Tides: curvature you can measure

10.1 Accelerometers and tidal measurements#

An ideal accelerometer carried by a freely falling point particle reads zero. This is true in Minkowski space and beside a black hole, provided the particle follows a geodesic and is treated as an ideal test body.

Release a collection of particles, however, and their separations can accelerate. Near a gravitating body, one part of the collection may be pulled into a different geodesic than another. A sufficiently extended astronaut cannot follow every nearby geodesic simultaneously while maintaining an unchanged shape. Internal stresses arise because the body resists that relative motion.

An accelerometer measures proper acceleration: how its own motion departs from free fall. A gravity gradiometer compares nearby freely falling bodies to measure the spatial variation of their motion. A single laboratory can contain several test masses and make this comparison internally.

The governing equation is geodesic deviation, also called the Jacobi equation. We will derive it using only concepts already assembled.

10.2 Organize the experiment as a family of worldlines#

Consider a smooth local family of timelike geodesics

where $\tau$ is proper time along each geodesic and $s$ labels neighboring geodesics. Picture a smooth strip of paths: one direction runs along a particle’s history, and the other runs across the strip to a neighboring particle. Restrict attention to a patch where these two labels are independent and the strip has no crossings. Define

The first vector moves along one worldline. The second moves across the family at fixed $\tau$. More precisely, $\xi^\mu ds$ describes an infinitesimal connecting displacement. A finite separation between distant events does not automatically define a unique tangent vector; the family and the infinitesimal limit supply the meaning here.

Because $\tau$ and $s$ are commuting parameters,

With zero torsion, the definition of torsion therefore gives

Read this as a statement about the same smoothly labeled grid of paths. The rate at which the separation changes along a trajectory equals the change in velocity across neighboring trajectories. It is the curved-space version of exchanging $\partial_\tau\partial_sx$ with $\partial_s\partial_\tau x$.

Proper-time parameterization gives $u^\mu u_\mu=-c^2$. Geodesic motion gives

These assumptions have different jobs: normalization makes $\tau$ a physical clock reading; affine geodesic motion removes a tangential reparameterization term from the acceleration equation.

10.3 Deriving geodesic deviation#

The covariant relative acceleration is

Replace the inner derivative using $\nabla_u\xi=\nabla_\xi u$:

Now use the definition of curvature to interchange the two derivatives:

The first term vanishes because every worldline in the family is geodesic. The final term vanishes because the parameters commute. We are left with

In components, this is $R^\mu{}_{\alpha\beta\nu}u^\alpha u^\beta\xi^\nu$. Use antisymmetry in the last pair to place the separation index before the final velocity index:

The minus sign follows from the stated curvature convention and this index ordering. It is not an independently adjustable physical sign.

The equation is exact for a Jacobi field generated by an infinitesimal variation of geodesics. Using it for particles separated by a finite distance is a linear approximation in their separation. Higher-order separation effects involve additional geometric information, including curvature variation. The same geometric derivation works for null geodesics with an affine parameter, but a null worldline has no proper-time parameter or timelike rest frame.

To interpret $\xi$ as spatial separation for the reference observer, choose it initially orthogonal to $u$. This orthogonality persists:

The first equality uses geodesic motion; the last uses the equal normalization of the family. This is why the separation can consistently be discussed in the reference observer’s instantaneous rest space.

10.4 Put an instrument frame on the reference geodesic#

Choose a parallel-transported orthonormal frame along the reference trajectory, with $e_{\hat0}=u/c$. In this frame $u^{\hat0}=c$, $u^{\hat i}=0$, and the frame itself has zero covariant rate of change along the trajectory. Covariant differentiation of vector components along it becomes ordinary differentiation.

The spatial deviation equation reduces to

The matrix $c^2R_{\hat i\hat0\hat j\hat0}$ has units of inverse time squared. Its eigenvectors identify the principal tidal directions. With our sign convention, a positive eigenvalue produces relative acceleration toward the reference trajectory in that direction; a negative eigenvalue produces relative acceleration away.

This matrix is an observer-dependent projection of the invariant Riemann tensor. Another observer with a different four-velocity can obtain a different tidal matrix. That does not make the effect a coordinate illusion: they are physically different observers performing different local experiments. A complete reconstruction of curvature requires enough independent relative-motion measurements, not merely one three-dimensional tidal matrix.

