The Einstein Tensor

The Einstein tensor is the object that turns "spacetime tells matter how to move, matter tells spacetime how to curve" into an equation you can solve. Take the metric $g_{\mu\nu}$ to be the gravitational field, and ask a sharp question: what symmetric two-index tensor, built only from the curvature of that metric, is fit to stand opposite the matter source $T_{\mu\nu}$ ? The obvious candidate is the Ricci tensor $R_{\mu\nu}$ , which measures how gravity focuses a cloud of freely falling dust. But $\nabla^{\mu}R_{\mu\nu}\neq 0$ in general, so setting $R_{\mu\nu}\propto T_{\mu\nu}$ would demand that energy-momentum leak in and out of existence — a fatal contradiction with the local conservation any matter theory obeys. The repair is a single subtraction. Remove a multiple of $R\,g_{\mu\nu}$ and let the contracted Bianchi identity fix the coefficient: the combination $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 R g_{\mu\nu}$ is divergence-free for every metric, as an identity of Riemannian geometry. That is what makes $G_{\mu\nu}=\kappa T_{\mu\nu}$ consistent — the geometry on the left is conserved automatically, so it hands that conservation, $\nabla^{\mu}T_{\mu\nu}=0$ , to the matter on the right as a free corollary rather than an extra law. Three independent roads lead to the same tensor. Geometrically it is the unique divergence-free curvature combination. Variationally it drops out of extremizing the Einstein–Hilbert action, so the Bianchi identity is really diffeomorphism invariance wearing Noether's clothing. Axiomatically, Lovelock's theorem proves that in four dimensions $G_{\mu\nu}+\Lambda g_{\mu\nu}$ is the only symmetric, divergence-free, second-order tensor available. Master it on its own and the field equations stop looking like a guess and start looking forced.

Key takeaways

The Einstein tensor $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 R g_{\mu\nu}$ is the unique symmetric combination of Ricci tensor, Ricci scalar, and metric that is linear in the second derivatives of $g_{\mu\nu}$ and has vanishing covariant divergence.
Its defining property is the contracted Bianchi identity $\nabla^{\mu}G_{\mu\nu}=0$ , a geometric consequence of the second Bianchi identity for the Riemann tensor — true of every metric whatsoever.
Setting $G_{\mu\nu}=\kappa T_{\mu\nu}$ therefore forces $\nabla^{\mu}T_{\mu\nu}=0$ : local energy-momentum conservation is built into the geometry, not imposed by hand.
The trace is $G=g^{\mu\nu}G_{\mu\nu}=(1-D/2)R$ , equal to $-R$ in four dimensions, which lets the field equations be rewritten in trace-reversed form $R_{\mu\nu}=\kappa(T_{\mu\nu}-\tfrac12 T g_{\mu\nu})$ .
The same tensor arises by varying the Einstein–Hilbert action, and Lovelock's theorem makes it canonical: $G_{\mu\nu}+\Lambda g_{\mu\nu}$ is the most general admissible left-hand side in 4D, the cosmological constant being the only freedom.
Key idea: the factor $-\tfrac12$ in $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 R g_{\mu\nu}$ is not a fitting parameter — it is uniquely fixed by the contracted Bianchi identity so that $\nabla^{\mu}G_{\mu\nu}\equiv 0$ , making conservation of the source an automatic geometric identity rather than an extra physical law.

The core insight

The factor $-\tfrac12$ in $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 R g_{\mu\nu}$ is not a fitting parameter: it is uniquely fixed by the contracted Bianchi identity so that $\nabla^{\mu}G_{\mu\nu}\equiv 0$ . Conservation of the source is therefore an automatic geometric identity, not an extra physical law.

Where this lives today

The Einstein tensor is the working left-hand side of every quantitative test of gravity. In numerical relativity, the split of $G_{\mu\nu}=\kappa T_{\mu\nu}$ into constraints (the $G_{0\nu}$ components, fixed on each time slice) and evolution equations — a split guaranteed consistent by the contracted Bianchi identity — is the 3+1 (ADM, BSSN) machinery that let LIGO's modellers compute the waveform of two merging black holes and match GW150914 in 2015. In cosmology, $G_{\mu\nu}+\Lambda g_{\mu\nu}=\kappa T_{\mu\nu}$ applied to the FLRW metric gives the Friedmann equations, and the resurrected $\Lambda$ — measured as dark energy since 1998 — is precisely the one free term Lovelock's theorem permits, its tiny value ( $\sim10^{-52}\ \mathrm{m^{-2}}$ ) a sharp open problem. Every attempt to modify gravity — $f(R)$ , Gauss–Bonnet, massive gravity — is organized around what the Einstein tensor forbids: to change the left-hand side you must break one of Lovelock's hypotheses.

