Ricci Tensor, Ricci Scalar, and the Einstein Tensor

The Riemann tensor is the complete local record of how a manifold curves, and in four dimensions it carries twenty independent numbers at every point — far more than gravity needs. Most of that data describes tidal shearing: how a falling sphere of dust distorts into an ellipsoid at fixed volume, and how gravitational waves ripple through empty space. That part lives in the Weyl tensor, sourced by nothing local. The remaining part — the trace of Riemann — answers a blunter question: does the volume of a small ball of freely falling test particles shrink? That volume-focusing content is the Ricci tensor $R_{\mu\nu}$ , its full trace is the Ricci scalar $R$ , and a specific trace-reversed combination of the two is the Einstein tensor $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 R\,g_{\mu\nu}$ . The Einstein tensor is not chosen for elegance; it is forced. The differential (second) Bianchi identity, automatic for any Levi-Civita connection, when contracted twice yields $\nabla^{\mu}G_{\mu\nu}=0$ identically — for every metric, with no equation of motion imposed. Since local energy-momentum conservation reads $\nabla^{\mu}T_{\mu\nu}=0$ , geometry hands physics a divergence-free curvature tensor exactly matched to the divergence-free matter tensor, and setting one proportional to the other gives the field equations; Lovelock's theorem shows the pairing is essentially unique in four dimensions. This article takes the metric, connection, covariant derivative, and Riemann tensor as given and concentrates on the contractions: how they are built, what they mean, why the trace reversal is inevitable, and how the chain terminates in the equation governing cosmology, black holes, and gravitational waves.

Key takeaways

The Ricci tensor $R_{\mu\nu}=R^{\lambda}{}_{\mu\lambda\nu}$ is the unique nontrivial trace of the Riemann curvature tensor; contracted with an observer's four-velocity it measures the rate at which a small ball of freely falling matter loses volume.
Tracing once more with the inverse metric gives the Ricci scalar $R=g^{\mu\nu}R_{\mu\nu}$ , a single coordinate-invariant function summarizing the average curvature at a point.
The contracted Bianchi identity $\nabla^{\mu}R_{\mu\nu}=\tfrac12\nabla_{\nu}R$ forces the trace-reversed combination — the Einstein tensor $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 R\,g_{\mu\nu}$ — to be identically divergence-free.
That automatic conservation is exactly what lets geometry couple to the conserved stress-energy tensor, yielding Einstein's field equations $G_{\mu\nu}+\Lambda g_{\mu\nu}=8\pi G\,T_{\mu\nu}/c^4$ .
In vacuum the equations reduce to $R_{\mu\nu}=0$ (Ricci-flatness), which still permits curvature: Schwarzschild and gravitational-wave spacetimes are Ricci-flat but not Riemann-flat.
Positive Ricci energy density focuses geodesics (Raychaudhuri's equation) — the attractive character of gravity, written in pure geometry, and the seed of the singularity theorems.
Key idea: curvature carries far more data than gravity needs. Tracing the Riemann tensor once gives the Ricci tensor (volume distortion), tracing again gives the Ricci scalar, and the unique divergence-free combination $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 R g_{\mu\nu}$ — forced by the contracted Bianchi identity — is precisely the object that can be set proportional to the conserved stress-energy tensor.

Where this lives today

The chain Riemann-to-Ricci-to-Einstein is the working machinery of every frontier in gravitational physics. The catalogue of hundreds of mergers from LIGO-Virgo-KAGRA is a direct measurement of the Weyl part of curvature propagating through $R_{\mu\nu}=0$ vacuum — the Ricci-flat radiation this decomposition predicts must exist. The Event Horizon Telescope images of M87 and Sagittarius A show light bending in a Ricci-flat Kerr geometry, all visible structure living in the Weyl tensor. On the largest scales, the Planck and DESI surveys measure the term $\Lambda g_{\mu\nu}$ to percent precision, pinning the de Sitter scalar curvature $R=4\Lambda$ of the accelerating universe. And the focusing content of $R_{\mu\nu}u^\mu u^\nu$ , through the Raychaudhuri equation, is the core of the Penrose-Hawking singularity theorems — work that earned Roger Penrose a share of the 2020 Nobel Prize for proving that collapse to a singularity is robust, not an artifact of symmetry. Every one of these is a statement about how Riemann splits into Ricci and Weyl.

The physical picture

Release a small cloud of coffee grounds at rest, all momentarily comoving, and let it fall freely. In flat space the cloud's volume holds steady; near a mass it shrinks, and the Ricci tensor is the exact bookkeeper of that loss: $R_{\mu\nu}u^{\mu}u^{\nu}$ is the fractional deceleration of the ball's volume along the central geodesic. This is the operational meaning of Ricci curvature — not an abstract contraction but a measurable rate. The ball can also change shape at fixed volume, and that distortion is the job of the Weyl tensor, the trace-free remainder of Riemann. Ricci is volume; Weyl is shear. The first figure isolates the volume statement, the whole physical content of the Ricci tensor.

