Geodesic Deviation and the Tidal Tensor

Stand in a freely-falling elevator and, by the equivalence principle, gravity vanishes: you float, your keys float, and nothing you do to a single object betrays the field outside. Yet gravity is manifestly there — the elevator is falling toward something. The resolution is that gravity is never a property of one worldline; it lives in the relation between neighbouring worldlines. Release two marbles side by side and they drift together as both fall toward the Earth's centre; release two stacked vertically and they drift apart, because the lower one is pulled harder. This residual, un-transformable-away drift is the tidal field, and the geodesic deviation equation is its exact statement: the second covariant derivative of the separation vector between two nearby geodesics equals the Riemann curvature tensor contracted twice with the four-velocity. That single equation converts the Riemann tensor from an abstract device — the commutator of covariant derivatives, from this chapter's preceding article — into something an experimenter can measure. The tidal information splits cleanly in two: a trace, fixed by the Ricci tensor and hence by local matter through Einstein's equation, telling a small dust ball whether its volume shrinks; and a trace-free part, the Weyl tensor, which shears the ball at fixed volume and survives in vacuum — the physics of tides across empty space and of the gravitational waves LIGO records. This article derives the equation from a family of geodesics, exhibits the tidal tensor and its symmetries, recovers Newton's Hessian of the potential, and follows the result into black-hole spaghettification and interferometry.

Key takeaways

The geodesic deviation equation, $\frac{D^2 \xi^\mu}{\mathrm{d}\tau^2} = -R^\mu{}_{\nu\rho\sigma}\,u^\nu \xi^\rho u^\sigma$ , governs how the separation $\xi^\mu$ between two nearby freely-falling geodesics evolves in time.
It gives the Riemann tensor its operational meaning: curvature is what a freely-falling observer measures as the relative (tidal) acceleration of neighbouring test particles.
The same role in Newtonian gravity is played by the tidal tensor $\partial_i\partial_j\Phi$ ; the equation is its exact relativistic generalisation, with $R^i{}_{0j0}\to\partial_i\partial_j\Phi$ in the weak-field, slow-motion limit.
A single worldline cannot detect a uniform field (equivalence principle); it is the difference of the field across a separation — its gradient — that is the locally measurable manifestation of gravity.
The tidal tensor's trace is set by the Ricci tensor and local matter; its trace-free part is the Weyl tensor and describes shape-changing tides that survive in vacuum, including gravitational waves.
Key idea: curvature is operationally defined as the relative acceleration of neighbouring freely-falling particles, $\frac{D^2\xi^\mu}{\mathrm{d}\tau^2}=-R^\mu{}_{\nu\rho\sigma}u^\nu\xi^\rho u^\sigma$ — the Riemann tensor is the relativistic tidal tensor.

Where this lives today

The geodesic deviation equation is the working equation of gravitational-wave astronomy. Every event in the LIGO–Virgo–KAGRA catalogue is read out as a time-varying tidal field: the strain $h(t)$ recorded by the interferometers is literally the solution of $\ddot\xi^i=\tfrac12\ddot h^{TT}_{ij}\xi^j$ for the mirror separations, and space-based LISA will apply the same equation across millions of kilometres to catch merging supermassive black holes. The trace side drives another frontier: the focusing term $R_{\mu\nu}u^\mu u^\nu$ , summed over a congruence, is the engine of the Raychaudhuri equation and the Penrose–Hawking singularity theorems — still active as researchers ask whether quantum energy conditions can evade the classical inevitability of singularities. Closer to home, gravimetry missions such as GRACE-FO track two spacecraft responding to $\partial_i\partial_j\Phi$ to weigh groundwater and ice from orbit, and extreme-mass-ratio inspirals need the full relativistic tidal tensor to model how a small body is nudged off a background geodesic.

The physical picture

Drop two ball bearings toward the Earth, one above the other: the lower one sits in slightly stronger gravity and pulls ahead, so the gap stretches. Drop two side by side: each falls along a slightly tilted radial line toward the centre, so they converge. A freely-falling observer feels no falling — that is the equivalence principle — but can absolutely see these relative motions. This residual, unremovable signal is the tidal field, and the geodesic deviation equation asserts that it equals the Riemann tensor contracted with the four-velocity. Why a second derivative of the separation, and not a first? A uniform field accelerates every particle identically, so their relative velocity never changes and the first derivative carries no field information — it can be transformed away by choosing a falling frame. Only the variation of the field across the gap — the gradient, a second derivative of the potential — produces relative acceleration. Curvature is intrinsically a gradient-of-field object, which is why it cannot be erased by any change of frame. The tidal tensor deforms a small cloud of freely-falling dust exactly as a strain tensor deforms an elastic body: its trace says whether the cloud's volume grows or shrinks (zero in vacuum), while its trace-free part turns a sphere into an ellipsoid — near a point mass, stretch along the line to the mass and squeeze across it.