The physical picture

Start with what the Ricci tensor does. Release a small ball of dust — coffee grounds, in Wheeler's image — at momentary rest and let each grain fall freely. The ball's volume does not sit still: Ricci curvature focuses it, and the fractional acceleration of the volume is exactly $-R_{\hat t\hat t}$ , the time-time Ricci component in the ball's own frame. The field equations then tie that geometric focusing to the physical source, so a real density produces a real, computable contraction rate. The first figure makes "Ricci = focusing" a picture with a number attached.

Figure

A ball of freely falling dust, momentarily at rest, is focused by Ricci curvature. Its fractional volume acceleration equals minus R_tt, which the field equations tie to the source density: for ordinary matter at ρ = 1000 kg per cubic metre the ball contracts at −8.4×10⁻⁷ per second squared.

Now the deeper point, and the reason the Einstein tensor exists at all. Raw Ricci is not conserved: its divergence picks up the change of scalar curvature, $\nabla^{\mu}R_{\mu\nu}=\tfrac12\nabla_{\nu}R$ . That $\tfrac12\nabla_{\nu}R$ is a leak — a source term that would let energy appear from nowhere if $R_{\mu\nu}$ alone sat opposite $T_{\mu\nu}$ . Subtracting $-\tfrac12 R g_{\mu\nu}$ contributes a divergence of exactly $-\tfrac12\nabla_{\nu}R$ , and the two cancel term for term. The Einstein tensor is the plugged version of Ricci; its divergence is forced to zero. The second figure shows that cancellation and its consequence for how many equations the theory contains.

Figure

The factor minus one-half makes the divergence cancel: the plus one-half grad-R that leaks out of the Ricci tensor is exactly undone by the minus one-half grad-R from the trace term, so grad-mu G-mu-nu equals zero identically. That identity removes 4 of the 10 symmetric components, leaving 6 genuine evolution equations for the 6 physical degrees of freedom of the metric.

Counting drives the last point home. The symmetric $G_{\mu\nu}$ has ten independent components in four dimensions, but the four identities $\nabla^{\mu}G_{\mu\nu}=0$ mean four of the field equations are constraints that propagate, not independent evolution equations. Ten minus four leaves six, exactly matching the six physical degrees of freedom of the metric once the four-fold coordinate freedom is fixed — the Bianchi identity is the bookkeeping that stops the equations over-determining the metric. Lovelock's theorem then supplies the inevitability: demand a symmetric, divergence-free tensor built from $g_{\mu\nu}$ and at most its second derivatives, linear in those, in four dimensions, and you are forced to $G_{\mu\nu}+\Lambda g_{\mu\nu}$ .

The core equations

The Einstein tensor is defined by a single algebraic combination of quantities the earlier curvature articles own. It is symmetric because $R_{\mu\nu}$ and $g_{\mu\nu}$ are, and it is built to be linear in the second derivatives of the metric so the resulting field equations are second-order.

G_{\mu\nu} \equiv R_{\mu\nu} - \tfrac{1}{2} R\, g_{\mu\nu} \qquad (1)

where: $G_{\mu\nu}$ — Einstein tensor, symmetric rank-2; $R_{\mu\nu}$ — Ricci tensor, $R_{\mu\nu}=R^{\alpha}{}_{\mu\alpha\nu}$ ; $R$ — Ricci scalar, $R=g^{\mu\nu}R_{\mu\nu}$ ; $g_{\mu\nu}$ — metric tensor. Read it as "Ricci, with just enough of its own trace subtracted to make it conserved." The defining property — the whole reason this particular combination is singled out from all others — is that its covariant divergence vanishes identically, for every metric, with no equation of motion imposed.

\nabla^{\mu} G_{\mu\nu} = 0 \qquad (2)

where: $\nabla_{\mu}$ — Levi-Civita covariant derivative (metric-compatible, torsion-free). This is a kinematic identity of Riemannian geometry, not a dynamical statement; it holds off-shell, before any field equation is solved. Contracting the definition with the inverse metric gives the trace, which depends on the spacetime dimension in a way that quietly excludes the two-dimensional case. It is the relation that makes the trace-reversed form possible.

