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Use units $c=1$, signature $(+---)$, and the weak-field metric $g_{ab}={\eta }_{ab}+h_{ab}$. Define the trace-reversed metric perturbation by ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$, with $h={\eta }^{ab}{h}_{ab}$. In the Lorenz gauge in linearized gravity, ${\partial }^a{\bar{h}}_{ab}=0$, the sign convention of Q2 givesFor completeness, the linearized Ricci tensor in this convention isSubstituting the Lorenz condition ${\partial }_c{h}^c{}_b={\partial }_{b}h/2$ and trace reversing gives the stated Einstein tensor. For a stationary source, $\square =-{\nabla }^2$, so ${\nabla }^2{\bar{h}}_{ab}=16\pi G{T}_{ab}$.

Keep the monopole terms of order $ϵ=GM/R$ and the leading term linear in the rotation rate. Since $\Omega R=O({ϵ}^{1/2})$, the latter metric term is of order $ϵ\Omega R=O({ϵ}^{3/2})$ on the shell. Thus it would disappear from a strictly order-$ϵ$ truncation; the requested approximation must retain the leading rotation correction as well. Terms $T_{ij}=\rho v_i{v}_j$ generate metric corrections of order $ϵ(\Omega R{)}^2=O({ϵ}^2)$ and are neglected. Corrections to four-velocity normalization and the supporting shell stresses affect the omitted higher orders. To retained order,The minus sign in $T_{0i}$ comes from lowering its spatial index; using $T^{0i}$ in its place would reverse the dragging direction.

The scalar Poisson solution regular at the centre and vanishing at spatial infinity is ${\bar{h}}_{00}=-4GM/r$ outside the shell and ${\bar{h}}_{00}=-4GM/R=-4ϵ$ inside. Continuity and the derivative jump ${\bar{h}}_{00}^{\prime }(R^{+})-{\bar{h}}_{00}^{\prime }(R^{-})=4GM/R^2$ verify the density $M\delta (r-R)/(4\pi R^2)$. Also ${\bar{h}}_{ij}=0$ at this order. Undoing trace reversal givesComparing with the scalar parts of the interior line element yields $A=-ϵ$ and $C=ϵ$.

For the mass-current contribution writeEach nonzero angular component is a degree-one spherical harmonic. The supplied Laplacian eigenvalue therefore givesAway from the shell, the two radial solutions are $r$ and $r^{-2}$. Regularity at $r=0$ and decay at infinity select $F_{\mathrm{in}}=ar$ and $F_{\mathrm{out}}=b/r^2$. Continuity at $R$ requires $b=a{R}^3$, and integrating the differential equation across the shell givesThus $a=4GM\Omega /(3R)=\omega$, with $F_{\mathrm{in}}=\omega r$. Trace reversal leaves mixed components unchanged, so $h_{0i}=\omega (\hat{\mathbf{z}}\times \mathbf{x}{)}_i=\omega (-y,x,0{)}_i$. The paper writes $g_{0i}=-B_i$, giving the frame dragging inside a slowly rotating spherical shellThis dipole matching is valid throughout $r<R$; it does not assume the observation point is very close to the centre. The stationary current is divergence-free, so this solution also satisfies the Lorenz gauge. The exterior result is $h_{0i}=2G(\mathbf{J}\times \mathbf{x}{)}_i/r^3$, where the thin shell has $\mathbf{J}=(2/3)M{R}^2\mathbf{\Omega }$, providing a check on the coefficient.

To identify the freely falling frames inside, let ${\mathcal{R}}_z(\theta )$ be the ordinary spatial rotation by angle $\theta$. Define the inertial coordinates inside a slowly rotating spherical shell byDifferentiating and dropping ${ϵ}^2$, products beyond the retained rotation order, and ${\omega }^2$ giveswhich is precisely the interior metric. Thus the retained interior curvature is zero, despite its nontrivial relation to the nonrotating coordinates anchored at infinity. Timelike geodesics in these local inertial coordinates are straight lines $\mathbf{X}={\mathbf{X}}_0+\mathbf{V}T$, with $|\mathbf{V}|<1$. In the original coordinates this isunderstood to the same perturbative order. Their coordinate equation is $d^2\mathbf{x}/d{t}^2=2\mathbf{\omega }\times d\mathbf{x}/dt$ to first rotation order. There is no interior Newtonian acceleration from the constant potential; the velocity-dependent term is the apparent Coriolis term relative to the locally dragged inertial axes.

