Questions: Ricci Tensor and Scalar Curvature

Question 1

In vacuum (T_μν = 0), the Einstein field equations reduce to R_μν = 0. Which of the following is a correct consequence?

• A. Spacetime must be flat (Riemann tensor vanishes)
• B. The Riemann tensor can be nonzero — the Weyl tensor carries curvature information not captured by the Ricci tensor
• C. Gravitational waves cannot propagate in vacuum
• D. The scalar curvature R is nonzero but the Ricci tensor vanishes

Answer: (B)

R_μν = 0 means the Ricci tensor vanishes, but the full Riemann tensor has 20 independent components — only 10 are in the Ricci tensor. The remaining 10 are in the Weyl tensor C_{ρσμν}, which can be nonzero even when R_μν = 0. This is why the Schwarzschild solution is curved (tidal forces exist) despite being a vacuum solution. Gravitational waves also propagate in vacuum as Weyl curvature. If R_μν = 0, then R = g^{μν}R_μν = 0 as well, so option D is wrong.

Question 2

The Ricci scalar R is always positive in a region where matter is present.

• T. True
• F. False

Answer: False

The sign of R depends on the equation of state of the matter. For a perfect fluid with energy density ρ and pressure p, the trace of the Einstein equations gives R = -8πG(ρ - 3p)/c⁴ (in the +−−− convention) or similar expressions depending on sign convention. For ordinary matter (ρ > 0, p ≥ 0), R can be negative. For a cosmological constant (dark energy with p = -ρc²), R can be positive. The sign of R is not determined by the presence of matter alone.

Question 3

Explain the physical meaning of the Ricci tensor in terms of the volume of a small ball of freely falling test particles.

Model answer: Consider a small ball of test particles initially at rest relative to each other in a local inertial frame. As they fall freely, the ball deforms due to tidal forces. The Ricci tensor determines the rate of change of the volume of this ball: the fractional rate of volume decrease (convergence of geodesics) is proportional to R_μν u^μ u^ν, where u^μ is the four-velocity. In the presence of matter with positive energy density, R_μν u^μ u^ν > 0, so the volume decreases — gravity is attractive. In vacuum, R_μν = 0, so the volume of the ball is preserved to first order, even though its shape changes (tidal distortion from the Weyl tensor stretches it in some directions and compresses it in others).

This geometric interpretation makes clear the distinction between Ricci and Weyl curvature. The Ricci tensor controls volume changes (focusing of geodesics, directly sourced by matter), while the Weyl tensor controls shape changes (tidal distortion, present even in vacuum). A black hole's vacuum exterior has zero Ricci curvature but strong Weyl curvature — volumes are preserved but shapes are dramatically distorted.

Question 4

Why is the Einstein tensor G_μν = R_μν - (1/2)g_μν R preferred over the Ricci tensor alone on the left side of the field equations?

Model answer: The Einstein tensor is divergence-free: ∇^μ G_μν = 0 identically, as a consequence of the contracted Bianchi identity. This is essential because the right side of the field equations, 8πG T_μν / c⁴, must also be divergence-free (∇^μ T_μν = 0) to ensure local conservation of energy and momentum. The Ricci tensor alone does not satisfy ∇^μ R_μν = 0 in general, so using R_μν = 8πG T_μν / c⁴ alone would be inconsistent with energy-momentum conservation. The combination G_μν = R_μν - (1/2)g_μν R is the unique symmetric, divergence-free tensor built from the metric and its first two derivatives (up to the cosmological constant term).

The contracted Bianchi identity is the mathematical guarantee that Einstein's field equations are self-consistent: the geometry automatically conserves energy-momentum. This identity also reduces the number of truly independent field equations from 10 to 6, which is exactly right given the 4 coordinate degrees of freedom.