G \equiv g^{\mu\nu}G_{\mu\nu} = R - \tfrac{1}{2} R\,D = \Big(1-\tfrac{D}{2}\Big) R \qquad (3)

where: $D$ — spacetime dimension, with $g^{\mu\nu}g_{\mu\nu}=\delta^{\mu}{}_{\mu}=D$ . In $D=4$ the trace is $G=-R$ ; in $D=2$ it vanishes, a signal that $G_{\mu\nu}\equiv 0$ there and gravity has no dynamics. Feeding $G=-R$ back through $G_{\mu\nu}=\kappa T_{\mu\nu}$ lets you solve algebraically for the Ricci tensor. The result is the form most convenient in vacuum and in the Newtonian-limit matching, because its source side collapses for simple matter.

R_{\mu\nu} = \kappa\Big(T_{\mu\nu} - \tfrac{1}{2} T\, g_{\mu\nu}\Big) \qquad (4)

where: $T_{\mu\nu}$ — stress-energy tensor; $T=g^{\mu\nu}T_{\mu\nu}$ — its trace; $\kappa=8\pi G/c^4$ — the coupling. In vacuum $T_{\mu\nu}=0$ gives $R_{\mu\nu}=0$ directly, the equation obeyed by Schwarzschild and Kerr. The same tensor is not just handed down geometrically — it is what a variational principle produces. Extremizing the Einstein–Hilbert action with respect to the inverse metric yields $G_{\mu\nu}$ as the functional derivative, tying geometry to an action and the Bianchi identity to diffeomorphism invariance.

\frac{1}{\sqrt{-g}}\frac{\delta S_{\mathrm{EH}}}{\delta g^{\mu\nu}} = \frac{1}{2\kappa}G_{\mu\nu},\qquad S_{\mathrm{EH}}=\frac{1}{2\kappa}\int \mathrm{d}^4x\,\sqrt{-g}\,R \qquad (5)

where: $S_{\mathrm{EH}}$ — Einstein–Hilbert action; $g$ — determinant of $g_{\mu\nu}$ , negative for Lorentzian signature. Adding a matter action supplies $T_{\mu\nu}$ and reproduces $G_{\mu\nu}=\kappa T_{\mu\nu}$ . Lovelock's theorem answers "could it have been anything else?" with a flat no. In four dimensions the only symmetric, divergence-free, second-order tensor available is the Einstein tensor plus a term proportional to the metric itself.

G_{\mu\nu}+\Lambda g_{\mu\nu} \quad \text{— the most general admissible left-hand side} \qquad (6)

where: $\Lambda$ — cosmological constant, and $\nabla_{\mu}(\Lambda g_{\mu\nu})=0$ because $\nabla_{\mu}g_{\alpha\beta}=0$ . The metric term is the single free addition; it is why the cosmological constant is natural, not ad hoc. For a concrete handle, the time-time component of the mixed Einstein tensor for a static, spherically symmetric metric takes a compact form — the expression that, set to zero, delivers the Schwarzschild solution in Worked example 2.

G^{0}{}_{0} = -\frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\big[r\big(1-e^{-2\beta}\big)\big] \qquad (7)

where: $\beta(r)$ — metric function in $g_{rr}=e^{2\beta(r)}$ for the static spherically symmetric line element. Setting it to zero is a first-order ODE for $\beta$ , integrable in one line.

Derivation AThe minus one-half is uniquely fixed by conservation

The goal is to prove that the coefficient in front of $R g_{\mu\nu}$ is not a choice but a consequence: demanding that the geometric left-hand side be divergence-free for every metric forces it to be exactly $-\tfrac12$ , so the Einstein tensor is the only curvature combination compatible with local conservation of the source. We start from the most general candidate consistent with the symmetries, take its covariant divergence, feed in the once-contracted Bianchi identity, and read off the coefficient — the argument is short because the Bianchi identity does all the work.