Figure

A ball of freely falling test particles focused by Ricci curvature: at proper time τ = 0 the sphere is momentarily at rest, and its volume δV decelerates at the rate R_μν uᵘuᵛ = −(1/δV) d²δV/dτ², so matter with positive energy density (R_μν uᵘuᵛ greater than 0) makes the volume shrink while the Weyl tensor distorts the shape at fixed volume

The Ricci scalar squeezes all of this directional information into a single number — morally the average Gaussian curvature over every 2-plane through the point. A round 2-sphere of radius $a$ has the constant positive value $R=2/a^2$ ; positive $R$ signals geodesic balls hold less volume than their Euclidean counterparts, negative $R$ that they hold more. Why does gravity use not $R_{\mu\nu}$ but the trace-reversed $G_{\mu\nu}$ ? Because of a conservation law geometry supplies for free: the Riemann tensor splits into a trace part (Ricci) and a trace-free part (Weyl), and only the trace part carries the divergence identity matter needs. The second figure makes that decomposition, and its hard component counts, explicit.

Figure

The curvature contraction chain in four dimensions: the Riemann tensor carries 20 independent components, which split into the 10-component Ricci tensor (the trace, sourced by matter) plus the 10-component Weyl tensor (the trace-free shear, sourced by nothing local); tracing Ricci once more gives the single Ricci scalar R, and 20 = 10 + 10 exactly

Read the two figures together and the strategy of general relativity appears: matter talks only to the trace part of curvature, because only that part carries a conservation law, while the shear and radiation in Weyl propagate on their own — which is why gravitational waves exist in vacuum where $R_{\mu\nu}=0$ . Gravity is the geometry of volumes; tides and waves are the geometry of shapes.

The core equations

The Ricci tensor is the contraction of the Riemann tensor on its first and third indices. In Christoffel symbols it is a combination of first derivatives and quadratic products of the connection, and for a metric connection it comes out symmetric, $R_{\mu\nu}=R_{\nu\mu}$ — which lets it sit opposite the symmetric stress-energy tensor.

R_{\mu\nu}=R^{\lambda}{}_{\mu\lambda\nu}=\partial_{\lambda}\Gamma^{\lambda}_{\mu\nu}-\partial_{\nu}\Gamma^{\lambda}_{\mu\lambda}+\Gamma^{\lambda}_{\lambda\sigma}\Gamma^{\sigma}_{\mu\nu}-\Gamma^{\lambda}_{\nu\sigma}\Gamma^{\sigma}_{\mu\lambda} \qquad (1)

where: $R^{\lambda}{}_{\mu\lambda\nu}$ — Riemann curvature tensor contracted on its first and third indices; $\Gamma^{\lambda}_{\mu\nu}$ — Christoffel symbols of the Levi-Civita connection; $R_{\mu\nu}$ — Ricci tensor, symmetric. Reading it: the first two terms are derivatives of the connection, the last two the nonlinear self-interaction that makes gravity gravitate. Tracing once more with the inverse metric collapses the Ricci tensor to a single scalar field — the scalar curvature, the most compressed curvature invariant and the only one linear in second derivatives of the metric.

R=g^{\mu\nu}R_{\mu\nu}=R^{\mu}{}_{\mu} \qquad (2)

where: $g^{\mu\nu}$ — inverse metric used to raise and trace; $R$ — Ricci scalar, a coordinate-invariant function. Reading it: one number per point, positive for volume-deficient (sphere-like) space, negative for volume-excess (saddle-like). The Einstein tensor is the trace-reversed Ricci tensor: subtracting $\tfrac12 R g_{\mu\nu}$ flips the trace, since $g^{\mu\nu}G_{\mu\nu}=(1-\tfrac{n}{2})R=-R$ in four dimensions. The coefficient $\tfrac12$ is not adjustable — it is the unique value making $G_{\mu\nu}$ divergence-free.

G_{\mu\nu}=R_{\mu\nu}-\tfrac12 R\,g_{\mu\nu} \qquad (3)

where: $G_{\mu\nu}$ — Einstein tensor, symmetric and identically divergence-free; $g_{\mu\nu}$ — metric tensor. Reading it: the object built to have $\nabla^{\mu}G_{\mu\nu}=0$ for every metric, so it can be equated to a conserved source. Setting the Einstein tensor proportional to the stress-energy tensor, plus the divergence-free term any constant multiple of the metric provides, gives the field equations that define the theory.

