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Core numerical enquiries return a surface point and its derivatives through second order, together with the parameter domain, patch boundaries and regularity information. At a regular point these determine its tangent plane, normal vector, first and second fundamental forms and curvatures. Subdivision and certified enclosing volumes provide spatial search and intersection support. Singular charts or discontinuity boundaries require one-sided or alternative-chart information; a zero cross product does not define a unique normal vector.

For a stationary film of uniform thickness, thermal conduction gives $T_{zz}=0$, so $T=T_0+cz$. The cooling boundary condition determines $c$:Consequently the surface temperature isThis is the quasistatic temperature of a conductively cooled film, with a weak-cooling expansion when $\alpha h/\kappa \ll 1$.

Let the thickness and longitudinal scales be $h_{\ast },L$, with $ϵ=h_{\ast }/L\ll 1$. Incompressibility gives transverse speed of order $ϵU$, and film evolution has the kinematic time scale $L/U$. For the temperature variation scale $\Theta$, longitudinal diffusion divided by transverse diffusion is $h_{\ast }^2/L^2={ϵ}^2$. Each advective term, and the time derivative on that evolution scale, divided by transverse diffusion isThus ${ϵ}^2\ll 1$ and $Pe\ll 1$ make the leading thermal profile locally linear and quasistatic. Retaining the weak-cooling condition gives the same surface-temperature approximation with the local $h(x,t)$. For any independently imposed faster time variation one would also require its thermal relaxation ratio $h_{\ast }^2/(\kappa t_{\mathrm{evol}})$ to be small.

Define the positive surface-tension gradient coefficientThe weak-cooling approximation gives ${\gamma }_s\simeq {\gamma }_0+Bh$, so the Marangoni stress is ${\gamma }_{s,x}=B{h}_x$. The Young–Laplace equation gives the leading liquid pressure $p=p_a-{\gamma }_0{h}_{xx}$, with constant ambient pressure. In lubrication theory, $p_z=0$ and $\mu u_{zz}=p_x$. The no-slip boundary condition at the wall and the tangential surface condition $\mu u_z(h)={\gamma }_{s,x}$ giveHence the volume flux isConservation of mass gives the thermocapillary thin-film equationThe replacement of surface tension by ${\gamma }_0$ in the capillary term needs $B{h}_{\ast }/{\gamma }_0\ll 1$. For comparable capillary and thermocapillary fluxes,so the fractional tension change is only $O({ϵ}^2)$. Both the tension correction to $\gamma h_{xxx}$ and the differentiated-tension contribution ${\gamma }_x{h}_{xx}$ are then higher-order. Weak cooling alone would not justify that replacement if $|{\gamma }^{\prime }|\Delta T/{\gamma }_0$ were allowed to grow without bound.

The capillary term smooths the film: a crest has negative curvature and higher liquid pressure, causing flow away from it. The thermal term has the opposite effect. A thicker region has a colder surface and therefore higher surface tension; Marangoni stress pulls surface fluid toward that region, reinforcing the thickness disturbance. These mechanisms give fourth-order damping and second-order growth respectively.

Use $H=h/\hat{h}$, $X=ϵx/\hat{h}$, and $\tau =t/\hat{t}$. The two dimensionless flux-divergence coefficients are $B{ϵ}^2\hat{t}/(2\mu )$ and ${\gamma }_0{ϵ}^4\hat{t}/(3\mu \hat{h})$. Setting both equal to one gives the balanced scales for a thermocapillary film:These choices yield $H_{\tau }+(H^2{H}_X{)}_X+(H^3{H}_{XXX}{)}_X=0$. The predicted $ϵ$ must be small for the assumed long-wave approximation.

Linearize about $H=1$ using a normal mode of amplitude $\delta$. The resulting equation is $s-k^2+k^4=0$, so the dispersion relation isModes with $0<|k|<1$ grow, while $|k|>1$ modes decay. Differentiating $s$ gives the most unstable wavenumbersThe physical fastest wavenumber is $ϵ/(\sqrt{2}\hat{h})$.