Step 1. Write the most general symmetric tensor built from Ricci, the scalar, and the metric that is linear in second derivatives: $S_{\mu\nu}=R_{\mu\nu}+a\,R g_{\mu\nu}+b\,g_{\mu\nu}$ for constants $a,b$ . Take its divergence, $\nabla^{\mu}S_{\mu\nu}=\nabla^{\mu}R_{\mu\nu}+a\,\nabla_{\nu}R+b\,\nabla^{\mu}g_{\mu\nu}$ . Why: this is the widest net compatible with "second-order, symmetric, algebraic in curvature," so whatever survives is genuinely forced, not assumed.

Step 2. The last term dies immediately: $\nabla^{\mu}g_{\mu\nu}=0$ by metric compatibility of the Levi-Civita connection, whatever the value of $b$ . So the cosmological-constant piece never affects conservation. Why: this is why $\Lambda g_{\mu\nu}$ is always allowed — it is divergence-free on its own and cannot spoil the identity.

Step 3. Substitute the once-contracted Bianchi identity $\nabla^{\mu}R_{\mu\nu}=\tfrac12\nabla_{\nu}R$ , the double contraction of the second Bianchi identity. The divergence collapses to $\nabla^{\mu}S_{\mu\nu}=\big(\tfrac12+a\big)\nabla_{\nu}R$ . Why: everything now rides on a single generically nonzero vector, $\nabla_{\nu}R$ , so a single scalar equation decides the whole question.

Step 4. Since $\nabla_{\nu}R$ is nonzero for generic metrics (it vanishes only for special, e.g. constant-scalar-curvature, spacetimes), demanding $\nabla^{\mu}S_{\mu\nu}=0$ for all metrics requires $\tfrac12+a=0$ , hence $a=-\tfrac12$ . Then $S_{\mu\nu}=R_{\mu\nu}-\tfrac12 R g_{\mu\nu}+b\,g_{\mu\nu}=G_{\mu\nu}+\Lambda g_{\mu\nu}$ with $b=\Lambda$ . Why: this recovers exactly the Lovelock content by hand — the Einstein tensor plus a cosmological term, and nothing else.

Go deeper · Derivation B · 1 minThe Einstein tensor from the Einstein–Hilbert action

Derivation BThe Einstein tensor from the Einstein–Hilbert action

The goal is to show that the same tensor falls out unavoidably from a variational principle: extremizing $\int\sqrt{-g}\,R$ over the inverse metric produces $R_{\mu\nu}-\tfrac12 R g_{\mu\nu}$ in the bulk, with the awkward curvature-variation piece organizing itself into a pure boundary term. This is the deep reason the Bianchi identity holds — it is the Noether identity of the action's invariance under coordinate changes. We vary each factor of $\sqrt{-g}\,R$ and collect.

Step 1. Expand by the product rule, writing $R=g^{\mu\nu}R_{\mu\nu}$ : $\delta(\sqrt{-g}\,R)=\delta\sqrt{-g}\,R+\sqrt{-g}\,R_{\mu\nu}\,\delta g^{\mu\nu}+\sqrt{-g}\,g^{\mu\nu}\,\delta R_{\mu\nu}$ . Three pieces, one for each factor. Why: separating the variation isolates the metric-determinant term (which supplies the $-\tfrac12 R g_{\mu\nu}$ ) from the Ricci-variation term (which must be shown harmless).

Step 2. The determinant varies as $\delta\sqrt{-g}=-\tfrac12\sqrt{-g}\,g_{\mu\nu}\,\delta g^{\mu\nu}$ , from Jacobi's formula $\delta g=g\,g^{\mu\nu}\delta g_{\mu\nu}=-g\,g_{\mu\nu}\delta g^{\mu\nu}$ . The first term is therefore $-\tfrac12\sqrt{-g}\,R g_{\mu\nu}\,\delta g^{\mu\nu}$ . Why: this is the geometric origin of the trace term — the very same $-\tfrac12$ , arising here from how the volume element responds to a metric change.

Step 3. The Ricci variation is a total covariant divergence (the Palatini identity), $g^{\mu\nu}\delta R_{\mu\nu}=\nabla_{\sigma}\big(g^{\mu\nu}\delta\Gamma^{\sigma}_{\mu\nu}-g^{\sigma\nu}\delta\Gamma^{\mu}_{\mu\nu}\big)$ , which integrates to a surface term and drops out of the bulk equations. Why: it contributes only at the boundary, precisely the piece cancelled by the Gibbons–Hawking–York term, so it cannot affect the field equations in the interior.

