Why $R_{\mu\nu}=0$ does not imply $R_{\mu\nu\rho\sigma}=0$

Almost everyone makes this mistake once, and it’s a good one to make. You learn the Einstein field equations. You learn that in vacuum, with no matter and no energy and no cosmological constant, they collapse down to something very simple: $$R_{\mu\nu} = 0.$$ So you think: $R_{\mu\nu}$ is the curvature, it’s zero, so there’s no curvature out there, spacetime is flat, nothing is happening. Then somebody points out that the entire outside of a black hole is vacuum. Every single point out there satisfies $R_{\mu\nu}=0$ exactly. And things fall into it.

So the reasoning broke somewhere, and I want to find out where, because once you see which step failed you’ll understand what the Ricci tensor actually measures, what the rest of the curvature is doing, and why gravity has waves at all. Short version: $R_{\mu\nu}$ isn’t the curvature. It’s a contraction of the curvature, and contracting throws information away.

Before any tensors show up, though, I want to convince you that you already believe this and have believed it since your first mechanics course. Newtonian gravity obeys Poisson’s equation, $$\nabla^2 \Phi = 4\pi G \rho.$$ Where there’s no matter, $\rho = 0$, and you get Laplace’s equation, $\nabla^2\Phi = 0$. Now, does anybody think Laplace’s equation means the gravitational field is zero? No. The potential outside a spherical mass is $\Phi = -GM/r$, you can check for yourself that $\nabla^2(-GM/r) = 0$ everywhere except the origin, and $\Phi$ is obviously not zero, its gradient isn’t zero, and things fall. The air above your head is vacuum. Laplace’s equation holds up there. You’re still stuck to the floor.

That last bit is the whole answer already, so let me slow down on it. The second derivatives of the potential form a symmetric matrix, the Hessian $\partial_i \partial_j \Phi$, and in three dimensions that has six independent entries. Laplace’s equation sets exactly one number to zero: the trace, $\sum_i \partial_i\partial_i \Phi$. Five entries are left completely alone. Those five are the tidal field, the thing that stretches you along the line toward the Earth and squeezes you sideways, and they’re what gravity does in empty space.

General relativity does the same thing with better bookkeeping. The full curvature of spacetime is the Riemann tensor $R^\rho{}{\sigma \mu \nu}$, and the Ricci tensor isn’t some separate object, it’s what you get by contracting Riemann on a pair of indices: $$R{\mu\nu} = R^\alpha{}{\mu\alpha\nu}.$$ You’re summing over one direction. That’s an averaging operation, and averaging loses information. If I tell you a list of numbers averages to zero, you don’t conclude the numbers were all zero, because ${+1, -1}$ averages to zero too. So $R{\mu\nu} = 0$ says that certain averages of the curvature vanish. It doesn’t say the curvature vanishes.

You can make that exact by counting. Riemann has a lot of symmetry: antisymmetric in the first two indices, antisymmetric in the last two, and symmetric when you swap the two pairs. In four dimensions an antisymmetric pair takes $\binom{4}{2}=6$ values, so the pair symmetry makes Riemann a symmetric $6\times 6$ matrix with $\tfrac{6\cdot 7}{2}=21$ entries, and the first Bianchi identity $R_{\mu[\nu\rho\sigma]}=0$ knocks off one more. So the Riemann tensor of a four dimensional spacetime has $$20 \ \text{independent components.}$$ The Ricci tensor is symmetric, so it has $\tfrac{4\cdot 5}{2}=10$. Setting $R_{\mu\nu}=0$ puts ten conditions on twenty numbers.¹Worth doing this count yourself once. The first Bianchi identity $R_{\mu[\nu\rho\sigma]}=0$ looks like it should impose a lot of conditions, but nearly all of them already follow from the pair symmetries. The only new content is the totally antisymmetric part $R_{[\mu\nu\rho\sigma]}=0$, and in four dimensions there’s exactly $\binom{4}{4}=1$ way to pick four antisymmetrised indices, so exactly one new constraint. In general the answer is $\frac{n^2(n^2-1)}{12}$, which gives 1 for $n=2$, 6 for $n=3$, 20 for $n=4$, and 50 for $n=5$. Ten are left over. Whatever those ten are, they’re the curvature that empty space is allowed to have.

