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Adiabatic shielding suppresses the heating of an orbit when an external tidal tensor varies slowly compared with its orbital frequency. In a nonresonant slow encounter, adiabatic invariance of an orbital action lets the orbit adjust approximately reversibly, so a frozen-position impulse approximation overestimates the energy transfer. An orbital resonance can invalidate the simple slow-variation argument.

Use local axes radial, tangential in the orbital plane, and perpendicular to that plane around a circular Kepler orbit with ${\Omega }^2=GM/R^3$. The gravitational tidal tensor is ${\Omega }^2\mathrm{diag}(2,-1,-1)$. A uniformly corotating frame adds the gradient of centrifugal acceleration, ${\Omega }^2\mathrm{diag}(1,1,0)$, giving the displayed effective tensor for stationary offsets. Thus the radial differential acceleration is $3GMa/R^3$. These are accelerations, or forces per unit test mass; an actual differential force includes that mass. Moving offsets also experience Coriolis acceleration.

Stipulate a fixed source point mass $M=10^{12}\mathrm{kg}$, two equal test particles of mass $m=1\mathrm{kg}$, and $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$. This is a chosen model, not a measurement of a particular object. Place the test particles on the outward radial axis at $R-L/2$ and $R+L/2$, with midpoint distance $R=100\mathrm{m}$ and separation $L=2\mathrm{m}$. Neglect their mutual attraction, other sources and all supporting forces.

Each signed radial gravitational acceleration is inward:Their separation has outward relative accelerationThe exact value is $2.6702540\times 10^{-4}\mathrm{m}{\mathrm{s}}^{-2}$. The midpoint tidal tensor predicts $2\mu L/R^3=2.6697200\times 10^{-4}\mathrm{m}{\mathrm{s}}^{-2}$, an underestimate of $0.019999\%$ relative to the exact value. For these equal $1\mathrm{kg}$ masses, the difference of external forces has the same numerical value in the SI unit of force, the newton. It is not an additional mutual force between the test particles.

The exact formula requires $L<2R$ and describes the instantaneous relative acceleration. Because the two sample positions are symmetric about the midpoint, the Taylor expansion error relative to the linear result starts at order $(L/R{)}^2$; a one-sided comparison with a reference point generally has relative error of order $L/R$. If either test particle enters an extended source, use that source's interior Newtonian gravitational field instead of extrapolating the singular point mass model.

The differential Newtonian gravitational field of a galaxy across a smaller bound system. Its tidal tensor can slowly change the angular momentum of distant comets, changing their periapses and their access to planetary encounters.

Gravity gradiometry measures spatial differences of gravitational acceleration. Two simultaneous measurements along a unit vector $\mathbf{e}$, separated by $L\mathbf{e}$, giveThis measures one directional component of the tidal tensor, not its whole matrix. Any identical platform acceleration cancels in the difference. Unequal instrument responses, changing geometry and rotating reference frames can spoil that cancellation. For a large baseline, the quotient is an average of the directional tidal tensor along the baseline, rather than its exact value at one point.

For a fixed source point mass, the magnitude $g=\mu /R^2$ of the gravitational acceleration measures the local pull; the radial eigenvalue ${\lambda }_r=2\mu /R^3$ of the tidal tensor measures how that pull changes across space. Doubling $R$ divides $g$ by four but ${\lambda }_r$ by eight. For a fixed small radial separation $L$, the linear relative acceleration is $\Delta a\simeq {\lambda }_{r}L$, soA small body can therefore experience strong common gravitational acceleration with a comparatively small tidal force. A measurement of $g$ alone does not determine the tidal tensor: sources at different distances can have the same $g$ but different spatial variation.

The local stretching time scale is ${\lambda }_r^{-1/2}=\sqrt{R^3/(2\mu )}$: it marks when an initially resting separation could change by order itself if the tidal tensor stayed fixed. During free fall, $R$ changes, so this scale is not a complete trajectory or a universal disruption time. Tidal disruption additionally compares the tidal force with binding by self-gravity and material forces; $g$ by itself supplies no such comparison.

