Ricci, Weyl, and Einstein curvature

9.1 What a curvature trace leaves out#

The Riemann tensor is richly directional. Einstein’s equation uses a particular contraction of it, the Ricci tensor:

This identifies the output index with one curvature-direction index and sums. It is a trace of a curvature map. Because of the Riemann symmetries, $R_{\mu\nu}=R_{\nu\mu}$.

Contract again to obtain the Ricci scalar:

A trace adds selected entries. Consider $A=\operatorname{diag}(2,-1,-1)$. Its trace is $2-1-1=0$, but $A(1,0,0)=(2,0,0)$, so the linear map is not zero. On the other two axes it reverses arrows without shortening them. If such a matrix instead represents an acceleration law, its signs describe the directions of acceleration; that is a separate physical interpretation. A vanishing Ricci scalar is even less informative than a vanishing Ricci tensor, and a vanishing Ricci tensor is less informative than a vanishing Riemann tensor.

The hierarchy is

with neither reverse implication valid in general four-dimensional spacetime.

This is crucial for understanding gravity in empty space. For $\Lambda=0$, the vacuum Einstein equation gives $R_{\mu\nu}=0$. A black hole exterior can still have large tidal curvature, and gravitational waves can still propagate through vacuum. Ricci-flat does not mean Riemann-flat.

WORKED EXAMPLE

A trace loses information you can name

Can you exhibit nonzero curvature whose entire Ricci tensor vanishes?

Builds onCurvature

See the idea

A trace is a particular sum, and positive and negative contributions can cancel. To see what it forgets, work with a small supplied curvature table and calculate its contractions. Use an orthonormal frame with signature $(-+++)$.

Work it out

1. 01
   Supply six independent sectional entries

   Let the only independent nonzero entries be $R_{0101}=A$, $R_{0202}=B$, $R_{0303}=C$, $R_{1212}=D$, $R_{1313}=E$, and $R_{2323}=F$. Generate their partners using pair antisymmetry and pair exchange. Each listed index pair couples only to itself; the four-distinct-index entries in the algebraic Bianchi identity are all zero.

   $$R_{ab}=\eta^{cd}R_{cadb}.$$

   Why this step works The inverse metric contributes a minus sign when the contracted index is temporal.

2. 02
   Perform the contraction

   For example $R_{11}=-R_{0101}+R_{2121}+R_{3131}=-A+D+E$. Applying the same rule to the other diagonal entries gives the whole Ricci tensor for this table. Off-diagonal entries vanish.

   $$R_{00}=A+B+C,\quad R_{11}=-A+D+E,\quad R_{22}=-B+D+F,\quad R_{33}=-C+E+F.$$

   Why this step works Pair antisymmetry is used twice for entries such as R₂₁₂₁, leaving the original sign.

3. 03
   Build a vacuum counterexample

   Choose $(A,B,C,D,E,F)=q(-2,1,1,-1,-1,2)$ with nonzero q of inverse-length-squared dimension. Every displayed Ricci entry is zero, while the original curvature table is not. The scalar contraction also vanishes.

   $$R=2(-A-B-C+D+E+F)=0,\qquad R_{abcd}R^{abcd}=48q^2.$$

   Why this step works For this particular diagonal table each independent squared entry appears four times in the full invariant.

Go deeper

This is an algebraic curvature tensor at a point, with the pattern of Schwarzschild curvature in a suitably aligned orthonormal frame and a chosen curvature scale q. Section 9.3 will name the trace-free remainder the Weyl tensor. The full contraction $R_{abcd}R^{abcd}$ is called the Kretschmann scalar. It need not be positive in Lorentzian geometry: the positivity above follows from this special table. Other spacetimes can have nonzero curvature with vanishing polynomial scalar invariants.

Further readingDavid Tong · General relativity

Test the idea

FIRST, PREDICT

If every Ricci component is zero, what can the supplied table still contain?

Nonzero Weyl curvature and tidal effects.No curvature of any kind.Only a coordinate singularity.

Compare the reasoning

Nonzero Weyl curvature and tidal effects.

The explicit six-entry example leaves the Riemann tensor nonzero after all Ricci contractions cancel.

No curvature of any kind.

The original entries were not determined by their contracted sums.

Only a coordinate singularity.