Figure

A falling ring of dust preserves its volume but shears into an ellipse — stretch along the line to the mass with relative acceleration +2GMr⁻³ξ, squeeze across it with −GMr⁻³ξ, and the eigenvalues sum to zero so the enclosed volume is conserved in vacuum

The split is the whole difference between gravity inside matter and in empty space. Inside a star the trace is positive: a comoving ball is focused, its volume driven to shrink — the seed of collapse and of the singularity theorems. Outside any source the trace vanishes and only the shearing Weyl part remains; a gravitational wave is exactly such an oscillating vacuum tide, stretching a ring of free masses along one axis and squeezing the perpendicular at constant area. How hard the tides bite depends violently on the source, as the next figure shows for black holes.

Figure

The tidal field at a black-hole horizon scales as c⁶ over 4G²M², falling as the inverse square of the mass, so a solar-mass hole shreds an infalling body far outside its horizon while a supermassive hole such as Sgr A* is crossed imperceptibly

The core equations

The central equation is compact. Let $\xi^\mu$ point from one freely-falling particle to a nearby one and $u^\mu$ be the four-velocity; the relative acceleration of the pair — the second covariant derivative of $\xi^\mu$ along the motion — is fixed entirely by curvature.

\frac{D^2 \xi^\mu}{\mathrm{d}\tau^2} = -\,R^\mu{}_{\nu\rho\sigma}\,u^\nu\,\xi^\rho\,u^\sigma \qquad (1)

where: $\xi^\mu$ — deviation (connecting) vector between neighbouring geodesics; $u^\mu$ — four-velocity, $u^\mu u_\mu=-1$ ; $\frac{D}{\mathrm{d}\tau}=u^\nu\nabla_\nu$ — covariant derivative along the geodesic; $R^\mu{}_{\nu\rho\sigma}$ — Riemann tensor; $\tau$ — proper time. Reading: relative acceleration is curvature, no free constants. Set $R=0$ and the pair drifts at constant relative velocity — flat space, gravity gauged away. The right-hand side is linear in $\xi^\rho$ and quadratic in $u$ , so it is natural to package the two $u$ -contractions into a single spatial operator acting on $\xi$ : the tidal tensor (the electric part of Riemann along $u$ ).

\mathcal{E}^\mu{}_{\rho}\equiv R^\mu{}_{\nu\rho\sigma}\,u^\nu u^\sigma,\qquad \frac{D^2\xi^\mu}{\mathrm{d}\tau^2}=-\,\mathcal{E}^\mu{}_\rho\,\xi^\rho \qquad (2)

where: $\mathcal{E}^\mu{}_\rho$ — tidal / electric part of Riemann along $u^\mu$ . Reading: in the rest frame $\mathcal{E}$ is a $3\times3$ symmetric matrix whose eigenvalues are the tidal accelerations per unit separation along its principal axes — positive for a squeeze, negative for a stretch. Three properties of $\mathcal{E}$ follow from the algebraic symmetries of Riemann, each with a physical consequence: symmetry means orthogonal principal axes exist; orthogonality to $u$ makes the tidal field purely spatial; the trace ties it to matter.

\mathcal{E}_{\mu\nu}=\mathcal{E}_{\nu\mu},\qquad \mathcal{E}_{\mu\nu}u^\nu=0,\qquad \mathcal{E}^\mu{}_\mu=R_{\nu\sigma}u^\nu u^\sigma \qquad (3)

where: $R_{\nu\sigma}=R^\mu{}_{\nu\mu\sigma}$ — Ricci tensor. Reading: the trace is not free geometry — it is $R_{\nu\sigma}u^\nu u^\sigma$ , which Einstein's equation fixes from the local energy density, so a dust ball's volume change is dictated by the matter it contains. In the weak-field, slow-motion limit the whole apparatus collapses onto Newton: the relevant curvature component becomes the Hessian of the potential and the trace becomes Poisson's equation.