The ten leftovers have a name. There’s exactly one way to split the Riemann tensor into a piece built out of Ricci and a remainder that’s completely traceless, and the remainder is the Weyl tensor $C_{\mu\nu\rho\sigma}$. In $n$ dimensions the split reads $$R_{\mu\nu\rho\sigma} = C_{\mu\nu\rho\sigma} + \frac{2}{n-2}\Bigl(g_{\mu[\rho}R_{\sigma]\nu} – g_{\nu[\rho}R_{\sigma]\mu}\Bigr) – \frac{2R}{(n-1)(n-2)}\, g_{\mu[\rho}g_{\sigma]\nu},$$ where the square brackets mean antisymmetrise. Every trace of $C$ vanishes by construction, $C^\alpha{}{\mu\alpha\nu}=0$, and that’s exactly why the field equations can’t see it. So in vacuum, where $R{\mu\nu}=0$ and therefore $R=0$ as well, the whole thing collapses to $$R_{\mu\nu\rho\sigma} = C_{\mu\nu\rho\sigma}.$$ The curvature of empty space is pure Weyl. Not mostly Weyl, not approximately Weyl. Exactly and only Weyl.²The Weyl tensor has $\frac{n(n+1)(n+2)(n-3)}{12}$ independent components in $n$ dimensions. It’s also conformally invariant: rescale the metric by any positive function, $g_{\mu\nu} \to \Omega^2 g_{\mu\nu}$, and the Weyl tensor with one index raised doesn’t change. That’s why it sometimes gets called the conformal curvature tensor, and why a spacetime is conformally flat exactly when its Weyl tensor vanishes.
Now stare at that component formula for a second, because there’s something odd hiding in it. The factor $(n-3)$ means the Weyl tensor has zero components whenever $n \le 3$. Not zero for some particular metric, zero identically, always, for every three dimensional geometry there is. Which means the Riemann tensor in three dimensions is completely fixed by the Ricci tensor. Which means that in three dimensions, $R_{\mu\nu}=0$ really does force spacetime to be flat.

This still surprises me. The wrong intuition isn’t wrong because anyone reasoned badly. It’s wrong because it’s reasoning about a universe with one dimension too few. In three dimensional gravity there’s no curvature outside a source, no light bending around a star, no structure in a black hole exterior, and no gravitational radiation, because there’s simply nowhere for that information to sit. The theory is topological. Our fourth dimension is what opens up ten free slots in the curvature and turns gravity into a field with a life of its own.³Three dimensional gravity isn’t boring, and the qualification matters: with a negative cosmological constant you get the BTZ black hole, with a horizon and a temperature and an entropy. But that isn’t a counterexample. BTZ is locally identical to anti de Sitter space everywhere, with constant curvature set entirely by $\Lambda$; what makes it a black hole is a global identification of points, which is topology rather than local geometry. There are still no local gravitational degrees of freedom in three dimensions and no gravitational waves.

So Weyl is what survives. The question that matters is what it does, and here the cleanest statement in the whole subject is waiting. Take a small ball of test particles, freely falling, starting at rest relative to each other. Coffee grounds released in a spacecraft will do fine. Watch the ball. Its volume is controlled by the expansion $\theta = \nabla_\mu u^\mu$ of the congruence of worldlines, which obeys the Raychaudhuri equation $$\frac{d\theta}{d\tau} = -\frac{1}{3}\theta^2 – \sigma_{\mu\nu}\sigma^{\mu\nu} + \omega_{\mu\nu}\omega^{\mu\nu} – R_{\mu\nu}u^\mu u^\nu,$$ with $\sigma_{\mu\nu}$ the shear and $\omega_{\mu\nu}$ the vorticity. Start the ball at rest with no shear and no rotation, so at the first instant $\theta$, $\sigma$ and $\omega$ are all zero, and every term on the right dies except the last one. You’re left with $$\left.\frac{\ddot V}{V}\right|{\tau=0} = -R{\mu\nu}u^\mu u^\nu.$$ The Ricci tensor is the thing, and the only thing, that changes the volume of a ball of freely falling matter.⁴Getting from $\theta$ to volume: $\theta = \dot V/V$, so $\dot\theta = \ddot V/V – (\dot V/V)^2$, and at the initial instant where $\dot V=0$ that’s just $\ddot V / V$. Feed in Einstein’s equation and the geometric statement becomes a statement about matter: for a perfect fluid you get $\ddot V/V = -4\pi G(\rho + 3p/c^2)$, which is a very direct way to see why pressure gravitates in general relativity, and why no equation of state stiff enough can save a star from collapsing.