The nonzero invariant in this example is independent of coordinates.

A hint

Substitute the six values into each Ricci entry before concluding anything.

NOW CHANGE THE EXAMPLE

For $(A,B,C,D,E,F)=(-2,-2,-2,2,2,2)$ in a common curvature unit, calculate R in that unit.

A hint

Use $R=2(-A-B-C+D+E+F)$.

Work through the solution

$R=2(6+6)=24$. This is a constant-sectional-curvature pattern.

A vanishing contraction is a statement about specific sums, not about all of the original curvature.

9.2 Ricci curvature and an initially stationary cloud#

Choose a freely falling observer with unit orthonormal frame and $e_{\hat0}=u/c$. The spatial tidal matrix is

It is symmetric. The next chapter derives the measurable relative acceleration it produces:

The trace of this matrix is

Thus Ricci curvature evaluated on the observer’s time direction measures the sum of the three principal tidal effects. One direction can stretch while another compresses; the trace tells you the aggregate.

To turn that into a volume statement, release an infinitesimal ball of freely falling particles initially at rest relative to one another in the observer’s frame. Let $\lambda_i$ be the three eigenvalues of $\mathcal E$. An initial edge length $L_i$ has zero initial rate and acceleration $-c^2\lambda_iL_i$, so its short-time length is $L_i[1-\tfrac12c^2\lambda_i(\Delta\tau)^2]+O((\Delta\tau)^3)$. Multiply the three lengths. The leading fractional volume change contains their sum, $c^2\sum_i\lambda_i=R_{\mu\nu}u^\mu u^\nu$. Thus initially

and, for a short proper-time interval,

Why the trace? At first, each principal edge changes by its own small fractional amount. Multiplying the three edge lengths, the leading fractional volume change is the sum of those changes. Products of the small changes enter at higher order.

This result uses zero initial relative velocity. A cloud already expanding, changing shape, or rotating can have additional contributions to its volume change. Chapter 22 derives the more general volume-evolution law, called the Raychaudhuri equation.

9.3 Separating Ricci and Weyl curvature#

In four dimensions the Riemann tensor can be decomposed as

The tensor $C_{\rho\sigma\mu\nu}$ is the Weyl tensor. It has the Riemann symmetries and every metric trace vanishes. It is the curvature information left after removing Ricci’s traces.

The factors $1/2$ and $1/6$ can be derived rather than memorized. Write the four-term Ricci combination in parentheses as $Q_{\rho\sigma\mu\nu}$ and the two-term metric combination as $P_{\rho\sigma\mu\nu}$. Suppose

Contract the first and third indices. In $n$ dimensions,

The Weyl contribution must vanish under this contraction, and the left side must equal $R_{\sigma\nu}$. Therefore $A(n-2)=1$ and $A+B(n-1)=0$. With $n=4$, $A=1/2$ and $B=-1/6$. The coefficients are forced by trace removal.

At an event in four dimensions, Ricci carries ten independent components and Weyl carries the remaining ten. Again, this is algebraic information, not ten freely propagating fields.

Weyl curvature is often described as “shape-changing curvature.” Its contribution to the tidal matrix is trace-free, so it produces no leading initial volume acceleration for the initially comoving ball just described. It can stretch a sphere into an ellipsoid while preserving volume to that leading order.

Two qualifications prevent this useful slogan from becoming misleading:

• Ricci curvature can also contribute to anisotropic distortion; it is not generally a purely isotropic squeeze. The Weyl/Ricci split is a decomposition of the full spacetime curvature, while “shape versus volume” refers to a chosen observer’s spatial tidal problem.
• Weyl-induced shear can subsequently affect volume evolution. A Ricci-flat cloud is not guaranteed to maintain its initial volume forever. Trace-free initial acceleration is weaker than an exact volume-preservation theorem.

When $R_{\mu\nu}=0$, the whole Riemann tensor equals Weyl. This is the curvature of vacuum tidal fields and vacuum gravitational waves. If $\Lambda\ne0$, vacuum instead has $R_{\mu\nu}=\Lambda g_{\mu\nu}$, and the Ricci part need not vanish.

