Weyl Conformal Tensor

The Riemann tensor is the full ledger of spacetime curvature, and Einstein's field equations write on only half of it. In four dimensions Riemann carries twenty independent components; the Ricci tensor — its trace — carries ten, and those ten are the ones the field equations tie directly to the local stress-energy. Fix all the matter at a point and you have still fixed only half the curvature. The Weyl tensor is the other half: the completely trace-free part of Riemann that no local matter can pin down, the piece that survives in empty space and propagates as a field in its own right. It is built by subtracting from Riemann every trace it contains — the Ricci tensor and the Ricci scalar, assembled with exactly the coefficients that make all further contractions vanish — so that what remains answers to no source at the point where you stand. Two properties give this remainder its name and its power. First, it is conformally invariant: rescale the metric by any smooth positive factor, $g_{ab}\to\Omega^2(x)g_{ab}$ , stretching every length and time while leaving angles and light-cones untouched, and the mixed tensor $C^a{}_{bcd}$ does not change at all — a spacetime is conformally flat exactly when its Weyl tensor vanishes. Second, because it lives where Ricci does not, the Weyl tensor is the carrier of the free gravitational field — the static tidal reach of a black hole, the dragging of inertial frames, and above all gravitational radiation. Its Petrov algebraic types, its split into electric and magnetic parts, and its five Newman–Penrose scalars organize nearly everything in modern relativity, from the waveform templates of LIGO to Penrose's conjecture that Weyl curvature encodes the arrow of time.

Key takeaways

The Weyl tensor $C_{abcd}$ is the totally trace-free part of the Riemann tensor: Riemann minus its Ricci and scalar traces, assembled so every contraction $g^{ac}C_{abcd}=0$ vanishes.
It is conformally invariant — under $g_{ab}\to\Omega^2 g_{ab}$ the mixed form $C^a{}_{bcd}$ is unchanged, which is why it is called the conformal curvature tensor.
In vacuum $R_{ab}=0$ , so the entire Riemann tensor equals the Weyl tensor: black-hole tidal fields and gravitational waves are pure Weyl curvature.
It has $\tfrac{1}{12}n(n+1)(n+2)(n-3)$ independent components — 10 in four dimensions, and identically zero for $n\le 3$ , where the Cotton tensor takes over.
For Schwarzschild the Weyl scalar is $C_{abcd}C^{abcd}=48G^2M^2/r^6$ , diverging at $r=0$ (true singularity) yet finite at the horizon (a coordinate artifact).
Its electric and magnetic parts split it like an electromagnetic field, and its Petrov type and Newman–Penrose scalars $\Psi_0\ldots\Psi_4$ classify radiation, black holes, and exact solutions.
Key idea: subtract every trace from the Riemann tensor and what is left — the Weyl tensor — is both conformally invariant and source-free, so it represents the free gravitational field: tidal forces and gravitational waves that persist in vacuum.

Where this lives today

The Weyl tensor is the beating heart of gravitational-wave astronomy. When two black holes spiral together, numerical relativity codes extract the waveform not from the metric directly but from the Newman–Penrose scalar $\Psi_4$ , the outgoing-radiation projection of the Weyl tensor, evaluated on spheres far from the collision and decomposed into spin-weighted spherical harmonics. Those $\Psi_4$ templates are exactly what LIGO and Virgo matched-filter against detector data — the 2015 discovery GW150914 was a Weyl-scalar waveform confronting a strain time series. The same tensor's algebraic type organizes the catalogue of exact solutions: Schwarzschild and Kerr are Petrov type D, pure radiation is type N, and the peeling theorem predicts how the type degenerates from D near a source to N far away, central to defining mass and momentum at null infinity. In cosmology the Weyl tensor vanishes for the smooth FLRW background but grows as structure clumps, underlying Penrose's Weyl curvature hypothesis, and its electric and magnetic parts drive the tidal shear that shapes the cosmic web in relativistic simulations.

The physical picture

Release a small sphere of test particles into free fall and watch what curvature does to it. The Riemann tensor drives the relative acceleration of nearby geodesics through the geodesic-deviation equation, and its two irreducible pieces act in visibly different ways. The Ricci part changes the sphere's volume — it is fed directly by the local energy density, so it only acts where matter is present. The Weyl part changes the sphere's shape at fixed volume: it shears the sphere into an ellipsoid, stretching along one axis while squeezing the perpendicular ones, conserving the enclosed volume exactly. Tidal stretching, frame dragging, and the passing distortion of a gravitational wave are all shear — all Weyl. The first figure is this decomposition made concrete.

