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Take horizontal motion towards the second projectile as positive. In projectile motion, both bodies have the same downward gravitational acceleration, so it cancels from their relative motion. Their coordinates areFor a collision at $t>0$, equality of the horizontal coordinates requires $v_2\cos \theta =v_1$. Equality of the heights then requires $v_{2}t\sin \theta =h$. HenceThe upward solution requires $v_2>v_1$. A collision strictly above the ground further requires $t<\sqrt{2h/g}$, or $v_2^2-v_1^2>gh/2$. The horizontal target distance does not independently affect the interception geometry. If the first shot is also required to hit that ground target, its parameters satisfy $l=v_1\sqrt{2h/g}$.

The specified rocket fuel fraction gives $m(T)=2m_0/3$, so $\alpha T=m_0/3$. Imposing $v(T)=0$ yieldsThen $h(T)=0$ fixesNumerically,The associated constant mass loss rate is $\alpha =m_0/(3T)\simeq 88.78\mathrm{kg}{\mathrm{s}}^{-1}$. The initial upward thrust acceleration already exceeds $g$ and increases as mass falls, so the downward speed decreases monotonically to zero. This calculation assumes constant lunar gravitational acceleration, constant relative rocket exhaust speed, and negligible drag.

Take downward distance $s$ and downward speed $v=ds/dt\ge 0$, with constant gravitational acceleration $g>0$ and quadratic drag coefficient $k>0$. The printed vector square has no standard vector meaning; the directionally consistent drag force is ${\mathbf{F}}_d=-mk|\mathbf{v}|\mathbf{v}$. For this downward fall, Newton's second law becomesThe terminal velocity is obtained by setting the net force to zero:From rest, the speed increases monotonically towards this limiting value. In fact $v(t)=v_{\infty }\tanh (\sqrt{gk}t)$; thus it is a supremum approached at infinite time, and a tower of finite height need not permit it to be reached.

For the required distance, use the chain rule $dv/dt=vdv/ds$. Writing $w=v^2$ gives a linear ordinary differential equation:Consequently,Here $k$ has units of inverse length, so the result has the required units of distance and is independent of the falling mass.

Powered ascent. Use upward signed velocity, constant gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, constant relative rocket exhaust speed $u=2500\mathrm{m}{\mathrm{s}}^{-1}$, vertical flight from rest, no drag force, and no impulse at stage separation. The constant-$g$ approximation is an explicit modelling assumption; a varying Newtonian gravitational field would require integrating a different ascent equation. Put $m_0=540000\mathrm{kg}$ and $q=2500\mathrm{kg}{\mathrm{s}}^{-1}$. During ascent $m(t)=m_0-qt$, and momentum balance gives the rocket equation with gravity:The launch rocket thrust $uq=6.25\times 10^6\mathrm{N}$ exceeds $m_{0}g$, so immediate lift-off is consistent. At ascent burnout the entire attached vehicle still has mass $m_b=140000\mathrm{kg}$, not $40000\mathrm{kg}$; the latter is the separated first stage. The burn duration and endpoint speed are

Coasting height. Integrating the ascent velocity from zero height givesFor example this follows by substituting $m=m_0-qt$ in ${\int }_0^t\ln [m_0/(m_0-qs)]ds$. At separation,The separated stage then undergoes free fall in a constant gravitational field with initial upward speed $v_b$. Conservation of mechanical energy during this coast givesIts smaller mass after separation does not change that coasting height.

Reserve burn. On descending to $z_i=100000\mathrm{m}$, conservation of mechanical energy gives downward speedReset time to $s=0$ at reserve ignition. Now $m_i=40000\mathrm{kg}$, $m_f=14000\mathrm{kg}$ and $q_r=26000/92\mathrm{kg}{\mathrm{s}}^{-1}$. Downward ejection supplies upward rocket thrust. The upward signed velocity isThus the formal free-flight burnout value iscorresponding to a formal downward speed of $2.26\mathrm{m}{\mathrm{s}}^{-1}$. Keeping $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ throughout instead gives $3.82\mathrm{m}{\mathrm{s}}^{-1}$ downward; premature rounding is especially damaging because three large velocity terms almost cancel.