10.5 Recovering Newtonian tides#

Let $\Phi$ be a weak, slowly varying Newtonian potential. For this calculation use $x^0=ct$ and retain leading weak-field terms:

The spatial inverse metric can be replaced by $\delta^{ij}$ at this order in the particular expression we need. For an approximately static field,

Therefore

Time-derivative terms have been neglected by the static approximation, and products of weak-field connection coefficients are higher order. With $u^0\simeq c$ and slow reference motion, geodesic deviation gives

This is precisely Newtonian relative acceleration. If a nearby particle is at $\mathbf x+\boldsymbol\xi$, subtract the reference particle’s acceleration from the nearby particle’s acceleration and Taylor-expand:

The common acceleration disappears. Only the spatial gradient of acceleration remains. This is the mathematical content of the elevator argument: free fall removes a shared gravitational acceleration locally; it does not remove differences in gravitational acceleration across a finite laboratory.

For a point mass, let $r=\sqrt{x_ix_i}$ and $n_i=x_i/r$ be the components of the radial unit vector. Differentiation gives $\partial_jr=n_j$ and $\partial_jn_i=(\delta_{ij}-n_in_j)/r$. Outside the source, $\Phi=-G_NM/r$ has gradient $\partial_i\Phi=G_NM n_i/r^2$. Differentiate once more:

The Hessian has radial eigenvalue $-2G_NM/r^3$ and two tangential eigenvalues $+G_NM/r^3$. The actual relative accelerations carry the minus sign:

A falling cloud stretches radially and squeezes sideways. The eigenvalues of the acceleration map add to zero outside the source. This is the Newtonian shadow of vacuum Ricci-flatness, while the nonzero trace-free tidal field is the shadow of Weyl curvature.

The ratio $2:-1:-1$ compares radial stretching acceleration with the two transverse squeezing accelerations for equal initial separations. This pattern explains how strong tides can lengthen a falling body while narrowing it.

SPACETIME LAB / 05

Gravity changes a cloud’s shape

A small freely falling cloud stretches radially and squeezes in two independent sideways directions. Rotate to find the dimension a flat diagram hides.

ExploreDiagram

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Early-time deformation

Spherical cloudStretched cloud

$$\frac{\ddot\xi^i}{\xi^i}=\frac{GM}{r^3}(2,-1,-1)$$

Read the scene. The faint dashed circles retain the initial spherical shape. The local orthonormal axes are radial $\hat r$ and two transverse directions $\hat\theta,\hat\phi$; the three eigenvalues in the equation follow that order. In this local, early-time visualization with initially comoving particles, lengths change by $(1+2s,1-s,1-s)$. It preserves volume only to first order in the small parameter $s$.

WORKED EXAMPLE

Read the tidal matrix as an instrument

What can a tiny cloud determine—and what can it miss?

Builds onContractions and Weyl Spherical tidal calculation

See the idea

A freely falling observer releases nearby test masses with initially negligible relative velocity. Their relative accelerations determine a linear map from separation to acceleration. The observer’s clock and orthonormal axes make the map’s components physical measurements, within the small-cloud approximation.

Work it out

1. 01
   Calibrate the sign using spherical gravity

   At distance r from a spherical mass, set $q=G_NM/r^3>0$. In the Newtonian limit, differentiate the radial acceleration and compare neighboring directions. The acceleration map stretches radially and compresses in the two transverse directions.

   $$\ddot{\boldsymbol\xi}=\operatorname{diag}(2q,-q,-q)\boldsymbol\xi.$$

   Why this step works The inward field is weaker at the more distant radial test mass.

2. 02
   Infer the curvature measured by the cloud

   In the local orthonormal frame with $x^{\hat0}=c\tau$, geodesic deviation reads $\ddot\xi^{\hat i}=-c^2R_{\hat i\hat0\hat j\hat0}\xi^{\hat j}$ at leading order. Thus the curvature’s electric tidal entries are the negatives of the acceleration-map entries divided by $c^2$.

   $$R_{\hat i\hat0\hat j\hat0}=\frac1{c^2}\operatorname{diag}(-2q,q,q).$$

   Why this step works The curvature convention is calibrated against a physical relative acceleration.

3. 03
   Distinguish shape from volume

   The trace of the acceleration map is $2q-q-q=0$. An initially comoving infinitesimal cloud has no leading $\tau^2$ fractional volume change in this vacuum calibration, while its shape immediately begins to distort. Later volume changes can arise through the shear that has developed.

   $$\Delta\xi_i\simeq\tfrac12 a_i\tau^2,\qquad \sum_i a_i/\xi_i=0.$$

   Why this step works Trace-free initial acceleration does not imply every axis stays fixed or the volume stays constant forever.

