Is there a version of the Einstein field equations that uses the Riemann curvature tensor instead of the Ricci curvature tensor?

Question

I understand that the Einstein field equation uses the Ricci curvature tensor, Ricci curvature scalar, and stress-energy momentum tensor. But is there a way to form an equation that uses the Riemann curvature tensor and the stress-energy momentum tensor?

Answer 1

Well, as you know the Ricci tensor/scalar are built from the Riemann tensor, so it's not entirely clear to me what you're asking. You can of course write $$G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}R g_{\mu\nu} = R^\alpha_{\ \ \mu\alpha\nu} - \frac{1}{2} g^{\beta\gamma}R^\alpha_{\ \ \beta \alpha \gamma} g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$$

if you wish, but there is no way to e.g. write an equation for the components of the Riemann tensor in terms of the stress-energy tensor because the stress-energy tensor does not uniquely determine the Riemann tensor — in particular, even if $T_{\mu\nu}=0$, it is possible that $R^\alpha_{\ \ \beta \mu\nu} \neq 0$ as is the case in e.g. the Schwarzschild spacetime.

Answer 2

Here is the Lichnerowicz' form of Einstein's equations: \begin{eqnarray} R^μ{}_{νρσ;μ}=J_{νρσ},\\ R_{μνρσ;α}+R_{μναρ;σ}+R_{μνσα;ρ}=0,\\ \end{eqnarray} where the “current” tensor is $$J_{νρσ}=\left(T_{νσ;ρ}-\frac12 g_{νσ}T_{;ρ}\right)-\left(T_{νρ;σ}-\frac12 g_{νρ}T_{;σ}\right) .$$

The second equation is just (differential) Bianchi identity, while the first could be obtained by combining Bianchi identity with the standard form of Einstein equations.

Compare this system with Maxwell's equations: \begin{eqnarray} F^μ{}_{ν;μ}=J_ν,\\ F_{μν;ρ}+F_{ρμ;ν}+F_{νρ;μ}=0. \end{eqnarray}

Just from the formal analogy between gravitational and electromagnetic equations we could guess that free gravitational field propagates at the speed of light (at least when the linearized approximation is appplicable) and that derivatives of stess-energy tensor serve as sources for that free gravitational field.

Answer 3

If you mean an equation such that the stress energy tensor at a point of the spacetime determines the Riemann tensor at that point, then the answer is negative for physical reasons. Such an equation would imply that the spacetime is flat outside the gravitational sources. This makes no sense because the gravitational field propagates outside its sources.

Answer 4

You can start with Ricci decomposition in a $n$ dimensional Riemannian manifold:

$$R_{abcd}=C_{abcd}+\frac{1}{n-2}(S_{ad}g_{bc}+S_{bc}g_{ad}-S_{ab}g_{cd}-S_{ac}g_{bd})+\frac{R}{n(n-1)}(g_{ad}g_{bc}-g_{ac}g_{bd})$$ where $C_{abcd}$ is the Weyl tensor and $S_{ab}=R_{ab}-\frac{1}{n}g_{ab}R$ defines the trace free part of Ricci tensor. You can substitute $R_{ab}$ and $R$ in terms of $T_{ab}$ and $T$ from Einstein's field equations. $C_{abcd}$ is not completely independent though: you have the 2nd Bianchi identity: $$\nabla_{[a}R_{bc]de}=0$$ which will give some relations between divergence of the Weyl tensor in terms of derivatives of $T_{ab}$ and $T$