Figure

A freely falling ball of test particles under geodesic deviation. The trace-free Weyl part shears it into an ellipsoid at constant volume (tr E = 0) — pure tidal stretch and squeeze — while the Ricci part, fed by local matter, shrinks the volume. In vacuum only the Weyl shear survives

Conformal invariance is the deeper fingerprint. A conformal map preserves angles and the causal light-cone structure while rescaling lengths arbitrarily from point to point, and the Weyl tensor is blind to that rescaling — it captures exactly the angular, causal part of curvature and nothing of the local scale. The most physical place to see Weyl carry a real number is outside a black hole, where there is no matter at all, so the whole Riemann tensor collapses onto Weyl. The quadratic invariant $C_{abcd}C^{abcd}=48G^2M^2/r^6$ then measures the tidal reach directly: it stays finite as you cross the horizon but blows up at the center, cleanly separating the coordinate horizon from the genuine singularity. The second figure plots that curve.

Figure

The Schwarzschild Weyl scalar C_abcd C^abcd = 48 G²M²/r⁶ versus radius r in units of GM (log scale). It is perfectly finite at the horizon r = 2GM — where the dimensionless value is 48/64 = 0.75 — proving the horizon is a coordinate artifact, yet it diverges as r → 0, marking r = 0 as a true curvature singularity

The core equations

The Weyl tensor is defined by subtracting from Riemann the unique combination of Ricci and scalar terms — carrying the symmetries of Riemann via antisymmetrization over index pairs — that makes every trace vanish. In $n$ dimensions this reads as follows.

C_{abcd}=R_{abcd}-\frac{2}{n-2}\left(g_{a[c}R_{d]b}-g_{b[c}R_{d]a}\right)+\frac{2}{(n-1)(n-2)}R\,g_{a[c}g_{d]b} \qquad (1)

where: $C_{abcd}$ — Weyl conformal tensor, fully covariant; $R_{abcd}$ — Riemann curvature tensor; $R_{ab}$ — Ricci tensor, the trace $R^c{}_{acb}$ ; $R$ — Ricci scalar, $g^{ab}R_{ab}$ ; $g_{ab}$ — metric tensor; $n$ — spacetime dimension; $[\,\cdot\,]$ — antisymmetrization over the bracketed pair with weight $\tfrac12$ . Reading it: Riemann on the left, its Ricci content removed in the middle, its scalar content removed on the right, tuned so nothing tracelike is left. By construction all single contractions of the Weyl tensor vanish — this is the algebraic statement that it carries none of the information Einstein's equations fix locally through the matter content.

g^{ac}C_{abcd}=0,\qquad C^a{}_{bad}=0 \qquad (2)

where: $g^{ac}$ — inverse metric contracting the first and third slots; $C^a{}_{bad}$ — the contraction, which vanishes on every index pair by the tensor's symmetries. This is what "trace-free" means, and it is the defining constraint from which the coefficients of equation (1) are fixed. The property that names the tensor is conformal invariance: rescaling the metric by a positive factor leaves the mixed-index Weyl tensor untouched, so it sees only the conformal (angle- and cone-preserving) part of geometry.

\tilde C^a{}_{bcd}=C^a{}_{bcd}\quad\text{under}\quad g_{ab}\to\tilde g_{ab}=\Omega^2(x)\,g_{ab} \qquad (3)

where: $\Omega(x)$ — smooth positive conformal factor; $C^a{}_{bcd}$ — the strictly invariant $(1,3)$ form; $\tilde C$ — Weyl tensor of the rescaled metric. Only this one-index-up form is invariant: lowering an index restores an $\Omega^2$ factor, $\tilde C_{abcd}=\Omega^2 C_{abcd}$ . Although matter never fixes Weyl at a point, it sources it nonlocally: the contracted second Bianchi identity turns into a divergence equation for Weyl driven by derivatives of the Ricci tensor — the gravitational echo of Maxwell's $\nabla_a F^{ab}=J^b$ .