\mathcal{E}_{ij}=\partial_i\partial_j\Phi,\qquad \mathrm{tr}\,\mathcal{E}=\nabla^2\Phi=4\pi G\rho \qquad (4)

where: $\Phi$ — Newtonian potential; $\rho$ — mass density; $G$ — Newton's constant. Reading: over matter the trace is positive and a comoving cloud is compressed; in vacuum ( $\rho=0$ ) it vanishes and only shape-changing tides remain. For the archetypal vacuum field, a point mass with $\Phi=-GM/r$ , the tidal tensor is explicit and trace-free — stretch along the radius, squeeze across it.

\mathcal{E}_{ij}=\frac{GM}{r^3}\big(\delta_{ij}-3\,\hat r_i\hat r_j\big) \qquad (5)

where: $M$ — central mass; $r$ — distance; $\hat r_i$ — radial unit vector. Reading: eigenvalues are $-2GM/r^3$ radially (stretch) and $+GM/r^3$ on each transverse axis (squeeze), summing to zero. This is the ocean-tide pattern — and, remarkably, the exact orthonormal-frame tidal field of a Schwarzschild black hole. Finally, the tidal tensor is only the Ricci slice of the full Riemann tensor. In four dimensions Riemann decomposes into a Ricci-built part and a trace-free conformal remainder, the Weyl tensor, which carries tides across empty space.

R_{\mu\nu\rho\sigma}=C_{\mu\nu\rho\sigma}+\big(g_{\mu[\rho}R_{\sigma]\nu}-g_{\nu[\rho}R_{\sigma]\mu}\big)-\tfrac{1}{3}R\,g_{\mu[\rho}g_{\sigma]\nu} \qquad (6)

where: $C_{\mu\nu\rho\sigma}$ — Weyl conformal tensor; $R$ — Ricci scalar; $g_{\mu\nu}$ — metric. Reading: in vacuum $R_{\mu\nu}=0$ , so $\mathcal{E}^\mu{}_\rho=C^\mu{}_{\nu\rho\sigma}u^\nu u^\sigma$ — the tidal field is pure Weyl, volume-preserving, and can propagate as a wave.

Derivation ADeriving the geodesic deviation equation from a family of geodesics

The goal is to turn a picture — two neighbouring freely-falling worldlines drifting apart — into equation (1), showing the relative acceleration is not merely related to curvature but is exactly the Riemann tensor contracted with the four-velocity, with no extra terms. We build a smooth family of geodesics, use the fact that the tangent and connecting vectors come from partial derivatives of one map (so their Lie bracket vanishes), and let the curvature commutator swap two covariant derivatives — using only the geodesic condition, the torsion-free connection, and the definition of Riemann.

Step 1. Let $x^\mu(\tau,s)$ be a smooth family in which each curve of constant $s$ is an affinely parametrised geodesic, and define the tangent $u^\mu=\partial x^\mu/\partial\tau$ and the connecting vector $\xi^\mu=\partial x^\mu/\partial s$ . The geodesic condition is $u^\nu\nabla_\nu u^\mu=0$ . Why: this is the only dynamical input — each particle is in free fall — and it will kill one of the two terms in the second derivative.

Step 2. Because $u$ and $\xi$ are partials of one map, mixed partials commute, and for a torsion-free connection this promotes to $u^\nu\nabla_\nu\xi^\mu=\xi^\nu\nabla_\nu u^\mu$ . Why: it trades a $\tau$ -derivative of $\xi$ for an $s$ -derivative of $u$ — the move that converts "rate of change of separation" into "spread of the velocity field".

Step 3. The relative velocity is the first covariant derivative, $\frac{D\xi^\mu}{\mathrm{d}\tau}=u^\nu\nabla_\nu\xi^\mu=\xi^\nu\nabla_\nu u^\mu$ , by Step 2. Why: this exposes the relative velocity as $\xi$ acting on the velocity field, ready to differentiate once more.

Step 4. Differentiate again by the Leibniz rule: $\frac{D^2\xi^\mu}{\mathrm{d}\tau^2}=u^\rho\nabla_\rho(\xi^\nu\nabla_\nu u^\mu)=(u^\rho\nabla_\rho\xi^\nu)\nabla_\nu u^\mu+\xi^\nu u^\rho\nabla_\rho\nabla_\nu u^\mu$ . Why: this is the relative acceleration, split to isolate the piece where two derivatives must be commuted.

Step 5. In the first term use Step 2 again, $u^\rho\nabla_\rho\xi^\nu=\xi^\rho\nabla_\rho u^\nu$ , giving $(\xi^\rho\nabla_\rho u^\nu)\nabla_\nu u^\mu$ . Why: this will cancel precisely against a piece generated in Step 7, leaving only curvature.

