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For conservative forces and ideal constraints that do no work, mechanical energy $E=T+U$ is constant. Work-energy balance remains useful with nonconservative forces: their power gives $dE/dt$. In one dimension,locates accessible regions and turning points, and reduces the trajectory to a quadrature $dt=dq/\sqrt{2(E-U)/m}$. A local minimum of potential is a stable equilibrium; a local maximum is unstable. Near a nondegenerate minimum, $U=U_0+K(q-q_0{)}^2/2+\cdots$ gives ${\omega }^2=K/m$.

For a central force, conserved angular momentum reduces radial motion to an effective potential $U_{\mathrm{eff}}=U(r)+L^2/(2m{r}^2)$. Its extrema determine circular orbits and their radial stability. In several coordinates, small oscillations follow from the kinetic-energy mass matrix $M$ and potential Hessian $K$ through $\det (K-{\omega }^{2}M)=0$.

Energy is also central to the Lagrangian $\mathcal{L}=T-U$ and the Euler-Lagrange equations. A time-independent natural mechanical system has conserved energy; cyclic coordinates yield additional momenta. Energy alone does not normally determine every coordinate of a multi-degree-of-freedom motion. Its value is to expose allowed motion, stability and reductions before solving the full differential equations.

Generalized coordinates incorporate holonomic constraints without solving separately for every constraint force. For conservative mechanics use $L=T-V$. The principle of stationary action makes the action $\int Ldt$ stationary for variations vanishing at the endpoints. Integration by parts givesThe canonical momentum is $p_i=\partial L/\partial {\dot{q}}_i$. A cyclic coordinate gives conserved $p_i$; absent explicit time dependence, ${\sum }_i{p}_i{\dot{q}}_i-L$ is conserved. These are symmetry–conservation relationships, not assumptions that all momenta are constant. Nonconservative generalized forces enter the right side.

For example, a pendulum has $L=m{\ell }^2{\dot{\theta }}^2/2+mg\ell \cos \theta$, giving $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0$. The method extends directly to coupled coordinates and velocity-dependent interactions when the correct Lagrangian is used; $T-V$ with a position-only potential is not the most general form.

The Einstein tensor and its contractions areTo match the supplied Ricci tensor components, use the Riemann curvature tensor conventionThis convention also fixes the sign used for geodesic deviation above. In vacuum, tracing the Einstein field equations gives $-R+4\Lambda =0$. Substitution therefore yields $R_{\mu \nu }=\Lambda g_{\mu \nu }$.

Differentiate the angular field equation to obtainConsequently $R_{00}=\Lambda A=\Lambda g_{00}$ and $R_{11}=-\Lambda /A=\Lambda g_{11}$. The other angular component follows from $R_{33}={\sin }^2\theta R_{22}$. Thus the single angular equation implies all the displayed vacuum equations wherever this coordinate chart is regular.

Integrating $(rA{)}^{\prime }=1-\Lambda r^2$ gives $A=1+C/r-\Lambda r^2/3$. Match $g_{tt}\simeq 1+2\Phi$ with the Newtonian gravitational potential $\Phi =-GM/r$ in the weak-field overlap region to identify $C=-2GM$. Hence the Schwarzschild-de Sitter spacetime hasNegative $\Lambda$ instead gives the Schwarzschild--anti-de Sitter black hole case. This matching fixes the normalization locally; a spacetime with nonzero $\Lambda$ is not asymptotically flat.

For timelike geodesics in the orbital plane, take proper time as parameter and use $L=(A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2)/2$. The cyclic coordinates give $A\dot{t}=k$ and $r^2\dot{\varphi }=h$; the angular canonical momentum is $-h$ in this signature. The four-velocity normalization $2L=1$ then givesHere $k$ is conserved Killing energy per unit rest mass in the chosen time normalization, and $h$ is conserved angular momentum per unit rest mass. There is no general interpretation of $k$ as energy measured at an asymptotically flat infinity when $\Lambda \ne 0$.

For radial motion $h=0$, differentiation gives $2\dot{r}\ddot{r}=-A^{\prime }\dot{r}$. The radial Euler-Lagrange equation extends the resulting expression through a turning point where division by $\dot{r}$ would fail. ThusThis is the proper-time second derivative of the areal radius, not a nonzero four-acceleration of the freely falling particle. Positive $\Lambda$ contributes an outward term; negative $\Lambda$ reinforces inward attraction. Restoring units in the slow-motion, weak-field Newtonian limit givesThe extra term describes the gravitational effect of a cosmological constant, not drag or a contact force. The approximation requires $GM/(r{c}^2)\ll 1$, $|\Lambda |r^2\ll 1$ and speed much less than $c$.