A freely transported gyroscope at rest in the interior to the retained order has constant spatial direction in the $\mathbf{X}$ frame, hence its direction in the original axes obeys $d\mathbf{S}/dt=\mathbf{\omega }\times \mathbf{S}$. Local inertial axes precess in the same sense as the shell, at $\omega =(4ϵ/3)\Omega$ relative to distant nonrotating axes. This is the interior Lense-Thirring precession. For a weak shell the dragging is only a small fraction of its angular velocity, not perfect corotation. It is compatible with the equivalence principle: rotation relative to distant axes is a comparison between frames, and local flatness at the retained order does not make that comparison vanish. Quadratic rotation terms, which would include centrifugal effects and additional curvature, are outside the requested leading approximation.

Both metrics have curvature radius one and signature $(+-)$. Their global topology must be specified as well as their local metric. The usual two-dimensional de Sitter spacetime is the hyperboloid $(X^0{)}^2-(X^1{)}^2-(X^2{)}^2=-1$ in three-dimensional Minkowski spacetime, parametrized byThus $t\in \mathbb{R}$ and $\chi$ is periodic modulo $2\pi$, for example $0\le \chi <2\pi$. Spatial slices are circles. Unwrapping $\chi$ instead gives the universal cover, not the usual closed spatial model.

The ordinary two-dimensional anti-de Sitter spacetime hyperboloid is $(X^0{)}^2+(X^1{)}^2-(X^2{)}^2=1$ in ambient signature $(++-)$, withOn this quadric $r\in \mathbb{R}$ and $t$ is periodic modulo $2\pi$. The closed time circles are timelike, so it contains closed timelike curves. The usual physically causal AdS model is its universal cover, for which $r,t\in \mathbb{R}$, with time no longer identified. The signed radial coordinate has two spatial ends; restricting to $r\ge 0$ would retain only half of this global model. The metric by itself cannot distinguish periodic time from the cover. The scalar curvature is $+2$ for de Sitter and $-2$ for AdS in the Ricci convention of Q2. Both are maximally symmetric and have no curvature singularity; these toy two-dimensional metrics illustrate causal geometry rather than the ordinary four-dimensional matter field equations.

For the conformal cylinder of two-dimensional de Sitter spacetime, introduceThenThe conformal boundaries $\eta =\pm \pi /2$ are spacelike circles, representing past and future infinity. Null geodesics are straight lines $\chi ={\chi }_0\pm (\eta -{\eta }_0)$, with the periodic spatial identification. The finite total conformal-time interval explains the cosmological horizons: light cannot complete a full trip around the circle during the entire history.

For the conformal strip of two-dimensional anti-de Sitter spacetime, useso on the universal coverIts two conformal boundaries at $\psi =\pm \pi /2$ are timelike. Null rays are $t\pm \psi =\text{constant}$, and the strip extends for arbitrarily long coordinate time. The distances in these conformal diagrams are not physical lengths: a conformal boundary may be reached in finite coordinate time while lying infinitely far away in physical affine parameter.

The sides of the de Sitter diagram are identified; its top and bottom are spacelike infinities. The AdS diagram shows a finite window of an infinite time strip, with no identification of top and bottom. Null rays have slope one in both diagrams. The blue curves are timelike geodesics, while the colored de Sitter lines bound the central observer's causal domains.

