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A smooth wire has instantaneous shape $z=f(r)$ in cylindrical coordinates and rotates at constant angular velocity $\dot{\theta }=\omega$. The tangent vector at fixed time is ${\mathbf{e}}_r+f^{\prime }(r){\mathbf{e}}_z$. Its dot product with the constraint force vanishes. Projecting Newton's second law along this tangent vector gives the equation in the title.

Multiplying that equation by $\dot{r}$ proves conservation ofThis is a conserved quantity in the rotating description; the motor can exchange ordinary mechanical energy with the bead. A relative equilibrium at $a>0$ satisfies $U^{\prime }(a)=g{f}^{\prime }(a)-{\omega }^{2}a=0$. Its linearized equation is $(1+f^{\prime }(a{)}^2)\ddot{ϵ}=-U^{\prime \prime }(a)ϵ$. A strict local minimum with $U^{\prime \prime }(a)>0$ gives Lyapunov stability, while $U^{\prime \prime }(a)<0$ gives an unstable equilibrium. If $U^{\prime \prime }(a)=0$, higher orders are needed.

For a massive test particle with electric charge $e$, choose the electromagnetic four-potential $A_t=-Q/r$ and proper time $\tau$. The cyclic coordinates $t,\varphi$ in the equatorial Lagrangian give the conserved quantities $E=mF\dot{t}+eQ/r$ and $l=r^2\dot{\varphi }$. Here $l$ is angular momentum per unit mass. Substituting these quantities into the four-velocity normalization $g(U,U)=-1$ givesThe energy includes the gauge-dependent electrostatic term; the displayed convention fixes that gauge by taking the potential to vanish at infinity.

For a finite-dimensional Hamiltonian system, integrability is normally expressed by enough independent first integrals in mutual Poisson bracket involution. In classical field theories, an infinite hierarchy of compatible conserved quantities is the corresponding structure. For Sine-Gordon theory, Bäcklund transformations generate local conservation laws and explicit soliton solutions. Conservation of arbitrary quantities alone should not be confused with proof of their Hamiltonian involution.

Let $X$ be a Conformal Killing vector field with conformal Killing factor $\lambda$, and let $U$ be the tangent vector of a geodesic with affine parameter $s$. The product rule, metric compatibility and geodesic equation giveEvery null geodesic therefore has the conserved quantity $g(X,U)$. On a timelike geodesic with proper time as parameter, the derivative is $-\lambda$ in metric signature $(-,+,+,+)$. With a nonaffine parameter satisfying ${\nabla }_{U}U=fU$, an extra term $fg(X,U)$ appears; the conserved expression for a null geodesic is tied to its affine parameter normalization.

With periodic boundary conditions, the spatial mean of $B$ is a conserved quantity. For zero mean write $B=H_X$ with periodic zero-mean $H$. The functional $V=⟨\mu A^2/2+\alpha A^3/3-A^4/4-A_X^2/2-A^2{H}_X/2-H_X^2/(4\delta )⟩$ generates $A_T=\delta V/\delta A$, $H_T=2\delta \sigma \delta V/\delta H$. Thus $V_T=⟨A_T^2+H_T^2/(2\delta \sigma )⟩\ge 0$. Completing the square in $H_X$ bounds $V$ above by a coercive quartic polynomial when $0<\delta <1$, ruling out nonconstant recurrent motion on regular precompact trajectories.

For the linear ordinary differential equation $c^{\prime }=Ac-b$, a fixed left null space vector gives $(w^{T}c{)}^{\prime }=-w^{T}b$. Thus $w^{T}c$ is a conserved quantity when $w^{T}b=0$. In a box model written for concentrations per unit volume, physical tracer inventory uses the box volumes as weights. Zero column sums make $w=1$ possible, but that property must be checked rather than assumed.

Each Killing vector field supplies a conserved quantity for freely moving test particles: it turns a geometric symmetry into a first integral of the geodesic equation. Let $\gamma (\lambda )$ be a geodesic with affine parameter $\lambda$, tangent vector $U^{\mu }=d{x}^{\mu }/d\lambda$, and ${\nabla }_{U}U=0$. Wherever a Killing vector field $K$ is defined along the geodesic,The statement applies to Riemannian manifolds and to timelike geodesics, null geodesics and spacelike geodesics in a Lorentzian manifold. It requires an affine parameter; for a massive test particle, proper time is one choice. See Tong, section 4.3.2, equation (4.32).

