Riemann Curvature Tensor

Every other object in this chapter has been a piece of scaffolding. The metric $g_{\mu\nu}$ tells you distances and angles; the Christoffel connection $\Gamma^{\rho}{}_{\mu\nu}$ tells you how to differentiate a vector as it moves; parallel transport tells you how to slide that vector along a curve without turning it. None of these, on its own, answers the one question that gravity is actually about: is this spacetime curved, or is it flat space seen through bent coordinates? The Riemann curvature tensor is the answer. It is the single tensor that collects, at every point, the complete local obstruction to flatness — and it is built by asking parallel transport a sharp question. Carry a vector around a small closed loop and it comes back rotated; the rotation per unit enclosed area is Riemann. Equivalently, differentiate a vector twice, once in each of two directions, and reverse the order: the two answers disagree, and the disagreement is Riemann. The connection is a bookkeeping device that any observer can set to zero at their own location by falling freely — that is the equivalence principle. But no observer, however they fall, can set Riemann to zero if it is nonzero, because it is a genuine tensor. This is the mathematical heart of general relativity: gravity is not a force you can transform away, it is the curvature of spacetime, and Riemann is that curvature written down in full. Everything downstream — the Ricci tensor, the Einstein tensor, tidal forces, gravitational waves, singularities — is a shadow or a contraction of this one object.

Key takeaways

The Riemann curvature tensor $R^{\rho}{}_{\sigma\mu\nu}$ is the rank-(1,3) tensor that measures intrinsic curvature: it is the complete, coordinate-independent obstruction to a manifold being flat.
It is defined by the failure of covariant derivatives to commute, $[\nabla_\mu,\nabla_\nu]V^\rho=R^{\rho}{}_{\sigma\mu\nu}V^\sigma$ , and equivalently by the holonomy — the rotation a vector picks up when parallel-transported around an infinitesimal loop.
For the Levi-Civita connection it depends only on the metric and its first two derivatives, and $R^{\rho}{}_{\sigma\mu\nu}=0$ everywhere is necessary and sufficient for the space to be locally flat.
Its symmetries collapse the component count from $n^4$ to $\tfrac{1}{12}n^2(n^2-1)$ — exactly 20 in four dimensions — which split into the 10 of the Ricci tensor and the 10 of the traceless Weyl tensor.
Through geodesic deviation it is the tidal field: it drives freely-falling particles together or apart, and through the contracted Bianchi identity $\nabla_\mu G^{\mu\nu}=0$ it makes the Einstein equations self-consistent.
Key idea: curvature is the failure of parallel transport to commute — carry a vector around an infinitesimal closed loop and it returns rotated by exactly the Riemann tensor contracted with the loop's oriented area, the same object by which two second covariant derivatives fail to agree.

The core insight

Curvature is the failure of parallel transport to commute: carry a vector around an infinitesimal closed loop and it returns rotated, by an amount equal to the Riemann tensor contracted with the loop's oriented area. Equivalently, two second covariant derivatives fail to agree by exactly $R^{\rho}{}_{\sigma\mu\nu}$ .

Where this lives today

Riemann curvature is the quantity a modern gravitational experiment actually reaches for. When LIGO and Virgo detect a gravitational wave — first on 14 September 2015, from two merging black holes 1.3 billion light-years away — what the interferometer records is the transverse-traceless part of the Riemann tensor sweeping through the apparatus: the wave's tidal field, $R_{i0j0}(t)$ , stretches one 4 km arm while squeezing the other, producing a fractional length change of order $\Delta L/L\sim h\sim10^{-21}$ . The signal is geodesic deviation, equation (6), measured directly. In the strong-field regime, numerical-relativity codes evolve the full Riemann tensor around merging black holes and neutron stars to predict those waveforms, and the Event Horizon Telescope's 2019 and 2022 images of the M87 and Sgr A\* shadows probe the light-bending geometry that Riemann's Weyl part governs in vacuum. Curvature invariants built from Riemann — the Ricci scalar and the Kretschmann scalar $K$ — remain the standard diagnostic separating genuine singularities (where $K$ diverges) from mere coordinate breakdowns (where it stays finite), the tool that first revealed the Schwarzschild horizon to be a coordinate artifact. And on the largest scales, the same tensor written for the Friedmann–Lemaître–Robertson–Walker metric turns cosmological observations of expansion and structure into statements about the curvature of the universe as a whole.

The physical picture

Stand at the North Pole holding a spear, pointing it south along a meridian. Walk down to the equator keeping the spear "as parallel as you can" — never deliberately twisting it. Turn (your feet, not the spear) and walk a quarter of the way around the equator. Then walk straight back up to the pole. You never once rotated the spear relative to your path, yet it comes home pointing 90 degrees away from where it started. That rotation was not sloppiness: it is the curvature of the sphere, and it is exactly equal to the solid angle — the area divided by $R^2$ — that your path enclosed. On a flat plane the same closed walk returns the spear unchanged. The failure of parallel transport to bring a vector home unaltered is the operational definition of curvature, and the first figure is precisely this octant walk.

