advanced28 min studyLesson 807 in Physics

Curvature traces obey identities without containing every tidal component

Contract the Riemann tensor, derive the divergence-free Einstein tensor, and distinguish local identities from global flux statements.

On this page

What you’ll understand

  • Compute Ricci and scalar contractions with the stated curvature convention.
  • Use the contracted Bianchi identity to identify a divergence-free tensor.
  • Distinguish Ricci-flat geometry from fully flat geometry and handle volume boundaries.

The Riemann tensor contains more information than many gravitational equations need directly. Its Ricci contraction and scalar trace organize local volume focusing and enter Einstein’s equation, while a differential identity constrains their divergence. Tracing loses information, however: a four-dimensional vacuum region can have nonzero tidal curvature even when its Ricci tensor vanishes. We will compute a sphere example and explain how local covariant identities connect to integration only after the measure and boundary terms are specified.

Contractions keep some curvature and discard other parts

Use Rμν=Rρμρν with the Riemann convention already declared. Metric compatibility and the Levi-Civita curvature symmetries make this Ricci tensor symmetric. A further contraction with the inverse metric gives the Ricci scalar. The inverse matters: in angular coordinates its entries supply inverse length squared, so a dimensionless angular Ricci component can still contribute to a scalar in m⁻². The Einstein tensor subtracts one half the scalar times the metric. These are tensor constructions, not operations tied to one special coordinate chart.

On a round two-sphere, Rθθ=1 and Rφφ=sin²θ. Contracting with the inverse radius-squared metric gives R=2/a². In two dimensions the Einstein tensor vanishes identically despite this nonzero curvature; a four-dimensional interpretation must not be imported into that dimensional identity. In four dimensions, the Weyl part retains curvature not specified by Ricci traces. Schwarzschild vacuum has Rμν=0 but a nonzero Riemann-square invariant 48m²/r⁶, where m=GM/c² is a length. Thus even a vanishing Ricci scalar is far from a proof of flatness.

Rμν=Rρμρν,R=gμνRμν,Gμν=Rμν−12RgμνR_{\mu\nu}=R^\rho{}_{\mu\rho\nu},\qquad R=g^{\mu\nu}R_{\mu\nu},\qquad G_{\mu\nu}=R_{\mu\nu}-\frac12Rg_{\mu\nu}
The spacetime signature is −+++ and curvature sign is inherited from the previous lesson. R and orthonormal Ricci components have units m⁻². Coordinate components depend additionally on coordinate units; the sphere example uses dimensionless angular coordinates.↔ Scroll wide equations horizontally.

The Bianchi identity is geometric rather than a field equation

The differential Bianchi identity is a cyclic relation among covariant derivatives of Riemann curvature. Contract it twice and use metric compatibility to obtain ∇μRμν equal to one half the gradient of R. The Einstein tensor therefore has zero covariant divergence for every sufficiently smooth metric with its Levi-Civita connection. No matter equation or Einstein equation was required. When the gravitational field equation relates this tensor to matter stress, the identity provides a consistency condition with covariant matter conservation.

Covariant conservation does not automatically provide a globally conserved ordinary energy in an arbitrary spacetime. The extra free tensor index means one cannot integrate components at different points as though their bases were the same. Symmetry, boundary conditions, and suitable currents are needed for a global charge. A cosmological term Λgμν has vanishing divergence when Λ is constant. Treating Λ as an arbitrary position-dependent scalar while keeping an otherwise unchanged conserved source would violate the same local consistency condition unless additional dynamical contributions are included.

∇λRρσμν+∇μRρσνλ+∇νRρσλμ=0,∇μRμν=12∇νR,∇μGμν=0\nabla_\lambda R^\rho{}_{\sigma\mu\nu}+\nabla_\mu R^\rho{}_{\sigma\nu\lambda}+\nabla_\nu R^\rho{}_{\sigma\lambda\mu}=0,\qquad \nabla^\mu R_{\mu\nu}=\frac12\nabla_\nu R,\qquad \nabla^\mu G_{\mu\nu}=0
The cyclic identity assumes the torsion-free Levi-Civita connection. Raising the derivative index uses the inverse metric. These are local differential identities in any dimension where the stated contractions exist, with no assumption of Einstein’s dynamical equation.↔ Scroll wide equations horizontally.

A volume identity requires a directed boundary flux

For a vector current the covariant divergence is the coordinate derivative of √|g| times the current, divided by √|g|. Integrating with the invariant density converts it to a directed boundary flux by Stokes’ theorem. Orientation fixes the signs of initial and final pieces, and induced measures must be used. A non-null boundary can be expressed using a normalized normal with the appropriate Lorentzian orientation sign. A null boundary needs a separate surface-element construction because its normal is also tangent; setting a unit normal there is invalid. Boundary terms vanish only under stated boundary conditions.

The function lab illustrates the sphere scalar’s inverse-radius scaling. Reset and set both controls to a=2,x=0.5, predicting y=0.5. Reset and set a=2,x=1, predicting y=2. Assign x to an inverse radius in chosen reference-length units and y to the corresponding dimensionless scalar 2x². Halving radius quadruples this curvature. The lab computes neither tensor traces nor a boundary flux; the interpretation depends on the analytic sphere metric, not on the parabola alone.

Trace the curvature of a two-sphere

A round sphere has radius a=2 m. At its equator the metric is diag(4,4) m² and Ricci components are diag(1,1). Find R and the two Einstein-tensor components, and explain why their values do not imply flatness.

  1. The inverse metric at the equator is diag(1/4,1/4) m⁻². The scalar is R=(1/4)×1+(1/4)×1=0.50 m⁻².

  2. Each Einstein component is 1−(1/2)×0.50×4=0. The vanishing occurs because of the two-dimensional trace identity.

  3. The sphere still has sectional curvature 1/a²=0.25 m⁻² and nonzero Riemann curvature. The two-dimensional Einstein tensor has lost that information through its special identity.

The result

The scalar is R=0.50 m⁻² and both diagonal Einstein components vanish, while the sphere remains intrinsically curved.

Explore the model

Interactive lab

Interactive lab · function

Test curvature versus inverse radius

Reset and set BOTH controls to a=2,x=0.5, predicting y=0.5. Reset and set a=2,x=1, predicting y=2. Assign x to normalized inverse sphere radius and y=2x². This is scalar scaling only, with no Ricci contraction or Bianchi solver.

Test curvature versus inverse radiusThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forSphere curvature scales as inverse radius squared; a vanishing traced combination need not mean full curvature vanishes.

Check your understanding

What follows from Rμν=0 in a four-dimensional vacuum region?

Try it yourself

In a four-dimensional Einstein space let Rμν=Λgμν with constant Λ=0.020 m⁻². Find R, Gμν, and G₀₀ in an orthonormal −+++ frame. Verify the vacuum equation Gμν+Λgμν=0.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The mixed trace gμνgμν is the dimension four. The orthonormal time-time metric entry is −1.

Compare with the solution

R=4Λ=0.080 m⁻². Then Gμν=Λgμν−(1/2)(4Λ)gμν=−Λgμν. In the orthonormal frame G₀₀=+0.020 m⁻². Adding Λgμν cancels every component. Constancy of Λ also makes its covariant divergence vanish by metric compatibility.

Take it with you

  • Ricci contractions discard curvature information that can remain in Weyl tides.
  • The contracted Bianchi identity holds geometrically before a field equation is imposed.
  • Global flux claims require invariant measures, orientation and boundary conditions.

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