The free-flight endpoint is below ground. Integrating the same reserve-burn velocity givesAt the stated endpoint this gives $z_r(92)=-167.27\mathrm{m}$ for $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. The surface is crossed at approximately $s=89.15\mathrm{s}$, before reserve burnout. The stage has already hit the surface, so a physical post-impact burnout speed cannot be inferred without a ground-contact model. The boxed velocity is the intended ideal continuation, not an achievable free-flight landing result under these exact assumptions. In particular, the given numbers do not describe an exact soft landing. The constant-rate retro-rocket landing must satisfy both its velocity and height conditions.

For one wire, Ampère's circuital law on a concentric circle of radius $r$ givesThe other wire experiences the Lorentz force on its electric current, $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$. Its direction is perpendicular to this magnetic field, so the force per unit length isThe right-hand rule gives attraction for parallel electric currents in the same direction and repulsion for opposite directions.

For the upper wire, the upward magnetic force per unit length balances its weight per unit length, $\lambda g$, where $g$ is the gravitational acceleration. ThereforeThis assumes the lower wire is held fixed and both wires remain straight. Within the vertical-motion model, increasing the separation weakens the upward magnetic force, so this equilibrium is vertically restoring.

A tidal force measures the difference of gravitational acceleration across an extended object. A freely falling reference frame removes a spatially uniform acceleration, but not its gradient. For a distant point mass $M$ at distance $R$ and direction $\hat{\mathbf{n}}$ from the reference centre, a small internal displacement $\mathbf{x}$ givesThe tidal tensor has eigenvalues $2GM/R^3$ radially and $-GM/R^3$ in each transverse direction: radial stretching and transverse squeezing. Its trace vanishes in the source-free region, consistent with the gravitational Gauss law. An associated relative potential is ${\Phi }_{\mathrm{tide}}=-GM[3(\hat{\mathbf{n}}\cdot \mathbf{x}{)}^2-x^2]/(2R^3)$.

The scale is $GMl/R^3$ across size $l$, not the centre's absolute gravity $GM/R^2$. This distinction explains how a more distant massive object can exert a large force but a modest tide. Opposite-side stretching supports the elementary two-bulge picture; real ocean tides also involve rotation, basin geometry and fluid dynamics, so the static gradient alone does not predict coastal heights or timing. The same gradient stresses satellites and can disrupt a self-gravitating body when it competes with the body's own gravitational restoring acceleration.

Take upward as positive and interpret the stated $97\mathrm{kg}$ as the total mass immediately before launch, including the harpoon. The remaining probe mass is $M=95.6\mathrm{kg}$. Conservation of momentum during the short firing interval givesThe harpoon speed used here is relative to the comet, as printed. From $v_{\mathrm{esc}}^2=2G{M}_c/R$, the near-surface gravitational acceleration isFor the near-surface recoil height with constant downward thrust, the upward motion has downward acceleration $a=g_c+F/M$. ThereforeSince $H/R\simeq 1.8\times 10^{-4}$, treating gravity as constant is consistent. This uses only the stated gravity and thrust after firing; a taut tether could reduce this free-flight height, but no tether length or elasticity is specified. If instead $97\mathrm{kg}$ meant the probe mass after firing, the pre-firing mass would be $98.4\mathrm{kg}$ and its momentum would leave the probe still moving downward, with no upward excursion. This illustrates nearly cancelling recoil momenta in payload ejection. The mass convention matters because the launch momenta nearly cancel.

Archimedes' principle states that the resultant hydrostatic pressure force on a body equals the weight of the fluid displaced, directed opposite to gravity. Let $V_d$ be the displaced fluid region, with outward normal $\hat{\mathbf{n}}$ from the body into the fluid. Since the fluid pressure force is inward,Hydrostatic balance is $\nabla p={\rho }_f\mathbf{g}$, with downward gravitational acceleration $\mathbf{g}$. Thereforefor uniform fluid density and gravity. Uniform atmospheric pressure gives no net force on a closed surface. For a floating body, the same displaced-volume construction includes the free surface; any significant air buoyancy can be included as a second displaced fluid.