Now put the ball in vacuum. There $R_{\mu\nu}u^\mu u^\nu = 0$ identically, so the volume doesn’t change at all. But something obviously does happen to coffee grounds falling toward the Earth, because the ones nearer the Earth fall faster and the ones off to the sides fall along slightly converging lines. The ball deforms. It just deforms at fixed volume. In vacuum the geodesic deviation equation reads $$\frac{D^2 \xi^\mu}{d\tau^2} = -R^\mu{}{\alpha\nu\beta},u^\alpha \xi^\nu u^\beta = -E^\mu{}\nu, \xi^\nu, \qquad E_{\mu\nu} \equiv C_{\mu\alpha\nu\beta}u^\alpha u^\beta,$$ where $E_{\mu\nu}$ is called the electric part of the Weyl tensor. It’s symmetric, and since every trace of Weyl vanishes it’s traceless too: $E^\mu{}\mu = 0$. A traceless matrix of relative accelerations is exactly the statement that volume is preserved while shape isn’t. That’s the division of labour, and it’s a theorem, not a slogan. Ricci changes volume. Weyl changes shape.⁵Weyl also has a magnetic part, $B{\mu\nu} = \tfrac{1}{2}\epsilon_{\mu\alpha\beta\gamma}C^{\alpha\beta}{}_{\nu\delta}u^\gamma u^\delta$, and the two together account for all ten components, five each, both symmetric and traceless. The electric and magnetic language isn’t a loose analogy either: $E$ and $B$ obey equations closely parallel to Maxwell’s, and it’s the two of them oscillating into each other that makes a gravitational wave. A purely electric Weyl field is just the static tidal field of a mass. You need the magnetic part to get radiation.

You can check the whole story with nothing but calculus, and I’d recommend actually doing it, because it makes the abstraction concrete. Take $\Phi = -GM/r$ and compute the Hessian: $$\partial_i \partial_j \Phi = GM\left(\frac{\delta_{ij}}{r^3} – \frac{3 x_i x_j}{r^5}\right).$$ The trace is $GM(3/r^3 – 3r^2/r^5) = 0$, which is Laplace’s equation turning up as promised. The eigenvalues are easy to read off: along the radial direction you get $-2GM/r^3$, and in each of the two transverse directions you get $+GM/r^3$. Since the relative acceleration of nearby particles is $\ddot\xi^i = -\partial_i\partial_j\Phi\,\xi^j$, the ball gets pulled apart radially at a rate $2GM/r^3$ and squeezed inward in each transverse direction at $GM/r^3$. One stretch, two squeezes, and because $2 = 1+1$ the volume stays put. That’s the traceless condition, sitting there in numbers you can work out in a minute.

Some numbers help here. Near the Earth’s surface the tidal rate is $2GM_\oplus/R_\oplus^3 \approx 3.1\times 10^{-6}\ \mathrm{s^{-2}}$, so across the two metres of a standing person the difference in acceleration between head and feet is about $6\times 10^{-6}\ \mathrm{m\,s^{-2}}$, which is why you’ve never noticed it. Go to a neutron star of $1.4$ solar masses and stand $100$ kilometres away and the same expression gives $3.7\times 10^{5}\ \mathrm{s^{-2}}$, so the head to foot difference is around seventy five thousand times Earth gravity and you get pulled into a thread. Both of those places are perfect vacuum. Both have $R_{\mu\nu}=0$ exactly. The difference between an effect you can’t feel and one that kills you is entirely a difference in the Weyl tensor, which the field equations never mention.