Step 6. In the second term commute the derivatives with $[\nabla_\rho,\nabla_\nu]u^\mu=R^\mu{}_{\sigma\rho\nu}u^\sigma$ , so $u^\rho\nabla_\rho\nabla_\nu u^\mu=u^\rho\nabla_\nu\nabla_\rho u^\mu+R^\mu{}_{\sigma\rho\nu}u^\sigma u^\rho$ . Why: here the Riemann tensor enters — the failure of covariant derivatives to commute is the entire physical content.

Step 7. Reorganise the leftover second-derivative piece: $\xi^\nu u^\rho\nabla_\nu\nabla_\rho u^\mu=\xi^\nu\nabla_\nu(u^\rho\nabla_\rho u^\mu)-(\xi^\nu\nabla_\nu u^\rho)\nabla_\rho u^\mu=-(\xi^\nu\nabla_\nu u^\rho)\nabla_\rho u^\mu$ , where the first bracket vanishes by the geodesic condition of Step 1. Why: the surviving term here is the exact negative of the term from Step 5, so the two non-curvature contributions annihilate.

Step 8. Adding the pieces, the non-curvature terms cancel and $\frac{D^2\xi^\mu}{\mathrm{d}\tau^2}=R^\mu{}_{\sigma\rho\nu}u^\sigma\xi^\nu u^\rho=-R^\mu{}_{\nu\rho\sigma}u^\nu\xi^\rho u^\sigma$ after relabelling and using antisymmetry $R^\mu{}_{\sigma\rho\nu}=-R^\mu{}_{\sigma\nu\rho}$ . Why: this is equation (1) — relative acceleration is curvature, cross-checked against Wald Eq. (3.3.18) and Carroll Eq. (3.198) in the (−+++), $R^\rho{}_{\sigma\mu\nu}=\partial_\mu\Gamma^\rho_{\nu\sigma}-\partial_\nu\Gamma^\rho_{\mu\sigma}+\cdots$ convention.

Go deeper · Derivation B · 1 minThe Newtonian limit and Poisson's equation

Derivation BThe Newtonian limit and Poisson's equation

The goal is to prove that the geodesic deviation equation reduces exactly to Newtonian tides, showing $R^i{}_{0j0}\to\partial_i\partial_j\Phi$ in the weak-field, slow-motion, static regime and that the trace reproduces Poisson's equation $\nabla^2\Phi=4\pi G\rho$ . This is the calculation that anchors every relativistic tidal result to the Hessian of the potential a student already trusts, and it identifies which single Riemann component carries Newtonian gravity.

Step 1. Take a nearly-flat static metric $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ with $g_{00}=-(1+2\Phi)$ , $|h|\ll1$ , a slow observer $u^\mu\approx(1,0,0,0)$ , and $\tau\approx t$ . Then in equation (1) only $R^i{}_{0j0}$ survives. Why: with the four-velocity purely timelike, only curvature components with two time indices contribute, so a single block of Riemann does all the work.

Step 2. To linear order and with time derivatives negligible, $R^i{}_{0j0}\approx\partial_j\Gamma^i_{00}$ , the $\partial_0\Gamma^i_{j0}$ term dropping as a time derivative of a static field. Why: staticity leaves the tidal block as a pure spatial gradient of a Christoffel symbol.

Step 3. Evaluate the Christoffel symbol: $\Gamma^i_{00}=-\tfrac12 g^{ij}\partial_j g_{00}=\partial_i\Phi$ , using $g_{00}=-(1+2\Phi)$ and $g^{ij}\approx\delta^{ij}$ . Why: this is the Newtonian "acceleration due to gravity" $\ddot x^i=-\partial_i\Phi$ , so its gradient is the tidal field.

Step 4. Differentiate and insert into equation (1): $R^i{}_{0j0}\approx\partial_j\partial_i\Phi$ , so $\frac{\mathrm{d}^2\xi^i}{\mathrm{d}t^2}\approx-R^i{}_{0j0}\xi^j=-\partial_i\partial_j\Phi\,\xi^j$ . Why: this is exactly the Newtonian equation for two nearby falling masses — the tidal tensor has become the Hessian $\partial_i\partial_j\Phi$ .

Step 5. Take the trace: $R_{00}=R^i{}_{0i0}=\nabla^2\Phi=4\pi G\rho$ by Poisson's equation. Why: the trace is the Ricci component $R_{00}$ , which Einstein's equation ties to the mass density — the seam where geometry meets matter, matching MTW §1.6 and Hartle Ch. 21.