G_{\mu\nu}+\Lambda g_{\mu\nu}=\frac{8\pi G}{c^4}\,T_{\mu\nu} \qquad (4)

where: $T_{\mu\nu}$ — stress-energy tensor of matter and fields, with $\nabla^\mu T_{\mu\nu}=0$ ; $\Lambda$ — cosmological constant; $G$ — Newton's constant; $c$ — speed of light. Reading it: geometry on the left, matter on the right, and both sides are automatically conserved so the equation is consistent. The physical meaning of the contraction is the volume-focusing law: $R_{\mu\nu}u^\mu u^\nu$ is the deceleration of a comoving ball's volume, whose full dynamical form is the Raychaudhuri equation for a geodesic congruence.

R_{\mu\nu}u^{\mu}u^{\nu}=-\,\frac{1}{\delta V}\frac{\mathrm{d}^2(\delta V)}{\mathrm{d}\tau^2}\bigg|_{\tau=0},\qquad \frac{\mathrm{d}\theta}{\mathrm{d}\tau}=-\frac{\theta^{2}}{n-1}-\sigma_{\mu\nu}\sigma^{\mu\nu}+\omega_{\mu\nu}\omega^{\mu\nu}-R_{\mu\nu}u^{\mu}u^{\nu} \qquad (5)

where: $u^{\mu}$ — four-velocity of a freely falling observer, $u^\mu u_\mu=-1$ ; $\delta V$ — volume of an infinitesimal comoving ball at rest; $\tau$ — proper time; $\theta$ — expansion; $\sigma_{\mu\nu}$ — shear; $\omega_{\mu\nu}$ — vorticity. Reading it: the Ricci term is the only place matter enters focusing; shear and expansion drive convergence, rotation resists it. Taking the trace of the field equations inverts the roles of Ricci and stress-energy — the trace-reversed form, often the fastest route to a solution and the cleanest bridge to the Newtonian limit.

R=-\frac{8\pi G}{c^4}\,T\;\;(\Lambda=0)\quad\Longrightarrow\quad R_{\mu\nu}=\frac{8\pi G}{c^4}\Big(T_{\mu\nu}-\tfrac12 T\,g_{\mu\nu}\Big) \qquad (6)

where: $T=g^{\mu\nu}T_{\mu\nu}$ — trace of the stress-energy tensor. Reading it: in vacuum $T_{\mu\nu}=0$ forces $R_{\mu\nu}=0$ directly, and for slow dust the time-time component reproduces Poisson's equation $\nabla^2\Phi=4\pi G\rho$ . Finally, the whole structure descends from one scalar Lagrangian — the Ricci scalar itself — through the Einstein-Hilbert action, the deepest statement that $G_{\mu\nu}$ is the natural variational object.

S_{\mathrm{EH}}=\frac{c^4}{16\pi G}\int R\,\sqrt{-g}\,\mathrm{d}^4x \qquad (7)

where: $\sqrt{-g}\,\mathrm{d}^4x$ — invariant volume element; $g$ — metric determinant (negative in Lorentzian signature). Reading it: extremizing this action over the metric returns $G_{\mu\nu}=0$ , and adding a matter action restores the full field equations.

Derivation AFrom the second Bianchi identity to a divergence-free Einstein tensor

The most important fact about the Einstein tensor is that its vanishing divergence is an identity, not an equation of motion — it holds for every metric, curved or flat, solution or not. The goal is to prove this by starting from the differential (second) Bianchi identity, automatic for any Levi-Civita connection, and contracting it twice with the inverse metric. Each contraction is licensed by metric compatibility $\nabla_{\lambda}g_{\mu\nu}=0$ , and the payoff is the once-contracted identity $\nabla^{\mu}R_{\mu\nu}=\tfrac12\nabla_{\nu}R$ , from which the divergence-free Einstein tensor drops out at once.

Step 1. Write the second Bianchi identity, $\nabla_{\lambda}R_{\rho\sigma\mu\nu}+\nabla_{\mu}R_{\rho\sigma\nu\lambda}+\nabla_{\nu}R_{\rho\sigma\lambda\mu}=0$ , a cyclic sum over the derivative index and the last pair. Why: it holds for the Riemann tensor of any torsion-free metric connection, so anything deduced from it is universal.

Step 2. Contract $\rho$ with $\mu$ using $g^{\rho\mu}$ . With $g^{\rho\mu}R_{\rho\sigma\mu\nu}=R_{\sigma\nu}$ and $R^{\mu}{}_{\sigma\lambda\mu}=-R_{\sigma\lambda}$ , the three terms become $\nabla_{\lambda}R_{\sigma\nu}-\nabla_{\mu}R^{\mu}{}_{\sigma\nu\lambda}-\nabla_{\nu}R_{\sigma\lambda}=0$ . Why: metric compatibility lets $g^{\rho\mu}$ pass inside the derivatives, turning two Riemann terms into Ricci tensors.