The geodesics of two-dimensional de Sitter spacetime can be derived directly. Let an overdot denote an affine parameter normalized so that the tangent norm is $\kappa =1,0,-1$ for timelike, null and spacelike geodesics. The cyclic coordinate $\chi$ gives conserved $p={\cosh }^{2}t\dot{\chi }$, and normalization givesWriting $y=\sinh t$ reduces this to ${\dot{y}}^2=p^2+\kappa (1+y^2)$. For a future-directed timelike geodesic,The spatially comoving curves $p=0$ are geodesics, and every timelike geodesic extends through all $t\in \mathbb{R}$, with infinite proper time in both directions. Initially comoving nearby observers have physical separation proportional to $\cosh t$, displaying de Sitter's tidal defocusing. Relative separation can initially decrease for other initial velocities; not every pair is monotonically separating.

For a nontrivial null geodesic, $p\ne 0$ and $\sinh t=\pm |p|\lambda +\text{constant}$. Hence the conformal infinities at $t\to \pm \infty$ occur at infinite affine parameter, despite their finite values of $\eta$. For a spacelike geodesic, $|p|\ge 1$ and $y=\sqrt{p^2-1}\sin (s-s_0)$. Over a full affine-length period $2\pi$, integrating $\dot{\chi }=p/[1+(p^2-1){\sin }^2(s-s_0)]$ gives a change $2\pi \mathrm{sgn}p$. Thus the spacelike geodesics close on the usual cylinder. The special case $|p|=1$ is the equatorial circle $t=0$. These closed spacelike curves do not threaten causality.

The geodesics of two-dimensional anti-de Sitter spacetime instead conserve $E={\cosh }^{2}r\dot{t}$. Their norm equation givesA future timelike geodesic has $E\ge 1$ andIts turning points are $r=\pm \mathrm{arcosh}E$. Timelike geodesics oscillate rather than escaping to the spatial boundary. For $E=1$, the central curve $r=0$ is geodesic. Integrating $\dot{t}=E/[1+(E^2-1){\sin }^2(\tau -{\tau }_0)]$ shows that a full proper-time period $2\pi$ advances coordinate time by $2\pi$. They close on the periodically identified hyperboloid but remain complete nonclosed curves on the universal cover. Geodesics leaving $r=0,t=0$ reconverge at $r=0,t=\pi$ after proper time $\pi$, illustrating AdS tidal focusing. A static observer at nonzero fixed $r$ is accelerated, not one of these oscillating free trajectories.

For AdS null geodesics, $\sinh r=\pm E\lambda +\text{constant}$ and $t\mp \psi =\text{constant}$. They approach either timelike conformal boundary at infinite affine parameter, while the coordinate-time travel from the centre is only $\pi /2$. A spacelike geodesic has $\sinh r=\sqrt{E^2+1}\sinh (s-s_0)$ and connects the two spatial ends with infinite affine length. Both global models are geodesically complete. Reflecting a signal at the AdS conformal boundary is a choice of boundary condition for fields; it is not an automatic continuation of a geodesic at a finite affine endpoint.

For the complete de Sitter observer at $\chi =0$, let $d(\chi ,0)\in [0,\pi ]$ be the shortest angular distance. An event at $(\eta ,\chi )$ can send a signal reaching the observer at a finite future time exactly whenIt can receive a signal sent by the observer at a finite past time exactly when $d(\chi ,0)<\eta +\pi /2$. Equality gives the future and past observer horizons in two-dimensional de Sitter spacetime; their null branches are respectively $\chi =\pm (\pi /2-\eta )$ and $\chi =\pm (\eta +\pi /2)$ on the displayed cylinder. Their intersection is the central static patch, $d(\chi ,0)+|\eta |<\pi /2$. All complete timelike geodesic observers have analogous horizons by de Sitter symmetry. These are observer-dependent cosmological horizons, not singularities of the global coordinates or Cauchy horizons.

In contrast, the complete central AdS observer can exchange signals with every event at finite $r$: the conformal distance $|\psi |$ is finite and the observer exists for all $t\in \mathbb{R}$. Its static Killing vector has squared norm ${\cosh }^{2}r>0$ everywhere. Thus global AdS on its cover has no cosmological event horizon for this observer, although other restricted patches or accelerated observers can have horizons.