For a worked example in the Euclidean plane, take $\gamma (t)=(1+t,2-t)$ with affine parameter $t$ and velocity $U=(1,-1)$. The Euclidean Killing vector fields for translation and rotation giveThe rotational field along the geodesic is $K_{\mathrm{rot}}(\gamma (t))=(t-2,1+t)$: it changes from $(-2,1)$ at $t=0$ to $(-1,2)$ at $t=1$, yet its pairing with $U$ stays $-3$. With toy mass $m=1$ and $t$ interpreted as time, the first constant is the $x$ component of linear momentum, and the second is the signed $z$ component of angular momentum about the origin. Their numerical values use toy units. The geodesic is straight; the rotation flow orbits are circles.

This conservation law follows from metric compatibility and the Killing equation:The geodesic equation kills the acceleration term, and the symmetric product $U^{\mu }{U}^{\nu }$ selects the vanishing symmetric derivative. This tensor contraction is the whole proof outline; see Carroll, equation (5.43).

In a stationary spacetime, a chosen time-translation Killing vector field $K$ gives Killing energy. With metric signature $(-+++)$, $c=1$, and a massive test particle's four-momentum $p^{\mu }=m{U}^{\mu }$, the physical normalization isHere $\Phi$ generates the axial rotational symmetry when one exists. In coordinates with $K={\partial }_t$ and $\Phi ={\partial }_{\varphi }$, these are $E=-p_t$ and $L_z=p_{\varphi }$. Both are conserved quantities along the geodesic, even if the metric has a time-angle cross term. A future timelike $K$ makes $E$ positive for a future massive particle; the scale of $E$ depends on the normalization of $K$. Not every spacetime has a global timelike Killing vector field, and a stationary field can become spacelike in an ergoregion of a stationary spacetime. Conservation on the field's domain does not imply a universal global positive energy; see Tong, sections 4.3.1-4.3.2 on the timelike-domain qualification and conserved charges.

For a field evolving by a Hamiltonian partial differential equation, infinite-dimensional integrability combines sufficiently rich commuting conservation laws with a mechanism such as the inverse scattering transform or a spectral coordinate system that linearizes evolution. Lax equations supply spectral invariants; Lenard-Magri recursions supply commuting Hamiltonians when their solvability conditions hold. Infinite many conserved quantities alone do not prove completeness or an infinite-dimensional action-angle theorem: the function space, boundary conditions and analytic independence all matter.

An integrable partial differential equation has enough exact structure, such as a Lax pair, infinitely many conserved quantities, or an inverse scattering transform, to reduce its evolution to auxiliary linear or finite-dimensional data.

For the symmetric top, set $I_1=I_2=J$. The third of Euler's equations reduces to $I_3{\dot{\omega }}_3={\tau }_3=0$. Thus the axial angular velocity component is a conserved quantity:The transverse angular velocity components are generally variable. A torque constant in the principal body frame generally rotates in the inertial frame; constancy in the two frames is different.

The Poisson bracket is antisymmetric, so $\{H,H\}=0$. The Hamiltonian energy balance consequently givesFor a regular Lagrangian the Legendre transform in mechanics gives $H_t=-L_t$, evaluated at the corresponding velocity. In particular, a Lagrangian with no explicit time dependence has time-translation symmetry. Noether theorem associates this continuous symmetry of the action with the conserved quantity ${\sum }_i{\dot{q}}_i\partial L/\partial {\dot{q}}_i-L=H$. For an ordinary natural mechanical Lagrangian $L=T-V$, with $T$ quadratic in the velocities, this quantity is the total energy $T+V$.

Since $H$ is a conserved quantity, the Hamiltonian energy balance gives $H_t=0$. Since $f$ is a conserved quantity, $f_t+\{f,H\}=0$. Take an explicit partial derivative with respect to time, holding the phase space coordinates fixed:The last Poisson bracket vanishes because $H_t=0$. ThusThis is the time derivative of a conserved quantity for an autonomous Hamiltonian. The conservation hypotheses are understood to hold for every trajectory, as identities on phase space.

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 $I=H_0/\omega$.

Set $m=GM/c^2$ and $f(r)=1-2m/r$. A photon follows a null geodesic, with an affine parameter $\lambda$ satisfyingThe Christoffel symbols follow from the supplied Schwarzschild spacetime metric tensor. An efficient equivalent procedure is to use the geodesic Lagrangianwhere dots denote $d/d\lambda$. Its Euler-Lagrange equations give the affinely parametrized geodesic equations. The condition $\mathcal{L}=0$ is imposed after variation; setting the Lagrangian identically to zero before varying would discard the equations.