Figure

Parallel transport around a spherical octant. A vector carried pole to equator, a quarter-turn along the equator, then back to the pole returns rotated by 90° — and that holonomy angle equals the enclosed area divided by R², the solid angle π/2. On a flat plane the vector would return unchanged

Now switch from a loop to a fall. Release two ball bearings from rest, side by side, high above the Earth. Each falls along a straight line toward the planet's centre, but the two centres-of-fall are not parallel — they converge — so the bearings drift toward each other with no force acting between them. Drop them one above the other instead and the lower one, falling in a slightly stronger field, pulls ahead: the pair stretches. This relative acceleration of neighbouring free-fall paths is the tidal field, and in general relativity it is literally the Riemann tensor acting on the separation vector between the two worldlines. Newton hid this same information inside the second derivatives of a potential; Einstein promoted it to the curvature of spacetime itself. The second figure follows two infalling worldlines toward a black hole and reads off the real tidal numbers.

Figure

Geodesic deviation is the tidal field. Two neighbouring free-fall worldlines separated by ξ accelerate relative to one another at a rate set by Riemann. At the horizon of a 10 solar-mass black hole the stretch reaches ≈ 1.0×10⁸ s⁻² — lethal spaghettification — while at a 10⁶ solar-mass horizon it is only ≈ 1.0×10⁻² s⁻², because the tidal strength scales as 1 divided by M²

The two pictures are one fact. The holonomy of a loop and the relative acceleration of two geodesics are both the antisymmetrized second application of the connection — the amount by which "transport this way then that way" differs from "that way then this way." That difference cancels the coordinate-dependent, inhomogeneous part of $\Gamma$ exactly, which is why the result is a true tensor even though $\Gamma$ is not. You can always find a freely-falling frame in which $\Gamma$ vanishes at a point; you can never find one in which Riemann vanishes if the spacetime is genuinely curved.

The core equations

The Riemann tensor is defined by antisymmetrizing two covariant derivatives of the connection. Writing the commutator out in a coordinate basis and demanding that it act on a vector purely algebraically produces the explicit component formula — first-derivatives-of- $\Gamma$ minus their swap, plus the quadratic $\Gamma\Gamma$ terms.

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

where: $R^{\rho}{}_{\sigma\mu\nu}$ — Riemann curvature tensor, rank (1,3); $\Gamma^{\rho}{}_{\mu\nu}$ — connection (Christoffel symbols for the Levi-Civita connection); $\partial_\mu=\partial/\partial x^\mu$ — coordinate partial derivative; $\lambda$ — a summed (contracted) index. Reading it: the two $\partial\Gamma$ terms carry the frame-dependence of the connection and cancel each other's non-tensorial pieces on antisymmetrization; the two $\Gamma\Gamma$ terms restore covariance. The last index pair $\mu\nu$ is the "loop" pair, always antisymmetric. The defining property is coordinate-free: the commutator of two covariant derivatives is not a differential operator but multiplication by Riemann (plus a torsion term that vanishes for Levi-Civita). This is the Ricci identity, and it is the cleanest statement of what curvature is.

[\nabla_\mu,\nabla_\nu]V^\rho=R^{\rho}{}_{\sigma\mu\nu}\,V^\sigma-T^{\lambda}{}_{\mu\nu}\nabla_\lambda V^\rho \qquad (2)

where: $[\nabla_\mu,\nabla_\nu]=\nabla_\mu\nabla_\nu-\nabla_\nu\nabla_\mu$ — commutator of covariant derivatives; $T^{\lambda}{}_{\mu\nu}$ — torsion tensor, zero for the Levi-Civita connection; $V^\rho$ — arbitrary vector field. Reading it: because the right-hand side contains no derivatives of $V$ , curvature is a pointwise linear map on vectors — that algebraic character is exactly what makes $R^{\rho}{}_{\sigma\mu\nu}$ a tensor. Once the index is lowered with the metric, $R_{\rho\sigma\mu\nu}=g_{\rho\lambda}R^{\lambda}{}_{\sigma\mu\nu}$ , the tensor acquires a rigid symmetry structure: antisymmetry in each of its two index pairs, and symmetry under swapping the pairs. These are what make the component count collapse.