A freely floating body of mass $m$ satisfies $mg={\rho }_{f}g{V}_d$, so its displaced volume is $m/{\rho }_f$. A homogeneous body of density ${\rho }_b<{\rho }_f$ floats with submerged fraction ${\rho }_b/{\rho }_f$; if ${\rho }_b>{\rho }_f$, hydrostatic buoyancy alone cannot support it. The buoyant force acts through the centre of buoyancy, the centroid of the displaced volume for constant density, while weight acts through the centre of mass. Their relative positions determine rotational equilibrium and stability. For a fully submerged body, a centre of mass below the centre of buoyancy gives a restoring torque; a floating body's changing displaced shape additionally requires the metacentric height criterion. Surface tension, externally driven flow and contact forces can alter the simple force balance.

The differential gravitational acceleration responsible for a tidal force is the difference between accelerations at separated points of a body. Relative to a freely falling reference point $\mathbf{R}$, a small offset $\mathbf{\xi }$ experiencesThe tidal tensor has units ${\mathrm{s}}^{-2}$; the vector called $\mathbf{T}(\mathbf{\xi })$ in the source is the resulting differential acceleration, in $\mathrm{m}{\mathrm{s}}^{-2}$. An actual differential force is this vector multiplied by the test mass. Measurable effects include relative motion of separated freely falling masses, radial stretching, transverse squeezing, deformation and, when self-gravity or strength is insufficient, disruption. A spatially uniform acceleration alone produces no tidal deformation.

A spatially uniform external gravitational acceleration moves a body's centre of mass without stretching it. Tidal forces arise from the variation of that acceleration across the body. Let a distant mass $M$ lie at distance $D$ along unit vector $\mathbf{n}$, and let $|\mathbf{r}|\ll D$ denote position relative to the body's centre. Subtract the centre's acceleration. To first order,The tidal tensor is $T_{ij}=GM(3n_i{n}_j-{\delta }_{ij})/D^3$. It stretches along the line to the source with eigenvalue $2GM/D^3$ and compresses in the two perpendicular directions with eigenvalues $-GM/D^3$. Its trace vanishes outside the source, consistently with Laplace's equation for the gravitational potential.

After removal of the constant and uniform-acceleration terms, the tidal potential isAn ideal equilibrium ocean on a sphere of radius $R$ follows an equipotential of the body's gravity plus this perturbation. For surface gravity $g$, the leading displacement isup to a constant fixing total water volume. The quadrupolar form produces bulges on both the near and far sides; the far-side bulge follows from differential acceleration, without postulating a repulsive gravitational force.

The Moon and Sun both raise terrestrial tides, with relative strength set mainly by $M/D^3$, rather than by the total gravitational pull $M/D^2$. Their contributions reinforce near new and full Moon to produce spring tides, and partly cancel near quadrature to produce neap tides. Earth's rotation relative to the tidal pattern gives a strong semidiurnal component in the simplest geometry. Real tide heights and timing depend on ocean depth, coastlines, basin resonances, Coriolis acceleration, friction and the declination of the forcing bodies. Tidal dissipation produces phase lags and torques, exchanging rotational and orbital angular momentum and gradually altering the spin and orbit. The equilibrium bulge is therefore a useful starting model, not an exact local tide forecast.

The uniform weight per unit length is $q=\rho abg$, where $\rho$ is the mass density and $g$ the gravitational acceleration. The total weight is $qL$, so each support provides an upward force $qL/2$. The weight on the portion from $0$ to $x$ is $qx$ acting at that portion's midpoint, a distance $x/2$ from the cut. In the same bending moment convention as part (c),The units are $(\mathrm{N}{\mathrm{m}}^{-1}){\mathrm{m}}^2=\mathrm{N}\mathrm{m}$. Its greatest magnitude occurs at $x=L/2$, where $B=-q{L}^2/8$.

Far from its ends, approximate the body by an infinite uniform cylinder. By cylindrical symmetry, the Newtonian gravitational field is radial. Applying Gauss's law for gravity to a coaxial cylindrical surface of radius $s$ and length $\ell$ outside the material givesAt the surface, $R=50\mathrm{m}$, soThe gravitational acceleration points inward toward the axis. Its small magnitude reflects both the small size and the modest density. The infinite-cylinder approximation uses the specified distance from both ends; a spherical approximation would discard the relevant geometry. No rotational contribution is included, since no rotation rate is supplied.