Go deeper

A symmetric three-by-three tidal matrix supplies at most six numbers for one observer, fewer than the twenty independent components of a general four-dimensional Riemann tensor. In vacuum its electric Weyl part is trace free and has five. Define the magnetic Weyl part by $B_{ij}=\tfrac12\varepsilon_i{}^{kl}C_{klj0}$ in an oriented orthonormal frame, where $\varepsilon_{123}=1$, permutations change sign, and repeated indices give zero. It is another symmetric trace-free spatial matrix in vacuum. E and B together encode the ten vacuum Weyl components. This definition does not make B an ordinary magnetic field; measurements involving relative velocities or different observers probe information absent from one initially comoving cloud. Finite apparatus size also introduces curvature-gradient corrections.

Further readingDavid Tong · General relativity

Test the idea

FIRST, PREDICT

An initially comoving cloud has zero leading volume acceleration. Must its shape remain unchanged?

Yes. Volume determines all three axis lengths.It proves the entire Riemann tensor is zero.No. Opposite axis changes can cancel in the trace.

Compare the reasoning

Yes. Volume determines all three axis lengths.

One scalar cannot determine the independent changes of three axes.

It proves the entire Riemann tensor is zero.

The tidal map itself already contains nonzero measured components.

No. Opposite axis changes can cancel in the trace.

The radial stretching and transverse compression in the explicit vacuum example demonstrate this.

A hint

Add the axis rates only after looking at them separately.

NOW CHANGE THE EXAMPLE

With $q=0.01\,\mathrm{s^{-2}}$ and radial separation $3\,\mathrm m$, find the initial radial relative acceleration.

A hint

Use $2q\xi_r$.

Work through the solution

$2(0.01)(3)=0.06\,\mathrm{m/s^2}$.

State the observer and apparatus before saying which part of curvature has been measured.

10.6 Normal coordinates and the size of a laboratory#

At any regular event $p$, choose Riemann normal coordinates with an orthonormal basis at the origin. Then

Section 8.5 showed how a quadratic coordinate change cancels the connection at one event. Riemann normal coordinates have a further geometric definition. Choose an initial tangent $X$ at $p$, follow the geodesic with that tangent from affine parameter 0 to 1, and assign its endpoint the coordinate list $X^\mu$. This endpoint rule is called the exponential map, written $\exp_p(X)$. It is a smooth invertible map sufficiently near $X=0$, where its derivative is the identity. The resulting normal neighborhood is a region small enough that these geodesic labels remain unique.

But the derivatives of $\Gamma$ generally survive. The metric expansion is

Further calculation: the factor one third in the normal-coordinate metric

Radial geodesics have $X^\mu(\lambda)=\lambda v^\mu$, so their equation requires $\Gamma^\rho{}_{\mu\nu}(X)X^\mu X^\nu=0$. Write $C^\rho{}_{\mu\nu\alpha}=\partial_\alpha\Gamma^\rho{}_{\mu\nu}(0)$. Torsion-free symmetry and the cubic term of the radial condition give

$$C^\rho{}_{\mu\nu\alpha}=C^\rho{}_{\nu\mu\alpha}, \qquad C^\rho{}_{\mu\nu\alpha} +C^\rho{}_{\nu\alpha\mu}+C^\rho{}_{\alpha\mu\nu}=0.$$

At the origin, curvature is $R^\rho{}_{\sigma\mu\nu}=C^\rho{}_{\nu\sigma\mu}-C^\rho{}_{\mu\sigma\nu}$. Solving these linear relations gives

$$C^\rho{}_{\mu\nu\alpha} =-\frac13\left(R^\rho{}_{\mu\nu\alpha} +R^\rho{}_{\nu\mu\alpha}\right).$$

Substitution checks the symmetry and cyclic relation directly. Differentiate metric compatibility once, using $\partial g=\Gamma=0$ at the origin, and substitute this value of $C$:

$$\partial_\alpha\partial_\beta g_{\mu\nu}(0) =-\frac13\left(R_{\mu\alpha\nu\beta} +R_{\mu\beta\nu\alpha}\right).$$

Taylor’s formula multiplies this by $X^\alpha X^\beta/2$. The two curvature terms contribute equally after relabeling the summed indices, producing the displayed coefficient $-1/3$.