For a positive steady film with zero volume flux, $H^2{H}_X+H^3{H}_{XXX}=0$. Divide by $H^3$ and integrate once:Choosing the thickness scale so that $H=1$ when $H_{XX}=0$ sets $C=0$. Multiply $H_{XX}=-\ln H$ by $H_X$ and integrate to obtain the zero-flux thermocapillary film profiles relationThus the stated potential is $V(H)=H(\ln H-1)$ and the integration constant $E$ is the corresponding first integral.

Use $c=1$, background Minkowski spacetime metric tensor ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, and $g_{ab}={\eta }_{ab}+h_{ab}$ with small metric perturbation. All indices and derivatives in this linear approximation use $\eta$. The linearized inverse metric is ${\eta }^{ab}-h^{ab}$, and the linearized Levi-Civita connection isDefine the trace-reversed metric perturbation ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$, with $h={\eta }^{ab}{h}_{ab}$.

An infinitesimal coordinate change $x^{\prime a}=x^a+{\xi }^a$ gives the linearized coordinate gauge transformationThe metric perturbation therefore has redundant components. The gauge invariance of the linearized Riemann tensor follows because its change contains third derivatives of $\xi$ cancelling by commutation of partial derivatives. Conversely, flat linearized metric perturbations are locally pure gauge on a contractible patch. Thus linearized curvature, rather than a particular coordinate value of $h$, detects the physical disturbance. Gauge freedom can also be restricted by global topology or boundary conditions.

Keeping the paper's curvature sign convention, its linearized Ricci tensor and scalar areThese are the negatives of the alternative frequently used curvature convention. Put $v_b={\partial }^a{\bar{h}}_{ab}$. The Linearized Einstein equations in vacuum areUnder a gauge change $v_b\mapsto v_b-\square {\xi }_b$, so solving $\square {\xi }_b=v_b$ imposes the Lorenz gauge in linearized gravity. The field equations then become $\square {\bar{h}}_{ab}=0$. The remaining residual gauge symmetry of linearized gravity has $\square {\xi }_b=0$; Lorenz gauge does not exhaust the coordinate freedom.

A Fourier mode ${\bar{h}}_{ab}=A_{ab}{e}^{i{k}_c{x}^c}$ obeys $k^a{k}_a=0$ and $k^a{A}_{ab}=0$. Hence the plane gravitational waves in linearized gravity propagate on the background light cones. Four transversality conditions and four residual gauge amplitudes leave $10-4-4=2$ physical degrees of freedom. The explicit plane-wave reduction to transverse-traceless gauge can set $h_{0a}=0$, spatial trace zero and $k^i{h}_{ij}=0$. For a wave in the $z$ direction, write its physical spatial strain as $H_{ij}=-h_{ij}$, so the spatial metric tensor is $-({\delta }_{ij}+H_{ij})$:There are two gravitational wave polarizations, plus polarization and cross polarization. Rotating transverse axes through $\theta$ rotates their amplitude pair through $2\theta$. Nonconstant profiles with nonzero second derivative have nonzero linearized Riemann curvature tensor and cannot be removed as pure gauge.

The TT coordinate separation and measured separation distinction explains a detector's response. Freely falling particles initially at rest can keep fixed TT coordinates, since ${\Gamma }^a{}_{00}=0$, while their proper separation changes. For a short arm along a unit vector $n$,This is the tidal response described by geodesic deviation; coordinate motion by itself is not the measured signal.

Finally, gravitational-wave energy is second order in the perturbation, so it is absent from a first-order vacuum equation. In a short-wavelength averaging regime the averaged stress-energy of transverse gravitational waves isFor a plane wave its energy flux is $(16\pi G{)}^{-1}⟨{\dot{H}}_{+}^2+{\dot{H}}_{\times }^2⟩$. The averaging scale and weak-field assumptions matter: this does not assign a coordinate-independent local gravitational energy density to an arbitrary first-order perturbation.

The Brans-Dicke theory couples a scalar field $\Phi$ to the Ricci scalar. Assume $\Phi \ne 0$, constant $\omega$, matter independent of $\Phi$, and variations of compact support so boundary terms may be discarded. Use the paper's curvature sign convention and signature $(+---)$. Set $X=g^{ab}{\partial }_a\Phi {\partial }_b\Phi$ and $\square \Phi ={\nabla }_a{\nabla }^a\Phi$. Define the stress-energy tensor consistently byThis convention fixes the sign of the matter term.