If you want a proof rather than an argument that Schwarzschild isn’t flat, compute a curvature invariant, because scalars can’t be changed by any coordinate transformation. The Kretschmann scalar is $$K = R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma} = \frac{48G^2M^2}{c^4 r^6}.$$ That’s nonzero at every finite radius. If spacetime were flat then $R_{\mu\nu\rho\sigma}=0$ and $K$ would have to vanish, and it doesn’t. There’s no clever choice of coordinates that turns a nonzero scalar into a zero one. So the outside of a star is curved, curved despite being empty, and since Riemann and Weyl coincide there we can also write $K = C_{\mu\nu\rho\sigma}C^{\mu\nu\rho\sigma}$.⁶The same scalar settles the other classic Schwarzschild confusion. At the horizon $r = r_S$ the metric components go haywire, but $K = 48G^2M^2/(c^4 r_S^6)$ is perfectly finite there, which tells you the horizon is a problem with the coordinates and not with the geometry. As $r \to 0$ the scalar blows up, and that singularity is real and no choice of coordinates removes it.

The most extreme version of all this is gravitational radiation. A gravitational wave crossing interstellar space solves $R_{\mu\nu}=0$ at every point along its path. No matter anywhere in it, nothing on the right hand side of the field equations, nothing at all in the naive sense. It’s pure Weyl curvature moving at the speed of light, made entirely out of the components the vacuum equations decline to constrain. And when one arrives it does exactly what we derived above: changes the shape of things while leaving their volume alone, stretching along one axis and squeezing along the perpendicular one. That’s precisely what LIGO measures, a difference in the lengths of two perpendicular arms, and in September 2015 that difference was about $4\times 10^{-18}$ metres over a four kilometre baseline, roughly a thousandth of the width of a proton. Empty space, curved, and curved strongly enough to detect.

So the corrected statement is this. $R_{\mu\nu}=0$ doesn’t say spacetime is flat. It says spacetime is Ricci flat, which is a real and restrictive condition, but one that leaves half the curvature untouched in four dimensions. The half it kills is the half that responds directly to local matter and changes the volumes of falling objects. The half it leaves is the tidal, shape distorting, freely propagating half, and that half isn’t sourced locally at all. It’s set by matter somewhere else and carried through the vacuum by the Bianchi identities, in exactly the way that $\nabla^2\Phi=0$ in the air above your head is perfectly compatible with the Earth underneath pulling on you.⁷The mechanism by which distant matter fixes Weyl in a vacuum region is the second Bianchi identity, $\nabla_{[\lambda}R_{\mu\nu]\rho\sigma}=0$. Contract it in vacuum and you get $\nabla^\mu C_{\mu\nu\rho\sigma}=0$, a wave equation for the Weyl tensor where the matter enters only through boundary conditions. That’s the precise sense in which Weyl curvature is the free gravitational field, and the parallel with the source free Maxwell equations $\nabla^\mu F_{\mu\nu}=0$ is very close.

One last thing, for anyone drifting toward pure mathematics. Ricci flat manifolds aren’t something physicists invented to describe black holes, they’re a central object in Riemannian geometry, and $R_{\mu\nu}=0$ is a hard nonlinear equation with a big space of solutions. Calabi conjectured and Yau proved that every compact Kähler manifold with vanishing first Chern class carries a Ricci flat metric, and the resulting Calabi-Yau manifolds are highly curved things that happen to have no Ricci curvature anywhere. The K3 surface is a four dimensional example: compact, Ricci flat, and very much not flat. When a geometer says Ricci flat they never mean flat, and it wouldn’t occur to them that anyone might. It only occurs to us because we met the Ricci tensor for the first time in an equation with the word vacuum sitting next to it.

