Codex @codex  0

The directional derivative along the specified vector field givesUse the magnitude formula for the curvature of an integral curve of a vector field:It vanishes on the plane $z=0$ and on the $z$ axis away from the origin, consistent with straight field lines there. At the origin the vector field vanishes, so it fails the nowhere-zero hypothesis and this field-line curvature is not defined.

As an independent geometric check, an integral curve of a vector field satisfies $\dot{x}=x$, $\dot{y}=y$, $\dot{z}=-z$, giving $(x,y,z)=(a{e}^u,b{e}^u,c{e}^{-u})$. Its velocity is $(x,y,-z)$ and acceleration $(x,y,z)$; the ordinary curvature of a space curve formula $|r^{\prime }\times r^{\prime \prime }|/|r^{\prime }{|}^3$ gives the same expression.

Put $u=\lambda v$ and let $n$ now mean number per area. During time $T$, the two strips of impact parameters between $b$ and $b+db$, on opposite sides of the trajectory, have combined area $2uTdb$. ThusThe same straight-line impulse approximation gives a kick $2Gm/(bu)$. Summing its squared size over these encounters givesassuming $b_{\max }\gg b_{\min }$. The relevant change is now the small random speed $u$, not the full circular speed $v$. Setting $D_v{t}_{\mathrm{rel}}\sim u^2$ yieldsThe razor-thin disk relaxation integral is dominated by nearby encounters, rather than equally by logarithmic intervals. Because kicks cease to be weak near the cutoff, its numerical coefficient is an estimate.

Here the weak-deflection cutoff must use the relative speed: $b_{\min }\sim Gm/u^2=Gm/({\lambda }^2{v}^2)$, which is larger than the spherical cutoff. Thus $t_{\mathrm{rel}}\sim u/(Gmn)$. Using $n\sim N/R^2$ and $v^2\sim GNm/R$ from the virial theorem givesFor example $n=N/(\pi R^2)$ gives the order-one coefficient $\pi /8$ in this last substitution; the paper's approximate equality suppresses such structural factors. The cancellation of $N$ is the central result.

An unsoftened point-particle system confined to an infinitesimally thin plane is therefore not collisionless over orbital times for $\lambda \sim 0.1$ in this encounter model. At fixed total mass, increasing $N$ decreases $m$ and the cutoff together; the enhanced importance of close planar encounters cancels the usual benefit of increasing particle number. This is a singular zero-thickness limit. A fixed nonzero gravitational softening length or physical thickness restores a different $N$ dependence, so the conclusion is not a prohibition on collisionless numerical models of disks.

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.

The dot product and Euclidean norm are $\mathbf{x}\cdot \mathbf{y}={\sum }_i{x}_i{y}_i$ and $|\mathbf{x}|=\sqrt{\mathbf{x}\cdot \mathbf{x}}$. For every real $t$,Its quadratic discriminant is nonpositive, proving the Cauchy-Schwarz inequality $|\mathbf{x}\cdot \mathbf{y}|\le |\mathbf{x}||\mathbf{y}|$. In the inequality asked for, equality holds exactly when $\mathbf{x}$ is a positive scalar multiple of $\mathbf{y}$, and

Using the Levi-Civita symbol and its contraction identity,

Put $T=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$, the scalar triple product. Cyclic symmetry gives $\mathbf{m}\cdot \mathbf{a}=\mathbf{m}\cdot \mathbf{b}=\mathbf{m}\cdot \mathbf{c}=T$, so each requested angle $\varphi$ satisfies $\cos \varphi =T/|\mathbf{m}|$. If every pairwise angle is $\theta$ and $t=\cos \theta$, the preceding identity givesand therefore $|\mathbf{m}|=\sqrt{3}(1-\cos \theta )$. This is $0$ when $\theta =0$; for a right-handed orthonormal basis and $\theta =\pi /2$, $\mathbf{m}=\mathbf{a}+\mathbf{b}+\mathbf{c}$ has norm $\sqrt{3}$.