Vary with respect to $g^{ab}$. The variation of metric volume density contributes $-g_{ab}/2$ times each scalar Lagrangian, while $\delta R=R_{ab}\delta g^{ab}+g^{ab}\delta R_{ab}$. Applying the Palatini identity followed by two rounds of integration by parts gives the metric variation of a scalar-curvature coupling,The differentiated $\Phi$ terms remain because the curvature multiplier is not constant. Varying the kinetic term at fixed $\Phi$ givesCombining these terms with the matter variation, the Euler-Lagrange equation isRaising both free indices reproduces the contravariant form, because ${\nabla }_a{g}_{bc}=0$.

The independent scalar variation givesor $R+\omega {\Phi }^{-2}X-2\omega {\Phi }^{-1}\square \Phi =0$. The trace of the metric tensor equation in four dimensions isMultiplying the scalar equation by $\Phi$ eliminates the same combination $\Phi R+\omega {\Phi }^{-1}X$. Thus the scalar equation of Brans-Dicke theory isAt the degenerate coupling of Brans-Dicke theory, $\omega =-3/2$, the combined equations impose $T=0$ instead; division by $3+2\omega$ is unavailable. If matter depended directly on $\Phi$, its scalar variation would add a source and the displayed scalar equation would change.

The following lettered sections justify the supplied variation identities; they are supporting identities in the PDF, rather than three independent field-equation problems.

The local metric tensors do not determine the global coordinate identifications. The usual complete two-dimensional de Sitter spacetime is the hyperboloid $(X^0{)}^2-(X^1{)}^2-(X^2{)}^2=-1$ in ambient signature $(+--)$, withPulling back the ambient metric tensor gives $d{t}^2-{\cosh }^{2}td{\chi }^2$, withUnwrapping $\chi$ gives the universal cover of two-dimensional de Sitter spacetime instead. The usual embedded two-dimensional anti-de Sitter spacetime has $(X^0{)}^2+(X^1{)}^2-(X^2{)}^2=1$ in signature $(++-)$ andHere $r\in \mathbb{R}$ and $t$ is initially periodic modulo $2\pi$. The time circles are closed timelike curves. Its physically standard universal cover removes that identification:The two signs of $r$ describe two spatial ends; imposing $r\ge 0$ would retain only half of this complete model.

Both spacetimes have constant curvature and are geodesically complete. In the paper's convention their Ricci scalars are respectively $+2$ and $-2$. Their embeddings show that geodesics are intersections with planes through the ambient origin. They nevertheless have very different causal structures.

For the geodesics of two-dimensional de Sitter spacetime, use an affine parameter $\lambda$, normalization $\kappa ={\dot{t}}^2-{\cosh }^{2}t{\dot{\chi }}^2\in \{1,0,-1\}$ and conserved momentum $L={\cosh }^{2}t\dot{\chi }$. ThenTimelike geodesics extend to both infinite proper times; $\chi =\text{constant}$ is a simple example. Null geodesics obey $\sinh t=\pm |L|(\lambda -{\lambda }_0)$, so their affine parameter is also infinite at either temporal end. Spacelike geodesics have $|L|\ge 1$, obey $\sinh t=\sqrt{L^2-1}\sin (\lambda -{\lambda }_0)$ and are closed curves on the embedded cylinder; $t=0$ is its simplest spacelike geodesic circle. Unwrapping the spatial circle removes closure but not completeness.

For the geodesics of two-dimensional anti-de Sitter spacetime, the timelike Killing vector ${\partial }_t$ gives $E={\cosh }^{2}r\dot{t}$. The normalization is $\kappa ={\cosh }^{2}r{\dot{t}}^2-{\dot{r}}^2$, henceChoose a future-directed timelike geodesic, so $E\ge 1$ andFor $E>1$, it oscillates radially with proper period $2\pi$; $E=1$ gives the central geodesic $r=0$. In both cases, global time advances by $2\pi$ over a proper time interval $2\pi$. The curve is closed on the original hyperboloid, while its lift on the universal cover is nonclosed and future-directed. Null geodesics have $\sinh r=\pm E(\lambda -{\lambda }_0)$ and reach either spatial end only at infinite affine parameter. Spacelike geodesics satisfy $\sinh r=\sqrt{E^2+1}\sinh (\lambda -{\lambda }_0)$ and also have infinite proper length toward either end. Thus the finite coordinate time discussed below does not imply physical geodesic incompleteness.