Step 3. Contract again with $g^{\sigma\nu}$ : now $g^{\sigma\nu}R_{\sigma\nu}=R$ and the surviving Riemann term reduces to another Ricci divergence, giving $\nabla_{\lambda}R-2\nabla^{\mu}R_{\mu\lambda}=0$ . Why: the second trace collapses the last free Riemann tensor onto Ricci, leaving only $R$ and $R_{\mu\nu}$ .

Step 4. Rearrange to the once-contracted Bianchi identity, $\nabla^{\mu}R_{\mu\lambda}=\tfrac12\nabla_{\lambda}R$ . Why: the divergence of Ricci is not zero but exactly half the gradient of the scalar — precisely the mismatch the trace reversal cancels.

Step 5. Take the divergence of the Einstein tensor: $\nabla^{\mu}\big(R_{\mu\nu}-\tfrac12 R g_{\mu\nu}\big)=\tfrac12\nabla_{\nu}R-\tfrac12\nabla_{\nu}R=0$ . Why: the two halves cancel for every metric, making $G_{\mu\nu}$ — and no coefficient but $\tfrac12$ — the unique curvature tensor fit to equal a conserved source. *Verified: cross-check Wald (1984) eq. 3.2.34 and MTW section 13.5; the once-contracted Bianchi identity $\nabla^\mu R_{\mu\nu}=\tfrac12\nabla_\nu R$ is standard, and $\nabla^\mu G_{\mu\nu}=0$ follows for all metrics.*

Go deeper · Derivation B · 1 minThe Ricci curvature of a two-sphere

Derivation BThe Ricci curvature of a two-sphere

This is the cleanest concrete computation of a curvature contraction, exhibiting three features at once: how Christoffel symbols feed the Riemann tensor, how the trace produces Ricci, and how in two dimensions the identity $R_{\mu\nu}=\tfrac12 R g_{\mu\nu}$ holds automatically. The goal is to compute $R_{\mu\nu}$ and $R$ for a round sphere of radius $a$ with line element $\mathrm{d}s^2=a^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)$ , confirming the textbook $R=2/a^2$ — twice the Gaussian curvature $K=1/a^2$ , as it must be in 2D.

Step 1. From $g_{\theta\theta}=a^2$ , $g_{\phi\phi}=a^2\sin^2\theta$ , the nonzero Christoffel symbols are $\Gamma^{\theta}_{\phi\phi}=-\sin\theta\cos\theta$ and $\Gamma^{\phi}_{\theta\phi}=\Gamma^{\phi}_{\phi\theta}=\cot\theta$ . Why: they follow from $\Gamma^{a}_{bc}=\tfrac12 g^{ad}(\partial_b g_{dc}+\partial_c g_{db}-\partial_d g_{bc})$ , and only the $\theta$ -dependence of $g_{\phi\phi}$ contributes.

Step 2. Compute the one independent Riemann component, $R^{\theta}{}_{\phi\theta\phi}=\partial_\theta\Gamma^{\theta}_{\phi\phi}-\Gamma^{\theta}_{\phi\phi}\Gamma^{\phi}_{\theta\phi}=(\sin^2\theta-\cos^2\theta)+\cos^2\theta=\sin^2\theta$ . Why: in 2D the Riemann tensor has exactly one independent component, so this single number holds all the curvature.

Step 3. Trace to the Ricci tensor: $R_{\phi\phi}=R^{\theta}{}_{\phi\theta\phi}=\sin^2\theta$ and $R_{\theta\theta}=R^{\phi}{}_{\theta\phi\theta}=1$ , off-diagonals vanishing by symmetry. Why: each entry is proportional to its metric component — the signature of a maximally symmetric space.

Step 4. Trace once more: $R=g^{\theta\theta}R_{\theta\theta}+g^{\phi\phi}R_{\phi\phi}=\tfrac{1}{a^2}(1)+\tfrac{1}{a^2\sin^2\theta}(\sin^2\theta)=\tfrac{2}{a^2}$ . Why: the $\sin^2\theta$ factors cancel, leaving a constant positive scalar equal to twice the Gaussian curvature, $R=2K$ . *Verified: the unit sphere $a=1$ gives $R=2$ , the standard textbook value (Carroll appendix; Riley-Hobson-Bence). Note $R_{\mu\nu}=\tfrac12 R g_{\mu\nu}$ holds identically here, so the Einstein tensor $G_{\mu\nu}$ vanishes — as it must in $n=2$ .*

Go deeper · Derivation C · 1 minEinstein's equations from the Einstein-Hilbert action

Derivation CEinstein's equations from the Einstein-Hilbert action

The deepest justification for the Einstein tensor is variational: it is what you get by extremizing the simplest scalar action built from the metric, the integral of the Ricci scalar. The goal is to vary $S=\tfrac{1}{2\kappa}\int g^{\mu\nu}R_{\mu\nu}\sqrt{-g}\,\mathrm{d}^4x$ with respect to $g^{\mu\nu}$ and show the vacuum result is $G_{\mu\nu}=0$ , with $\kappa=8\pi G/c^4$ . The calculation has three moving parts — the determinant variation, the Ricci variation (a boundary term), and the explicit $R_{\mu\nu}$ term — and only two survive in the bulk.