Further calculation: rescaling the metric and the Weyl tensor

A conformal rescaling multiplies the metric by a smooth positive function squared: $\widetilde g_{ab}=\Omega^2g_{ab}$. Local lengths and proper times acquire the factor $\Omega$. Null directions remain null because multiplying zero by $\Omega^2$ still gives zero.

Put $f=\ln\Omega$, taking $\Omega$ dimensionless. Substituting $\widetilde g$ into the Christoffel formula and differentiating the product gives

$$\widetilde\Gamma^a{}_{bc}-\Gamma^a{}_{bc} =\delta^a_b\partial_cf+\delta^a_c\partial_bf -g_{bc}g^{ad}\partial_df.$$

Define the symmetric tensor

$$B_{ab}=\nabla_a\partial_bf-\partial_af\,\partial_bf +\frac12g_{ab}g^{cd}\partial_cf\,\partial_df.$$

Using the connection difference in the curvature formula, the new first derivatives give the Hessian terms in $B$ and the connection products give its squared-gradient terms. Collecting them yields

$$\begin{aligned} \widetilde R_{abcd}=\Omega^2\big(&R_{abcd} +g_{ad}B_{bc}+g_{bc}B_{ad}\\ &-g_{ac}B_{bd}-g_{bd}B_{ac}\big). \end{aligned}$$

Every added term has exactly the metric-times-symmetric-tensor form removed by the Ricci subtraction. For a direct check in $n$ dimensions, contracting the added four-term combination gives $-(n-2)B_{bd}-g_{bd}B^a{}_a$. The same trace removal that fixed the decomposition coefficients therefore leaves

$$\widetilde C_{abcd}=\Omega^2 C_{abcd},\qquad \widetilde C^a{}_{bcd}=C^a{}_{bcd}.$$

The inverse metric used to raise the first index supplies $\Omega^{-2}$. Thus the mixed-index Weyl tensor is unchanged. In particular, a metric of the form $g_{ab}=\Omega^2\eta_{ab}$ has zero Weyl curvature. Such a metric is called conformally flat in that coordinate region. For further conformal geometry, see Tong’s curvature notes.

A FREELY FALLING CLOUD

Watch shape and volume part company.

Release a small ball of particles with no relative motion. Compare the evolving cloud with its gray, undeformed reference.

Isotropic focusingTrace-free tide

Drag to turn · arrows to orbit

$\ell_1/\ell_0=1$$\ell_2/\ell_0=1$$\ell_3/\ell_0=1$

Release the cloudReset$s=0$

$s=0$$s=0.3$$s=0.6$$s=0.9$

Three-dimensional volume$\mathcal V/\mathcal V_0=1$

Longest / shortest axis$\ell_{\max}/\ell_{\min}=1$

At the instant of release$$\frac{\ddot{\mathcal V}(0)}{\omega^2\mathcal V_0}=0$$

Stretching in one direction initially cancels squeezing in the other two. It does not cancel the later volume change.

$$\frac{d^2\boldsymbol\xi}{d\tau^2}=\mathsf A\boldsymbol\xi,\qquad \frac{\mathsf A}{\omega^2}=\operatorname{diag}(2,-1,-1)$$

$s=\omega\tau$, with $\omega>0$ setting the time scale. The displayed volume is the product of all three principal lengths, not the area of the projection.

Why the model moves this way

In the chosen orthonormal frame, $\mathsf A=-c^2\mathcal E$, where $\mathcal E_{ij}=R_{\hat i\hat0\hat j\hat0}$. Each normalized principal length starts at one with zero derivative and solves $a_i^{\prime\prime}(s)=\lambda_i a_i(s)$. Negative eigenvalues give cosine; positive ones give hyperbolic cosine.

$$a_i(s)=\begin{cases}\cos(\sqrt{-\lambda_i}\,s),&\lambda_i<0,\\\cosh(\sqrt{\lambda_i}\,s),&\lambda_i>0.\end{cases}$$

$\mathcal V/\mathcal V_0=a_1a_2a_3$. For the trace-free case, $\cosh(\sqrt2s)\cos^2s=1-\tfrac12s^4+O(s^6)$: the quadratic volume term vanishes, but the later change does not.

This solves a constant tidal-matrix model exactly. It illustrates the local geodesic-deviation argument, not an entire black-hole spacetime. In vacuum the trace-free tide is Weyl curvature. In general, Ricci curvature can also produce anisotropic distortion; this two-mode comparison is not a universal Ricci/Weyl decomposition.