Step 4. Assemble the surviving bulk: $\delta S_{\mathrm{EH}}=\tfrac{1}{2\kappa}\int\sqrt{-g}\,\big(R_{\mu\nu}-\tfrac12 R g_{\mu\nu}\big)\delta g^{\mu\nu}\,\mathrm{d}^4x$ , and the bracket is $G_{\mu\nu}$ . Adding a matter action with $T_{\mu\nu}=-\tfrac{2}{\sqrt{-g}}\,\delta S_m/\delta g^{\mu\nu}$ and demanding $\delta S=0$ for arbitrary $\delta g^{\mu\nu}$ gives $G_{\mu\nu}=\kappa T_{\mu\nu}$ . Why: stationarity of the total action is the field equation, and the geometric side is automatically $G_{\mu\nu}$ — its divergence-freedom now reads as diffeomorphism invariance of $S_{\mathrm{EH}}$ .

Go deeper · Derivation C · 1 minThe Newtonian limit fixes the coupling constant

Derivation CThe Newtonian limit fixes the coupling constant

The goal is to pin the number $\kappa$ by insisting that general relativity reproduce Newton where Newton works: in a weak, static, slowly varying field sourced by ordinary matter, the field equations must collapse to Poisson's equation $\nabla^2\Phi=4\pi G\rho$ , and matching determines $\kappa=8\pi G/c^4$ with no remaining freedom. We linearize the metric, simplify the source for dust, and use the trace-reversed form so the source side is easy to compare with the linearized Ricci tensor.

Step 1. Write $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ with $|h_{\mu\nu}|\ll 1$ and signature $(-,+,+,+)$ , static fields, and dust $T_{\mu\nu}=\rho\,u_{\mu}u_{\nu}$ . Use the trace-reversed equation $R_{00}=\kappa\big(T_{00}-\tfrac12 T\,\eta_{00}\big)$ because its right side is trivial for dust. Why: the trace-reversed form moves the metric-trace complication to the source, where dust makes it disappear.

Step 2. For dust at rest, $T_{00}\approx\rho c^2$ and the trace $T\approx-\rho c^2$ , so with $\eta_{00}=-1$ the source combination is $T_{00}-\tfrac12 T\,\eta_{00}\approx\rho c^2-\tfrac12\rho c^2=\tfrac12\rho c^2$ . Why: rest-mass energy dominates over pressure for non-relativistic matter, so only the $\tfrac12\rho c^2$ combination survives.

Step 3. The linearized time-time Ricci tensor for a static field keeps only spatial second derivatives, $R_{00}\approx-\tfrac12\nabla^2 h_{00}$ , and the Newtonian potential enters through $h_{00}=-2\Phi/c^2$ , giving $R_{00}\approx\tfrac{1}{c^2}\nabla^2\Phi$ . Why: identifying $h_{00}$ with the potential is what connects the metric to the familiar $\Phi$ , and staticity kills the time derivatives.

Step 4. Equate the two expressions for $R_{00}$ : $\tfrac{1}{c^2}\nabla^2\Phi=\kappa\cdot\tfrac12\rho c^2$ , i.e. $\nabla^2\Phi=\tfrac12\kappa c^4\rho$ . Demanding this be Poisson's $\nabla^2\Phi=4\pi G\rho$ forces $\tfrac12\kappa c^4=4\pi G$ , hence $\kappa=8\pi G/c^4$ . Why: the single free constant in the theory is fixed by one experimental anchor — the correspondence with Newtonian gravity — and nothing about GR's structure was adjustable beyond it. *The Newtonian-limit matching yields $\kappa=8\pi G/c^4\approx 2.08\times10^{-43}\ \mathrm{s^2\,kg^{-1}\,m^{-1}}$ .*

Limiting cases

The Einstein tensor specializes cleanly across the physically important regimes, and each specialization is a check on the definition.

