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In a parallel-propagated orthonormal frame, orthogonal geodesic deviation in constant sectional curvature reduces to three scalar equations ${\ddot{S}}^i=\kappa S^i$. For $\kappa >0$, $S=A{e}^{\sqrt{\kappa }\tau }+B{e}^{-\sqrt{\kappa }\tau }$: generic initial data grow exponentially, but the pure decaying mode is a counterexample to universal growth. Initially comoving neighbors have $S=S(0)\cosh (\sqrt{\kappa }\tau )$. For $\kappa <0$, $S=A\cos (\sqrt{-\kappa }\tau )+B\sin (\sqrt{-\kappa }\tau )$, so the deviation is bounded and oscillatory. Simultaneous focusing requires compatible vector initial data and is not automatic for every oscillatory solution.

For a scalar, the Lie derivative of a function is its directional derivative:The Lie derivative of a vector field is their Lie bracket of vector fields:For a torsion-free connection,The last term vanishes because the connection coefficients are symmetric in $b,c$. This proves ${\mathcal{L}}_{\xi }X={\nabla }_{\xi }X-{\nabla }_X\xi$.

For the metric tensor statements use the Levi-Civita connection, which also has metric compatibility. The tensor derivation property of the Lie derivative of a tensor field givesExpanding the first term using $\nabla g=0$ and using the vector identity cancels the derivatives of $X$ and $Y$. What remains isSince $X,Y$ are arbitrary, the Killing equation isMetric compatibility is needed here, even though symmetry alone sufficed for the vector identity. In general the Lie derivative of a metric with nonmetricity contains a term ${\xi }^c{\nabla }_c{g}_{ab}$. For example, with the metric tensor $g=e^{2x}d{x}^2$, zero connection coefficients and $\xi ={\partial }_x$, the Lie derivative of a tensor field is $2e^{2x}$, whereas $2{\nabla }_x{\xi }_x=4e^{2x}$. The usual metric connection interpretation of the question supplies the required compatibility.

For the Riemann curvature tensor calculation, keep the original PDF's convention. With $R_{abcd}=g_{ae}{R}^e{}_{bcd}$ and $R_{abc}{}^d=g^{de}{R}_{abce}$, it meansThus the commutator on a covector has the opposite sign. Write $K_{abc}={\xi }_{a;bc}={\nabla }_c{\nabla }_b{\xi }_a$. Taking a covariant derivative of the Killing equation makes $K$ antisymmetric in its first two indices, while the Ricci identity givesCombining this with the first-two-index antisymmetry gives the three relationsAdd the first and third and subtract the second. Pair symmetry and the first Bianchi identity imply $R_{adbc}+R_{bdca}-R_{cdab}=-2R_{abcd}$. ConsequentlyThis is the second covariant derivative of a Killing vector in the paper's convention; its sign is tied to the declared curvature-sign convention in Killing derivative identities, not to raising a particular index.

Set $L_{ab}={\xi }_{a;b}$. Along any smooth curve with tangent vector $V$, the chain rule and the just-proved identity give the Killing transport equationsChoose a parallel-propagated orthonormal frame along the curve. These become a homogeneous linear system of ordinary differential equations for the components of $(\xi ,L)$. Uniqueness of its initial value problem implies that zero initial data remain zero. Any point in the same connected component as $P$ can be joined to $P$ by a piecewise smooth curve. Therefore zero value and zero first derivative at $P$ force the Killing field and its derivative to vanish throughout that component, with no completeness assumption. On a manifold with one connected component this is the requested global result. Connectedness cannot be dropped: on two disjoint flat spacetimes, take the zero field on one and a nonzero translation on the other.

On a pseudo-Riemannian manifold of dimension $n$ with one connected component, the map $\xi \mapsto ({\xi }_a(P),L_{ab}(P))$ is therefore injective. There are $n$ independent components of the value and $n(n-1)/2$ of its antisymmetric derivative. The dimension bound for Killing vector fields isIt is attained in flat space by ${\xi }_a=A_a+B_{ab}{x}^b$ with constant antisymmetric $B$, giving translations and infinitesimal rotations or Lorentz transformations. The four-dimensional maximum is 10. For a disconnected manifold the bound applies componentwise; there is no uniform global bound depending only on $n$ if the number of components is unrestricted.