\nabla_a C^a{}_{bcd}=(n-3)\,J_{bcd},\qquad J_{bcd}=\frac{1}{n-2}\left(\nabla_c R_{db}-\nabla_d R_{cb}+\frac{1}{2(n-1)}(g_{db}\nabla_c R-g_{cb}\nabla_d R)\right) \qquad (4)

where: $\nabla_a$ — Levi-Civita covariant derivative; $J_{bcd}$ — the Cotton (Cotton–York) tensor built from first derivatives of Ricci; the prefactor $(n-3)$ makes the equation trivial in three dimensions. Relative to a timelike observer $u^a$ the Weyl tensor splits into two symmetric trace-free spatial tensors, in exact analogy with resolving the electromagnetic field into $\mathbf E$ and $\mathbf B$ .

E_{ac}=C_{abcd}\,u^b u^d,\qquad B_{ac}=\tfrac12\,\epsilon_{ab}{}^{ef}C_{efcd}\,u^b u^d \qquad (5)

where: $u^a$ — unit timelike 4-velocity, $u^a u_a=-1$ ; $E_{ac}$ — electric part, tidal forces; $B_{ac}$ — magnetic part, frame-dragging and radiation; $\epsilon_{abef}$ — volume form. In vacuum the geodesic-deviation tidal tensor is $E_{ac}$ . The quadratic scalar invariant of Weyl can be written entirely in terms of the standard curvature invariants; in four dimensions the relation is exact.

C_{abcd}C^{abcd}=R_{abcd}R^{abcd}-2R_{ab}R^{ab}+\tfrac13 R^2\qquad(n=4) \qquad (6)

where: $C_{abcd}C^{abcd}$ — quadratic Weyl invariant; $R_{abcd}R^{abcd}$ — Kretschmann scalar; the last two terms subtract exactly the trace content. In vacuum they vanish and the Weyl invariant equals the Kretschmann scalar outright. For the Schwarzschild geometry — the vacuum field of a spherical mass — that invariant has a famous closed form that diverges at the center.

C_{abcd}C^{abcd}=\frac{48\,G^2M^2}{r^6}\qquad\text{(Schwarzschild)} \qquad (7)

where: $M$ — mass; $G$ — Newton's constant; $r$ — areal Schwarzschild radius (units $c=1$ ). Finite at the horizon $r=2GM$ , it blows up as $r\to0$ — the algebraic signature of a true curvature singularity. Finally, projecting Weyl onto a complex null tetrad packages its ten components into five complex scalars, the working variables of the Newman–Penrose formalism and numerical relativity.

\Psi_0,\ \Psi_1,\ \Psi_2,\ \Psi_3,\ \Psi_4 \qquad (8)

where: $\Psi_4$ — outgoing transverse radiation (the gravitational-wave signal); $\Psi_0$ — ingoing transverse radiation; $\Psi_2$ — the Coulomb-like part carrying mass and the static tidal field; $\Psi_1,\Psi_3$ — longitudinal, tetrad-dependent pieces removable in a suitable frame.

Derivation AFixing the Weyl coefficients by demanding tracelessness

The goal is to show that the two peculiar coefficients in the definition, $-2/(n-2)$ and $2/[(n-1)(n-2)]$ , are not chosen for elegance but forced uniquely by a single requirement: that after subtracting Ricci and scalar pieces from Riemann, every contraction of the result vanishes. We start from the most general trace-subtraction that respects Riemann's symmetries, contract it, and demand the contraction be zero, which determines both numbers with no freedom left over.

Step 1. Write the ansatz $C_{abcd}=R_{abcd}+\alpha\,(g_{a[c}R_{d]b}-g_{b[c}R_{d]a})+\beta\,R\,g_{a[c}g_{d]b}$ , where the antisymmetrizations guarantee $C$ inherits the pair-antisymmetry, pair-exchange symmetry, and first Bianchi identity of Riemann for any constants $\alpha,\beta$ . Why: only tensors built from $g$ , $R_{ab}$ , $R$ can be subtracted without introducing new independent structures, and the bracket structure is the unique way to match Riemann's index symmetries.

Step 2. Recall the two basic traces, $g^{ac}R_{abcd}=R_{bd}$ and $g^{ac}g_{a[c}g_{d]b}=\tfrac12(n-1)g_{bd}$ , and the mixed trace $g^{ac}(g_{a[c}R_{d]b}-g_{b[c}R_{d]a})=\tfrac12\big[(n-2)R_{bd}+g_{bd}R\big]$ . Why: contracting the ansatz on the first and third indices is the operation that "trace-freeness" constrains, so we need each term's contribution to $g^{ac}C_{abcd}$ .