Use arc length $s$ as the affine parameter, and put a dot over derivatives with respect to $s$. The geodesic Lagrangian on the sphere isThe cyclic coordinate $\varphi$ gives $\partial L/\partial \dot{\varphi }=a^2{\sin }^2\theta \dot{\varphi }=\text{constant}$. With the constant normalized as in the paper,Here $h$ has units of inverse length, because $s$ is physical arc length. Unit-speed normalization givesFor $h\ne 0$, division by ${\dot{\varphi }}^2$ yields the spherical-geodesic first integralThe turning latitudes satisfy $\sin \theta =|ah|$. In the usual polar-coordinate range $0\le \theta \le \pi$,The maximum departure from the equator, often the geometric quantity intended by an angular amplitude, is insteadThus polar angle and signed latitude must not be interchanged. At $|ah|=1$ the geodesic is the equator. At $h=0$ it is a meridian through the poles; $\varphi$ cannot serve as a global parameter and the divided equation is inapplicable. The limiting polar range is then $[0,\pi ]$.

The original PDF prints an ambiguous derivative with the square on $\theta$ rather than a conventional derivative notation. If the intended equation was second order, differentiating the first integral, or eliminating $\dot{\theta }$ from the Euler-Lagrange equation, gives the equally useful formIt has the same normalized geodesics and the same turning angles. Giving both forms resolves the typographical ambiguity without pretending that a second derivative is the square of a first derivative. Geometrically these trajectories are great circles, with inclination $\arccos |ah|$ to the equator.

Take $\tau$ as particle proper time, dots as $d/d\tau$, and $f(r)=1-2\mu /r$ with $\mu =GM/c^2$. In the equatorial plane the geodesic Lagrangian isThe cyclic coordinates give $d(c^{2}f\dot{t})/d\tau =0$ and $d(-r^2\dot{\varphi })/d\tau =0$. Set the resulting constants to $c^{2}k$ and $-h$ respectively. The explicit second-order equations areThe polar equation is satisfied by $\theta =\pi /2$, $\dot{\theta }=0$, consistently with spherical symmetry. The proper-time normalization $d{s}^2=c^{2}d{\tau }^2$ supplies the third first integral:The normalization is $2\mathcal{L}=c^2$; it is distinct from choosing an arbitrary affine tangent's norm.

Stationarity and axial symmetry make the covariant time and azimuthal momenta constant. For physical momentum and coordinate $t$ in seconds, $P_t=m{c}^{2}f\dot{t}=m{c}^{2}k$ and $P_{\varphi }=-m{r}^2\dot{\varphi }=-mh$. Thus $m{c}^{2}k$ is the conserved energy normalized at infinity and $mh$ is the signed axial angular momentum. This is Killing energy with positive time signature; a convention with negative timelike norm uses the opposite contraction sign. The supplied covariant-momentum form of the geodesic equation makes these cyclic constants immediate, but its parametrization must be understood: for $p^{\mu }=d{x}^{\mu }/d\lambda$ it is exactly as written. For physical $P^{\mu }=md{x}^{\mu }/d\tau$, proper-time differentiation instead gives $d{P}_{\mu }/d\tau =({\partial }_{\mu }{g}_{\nu \sigma })P^{\nu }{P}^{\sigma }/(2m)$; equivalently, the displayed mass-free form uses $\lambda =\tau /m$. This affine momentum normalization does not change either conservation law.

Multiply the normalization equation by $f$ and substitute the two cyclic integrals. Rearrangement gives the radial energy equationFor radial marginal escape $h=0$ and the future-directed constant is $k=1$, because the speed at infinity vanishes. Choose the outward root:Hence the marginal escape geodesic in Schwarzschild spacetime hasThe constant $a$ has units of length to the power $3/2$ divided by time. This radius is an areal coordinate, not accumulated local proper distance.