The conformal cylinder of two-dimensional de Sitter spacetime follows by setting $\eta =\arctan (\sinh t)$:Its past and future conformal boundaries are spacelike circles. Null geodesics have $d\chi /d\eta =\pm 1$ and can travel only a finite angular distance over the entire infinite proper-time history. For the complete observer $\chi =0$, an event can send a signal to that observer precisely when its shortest angular distance $d(\chi ,0)$ is less than $\pi /2-\eta$. The observer horizons in two-dimensional de Sitter spacetime are therefore the null curves $d(\chi ,0)=\pi /2-\eta$; its past signal horizon has $d(\chi ,0)=\eta +\pi /2$. These are observer horizons, without curvature singularities. Their intersection bounds a static patch with metric tensorThe horizons at $\rho =\pm 1$ are regular Killing horizons beyond which this static chart fails, while the global coordinates remain smooth.

For the conformal strip of two-dimensional anti-de Sitter spacetime, set $\psi =\arctan (\sinh r)$. On its universal cover,The two conformal boundaries are timelike. A null ray from the center approaches a boundary after coordinate time $\pi /2$, though its affine parameter diverges. Every interior event can signal to the complete central static observer in finite global time, so that observer has no event horizon. A Poincaré horizon concerns a restricted coordinate patch, and accelerated observers can have different causal horizons.

Finally, the global de Sitter cylinder and its spatial cover are globally hyperbolic: constant-time slices are Cauchy hypersurfaces. The anti-de Sitter universal cover is not globally hyperbolic, because timelike infinity admits incoming signals within finite global time. A field's evolution therefore requires boundary conditions at its two conformal boundaries in addition to initial data. The opposite placement of the conformal boundaries explains why de Sitter observers have cosmological horizons despite global hyperbolicity, while the central anti-de Sitter observer has no event horizon despite the failure of global hyperbolicity.

The material derivative follows surfactant carried by the moving interface. Tangential surface divergence changes the local surface area and hence dilutes or concentrates insoluble material. Normal motion changes area through the curvature: an expanding sphere has positive $u_n{\nabla }_s\cdot \mathbf{n}$ and its concentration decreases. The last term is surface diffusion, which smooths concentration gradients. There is no exchange with the bulk; this is conservation of insoluble surfactant on a moving interface.

The interfacial speed is of order $a|\mathbf{F}|$. Diffusion dominates redistribution when the surface Péclet number satisfiesThen $C^{\prime }/C_0=O({\mathrm{Pe}}_s)$, and advection of $C^{\prime }$ is a second-order effect. Write $S=\mathbf{n}\cdot \mathbf{Fn}$. Since ${\mathbf{u}}_s=\mathbf{Fx}-S\mathbf{x}$, the projected divergence of the first term is $\mathrm{tr}\mathbf{F}-S=-S$, whereas the second has divergence $2S$; its gradient contracted with $\mathbf{x}$ is zero. Hence ${\nabla }_s\cdot {\mathbf{u}}_s=-3S$.

The leading steady transport equation is $D_s{\Delta }_s{C}^{\prime }=C_0{\nabla }_s\cdot {\mathbf{u}}_s=-3C_{0}S$. Trace-freeness and the supplied surface Laplacian identity give ${\Delta }_{s}S=-6S/a^2$. Fixing the conserved total surfactant removes the uniform null mode and yieldsThis is quadrupolar surfactant distribution on a spherical interface; its spherical mean is zero because $⟨n_i{n}_j⟩={\delta }_{ij}/3$.

With $+$ outside, $-$ inside, and the normal directed out of the bubble, the interfacial stress balance with variable surface tension isNow ${\nabla }_{s}S=2I_s\mathbf{Fn}/a$, so $-{\nabla }_s\gamma =2A{\gamma }_1{I}_s\mathbf{Fn}/a$. Multiplying the curvature by the tension and retaining linear terms givesThe undeformed isotropic pressure jump is kept with its normal; the perturbation terms are evaluated on the reference sphere. Products of deformation and concentration perturbations are beyond the approximation.

