilde{\mathbf{p}}-q\tilde{\mathbf{A}}{|}^2/(2m)+q\tilde{\varphi }$, so the gauge change is a time-dependent canonical transformation.

In polar coordinates,With the change of radial coordinate $u=\log r$, the detection functional becomesIt is therefore $\lambda$ times ordinary path length in the $(u,\theta )$-plane. The shortest path is the line segment from $(\log A,0)$ to $(\log B,\alpha )$, soReturning to $r$ gives the logarithmic spiralIts minimum detection probability isThe functional depends only on the geometric path and not its parametrization, so tiptoeing and running give the same probability.

For the improved sensor, omit the irrelevant positive factor $\lambda$ and use the LagrangianThe coordinate $\theta$ is cyclic, so its conjugate momentum is conserved. Equivalently,is constant. Because $L$ has no explicit time dependence, the associated conservation of energy gives the constantMultiplying by $r$ and using $h=r\dot{\theta }$ gives the required equation

Let each mass be $m$ and each spring constant be $k$, with the $x_i$ denoting longitudinal displacements from equilibrium. The Lagrangian isThe inverse coordinate transformation isSubstitution givesThus the Euler-Lagrange equations separate into a symmetric pair,and an antisymmetric pair,

The symmetric normal modes are:

For antisymmetric modes, the squared dimensionless frequencies are the eigenvalues ofnamely $2\mp \sqrt{2}$. They giveandIn the original coordinates their mode shapes are respectivelyand

The coordinates $\varphi$ and $\psi$ are cyclic, so their canonical momenta are conserved:Because the Lagrangian has no explicit time dependence, the conserved energy isThese are the three integrals of motion requested.

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

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

The kinetic energy of the two equal masses isThe extensions of the left, coupling, and right springs are respectively $x_1$, $x_2-x_1$, and $x_2$. Their potential energy is thereforeThus the mechanical Lagrangian $L=T-U$ has the requested quadratic form with the symmetric small-oscillation mass and stiffness matrices