A Schwarzschild static observer at exterior radius $r$ uses local radial rulers $d\ell =dr/\sqrt{f}$ and a local clock $dT=\sqrt{f}dt$. Therefore the local radial speed of a Schwarzschild geodesic isFor the marginal trajectory $k=1$ this givesIts equality to $\dot{r}$ is particular to $k=1$; the local definition, rather than that accidental equality, is the required observer measurement. The Schwarzschild-coordinate velocity is instead $dr/dt=f\sqrt{2GM/r}$.

For nonzero asymptotic speed $0\le u<c$, the same energy constant is $k={\gamma }_{\infty }=(1-u^2/c^2{)}^{-1/2}$. Combining the radial equation with the local-ruler relation gives radial Schwarzschild escape with asymptotic speedThe positive root describes the outward particle, at radii on its trajectory with $r\ge r_0$. Independently, a static observer has four-velocity $U=(1/\sqrt{f},0,0,0)$ in these coordinates, giving relative Lorentz factor $g(U,u_{\mathrm{particle}})/c^2=\sqrt{f}\dot{t}=k/\sqrt{f}$ and the same speed. As $r\to \infty$ it tends to $u$; for $u=0$ it gives the preceding result. Its limiting value approaches $c$ as $r\downarrow 2GM/c^2$, but that is a limit through exterior observers: no static timelike observer exists on or inside the horizon.

Let $x^{\mu }(\tau )$ denote the spacetime coordinates, with dots meaning differentiation with respect to an affine parameter; for a massive particle take proper time $\tau$. The metric tensor determines the quadratic geodesic Lagrangian $L=g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }$. The Euler-Lagrange equations areSymmetry of the metric tensor gives $\partial L/\partial {\dot{x}}^{\mu }=2g_{\mu \nu }{\dot{x}}^{\nu }$. Hence the covariant four-momentum is $p_{\mu }=m{g}_{\mu \nu }{\dot{x}}^{\nu }=(m/2)\partial L/\partial {\dot{x}}^{\mu }$. If $x^{\mu }$ is a cyclic coordinate, its Euler-Lagrange equation immediately gives ${\dot{p}}_{\mu }=0$.

For the rotating metric, $t,\varphi ,z$ are all cyclic. ThusCombining the first two equations cancels their $r$ dependence:The radial Euler-Lagrange equation is $-2\ddot{r}=-2r(\dot{\varphi }+\omega \dot{t}{)}^2$, soThis is the geodesic equation in a uniformly rotating Minkowski chart. With inertial angle $\Phi =\varphi +\omega t$, $-p_{\varphi }$ is the inertial axial angular momentum. The outward radial-coordinate acceleration expresses the familiar centrifugal term associated with angular motion: the underlying free trajectory is a straight line in Minkowski spacetime. It does not imply an external radial force. Holding a particle at rest in the rotating reference frame instead requires a real inward force.

Curvature contractions. Use the index contraction stipulated in the question. In three dimensions,The same contraction in four dimensions givesThen the Ricci scalar is $R=4\Lambda$ and the Einstein tensor is $G_{\mu \nu }=-\Lambda g_{\mu \nu }$. Thus $R_{\mu \nu }=\Lambda g_{\mu \nu }$ is precisely the vacuum Einstein field equations $G_{\mu \nu }+\Lambda g_{\mu \nu }=0$.

The static solution. For $A=1+{\lambda }^2{r}^2$, $A^{\prime }=2{\lambda }^{2}r$ and $A^{\prime \prime }=2{\lambda }^2$. The given Ricci tensor components becomeEach equals $-3{\lambda }^2$ times the corresponding metric tensor component. ThereforeThis is Anti-de Sitter spacetime in static coordinates on anti-de Sitter spacetime, with curvature radius $1/\lambda$. The overall metric signature is opposite to the convention used in some descriptions of those coordinates; the geodesic constants below use the question's $(+,-,-,-)$ convention.

Conserved quantities and radial motion. In the equatorial plane, the geodesic Lagrangian is $L=A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2$. Its cyclic coordinates giveFor rest mass $m$, $mk=p_t$ is the conserved Killing energy and $mh=-p_{\varphi }$ is the axial angular momentum. Timelike normalization $L=1$ givesFor purely radial motion $h=0$, differentiating away from a turning point gives $\ddot{r}=-{\lambda }^{2}r$, and continuity extends that equation through the turning point. An outward launch from the origin at $\tau =0$ hasThis is the radial timelike geodesic in anti-de Sitter spacetime. Its first return to the origin isindependent of the outward launch speed for every nontrivial launch $k>1$. The coordinate $r$ is nonnegative, so this sine expression describes the excursion up to that return; subsequent passage through the origin requires changing the radial direction or using a signed diameter coordinate. The speed changes the amplitude, not the harmonic frequency.