There’s a general lesson in here somewhere, and it’s the thing I’d want you to keep. When a theory hands you an equation of the form “some contraction of the interesting object is zero,” the first question isn’t what the equation forbids rather it’s what it leaves alone. Count the components. Find the part the equation never touches. In general relativity that part is the Weyl tensor, it has ten components, it exists only because we live in four dimensions instead of three, and it’s where black holes and tides and gravitational waves all live.

References and Footnotes

• 1
  Worth doing this count yourself once. The first Bianchi identity $R_{\mu[\nu\rho\sigma]}=0$ looks like it should impose a lot of conditions, but nearly all of them already follow from the pair symmetries. The only new content is the totally antisymmetric part $R_{[\mu\nu\rho\sigma]}=0$, and in four dimensions there’s exactly $\binom{4}{4}=1$ way to pick four antisymmetrised indices, so exactly one new constraint. In general the answer is $\frac{n^2(n^2-1)}{12}$, which gives 1 for $n=2$, 6 for $n=3$, 20 for $n=4$, and 50 for $n=5$. ↩︎
• 2
  The Weyl tensor has $\frac{n(n+1)(n+2)(n-3)}{12}$ independent components in $n$ dimensions. It’s also conformally invariant: rescale the metric by any positive function, $g_{\mu\nu} \to \Omega^2 g_{\mu\nu}$, and the Weyl tensor with one index raised doesn’t change. That’s why it sometimes gets called the conformal curvature tensor, and why a spacetime is conformally flat exactly when its Weyl tensor vanishes. ↩︎
• 3
  Three dimensional gravity isn’t boring, and the qualification matters: with a negative cosmological constant you get the BTZ black hole, with a horizon and a temperature and an entropy. But that isn’t a counterexample. BTZ is locally identical to anti de Sitter space everywhere, with constant curvature set entirely by $\Lambda$; what makes it a black hole is a global identification of points, which is topology rather than local geometry. There are still no local gravitational degrees of freedom in three dimensions and no gravitational waves. ↩︎
• 4
  Getting from $\theta$ to volume: $\theta = \dot V/V$, so $\dot\theta = \ddot V/V – (\dot V/V)^2$, and at the initial instant where $\dot V=0$ that’s just $\ddot V / V$. Feed in Einstein’s equation and the geometric statement becomes a statement about matter: for a perfect fluid you get $\ddot V/V = -4\pi G(\rho + 3p/c^2)$, which is a very direct way to see why pressure gravitates in general relativity, and why no equation of state stiff enough can save a star from collapsing. ↩︎
• 5
  Weyl also has a magnetic part, $B{\mu\nu} = \tfrac{1}{2}\epsilon_{\mu\alpha\beta\gamma}C^{\alpha\beta}{}_{\nu\delta}u^\gamma u^\delta$, and the two together account for all ten components, five each, both symmetric and traceless. The electric and magnetic language isn’t a loose analogy either: $E$ and $B$ obey equations closely parallel to Maxwell’s, and it’s the two of them oscillating into each other that makes a gravitational wave. A purely electric Weyl field is just the static tidal field of a mass. You need the magnetic part to get radiation. ↩︎
• 6
  The same scalar settles the other classic Schwarzschild confusion. At the horizon $r = r_S$ the metric components go haywire, but $K = 48G^2M^2/(c^4 r_S^6)$ is perfectly finite there, which tells you the horizon is a problem with the coordinates and not with the geometry. As $r \to 0$ the scalar blows up, and that singularity is real and no choice of coordinates removes it. ↩︎
• 7
  The mechanism by which distant matter fixes Weyl in a vacuum region is the second Bianchi identity, $\nabla_{[\lambda}R_{\mu\nu]\rho\sigma}=0$. Contract it in vacuum and you get $\nabla^\mu C_{\mu\nu\rho\sigma}=0$, a wave equation for the Weyl tensor where the matter enters only through boundary conditions. That’s the precise sense in which Weyl curvature is the free gravitational field, and the parallel with the source free Maxwell equations $\nabla^\mu F_{\mu\nu}=0$ is very close. ↩︎