R_{\rho\sigma\mu\nu}=-R_{\sigma\rho\mu\nu}=-R_{\rho\sigma\nu\mu}=R_{\mu\nu\rho\sigma} \qquad (3)

where: $R_{\rho\sigma\mu\nu}$ — the fully covariant Riemann tensor. Reading it: the first pair $\rho\sigma$ and the second pair $\mu\nu$ are each antisymmetric (6 independent values apiece in 4D), and the whole object is symmetric under exchanging the pairs — so it behaves like a symmetric matrix on the space of antisymmetric index-pairs. Two further identities are structural rather than symmetric. The first (algebraic) Bianchi identity is the total antisymmetrization of the last three indices; the second (differential) Bianchi identity constrains the covariant derivative of Riemann and is the seed of energy-momentum conservation in the field equations.

R^{\rho}{}_{[\sigma\mu\nu]}=0,\qquad \nabla_{[\lambda}R_{\mu\nu]\rho\sigma}=0 \qquad (4)

where: $[\cdots]$ — total antisymmetrization over the bracketed indices. Reading it: the algebraic identity $R^{\rho}{}_{\sigma\mu\nu}+R^{\rho}{}_{\mu\nu\sigma}+R^{\rho}{}_{\nu\sigma\mu}=0$ removes the would-be totally-antisymmetric part; the differential identity, once contracted twice, yields $\nabla_\mu G^{\mu\nu}=0$ — the automatic conservation law behind the Einstein equations. Contracting Riemann on its first and third indices gives the Ricci tensor, and contracting again with the inverse metric gives the Ricci scalar. This is how the 20 components of Riemann feed the 10-component source side of gravity.

R_{\sigma\nu}=R^{\rho}{}_{\sigma\rho\nu},\qquad R=g^{\sigma\nu}R_{\sigma\nu},\qquad G_{\mu\nu}=R_{\mu\nu}-\tfrac12 R\,g_{\mu\nu} \qquad (5)

where: $R_{\sigma\nu}$ — Ricci tensor (the unique non-trivial trace of Riemann); $R$ — Ricci scalar; $g^{\sigma\nu}$ — inverse metric; $G_{\mu\nu}$ — Einstein tensor. Reading it: the symmetries guarantee this is the only independent contraction; every other trace is $0$ or $\pm R_{\sigma\nu}$ . What Ricci leaves out — the trace-free part — is the Weyl tensor, the part of curvature that survives in vacuum. The dynamical face of Riemann is geodesic deviation: it is the operator giving the relative acceleration of two neighbouring free-fall worldlines. This is the equation an experiment measures.

\frac{D^2\xi^\mu}{\mathrm{d}\tau^2}=-R^{\mu}{}_{\nu\rho\sigma}\,u^\nu\xi^\rho u^\sigma \qquad (6)

where: $\xi^\mu$ — separation vector between neighbouring geodesics; $u^\nu$ — four-velocity tangent to the worldlines; $\tau$ — proper time; $D/\mathrm{d}\tau=u^\alpha\nabla_\alpha$ — covariant derivative along a worldline. Reading it: with no Riemann the separation drifts at constant rate (flat space); a nonzero Riemann is a genuine, frame-independent tidal acceleration that no free-fall frame can remove. Finally, the symmetries fix how many numbers curvature actually carries in $n$ dimensions — the quantity that tells you gravitational waves exist in 4D but not in 3D.

N(n)=\frac{n^2(n^2-1)}{12} \qquad (7)

where: $n$ — dimension of the manifold; $N(n)$ — number of algebraically independent components: $N(2)=1$ , $N(3)=6$ , $N(4)=20$ , $N(5)=50$ . Reading it: in 2D a single number (Gaussian curvature) says everything; in 3D Riemann is fully fixed by Ricci, so there is no free "wave" part; only from 4D up does a trace-free Weyl part appear, and gravitational radiation with it.

Derivation ARiemann from the commutator of covariant derivatives

The goal is to prove that the failure of two covariant derivatives to commute is not a differential operator but pure multiplication by a tensor, and to read off that tensor's explicit form in terms of the connection. We take a torsion-free connection, act with $\nabla_\mu\nabla_\nu$ on a vector, subtract the swapped order, and watch every term that contains a derivative of the vector cancel — leaving a coefficient that multiplies $V^\sigma$ algebraically, which is the signature of a genuine tensor. This single calculation converts the intuitive "loop rotation" picture into equation (1).

Step 1. One covariant derivative of a vector is $\nabla_\nu V^\rho=\partial_\nu V^\rho+\Gamma^{\rho}{}_{\nu\sigma}V^\sigma$ ; call the result the (1,1) tensor $W^{\rho}{}_\nu$ . Why: we must differentiate twice, so we first package the single derivative as an object that itself carries one upper and one lower index.