For a perturber following $(p,0,vt)$, the impulse approximation integrates its tidal tensor to $2G{m}_p\mathrm{diag}(1,-1,0)/(v{p}^2)$. A fixed target position therefore receives $\Delta \mathbf{v}=2G{m}_p(x,-y,0)/(v{p}^2)$. The longitudinal kick vanishes after integrating the symmetric trajectory.

For $\Phi =-GM/r$, the acceleration tidal tensor isIts radial eigenvalue is $2GM/r^3$ and its two transverse eigenvalues are $-GM/r^3$. This stretches a small freely falling cloud radially and compresses it transversely. The zero trace expresses the vacuum Poisson equation outside the source; within matter the trace is $-4\pi G\rho$.

For one target galaxy, an encounter with relative speed $u$ sweeps a cylinder of volume $\pi R^{2}udt$ in time $dt$. Multiplication by the number density gives the encounter rate $\Gamma =\pi R^{2}N⟨v_{\mathrm{rel}}⟩$. Thus the geometric galaxy-encounter rate givesThis is the expected number of mergers in the geometric model. For independent encounters, the Poisson process probability of at least one is $P_{\ge 1}=1-e^{-\mu }$, so the displayed linear probability is valid for rare encounters, $\mu \ll 1$. There is no factor of one-half for a single target; that factor would enter a count of distinct pairs across the whole population.

For an illustrative group environment, take $R=20\mathrm{kpc}=0.02\mathrm{Mpc}$, $N\sim 1{\mathrm{Mpc}}^{-3}$, $⟨v_{\mathrm{rel}}⟩\sim 300\mathrm{km}{\mathrm{s}}^{-1}$ and a Hubble time of order $10^{10}$ years, or $3\times 10^{17}\mathrm{s}$. The path length is about $9\times 10^{19}\mathrm{km}\simeq 3\mathrm{Mpc}$, using the stated distance conversion. HenceThe estimate scales linearly with environment density and quadratically with the adopted merger radius. It is an order-of-magnitude model estimate, not a universal observed merger fraction; reducing the illustrative density by a factor of one hundred reduces the estimate by the same factor.

For the tidal calculation, use the relative potential convention in which acceleration is $\nabla \psi$. Expand the point-mass potential about the target centre:The constant does not exert a force. The linear term accelerates the target centre and is subtracted in its freely falling frame. The remaining leading quadrupolar point-mass tidal potential isThe condition $r<r_p$ permits the convergent expansion; using only this term is the leading tidal approximation, accurate when the target radius is small compared with the closest separation.

The given impact and velocity directions imply ${\mathbf{r}}_p(t)=(p,0,vt)$. This trajectory is in the $x,z$ plane, correcting the incompatible printed $x,y$ plane. In the impulse approximation, hold the stellar position fixed while integrating the tidal tensor:For $s=vt/p$, the needed integrals arewith the mixed integral zero by oddness. Thus the integrated tidal tensor of a straight-line flyby is $2G{m}_p\mathrm{diag}(1,-1,0)/(v{p}^2)$ andThere is stretching along the impact direction, compression in the other transverse direction and no net kick along the path.

Write $\mathbf{u}$ for the star's initial velocity, distinguishing it from the perturber speed $v$. Its specific energy change is $\mathbf{u}\cdot \Delta \mathbf{v}+\frac{1}{2}|\Delta \mathbf{v}{|}^2$. The no-correlation assumption removes the first term on averaging; it is not an identity for every individual star. The mean specific heating at a given position is thereforeFor a spherical target, the mass-weighted averages satisfy $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩=⟨r^2⟩/3$. The tidal impulse heating of a spherical galaxy is consequentlyHere $⟨r^2⟩=m_g^{-1}\int r^2\rho d^{3}r$ is assumed finite. Notice the change from specific energy to total energy after multiplying by $m_g$.