The causal predictability distinction is also important. Global de Sitter is globally hyperbolic: constant-$t$ circles are Cauchy surfaces. A causal curve is monotone in $t$ and satisfies $|d\chi /dt|\le \mathrm{sech}t$; if it ended at finite $t$, compactness of the spatial circle would allow an endpoint and extension. Hence an inextendible causal curve crosses every such slice once. AdS on the cover has no closed timelike curves, but it is not globally hyperbolic. For two central events at $t=-T$ and $t=T$, with $T>\pi /2$, their causal diamond contains events at $t=0$ and arbitrarily large $r$, so it is not compact. Initial data alone do not determine field evolution without conditions at the timelike conformal boundaries. de Sitter has spacelike conformal infinities and cosmological observer horizons; covered AdS has timelike conformal infinities and requires boundary data. The distinction between the AdS quadric and its universal cover is also explained in David Tong's general relativity lectures.

Let ${\eta }_{ab}=g(e_a,e_b)$ be the constant diagonal metric matrix in a local pseudo-orthonormal frame, and define ${\theta }_{ab}={\eta }_{ac}{\theta }^c{}_b$. The metric compatibility of the Levi-Civita connection givesThus the lowered connection matrix is antisymmetric, including in indefinite signature. Ordinary matrix antisymmetry of ${\theta }^a{}_b$ itself is not the corresponding assertion in indefinite signature.

Let ${\omega }^a$ be the dual coframe, and write a vector field as $Y=e_b{\omega }^b(Y)$. ThenSubtract the same expression with $X,Y$ exchanged and subtract ${\omega }^a([X,Y])$. By the definition of the exterior derivative, this gives the components of the torsion tensor:The Levi-Civita connection is torsion-free, so Cartan's first structure equation is

Use $R(X,Y)={\nabla }_X{\nabla }_Y-{\nabla }_Y{\nabla }_X-{\nabla }_{[X,Y]}$ and define its matrix of curvature two-forms by $R(X,Y)e_b=e_a{R}^a{}_b(X,Y)$. Applying two covariant derivatives to $e_b$ yieldsThe first three terms are $d{\theta }^a{}_b(X,Y)$ and the last two are $({\theta }^a{}_c\wedge {\theta }^c{}_b)(X,Y)$. Therefore Cartan's second structure equation isThe associated Riemann curvature tensor components satisfy $R^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{\omega }^c\wedge {\omega }^d$ with this convention.

For a position-dependent change of pseudo-orthonormal frame, write the frame row as $E^{\prime }=E\Lambda$, with ${\Lambda }^T\eta \Lambda =\eta$. The dual coframe obeys ${\omega }^{\prime }={\Lambda }^{-1}\omega$. Applying the Leibniz rule to $\nabla (E\Lambda )$ givesThe connection one-form has an inhomogeneous term because differentiating the moving basis produces $d\Lambda$. The curvature two-form instead transforms homogeneously. Indeed, curvature is tensorial in the vector to which it is applied, so $R(X,Y)(E\Lambda )=(R(X,Y)E)\Lambda =E^{\prime }{\Lambda }^{-1}R(X,Y)\Lambda$. ConsequentlyEquivalently, substituting the transformed connection into $d{\theta }^{\prime }+{\theta }^{\prime }\wedge {\theta }^{\prime }$ and using $d{\Lambda }^{-1}=-{\Lambda }^{-1}(d\Lambda ){\Lambda }^{-1}$ cancels all derivative terms and gives the same result. This transformation law holds for a general invertible frame change as well; the orthogonality condition ensures both frames remain pseudo-orthonormal.

Evaluate all quantities at $(u_0,v_0)$ and assume a regular surface point, $S_u\times S_v\ne 0$. The tangent plane passes through $S_0=S(u_0,v_0)$ and is spanned by $S_u,S_v$. ThusChoose the unit normal $n=(S_u\times S_v)/\Vert S_u\times S_v\Vert$. The first fundamental form has coefficientswhile the second fundamental form has coefficientsTo derive the curvature formula, differentiate $n\cdot S_u=n\cdot S_v=0$. This gives $-n_u\cdot S_u=e$, $-n_u\cdot S_v=f$, $-n_v\cdot S_u=f$ and $-n_v\cdot S_v=g$. Therefore the shape operator $-dn$, expressed in the tangent basis, isIts eigenvalues are the principal curvatures. Their product, the Gaussian curvature, is its determinant:Regularity ensures $EG-F^2=\Vert S_u\times S_v{\Vert }^2>0$. Reversing $n$ changes the signs of $e,f,g$ and both principal curvatures but leaves $K$ unchanged. At a singular parameter point this quotient and the cross-product normal vector are undefined; an alternative regular chart or separate geometric analysis is required.

