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Differentiate $f=g(S,S)$ twice, use geodesic deviation, and eliminate $g(DS,DS)$ with the conserved $K$:The most informative way to interpret this linear differential equation is to express $S$ in a parallel-propagated orthonormal frame along the timelike geodesic. Its three spatial components satisfy ${\ddot{S}}^i=\kappa S^i$.

If $R>0$, put $H=\sqrt{R/12}$. With constant spatial vectors $A,B$ in that frame,Thus generic deviations grow in length like $e^{H\tau }$, and their squared separation grows like $e^{2H\tau }$. Initially comoving neighbors, $DS(0)=0$, have $S(\tau )=S(0)\cosh (H\tau )$ and exhibit this growth. The unqualified printed assertion is too strong: the nonzero Jacobi field $S=B{e}^{-H\tau }$ has $K=0$ and shrinks exponentially. Such a Jacobi field is realizable by varying initial position and initial unit timelike velocity, with initial relative velocity $DS(0)=-HB$.

If $R<0$, put $\omega =\sqrt{-R/12}$. ThenThe deviations are bounded and oscillatory, rather than exponentially growing. The squared separation is periodic with period $\pi /\omega$ or is constant. Initially comoving neighbors have $S(\tau )=S(0)\cos (\omega \tau )$ and focus at $\tau =\pi /(2\omega )$; arbitrary vector initial data need not focus together. For completeness, $R=0$ gives $S=A+B\tau$ and $f=|A+B\tau {|}^2$. These conclusions concern infinitesimal geodesic deviation on the domain of the variation, not arbitrarily large finite separations.

Put $A,B\in \{1,2\}$ and write $h_{AB}=H_{AB}{e}^{ik\cdot x}$, where $H_{11}=H_{+}$, $H_{22}=-H_{+}$ and $H_{12}=H_{21}=H_{\times }$. Physical fields are the real parts. Use $c=1$ here, and take the initial separation at equal time, so $L^0=0$. For propagation in the positive $x^3$ direction one may set $k_{\mu }=(-\omega ,0,0,\omega )$; only $k_0^2={\omega }^2$ enters the answer.

The linearized Levi-Civita connection in this transverse-traceless gauge has the potentially nonzero componentswith all others zero to this order. In particular ${\Gamma }^{\mu }{}_{00}=0$, so a particle initially at rest has fixed spatial coordinates in transverse-traceless gauge, and $t$ equals its proper time. Expanding the Riemann curvature tensor givesThe plus sign in the last expression agrees with $R(T,S)T$ in the geodesic deviation convention above.

For the separation measured in a parallel-propagated orthonormal frame, write $S^{\hat{\mu }}=L^{\hat{\mu }}+{\xi }^{\hat{\mu }}$. The frame removes the connection terms from the time derivative, so the geodesic deviation equation reduces to ${\ddot{\xi }}^{\hat{A}}=R^{\hat{A}}{}_{\hat{0}\hat{0}\hat{B}}{L}^{\hat{B}}={\ddot{h}}_{AB}{L}^{\hat{B}}/2$ at first order. The observable transverse relative acceleration isThese are the plus polarization and cross polarization tidal patterns; there is no longitudinal acceleration. The curvature is evaluated along the reference worldline, and replacing $S$ by $L$ on the right changes only second-order terms. This is the infinitesimal-separation approximation intrinsic to geodesic deviation.

There is an important component distinction if the symbols $L^{\mu }+{\xi }^{\mu }$ are instead interpreted literally in the transverse-traceless gauge coordinate basis. Expanding the left side, rather than replacing a covariant derivative by an ordinary derivative, givesThus in TT coordinatesWith zero initial coordinate displacement and velocity, ${\xi }_{\mathrm{coord}}=0$ throughout. This does not remove the physical gravitational wave strain. To first order an orthonormal coframe in spacetime along the reference worldline is ${\vartheta }^{\hat{A}}=({\delta }^A{}_B+h_{AB}/2)d{x}^B$; its dual frame is parallel propagated because ${\partial }_0({\delta }^A{}_B-h_{AB}/2)+{\Gamma }^A{}_{0B}=0$. Henceafter choosing the initial flat-frame reference. The TT coordinate separation and measured separation distinction accounts exactly for the apparently different accelerations. In a region initially free of the wave, the measured displacement is ${\xi }^{\hat{A}}=h_{AB}{L}^{\hat{B}}/2$ with the corresponding initial conditions.

Use linearization around Minkowski spacetime after constant rescaling of $x,y$: set $A=1+\epsilon a(u)$, $B=1+\epsilon b(u)$, $u=t-z$. To first order,Hence $s=a+b$ is affine in $u$, while $h=a-b$ is the freely propagating profile. The affine transverse trace in a diagonal linearized plane wave is pure gauge: it has zero linearized curvature and is locally a coordinate artifact. To see this without imposing extra boundary conditions, use a first-order linearized coordinate gauge transformation $h_{\mu \nu }\mapsto h_{\mu \nu }-{\partial }_{\mu }{\xi }_{\nu }-{\partial }_{\nu }{\xi }_{\mu }$, withThe $tx,ty,zx,zy$ changes cancel and the remaining time/longitudinal changes are proportional to $s^{\prime \prime }=0$. The transverse perturbation becomeswith all time and longitudinal components zero. It is a transverse-traceless gauge plane gravitational wave travelling in the positive $z$-direction, with plus polarization and no cross polarization in these transverse axes.