For two equal targets, each gains the same internal energy, so $\Delta U_{\mathrm{tot}}=8G^2{m}_g^3⟨r^2⟩/(3v^2{p}^4)$. Their relative-motion reduced mass is $m_g/2$. If $v$ denotes their relative speed at infinity, their initial orbital energy is $E_{\mathrm{orb}}=\frac{1}{2}(m_g/2)v^2=m_g{v}^2/4$. Internal tidal heating comes from this orbital energy. In the tidal capture of galaxies model, capture occurs if $\Delta U_{\mathrm{tot}}>E_{\mathrm{orb}}$, which gives the equal-mass tidal-capture thresholdCapture allows further passages and eventual merger in this simplified picture.

The encounter lasts roughly $p/v$. The impulse approximation needs $\Omega p/v\ll 1$, so that a star barely moves during the tide. When $v/p<\Omega$, stellar orbits respond during the perturbation and the kicks can cancel. Adiabatic invariance of an orbital action produces adiabatic shielding of tidal encounters, rather than the impulsive heating used above. The capture inequality cannot be extrapolated into that slow-encounter regime.

Ignoring gravitational focusing, a galaxy sweeps a cylinder of volume $\pi R^2{v}_{\mathrm{rel}}T$. Multiplying this by the number density $N$ of target galaxies and averaging the relative speed gives the mean encounter countFor independent encounters modeled by a Poisson process, the probability of at least one galaxy merger is $1-e^{-\mu }$. The printed expression $P=\mu$ is its rare-encounter approximation $\mu \ll 1$; it is not an exact probability for arbitrary $T$.

For an illustrative present-day field population take an effective merger impact parameter $R=20\mathrm{kpc}=0.02\mathrm{Mpc}$, $N=10^{-2}{\mathrm{Mpc}}^{-3}$, and $⟨v_{\mathrm{rel}}⟩=200\mathrm{km}{\mathrm{s}}^{-1}$. These are assumed order-of-magnitude inputs, not precise observational measurements. Taking a Hubble time of $T\simeq 1.4\times 10^{10}\mathrm{yr}\simeq 4.4\times 10^{17}\mathrm{s}$, the given distance conversion yieldsand henceThus the geometric field estimate is of order $10^{-5}$ to $10^{-4}$ per Hubble time. It scales as $R^{2}N⟨v_{\mathrm{rel}}⟩T$ and is very sensitive to environment and the adopted effective merger radius. Enhanced density in groups, gravitational focusing, and the evolution of the galaxy population are all omitted; this number is not a prediction of the full cosmological merger fraction.

Now use the stipulated rapid, distant encounter. The consistent rectilinear trajectory isWith the stated impact parameter along $x$ and velocity along $z$, this trajectory lies in the $xz$-plane. The original PDF's reference to the $xy$-plane is a typo; the local TeX also corrupts the impact-parameter direction. The original PDF fixes that direction as $x$.

Use the positive potential $G{m}_p/|\mathbf{R}-\mathbf{r}|$, whose gradient is the attractive acceleration under the question's sign convention. Its multipole expansion isThe first term has no force; the linear term accelerates the entire galaxy and disappears in the frame following its centre of mass. The leading internal tidal potential is thereforeThis is a quadrupole approximation valid for $r\ll p$, rather than an exact equality for every $r<p$. Its acceleration tidal tensor acts on $\mathbf{r}$ asIn the impulse approximation, each stellar position is held fixed during the flyby and the velocity kick is the integral of this acceleration. Put $s=vt$; terms odd in $s$ integrate to zero, whileThe integrated diagonal coefficients of the tidal tensor are consequently $(2,-2,0)/(v{p}^2)$, givingThe encounter stretches the galaxy along the impact parameter, compresses it along $y$, and gives no net leading kick along the flyby direction.