The core parametric surface interrogation is evaluation of the position and its derivatives through second order:Position answers point queries; first derivatives give the differential, tangent directions and normal vector; second derivatives supply curvature. An evaluator should also report the parameter domain, patch adjacency, boundary curves and knot or crease locations, and whether these derivatives exist at the queried point. At a discontinuity boundary it should provide the relevant one-sided limits rather than silently average them.

For spatial searches, provide restriction or subdivision into equivalent smaller patches and certified bounding volumes, with error estimates when an approximate mesh is used. These allow intersection and closest-point algorithms to reject remote patches and refine the remaining candidates. A useful interface therefore combines differential evaluation with domain/topology and enclosure information. Inverse parameter lookup, ray intersection and nearest-point search can be built from these primitives using subdivision and local root or minimization solves; they need not be presumed exact basic evaluations of every representation.

A useful subdivision curve interrogation supplies the following connected capabilities. It describes the parameter interval, open or closed topology, endpoints and break points; it evaluates the limit point $C(t)$ and, where the scheme guarantees them, $C^{\prime }(t)$ and $C^{\prime \prime }(t)$; and it refines or restricts an interval while retaining the mapping to the original parameter. It also provides enclosing bounds and a certified curve-to-chord error on each restricted piece. These are enough to build clipping, intersections, closest-point searches, tangents, curvature and adaptive sampling.

The enquiries concern the subdivision limit, not just a finite refined control polygon. For a stationary local refinement matrix $A$, a limit vertex can be obtained from a normalized left eigenvector $w^{T}A=w^T$, $w^{T}1=1$, as $w^{T}P$. Derivatives use the appropriately scaled difference schemes or equivalent spline evaluation on regular intervals. The convergence and derivative hypotheses must actually hold for the scheme; no second derivative can be requested at an intrinsic crease.

If the subdivision mask is nonnegative and preserves constants, descendants and limit points lie in the relevant control-point convex hull, giving a useful enclosing volume. For subdivision masks with negative weights, that hull need not be an enclosure: the interface must supply another valid bound, for example from a proven refinement-error estimate. This distinction is important for safe spatial rejection in an algorithm.

Measure $z$ from the rigid tube wall towards the air core, so $0<z<h$. To leading order in $h_0/a$ and small interface slope, the curvature entering the Young–Laplace equation isThe air pressure is spatially uniform at the order retained. Thus the linear liquid pressure perturbation associated with thickness amplitude $H$ is $p_1=-{\sigma }_0(a^{-2}-k^2)H$. The circumferential curvature favors growth at long wavelength; the curvature opposes it at short wavelength.

The lubrication theory momentum equation is $\mu u_{zz}=p_x$. With the wall no-slip boundary condition $u(0)=0$ and zero surface shear stress, $u_z(h)=0$, its solution and flux areLeading thin-film mass conservation is $h_t+q_x=0$. For a mode $e^{\beta t+ikx}$, this gives the capillary instability of an annular liquid liningTherefore the unstable band is $\boxed{0<k^2<a^{-2}}$. Maximization with respect to $k^2$ gives $k_{\ast }^2=1/(2a^2)$ andThese curvature coefficients use $a$ instead of $a-h_0$ consistently at leading order in the thin-lining approximation.

The plane planar isoperimetric inequality says that a simple closed curve of length $L$ enclosing area $A$ satisfies $L^2\ge 4\pi A$, with equality for circles.