Powered vertical flight. Take upward velocity $v$ as positive, and let $v_e>0$ be the downward exhaust speed relative to the rocket. During $dt$ the rocket's mass changes by $dm<0$, and exhaust mass $-dm$ leaves with Earth-frame velocity $v-v_e$. Compare the momentum of the rocket plus this newly expelled exhaust before and after the interval:Discard the second-order product $dmdv$. The result is $mdv+v_{e}dm=-mgdt$, orThe positive rocket thrust is $-v_e\dot{m}$. Integrating with $v(0)=0$ and $m(0)=m_0$ gives the rocket equation with gravitational loss:Using $d(mv)/dt$ for the rocket alone without the exhaust momentum flux would give the wrong equation for this open system.

Orbital speed under the stated gravity assumption. A circular orbit of radius $r_o=R_E+h_o=7.90\times 10^6\mathrm{m}$ requires centripetal acceleration $v_o^2/r_o$. If the magnitude of gravitational acceleration is the stipulated constant $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ even at the orbit,This treats gravity as centrally directed for the orbital calculation, while retaining its constant magnitude. The subsequent flat-surface approximation treats its direction as locally vertical.

Burnout height. During the unpowered projectile motion, horizontal velocity is constant. At the top of the trajectory the vertical component vanishes, so the horizontal component must equal the orbital speed $v_o$. At burnout, therefore,These follow either by integrating $\ddot{y}=-g$ or by the vertical work-energy theorem. Substituting the preceding constant-$g$ orbital result givesThe vertical component is initially positive, as required to reach the orbit after burnout.

Alternative inverse-square interpretation. A real Earth's exterior Newtonian gravitational field is not constant over a $1500\mathrm{km}$ altitude change. If $9.8\mathrm{m}{\mathrm{s}}^{-2}$ instead means the surface value $g_0$, then Newton's law of universal gravitation gives $GM=g_0{R}_E^2$ andUsing this orbital speed but the paper's constant-$g$ flat projectile motion approximation for the coast would giveThis alternative changes the global constant-gravity assumption; it is not the same model as the $183\mathrm{km}$ result. An exact curved Kepler orbit coast would also abandon the prescribed flat trajectory. The printed assumptions support the constant-magnitude answers above, while the inverse-square values explain the different answer obtained from the usual physical Earth-orbit formula.

The tunnel is taken thin enough that removal of its material does not significantly disturb spherical symmetry. For $a\le r\le 2a$, the enclosed mass isBy the spherical shell theorem, the inward gravitational acceleration has magnitudeThe released gravitational potential energy is the positive work done by gravity while the mass falls from radius $2a$ to $a$:By conservation of mechanical energy,Inside the cavity, the spherical shell theorem gives zero gravitational field, so the velocity stays constant. The original figure places B and C at opposite ends of a cavity diameter, a distance $2a$ apart. ThereforeThe quantity $G\rho$ has units ${\mathrm{s}}^{-2}$, so this is a time. There are no numerical values for $a$ or $\rho$ in the paper. The transit time is independent of $a$ and the falling mass $m$. The cavity's gravitational potential is constant, not necessarily zero; only its gradient vanishes. The calculation is a hollow-sphere gravitational transit.

For a prograde equatorial synchronous circular orbit, equate gravitational acceleration and centripetal acceleration. With planetary radius $R_M$, $GM=g_s{R}_M^2$, and rotation period $P$,The tabulated $g_s$ is a gravitational acceleration; its printed unit must be read as $\mathrm{m}{\mathrm{s}}^{-2}$, rather than $\mathrm{m}{\mathrm{s}}^{-1}$. Using $P=88642\mathrm{s}$ givesThe orbital altitude is measured from the surface, rather than the centre.

At the visible horizon, the line of sight is tangent to the spherical surface. The limiting central angle ${\theta }_h$ satisfies $\cos {\theta }_h=R_M/r_o$. The visible spherical cap has area $2\pi R_M^2(1-\cos {\theta }_h)$, givingAbout $41.7\%$ of the whole surface is geometrically visible, with ${\theta }_h\simeq 80.4^{\circ }$. This is a surface-area fraction, not an apparent-disc area ratio. Useful imaging near the horizon is further limited by foreshortening and the long atmospheric path.