Circular periods. The effective radial potential is $V_{\mathrm{eff}}=A(1+h^2/r^2)$. For a circle at $r_0$, $V_{\mathrm{eff}}^{\prime }=0$ gives $h^2={\lambda }^2{r}_0^4$, and normalization gives $k=A(r_0)$. Thus $\dot{t}=1$ and $\dot{\varphi }=\pm \lambda$. A stable circular timelike geodesic in anti-de Sitter spacetime therefore hasindependently of radius. This is the same isochronous behavior as the central harmonic radial equation. A static clock placed at the orbit's radius instead measures $\sqrt{1+{\lambda }^2{r}_0^2}\Delta t$, which does depend on radius; it is a different clock from the orbiting particle's proper time.

Use the $+---$ metric signature and abbreviate $f(r)=1-2\mu /r$, with $\mu =GM/c^2$ a length. Dots initially mean differentiation with respect to an affine parameter $\lambda$. The geodesic Lagrangian in the equatorial plane isThe planarity of geodesics in spherical symmetry permits this plane to be chosen. More explicitly, the full angular Euler-Lagrange equation is $\ddot{\theta }+2\dot{r}\dot{\theta }/r-\sin \theta \cos \theta {\dot{\varphi }}^2=0$, so initial data $\theta =\pi /2$, $\dot{\theta }=0$ preserve it.

Both $t$ and $\varphi$ are cyclic coordinates. Their Euler-Lagrange equations give conserved conjugate momenta $p_t=c^{2}f\dot{t}$ and $p_{\varphi }=-r^2\dot{\varphi }$. Defining constants $k,h$ therefore givesThe corresponding second-order geodesic equations areand variation of $r$ givesThe geodesic Lagrangian has no explicit $\lambda$ dependence. Its conserved Lagrangian energy function is ${\dot{x}}^{\alpha }\partial \mathcal{L}/\partial {\dot{x}}^{\alpha }-\mathcal{L}=\mathcal{L}$, since it is quadratic in the tangent components. Thus twice this function is a constant:This is the constant squared tangent norm of an affinely parametrized geodesic.

For a massive particle choose $\lambda =\tau$, its proper time, and obtain $\kappa =c^2$. For a photon, the tangent is null and $\kappa =0$. A general rescaling of a massive particle's affine parameter rescales a positive $\kappa$; the numerical value $c^2$ requires the proper time choice. With that choice, $h=L_z/m$ is signed specific angular momentum, of units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, and $k=E/(m{c}^2)$ is dimensionless conserved specific Killing energy. The time coordinate is normalized to proper time at infinity, so this $E$ is the particle's conserved energy referred to infinity. A photon's constants scale with the normalization of its null tangent; it has no rest frame or proper time parametrization.

Substitution of $\dot{t}=k/f$ and $\dot{\varphi }=h/r^2$ into the norm gives the general radial first integralFor a massive particle this becomesAdopting the usual half-normalization for the effective potential for timelike Schwarzschild geodesics, defineAll terms have units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$, or energy per unit mass. The first two terms are the Newtonian gravitational and centrifugal contributions; the last is the relativistic correction.

A circular orbit requires both $\dot{r}=0$ and $V_{\mathrm{eff}}^{\prime }(R)=0$. Differentiating at fixed $h$ givesThe radial first integral also givesFor the future-directed circular orbit at $R=4\mu$, these constants are $|h|=4\mu c$, $k=1$, and $f=1/2$. ThereforeSince Schwarzschild time agrees with the clock time of a static observer at infinity, the intended infinity-referred tangential orbital speed isThis uses the circumferential areal radius $R$ with the clock at infinity; it is not a local measurement made at the orbit. A local static observer's clock instead has $d{\tau }_{\mathrm{stat}}=\sqrt{f}dt$ and measures $v_{\mathrm{local}}=v_{\infty }/\sqrt{f}=\boxed{c/\sqrt{2}}$. These two operationally different speeds explain the significance of the phrase “at infinity”.

The effective-potential stability of a circular orbit follows fromConsequently the $R=4\mu$ orbit exists but is unstable to radial perturbations. The original plot below shows its position at a maximum, for $h=4\mu c$.