For sufficiently small $r$, below the injectivity radius at $p$, geodesic polar coordinates give the surface metric $d{r}^2+j(r,\theta {)}^{2}d{\theta }^2$. The radial variation field is a Jacobi field, soIts integral equation is $j=r-{\int }_0^r(r-u)K({\gamma }_{\theta }(u))j(u,\theta )du$. Continuity of curvature and $j(u,\theta )=u+o(u)$, uniformly in direction, giveThe Gauss lemma makes the angular metric factor both the circle line element and the polar area factor. ThereforeSquaring the length expansion and subtracting givesThis identifies the requested remainder and proves its ratio to $r^4$ tends to zero. For $K(p)>0$, sufficiently small geodesic disks have $L^2<4\pi A$, the opposite strict inequality to the planar bound. Their positive curvature allows more area for the same perimeter; the Euclidean inequality is not a universal inequality for curved surfaces.

At a chosen event $p$, local inertial coordinates have $g_{ab}(p)={\eta }_{ab}$, ${\partial }_c{g}_{ab}(p)=0$ and ${\Gamma }^a{}_{bc}(p)=0$. The metric agrees with Minkowski space to first order and freely falling trajectories are momentarily straight, expressing the local equivalence principle. Second derivatives generally remain, recording curvature and tidal effects.

Differentiate the Levi-Civita connection. At $p$, the terms from derivatives of the inverse metric multiply first derivatives of $g$ and vanish, givingFor the convention $R^a{}_{bcd}={\partial }_c{\Gamma }^a{}_{bd}-{\partial }_d{\Gamma }^a{}_{bc}+\Gamma \Gamma -\Gamma \Gamma$, the connection products vanish at $p$. Lowering the first index and subtracting these derivatives cancels the ${\partial }_c{\partial }_d{g}_{ab}$ terms, yieldingSymmetry of $g$ and commutation of second derivatives show $R_{abcd}=R_{cdab}$ in these coordinates. Both sides are tensors, so equality in an inertial chart at each event implies equality at any point in every chart.

For the specified conformal metric, let $s=L^{-2}{\eta }_{ef}{x}^e{x}^f$. Its expansion is $g_{ab}={\eta }_{ab}(1-2s+O(|x{|}^4))$. Thus the original coordinates are already inertial at zero, andSubstitution gives $R_{abcd}(0)=4({\eta }_{ac}{\eta }_{bd}-{\eta }_{ad}{\eta }_{bc})/L^2$. In four dimensions the indicated contraction isThe sign is fixed by the Riemann-tensor formula in the question; reversing the Riemann convention would reverse the scalar as well.

In the upper half-plane model $H=\{(x,y):y>0\}$ the hyperbolic metric isThe length of a differentiable curve is $\int \sqrt{{\dot{x}}^2+{\dot{y}}^2}dt/y$. The area element is the square root of the metric determinant times $dxdy$, so the hyperbolic area of a measurable region $S$ is ${\int }_S{y}^{-2}dxdy$.

The Gauss-Bonnet theorem for a geodesic hyperbolic triangle of angles $\alpha ,\beta ,\gamma$ in curvature $-1$ gives area $\pi -\alpha -\beta -\gamma$; an ideal vertex has angle zero. The horizontal edge in the present region is not a geodesic, so it is simpler to use the area element directly:The open boundary does not affect the integral because its area is zero.

Work first in the parameter rectangle of the parametric surface. Set $g(u,v)=f(F(u,v))$. With $F_u,F_v$ linearly independent, the chain rule givesThe intersection is transversal precisely when the implicit normal vector $\nabla f(F)$ and the parametric normal vector $F_u\times F_v$ are not parallel. Consequently $(g_u,g_v)\ne (0,0)$, and the implicit function theorem makes $g=0$ a locally regular parameter curve. This is the setting for transversal surface intersection tracing by a predictor-corrector continuation method.

Find one seed on the required component. Subdivide parameter patches, enclosing their images by bounding volumes and testing whether a certified range of $f$ excludes zero. For a positive-weight NURBS, the active control-point convex hull supplies a convenient enclosure; interval arithmetic can tighten the range of the composed function. Search the surviving patches for a bracketed root or a local corrected seed. Merely checking signs at four corners is inadequate: a small closed intersection can lie wholly inside a patch with the same corner signs. An adaptive or certified patch search, or a supplied seed, avoids this failure. For a sphere centered at $C$, use $f(P)=\Vert P-C{\Vert }^2-r^2$ and $\nabla f=2(P-C)$; for a torus, use its implicit polynomial and its differentiated polynomial in the same algorithm.