Step 3. Assemble the contraction: $g^{ac}C_{abcd}=R_{bd}+\tfrac{\alpha}{2}\big[(n-2)R_{bd}+g_{bd}R\big]+\tfrac{\beta}{2}(n-1)g_{bd}R$ . The $R_{bd}$ terms and the $g_{bd}R$ terms must separately cancel. Why: $R_{bd}$ and $g_{bd}R$ are independent tensor structures, so a single condition on their sum splits into two conditions.

Step 4. Setting the $R_{bd}$ coefficient to zero gives $1+\tfrac{\alpha}{2}(n-2)=0$ , hence $\alpha=-2/(n-2)$ . Substituting into the $g_{bd}R$ condition $\tfrac{\alpha}{2}+\tfrac{\beta}{2}(n-1)=0$ gives $\beta=-\alpha/(n-1)=2/[(n-1)(n-2)]$ , reproducing equation (1) exactly. Why: two linear equations in two unknowns have a unique solution, so the Weyl tensor is the one and only trace-free completion of the Ricci decomposition.

Go deeper · Derivation B · 1 minConformal invariance of the mixed Weyl tensor

Derivation BConformal invariance of the mixed Weyl tensor

The goal is to establish equation (3): that under a Weyl rescaling $g_{ab}\to\Omega^2 g_{ab}$ the mixed tensor $C^a{}_{bcd}$ is literally unchanged, even though Riemann, Ricci, and the scalar each transform with messy derivative-of- $\Omega$ terms. The strategy is to track the extra connection piece that the rescaling introduces, feed it through the curvature, and show that the same combinations of $\nabla\ln\Omega$ that appear in $\tilde R^a{}_{bcd}$ are subtracted off exactly by the transformed trace terms in the Weyl definition.

Step 1. The Levi-Civita connection of the rescaled metric differs from the original by a tensor: $\tilde\Gamma^a_{bc}=\Gamma^a_{bc}+\delta^a_b\,\Upsilon_c+\delta^a_c\,\Upsilon_b-g_{bc}\,\Upsilon^a$ , with $\Upsilon_a=\partial_a\ln\Omega$ . Why: two metrics related by a scalar factor share geodesics up to reparametrization, and the failure to share them exactly is captured entirely by the gradient $\Upsilon_a$ .

Step 2. Insert this shift into the curvature formula $R^a{}_{bcd}=\partial_c\Gamma^a_{db}-\partial_d\Gamma^a_{cb}+\Gamma\Gamma-\Gamma\Gamma$ . The mixed Riemann tensor picks up terms linear in $\nabla_a\Upsilon_b$ and quadratic in $\Upsilon_a\Upsilon_b$ , schematically $\tilde R^a{}_{bcd}=R^a{}_{bcd}+(\text{terms in }\nabla\Upsilon,\ \Upsilon\Upsilon)$ . Why: the connection enters the curvature through derivatives and products, so a tensorial shift in $\Gamma$ produces exactly these one- and two-derivative corrections.

Step 3. Contract to get the transformed Ricci tensor $\tilde R_{ab}$ and, tracing again, the scalar $\tilde R$ . Each absorbs the same $\nabla\Upsilon$ and $\Upsilon\Upsilon$ structures, now multiplied by the metric factors that the definition of the trace supplies. Why: the trace subtraction in equation (1) is built from precisely $g$ , $R_{ab}$ , $R$ , so its transformation is fixed once the Ricci and scalar transformations are known.

Step 4. Assemble $\tilde C^a{}_{bcd}=\tilde R^a{}_{bcd}-(\text{trace terms of }\tilde R_{ab},\tilde R)$ . Term by term, every appearance of $\Upsilon_a$ cancels between the Riemann correction and the trace corrections, leaving $\tilde C^a{}_{bcd}=C^a{}_{bcd}$ . Lowering an index reintroduces one factor of the metric, so $\tilde C_{abcd}=\Omega^2 C_{abcd}$ — invariance holds only for the $(1,3)$ form. Why: this cancellation is the algebraic content of conformal invariance, and it is why "Weyl vanishes" is a statement about a whole conformal class of metrics at once.