For constant gravitational acceleration, hydrostatic balance gives the atmospheric column mass $m=p_{\mathrm{sfc}}/g$, with negligible pressure at the top. For the same surface temperature $T_{\ast }<T_{\mathrm{sfc}}<T_c$, the variable latent heat gives a smaller integrated $\ln (p_{\mathrm{sfc}}/p_{\ast })$, and henceMultiplying by the same planetary surface area gives the same comparison for total mass. The models agree for a surface at or below the triple point, where they use the same sublimation branch. The comparison assumes the same surface temperature, rather than the same surface pressure.

Choose a common horizontal compensation depth beneath the deepest cooling-induced density anomaly, within mantle that can relax pressure differences over the timescale considered. In a normal oceanic plate model, this means below the old plate's thermal lithosphere, not at the base of the approximately 7 km thick oceanic crust.

The reason is that cooling changes density through both crust and lithospheric mantle. A datum at the Mohorovičić discontinuity would omit most of the changing column mass. A different datum for each age would also compare unlike columns. Once the common datum is sufficiently deep, extending it farther into laterally uniform mantle adds the same mass per unit area to each column and does not change the predicted relative subsidence.

There is no unique number fixed by the question. For an illustrative oceanic plate cooling model with a thermal thickness about 100–125 km, a common datum beneath that layer, of order 150–200 km below sea level, is a defensible choice provided the model has no substantial lateral density anomaly there. These are model choices, not measured universal boundaries. In a half-space cooling model, use the oldest age $t_{\max }$ to choose a practical cutoff: $z_c\gtrsim 4\sqrt{\kappa t_{\max }}$ makes the remaining temperature anomaly less than about 0.5% of its surface contrast. For $\kappa =10^{-6}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ and $t_{\max }=100$ million years, this is about 225 km. A half-space model has no mathematically sharp finite thermal base.

Finally, equal column mass is the local, approximately constant-gravitational acceleration representation of equal pressure at the datum. A weak asthenosphere permits pressure relaxation through slow solid-state material creep; it need not be liquid. Short-wavelength loads, lithospheric flexure or dynamic topography can violate local column-by-column isostasy, requiring a more complete mechanical model.

Specify what “onset” means. I would distinguish the first local, perhaps short-lived subduction event from persistent, self-sustaining recycling in a connected global network of tectonic plates. Modern plate tectonics combines relatively rigid moving lithosphere, localized boundary deformation, seafloor spreading and sustained convergence. A sinking drip or a single imposed descending sheet is not sufficient evidence of that global regime. Different processes may become established at different times, so a transition interval may be a better hypothesis than an instantaneous worldwide switch.

Construct competing physical models. Solve conservation of mass, momentum and energy for evolving mantle convection and lithosphere, including temperature- and stress-dependent rheology, brittle yielding, plastic deformation, melting, compositional buoyancy and relevant phase changes. Use geodynamic modelling to compare a stagnant lid, episodic overturn, drip-dominated recycling and a sustained mobile lid. In a simplified incompressible creeping-flow framework,Here $\mathbf{u}$ is velocity, $p$ pressure, $\eta$ effective viscosity, $\dot{\mathbf{ϵ}}$ the strain-rate tensor, $\rho$ density, $\mathbf{g}$ gravitational acceleration, $c_p$ specific heat capacity, $k$ thermal conductivity, $H$ radiogenic power per unit mass, and $\Phi$ dissipative heating per unit volume. These equations are a framework: compressibility, latent heat and chemical transport may require additional terms. A simple yield stress law ${\tau }_y=C+\mu (\rho gz-P_f)$ represents cohesion $C$, friction coefficient $\mu$ and pore pressure $P_f$, illustrating why hydration and interface weakening can change the outcome.

Vary the early-planet controls rather than assuming the answer. Explore plausible initial mantle potential temperature, radiogenic heating, crust thickness and composition, water inventory, grain-size evolution, surface temperature, and boundary strength. Hotter mantle lowers viscosity but can produce thicker buoyant basaltic crust; it does not automatically make sustained subduction easier. A cold strong plate can transmit slab pull yet resist bending and initiation. Water weakening of rocks, melt-induced weakening, inherited faults and mechanically weak sediments can change this balance. Test proposed triggers such as mantle plume impingement or gravitational instability without presupposing an established global plate-force system.