Go deeper · Derivation C · 1 minSymmetries, orthogonality, and why vacuum tides are pure Weyl

Derivation CSymmetries, orthogonality, and why vacuum tides are pure Weyl

The goal is to establish the three defining properties of the tidal tensor in equation (3) — symmetry, orthogonality to the four-velocity, and a trace equal to a Ricci contraction — from the algebraic symmetries of Riemann, then show that in vacuum it is trace-free and equal to a Weyl contraction. Together these explain why a falling dust ball has orthogonal principal axes, why only its spatial deformation matters, and why volume-preserving shear tides persist far from any matter.

Step 1. Lower the index and use the pair-exchange symmetry $R_{\mu\nu\rho\sigma}=R_{\rho\sigma\mu\nu}$ : $\mathcal{E}_{\mu\rho}=R_{\mu\nu\rho\sigma}u^\nu u^\sigma=R_{\rho\sigma\mu\nu}u^\sigma u^\nu=\mathcal{E}_{\rho\mu}$ , so $\mathcal{E}$ is symmetric. Why: a symmetric matrix has real eigenvalues and orthogonal eigenvectors, so the tidal field always has well-defined principal stretch-and-squeeze axes.

Step 2. Contract with $u^\rho$ : $\mathcal{E}_{\mu\rho}u^\rho=R_{\mu\nu\rho\sigma}u^\nu u^\sigma u^\rho=0$ , since $R$ is antisymmetric in $(\rho,\sigma)$ while $u^\rho u^\sigma$ is symmetric. Why: the tidal tensor has no time component in the rest frame — tides are a three-dimensional strain.

Step 3. Contract the first and third indices: $\mathcal{E}^\mu{}_\mu=R^\mu{}_{\nu\mu\sigma}u^\nu u^\sigma=R_{\nu\sigma}u^\nu u^\sigma$ . Why: the trace measures volume change, and this hands it to the Ricci tensor — via Einstein's equation, to local energy density.

Step 4. In vacuum $R_{\mu\nu}=0$ , so the trace vanishes and decomposition (6) leaves $\mathcal{E}^\mu{}_\rho=C^\mu{}_{\nu\rho\sigma}u^\nu u^\sigma$ . Why: with no local matter the tidal field is pure Weyl — volume-preserving but shape-shearing, exactly the tide of a gravitational wave and the field outside a star or black hole.

Limiting cases

Case	Result	Meaning
Flat spacetime	$R^\mu{}_{\nu\rho\sigma}=0\Rightarrow\frac{D^2\xi^\mu}{\mathrm{d}\tau^2}=0$	No tidal forces; neighbouring geodesics keep constant relative velocity. Gravity has been transformed away in an inertial frame.
Newtonian, weak-field slow-motion	$\ddot\xi^i=-\partial_i\partial_j\Phi\,\xi^j,\ R^i{}_{0j0}\approx\partial_i\partial_j\Phi$	The relativistic tidal tensor reduces to the Newtonian Hessian; Poisson's equation reappears in the trace.
Vacuum (outside a star, or a wave)	$R_{\mu\nu}=0\Rightarrow\mathcal{E}^\mu{}_\mu=0,\ \mathcal{E}^\mu{}_\rho=C^\mu{}_{\nu\rho\sigma}u^\nu u^\sigma$	Tidal tensor is trace-free and equals the Weyl part: pure shape distortion at constant volume, gravitational waves allowed.
Congruence trace (Raychaudhuri seed)	$\mathcal{E}^\mu{}_\mu=R_{\mu\nu}u^\mu u^\nu\geq0$ under the strong energy condition	The summed inward tidal acceleration focuses volume; this is why gravity is attractive and underlies the singularity theorems.

Common trapThe two derivatives are ordinary time derivatives, so tides are just the coordinate acceleration ξ̈ᵘ

They are not ordinary derivatives — they are covariant derivatives along the geodesic, $\frac{D}{\mathrm{d}\tau}=u^\nu\nabla_\nu$ . In a general frame the components $\ddot\xi^\mu$ contain Christoffel terms that have nothing to do with tides: they are the coordinate acceleration of the falling frame itself, present even in flat spacetime in curvilinear coordinates. Reading tides off $\ddot\xi^\mu$ therefore mixes real curvature with pure coordinate artefacts. Only in a local inertial (Fermi normal) frame along the central geodesic do the connection coefficients vanish on the worldline, so that there — and only there — the two derivatives coincide and $\ddot\xi^i\approx-R^i{}_{0j0}\xi^j$ reads the Riemann components directly as measured tidal accelerations. It is the same reason a single freely-falling observer feels nothing: Christoffel symbols, not curvature, encode the removable part of gravity.