Case	Result	Meaning
Vacuum	$G_{\mu\nu}=0\;\Longleftrightarrow\;R_{\mu\nu}=0$	The trace of $G_{\mu\nu}=0$ gives $R=0$ (in $D\neq 2$ ), so vacuum spacetimes are Ricci-flat. They need not be flat: Weyl curvature survives, e.g. Schwarzschild — which is how a star's exterior field, and gravitational waves, exist in vacuum.
Maximally symmetric	$G_{\mu\nu}=-\Lambda g_{\mu\nu}$	For de Sitter or anti-de Sitter space, constant curvature makes $G_{\mu\nu}$ proportional to the metric, reproducing a pure cosmological constant $\Lambda$ .
Weak field / linearized	$G_{\mu\nu}^{(1)}=-\tfrac12\,\Box\bar h_{\mu\nu}$ (Lorenz gauge)	To first order in $h_{\mu\nu}=g_{\mu\nu}-\eta_{\mu\nu}$ , with trace-reversed $\bar h_{\mu\nu}=h_{\mu\nu}-\tfrac12\eta_{\mu\nu}h$ , the Einstein tensor becomes a wave operator — the seed of gravitational radiation.
Two dimensions	$G_{\mu\nu}\equiv 0$	In $D=2$ , $R_{\mu\nu}=\tfrac12 R g_{\mu\nu}$ identically, so the Einstein tensor vanishes for every metric. Ordinary GR has no dynamics in 2D; gravity must be reformulated (dilaton gravity).

Common trapThinking the divergence in the Bianchi identity is the ordinary partial derivative

The identity $\nabla^{\mu}G_{\mu\nu}=0$ uses the covariant divergence with the metric-compatible Levi-Civita connection — it is emphatically not the partial-derivative divergence $\partial^{\mu}G_{\mu\nu}$ , which does not vanish in general. The Christoffel terms are the whole point. The same caution applies to $\nabla^{\mu}T_{\mu\nu}=0$ : it is a local statement, and because of those connection terms it does not integrate to a globally conserved total energy without extra structure. A genuine conserved charge needs a Killing vector $\xi^{\mu}$ , giving the current $J^{\mu}=T^{\mu}{}_{\nu}\xi^{\nu}$ with an honest $\partial_\mu(\sqrt{-g}\,J^\mu)=0$ .

Worked examples

1 · The coupling constant and how fast matter focuses (fully worked)

Evaluate $\kappa=8\pi G/c^4$ in SI units, and use the Ricci-focusing relation $(1/V)\,\mathrm{d}^2V/\mathrm{d}\tau^2=-4\pi G\rho$ to find how fast a small ball of water-density dust begins to contract. Comment on why $\kappa$ is so tiny.

Step 1. Assemble the constants: $G=6.674\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}$ and $c=2.998\times10^8\ \mathrm{m/s}$ , so $c^4=(2.998\times10^8)^4=8.08\times10^{33}\ \mathrm{m^4\,s^{-4}}$ . The numerator is $8\pi G=8\pi(6.674\times10^{-11})=1.677\times10^{-9}$ .

Step 2. Divide: $\kappa=\dfrac{1.677\times10^{-9}}{8.08\times10^{33}}=\mathbf{2.08\times10^{-43}\ s^2\,kg^{-1}\,m^{-1}}$ . The colossal $c^4$ in the denominator is why spacetime is so stiff — it takes a planet's worth of stress-energy to bend geometry appreciably.

Step 3. Now the focusing rate for dust at $\rho=1000\ \mathrm{kg/m^3}$ : $(1/V)\,\mathrm{d}^2V/\mathrm{d}\tau^2=-4\pi G\rho=-4\pi(6.674\times10^{-11})(1000)=\mathbf{-8.39\times10^{-7}\ s^{-2}}$ . The ball's volume starts curving inward at that fractional acceleration.

Step 4. The two numbers cross-check: the focusing relation is the trace-reversed equation in disguise, $R_{\hat t\hat t}=\kappa(T_{\hat t\hat t}-\tfrac12 T)=\kappa\cdot\tfrac12\rho c^2$ , and $\kappa\cdot\tfrac12 c^2\rho=\tfrac12(2.08\times10^{-43})(8.99\times10^{16})(1000)\approx4\pi G\rho=8.4\times10^{-7}\ \mathrm{s^{-2}}$ . Independently, $\kappa\approx2.08\times10^{-43}$ SI and the ball contracts at $-8.4\times10^{-7}\ \mathrm{s^{-2}}$ .

2 · Schwarzschild passes the vacuum test

Show that the exterior Schwarzschild geometry, with $e^{-2\beta}=1-r_s/r$ and $r_s=2GM/c^2$ , satisfies the vacuum equation $G^{0}{}_{0}=0$ using the compact form in equation (7).