At a corrected seed ${\eta }_k=(u_k,v_k)$, an unnormalized parameter tangent vector and a unit spatial tangent vector areChoose the sign consistently with the previous tangent vector. Normalize $q=q_0/\Vert DF{q}_0\Vert$ and predict ${\eta }_p={\eta }_k+hq$, so $h$ is approximately a physical arc length step. Correct the predictor by Newton method applied toThe Jacobian has rows $(g_u,g_v)$ and $(T_k\cdot F_u,T_k\cdot F_v)$. At the starting point its first row vanishes along $q$, whereas its second row applied to $q$ is one. The two rows are therefore independent. The transverse-plane condition prevents the corrector from sliding arbitrarily along the intersection. Restrict correction to a neighborhood of the predictor, use damping, and reduce $h$ when correction fails or its displacement becomes excessive; this prevents jumps to another nearby branch.

To obtain a sufficiently dense sequence, impose a physical point-spacing bound, a bound on tangent turning, and a chord-error estimate. For local curvature $\kappa$, the small-step sagitta is approximately $\kappa h^2/8$; estimate curvature from successive tangent vectors and reduce $h$ until this and the desired spacing are acceptable. The correction residual should be small relative to the drawing tolerance. Uniform steps in $u$ or $v$ alone do not ensure spatial density because the surface metric can vary considerably.

For an open component, trace in both tangent directions from the seed. On reaching a parameter boundary, locate the endpoint by solving $g=0$ together with that boundary's parameter equation, rather than accepting an overshoot. For a closed component, stop only after nontrivial travel and return to the initial parameter location with compatible tangent orientation; near-coincidence in three-dimensional space alone can mistake a self-intersection or neighboring branch for closure. Cross periodic seams using the appropriate chart identification. Adaptive physical stepping, local correction, and component-aware termination give the required point sequence. A vanishing parameter gradient signals a tangency or singularity outside the transversal hypothesis and must be reported rather than passed through by this regular algorithm.

Use the equal-time points to estimate velocity and acceleration with centered finite differences:At the endpoints use one-sided formulas of the desired order. A differentiated path and its time parametrization can instead supply these quantities. Let $g_k$ denote the gravitational acceleration, directed downwards. In the accelerating camera, the effective downward vector is $g_k-a_k$, so the locally felt upward vector is $U_k=a_k-g_k$. For a stationary camera this gives $-g_k$, fixing the sign.

The apparent-gravity camera frame is obtained by orthogonal projection and a cross product:When $W_k\ne 0$, $X_k,Z_k$ are perpendicular unit vectors, and $Y_k$ completes a right-handed orthonormal basis: $X_k\times Y_k=Z_k$. Also $U_k=(U_k\cdot X_k)X_k+\Vert W_k\Vert Z_k$, so the felt-up vector lies in the $XZ$ plane, with its transverse component pointing towards positive $Z$.

The stated exclusion of $a_k=g_k$ ensures $U_k\ne 0$, but it does not ensure $W_k\ne 0$. If felt-up is parallel to travel, the roll angle is undetermined rather than the construction being impossible. Every transverse orthonormal basis then has the required plane property. For continuity, project the previous $Z$ into $X_k^{\perp }$ and normalize it, using $Y_k=Z_k\times X_k$. If this projection is also too small, choose a coordinate unit vector least aligned with $X_k$ and project that vector instead. Switch to this transported choice near the parallel case to avoid numerical amplification. A stopped vehicle likewise needs a chosen limiting travel tangent vector, since its zero velocity itself has no direction. None of this requires a Frenet frame, which can fail where curvature vanishes.