Denote the pre-encounter stellar velocity by $\mathbf{u}$, to distinguish it from the relative flyby speed $v$. The instantaneous change of specific kinetic energy isUncorrelated kicks with zero mean cross term imply $⟨\mathbf{u}\cdot \Delta \mathbf{v}⟩=0$, so the phase-averaged heating at a specified position isFor an individual star the cross term need not vanish: the formula is an ensemble or orbital-phase average. Integrating over a spherical galaxy of mass $m_g$, symmetry gives $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩=⟨r^2⟩/3$. Therefore the total tidal heating isFor two identical galaxies, each receives this heating with $m_p=m_g$. Adding both contributions gives

In the centre of mass frame, equal masses approach with speeds $v/2$ if $v$ is their relative speed at infinity. The initial orbital energy iswhere $m_g/2$ is the reduced mass. The orbital potential energy vanishes at infinite separation. Within the weak-deflection approximation this asymptotic speed is also the nearly constant speed used in the flyby calculation.

Conservation of energy transfers the positive internal heating out of the relative orbit. Tidal capture of galaxies occurs in this model if $E_{\mathrm{orb}}-\Delta E_{\mathrm{int}}<0$, which givesThe equality is marginal capture. A bound pair still needs subsequent evolution to coalesce, so the result is an approximate capture criterion used here as a merger criterion. Its right-hand side has dimensions of length times speed, as required.

The impulse approximation requires the flyby duration $\tau \sim p/v$ to be short compared with the stellar dynamical time ${\Omega }^{-1}$, namely $v/p\gg \Omega$. If $v/p<\Omega$, a star moves substantially during the encounter. Its orbital adiabatic invariance of an orbital action suppresses net heating by a slowly varying tidal field: adiabatic shielding of tidal encounters replaces the frozen-position calculation. The divergent $v^{-2}$ heating predicted by extrapolating the impulsive formula to slow encounters is therefore spurious. Close passages with $p$ comparable to the galaxy size or strong gravitational focusing also lie outside the derivation.

Assume nonrelativistic pressureless matter, an initial irrotational vector field of growing-mode linear cosmological density perturbations, a homogeneous expanding background, and no shell crossing. All gradients below refer to initial comoving coordinates. Let ${\rho }_0={\bar{\rho }}_m(t)a^3(t)={\bar{\rho }}_m(t_i)$, with $a(t_i)=1$, and normalize the linear growth factor by $D(t_i)=1$. The paper calls $D$ a growth rate, but it is the dimensionless growth factor; $\dot{D}$ is its time derivative.

Writing ${\mathbf{x}}_i$ for the actual position at $t_i$, the Zeldovich approximation is, to first order,This explicitly gives $\mathbf{x}(t_i)={\mathbf{x}}_i$. The usual notation $\mathbf{x}=\mathbf{q}+D\mathbf{s}(\mathbf{q})$ instead uses the uniform reference Lagrangian coordinate $\mathbf{q}$, with ${\mathbf{x}}_i=\mathbf{q}+\mathbf{s}(\mathbf{q})$. The two labels can be interchanged inside a first-order displacement, but not in the unperturbed position without the initial displacement correction.

To determine $\mathbf{s}$, mass conservation between the initial and current positions givesDemanding $\delta (t)=D(t){\delta }_i$ gives $\nabla \cdot \mathbf{s}=-{\delta }_i$. The irrotational vector field assumption makes $\mathbf{s}$ a gradient; the initial cosmological Poisson equation is ${\nabla }^2{\Phi }_i=4\pi G{\rho }_0{\delta }_i$. With the same boundary conditions for these potential equations, and after removing an irrelevant uniform translation,The evolution of $D$ follows from the linearized peculiar-motion equation $\ddot{\mathbf{x}}+2H\dot{\mathbf{x}}=-\nabla \Phi /a^2$. Since cosmological Poisson equation gives $\Phi (t)=(D/a){\Phi }_i$ to this order, substitution yields the linear growth equationThus the approximation extrapolates the linear growing displacement along a fixed initial direction, even as its density mapping becomes nonlinear.

The early spin of a dark matter halo comes from an external gravitational torque on its nonspherical initial mass region. A uniform external acceleration moves its centre of mass without spinning it; the spatially varying tidal tensor exerts different forces on different parts. The tidal torque theory requires misalignment of the region's shape and the surrounding tidal field. A local irrotational vector field of velocity does not imply zero integrated angular momentum for a nonspherical region.