Step 1. Insert the metric function into $r(1-e^{-2\beta})$ : with $e^{-2\beta}=1-r_s/r$ , the bracket is $r\big(1-(1-r_s/r)\big)=r\cdot\dfrac{r_s}{r}=r_s$ , a constant independent of $r$ .

Step 2. Differentiate: $\dfrac{\mathrm{d}}{\mathrm{d}r}\big[r(1-e^{-2\beta})\big]=\dfrac{\mathrm{d}}{\mathrm{d}r}\,r_s=0$ , so from equation (7) $G^{0}{}_{0}=-\dfrac{1}{r^2}\cdot 0=\mathbf{0}$ .

Step 3. Interpret. The vanishing of $G^{0}{}_{0}$ (and, working the other components, all of $G_{\mu\nu}$ ) says the Schwarzschild exterior is a vacuum solution, $R_{\mu\nu}=0$ , consistent with there being no matter outside the mass. Yet the spacetime is curved — the Weyl tensor is nonzero and encodes the tidal field a falling astronaut feels. Ricci-flat is not flat, which is what lets a black hole's gravity live in empty space. $G^{0}{}_{0}=0$ because $r(1-r_s/r)=r_s$ is constant; Schwarzschild is vacuum but not flat.

Physics of the portrait

The Einstein tensor sits at a crossroads of nineteenth-century geometry, Einstein's insistence on conservation, and a twentieth-century uniqueness theorem — many hands shaped it, but one name is attached because one man forced geometry and conservation to meet.

Origins · 1 minAlbert Einstein and how this was discovered

Albert Einstein

1879–1955 · the geometric left-hand side of the field equations

Einstein's road to $G_{\mu\nu}$ was not a derivation but a three-year struggle. By 1913, working with Marcel Grossmann, he had the idea that gravity is spacetime curvature and that the Ricci tensor should sit opposite the matter source — but his "Entwurf" equations were not generally covariant. The missing ingredient was conservation: whatever stood opposite $T_{\mu\nu}$ had to be divergence-free, because $\nabla^{\mu}T_{\mu\nu}=0$ is non-negotiable for matter. The bare Ricci tensor fails that test, and through the autumn of 1915 Einstein circled the correct combination, presenting $G_{\mu\nu}=\kappa T_{\mu\nu}$ to the Prussian Academy on 25 November 1915 — the same month David Hilbert obtained the equations from a variational principle. The differential identities guaranteeing $\nabla^{\mu}G_{\mu\nu}=0$ had been published by Luigi Bianchi in 1902 (and glimpsed earlier by Voss and Ricci), but Einstein did not know them by name; he verified conservation the hard way. In 1917 he added the cosmological term $\Lambda g_{\mu\nu}$ to force a static universe, later calling it his "greatest blunder" — though Lovelock's 1971 theorem would show it was never optional but always allowed, the one free addition the geometry permits. The tensor that bears his name is thus less an invention than a discovery of the only consistent possibility.

History

1902

— Bianchi publishes the differential curvature identities, earlier found by Voss and Ricci

1915

— Einstein presents Gμν = κTμν to the Prussian Academy on 25 November

1915

— Hilbert derives the same field equations from a variational principle

1917

— Einstein introduces the cosmological term Λgμν to model a static universe

1971

— Lovelock proves Gμν + Λgμν is the unique divergence-free second-order tensor in four dimensions

Going further

References. Misner, Thorne & Wheeler, Gravitation, §13.5, §17.1 (Bianchi identity and the Einstein tensor) · Wald, General Relativity, §3.2, §4.3 (curvature identities and field equations) · Carroll, Spacetime and Geometry, §3.7–3.10, §4.1–4.3 · Hartle, Gravity: An Introduction to Einstein's General Relativity, Ch. 22 · Schutz, A First Course in General Relativity, §8.1–8.4 · Lovelock, J. Math. Phys. 12, 498 (1971), "The Einstein Tensor and Its Generalizations". The key results used throughout this article are: the contracted Bianchi identity $\nabla^{\mu}R_{\mu\nu}=\tfrac12\nabla_{\nu}R$ ; the trace relation $G=(1-D/2)R$ ; the trace-reversed field equations; the coupling $\kappa=8\pi G/c^4\approx 2.08\times10^{-43}\ \mathrm{s^2\,kg^{-1}\,m^{-1}}$ ; and the Schwarzschild vacuum check $G^{0}{}_{0}=0$ .