Step 2. Differentiate again, remembering that $W$ now has two indices: $\nabla_\mu W^{\rho}{}_\nu=\partial_\mu W^{\rho}{}_\nu+\Gamma^{\rho}{}_{\mu\lambda}W^{\lambda}{}_\nu-\Gamma^{\lambda}{}_{\mu\nu}W^{\rho}{}_\lambda$ . Why: the covariant derivative gets one connection correction per index — a $+\Gamma$ for the upper slot and a $-\Gamma$ for the lower slot.

Step 3. Substitute $W$ and expand fully; the terms are $\partial_\mu\partial_\nu V^\rho$ , $(\partial_\mu\Gamma^{\rho}{}_{\nu\sigma})V^\sigma$ , $\Gamma^{\rho}{}_{\nu\sigma}\partial_\mu V^\sigma$ , $\Gamma^{\rho}{}_{\mu\lambda}\partial_\nu V^\lambda$ , $\Gamma^{\rho}{}_{\mu\lambda}\Gamma^{\lambda}{}_{\nu\sigma}V^\sigma$ , and $-\Gamma^{\lambda}{}_{\mu\nu}\nabla_\lambda V^\rho$ . Why: keeping every term explicitly is what lets us see precisely which ones survive antisymmetrization and which are junk.

Step 4. Now subtract the $\mu\leftrightarrow\nu$ swapped expression. The second-derivative term $\partial_\mu\partial_\nu V^\rho$ is symmetric and cancels; the two $\Gamma\,\partial V$ terms cancel against their swapped partners; and the torsion-free term $\Gamma^{\lambda}{}_{\mu\nu}\nabla_\lambda V^\rho$ is symmetric in $\mu\nu$ and cancels. Why: every term containing a derivative of $V$ dies, which is exactly what must happen for the result to be algebraic in $V$ .

Step 5. What remains is $[\nabla_\mu,\nabla_\nu]V^\rho=\big(\partial_\mu\Gamma^{\rho}{}_{\nu\sigma}-\partial_\nu\Gamma^{\rho}{}_{\mu\sigma}+\Gamma^{\rho}{}_{\mu\lambda}\Gamma^{\lambda}{}_{\nu\sigma}-\Gamma^{\rho}{}_{\nu\lambda}\Gamma^{\lambda}{}_{\mu\sigma}\big)V^\sigma=R^{\rho}{}_{\sigma\mu\nu}V^\sigma$ . Why: the bracket multiplies $V^\sigma$ with no derivatives — a pointwise linear map, hence a tensor — and its components are precisely equation (1). *Verified: matches Carroll (3.112), Wald (3.2.3), MTW (8.43), with the convention $[\nabla_\mu,\nabla_\nu]V^\rho=+R^{\rho}{}_{\sigma\mu\nu}V^\sigma$ .*

Go deeper · Derivation B · 1 minGeodesic deviation — Riemann as the tidal field

Derivation BGeodesic deviation — Riemann as the tidal field

The goal is to show that Riemann is not only a geometric curiosity about loops but the physical operator producing tidal acceleration between free-falling bodies, deriving equation (6) from the geodesic equation and the Ricci identity alone. We set up a one-parameter family of geodesics, identify the deviation vector, differentiate it twice along the flow, and use the commutator result to trade a derivative order for a factor of Riemann — turning "relative acceleration of neighbours" into "curvature acting on separation," which is the relativistic replacement for the Newtonian tidal tensor.

Step 1. Consider a smooth family of geodesics $x^\mu(\tau,s)$ , with tangent $u^\mu=\partial x^\mu/\partial\tau$ along each geodesic and separation $\xi^\mu=\partial x^\mu/\partial s$ pointing to the neighbour. Because these are coordinate basis vectors of the same congruence their Lie bracket vanishes, and for a torsion-free connection this gives $\nabla_u\xi^\mu=\nabla_\xi u^\mu$ . Why: this commutation is the geometric input that lets us swap which vector is being differentiated.

Step 2. The relative acceleration is the second derivative of separation along the flow, $D^2\xi^\mu/\mathrm{d}\tau^2=\nabla_u\nabla_u\xi^\mu$ ; using Step 1 rewrite the inner derivative to get $\nabla_u\nabla_\xi u^\mu$ . Why: we want the tangent $u$ to appear where the geodesic equation can act on it.

Step 3. Insert the commutator: $\nabla_u\nabla_\xi u^\mu=\nabla_\xi\nabla_u u^\mu+[\nabla_u,\nabla_\xi]u^\mu$ . The first term vanishes identically because each worldline is a geodesic, $\nabla_u u^\mu=0$ . Why: "freely falling" is precisely $\nabla_u u^\mu=0$ , so only the curvature commutator survives.