Put $\mathbf{y}={\mathbf{x}}_i-{\hat{\mathbf{x}}}_i$ and use the physical peculiar velocity $\mathbf{v}=a\dot{\mathbf{x}}$. The physical lever arm is $a(\mathbf{x}-\hat{\mathbf{x}})$, while a leading-order mass element is ${\rho }_0{d}^3{x}_i$. Substituting the velocity formula into the supplied definition of angular momentum about the centre of mass givesTerms from the perturbed mass measure are higher order. Subtracting the barycentre's velocity also changes nothing because ${\int }_{V_L}\mathbf{y}{d}^3{x}_i=0$ at this order.

Apply the divergence theorem componentwise:The antisymmetry of the Levi-Civita symbol kills the last term. Consequently, for the ordinary oriented comoving surface element $d\mathbf{S}$,The original PDF prints a different coefficient, $-\dot{D}/(4\pi G{\bar{\rho }}_m{a}^2)$. That coefficient does not follow from its own mass measure and velocity equation and has the wrong dimensions for total angular momentum. The corrected coefficient above also gives the time dependence consistent with the paper's later tensor expression; this is a source typo, not a TeX transcription issue.

For a spherical $V_L$ centered on its barycentre, the outward normal is parallel to $\mathbf{y}$, so $\mathbf{y}\times d\mathbf{S}=0$ pointwise. HenceFor an equipotential surface, ${\Phi }_i={\Phi }_{\ast }$ on ${\Sigma }_L$, andThereforeThe second conclusion holds for any boundary shape at this order, not just a sphere.

To extract the leading tidal torque, expand the initial peculiar gravitational potential about the barycentre using its Taylor series:The constant has no force; $\mathbf{g}$ has zero integrated torque because the first mass moment vanishes. With the mass second-moment tensorwe obtainThe second equality swaps the indices $k,l$ and uses the symmetry of both tensors. Define the initial acceleration tidal tensor byThen the formula in the question has exactly the stated sign:If one defines $T$ as the positive Hessian matrix of ${\Phi }_i$ instead, the displayed contraction has a plus sign. Defining the convention is essential. Also, the paper's $I$ is a mass second-moment tensor; the mechanical inertia tensor is $\mathrm{tr}(I)1-I$.

Only the anisotropic parts contribute: a multiple of the identity produces zero contraction with the Levi-Civita symbol. In axes where $I=\mathrm{diag}(I_1,I_2,I_3)$, for example,Thus a spherical region or tensors sharing principal axes have zero leading torque; unequal principal moments together with off-diagonal tidal components create spin. The Taylor series truncation assumes that higher spatial derivatives of the tidal field are sufficiently small across the region.

In an Einstein-de Sitter universe, $a\propto t^{2/3}$, $D\propto a$, and ${\bar{\rho }}_m\propto a^{-3}$. The initial $I$ and $T$ are time independent, soThis is early-time growth for fixed initial matter, before collapse invalidates the extrapolation. It does not predict indefinite linear spin growth for a virialized halo.

Two reasons for only approximate agreement with simulations are that nonlinear collapse and shell crossing change the trajectories and the tidal field, invalidating the first-order displacement and fixed leading tidal tensor; and that real halos undergo dark-matter halo mergers, anisotropic accretion, and exchange of material and angular momentum, so the halo identified at a later time need not be the same isolated collection of initial particles. These processes can change both the magnitude and the direction of the spin.

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.

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.

Nearby freely falling particles can accelerate relative to one another even though each follows a geodesic. A local inertial frame removes the common gravitational acceleration at one event; it cannot remove the variation across an extended laboratory. That variation is a tidal force, measured covariantly by geodesic deviation.

Let $u^{\mu }$ be the reference four-velocity and ${\xi }^{\mu }$ an infinitesimal separation to a neighboring geodesic, with matching affine parameters. With the convention $[{\nabla }_{\beta },{\nabla }_{\gamma }]V^{\mu }=R^{\mu }{}_{\nu \beta \gamma }{V}^{\nu }$,This relates measured relative acceleration to the Riemann curvature tensor. Reversing the curvature-tensor convention reverses the displayed curvature sign, without changing the physical relative motion.