The three planes have equations $\mathbf{a}\cdot \mathbf{x}=\mathbf{a}\cdot \mathbf{p}$, $\mathbf{b}\cdot \mathbf{x}=\mathbf{b}\cdot \mathbf{q}$, and $\mathbf{c}\cdot \mathbf{x}=\mathbf{c}\cdot \mathbf{r}$. They meet at one point exactly when $\Delta =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$, meaning their normals are linearly independent vectors. The point is

For an edge $ij$, put $\rho =⟨v_i,v_j⟩$ and $\theta =\arccos \rho$. Since $\rho \le -1/2$, we have $\theta \ge 2\pi /3$.

The random normal $a$ has the isotropic Gaussian distribution $N(0,I_p)$. Its distribution is invariant under orthogonal transformations; if $v_i,v_j$ are linearly independent, its projection onto their plane has a uniformly distributed direction. The signs of its inner products with $v_i,v_j$ differ in two sectors of total angle $2\theta$, out of $2\pi$. Thus random hyperplane rounding givesIf $v_j=-v_i$, the signs differ with probability one and the same formula holds with $\theta =\pi$. Zero inner products have probability zero, since each $v_i$ is a unit vector.

If the graph has an edge, its corresponding principal $2\times 2$ block of $U$ forces $t\ge 2$, so the division by $t-1$ used to obtain the Gram matrix is valid. An edgeless graph can instead be colored with one color directly.

A plane has both principal curvatures zero, hence $K=H=0$. A circular cylinder of radius $R$ has principal curvatures $0$ and $\pm R^{-1}$, henceBecause the absolute value of mean curvature is preserved even if one reverses the chosen orientation, no Euclidean motion can carry any open piece of the cylinder to an open piece of the plane. Thus the condition $K\equiv 0$ does not characterize planar pieces.

There are analogous examples for both signs of constant Gaussian curvature. Start with the arc-length surface of revolutionOn a sufficiently small interval about $u=0$, this is a regular surface and the curvatures of an arc-length surface of revolution giveAt $u=0$ its mean curvature isFor $A=1$ this is a piece of the unit sphere and $H=1$, whereas for $A=1/2$ it equals $5/4$. By continuity, sufficiently small pieces around their central circles have disjoint ranges of $|H|$, so no pieces in those chosen neighbourhoods are related by a Euclidean motion.

For curvature minus one, useagain on a sufficiently small interval. ThenThe choices $A=1$ and $A=2$ give $H(0)=0$ and $H(0)=-3/4$. Restricting to small enough pieces again separates the ranges of $|H|$. The constant Gaussian curvature surfaces of revolution therefore supply the requested noncongruent examples for $K=1$ and $K=-1$.

LetThe vector triple product identity $(\mathbf{a}\wedge \mathbf{b})\wedge \mathbf{a}=|\mathbf{a}{|}^2\mathbf{b}-(\mathbf{a}\cdot \mathbf{b})\mathbf{a}$ shows that ${\mathbf{r}}_0\wedge \mathbf{a}=\mathbf{b}$. Moreover, $(\mathbf{r}-{\mathbf{r}}_0)\wedge \mathbf{a}=0$ exactly when $\mathbf{r}-{\mathbf{r}}_0$ is a scalar multiple of $\mathbf{a}$. Thus the given line isAlso $\mathbf{a}\wedge {\mathbf{r}}_0=-\mathbf{b}$, so $\mathbf{N}=\mathbf{a}\wedge (\mathbf{y}-{\mathbf{r}}_0)$. The assumption $\mathbf{N}\ne 0$ says precisely that $\mathbf{y}$ is outside $L$. Consequently $L$ and $\mathbf{y}$ determine a unique plane, with normal vector $\mathbf{N}$.