The imposed tensor is symmetric and traceless, so $\mathbf{x}\cdot \mathbf{Ex}$ is a harmonic scalar of degree two. Its decaying partner is $\mathbf{E}:\nabla \nabla (1/r)$; the corresponding harmonic vector uses $\mathbf{E}\nabla (1/r)$. Linearity and rotational symmetry give precisely these tensorial modes, with the growing scalar chosen to reproduce the imposed strain. Thus in the unscaled convention one can takeThe coefficients multiply distinct decaying modes. No other angular harmonic is forced at this order.

No penetration in the steady state gives $P=3Q-1$. The given tangential surface velocity implies $\alpha =1+2Q$. The tangential stress balance, with $M=A{\gamma }_1/(\mu a)$, gives $1+P-8Q=M\alpha$, or $-5Q=M(1+2Q)$. ThereforeThe nonuniform normal stress is unaffected by the uniform inviscid interior pressure. Its balance isSubstitution and rearrangement giveAs $M\to \infty$, $\alpha \to 0$: the concentration-induced Marangoni stress suppresses tangential surface motion, making the exterior boundary behave like an immobile surface despite the inviscid interior. The deformation coefficient tends to $5\mu a/(2{\gamma }_0)$; the clean limit is $2\mu a/{\gamma }_0$. Both small surface Péclet number and small deformation, of order $\mu a|\mathbf{E}|/{\gamma }_0$, are required. This is Marangoni immobilization of a bubble in straining flow.

Measure pressure relative to its hydrostatic value and let $\Delta P$ be the lower-side excess pressure minus the upper-side excess pressure. The long unobstructed parts of the tube act as Poiseuille resistances in series. Their lengths add to $L$ at leading order because $l\ll L$; entrance and exterior resistances are smaller for $L\gg a$. ConsequentlyHere $Q$ is downward laboratory-frame volume flux.

Take $z$ upwards in the particle frame and let $s$ measure distance from the particle surface to the tube wall. In this frame the particle is stationary and the wall moves upwards at speed $U$. In the thin, slowly varying gap, pressure is uniform across $s$ and the leading axial Stokes equation is $\mu w_{ss}=p_z$. No slip givesThe gap's upward relative flux is constant, while conservation of displaced volume gives $2\pi aq=\pi a^{2}U-Q$. Thus $p_z=6\mu U/h^2-12\mu q/h^3$, and integration from bottom to top yieldsCurvature corrections to the gap circumference are relative order $h/a$. Rapid end regions do not contribute to the leading singular neck resistance.

For the force use a cylindrical fluid control volume containing the particle, with full-area end caps just beyond it and the particle surface as an inner boundary. The upward pressure force on the caps is $\pi a^2\Delta P$. The outer wall exerts axial traction $\mu w_s(h)$ on the fluid, whereas the particle's traction on the fluid integrates to minus its upward hydrodynamic resistance $F$. The zero-inertia momentum balance is thereforeThe velocity profile gives $w_s(h)=4U/h-6q/h^2$, provingUsing the outer-wall traction is essential; directly using the particle-side shear and a flat-end pressure force would omit the pressure on the sloping particle surface. This is control-volume drag formula for a sphere in a tube.

The four dimensionless relations areSolving them givesWithin the preceding lubrication and tube-resistance approximations, the force isSince $h/a\ll 1$, $I_3^{\ast }\gg I_2^{\ast }\gg I_1^{\ast }$. Dropping the smaller additive $-2I_2^{\ast }+4I_1^{\ast }/3$ terms in the numerator yields the requested leading formulaThe Cauchy-Schwarz inequality gives $(I_2^{\ast }{)}^2\le I_1^{\ast }{I}_3^{\ast }$, so the retained $\delta$ term is positive. The term multiplied by $\delta$ must be retained: it can compete with $I_3^{\ast }$ even though the unmultiplied lower integrals cannot.