The metric is Minkowskian through first order in displacement, while curvature enters at second order. This is the precise limitation of a local inertial frame. Making the connection vanish at one point does not make all second derivatives of the metric vanish there, and does not make the surrounding neighborhood flat.

Riemann normal coordinates make the geodesics launched from the origin straight coordinate rays. They do not make every geodesic in the neighborhood a straight coordinate line. If they did, they would have erased curvature rather than merely chosen convenient labels.

For a moving freely falling laboratory, Fermi normal coordinates extend the construction along a reference timelike geodesic. Parallel transport a nonrotating orthonormal frame along it, then use short spacelike geodesics orthogonal to the reference worldline to label nearby events. Locally, the metric is Minkowskian and its first derivatives vanish on the entire reference geodesic, while transverse second-order terms contain curvature. This construction and its quadratic expansion are developed in Manasse and Misner’s original Fermi-coordinate paper.

For example, with $X^0=c\tau$ and spatial distances $X^i$,

At the reference worldline, the time coordinate is the observer’s clock and the first-order gravitational terms are absent. Away from it, a quadratic tidal potential remains. The coefficient differs from the Riemann-normal expansion because these are different coordinate constructions, one centered on an event and the other on a worldline.

If the laboratory accelerates, its comoving nonrotating coordinates acquire acceleration terms already at first order in distance. A rocket can keep itself fixed in its own coordinates, but it cannot make its accelerometer reading disappear by relabeling events.

The size of an approximately inertial laboratory is controlled by quantities such as $|R_{\hat a\hat b\hat c\hat d}|L^2\ll1$, together with sufficiently small curvature-variation effects across the region. A nominal curvature radius alone is not enough if the curvature changes rapidly. The length and time scales of a measurement must therefore be compared with both the curvature and its variation.

10.7 Curvature scalars and their limits#

Coordinates can become singular while geometry remains regular. To distinguish a coordinate problem from a physical curvature problem, form scalars such as

The last is the Kretschmann scalar. In Schwarzschild spacetime,

It stays finite at $r=2G_NM/c^2$ and diverges as $r\to0$. This helps show why the standard Schwarzschild-coordinate problem at the horizon is removable, while the central curvature divergence is not. Establishing smooth extension through a horizon still requires regular coordinates; finiteness of one scalar by itself is not an extension theorem.

A limitation of these scalar tests is that even all polynomial scalar curvature invariants can vanish while the Riemann tensor is nonzero. Lorentzian contractions are not positive sums of squares. A nonzero null vector already demonstrates the basic logic: its norm can be zero without the vector being zero.

Further example: a curved wave spacetime with zero scalar contractions

Start with flat coordinates $U=(ct-z)/\sqrt2$, $V=(ct+z)/\sqrt2$, both with length units. Then $-c^2dt^2+dz^2=-2\,dU\,dV$. To construct a wave geometry, add a position-dependent term:

$$ds^2=-2\,dU\,dV+dx^2+dy^2+H(U,x,y)\,dU^2,$$

where

$$H=A(U)(x^2-y^2)+2B(U)xy.$$

The profiles $A(U)$ and $B(U)$ have units of inverse length squared, so $H$ is dimensionless. The inverse of the $U,V$ metric block is

$$\begin{pmatrix}H&-1\\-1&0\end{pmatrix}^{-1} =\begin{pmatrix}0&-1\\-1&-H\end{pmatrix}.$$

Thus $g^{UU}=0$ throughout the wave geometry, while $g^{VV}$ need not vanish. The nonzero Christoffel coefficients are, apart from their lower-index symmetry,

$$\Gamma^V{}_{UU}=-\frac12\partial_UH, \qquad\Gamma^V{}_{Ui}=-\frac12\partial_iH, \qquad\Gamma^i{}_{UU}=-\frac12\partial_iH.$$

There are no coefficients with upper $U$, and $H$ is independent of $V$. In the curvature formula, $R^V{}_{iUj}=-\partial_j\Gamma^V{}_{Ui}=\tfrac12\partial_i\partial_jH$; the other terms vanish. Lowering the first index supplies $g_{UV}=-1$, giving

$$R_{UiUj}=-\frac12\partial_i\partial_jH, \qquad i,j\in\{x,y\},$$

so it is nonzero whenever the profiles $A$ or $B$ are nonzero. Yet

$$R_{UU}=-\frac12(\partial_x^2H+\partial_y^2H)=0,$$

and the other Ricci components vanish: this is a vacuum plane gravitational wave.