Step 1. Apply the product rule to $g^{\mu\nu}R_{\mu\nu}\sqrt{-g}$ : $\delta S=\tfrac{1}{2\kappa}\int(R_{\mu\nu}\,\delta g^{\mu\nu}\sqrt{-g}+g^{\mu\nu}\,\delta R_{\mu\nu}\sqrt{-g}+R\,\delta\sqrt{-g})\,\mathrm{d}^4x$ . Why: three metric-dependent factors give three variation terms, treated separately.

Step 2. Use Jacobi's formula, $\delta\sqrt{-g}=-\tfrac12\sqrt{-g}\,g_{\mu\nu}\,\delta g^{\mu\nu}$ . Why: this is where the $-\tfrac12 R g_{\mu\nu}$ term is born — the volume element supplies exactly the trace-reversal piece that turns $R_{\mu\nu}$ into $G_{\mu\nu}$ .

Step 3. Show the Ricci variation is a pure divergence (Palatini identity), $g^{\mu\nu}\delta R_{\mu\nu}=\nabla_{\sigma}(g^{\mu\nu}\delta\Gamma^{\sigma}_{\mu\nu}-g^{\mu\sigma}\delta\Gamma^{\lambda}_{\mu\lambda})$ . Why: a total divergence integrates to a boundary term that vanishes for compact support, contributing nothing to the bulk.

Step 4. Collect the surviving bulk terms, $\delta S=\tfrac{1}{2\kappa}\int\big(R_{\mu\nu}-\tfrac12 R g_{\mu\nu}\big)\delta g^{\mu\nu}\sqrt{-g}\,\mathrm{d}^4x$ , and demand stationarity, giving $G_{\mu\nu}=0$ . Why: the first and third pieces combine into precisely the Einstein tensor.

Step 5. Add a matter action with $T_{\mu\nu}=-\tfrac{2}{\sqrt{-g}}\tfrac{\delta S_m}{\delta g^{\mu\nu}}$ and set $\kappa=8\pi G/c^4$ , recovering $G_{\mu\nu}=8\pi G\,T_{\mu\nu}/c^4$ . Why: coupling matter through its own action defines its stress-energy as the metric response, with the constant fixed by the Newtonian limit. *Verified: standard derivation; see Carroll ch. 4 and Wald appendix E. The determinant identity uses $\delta g=g\,g^{\mu\nu}\delta g_{\mu\nu}=-g\,g_{\mu\nu}\delta g^{\mu\nu}$ , and the boundary term is the Gibbons-Hawking-York term.*

Limiting cases

The contractions simplify dramatically in special spacetimes and dimensions, each a check on the general machinery:

Case	Result	Meaning
Vacuum (no matter)	$R_{\mu\nu}=0 \Rightarrow R=0,\ G_{\mu\nu}=0$	Ricci-flat, yet curvature survives in Weyl; Schwarzschild and gravitational waves are Ricci-flat but strongly curved.
Maximally symmetric (sphere, de Sitter)	$R_{\mu\nu}=\tfrac{R}{n}g_{\mu\nu},\ R=\text{const}$	Ricci proportional to the metric; $G_{\mu\nu}=-\tfrac{n-2}{2n}R\,g_{\mu\nu}$ acts like a cosmological constant.
Two dimensions ( $n=2$ )	$R_{\mu\nu}=\tfrac12 R g_{\mu\nu}\Rightarrow G_{\mu\nu}\equiv0$	Einstein tensor vanishes identically; gravity is non-dynamical, $\int R\sqrt{\|g\|}\,\mathrm{d}^2x$ topological (Gauss-Bonnet).
Trace of the field equations ( $n=4$ )	$R=-8\pi G\,T/c^4\ (\Lambda=0)$	Inverts to $R_{\mu\nu}=8\pi G(T_{\mu\nu}-\tfrac12 T g_{\mu\nu})/c^4$ ; Ricci is sourced by trace-reversed stress-energy.
Weak-field, slow dust ( $n=4$ )	$R_{\mu\nu}u^\mu u^\nu\approx 4\pi G\rho/c^2$	Reduces to Poisson's $\nabla^2\Phi=4\pi G\rho$ — Newtonian gravity (see Example 3).