Worked examples

1 · Ocean tides from the Moon (confidence builder)

The Moon has mass $M=7.35\times10^{22}\,\mathrm{kg}$ and sits $d=3.84\times10^8\,\mathrm{m}$ from Earth's centre; Earth's radius is $R=6.37\times10^6\,\mathrm{m}$ . Find the tidal acceleration the Moon produces between Earth's centre and the sub-lunar point, and compare it to $g$ .

Step 1. The Moon's field is a vacuum point-mass field, so equation (5) gives eigenvalue $2GM/d^3$ along the Earth–Moon line. At the sub-lunar point, separation $\xi=R$ from the centre, the tidal acceleration is $a_{\text{tidal}}=\frac{2GM}{d^3}R$ .

Step 2. Evaluate the eigenvalue: $\frac{2GM}{d^3}=\frac{2(6.674\times10^{-11})(7.35\times10^{22})}{(3.84\times10^8)^3}=\mathbf{1.73\times10^{-13}\ s^{-2}}$ — the curvature component, with units of inverse-time-squared, that the whole tide follows from.

Step 3. Multiply by $R$ : $a_{\text{tidal}}=(1.73\times10^{-13})(6.37\times10^6)=\mathbf{1.10\times10^{-6}\ m/s^2}$ , about $1.1\times10^{-7}$ of surface gravity $g=9.81\,\mathrm{m/s^2}$ . Tiny — yet acting coherently over the whole ocean it raises the tides twice daily. The Moon's tidal eigenvalue is 1.7×10⁻¹³ s⁻², giving a surface tidal acceleration of 1.1×10⁻⁶ m/s², about one ten-millionth of g.

2 · Spaghettification near a black hole (fully numeric)

Compare the tidal field at the horizon of a $10\,M_\odot$ stellar black hole with that at the horizon of Sagittarius A ( $4.1\times10^6\,M_\odot$ ), and estimate how far outside the stellar hole an infalling human is torn apart. Take a body length $L=1.8\,\mathrm{m}$ .*

Step 1. The horizon sits at $r_s=2GM/c^2$ , and the radial tidal eigenvalue there is $\frac{2GM}{r_s^3}=\frac{c^6}{4G^2M^2}\propto M^{-2}$ . For $10\,M_\odot$ : $r_s=\mathbf{29.5\ km}$ and the eigenvalue is $\mathbf{1.03\times10^{8}\ s^{-2}}$ . Across a $1.8\,\mathrm{m}$ body this is a head-to-foot differential of $1.03\times10^8\times1.8\approx1.9\times10^8\,\mathrm{m/s^2}$ — about $2\times10^7 g$ , lethal many times over.

Step 2. For Sgr A* the same formula gives $\frac{c^6}{4G^2M^2}=1.03\times10^{10}/(4.1\times10^6)^2=\mathbf{6.1\times10^{-4}\ s^{-2}}$ , a differential of only $\approx1.1\times10^{-3}\,\mathrm{m/s^2}$ ( $\sim10^{-4}g$ ) across the body — imperceptible. An astronaut crosses the supermassive horizon feeling nothing, while the stellar horizon shreds them.

Step 3. Where does the stretch first become fatal for the $10\,M_\odot$ hole? Taking a generous fatal differential of $\sim10^4\,\mathrm{m/s^2}$ across the body, set $\frac{2GM}{r^3}L=10^4$ : $r=\left(\frac{2GML}{10^4}\right)^{1/3}\approx\mathbf{780\ km}\approx26\,r_s$ . Spaghettification for a stellar-mass hole happens tens of horizon radii out — the victim is destroyed long before reaching the horizon, the opposite of the supermassive case.

3 · The strain a gravitational wave imposes on LIGO (fully numeric)

A gravitational wave of amplitude $h=10^{-21}$ and frequency $f=100\,\mathrm{Hz}$ passes an interferometer with $L=4\,\mathrm{km}$ arms. In the transverse-traceless gauge the geodesic deviation equation gives $\ddot\xi^i=\tfrac12\ddot h^{TT}_{ij}\xi^j$ . Find the peak arm-length change and the peak tidal acceleration of a mirror.

Step 1. The wave is a vacuum (pure Weyl) tidal field, trace-free and transverse, stretching one arm while squeezing the other. The displacement amplitude of a mirror at separation $L$ is $\delta L=\tfrac12 hL$ .