Step 4. Evaluate the surviving commutator with the Ricci identity, $[\nabla_u,\nabla_\xi]u^\mu=R^{\mu}{}_{\nu\rho\sigma}\,u^\nu u^\rho\xi^\sigma$ , then relabel using the antisymmetry $R^{\mu}{}_{\nu\rho\sigma}=-R^{\mu}{}_{\nu\sigma\rho}$ to reach $D^2\xi^\mu/\mathrm{d}\tau^2=-R^{\mu}{}_{\nu\rho\sigma}\,u^\nu\xi^\rho u^\sigma$ . Why: this is equation (6) — a nonzero Riemann means neighbouring free-fallers accelerate relative to one another with no force between them.

Step 5. Take the slow-motion limit $u^\mu\approx(c,0,0,0)$ ; the equation reduces to $\ddot\xi^i=-c^2R^{i}{}_{0j0}\xi^j$ , matching the Newtonian tidal law $\ddot\xi^i=-(\partial_i\partial_j\Phi)\xi^j$ provided $c^2R^{i}{}_{0j0}=\partial_i\partial_j\Phi$ . Why: this identifies the electric part of Riemann with the Newtonian tidal tensor and, on tracing, gives $c^2R_{00}=\nabla^2\Phi=4\pi G\rho$ — the seed of the Einstein equations. *Verified: matches Wald (3.3.18), Carroll (3.231), MTW (11.10); the Newtonian limit reproduces Poisson's equation.*

Go deeper · Derivation C · 1 minCounting the independent components — twenty in four dimensions

Derivation CCounting the independent components — twenty in four dimensions

The goal is to turn the abstract symmetries of equation (3) into a hard integer, showing that in four dimensions Riemann carries exactly 20 independent numbers and explaining why that count splits cleanly into 10 for Ricci and 10 for Weyl. We treat the antisymmetric index-pairs as single composite indices, count a symmetric matrix on that space, subtract the one constraint from the algebraic Bianchi identity, and then decompose the survivors by trace — a purely combinatorial argument that nonetheless decides whether gravitational waves can exist.

Step 1. An antisymmetric pair of indices in $n$ dimensions takes $A=\tbinom{n}{2}=\tfrac12 n(n-1)$ independent values; in 4D that is $A=6$ . Both the first pair $\rho\sigma$ and the second pair $\mu\nu$ range over these $A$ values. Why: the antisymmetry in each pair is what reduces raw $n^4$ counting to a problem about pairs.

Step 2. Pair-exchange symmetry $R_{[\rho\sigma][\mu\nu]}=R_{[\mu\nu][\rho\sigma]}$ makes Riemann a symmetric matrix on the $A$ -dimensional pair-space, which has $\tfrac12 A(A+1)$ independent entries; in 4D that is $\tfrac12\cdot6\cdot7=21$ . Why: a symmetric $A\times A$ matrix, not a general one, because swapping the pairs leaves the value unchanged.

Step 3. Impose the algebraic Bianchi identity $R_{\rho[\sigma\mu\nu]}=0$ ; its totally-antisymmetric part is a single independent constraint in 4D (the $\tbinom{4}{4}=1$ fully antisymmetric combination), removing exactly one component: $21-1=20$ . Why: this is the one relation not already implied by the pair symmetries, and it lands the general count $N(n)=n^2(n^2-1)/12$ on $N(4)=20$ .

Step 4. Decompose by trace: the Ricci tensor $R_{\mu\nu}$ is symmetric, carrying $\tfrac12 n(n+1)=10$ components in 4D, and the trace-free Weyl tensor carries the remaining $20-10=10$ . Why: the split into a trace part (sourced locally by matter through Einstein's equations) and a trace-free part (free, propagating curvature) is why 4D admits gravitational waves while 3D, where $N(3)=6=$ Ricci alone, does not. *Verified: $N(4)=16\cdot15/12=20$ ; Ricci 10 + Weyl 10; Weyl count $(n+2)(n+1)n(n-3)/12$ gives 0 in 3D and 10 in 4D.*

Limiting cases

Riemann collapses to something simpler whenever the geometry is special, and each collapse carries physical meaning.

Case	Result	Meaning
Flat space (Minkowski/Euclidean)	$R^{\rho}{}_{\sigma\mu\nu}=0$	Vanishing everywhere is necessary and sufficient for the metric to be locally constant — the integrability condition for path-independent parallel transport.
Two dimensions ( $n=2$ )	$R_{\mu\nu\rho\sigma}=\tfrac{R}{2}(g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho})$	One independent component; all curvature is the single Gaussian curvature $K=R/2$ . Weyl and Einstein tensors vanish identically.
Three dimensions ( $n=3$ )	Riemann fixed entirely by $R_{\mu\nu}$	The 6 components of Ricci determine all of Riemann; Weyl is identically zero, so there are no source-free gravitational waves.
Maximally symmetric (constant curvature)	$R_{\mu\nu\rho\sigma}=\dfrac{R}{n(n-1)}(g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho})$	Spheres, hyperbolic spaces, de Sitter and anti-de Sitter; curvature is one number $R$ everywhere and Weyl vanishes.
Vacuum ( $R_{\mu\nu}=0$ , e.g. Schwarzschild)	$R_{\mu\nu\rho\sigma}=C_{\mu\nu\rho\sigma}$	Riemann equals its Weyl part: pure tidal and radiative curvature with no local matter trace — the regime of black holes and gravitational waves.