Source: Tong, General Relativity, geodesic deviation and curvature · Weyl tensor and its vanishing traces

Reference diagram

9.4 The differential Bianchi identity#

A metric can vary from point to point, but the resulting curvature is not an arbitrary tensor field with 20 freely assignable functions. Because it comes from a connection, it obeys a differential identity:

This is the differential or second Bianchi identity. It is an identity of Levi-Civita geometry, before Einstein’s equation or any matter model is introduced.

Here is a local proof that shows why it exists. At an arbitrary point choose normal coordinates so that $\Gamma=0$ there. At that point $\nabla R=\partial R$. Differentiating $R=\partial\Gamma-\partial\Gamma+\Gamma\Gamma-\Gamma\Gamma$ gives second derivatives of $\Gamma$; the derivatives of its quadratic products contain an undifferentiated $\Gamma$ and vanish at the point. Add the three cyclic terms. Every second derivative appears twice with opposite sign, so commutation of ordinary mixed partial derivatives makes the sum zero.

The expression is tensorial, so if it vanishes in normal coordinates it vanishes in every chart at that point. The point was arbitrary, so the identity holds throughout the smooth region. The convenient chart simplified the calculation without restricting the metric.

The related algebraic consistency rule is the Jacobi identity. For operators $A,B,C$ and commutator $[A,B]=AB-BA$,

To verify it, expand each bracket. For example, the first contributes $ABC-ACB-BCA+CBA$; each of these ordered products appears with the opposite sign in one of the other two brackets. Thus failures of pairwise commutation cannot be assigned independently. The direct normal-coordinate calculation above establishes the corresponding differential identity for curvature.

9.5 A divergence-free curvature tensor#

Set the differentiation index $\lambda$ equal to the upper curvature index $\rho$ and sum in the identity above. The second term contains $R^\rho{}_{\sigma\nu\rho}=-R_{\sigma\nu}$, while the third contains $R^\rho{}_{\sigma\rho\mu}=R_{\sigma\mu}$. Moving those two terms to the right gives

Now contract with $g^{\sigma\nu}$. Compatibility allows the metric to move through the derivative. The left side becomes $\nabla_\rho R^\rho{}_{\mu}$; the right side becomes $\nabla_\mu R-\nabla_\nu R^\nu{}_{\mu}$. Thus

The two Ricci-divergence expressions differ only by the name of their summed index. Bring them together:

This tells us exactly how to repair the Ricci tensor’s divergence. Define

Then

The coefficient $1/2$ is exactly what cancels the Ricci divergence. The remaining derivative of the metric is zero by compatibility.

Divergence-free does not mean covariantly constant. The identity is $\nabla^\mu G_{\mu\nu}=0$, involving a contraction. It does not say $\nabla_\lambda G_{\mu\nu}=0$ for every choice of indices. A fluid can have zero net outflow from each small region while still varying across the room; similarly, divergence-free geometry can vary.

The result prepares the field equation. A consistent geometric left side can be matched to a covariantly conserved stress-energy tensor on the right. A constant multiple of $g_{\mu\nu}$ is also divergence-free, allowing the cosmological term $\Lambda g_{\mu\nu}$. Geometry alone has not yet fixed the physical coupling, matter content, or the theory’s full action. Those are the next stage.

9.6 Why dimension matters#

For the two-sphere we found $R_{ab}=a^{-2}g_{ab}$ and $R=2/a^2$. Consequently,

for that intrinsically curved surface. The cancellation holds for every two-dimensional metric. There is only one independent curvature entry, so the tensor has the form

Contracting gives $R_{ab}=Kg_{ab}$ and $R=2K$, hence $G_{ab}=Kg_{ab}-Kg_{ab}=0$, whether or not $K$ vanishes.

This is not a counterexample to the usefulness of Einstein’s equation in four dimensions. It shows that a tensor’s information content depends on dimension. In two dimensions all intrinsic curvature is summarized by one scalar, and the Einstein combination cancels it. In three dimensions Weyl vanishes identically and Ricci determines the full Riemann tensor. Four dimensions are the first in which a nonzero Weyl tensor carries local curvature information independent of Ricci.