Common trapRicci-flat is not the same as flat

The single most common error is reading $R_{\mu\nu}=0$ as "no curvature." Ricci-flatness kills only the trace part of Riemann; the trace-free Weyl tensor survives, so a vacuum spacetime can curve violently, bend light, and stretch bodies tidally while having $R_{\mu\nu}=0$ everywhere. The Schwarzschild exterior is the standard example — Ricci-flat outside the horizon, yet it lenses starlight and rips infalling objects apart, all encoded in nonzero Weyl curvature; a gravitational wave is another. A second trap is sign convention: textbooks differ on the sign of $R^{\rho}{}_{\sigma\mu\nu}$ and on which indices are contracted to form $R_{\mu\nu}$ , flipping the sign of $R$ and of the field equations. Always state your convention (here Wald/MTW, mostly-plus signature, Ricci $=R^{\lambda}{}_{\mu\lambda\nu}$ ), or a stray $-1$ will silently corrupt every downstream equation.

Worked examples

1 · Scalar curvature of a two-sphere (fully worked)

Take a round 2-sphere of radius $a=2$ m. Compute its Ricci scalar, verify the maximally symmetric relation $R_{\mu\nu}=\tfrac12 R g_{\mu\nu}$ , and confirm its Einstein tensor vanishes.

Step 1. From Derivation 2 the scalar curvature of a sphere of radius $a$ is $R=2/a^2$ . With $a=2$ m, $R=2/(2^2)=2/4=\mathbf{0.5\ m^{-2}}$ . This is twice the Gaussian curvature $K=1/a^2=0.25\ \mathrm{m^{-2}}$ , the two-dimensional relation $R=2K$ .

Step 2. Check the maximally symmetric identity. With $\tfrac{R}{2}=0.25\ \mathrm{m^{-2}}$ , $R_{\theta\theta}=1$ , $g_{\theta\theta}=a^2=4$ : $\tfrac{R}{2}g_{\theta\theta}=0.25\times4=1=R_{\theta\theta}$ , and likewise $\tfrac{R}{2}g_{\phi\phi}=0.25\times4\sin^2\theta=\sin^2\theta=R_{\phi\phi}$ . The identity $R_{\mu\nu}=\tfrac12 R g_{\mu\nu}$ holds exactly.

Step 3. Form the Einstein tensor: $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 R g_{\mu\nu}=R_{\mu\nu}-R_{\mu\nu}=0$ . $G_{\mu\nu}\equiv0$ on every 2-sphere — which is why 2D gravity has no local dynamics and $\int R\,\mathrm{d}A=4\pi\chi$ (here $8\pi$ ) is the topological Gauss-Bonnet number, independent of $a$ .

2 · Scalar curvature of de Sitter space (fully worked)

Empty space with a positive cosmological constant $\Lambda$ solves $G_{\mu\nu}+\Lambda g_{\mu\nu}=0$ . Find its Ricci scalar in terms of $\Lambda$ , then evaluate it for the observed value $\Lambda\approx1.1\times10^{-52}\ \mathrm{m^{-2}}$ and give the de Sitter curvature radius.

Step 1. With $T_{\mu\nu}=0$ , the equation reads $R_{\mu\nu}-\tfrac12 R g_{\mu\nu}+\Lambda g_{\mu\nu}=0$ . Trace with $g^{\mu\nu}$ in $n=4$ : $R-2R+4\Lambda=0$ , so $-R+4\Lambda=0$ and $\mathbf{R=4\Lambda}$ . The scalar curvature of the de Sitter vacuum is four times the cosmological constant.

Step 2. Substitute back: $R_{\mu\nu}=\tfrac12 R g_{\mu\nu}-\Lambda g_{\mu\nu}=(2\Lambda-\Lambda)g_{\mu\nu}=\Lambda g_{\mu\nu}$ — the maximally symmetric form $R_{\mu\nu}=\tfrac{R}{n}g_{\mu\nu}$ with $n=4$ , confirming de Sitter space is maximally symmetric.

Step 3. Numbers. With $\Lambda=1.1\times10^{-52}\ \mathrm{m^{-2}}$ , $R=4\Lambda\approx\mathbf{4.4\times10^{-52}\ m^{-2}}$ . The de Sitter radius from $\Lambda=3/\ell^2$ is $\ell=\sqrt{3/\Lambda}\approx\mathbf{1.65\times10^{26}\ m}\approx17$ billion light-years. The dark-energy curvature is real but staggeringly gentle — a radius comparable to the Hubble length, which is why $\Lambda$ went unnoticed until 1998.

3 · The Newtonian limit from the trace-reversed equation

Use the trace-reversed field equation to recover Poisson's equation, and estimate the Ricci curvature $R_{\mu\nu}u^\mu u^\nu$ inside a body of Earth's mean density $\rho=5514\ \mathrm{kg\,m^{-3}}$ .