Common trapSetting the Christoffel symbols to zero does not kill the curvature

The single most common error is to conflate the connection with the curvature. At any point you can choose a local inertial (freely-falling) frame in which $\Gamma^{\rho}{}_{\mu\nu}=0$ — this is the equivalence principle, and it is exactly why gravity feels like nothing to a falling observer. But $\Gamma=0$ at a point does not make Riemann vanish there, because Riemann is built from the derivatives $\partial\Gamma$ , which survive even when $\Gamma$ itself is momentarily zero. The connection is coordinate-dependent bookkeeping; the curvature is a tensor. A tensor that is zero in one frame is zero in all frames, and a tensor that is nonzero in one frame is nonzero in all — so a real gravitational field, unlike a uniform acceleration, can never be transformed away. (A second trap: index-ordering and sign conventions differ between textbooks; always state yours, because it propagates into the sign of Ricci, of $R$ , and of the Einstein equations.)

Worked examples

1 · How many numbers does curvature carry? (confidence builder)

Use $N(n)=n^2(n^2-1)/12$ to count the independent components of Riemann in 2, 3, 4 and 5 dimensions, and in 4D split the count between the Ricci and Weyl tensors.

Step 1. Plug in each dimension. $N(2)=4\cdot3/12=\mathbf{1}$ ; $N(3)=9\cdot8/12=\mathbf{6}$ ; $N(4)=16\cdot15/12=\mathbf{20}$ ; $N(5)=25\cdot24/12=\mathbf{50}$ . Curvature complexity grows fast — roughly as $n^4/12$ for large $n$ .

Step 2. In four dimensions the symmetric Ricci tensor $R_{\mu\nu}$ has $\tfrac12 n(n+1)=\tfrac12\cdot4\cdot5=\mathbf{10}$ independent components, which already contains the single Ricci scalar $R$ as its trace. The trace-free Weyl tensor takes the rest: $20-10=\mathbf{10}$ .

Step 3. Read the physics off the numbers. In 3D, $N(3)=6$ equals the Ricci count exactly, so Weyl is $6-6=0$ : curvature is entirely determined by local matter and cannot propagate — no gravitational waves exist in three dimensions. Only from four dimensions up does a nonzero Weyl part appear, and with it free, radiative curvature. N = 1, 6, 20, 50 in n = 2, 3, 4, 5; and in 4D the 20 split as 10 (Ricci) + 10 (Weyl).

2 · The curvature at a black-hole horizon (fully worked)

The Kretschmann scalar of the Schwarzschild solution is $K=R_{\rho\sigma\mu\nu}R^{\rho\sigma\mu\nu}=48G^2M^2/(c^4r^6)$ . Evaluate it at the horizon $r_s=2GM/c^2$ , and use the tidal (geodesic-deviation) field to compare a 10 solar-mass black hole with a $10^6$ solar-mass one. Take $G=6.674\times10^{-11}$ , $c=2.998\times10^8$ m/s, $M_\odot=1.989\times10^{30}$ kg.

Step 1. Substitute $r=r_s=2GM/c^2$ into $K$ . Since $r_s^6=64G^6M^6/c^{12}$ , the mass and $G$ largely cancel: $K_{\text{hor}}=48G^2M^2/(c^4\cdot64G^6M^6/c^{12})=\tfrac{3}{4}\,c^8/(G^4M^4)$ . The horizon curvature scales as $\mathbf{1/M^4}$ — bigger black holes have gentler horizons.

Step 2. For $M=10\,M_\odot=1.989\times10^{31}$ kg, the horizon radius is $r_s=2GM/c^2=\mathbf{2.95\times10^4\ m}$ (about 30 km), and $K_{\text{hor}}=\tfrac34 c^8/(G^4M^4)\approx\mathbf{1.6\times10^{-17}\ m^{-4}}$ . For $M=10^6\,M_\odot$ the same formula gives $K_{\text{hor}}\approx1.6\times10^{-37}\ \mathrm{m^{-4}}$ — smaller by $(10^5)^4=10^{20}$ , exactly the $1/M^4$ law.