For a sphere of radius $a(1-ϵ)$, the neck has $h=aϵ+z^2/(2a)$ at leading order. With $z=a\sqrt{2ϵ}t$, the integrals reduce to $\sqrt{2}{ϵ}^{1/2-n}{\int }_{-\infty }^{\infty }(1+t^2{)}^{-n}dt$. For $n=1,2,3$ these last integrals are $\pi ,\pi /2,3\pi /8$, giving the full two-sided integrals. PutThe printed sphere integrals are half these values: they correspond to integrating over one side of the neck rather than the stated interval spanning both sides. The substitution above establishes the radius-deficit normalization of sphere-tube lubrication integrals. Using the two-sided values, the force becomesso the three answers areThese are the corrected sphere-tube drag regimes with a radius-deficit gap. If the supplied half-integrals are treated as stipulated data instead, the same algebra gives $3\pi {ϵ}^{-5/2}/(8\sqrt{2})$, $1/\delta$ and $\pi \sqrt{2}{ϵ}^{-1/2}/3$ in the three regimes. Thus the first and third numerical drag coefficients double when the geometrically correct integrals are used; the regime boundaries and dominant flow patterns are unchanged.

In the first regime, $Q^{\ast }\sim \delta c_3{ϵ}^{-5/2}\ll 1$, $q^{\ast }\sim 1/4$, and $\Delta P^{\ast }\sim c_3{ϵ}^{-5/2}\sim F^{\ast }$. Almost no displaced fluid escapes through the long tube; it bypasses upwards through the thin annulus. The large pressure-driven bypass resistance sets the drag, with shear smaller.

In the second regime, $Q^{\ast }\sim 1$, $\Delta P^{\ast }\sim F^{\ast }\sim 1/\delta$, andAlmost all displaced volume travels through the unobstructed tube, whose pressure resistance dominates the force. Both terms in this small gap flux are worth displaying: their crossover at $\delta \sim {ϵ}^{3/2}$ changes the balance between pressure-driven and wall-driven gap transport but does not create a fourth leading force regime.

In the third regime, $1-Q^{\ast }\sim 4ϵ/3$, $q^{\ast }\sim ϵ/3$, and $\Delta P^{\ast }\sim 1/\delta \ll F^{\ast }$. Tube throughflow still carries almost all the displaced volume, but the dominant resistance is local annular lubrication rather than the net tube pressure drop. The relative gap flux is now of order $Uh$, so wall-driven and pressure-driven components both matter locally. The formulas describe hydrodynamic resistance after subtracting hydrostatic buoyancy; terminal settling would balance it against the particle's excess weight.

An efficient subdivision surface interrogation interface needs the mesh topology and patch adjacency, boundary and trimming information, patch-local parameter charts and their transition maps, and local restriction/refinement with the complete influencing control net. Numerical enquiries should supply limit positions $S$, first derivatives $S_u,S_v$, and second derivatives where defined; these determine normal vectors and curvatures at regular points. It must also return certified patch bounding volumes and convergent geometric/derivative error bounds. Near an extraordinary subdivision vertex, use the local subdivision matrix and limit/tangent evaluation appropriate to that scheme; second derivatives need not exist merely because a tangent plane does. Finite refined mesh vertices are not exact limit positions.

Represent the cutting plane as $n\cdot x=d$, with $n$ normalized. On each chart setThe desired plane section of a subdivision surface is the image of the zero set $g=0$. For a bounding volume $B$, compute a certified interval $[g_{\min },g_{\max }]$ for the plane functional over $B$. If that interval is strictly positive or strictly negative, reject the patch. With positive partition-of-unity weights, the minimum and maximum of $n\cdot p_i-d$ over all influencing control points already bound its limit patch. For masks without this convex hull property, the interface must provide a different valid enclosure.

An adaptive algorithm that isolates components before tracing is:

This is exhaustive to a chosen tolerance when certified subdivision isolates finitely many regular transverse branches and enclosures converge. If $\Vert \nabla g\Vert \ge \gamma >0$ locally and $\Vert DS\Vert \le M$, a residual tolerance $|g|\le \delta$ corresponds locally to a geometric correction bounded by $M\delta /\gamma$, using the gradient flow toward the level set. Bounds on second derivatives similarly control curve-segment chord error bound. As $\gamma$ approaches zero, a fixed residual tolerance no longer gives a fixed geometric error, so the step and stopping rules must reflect conditioning.