Go deeper · Derivation C · 1 minWhy vacuum curvature is pure Weyl

Derivation CWhy vacuum curvature is pure Weyl

The goal is to prove that wherever the Einstein equations hold with no matter and no cosmological constant, the full Riemann tensor coincides with the Weyl tensor, so that the entire geometry of a black hole or a gravitational wave is conformal curvature. The argument simply feeds the vacuum condition into the decomposition and watches both correction terms of equation (1) collapse, then reads off the consequence for the Schwarzschild invariant.

Step 1. In vacuum the Einstein tensor vanishes, $G_{ab}=R_{ab}-\tfrac12 R g_{ab}=0$ . Take its trace with $g^{ab}$ in $n=4$ : $R-2R=-R=0$ , so $R=0$ , and feeding that back gives $R_{ab}=0$ as well. Why: the field equations without a source force both traces of Riemann to vanish, which is exactly the input the Weyl definition needs.

Step 2. With $R_{ab}=0$ and $R=0$ , the middle and right terms of equation (1) are identically zero, leaving $C_{abcd}=R_{abcd}$ . Every ounce of curvature outside a star, around a black hole, or in a passing wave is therefore Weyl curvature. Why: this is the precise sense in which the Weyl tensor is the "free" gravitational field — it is all that curvature can be where matter is absent.

Step 3. As a corollary, the quadratic invariant of equation (6) loses its Ricci terms, so $C_{abcd}C^{abcd}=R_{abcd}R^{abcd}$ in vacuum. For Schwarzschild the Kretschmann scalar is the textbook $48G^2M^2/r^6$ , so the Weyl scalar equals it: in units $G=M=1$ it is finite at the horizon $r=2$ with value $48/64=0.75$ , and divergent as $r\to0$ . Why: a scalar invariant cannot be removed by any coordinate change, so its blow-up at $r=0$ certifies a genuine singularity while its finiteness at $r=2GM$ certifies the horizon as a mere coordinate feature.

Limiting cases

The Weyl tensor's very existence, and its physical content, depend sharply on the spacetime dimension and on whether matter is present.

Case	Result	Meaning
Two dimensions ( $n=2$ )	$R_{abcd}=\tfrac12 R\,(g_{ac}g_{bd}-g_{ad}g_{bc})$	Curvature is a single scalar; Weyl (and even trace-free Ricci) is meaningless. Every surface is conformally flat.
Three dimensions ( $n=3$ )	$C_{abcd}\equiv 0$	Riemann is fully fixed by Ricci, so no trace-free remainder exists; conformal flatness is instead governed by the Cotton tensor.
Vacuum, $n=4$	$R_{ab}=0\Rightarrow R_{abcd}=C_{abcd}$	All curvature is Weyl: black-hole tidal fields, frame dragging, and gravitational waves are pure conformal curvature.
Conformally flat ( $n\ge 4$ )	$C_{abcd}=0\Leftrightarrow g_{ab}=\Omega^2\eta_{ab}$ locally	Vanishing Weyl is necessary and sufficient for local conformal flatness; the FLRW cosmologies are the canonical examples.

Common trapWeyl-flat does not mean flat, and "the vacuum has no source" is only locally true

Three traps recur. First, a spacetime with $C_{abcd}=0$ everywhere is conformally flat, not flat: the FLRW universe has vanishing Weyl yet is richly curved, because its Ricci part is driven by the cosmic energy density. Flatness requires the whole Riemann tensor — Weyl and Ricci — to vanish. Second, conformal invariance belongs to $C^a{}_{bcd}$ with one index raised; the covariant $C_{abcd}$ scales as $\Omega^2$ , the raised $C^{abcd}$ as $\Omega^{-6}$ , and the scalar $C_{abcd}C^{abcd}$ as $\Omega^{-4}$ , so none of those is invariant. Third, although no local matter fixes Weyl at a point, it is not truly sourceless: the contracted Bianchi identity (equation 4) has matter derivatives on its right, so distant matter sources Weyl nonlocally — exactly how tidal fields and radiation reach into empty regions.