Take $z$ upward from the unperturbed interface in the apparatus frame, with liquid at $z>0$ and pulling velocity $-V{\mathbf{e}}_z$. The primary ice contains no salt. We use the dilute, equal-density approximation and neglect fluid motion relative to the mould. The initial salt inventory per unit horizontal area, with a common density factor suppressed, is $N=h{C}_0$. Its conservation is essential: the far liquid has zero concentration, so an imposed nonzero far-field composition would describe a different experiment.

The steady advection-diffusion equation is $-V{C}^{\prime }=D{C}^{\prime \prime }$. Decay into the pure water and interfacial salt rejection giveSince ${\int }_0^{\infty }C(z)dz=D{C}_I/V=N$, the finite-inventory directional salt rejection solution isThe decay length is $D/V$. Local liquidus equilibrium givesThus the thermal apparatus selects the interface position at this isotherm. The stated speed inequality is precisely $C_I<C_E$, so the interface is warmer than the eutectic temperature, $T_I>T_E=-m{C}_E$. These are steady values after the initial finite saline layer has rearranged; the final exponential distribution is not confined to the original depth $h$.

For $Vh{C}_0/D>C_E$, a steady pure-ice front would require an interfacial composition beyond the eutectic composition. That branch is unavailable. During the transient, rejected brine concentrates until it reaches $C_E$; secondary salt-bearing crystals can then form and the residual brine freezes as an eutectic system. Salt is incorporated into the product rather than remaining indefinitely in a super-eutectic exponential layer. The detailed pattern can include brine pockets and a nonplanar freezing region; it depends on the thermal history and the morphological stability. The speed inequality alone does not decide that latter stability.

For the stability calculation use an interface displacement $\eta =\hat{\eta }{e}^{ikx+st}$ and a liquid perturbation $c=\hat{c}{e}^{ikx+st-qz}$. Here $k>0$ is the wavenumber, $s$ is the temporal rate and $\Re q>0$ gives spatial decay. The frozen thermal field is $T_I+Gz$. Define $\Gamma ={\gamma }_{sl}{T}_m/(\rho L)>0$, where ${\gamma }_{sl}$ is solid-liquid surface energy, $T_m$ is absolute pure-solvent melting temperature, $\rho$ is mass density and $L$ is specific latent heat. For a solid protrusion into liquid the curvature is $\mathcal{K}=-{\eta }_{xx}=k^2\eta$, and the Gibbs-Thomson effect gives local equilibrium $T=-mC-\Gamma \mathcal{K}$. We assume isotropic surface energy and instantaneous interface kinetics, so no kinetic undercooling term is present.

Linearizing the equilibrium condition at the displaced interface givesThe perturbation advection-diffusion equation givesThe local solute balance at a moving graph, to first order, isIts perturbation is $(V-Dq)\hat{c}+s{C}_I\hat{\eta }=0$, because the terms $D\eta C^{\prime \prime }(0)$ and $V\eta C^{\prime }(0)$ cancel. Consequently the full frozen-temperature solutal interface dispersion isNo quasistatic approximation to solute diffusion was made here. For a growing mode the unique admissible liquid root isWriting $b=(G+\Gamma k^2)/(m{C}_I)$ and eliminating $s$ gives $(q-V/D)(q-V/D+b)=k^2$. The root continuously connected to the growing interface branch is thereforeIn particular, unstable wavelengths exist exactly when $mV{C}_I/D>G$, and their band isThis is the constitutional supercooling criterion: immediately above the plane, the local liquidus rises at rate $-m{C}^{\prime }(0)=mV{C}_I/D$, so it overtakes the imposed thermal gradient when that rate exceeds $G$. The Gibbs-Thomson effect cuts off short waves. At long wavelength $s\sim D(V/D-b)k^2/b$ for positive $G$. The exactly uniform inventory-changing displacement is not an instability of a fixed-inventory experiment. On stable branches the full diffusion problem also admits relaxation transients; the root selection above is sufficient to identify every exponentially growing interface mode.

If one additionally neglects the perturbation time derivative in the liquid diffusion equation, the alternative reduced rate isIt has the same onset and cutoff, but need not have the same rate away from the slow-growth regime. The frozen-temperature approximation by itself does not justify this extra reduction.