For a weak-field elliptic orbit, the relativistic correction makes the direction of closest approach advance after each radial cycle. The derivations under i and ii below giveper orbit to first order, where $a_{\mathrm{orb}}$ is the semi-major axis and $e$ the orbital eccentricity. For Mercury, the additional Schwarzschild perihelion precession is approximately 43 arcseconds per century, over and above the much larger Newtonian precession produced primarily by other planets. It is the unexplained residual after those effects have been included that provides the classic quantitative test of general relativity; one should not identify the whole observed precession with this relativistic correction.^[ref]

The labels i and ii in the original paper introduce optional orbit-equation aids, rather than separate mandatory subquestions. They are retained here for derivations of both aids and of the perihelion result.

Take $\alpha >0$, $t\ne 0$, and the negative-norm convention for a timelike curve used by the printed metric. Writing $v=dx/dt$ and $d{s}^2=-c^{2}d{\tau }^2$ givesSpatial translation is a Killing vector field. Equivalently, the geodesic Lagrangian has the cyclic coordinate $x$, so its conserved momentum isTherefore the correct first integral for this metric isThe first-integral assertion in the PDF is false. Its dimensionless square root requires $v/c$, and even after setting $c=1$ it is missing the factor $|t|$. For example, the nonvertical geodesics below have $v/\sqrt{1-v^2/c^2}$ proportional to $1/|t|$.

Set $b=\sqrt{\alpha }P/c$. Solving for the velocity, with increasing $t$ as the time orientation, gives $v=cb/\sqrt{t^2+b^2}$. The complete unparametrized family of timelike geodesics of a quadratic-time conformal metric isEach curve is restricted to a connected positive- or negative-$t$ patch; $t=0$ is excluded because the metric becomes degenerate. For $t\ne 0$, $|v|<c$, and the two independent constants are $x_c$ and $b$. If an explicit affine parametrization is desired,while the vertical family has $d\tau =|t|dt/\sqrt{\alpha }$. A reversal of the time orientation reverses the parameter without changing these paths.

Let a dot denote differentiation with respect to the five-dimensional affine parameter $\lambda =\tilde{u}$, and define $W={\dot{x}}^4+A_a{\dot{x}}^a$. Use the quadratic geodesic LagrangianThe assumed independence of $x^4$ makes it a cyclic coordinate, so the corresponding Euler-Lagrange equation isSince $\partial L/\partial {\dot{x}}^4=W$, this givesGeometrically, this is the conserved momentum associated with the fifth-direction Killing vector field ${\partial }_4$. It is the conserved fibre momentum underlying the Kaluza-Klein geodesic Lorentz force.

For radial motion, use the geodesic LagrangianThe Euler-Lagrange equation for $t$ gives $c^2\ddot{t}+a(da/dt){\dot{\chi }}^2=0$. Since $\chi$ is a cyclic coordinate, its conjugate momentum is constant. HenceFor a massive particle, take $\tau$ as proper time; the four-velocity normalization is $c^2{\dot{t}}^2-a^2{\dot{\chi }}^2=c^2$. For a photon, use an affine parameter rather than proper time, and the corresponding tangent is a null vector. ThereforeThe normalization of the null affine parameter remains arbitrary.

Write $f(r)=1-2\mu /r$. In the equatorial plane, the geodesic Lagrangian is $\mathcal{L}=\frac{1}{2}(f{c}^2{\dot{t}}^2-f^{-1}{\dot{r}}^2-r^2{\dot{\varphi }}^2)$. The cyclic coordinates $t$ and $\varphi$ give the conserved quantitiesThe squared tangent norm is constant along an affinely parametrized geodesic, since metric compatibility gives $d[g(\dot{X},\dot{X})]/d\tau =2g(D\dot{X}/d\tau ,\dot{X})=0$. HenceFor a massive particle parametrized by proper time, $E=c^2$; a photon has a null tangent, so $E=0$. This $E$ denotes the tangent norm, not the photon's physical energy. A future-directed exterior photon has $k>0$, proportional to its positive Killing energy. Its affine parameter can be rescaled, so there is no mass-shell normalization fixing $k$ to a particular positive value.