In a small freely falling frame outside a spherical mass, the weak-field tidal tensor has radial and transverse components proportional to $2,-1,-1$. ThusAn initially spherical cloud stretches radially and compresses in both transverse directions. Two separated test masses, with their changing separation measured by light, form the basic idea of a gravity gradiometer; the common falling motion drops out. A uniform apparent acceleration in an accelerated laboratory can occur even in flat Minkowski spacetime, so it is the separation-dependent tidal response that diagnoses curvature.

Local equivalence and tides. In Newtonian mechanics, inertial mass $m_{\mathrm{i}}$ measures the response to a force, whereas passive gravitational mass $m_{\mathrm{g}}$ couples to an imposed Newtonian gravitational field. A freely falling body obeysThe observed composition independence of free fall says that this ratio is universal. With a common choice of units it is one. This is the Newtonian content of the weak equivalence principle for test particles; it does not by itself specify all nongravitational physics.

The Einstein equivalence principle further says that local nongravitational experiments in a sufficiently small freely falling laboratory have the form predicted by special relativity, independently of the laboratory's location and motion. In such a local inertial frame, the metric tensor is locally the Minkowski metric and the Christoffel symbols vanish at the selected event. The strong equivalence principle extends the universality claim to local gravitational experiments and bodies with significant gravitational self-energy. None of these statements removes measurable tidal forces across a finite apparatus.

In an ideal small freely falling lift, a released ball initially floats relative to the cabin, because both undergo the same free fall. Two separated balls, however, need not retain their separation: they sample different parts of the gravitational field. For a point source, linearizing $\mathbf{g}=-GM\mathbf{r}/r^3$ across a small separation $\mathbf{\xi }$ gives the tidal tensor predictionA radial separation stretches with coefficient $2GM/r^3$, whereas a transverse separation is compressed with coefficient $-GM/r^3$. Initially parallel velocities therefore develop relative motion even though the individual bodies experience zero proper acceleration.

The relativistic statement is geodesic deviation. With the Riemann curvature tensor conventiona connecting vector between neighboring freely falling worldlines satisfiesHere $D/d\tau$ is the covariant derivative along the reference geodesic, and $U^{\mu }$ its four-velocity. Curvature can be nonzero where the Christoffel symbols have been made zero. In the weak, static, slow-motion limit this equation reduces to ${\ddot{\xi }}^i=-{\partial }_i{\partial }_j\Phi {\xi }^j$, recovering the tidal tensor above. A finite falling lift can detect gravity through these relative accelerations; local equivalence does not imply a globally uniform field.

Weak-field connection and potential. Take $x^0=ct$ and the $+---$ metric signature. There is an index error in the printed formula for the Christoffel symbols: its first differentiated metric component must be $g_{\nu 0}$, not $g_{\mu 0}$. The correctly contracted expression isThe offending term vanishes under the stipulated time independence, so this correction does not change the intended calculation. Write $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ and treat both $h$ and its derivatives as first order. To that order $g^{00}=1-h_{00}$, $g^{0i}=O(h)$, and $g^{ij}=-{\delta }^{ij}+O(h)$. A stationary metric tensor givesFor the temporal component the inverse-metric coefficient is already first order, soFor a spatial component the zeroth-order inverse metric contributes:The time component of the approximate geodesic equation says that $dt/d\tau$ is constant to the retained order. In its spatial component, convert the proper time derivatives to coordinate-time derivatives and cancel $(dt/d\tau {)}^2$:Comparison with the Newtonian equation $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$ identifies $h_{00}=2\Phi /c^2+C$. For an isolated source, choose the gravitational potential to tend to zero at infinity and normalize the time coordinate there, giving $C=0$. ThusThe quantity $\Phi$ is the Newtonian gravitational potential, or gravitational potential energy per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For a point source, $\Phi =-GM/r$. A static clock has $d\tau =\sqrt{g_{00}}dt\approx (1+\Phi /c^2)dt$, connecting the Newtonian limit with gravitational time dilation.