Seeding must not rely just on corner signs. For example $S(u,v)=(u,v,u^2+v^2-r^2)$ cut by $z=0$ has a closed circle wholly inside a patch whose corner values are all positive. Certified range tests plus subdivision retain it; a corner-only marching rule can miss it. Tangencies also change the nature of the output: $S=(u,v,u^2+v^2)$ meets $z=0$ in one isolated point, while a coplanar patch contributes an area rather than a curve. Thus a general plane section is not necessarily a union of regular curves; the algorithm should classify these cases instead of silently fabricating or deleting a branch.

For efficiency, cache subdivision stencils, patch extraction, limit-evaluation coefficients and bounds; reuse child data instead of reconstructing the entire refined mesh. Use local influencing neighborhoods, with the halo dictated by mask support, and keep adjacency/boundary topology separate from changing geometric coordinates. Plane-functional bounds can be computed by refining scalar signed-distance controls, which is cheaper than refining all three coordinates. Combine coarse inexpensive boxes with tighter local convex hull or interval arithmetic tests only where needed. Adapt steps to curvature and transversality, reuse branch seeds, and use regular spline evaluation away from extraordinary points. Stable edge ownership, outward-rounded bounds, scale-aware tolerances, and explicit degeneracy flags are correctness requirements as well as performance issues. The result is a limit-surface intersection computation; merely intersecting a finitely refined control polyhedron with the plane supplies only an uncontrolled approximation.

For radial null propagation, $dr/\sqrt{1-k{r}^2}=|dt|/a$. Since $dz/dt=-(1+z)H(z)$ and $a=a_0/(1+z)$, its present-distance integral isThus the printed extra $a_0$ before the redshift integral is inappropriate when $H$ is the physical Hubble parameter; it disappears only if $a_0$ has been set to one.

Dust conservation and the curvature relation giveTo evaluate it, put $s=\sqrt{1-{\Omega }_0}$ and $v=\sqrt{1+{\Omega }_{0}z}$. The dimensionless integral $I=H_0{d}_S$ isInverting the proper-distance expression in part b gives $r=\sinh (sI)/(a_0{H}_{0}s)$. Substituting the logarithm, writing the hyperbolic sine as half the difference of its exponential and inverse, and using $v^2-s^2={\Omega }_0(1+z)$ yieldsThis is the open dust distance-redshift relation. The printed square root lacks the factor $z$; its literal form would give a negative, nonzero coordinate at $z=0$. The corrected expression has $r\sim z/(a_0{H}_0)$ near zero. At fixed positive ${\Omega }_0$ and $z\gg {\Omega }_0^{-1}$, the term linear in $z$ dominates, givingThe zero-density limit is nonuniform: a pure Milne model has no such finite limiting radial coordinate.

Separate conservation of a fluid with fixed $w=P/\rho$ gives $\dot{\rho }+3H(1+w)\rho =0$, hence $\rho \propto a^{-3(1+w)}$. The $w=-1/3$ component therefore has ${\rho }_Q\propto a^{-2}$. Flatness today fixes its normalization to ${\rho }_{Q0}=3H_0^2(1-{\Omega }_{M0})/(8\pi G)$, soThis is a coasting fluid: its contribution to $\rho +3P$ is zero, so it does not itself drive accelerated expansion. Its background Friedmann contribution resembles curvature, although the actual spatial geometry remains flat.

For a homogeneous canonical scalar, ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. The acceleration equation requires $V>{\dot{\varphi }}^2$ for cosmic inflation, equivalently ${ϵ}_H=-\dot{H}/H^2<1$ in the flat scalar-dominated limit. Inflation need not be exactly de Sitter.

A standard sufficient regime for a long period is positive potential energy dominating the kinetic energy, physical gradient energy and curvature. Homogeneity over an expanding approximately Hubble-sized patch allows the gradient term to redshift away. In comoving coordinates the gradient energy is $|{\nabla }_{\mathrm{com}}\varphi {|}^2/(2a^2)$; the gradient in the displayed energy formula must be interpreted physically if no $a^{-2}$ is written there. RequireThe slow-roll equations are then $3H\dot{\varphi }\simeq -V^{\prime }$ and $H^2\simeq 8\pi V/(3m_{\mathrm{pl}}^2)$. Sufficient potential conditions areThe potential must remain sufficiently flat over a large enough field interval to produce the required expansion. These slow-roll conditions are stronger than the bare necessary condition for acceleration.