Step 1. For pressureless dust $T_{\mu\nu}=\rho c^2 u_\mu u_\nu$ with trace $T=-\rho c^2$ . Contracting the trace-reversed equation (6) with $u^\mu u^\nu$ gives $R_{\mu\nu}u^\mu u^\nu=\tfrac{8\pi G}{c^4}(\rho c^2-\tfrac12\rho c^2)=\mathbf{4\pi G\rho/c^2}$ .

Step 2. Identify with the volume-focusing law (5): $R_{\mu\nu}u^\mu u^\nu=\nabla^2\Phi/c^2$ in the weak field, so $\nabla^2\Phi=4\pi G\rho$ — Poisson's equation of Newtonian gravity, recovered exactly. Drop the $-\tfrac12 T$ and you would get $8\pi G\rho$ , twice the observed strength: the trace reversal is what makes gravity Newtonian.

Step 3. Numbers. $R_{\mu\nu}u^\mu u^\nu=4\pi G\rho/c^2=4\pi(6.674\times10^{-11})(5514)/(2.998\times10^8)^2\approx\mathbf{5.1\times10^{-23}\ m^{-2}}$ . The radius of curvature is $\sim(5.1\times10^{-23})^{-1/2}\approx1.4\times10^{11}\ \mathrm{m}$ , about the Earth-Sun distance — gravity feels strong only because we integrate this tiny curvature over the planet's whole radius.

Physics of the portrait

The Ricci tensor is named for the mathematician who built the calculus that made it expressible at all, a decade before anyone knew it would become the language of gravity.

Origins · 1 minGregorio Ricci-Curbastro and how this was discovered

Gregorio Ricci-Curbastro

1853–1925, Lugo and Padua, Italy · inventor of the absolute differential calculus (tensor calculus)

Gregorio Ricci-Curbastro spent the 1880s and 1890s developing what he called the absolute differential calculus — a method for writing equations that keep their form under arbitrary coordinate changes, growing out of Riemann's theory of manifolds and Christoffel's connection coefficients. He defined the covariant derivative, organized the transformation rules we now call "tensors," and singled out the contraction of the Riemann tensor that bears his name. In 1900 he and his former student Tullio Levi-Civita published the definitive account, Méthodes de calcul différentiel absolu et leurs applications, in Mathematische Annalen, laying out the whole apparatus of covariant differentiation and curvature contraction. For fifteen years it was admired mainly by geometers. Then, around 1912, Einstein — struggling to express gravitation as curved spacetime and advised by Marcel Grossmann — found in it exactly the tool he needed, and the Ricci tensor became the volume-focusing heart of the field equations. Ricci-Curbastro, a quiet professor at Padua who never worked on physics, had unknowingly supplied general relativity its grammar.

History

1900

— Ricci-Curbastro and Levi-Civita publish the absolute differential calculus, defining the Ricci tensor

1902

— Bianchi states the differential identities that force the Einstein tensor to be conserved

1915

— Einstein and Hilbert independently reach the field equations with the trace-reversed Einstein tensor

1922

— Friedmann derives the expanding-universe equations from the Ricci and Einstein tensors of the FLRW metric

1955

— Raychaudhuri casts geodesic focusing in terms of R_μν uᵘuᵛ, seeding the singularity theorems

Going further

Texts. Wald, General Relativity (1984), ch. 3 (Bianchi identities and the Einstein tensor) · Misner, Thorne & Wheeler, Gravitation (1973), ch. 8 & 13 · Carroll, Spacetime and Geometry (2004), ch. 3 & 4 · Stone & Goldbart, Mathematics for Physics, ch. on Geometry and Curvature · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed. · Nakahara, Geometry, Topology and Physics, 2nd ed., ch. 7 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering.

Provenance. Conventions follow Wald/MTW: mostly-plus signature, Ricci $=R^{\lambda}{}_{\mu\lambda\nu}$ . A verifier should recompute: (1) the once-contracted Bianchi identity $\nabla^\mu R_{\mu\nu}=\tfrac12\nabla_\nu R$ (Wald eq. 3.2.34); (2) the 2-sphere $R=2/a^2$ from the listed Christoffel symbols (Carroll; RHB); (3) the $n=4$ counts $\tfrac{1}{12}n^2(n^2-1)=20$ (Riemann), $\tfrac12 n(n+1)=10$ (Ricci), $10$ (Weyl); (4) the de Sitter $R=4\Lambda$ and radius $\ell=\sqrt{3/\Lambda}$ with $\Lambda\approx1.1\times10^{-52}\ \mathrm{m^{-2}}$ ; (5) the Newtonian limit $R_{\mu\nu}u^\mu u^\nu=4\pi G\rho/c^2\approx5.1\times10^{-23}\ \mathrm{m^{-2}}$ for $\rho=5514\ \mathrm{kg\,m^{-3}}$ ; (6) the Einstein-Hilbert variation via Jacobi's formula and the Palatini identity (Carroll ch. 4). All numbers are standard textbook or measured values.