The Euler-Lagrange equations in cylindrical coordinates areThe last two follow because $\varphi$ and $z$ are cyclic coordinates. They give the conserved quantitiesSince the Lagrangian is autonomous, its energy function is conserved as well:The velocity-linear term cancels from this energy, but it still changes the equations of motion. A dimensionless logarithm means $\log (R/R_0)$ for some fixed reference length; replacing the printed $\log R$ by that expression changes the Lagrangian only by a total time derivative and shifts the numerical value of $k$, not the trajectory.

Time-translation symmetry conserves the energy $H$, and axial rotational symmetry makes $\varphi$ a cyclic coordinate, conservingThe radial Euler-Lagrange equation before elimination isInsert $\dot{\varphi }=J/(m{r}^2)-qB/(2m)$. The cross terms linear in $JqB$ cancel, leavingEquivalently use the reduced radial potentialThe last term is constant at fixed $J$ and has no radial derivative, but is part of the full conserved energy. Conservation of $J$ is stated in the rotationally symmetric gauge; the conserved mechanical combination can be written in terms of magnetic flux in a gauge-independent way.

The azimuth is a cyclic coordinate, so its canonical momentum is conserved:Initially the velocity is parallel to the axis in a field-free region, so $p_{\theta }=0$. Conservation of canonical angular momentum across a solenoid fringe therefore gives, inside the uniform field,This holds away from the coordinate singularity $\rho =0$; the Cartesian orbit will be smooth there.

Time independence also conserves the energy Hamiltonian,The velocity-linear magnetic term cancels in this Legendre transform. Thus $\gamma ={\gamma }_v=(1-v^2/c^2{)}^{-1/2}$ is constant, as expected because a static magnetic Lorentz force does no work. Mechanical angular momentum can change even though $p_{\theta }$ and the total speed are conserved.

Use the angular coordinate $\theta$ for the position on the ring. The kinetic energy is $T=m{a}^2{\dot{\theta }}^2/2$. The gravitational potential energy is constant because the ring is horizontal, and the ideal constraint force has no tangential component. Thus the Lagrangian has no dependence on $\theta$ itself: it is a cyclic coordinate.

The Euler-Lagrange equation givesEquivalently, the conserved conjugate momentum $p_{\theta }=m{a}^2\dot{\theta }$ is the vertical angular momentum, as required by rotational symmetry.

A common rotation changes neither the separation nor the spring energy. Its infinitesimal direction is $(1,1)$, exactly the zero-frequency normal mode. Thus the continuum of rotated equilibrium points requires that zero mode; a nonzero restoring frequency in that direction would contradict rotational symmetry.

The second mode has $I_1{\eta }_1+I_2{\eta }_2=0$, so it changes the relative angle while leaving the inertia-weighted common angle unchanged. It therefore carries zero perturbation in the conserved angular momentum associated with common motion. Indeed the coordinatesseparate the Lagrangian exactly:The cyclic coordinate $q$ is free, while $\delta$ has restoring or inverted curvature. This also explains why all members of either equilibrium family have the same frequencies. Reflection symmetry makes the potential even in $\delta$, and the stable oscillatory mode is consistent with time reversal in classical mechanics through its two opposite frequency signs.

Use the convention $R=L-{\sum }_i{p}_i{\dot{q}}_i$ for the eliminated cyclic coordinates, expressing their velocities in terms of their fixed conjugate momenta. Terms involving derivatives of those eliminated velocities cancel because $\partial L/\partial {\dot{q}}_i=p_i$. Thus the remaining coordinates obey the usual Euler-Lagrange equations for R. For a particle of fixed angular momentum $\ell$ in a central potential, this gives $R=m{\dot{r}}^2/2-V(r)-{\ell }^2/(2m{r}^2)$, rather than the expression obtained by direct substitution into L alone. An opposite overall sign convention is also used, with correspondingly signed mixed Hamiltonian equations.

A reduction of Lagrangian mechanics that fixes conserved momenta of cyclic coordinates while keeping an Euler-Lagrange description for the remaining coordinates. It is named for Edward Routh. The Routhian applies a partial Legendre transform to the eliminated velocities.

For a sphere with metric $a^2(d{\theta }^2+{\sin }^2\theta d{\varphi }^2)$, physical arc length $s$ gives $h={\sin }^2\theta d\varphi /ds$. The cyclic coordinate $\varphi$ proves conservation of $a^{2}h$, while unit-speed normalization givesThis azimuthal form requires $h\ne 0$. The solutions are great circles; meridians have $h=0$ and require another parameter.