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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?

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4 Answers 4

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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.

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  • $\begingroup$ Well ${R^{\alpha}}_{\mu\alpha\nu}$ is just $R_{\mu\nu}$ in disguise. How does that convey any extra information? $\endgroup$
    – KP99
    Jul 3 at 8:02
  • $\begingroup$ You could use the expression of the Riemann tensor in terms of Christoffel symbols and hence the metric, then reverse this huge formula to find the metric in terms of the Riemann tensor. Then use EFE on this gives equation for Riemann Tensor components. $\endgroup$
    – Cretin2
    Jul 3 at 12:26
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    $\begingroup$ @KP99 It doesn't, which is essentially my point. My reading of the question is, "can you reformulate Einstein's equations to use the Riemann tensor instead of the Ricci tensor/scalar," to which my answer would be no, other than to use the explicit definitions of the latter in terms of the former. $\endgroup$
    – J. Murray
    Jul 3 at 13:30
  • $\begingroup$ Ah I see. I thought OP is asking whether one can find some relationship b/w Riemann tensor and stress energy tensor given that Einstein field equations holds and given all possible identities involving curvature tensor. A relationship b/w these two quantities does exist but not the simple direct relationship. $\endgroup$
    – KP99
    Jul 3 at 17:20
  • $\begingroup$ @KP99 Well, it’s entirely possible that your understanding of the question is the correct one! +1 to your answer $\endgroup$
    – J. Murray
    Jul 3 at 18:31
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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.

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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.

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  • $\begingroup$ For void regions $T_{\mu\nu} = 0$, it seems that $J_{\mu\nu\sigma} = 0$ and, therefore, it seems that ${R^\mu}_{\nu\rho\sigma}$ is constant in such a connected region (because all its covariant derivatives are null). Is that correct? $\endgroup$
    – Davius
    Jul 4 at 13:33
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    $\begingroup$ @Davius No, $\nabla \mathrm{Riem}=0$ is a rather strong condition and is not satisfied in general by vacuum spacetimes (counterexample: Schwarzschild). It only follows that $\mathrm{Riem}$ is divergenceless in addition to satisfying the Bianchi identity. $\endgroup$ Jul 4 at 15:24
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    $\begingroup$ @Davius: Bence is correct. Note, that the first equation has summation over the $μ$ index, so all covariant derivatives will not be zero, just the divergence will. $\endgroup$
    – A.V.S.
    Jul 4 at 18:45
  • $\begingroup$ This formulation doesn't even seem to appear anywhere on the Internet save here. Can you point to somewhere it is discussed in more depth? $\endgroup$
    – Trebor
    Jul 5 at 14:22
  • $\begingroup$ @Trebor: See, for example, this paper and this paper. $\endgroup$
    – A.V.S.
    Jul 5 at 16:33
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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 $T$

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    $\begingroup$ @K.T. Thanks for the edit! $\endgroup$
    – KP99
    Jul 3 at 11:12

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