For freely falling nearby particles, geodesic deviation gives opposite transverse tidal accelerations, since $R_{txtx}=-\epsilon a^{\prime \prime }$ and $R_{tyty}=-\epsilon b^{\prime \prime }=+\epsilon a^{\prime \prime }$ in vacuum. Thus the transverse geodesic deviation eigenvalues have opposite signs; the stretching and compressing roles interchange when the curvature profile changes sign. If $h^{\prime \prime }=0$ as well, there is no linearized tidal wave: that special profile is also locally flat to this order. The exact diagonal metric describes a fixed transverse polarization through its opposite vacuum curvature eigenvalues; the linearized calculation identifies it as plus relative to the chosen axes.

The coordinate vector fields on the parameter surface commute, so their pushforwards satisfyBecause the Levi-Civita connection is torsion free, part (i) then givesEvery $t$-curve is an affinely parametrized geodesic, hence ${\nabla }_{T}T=0$. Applying the definition of the Riemann curvature tensor,Therefore the connecting vector is a Jacobi field and obeys the geodesic deviation equation

The fields $X$ and $S$ arise by pushing forward the two coordinate vectors on the two-parameter family of geodesics, so $[X,S]=0$. Part (i), or equivalently the torsion-free property, givesIn components,ThereforeThis says that the deformation tensor of a timelike geodesic congruence gives the relative velocity of neighboring geodesics represented by the geodesic deviation vector.

The vector $T^a$ is the tangent to a reference member of a one-parameter family of affinely parametrized geodesics. The Jacobi field $Y^a$ is the infinitesimal connecting vector from that geodesic to a neighboring one at equal parameter. The geodesic deviation equation states that curvature determines their relative acceleration.

In the propagated frame $T=e_0$, and the constant sectional curvature formula reduces geodesic deviation toThe stated temporal initial data give $y^0(t)=0$. Writing $\omega =\sqrt{-K}$ for $K<0$ and $\kappa =\sqrt{K}$ for $K>0$, the spatial displacement isThese formulas are valid to first order in the initial separation and relative velocity.

The curvature 2-forms show that orthonormal curvature components grow as $m/r^3$. The geodesic deviation equation converts them into the relative acceleration of nearby parts of an observer. As $r\to 0$, the Schwarzschild tidal force stretches radial separations and compresses transverse separations with unbounded magnitude. An extended body therefore undergoes destructive tidal deformation before reaching the Schwarzschild singularity.

The flow of a Killing vector field consists of local isometries. Isometries preserve both the metric and its unique torsion-free metric-compatible Levi-Civita connection, hence ${\mathcal{L}}_K\nabla =0$. Set $X=K$ and $Y=Z=U$ in part (b). Since ${\nabla }_{U}U=0$,and antisymmetry of the first two curvature arguments givesThis is the geodesic deviation equation with connecting field $K$: applying the isometry to the original geodesic produces a neighboring geodesic, and curvature determines their relative acceleration.

A gravitational wave of plus polarization has opposite transverse strain or geodesic deviation eigenvalues along two fixed perpendicular axes. For a sinusoidal periodic function profile, the stretching and compressing roles alternate every half-period.

The orthonormal-frame components of Schwarzschild curvature scale as $M/r^3$. The geodesic deviation equation therefore gives radial stretching and transverse compression whose magnitude diverges as $r\to 0$.

A short rigid nonrotating guide constrains two freely sliding beads to one axis $n$. In its parallel-propagated orthonormal frame, the axial geodesic deviation is $\ddot{L}=-R_{n0n0}L$. For a transverse gravitational wave, $R_{n0n0}=-ϵ{\ddot{h}}_{nn}/2$, so a wave incident after the beads are at rest changes the measured separation by $ϵh_{nn}{L}_0/2$ to first order. Transverse reaction forces do not supply an axial restoring force. Alignment with the propagation direction gives no first-order length change.

In transverse-traceless gauge, particles initially at rest have fixed spatial coordinates because ${\Gamma }^{\mu }{}_{00}=0$. Their separation measured in a parallel-propagated orthonormal frame changes: to first order $S^{\hat{A}}=S^A+h_{AB}{S}^B/2$. Hence the physical geodesic deviation is ${\ddot{\xi }}^{\hat{A}}={\ddot{h}}_{AB}{L}^B/2$, whereas the coordinate displacement obeys ${\ddot{\xi }}_{\mathrm{coord}}^A=0$. The extra ${\partial }_0{\Gamma }^A{}_{0B}{L}^B$ term in the coordinate expansion of the second covariant derivative reconciles the two results. This applies to infinitesimal separations and linear metric perturbations.