Frequency of the orbiting source. Treat the gas blob as a test particle on a timelike circular Schwarzschild geodesic, with pressure and other nongravitational forces negligible. Treat our very distant location as infinity, so the observer's own gravitational potential is negligible. These are the assumptions that make a unique answer in terms of $R_S/r$ possible.

Put $f=1-R_S/r$ and use a local static observer at the emission point as an intermediate measuring frame. The remote observer lies on the symmetry axis of the orbit. A ray arriving on that axis has zero conserved axial angular momentum, so its local direction is perpendicular to the emitter's azimuthal velocity. Bending within a meridional plane does not alter this orthogonality. The transverse relativistic Doppler effect therefore giveswhere $v$ is the orbital speed measured by that local static observer. The gravitational redshift between static Schwarzschild observers then givesA gravitational circular orbit has $v^2=GM/r+O((GM/r{)}^2/c^2)$ in the weak-field limit, so $v^2/c^2=R_S/(2r)+O((R_S/r{)}^2)$. Expanding both factors,Solving for the emitted frequency givesThe gravitational part is $R_S/(2r)$ and the transverse kinematic time dilation contributes another $R_S/(4r)$; the absence of line-of-sight motion does not remove this second effect.

As an exact check, the local speed on a timelike circular Schwarzschild geodesic satisfies $v^2/c^2=\mu /(r-2\mu )$, where $\mu =GM/c^2=R_S/2$. Therefore the face-on frequency shift of a circular Schwarzschild source iswhose expansion agrees with the boxed answer. If the gas is supported or accelerated by pressure or magnetic stresses rather than following a geodesic, its orbital speed is not fixed by the mass and radius alone; the appropriate first-order result is instead ${\nu }_{\mathrm{em}}\approx {\nu }_{\mathrm{ob}}[1+GM/(r{c}^2)+v^2/(2c^2)]$, with $v$ supplied independently.

For a source point mass $M$ at the origin, let $\mathbf{R}=R\mathbf{n}$, where $\mathbf{n}$ is an outward unit vector, and let $\mu =GM$ be the source's standard gravitational parameter. Differentiating $g_i=-\mu R_i/R^3$ gives the tidal tensorwhere ${\delta }_{ij}$ is the Kronecker delta and $I$ is the identity matrix. In an orthonormal basis with first axis radial, the tidal tensor isThe radial eigenvalue $2\mu /R^3$ separates nearby freely falling test particles; the two transverse eigenvalues $-\mu /R^3$ bring them together. The farther radial test particle falls inward less rapidly, so its acceleration relative to the nearer one points outward. Transverse gravitational accelerations converge toward the source. Thus an initially resting small cloud stretches radially and squeezes in both transverse directions. These are instantaneous relative accelerations; the subsequent shape also depends on initial velocities and the changing tidal tensor along the cloud's free fall.

Outside the source, the matrix trace is $2\mu /R^3-\mu /R^3-\mu /R^3=0$. This matches $\mathrm{tr}T=-{\nabla }^2\Phi =0$ in a region with zero mass density, by the Poisson equation for Newtonian gravity. Zero matrix trace means no leading change of a small initially resting cloud's volume at order $t^2$, not that each separation stays fixed. Inside matter, the same sign convention gives $\mathrm{tr}T=-4\pi G\rho$ instead.

Subtracting the constant and uniform-acceleration terms from the point-mass relative potential gives ${\psi }_{\mathrm{tid}}=G{m}_p[3(\mathbf{r}\cdot {\hat{\mathbf{r}}}_p{)}^2-r^2]/(2r_p^3)$, with force convention $\nabla \psi$. Its tidal tensor is $G{m}_p(3{\hat{\mathbf{r}}}_p{\hat{\mathbf{r}}}_p^T-I)/r_p^3$. This is the leading term for target size small compared with separation.