Why does $\mathcal K$ vanish too? Every nonzero curvature component has lower $U$ indices, but the inverse metric has $g^{UU}=0$. Raising a $U$ slot pairs it with a $V$ slot, and there are no corresponding nonzero curvature components containing $V$. The full contraction therefore vanishes even though the tidal tensor does not. These plane waves belong to the class of geometries with vanishing scalar polynomial invariants; the broader classification is given by Pravda, Pravdova, Coley, and Milson.

A vanishing scalar contraction does not establish that the full tensor vanishes. For difficult spacetime classification or singularity questions, curvature components in physically or geometrically specified frames, covariant derivatives, geodesic behavior, and extension properties may all matter.

10.8 From the metric to curvature#

The calculations in these chapters distinguish several kinds of measurement:

The next step is physical rather than merely geometrical: identify the tensor that describes matter’s energy, momentum, and stresses, then find the dynamical equation and action that relate it to the geometry you have just learned to measure.

10.9 Measuring tides near Earth#

The tidal equation lets us estimate an experiment near Earth’s surface. Outside a spherical Earth, the radial relative acceleration of two nearby freely falling particles separated by $\ell$ is approximately

Use $G_NM_\oplus=3.9860\times10^{14}\,\mathrm{m^3/s^2}$ and $R_\oplus=6.371\times10^6\,\mathrm m$. For $\ell=1\,\mathrm m$, this gives $\Delta a_r\simeq3.08\times10^{-6}\,\mathrm{m/s^2}$. It is a few millionths of a metre per second squared, not the approximately $9.8\,\mathrm{m/s^2}$ of a supported laboratory’s accelerometer.

In the static orthonormal frame outside Earth, the magnitude of the corresponding radial curvature component is

The associated scale $\mathcal R_{\rm tidal}^{-1/2}\simeq1.71\times10^{11}\,\mathrm m$ is about 1.14 astronomical units. One astronomical unit is approximately $1.496\times10^{11}\,\mathrm m$, the scale of the Earth–Sun distance. This is a scale constructed from one component, not a literal circle into which four-dimensional spacetime bends. Notice the units: restoring $c^2$ gives a relative acceleration per unit separation, and multiplying by $\ell$ gives the acceleration difference. Curvature by itself does not determine the weight of one supported object.

This symbol names a component magnitude, with units of inverse length squared. The Kretschmann scalar retains the symbol $\mathcal K$ and has units of inverse length to the fourth power. In Schwarzschild spacetime, $\mathcal K=12\mathcal R_{\rm tidal}^2$ for this radial component; the length $\mathcal K^{-1/4}$ differs from $\mathcal R_{\rm tidal}^{-1/2}$ by the factor $12^{-1/4}$.

Here is a second useful conversion. Earth’s geometrized mass is $G_NM_\oplus/c^2\simeq4.44\,\mathrm{mm}$, and its Schwarzschild radius is twice that, about $8.87\,\mathrm{mm}$. These are compactness scales. Earth is not a black hole: its actual radius is hundreds of millions of times larger.

10.10 Counting curvature through coordinate freedom#

Chapter 8 counted curvature components using tensor symmetries. There is a second useful perspective: compare the metric’s Taylor coefficients with the coordinate freedom that can change them. This is a count that supports the normal-coordinate construction, not a substitute for its existence proof.

At an event, a symmetric four-by-four metric has ten independent entries. A linear coordinate transformation has sixteen coefficients. Once it puts the metric into Minkowski form, six continuous freedoms remain: three spatial rotations and three boosts. The first derivatives of the metric then have $4\times10=40$ entries. The quadratic part of a coordinate transformation also has $4\times10=40$ coefficients, symmetric in its two lower coordinate labels; normal coordinates use this freedom to eliminate those first derivatives.

At the next order, the second derivatives have $10\times10=100$ entries. The cubic coordinate change has four choices of output component and twenty symmetric triples of input labels, giving $4\times20=80$ coefficients. We can count the unordered triples directly: four have all labels equal; $4\times3=12$ have a repeated label and a different third label; four have three distinct labels. The total is $4+12+4=20$. The remaining $100-80=20$ independent combinations are precisely the curvature information that cannot be eliminated at the event.

This count explains the hierarchy: the metric’s values supply local measuring units, its first derivatives can be simplified away at one event, and curvature survives in the quadratic spatial variation. It does not say that there are twenty propagating gravitational polarizations. Chapter 20 counts dynamical initial data and reaches a different answer to a different question.