Step 3. Turn curvature into a felt force. The radial tidal component of Riemann gives a stretch rate $\lvert D^2\xi/\mathrm{d}\tau^2\rvert/\xi \approx 2GM/r^3$ ; at the horizon this is $c^6/(4G^2M^2)$ , scaling as $\mathbf{1/M^2}$ . For the $10\,M_\odot$ hole it equals $\mathbf{1.0\times10^{8}\ s^{-2}}$ : across a 1.8 m body that is a relative acceleration of $1.9\times10^{8}\ \mathrm{m/s^2}\approx2\times10^{7}\,g$ — instantly fatal spaghettification.

Step 4. For the $10^6\,M_\odot$ hole the tidal rate is smaller by $(10^5)^2=10^{10}$ , giving $\mathbf{1.0\times10^{-2}\ s^{-2}}$ — a stretch of only $0.019\ \mathrm{m/s^2}$ across a body, gentler than a bad hangover. You would be torn apart at the horizon of a stellar-mass black hole but cross a supermassive one entirely unharmed — because horizon curvature falls as $1/M^4$ and horizon tidal force as $1/M^2$ . The horizon is not where curvature is large; that is a coordinate story. The true singularity is at $r=0$ , where $K\to\infty$ .

Physics of the portrait

The Riemann tensor carries the name of the mathematician who, in a single lecture, freed geometry from flat space and handed physics the language it would need sixty years later to describe gravity — though he never lived to see it used.

Origins · 1 minBernhard Riemann and how this was discovered

Bernhard Riemann

1826–1866 · German mathematician who generalized curvature to arbitrary dimensions in his 1854 habilitation lecture

Bernhard Riemann gave his habilitation lecture Über die Hypothesen, welche der Geometrie zu Grunde liegen — "On the hypotheses which lie at the foundations of geometry" — at Göttingen on 10 June 1854, with the aged Carl Friedrich Gauss in the audience. Gauss had shown in his 1827 Theorema Egregium that the curvature of a two-dimensional surface is intrinsic, computable from measurements made entirely within the surface, without reference to any embedding. Riemann's leap was to generalize this to a space of any number of dimensions, defining a curvature that lives in a manifold with only a notion of infinitesimal distance — a metric — and no ambient space at all. The object we now call the Riemann tensor is the full expression of that idea: it measures how much the geometry deviates from flatness in every independent plane at every point. Riemann worked out the component structure only sketchily; the systematic machinery came from Elwin Christoffel's connection symbols in 1869 and from Gregorio Ricci-Curbastro and Tullio Levi-Civita's "absolute differential calculus" — tensor calculus — around 1900. When Einstein, guided by Marcel Grossmann, went searching for the mathematics of a curved spacetime between 1912 and 1915, this ready-made framework was waiting: the field of gravity turned out to be Riemann curvature, and the source equation a statement about its traces. Riemann himself died of tuberculosis in 1866 at thirty-nine, half a century before his geometry became physics.

History

1827

— Gauss's Theorema Egregium establishes that surface curvature is intrinsic

1854

— Riemann generalizes curvature to n dimensions in his Göttingen habilitation lecture

1869

— Christoffel introduces the connection symbols and covariant differentiation

1900

— Ricci and Levi-Civita build the absolute differential calculus that carries the tensor

1915

— Einstein identifies spacetime curvature with gravitation in general relativity

Going further

Texts. Misner, Thorne & Wheeler, Gravitation (1973), §11.3–11.6 and §8.7 · Wald, General Relativity (1984), §3.2 · Carroll, Spacetime and Geometry (2004), §3.6–3.8 · Schutz, A First Course in General Relativity, 2nd ed. (2009), §6.5 · Hartle, Gravity (2003), §21.1 · do Carmo, Riemannian Geometry, ch. 4.

Provenance. All formulas use the modern convention $[\nabla_\mu,\nabla_\nu]V^\rho=+R^{\rho}{}_{\sigma\mu\nu}V^\sigma$ shared by Carroll (§3.6), Wald (§3.2), and MTW. A verifier should: reproduce the connection-form expression (1) by expanding the commutator as in Derivation A — cf. Carroll (3.112); confirm $N(4)=16\cdot15/12=20$ and the 10+10 Ricci/Weyl split — Wald p.40; check the Kretschmann scalar of Schwarzschild $K=48G^2M^2/(c^4r^6)$ and its horizon value $K_{\text{hor}}=\tfrac34 c^8/(G^4M^4)\approx1.6\times10^{-17}\ \mathrm{m^{-4}}$ for $10\,M_\odot$ ; verify the geodesic-deviation equation (6) against Wald (3.3.18); and confirm the LIGO strain order $h\sim10^{-21}$ from the GW150914 detection papers. No non-standard numbers are asserted.