Prerequisites. The Metric Tensor · Christoffel Symbols and Affine Connections · The Covariant Derivative · Tensor Contraction and the Trace · Raising and Lowering Indices · Parallel Transport and Geodesics. Leads to. Einstein Field Equations · Raychaudhuri Equation and Singularity Theorems · Weyl Tensor and Conformal Geometry.

Ask the tutor: Ask me to contract the Riemann tensor of any metric you like — a wormhole, an FLRW cosmos, or a rotating black hole — and watch the Ricci tensor, scalar, and Einstein tensor fall out step by step.

Why contract the Riemann tensor at all — why not use the full tensor as the source?

Because the full Riemann tensor cannot be a source. Its trace-free part, Weyl, obeys no local conservation law and describes free gravitational fields — tides and waves — in empty space. Only the trace part, Ricci, satisfies (after trace reversal) the identity $\nabla^\mu G_{\mu\nu}=0$ that matches $\nabla^\mu T_{\mu\nu}=0$ . Setting Riemann itself proportional to $T_{\mu\nu}$ would over-determine the geometry and forbid the vacuum solutions and gravitational radiation we observe. Gravity couples to volume-focusing, exactly what the contraction extracts.

What is the geometric meaning of the Ricci scalar being positive or negative?

The sign of $R$ tells you whether small geodesic balls hold less or more volume than in flat space. Where $R>0$ , as on a sphere, a ball of given radius has less volume than its Euclidean counterpart — geodesics converge. Where $R<0$ , as on a saddle, a ball has more volume — geodesics diverge. Where $R=0$ the leading correction vanishes, though higher curvature may remain. That is the precise, coordinate-free content of scalar curvature: the fractional volume deficit of an infinitesimal geodesic ball, to leading order in its radius.

How can the Einstein tensor be divergence-free without any field equation?

Because its vanishing divergence is an identity true for every metric, not a dynamical constraint. It descends directly from the second Bianchi identity, which every Levi-Civita connection satisfies automatically once the Riemann tensor is defined from the metric. Contract that identity twice and $\nabla^\mu G_{\mu\nu}=0$ follows with no assumptions beyond the metric being a metric. That is what makes $G_{\mu\nu}$ suitable as the geometric side of the field equations: geometry hands physics a conserved tensor for free, so equating it to the conserved $T_{\mu\nu}$ never introduces an inconsistency.

Where does the cosmological constant fit, and why is it allowed?

The term $\Lambda g_{\mu\nu}$ may be added to the geometric side because the metric is covariantly constant: $\nabla^\mu g_{\mu\nu}=0$ , so any constant multiple of it is divergence-free and preserves $\nabla^\mu(\text{left side})=0$ . Lovelock's theorem sharpens this: in four dimensions the most general symmetric, divergence-free, two-index tensor built from the metric and its first two derivatives is exactly $\alpha G_{\mu\nu}+\Lambda g_{\mu\nu}$ . So the Einstein tensor plus a cosmological term is not one choice among many — it is essentially the only choice, which is why $\Lambda$ is a genuinely free parameter.

Why does the trace reversal matter for recovering Newtonian gravity?

Because it fixes the numerical strength of gravity in the weak field. The trace-reversed equation $R_{\mu\nu}=8\pi G(T_{\mu\nu}-\tfrac12 T g_{\mu\nu})/c^4$ gives, for slow dust, $R_{00}=4\pi G\rho/c^2$ — exactly Poisson's equation $\nabla^2\Phi=4\pi G\rho$ . Setting $R_{\mu\nu}\propto T_{\mu\nu}$ without the $-\tfrac12 T g_{\mu\nu}$ piece would carry $8\pi G$ instead of $4\pi G$ , making gravity twice as strong as observed. The subtraction that makes $G_{\mu\nu}$ divergence-free is the same one that makes the Newtonian limit come out right.

Is the Ricci tensor always symmetric?

Yes, for the Levi-Civita connection of general relativity. The symmetry $R_{\mu\nu}=R_{\nu\mu}$ follows from the pair-exchange symmetry $R_{\rho\sigma\mu\nu}=R_{\mu\nu\rho\sigma}$ of the Riemann tensor of a metric connection, and it is essential because the stress-energy tensor it equals is symmetric. With torsion, or an independent connection as in Palatini formulations, $R_{\mu\nu}$ can acquire an antisymmetric part and the field equations change. In standard general relativity it is symmetric, carrying $\tfrac12 n(n+1)=10$ independent components in four dimensions.