By spherical symmetry the ray's angular momentum fixes an orbital plane through the center. Rotate the angular coordinates so that plane is $\theta =\pi /2$. The $\theta$ Euler-Lagrange equation isso $\theta =\pi /2$, $\dot{\theta }=0$ is preserved once imposed initially. A radial ray has no unique orbital plane, but any plane through that radial line can be chosen.

The coordinates $t$ and $\varphi$ are cyclic. Thus their canonical momenta obeyThe two conserved quantities can be chosen asThese are the geodesic conserved quantities from Killing vectors ${\partial }_t$ and ${\partial }_{\varphi }$. With the chosen parameter normalization the conserved energy is $E=c^2\epsilon$ and axial angular momentum is $\ell$; multiplying the affine tangent by a constant multiplies both quantities by that constant.

The additional invariant is the squared norm of the affine tangent. Metric compatibility and the geodesic equation make it constant, and a photon has the null conditionOn the equatorial plane this givesand the conserved quantities of part (a) yieldFor a nonradial ray $\ell \ne 0$, put $u=1/r$. Sincethe orbit equation becomesHere $b=|\ell |/(c\epsilon )$ is the positive impact parameter for a future-directed ray arriving from infinity. A strictly radial ray has $\ell =0$ and must be described by the radial equation rather than by $\varphi$ as parameter.

With $\tau$ the proper time of the massive test particle, the four-velocity has squared norm $-1$:The conserved quantities in part (i) giveMultiplying the normalization by $F$ and substituting gives the charged-particle radial equation in Reissner-Nordstrom spacetime:At a radial turning point $r_{\ast }>0$, the left side is zero, soThe parenthesis on the left is strictly positive. Hence $F(r_{\ast })\ge 0$. The factorization in part (b) now yieldsThere is no radial turning point between the horizons. Indeed, for $r_{-}<r<r_{+}$, the right-hand side of the radial equation is strictly positive, independently of the electric charge $e$ of the test particle. This is the charged-particle turning-point exclusion in Reissner-Nordstrom spacetime.

The inequalities are necessary, not a claim that a radial turning point exists for every $E,l,e$. At a regular horizon crossing a timelike vector cannot be tangent to a null hypersurface, so a massive test particle actually has $\dot{r}\ne 0$ there. Algebraic equality at $r_{\pm }$ would require $E=eQ/r_{\pm }$ and does not describe a regular finite-proper time radial turning point on the horizon. In the Schwarzschild spacetime limit $Q=0$, the positive-radius inner alternative disappears.

Reflection across $\theta =\pi /2$ is an isometry; a geodesic initially tangent to its fixed equatorial plane remains in that plane by uniqueness of the geodesic equation. Thus take $\theta =\pi /2$, $\dot{\theta }=0$, and $\Sigma =r^2$. Define the algebraic abbreviationsThe conserved quantities of part (b) becomeFor $\Delta \ne 0$, inversion givesThe null condition can be written without expanding a quadratic square:Substituting the preceding expressions yields the equatorial Kerr null geodesic radial equationAn equivalent useful polynomial form isThese expressions use an affine parameter $s$; a common rescaling of $E$ and $L$ accompanies its rescaling. The inversion was performed in the exterior $r>r_{+}$, where Boyer-Lindquist coordinates are regular. The polynomial radial identity has a smooth continuation to a horizon, but the formulas for $\dot{t}$ and $\dot{\varphi }$ there must be replaced by regular-coordinate expressions. In the limit $a=0$, the result reduces to the Schwarzschild spacetime equation ${\dot{r}}^2=E^2-(1-2M/r)L^2/r^2$.

The assertion about every exterior radius is false for geodesics. A circular orbit here means an equatorial null geodesic at constant $r$, rather than a genuinely radial trajectory. A zero of the radial effective potential alone gives a turning point, not necessarily a circular orbit. The radial geodesic equation must hold as well.

Write $q=L-aE$ andFor an exterior circular null geodesic, the two conditions are $f(r)=f^{\prime }(r)=0$, equivalently $\mathcal{R}(r)={\mathcal{R}}^{\prime }(r)=0$ with $E,L$ fixed. One way to justify the derivative condition without dividing by $\dot{r}=0$ is to use the radial Euler-Lagrange equation. At constant $r$ it givesTogether with $g_{AB}{\dot{x}}^A{\dot{x}}^B=0$, and the fixed conserved $p_A$, this is precisely the double-root condition for $\mathcal{R}$.

NowSetting this to zero and using $f=0$ givesA nontrivial exterior null geodesic cannot have $E=0$ here: the second equation would force $q=L=0$, and then every tangent component would vanish. ThereforeBoth equations, with the same constants $E,L$, are required. They give the requested radius relations in terms of the conserved quantities, while showing why an arbitrary exterior radius is not permitted.