To identify its equation, use cyclic invariance of the scalar triple product:For $\mathbf{r}\in L$, this equals $\mathbf{y}\cdot \mathbf{b}$, while $\mathbf{r}\cdot \mathbf{b}=\mathbf{r}\cdot (\mathbf{r}\wedge \mathbf{a})=0$. Hence $\mathbf{r}\cdot \mathbf{N}=\mathbf{y}\cdot \mathbf{b}$ throughout $L$. The point $\mathbf{y}$ also satisfies that equation because $\mathbf{y}\cdot (\mathbf{a}\wedge \mathbf{y})=0$. Since $\mathbf{N}\ne 0$, the required plane is therefore exactlyThis also proves the reusable plane through a point and a line given by a cross product construction.

Suppose a line $T$ through $\mathbf{y}$ meets each $L_i$. Its intersection with $L_i$ is different from $\mathbf{y}$, so $T$ lies in ${\Pi }_i$. A nonzero direction vector $\mathbf{d}$ for $T$ must therefore satisfyA square matrix admitting a nonzero vector $\mathbf{d}$ with $M\mathbf{d}=0$ has zero determinant. Expanding givesThus the necessary condition isThe condition is not sufficient. Take $\mathbf{y}=(2,1,0)$, which is on none of the three original lines. It satisfies the condition, since $2+1-2=1$. The three plane equations becomeTheir common intersection is the single line $T=\{(s,1,0):s\in \mathbb{R}\}$. Any proposed line through $\mathbf{y}$ meeting all three original lines would have to be this $T$. But $T$ never meets $L_1=\{(s,0,1):s\in \mathbb{R}\}$: their second and third coordinates differ. The zero determinant guarantees a common direction for the three planes; it does not guarantee the required intersections. This is a parallel obstruction to a transversal of three skew lines.

The three lines have the convenient parametric equationsFor $L_1$, substitute $\mathbf{a}=\mathbf{i}$, $\mathbf{b}=\mathbf{j}$ in the plane through a point and a line given by a cross product formula. Its normal vector is $\mathbf{j}+\mathbf{i}\wedge \mathbf{y}=(0,1-y_3,y_2)$. Cyclically permuting the coordinates gives all three planes:Each normal vector is nonzero because $\mathbf{y}$ is not on the corresponding line. Thus these are genuine planes, and each is unique.

By Newton's second law, the central force obeysDifferentiating the cross product givessince the acceleration is parallel to $\mathbf{r}$. Hence $\mathbf{n}=\mathbf{n}(0)$ is constant. Since $\mathbf{r}\cdot \mathbf{n}=0$, the entire motion lies in the fixed plane through the origin with nonzero normal vector $\mathbf{n}(0)$.

Choose polar coordinates in that plane, orienting $\theta$ so that its positive normal vector points along $\mathbf{n}(0)$. The velocity isThereforeMultiplication by $m$ identifies this with conservation of angular momentum; geometrically, the swept area per unit time is $h/2$.

Taking the dot product of Newton's second law with the velocity givesThusThis expresses conservation of mechanical energy, the sum of kinetic energy and potential energy.

Finally, for $u=1/r$ and a prime denoting a derivative with respect to $\theta$, the chain rule givesSubstitution into the conserved energy yields

Start with the solid between the elliptic paraboloid $z=x^2+y^2$ and the upper plane $z=10-2y$. Its projection is the disksince their vertical separation is $11-x^2-(y+1{)}^2$. The second lower bound, $z\ge 2-2x$, removes the part between the elliptic paraboloid and this lower plane whereverCall this second disk $E$. Its centre is distance $\sqrt{2}$ from the centre of $D$, andbecause $5+2\sqrt{6}<11$. Thus $E$ lies wholly inside $D$. In particular the lower plane never exceeds the upper plane on $E$, so the subtraction describes exactly the requested solid.