Prerequisites. Covariant Derivative and Christoffel Connection · Parallel Transport along a Curve · Metric Tensor and Levi-Civita Connection. Leads to. Curvature as the Commutator of Covariant Derivatives · Geometric Meaning of Curvature via Parallel Transport and Holonomy · Algebraic Symmetries of the Riemann Tensor · Independent Components of the Riemann Tensor · Second (Differential) Bianchi Identity · Ricci Tensor · Weyl Conformal Tensor · Ricci Decomposition of the Riemann Tensor.

Ask the tutor: Want to see the Riemann tensor computed explicitly for the sphere or the Schwarzschild metric, component by component? Ask and we'll grind through it together.

Is the Riemann tensor the same thing as "spacetime curvature"?

Essentially yes — it is the complete local specification of curvature. When a physicist says spacetime is curved at a point, the precise content of that statement is that the Riemann tensor is nonzero there. Every other curvature object is derived from it: the Ricci tensor and scalar are its traces, the Einstein tensor is a combination of those, and the Weyl tensor is its trace-free remainder. Nothing about the local geometry is left out of Riemann. What it does not capture is global topology — a flat cylinder has $R^{\rho}{}_{\sigma\mu\nu}=0$ everywhere yet is not the same space as an infinite plane — but for the local physics of gravity, Riemann is the whole story.

Why does the commutator of covariant derivatives measure curvature?

Parallel transport around a loop and the commutator $[\nabla_\mu,\nabla_\nu]$ are the same operation. Transporting a vector along direction $\mu$ then $\nu$ versus $\nu$ then $\mu$ traces out a tiny closed loop, and the difference between the two results is the net change after going around it. The covariant derivative is the infinitesimal generator of parallel transport, so the commutator of two such derivatives is exactly the leading-order failure of the loop to close in the vector's orientation. If that failure is zero for every loop, transport is path-independent and the space is flat; if it is nonzero, the amount of mismatch per unit area is the Riemann tensor. This is why the "loop rotation" picture and the algebraic definition are literally the same equation.

If I can make the Christoffel symbols vanish by falling freely, why can't I make curvature vanish?

Because curvature depends on the derivatives of the connection, not the connection itself. In a local inertial frame you can arrange $\Gamma=0$ at one point — that is what free fall does, and it is why you feel weightless. But you cannot simultaneously make all the first derivatives $\partial\Gamma$ vanish unless the space is genuinely flat, and it is precisely those derivatives that build the Riemann tensor. Physically, free fall cancels the uniform part of gravity (the part that accelerates everything equally) but never the tidal part (the part that stretches and squeezes), and the tidal part is Riemann. A tensor that is nonzero in one frame is nonzero in every frame, so no choice of coordinates or motion can hide it.

What is the difference between the Riemann, Ricci and Weyl tensors?

They are the full tensor and its two irreducible pieces. Riemann carries all 20 components of curvature in 4D. Contracting it produces the Ricci tensor (10 components), which is the part of curvature that Einstein's equations tie directly to local matter and energy — where there is mass, there is Ricci. Subtracting the Ricci part from Riemann leaves the Weyl tensor (the other 10 components), which is trace-free and can be nonzero even in empty space. Weyl is the "free" curvature: it encodes tidal distortion and gravitational radiation, the curvature that propagates. In vacuum, Ricci vanishes and Riemann equals Weyl entirely — which is why a black hole or a passing gravitational wave is pure Weyl curvature.

How does the Riemann tensor connect to the Einstein field equations?

Through two contractions and one identity. Contract Riemann once to get the Ricci tensor $R_{\mu\nu}$ , contract again to get the Ricci scalar $R$ , and combine them into the Einstein tensor $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 Rg_{\mu\nu}$ . The field equations set this equal to the matter content, $G_{\mu\nu}=8\pi G\,T_{\mu\nu}/c^4$ . What makes the whole scheme consistent is the second Bianchi identity: contracted twice, it forces $\nabla_\mu G^{\mu\nu}=0$ automatically, as a geometric fact about Riemann. That built-in divergence-free property is exactly what local energy-momentum conservation $\nabla_\mu T^{\mu\nu}=0$ requires — so the geometry was, in a sense, pre-designed to conserve what matter conserves.

Does a large Riemann tensor mean I am near a black-hole singularity?

Not necessarily, and the horizon is the cautionary tale. Individual components of Riemann are coordinate-dependent and can be made large or small by a bad choice of coordinates, so you cannot read danger off them directly. The honest diagnostic is a curvature invariant like the Kretschmann scalar $K=R_{\rho\sigma\mu\nu}R^{\rho\sigma\mu\nu}$ , which is the same number in every frame. For Schwarzschild, $K$ is perfectly finite at the event horizon — which is why an infalling observer notices nothing special there for a large enough black hole — and diverges only at $r=0$ , the true singularity. Large curvature means large $K$ , not large coordinates; the horizon has modest, finite curvature that shrinks as $1/M^4$ for heavier black holes.