Useful diagnostics include the Rayleigh number,with thermal expansion coefficient $\alpha$, layer thickness $D$, thermal diffusivity $\kappa$ and temperature contrast $\Delta T$, together with surface mobility, the localization of deformation, plate longevity, melt production and material recycling. A large Rayleigh number establishes vigorous convection under suitable conditions, not necessarily a mobile lid. A convincing model should sustain one-sided subduction, connected boundary evolution and seafloor spreading after any artificial trigger is removed. Compare initiation thresholds with continuation thresholds: hysteresis can allow an established regime to persist under conditions in which it would not spontaneously start.

As an example of the type of experiment, three-dimensional models have tested mantle plume-induced subduction initiation, finding sensitivity to plate buoyancy, magmatic weakening and hydration of the interface. The scientific use of such results here is to identify a feasible mechanism and its predicted signatures, not to adopt their success as a direct measurement of an ancient onset date.^[ref]

Connect a model regime to a date through independent evidence. Couple the experiments to candidate planetary cooling histories, then predict observable pressure-temperature-time paths, magmatic compositions, deformation patterns and recycling rates for each regime. If independently dated rocks constrain a thermal or compositional state, ask which regimes can reproduce that state and the observed geological associations. Repeat over uncertain initial conditions and constitutive parameters, check numerical resolution, and compare two- and three-dimensional results where geometry matters. Models that prescribe a descending slab, fix plate velocity, or insert a weak boundary may be excellent forward models of a working subduction zone but do not demonstrate spontaneous initiation.

The result is a conditional statement: persistent plate tectonics is physically feasible under a range of conditions, and dated geological observations may favor that range at a particular time. Uncertain cooling histories, several feasible regimes and uncertain rheology prevent numerical feasibility alone from dating onset. The strongest constraint comes when the same parameter range explains multiple independently dated observations and alternative regimes fail those tests.

tidal forces measure differences in gravitational acceleration across an extended region. Neighboring freely falling particles each have zero four-acceleration, but their separation need not accelerate uniformly. Let $u^{\mu }$ be the tangent to a reference timelike geodesic and ${\xi }^{\mu }$ its infinitesimal separation from a neighboring one, chosen from an affinely parametrized geodesic congruence. Using the Riemann curvature tensor convention specified in Solution 3,Changing the overall curvature convention changes this displayed sign. The covariant derivatives compare vectors by parallel transport, avoiding a coordinate-dependent subtraction of accelerations.

In the Newtonian limit, expand the force field at two nearby points:For $\Phi =-GM/r$, a local orthonormal frame aligned with the radius givesThere is radial stretching and transverse compression; the trace vanishes outside the mass. Two freely falling test masses and a measurement of their changing separation implement gravity gradiometry. Their common acceleration drops out, while a relative acceleration proportional to separation remains. A local inertial frame removes Christoffel symbols at a point, but cannot remove nonzero curvature over a finite region.

A tidal force is a difference of gravitational acceleration across an extended region. In Newtonian gravity, $g_i=-{\partial }_i\Phi$. Two neighboring freely falling particles separated by ${\xi }^j$ have, to first order,For a point mass, $\Phi =-GM/r$. In a local orthonormal radial/transverse frame,A small freely falling cloud stretches radially and compresses transversely. Outside the source the three eigenvalues sum to zero, consistently with ${\nabla }^2\Phi =0$; within matter the trace is $-4\pi G\rho$. These are accelerations of separations, not the much larger common acceleration of the cloud's centre.

In general relativity, freely falling test particles follow timelike geodesics. A freely falling frame can eliminate the connection at one event, but not the Riemann curvature tensor throughout a finite laboratory. Let $u$ be the geodesic tangent and $\xi$ a connecting vector for neighboring geodesics with $[u,\xi ]=0$. In the explicitly chosen curvature conventionthe geodesic deviation equation isIt follows by commuting ${\nabla }_u{\nabla }_{\xi }u$, using ${\nabla }_{u}u=0$ and zero torsion. In a parallel-propagated orthonormal frame, its spatial components directly describe measured relative acceleration. The weak-field, slow-motion limit recovers the Newtonian Hessian above. Reversing the curvature sign convention reverses the displayed curvature side, without changing the physical stretching. The equivalence principle removes locally uniform gravity, while geodesic deviation measures the residual tidal field. The connection between parallel transport, curvature and neighboring geodesics is developed systematically in the standard geometric construction.^[ref]