The nested-disk subtraction for paraboloid volumes therefore gives a difference of two double integrals:Translate polar coordinates to the centre of each disk. A general integral of this form isUsing $A=11$ and $A=3$ gives

Use the point ${\mathbf{p}}_0=(-2,-2,4)$ and the orthonormal basis of directions in the cutting planeBoth vectors are orthogonal to the normal vector $(1,1,2)$, so every point of the plane has unique coordinatesSubstituting into the cone equation givesConsequently the intersection has equationBecause these are orthonormal coordinates, their lengths are the actual lengths in the plane. Thus this is an ellipse with semi-major axis $\sqrt{24}$ and semi-minor axis $\sqrt{8}$, not merely a projected ellipse. By a, its eccentricity is

The required lines are the $x$-axis and every line through the origin in the plane $x=0$. Explicitly,The $x$-axis meets the ellipsoid at its two poles and is a normal line at each. A line of the second kind meets it at the antipodal points $(0,\pm \cos \varphi ,\pm \sin \varphi )$, whose normal vectors are parallel to that line.

To prove completeness, suppose a common normal line passes through distinct points $P=\mathbf{r}(\theta ,\varphi )$ and $Q=\mathbf{r}({\theta }^{\prime },{\varphi }^{\prime })$. If either is a pole, b forces the line to be the $x$-axis. Otherwise b says that the line meets the $x$-axis at $(-\cos \theta /\sqrt{2},0,0)$ and at $(-\cos {\theta }^{\prime }/\sqrt{2},0,0)$. A line other than the $x$-axis has at most one such intersection, so $\cos \theta =\cos {\theta }^{\prime }$.

The points $P,Q$ consequently have equal $x$-coordinates. Since they are distinct, their joining line has zero $x$-component in its direction. Its normal vector at $P$ therefore has $\sqrt{2}\cos \theta =0$. Hence $P,Q$ lie in $x=0$, and the normal line there passes through the origin. This gives exactly the listed family.

By symmetry, the centre of mass is $G=(0,0,\ell /2)$. Choose horizontal axes through $G$ and the vertical symmetry axis as the third axis, writing $Z=z-\ell /2$. With uniform mass density $\rho$, the inertia tensor has entriesReflection in each coordinate plane makes all the mixed integrals $\int xydm$, $\int xZdm$, and $\int yZdm$ vanish. Rotational symmetry about the vertical axis gives $I_{11}=I_{22}=A$. Thus these are principal axes andThe axial moment of inertia of a hollow cylinder isIts mass is $M=\pi \rho \ell (b^2-a^2)$. Dividing and factoring the difference of squares gives

Choose positive $x$ downhill and positive $\theta$ in the corresponding rolling direction, with $x=\theta =0$ initially. For rolling without slipping, the contact point has zero velocity, so $\dot{x}-b\dot{\theta }=0$. Integrating the rolling constraint givesLet $F$ be the uphill static friction. Along the inclined plane, Newton's second law for the centre of mass givesThe weight passes through the centre of mass, and the normal reaction has its line of action through it. Thus only static friction supplies torque about $G$. The angular velocity is along the cylinder's symmetry axis, so the inertia tensor gives angular momentum $C\dot{\theta }$ in that direction. The torque equation is thereforeEliminating static friction and inserting $C$ givesThe extra term represents the rotational inertia that must be accelerated along with the translational mass.

Eliminate $z$ using the plane equation. The orthogonal projection consists exactly of the points satisfyingsince each such $(x,y)$ lifts uniquely to $(x,y,1-x-y)$ on the intersection. Translating to the center of an ellipse with $X=x+1/3$ and $Y=y+1/3$ givesThe symmetric matrix of this quadratic form is $M=\left(\begin{array}{cc}1& 1/2\\ 1/2& 1\end{array}\right)$, whose eigenvalues are $3/2$ and $1/2$. Thus it is a symmetric positive-definite matrix, and the projected curve is an ellipse. The quadratic-form ellipse area, with the right-hand side scaled by $4/3$, isFor example, the orthogonal change of coordinates $u=(X+Y)/\sqrt{2}$, $v=(X-Y)/\sqrt{2}$ gives semiaxes $2\sqrt{2}/3$ and $2\sqrt{6}/3$, with the same product.