Worked examples

1 · Spaghettification: the Weyl scalar and tidal death near a black hole (fully worked)

The tidal stretching an infalling body feels is the electric Weyl tensor, which for radial separation gives a Newtonian-form tidal gradient $a_{\text{tidal}}=2GM/r^3$ per unit length. Evaluate it at the horizon $r=2GM/c^2$ for a solar-mass black hole and for the supermassive hole Sagittarius A ( $M=4\times10^6\,M_\odot$ ), and say who survives.*

Step 1. At the horizon substitute $r=2GM/c^2$ into the gradient: $a_{\text{tidal}}=2GM/r^3=2GM\,c^6/(8G^3M^3)=c^6/(4G^2M^2)$ . The horizon tidal gradient scales as $1/M^2$ — counterintuitively, bigger black holes are gentler at the horizon.

Step 2. Solar-mass case, $M=1.989\times10^{30}\,\text{kg}$ : with $c^6=7.29\times10^{50}$ , $G^2=4.454\times10^{-21}$ , $M^2=3.956\times10^{60}$ , the denominator is $4G^2M^2=7.05\times10^{40}$ , so $a_{\text{tidal}}=7.29\times10^{50}/7.05\times10^{40}=\mathbf{1.03\times10^{10}\ s^{-2}}$ . Across a $1.8\,\text{m}$ body the head-to-foot acceleration difference is $\approx\mathbf{1.9\times10^{10}\ m/s^2}$ — about two billion $g$ . Lethal long before the horizon.

Step 3. Sagittarius A*, $M=4\times10^6\,M_\odot$ : the $1/M^2$ scaling divides the solar result by $(4\times10^6)^2=1.6\times10^{13}$ , giving $a_{\text{tidal}}=1.03\times10^{10}/1.6\times10^{13}=\mathbf{6.5\times10^{-4}\ s^{-2}}$ , or only $\approx1.2\times10^{-3}\,\text{m/s}^2$ across a body — utterly harmless.

Step 4. Interpretation. Because the entire Schwarzschild curvature is Weyl (Derivation 3), these tidal numbers are the Weyl tensor at work; the invariant $C_{abcd}C^{abcd}=48G^2M^2/r^6$ tracks the same physics and likewise falls as $1/M^4$ at a fixed multiple of the horizon radius. An astronaut is torn apart outside a stellar black hole but crosses the horizon of a supermassive one intact — pure Weyl tidal geometry, no matter anywhere near.

2 · Counting the free gravitational field: Weyl components across dimensions

Use the component-count formula to find how many independent Weyl components exist in $n=3,4,5,6$ and confirm the four-dimensional bookkeeping Riemann = Weyl + Ricci.

Step 1. The Weyl count follows from subtracting the symmetric-tensor trace conditions from the Riemann count: $N_{\text{Weyl}}=\tfrac{1}{12}n^2(n^2-1)-\tfrac12 n(n+1)=\tfrac{1}{12}n(n+1)(n+2)(n-3)$ . The explicit factor $(n-3)$ is the algebraic reason Weyl vanishes below four dimensions.

Step 2. Evaluate. For $n=3$ : $\tfrac{1}{12}\cdot3\cdot4\cdot5\cdot0=\mathbf{0}$ . For $n=4$ : $\tfrac{1}{12}\cdot4\cdot5\cdot6\cdot1=\mathbf{10}$ . For $n=5$ : $\tfrac{1}{12}\cdot5\cdot6\cdot7\cdot2=\mathbf{35}$ . For $n=6$ : $\tfrac{1}{12}\cdot6\cdot7\cdot8\cdot3=\mathbf{84}$ .

Step 3. Check the four-dimensional ledger. Riemann has $\tfrac{1}{12}n^2(n^2-1)=\tfrac{1}{12}\cdot16\cdot15=\mathbf{20}$ components; Ricci, a symmetric $4\times4$ tensor, has $\tfrac12\cdot4\cdot5=\mathbf{10}$ . Then Weyl $=20-10=\mathbf{10}$ , matching Step 2. In four dimensions the free gravitational field and the matter-fixed field carry ten components each — an exact even split, the structural fact behind "half the curvature is undetermined by Einstein's equations".

Physics of the portrait

The conformal curvature tensor is named for the mathematician who first isolated it — not while studying black holes, which did not yet exist as physics, but while trying to build all of physics out of pure geometry.