For a more explicit classification let $b=L/E$ be the impact parameter. The second relation gives $b-a=s{r}^{3/2}/\sqrt{M}$ with $s=\pm 1$. Substituting into the first relation yieldsFor $0<a<M$, $s=+1$ gives the prograde equatorial circular Kerr photon orbit and $s=-1$ the retrograde one. The two allowed exterior radii areBoth are strictly greater than $r_{+}$ in the subextremal range. Squaring the signed radius equation gives $r(r-3M{)}^2=4M{a}^2$, but introduces an additional inner root and loses the sign distinguishing the two senses of rotation. At $a=0$ the two exterior radii coincide at the photon sphere $r=3M$, with $b=\pm 3\sqrt{3}M$.

For a direct counterexample to the all-radii claim take $a=0$ and $r_0=4M>r_{+}=2M$. Choose $E=1$ and $L=4\sqrt{2}M$ so that $f(r_0)=0$. The radial geodesic equation nevertheless givesIt is a turning point, not a circular orbit. Alternatively, constant-$r$ null directions exist throughout the exterior because the two angular-velocity roots of $g_{tt}+2\Omega g_{t\varphi }+{\Omega }^2{g}_{\varphi \varphi }=0$ are real; almost all of their integral curves require acceleration and fail the radial geodesic equation. This distinction explains how an all-radii assertion can arise.

Finally, $r=r_{+}$ itself has the null geodesic generators of its Killing horizon, proportional to ${\chi }_{+}={\partial }_v+{\Omega }_{+}{\partial }_{\psi }$. They are a separate horizon case. On the horizon ${\chi }_{+}$ is normal to every tangent, so its contractions with ${\partial }_v$ and ${\partial }_{\psi }$ both vanish: $E=L=0$ for a generator tangent. Its affine parameter can be chosen by rescaling that tangent. The exterior calculation dividing by $E$ does not include it. The presence of these generators does not supply circular null geodesics at the intervening arbitrary exterior radii.

Substitution of the weighted scaling and $\tau =\epsilon t$ givesThe derivative of the candidate Hamiltonian isThus $H$ is a conserved quantity at $\epsilon =0$. For $\alpha >0$ put $a=\sqrt{\alpha }$. The conservative equilibrium point $(-a,0)$ is a center with $H=-2a^3/3$, and $(a,0)$ is a saddle equilibrium with saddle-loop energyThe closed component of a level between these energies surrounds the center. At $H=H_s$,is the homoclinic orbit, with left turning point $u=-2a$. The upper branch is traversed to the right and the lower branch to the left. Levels above the barrier have escaping trajectories; the saddle level also has unbounded branches to the right. For $\alpha =0$ the two critical points coalesce; for $\alpha <0$ the potential is monotone and there is no center or saddle loop.

A field $u(x,t)$ has infinitely many degrees of freedom: its initial state is a function rather than a finite list of coordinates. For a Hamiltonian evolution, the analogue of the finite-dimensional construction seeks an infinite sequence of independent conserved functionals $H_j[u]$ whose Poisson brackets vanish. Their Hamiltonian vector fields then define commuting time evolutions. For example, a Poisson operator $J$ produces the field equation $u_t=J\delta H/\delta u$, where $\delta H/\delta u$ is the variational derivative.

A Lax equation packages many conservation laws into one operator identity. If $L_t=[A,L]$, cyclicity of an appropriate trace givesFor a formal pseudodifferential operator the relevant trace is the Adler trace; suitable fractional powers of a monic differential operator produce further conserved functionals. A Lenard-Magri recursion can prove that such functionals are Poisson-commuting functions, rather than merely conserved quantities. A Lax equation by itself does not establish all the independence and completeness properties needed for integrability.

The Korteweg-De Vries equation illustrates the mechanism. In one time normalization take $L={\partial }^2+u$ and $A={\partial }^3+\frac{3}{2}u\partial +\frac{3}{4}{u}_x$. Direct composition givesIts spectral problem, together with the inverse scattering transform for decaying data or suitable spectral coordinates for periodic data, turns nonlinear evolution into simple evolution of spectral data. Solitary waves arise from discrete spectral data, while continuous data describe dispersive radiation.

Thus infinite-dimensional Hamiltonian integrability is more than an infinite list of formulas: one wants sufficiently complete commuting invariants and a reconstruction mechanism that solves the evolution. Boundary conditions and the function space are essential. A formal hierarchy need not converge, and an infinite family of conserved quantities does not automatically give a compact infinite-dimensional torus or an unrestricted analogue of the Arnold-Liouville theorem.