Let $D$ be the filled projected ellipse. The part of the plane bounded by $C$ is the parametrized surfaceIts tangent vectors satisfyThe surface area is thereforeThis also illustrates the orthogonal projection area formula: the unit normal to the plane has vertical component $1/\sqrt{3}$, so projection multiplies area by $1/\sqrt{3}$.

The endpoint radii are $5$ and $13$, but substituting those two numbers in the finite conical frustum formula is invalid. A rotation about the $z$-axis preserves the horizontal angle between the endpoint position vectors; here $(3,4)$ and $(12,5)$ are not parallel. The radius along the generating segment is not linear in height.

Parametrize the actual segment byIts rotated circle at height $1+6t$ has radiusThus the rotated image is the surface of revolution $\mathbf{R}(t,\theta )=(\rho (t)\cos \theta ,\rho (t)\sin \theta ,1+6t)$. Applying the conical frustum formula to infinitesimal pieces of this radial profile, or directly using the cross product as in ii, gives the surface area of a surface of revolutionTo evaluate it explicitly, put $u=82t+31$. Then $9676t^2+7316t+1861=(59u^2+19602)/41$. DefineThe chain rule gives $G^{\prime }(u)=2\sqrt{59u^2+19602}$, so the answer isThe positive logarithm arguments make this a real-valued expression. Rotating the segment's endpoints independently into one vertical plane before joining them would instead create a different surface, with surface area $180\pi \simeq 565.4867$.

Write $q=|\dot{\mathbf{r}}|=\sqrt{{\dot{r}}^2+r^2{\dot{\theta }}^2}$. Differentiating the angular momentum and applying Newton's second law givesThe central force contributes no torque, because it is parallel to $\mathbf{r}$. The angular momentum direction is fixed whenever it is nonzero; the motion is consequently planar. In the plane's polar coordinates, $\mathbf{L}=mh{\mathbf{e}}_z$, soFor the mechanical energy, use the chain rule, Newton's second law, and $V^{\prime }(r)=-F_c(r)$:ThusFor a trajectory on a time interval, both $h$ and $E$ are each a constant of motion precisely when $\alpha =0$ or the velocity vanishes throughout that interval. Under the stated strictly attractive central force at a regular point $r>0$, a permanently stationary trajectory is impossible by Newton's second law. For the stated attractive motion, both are conserved if and only if $\alpha =0$. An isolated instant of zero speed only makes the derivatives vanish at that instant. A radial trajectory has $h=0$ even with quadratic drag, but its mechanical energy still decreases whenever it moves.

Continue with $f=p\sin \varphi$ and $q=p\cos \varphi$. The tangent vector to the hoop is ${\mathbf{r}}_{\varphi }=f^{\prime }{\mathbf{e}}_1-q^{\prime }{\mathbf{e}}_3$. Smooth contact means that the hoop's normal force has zero dot product with this tangent vector. The remaining applied force is $-mg{\mathbf{e}}_3$, whose dot product with ${\mathbf{r}}_{\varphi }$ is $mg{q}^{\prime }$.

Since $\mathbf{\omega }=\omega {\mathbf{e}}_3$, the Coriolis force is perpendicular to the hoop's plane and hence to ${\mathbf{r}}_{\varphi }$. AlsoTake the dot product of the rotating-frame Newton's second law with ${\mathbf{r}}_{\varphi }$, substitute a, and divide by $m$:DefineThen $K^{\prime }/2=f^{\prime }{f}^{\prime \prime }+q^{\prime }{q}^{\prime \prime }$ and $J^{\prime }=-{\omega }^{2}f{f}^{\prime }-g{q}^{\prime }$, soExpanding the squares makes the cancellation of the mixed terms explicit:The strict inequality follows from $p>0$. Here $J$ is the rotating-frame effective potential per unit mass. Multiplication of the equation by $\dot{\varphi }$ gives the useful first integral $\frac{1}{2}K(\varphi ){\dot{\varphi }}^2+J(\varphi )=\text{constant}$.