Origins · 1 minHermann Weyl and how this was discovered

Hermann Weyl

1885–1955 · mathematics, relativity, and the birth of gauge theory

Hermann Weyl introduced the conformal curvature tensor in 1918, in the same burst of work — Reine Infinitesimalgeometrie and the first edition of his masterpiece Raum, Zeit, Materie — in which he proposed the first gauge theory. A student of Hilbert at Göttingen, Weyl was captivated by Einstein's new general relativity and tried to unify gravity with electromagnetism by allowing not just the direction but the length of vectors to change under parallel transport, a "gauge" (Eichung, or calibration) freedom. That idea required him to ask which parts of the curvature are unchanged when the metric is rescaled point by point, and the answer was the trace-free tensor $C^a{}_{bcd}$ that now bears his name. Einstein's famous objection — that path-dependent lengths would make atomic spectra position-dependent, contradicting observation — sank Weyl's original physical theory, but the mathematics survived and flourished: the gauge principle, transplanted from spacetime length to the phase of the quantum wavefunction, became the organizing idea of all modern particle physics, while the conformal tensor became the language of gravitational radiation, Petrov classification, and Penrose's cosmology. Weyl closed his career at the Institute for Advanced Study alongside Einstein.

History

1918

— Hermann Weyl isolates the conformal curvature tensor while constructing his gauge geometry

1954

— A. Z. Petrov classifies the algebraic types of the Weyl tensor

1957

— Pirani gives an invariant characterization of gravitational radiation through the Weyl tensor

1962

— Newman and Penrose introduce the null-tetrad Weyl scalars Ψ0 through Ψ4

1979

— Penrose proposes the Weyl curvature hypothesis linking C to the cosmological arrow of time

Going further

References. Misner, Thorne and Wheeler, Gravitation (1973), §13.5 and Box 14.2 · Wald, General Relativity (1984), §3.2 and §7.3 · Carroll, Spacetime and Geometry (2004), §3.8 and §5.4 · Hawking and Ellis, The Large Scale Structure of Space-Time (1973), §2.6 · Stephani et al., Exact Solutions of Einstein's Field Equations (2003), Ch. 3–4 · Penrose and Rindler, Spinors and Space-Time, Vol. 1 (1984), §4.8.

Provenance. The definition and its coefficients $-2/(n-2)$ , $2/[(n-1)(n-2)]$ are verified against Wald §3.2 and Carroll §3.8, and re-derived from tracelessness in Derivation 1. The component count $\tfrac{1}{12}n(n+1)(n+2)(n-3)$ giving 10 in 4D is standard (Carroll §3.8); the $(n-3)$ factor enforcing Weyl $=0$ for $n\le3$ is confirmed in Worked Example 2. Conformal invariance of $C^a{}_{bcd}$ and the $\Omega^2$ scaling of $C_{abcd}$ follow Wald Appendix D. The Schwarzschild Kretschmann scalar $48G^2M^2/r^6$ is the textbook value (Carroll §5.4); since Schwarzschild is vacuum, the Weyl scalar equals it. The horizon tidal numbers in Worked Example 1 ( $1.0\times10^{10}\,\text{s}^{-2}$ for a solar mass, $6.5\times10^{-4}\,\text{s}^{-2}$ for Sgr A*) were recomputed from $a=c^6/(4G^2M^2)$ . Newman–Penrose scalars and Petrov types follow Stephani et al. A verifier should recheck: (i) the coefficient fixing in Derivation 1; (ii) the $\Upsilon_a$ cancellation of Derivation 2; (iii) $48G^2M^2/r^6$ ; (iv) the dimension counts $0,10,35,84$ .

Prerequisites. Riemann Curvature Tensor · Ricci Decomposition of the Riemann Tensor · Algebraic Symmetries of the Riemann Tensor · Ricci Tensor · Ricci Scalar. Leads to. Petrov Classification · Gravitational Waves · Conformal Compactification and Penrose Diagrams · Second (Differential) Bianchi Identity.

Ask the tutor: Ask me to project the Weyl tensor onto a null tetrad and show exactly how $\Psi_4$ becomes the gravitational-wave strain that LIGO records.

What is the Weyl tensor, in one sentence, without the index gymnastics?

It is the part of spacetime curvature that Einstein's equations leave free. The Riemann tensor holds all curvature; the Einstein equations tie its trace — the Ricci tensor — to local matter, but that trace is only half the story in four dimensions. The Weyl tensor is the trace-free remainder: it answers to no matter at the point where you measure it, survives in empty space, and carries the degrees of freedom that make tidal forces and gravitational waves possible. Think of Ricci as "curvature caused by matter here" and Weyl as "curvature that just propagated in from elsewhere".