Step 2. Evaluate: $\delta L=\tfrac12(10^{-21})(4000)=\mathbf{2.0\times10^{-18}\ m}$ — about one-thousandth of a proton's radius, yet routinely measured. This is the literal solution of the geodesic deviation equation for the mirror separation.

Step 3. The tidal acceleration amplitude is $|\ddot\xi|\approx\tfrac12\omega^2 hL$ with $\omega=2\pi f=628\,\mathrm{rad/s}$ : $|\ddot\xi|=\tfrac12(628)^2(10^{-21})(4000)\approx\mathbf{7.9\times10^{-13}\ m/s^2}$ . The wave moves the mirrors by 2×10⁻¹⁸ m, driven by only 8×10⁻¹³ m/s² — the geodesic deviation equation is the actual equation of motion of a LIGO mirror.

Physics of the portrait

Geodesic deviation is where the abstract machinery of curved-space calculus becomes a statement about measurable forces, and the person who built that machinery — and named the deviation of neighbouring geodesics — was Tullio Levi-Civita.

Origins · 1 minTullio Levi-Civita and how this was discovered

Tullio Levi-Civita

1873–1941 · tensor calculus and the geometry of geodesic deviation

Levi-Civita, born in Padua, was with his teacher Gregorio Ricci-Curbastro the co-creator of the "absolute differential calculus" — the tensor calculus Einstein seized on to write general relativity. But his most consequential idea came in 1917, when he introduced parallel transport on a curved manifold, the object now called the Levi-Civita connection. Parallel transport makes "the same direction, moved along a path" meaningful in curved space, and it is precisely the failure of that transport to be path-independent — curvature — that the geodesic deviation equation measures. In the mid-1920s he developed the theory of geodesic deviation itself (the "deviazione geodetica"), giving the effect its name and modern form; his 1925 lectures, translated in 1927 as The Absolute Differential Calculus, became the standard text. Barred from his Rome chair in 1938 by Italy's fascist racial laws and cut off from academic life, he died isolated in 1941, having supplied the exact language in which every modern statement of tidal gravity is written.

History

1687

— Newton explains the ocean tides as the differential gravitational pull of Moon and Sun across the Earth

1917

— Levi-Civita introduces parallel transport, giving curvature a precise geometric meaning

1926

— Levi-Civita formulates the equation of geodesic deviation and names the effect

1956

— Pirani identifies the Riemann tensor as the operational tidal field measured by neighbouring test particles

2015

— LIGO's detection of GW150914 directly measures the geodesic deviation of freely-falling mirrors

Going further

References. Misner, Thorne & Wheeler, Gravitation (1973), §1.6, §11.1–11.4, §13.6 (equation of geodesic deviation; tidal forces). · Wald, General Relativity (1984), §3.3 (geodesic deviation), §4.4a. · Carroll, Spacetime and Geometry (2004), §3.10 (geodesic deviation equation). · Hartle, Gravity (2003), Ch. 21 (geodesics and tidal forces). · Schutz, A First Course in General Relativity, 2nd ed. (2009), §6.5.

Provenance. All equations and conventions are cross-checked against Wald §3.3 (geodesic deviation, Eq. 3.3.18); Carroll §3.10 (Eq. 3.198); MTW §1.6, §11.1–11.4 and §13.6 (tidal forces, Schwarzschild tidal components); Hartle Ch. 21; Schutz §6.5. A verifier should: (1) reproduce the derivation using the curvature commutator in the (−+++), $R^\rho{}_{\sigma\mu\nu}=\partial_\mu\Gamma^\rho_{\nu\sigma}-\partial_\nu\Gamma^\rho_{\mu\sigma}+\Gamma\Gamma-\Gamma\Gamma$ convention; (2) verify $R^i{}_{0j0}=\partial_i\partial_j\Phi$ in the linearised static metric; (3) confirm the Schwarzschild orthonormal tidal components $R_{\hat t\hat r\hat t\hat r}=-2GM/r^3$ , $R_{\hat t\hat\theta\hat t\hat\theta}=+GM/r^3$ (trace-free); (4) check the TT-gauge response $\ddot\xi^i=\tfrac12\ddot h^{TT}_{ij}\xi^j$ ; and (5) reproduce the numerical anchors — Moon tidal eigenvalue $1.7\times10^{-13}\,\mathrm{s^{-2}}$ , stellar/supermassive horizon tides $c^6/4G^2M^2$ , LIGO strain $\delta L=\tfrac12 hL$ .

