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Take points $p=(3,0,4)$ and $q=(-2,3,3)$ with directions $u=(1,3,-1)$ and $v=(0,1,-1)$. A common normal is $u\times v=(-2,1,1)$, of length $\sqrt{6}$. The planes through the two lines with this common normal have separationThis is a lower bound for the distance between skew lines. It is attained at $s=-1$ and $t=-4$: the two points are $(2,-3,5)$ and $(-2,-1,7)$, whose difference $(4,-2,-2)$ is perpendicular to both directions and has length $2\sqrt{6}$. Thus the shortest distance is $\boxed{2\sqrt{6}}$.

The plane through $a_i$ has the constant normal coordinate $r\cdot \hat{n}=a_i\cdot \hat{n}$. Since $\hat{n}$ is a unit vector, the separation of these normal coordinates is a length. Every displacement joining the planes has this fixed normal component, so its length is at least its absolute value; a displacement parallel to $\hat{n}$ attains that bound. Hence the distance between parallel planes isThe absolute value, present in the PDF, is necessary because distance is nonnegative.

The first coordinate of the vector triple product isThe other two coordinates follow by cyclically permuting the indices, proving the identity.

A point on the line has the form $r=b+sm$. Substituting in the plane equation gives $s(m\cdot n)=(a-b)\cdot n$. When $m\cdot n\ne 0$ the parameter, and therefore the intersection point, is unique. The vector triple product gives $n\times (b\times m)=(n\cdot m)b-(n\cdot b)m$, so the unique intersection isFor the parallel case $m\cdot n=0$, the line is entirely in the plane if $(b-a)\cdot n=0$, and otherwise has no intersection. As usual, a genuine line and plane require nonzero direction and normal vectors. This is the line-plane intersection criterion.

The two displacement vectors $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$ lie in the required plane. Because the three points are non-collinear, their cross productis nonzero and is a normal vector to that plane. A point with position vector $\mathbf{r}$ lies in the plane exactly whenExpanding the cross product givesThe terms $\mathbf{a}\cdot (\mathbf{a}\times \mathbf{b})$ and $\mathbf{a}\cdot (\mathbf{c}\times \mathbf{a})$ vanish, so the equation of a plane through three points becomesNon-collinearity is important: it ensures this is a genuine plane equation with nonzero normal, rather than the vacuous equality $0=0$.

For the specified coordinates,Their sum is $(3,3,6)$, with Euclidean norm $3\sqrt{6}$. Consequently either orientation of the unit normal is valid:As a check, the scalar triple product is $12$, so the plane equation simplifies to $x+y+2z=4$, which all three supplied points satisfy. The normal direction is $(1,1,2)$.

The chosen unit normal points toward the plane from the origin, so the plane equation isTo locate the circle centre, project $\mathbf{p}$ onto this plane. PutThen $\mathbf{n}\cdot \mathbf{h}=d$, so $\mathbf{h}$ lies in the plane. For any other point $\mathbf{r}$ of the plane, $\mathbf{r}-\mathbf{h}$ is perpendicular to $\mathbf{n}$. The Pythagorean theorem therefore givesRestricting the sphere equation to the plane consequently gives a circle centred at $\mathbf{h}$. Thus $\mathbf{m}=\mathbf{h}$, proving that the displacement of its centre is purely normal:The same Pythagorean theorem gives $q^2={\rho }^2+{\gamma }^2$, and henceThe sign of $\gamma$ is fixed by which side of the plane contains $\mathbf{p}$; $q$ and $\rho$ alone determine only its magnitude. Substituting the signed offset yields the sphere-plane intersection radiusA genuine circle requires $|d-\mathbf{n}\cdot \mathbf{p}|<q$; equality gives tangency and $\rho =0$, while a larger distance gives no intersection. The signed distance from a point to a plane in the direction $\mathbf{n}$ is $\mathbf{n}\cdot \mathbf{p}-d=-\gamma$; the ordinary distance from a point to a plane is its absolute value.

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 $K=-1$.

For unit vectors $u,v$ and a random normal $a$ with independent standard normal distribution coordinates,The Gaussian distribution is invariant under orthogonal transformations. Projecting onto the plane spanned by $u,v$ therefore gives a uniformly distributed direction. If the angle between $u,v$ is $\theta$, the sign-disagreement directions form two sectors of total angle $2\theta$ out of $2\pi$. The endpoint cases $u=v$ and $u=-v$ give probabilities zero and one directly.

In an unsoftened plane with number surface density $n_2$ and relative speed $u$, an encounter strip has area $2udtdb$. Thus the velocity diffusion rate is $8G^2{m}^2{n}_2/u\int db/b^2$, dominated by the small-impact cutoff. Setting this cutoff to $Gm/u^2$ and requiring a change of order $u^2$ gives $t_{\mathrm{rel}}\sim u/(Gm{n}_2)$. With $u=\lambda v$, $n_2\sim N/R^2$ and $v^2\sim GNm/R$, this is of order $\lambda R/v$, independent of $N$ at fixed total mass. The point-particle, zero-thickness limit is singular; fixed softening or nonzero thickness changes it. Numerical coefficients are not controlled by extending weak kicks down to order-one deflections.

Orient the plane $\mathbf{n}\cdot \mathbf{r}=d$ with a unit normal $\mathbf{n}$. The signed distance of a point $\mathbf{p}$ from the plane isIts absolute value is the ordinary distance from a point to a plane. The closest point in the plane is $\mathbf{p}-s\mathbf{n}$, so the displacement from the point to its projection is $-s\mathbf{n}$. Reversing the normal and $d$ reverses $s$ without changing the geometric plane or the unsigned distance. A sphere of radius $q$ centred at $\mathbf{p}$ has sphere-plane intersection radius $\sqrt{q^2-s^2}$ when $|s|<q$.