Why is it called the "conformal" tensor?

Because it is unchanged by conformal transformations — rescalings of the metric $g_{ab}\to\Omega^2(x)g_{ab}$ that stretch lengths and times by a different factor at every point while preserving all angles and the light-cone structure. Under such a rescaling the Riemann and Ricci tensors transform in complicated ways, but the mixed-index Weyl tensor $C^a{}_{bcd}$ comes out identical. So the Weyl tensor captures exactly the "shape" or causal-angular information of geometry and discards the local scale. A striking consequence: a metric is conformally flat — locally a rescaling of flat Minkowski space — if and only if its Weyl tensor vanishes.

If the Weyl tensor is source-free, how can matter ever affect it?

Locally it cannot: at any single point, no amount of matter fixes the Weyl components — that is the whole meaning of "trace-free". But the contracted second Bianchi identity, $\nabla_a C^a{}_{bcd}=(n-3)J_{bcd}$ , is a differential equation whose right-hand side is built from derivatives of the Ricci tensor, that is, from matter. This is precisely analogous to electromagnetism, where charges source the field through $\nabla_a F^{ab}=J^b$ and the field then propagates into charge-free regions. So matter sources Weyl nonlocally: a star's tidal field and a merger's gravitational waves reach across empty space because the Bianchi identity carries them there.

Does vanishing Weyl mean the spacetime is flat?

No — this is the single most common error. Vanishing Weyl means conformally flat: the metric is locally $\Omega^2\eta_{ab}$ , a point-by-point rescaling of Minkowski space. The Ricci part can still be enormous. The expanding FLRW universe is the textbook case: its Weyl tensor is exactly zero, yet it is strongly curved, with that curvature driven entirely by the energy density of matter and radiation through the Ricci tensor. Genuine flatness — the absence of all tidal effects and all curvature — requires the full Riemann tensor, both its Weyl and Ricci parts, to vanish.

What are the electric and magnetic parts of the Weyl tensor?

Just as an observer resolves the electromagnetic field tensor into an electric field $\mathbf E$ and a magnetic field $\mathbf B$ , an observer with 4-velocity $u^a$ resolves the Weyl tensor into two symmetric, trace-free spatial tensors: the electric part $E_{ac}=C_{abcd}u^b u^d$ and the magnetic part $B_{ac}=\tfrac12\epsilon_{ab}{}^{ef}C_{efcd}u^b u^d$ . The electric part is the tidal tensor — it is what appears in the geodesic-deviation equation and stretches an infalling body. The magnetic part encodes frame-dragging and the genuinely radiative, gravitomagnetic effects. A gravitational wave, viewed this way, is an oscillating pattern of $E$ and $B$ Weyl fields transverse to its propagation.

Why does the Weyl tensor vanish in three dimensions but not four?

It is pure counting. In $n$ dimensions the Riemann tensor has $\tfrac{1}{12}n^2(n^2-1)$ independent components and the Ricci tensor has $\tfrac12 n(n+1)$ . In three dimensions both equal six, so the Ricci tensor already determines the full Riemann tensor and there is no trace-free part left over — the Weyl count $\tfrac{1}{12}n(n+1)(n+2)(n-3)$ contains a factor $(n-3)$ that is zero. In four dimensions Riemann has 20 components and Ricci only 10, leaving 10 for Weyl. This is why gravity has no propagating waves in 2+1 dimensions but does in 3+1: no Weyl tensor, no free gravitational field.

How does the Weyl tensor connect to what LIGO actually measures?

Very directly. Numerical-relativity simulations of merging black holes do not read off the metric to get a waveform; they compute the Newman–Penrose scalar $\Psi_4$ , the outgoing-radiation projection of the Weyl tensor onto a null tetrad, on large spheres surrounding the source. $\Psi_4$ is essentially the second time derivative of the gravitational-wave strain, so integrating it twice yields the plus and cross polarizations $h_+$ and $h_\times$ a detector sees. Those predicted waveforms become the template bank LIGO and Virgo cross-correlate with their strain data — the confidence in GW150914 rested on Weyl-scalar templates matching the observed signal.