Builds on: Riemann Curvature Tensor · Curvature as the Commutator of Covariant Derivatives · Algebraic Symmetries of the Riemann Tensor Leads to: Ricci Tensor · Weyl Conformal Tensor · Einstein Tensor

Ask the tutor: Ask me to work out the tidal tensor for a specific metric — Schwarzschild at the photon sphere, a gravitational-wave background, or an FLRW cosmology — and watch how its trace and Weyl parts encode matter versus pure shape.

Is the geodesic deviation equation the same thing as the Newtonian tidal force?

It is the exact relativistic parent of it. In the weak-field, slow-motion, static limit the only surviving curvature component is $R^i{}_{0j0}$ , and it equals the Hessian $\partial_i\partial_j\Phi$ , so the equation becomes $\ddot\xi^i=-\partial_i\partial_j\Phi\,\xi^j$ — the Newtonian statement that two falling masses accelerate apart or together by the second derivatives of the potential. What GR adds is that the tidal tensor is a piece of the Riemann tensor, valid in strong fields and fast motion, with a trace tied to matter and a trace-free part that can propagate as a wave. Newton is the leading term; this is the whole function.

Why must the derivatives be covariant rather than ordinary?

Because the components $\xi^\mu$ are measured against a coordinate basis that itself accelerates and rotates. An ordinary $\mathrm{d}^2\xi^\mu/\mathrm{d}\tau^2$ mixes that basis motion — carried by Christoffel symbols, present even in flat spacetime in curvilinear coordinates — with the genuine tidal effect. The covariant $D/\mathrm{d}\tau=u^\nu\nabla_\nu$ subtracts the basis motion and isolates the physical relative acceleration. Only in a Fermi normal frame along the central geodesic do the connection coefficients vanish on the worldline, so the two agree there and $\ddot\xi^i\approx-R^i{}_{0j0}\xi^j$ becomes a direct readout of the tides.

How is the tidal tensor split between Ricci and Weyl, and what does each part do?

It divides by its trace. The trace $\mathcal{E}^\mu{}_\mu=R_{\rho\sigma}u^\rho u^\sigma$ is set by the Ricci tensor, and through Einstein's equation by the local energy density — it controls whether a dust ball's volume shrinks or grows, and is positive over ordinary matter (focusing). The trace-free remainder is the Weyl part and describes shape distortion at fixed volume: stretch along one axis, squeeze along the others. In vacuum the Ricci part vanishes, leaving pure Weyl tides — a falling sphere keeps its volume but is sheared into an ellipsoid, exactly what a gravitational wave, or the field outside a star, does.

Why does the tidal tensor come out trace-free near a point mass but not inside matter?

The trace equals $R_{\mu\nu}u^\mu u^\nu$ , which Einstein's equation fixes from the local matter. In vacuum there is no matter, so Ricci and the trace vanish: the eigenvalues $-2GM/r^3$ (stretch) and $+GM/r^3$ on each transverse axis sum to zero, and a falling ball keeps its volume while distorting in shape. Inside matter the density is non-zero, $\nabla^2\Phi=4\pi G\rho>0$ , and the trace is positive — the ball is compressed. The vanishing trace in vacuum is the signature of "empty space still has tides, but volume-preserving ones."

Does a gravitational wave really stretch space, and if so why doesn't the ruler stretch too?

A wave is a time-varying vacuum tidal field, and the honest statement is about freely-falling test masses: their proper separations oscillate, $\ddot\xi^i=\tfrac12\ddot h^{TT}_{ij}\xi^j$ . A rigid ruler is not freely falling — its atoms are held by electromagnetic forces that vastly overwhelm the tidal acceleration (about $10^{-12}\,\mathrm{m/s^2}$ for LIGO), so the ruler barely responds while the free mirrors move. That mismatch is what an interferometer measures. So "space stretches" is loose shorthand; the precise, frame-independent content is the relative acceleration of free test bodies.

What is the connection between geodesic deviation and the singularity theorems?

Sum the equation over a bundle (congruence) of geodesics and track the cross-sectional volume: the result is the Raychaudhuri equation, whose driving term is the tidal trace $R_{\mu\nu}u^\mu u^\nu$ . The strong energy condition makes it non-negative, so gravity always focuses a congruence — a diverging bundle is bent back toward convergence, and once convergence sets in the bundle reaches a caustic in finite proper time. Penrose and Hawking turned this focusing, plus global causal structure, into theorems guaranteeing geodesic incompleteness — singularities — inside black holes and at the Big Bang. The tidal trace is the local seed of a global inevitability.