A linearized coordinate gauge transformation changes the identification of points of the perturbed spacetime with its Minkowski spacetime background. With $x^{\prime i}=x^i+ϵ{\xi }^i$, the same physical metric has perturbationThese changes are coordinate freedom, not extra physical gravitational wave polarizations. Raising indices with $\eta$, define $h={\eta }^{ij}{h}_{ij}$ and the trace-reversed metric perturbation ${\bar{h}}_{ij}=h_{ij}-{\eta }_{ij}h/2$. Then

The first-order Levi-Civita connection is ${\Gamma }^{(1)k}{}_{ij}=\frac{1}{2}{\eta }^{k\ell }({\partial }_i{h}_{j\ell }+{\partial }_j{h}_{i\ell }-{\partial }_{\ell }{h}_{ij})$. Contracting the Riemann curvature tensor in the stated convention givesPut $A_j={\partial }^i{\bar{h}}_{ij}$. The corresponding Einstein tensor isUnder the linearized coordinate gauge transformation, $A_j^{\prime }=A_j-\square {\xi }_j$. Choosing a solution of $\square {\xi }_j=A_j$ imposes Lorenz gauge in linearized gravity, $A_j^{\prime }=0$. Thus the vacuum Einstein field equations reduce toThere remains residual gauge symmetry of linearized gravity with $\square {\xi }_j=0$.

For the transverse profiles, write both symmetric off-diagonal entries as $h_{xy}=h_{yx}=h_{\times }(t-z)$ and the diagonal entries as $h_{xx}=-h_{yy}=h_{+}(t-z)$. The trace-reversed metric perturbation has zero trace and therefore equals $h$. Its divergence vanishes: its only nonzero index directions are $x,y$, while its coefficients depend only on $t-z$. Also $(-{\partial }_t^2+{\partial }_z^2)F(t-z)=0$ for every twice differentiable profile $F$. These are consequently plane gravitational waves in linearized gravity in transverse-traceless gauge, with the two independent gravitational wave polarizations $h_{+}$ and $h_{\times }$.

To calculate the detector response, use a parallel-propagated orthonormal frame along the centre of a short nonrotating stick. To first order its clock reads $t$, and the tidal curvature isFor a sliding-bead gravitational wave detector, the stick constrains transverse motion but supplies no axial restoring force. Projecting geodesic deviation along an $x$-directed stick givesHere $L$ is the proper separation and $L_0=L(t_0)$, with $\dot{L}(t_0)=0$. Integrating with those actual initial conditions yieldsFor beads at rest before an incident wave, both initial profile terms vanish and the familiar result is $\boxed{L(t)=L_0[1+ϵh_{+}(t-z_0)/2]+O({ϵ}^2)}$. The cross polarization produces a transverse tidal force, balanced by the guide, and no axial first-order displacement in this orientation. In transverse-traceless gauge the same plus response can be read from the proper line element, $d\ell =\sqrt{1+ϵh_{+}}dx$; coordinate distances alone are not measured distances. When a cross polarization is present, keeping the guide physically nonrotating may require transverse coordinate adjustments, which do not change the axial answer at this order.

For a $z$-directed stick, $R_{z0z0}=0$ and the wave is transverse. Therefore beads initially at rest retain their proper separation to first order: $L(t)=L_0+O({ϵ}^2)$. The tidal calculation is the usual linear detector limit, with bead spacing small compared with the wavelength; it does not treat a macroscopic rigid body as exactly rigid over arbitrary relativistic distances.

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.

Parallel propagation means ${\nabla }_T{e}_{\alpha }=0$, where $T=\dot{\lambda }$. By metric compatibility,The inner products therefore retain their initial values ${\eta }_{\alpha \beta }$, so the parallel-propagated orthonormal frame remains orthonormal.

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.