Prerequisites. The Stress-Energy Tensor · Local Conservation of Energy-Momentum · The Newtonian Limit. Leads to. The Einstein Field Equations · The Cosmological Constant · Variational Derivation of the Field Equations.

Ask the tutor: Ask me to walk through the double contraction of the second Bianchi identity, or to show step by step how $G_{\mu\nu}=\kappa T_{\mu\nu}$ becomes the Friedmann equations for an expanding universe.

Is the Einstein tensor an equation of motion or an identity?

Both, in different roles. The statement $\nabla^{\mu}G_{\mu\nu}=0$ is an identity — it holds for any metric whatsoever, before any physics, as a consequence of the Bianchi identities of geometry: kinematic, off-shell, always true. The statement $G_{\mu\nu}=\kappa T_{\mu\nu}$ is an equation of motion that selects which metrics are physically realized. The identity constrains the equation: because the left side is automatically divergence-free, the equation cannot hold unless the right side is too, which is why conservation of the source is guaranteed rather than assumed.

Why subtract exactly half the scalar curvature and not some other fraction?

Because the once-contracted Bianchi identity says $\nabla^{\mu}R_{\mu\nu}=\tfrac12\nabla_{\nu}R$ . Forming $R_{\mu\nu}+a R g_{\mu\nu}$ and taking its divergence gives $(\tfrac12+a)\nabla_{\nu}R$ . For this to vanish for every metric — not just the special ones with constant scalar curvature — the bracket must be identically zero, forcing $a=-\tfrac12$ . Any other coefficient leaves a term proportional to the generically nonzero $\nabla_{\nu}R$ , so the source would not be conserved. The $-\tfrac12$ is the unique fixed point, not a fit.

Does local conservation grad-mu T equals zero mean total energy is conserved?

Not by itself. The covariant divergence vanishing is purely local, and the Christoffel terms inside $\nabla_\mu$ prevent rewriting it as an ordinary divergence $\partial_\mu(\cdots)=0$ whose volume integral would give a conserved charge. In a general curved spacetime there is no globally conserved total energy. You recover one only when the spacetime has a symmetry: a Killing vector $\xi^{\mu}$ builds $J^{\mu}=T^{\mu}{}_{\nu}\xi^{\nu}$ , which satisfies an honest $\partial_\mu(\sqrt{-g}\,J^\mu)=0$ . Time-translation symmetry gives conserved energy; without it, energy in GR is genuinely subtle.

How can six equations determine a ten-component metric?

They do not have to — because four of the metric's ten components are gauge, mere labels you change by relabelling coordinates. The field equations $G_{\mu\nu}=\kappa T_{\mu\nu}$ are ten, but the four Bianchi identities $\nabla^{\mu}G_{\mu\nu}=0$ mean four of them are constraints on the initial data that the remaining equations automatically preserve, not independent evolution equations. That leaves six genuine dynamical equations for the six physical degrees of freedom — the Bianchi identity is what balances the count, and removing it would leave the theory over- or under-determined.

Why does the two-dimensional case break down?

In two dimensions the Riemann tensor has only one independent component, and the Ricci tensor is forced to obey $R_{\mu\nu}=\tfrac12 R g_{\mu\nu}$ for every metric. Substituting into the definition gives $G_{\mu\nu}\equiv 0$ identically — the Einstein tensor vanishes whatever the geometry, so the vacuum equation $G_{\mu\nu}=0$ is empty and carries no information about the metric. Two-dimensional gravity therefore has no propagating content in the ordinary formulation; to get dynamics one adds extra fields, as in dilaton gravity. The trace relation $G=(1-D/2)R$ signals it: the prefactor vanishes at $D=2$ .

What is the difference between the Einstein tensor being divergence-free and it being zero?

They are completely different statements. $\nabla^{\mu}G_{\mu\nu}=0$ (divergence-free) is an identity satisfied by every spacetime, curved or flat, with or without matter — a property of how the tensor is built. $G_{\mu\nu}=0$ (the tensor itself vanishing) is a vacuum field equation, a special condition picking out matter-free regions, equivalent to $R_{\mu\nu}=0$ . Schwarzschild has $\nabla^{\mu}G_{\mu\nu}=0$ everywhere but $G_{\mu\nu}=0$ only outside the star. Confusing the identity with the equation is the same slip as confusing "always true" with "constrains the solution."