# Is it possible to derive the Weyl tensor from the Ricci tensor or Ricci scalar? If so, how?

I understand that the Ricci tensor is derived from the Riemann tensor, but I know that the Weyl tensor is also derived from the Riemann tensor. So, is it possible to calculate the Weyl tensor from the Ricci tensor or Ricci scalar, and if so how?

• In general it is not possible. However, in two and three dimensions this is trivially possible, since there the Weyl tensor vanishes identically. Jul 1 at 4:29

No. The Weyl tensor and the Ricci tensor correspond to different "degrees of freedom" of the Riemann tensor. As an example, notice that there are many Ricci-flat solutions to the Einstein equations, such as

• Minkowski spacetime
• Schwarzschild spacetime
• gravitational waves in either of the previous spacetimes

and so on. Each of these solutions have different metrics and different Riemann tensors, but all of them have vanishing Ricci tensors. Hence, they show how knowledge of the Ricci tensor is not sufficient to reconstruct the Weyl tensor. Notice that if you know the Ricci tensor, you also know the Ricci scalar, so the Ricci scalar is also not enough to rebuild the Weyl tensor.

A common way of thinking about these tensors is that the Ricci tensor is the trace part of the Riemann tensor, while the Weyl tensor is the traceless part. Hence, to use one to determine the other would be similar to trying to use the trace of a matrix to figure out what is the actual matrix itself.

• So would that mean that it’s also impossible to use the stress energy tensor to derive the Weyl tensor? Jul 1 at 6:12
• @JungwoonSong Yes. Notice that all of the solutions I mentioned are vacuum solutions: they all share the vary same stress-energy tensor. The thing is that Einstein's equations are a system of PDEs, and you can't uniquely determine the solution to a PDE without some boundary/initial conditions. Hence, you need more information than just the stress tensor Jul 1 at 7:21

To add up to the above answer, even though the Weyl tensor contribution to the Riemann curvature tensor is naturally an independent degree of freedom from the Ricci tensor, it cannot be fully arbitrary - there are constraints that it has to satisfy, in the form of a divergence equation.

From the once-contraction of the differential Bianchi identity (also here)

$$R_{ab[cd;e]} = \nabla_{[e}R_{|ab|cd]} = 0$$

and the decomposition of the Riemann tensor into irreducible components, one can obtain the Weyl tensor propagation equation, which is formally satisfied by the very virtue of the properties of the Riemann curvature tensor (provable by assuming (I think, sufficiently) metricity, torsionlessness):

$$\nabla^{d}C_{abcd} = -\nabla_{[a}(R_{b]c} - \frac{1}{6}g_{b]c}R) = -\nabla_{[a}(T_{b]c} - \frac{1}{3}g_{b]c}T).$$

Such an equation can be used to help classify the types of spacetimes and investigate the tidal forces acting on matter; recently there appeared an article that does exactly that.

The study is based on the decomposition of the Weyl tensor into a gravito-electric and gravito-magnetic part along some vector field (one can think of observers' worldlines, or, in the context of 3+1 relativity (and here), coordinate time or normal vector field lines), and deriving from the above equation the evolution and constraint equations for these parts; said equations are called nonlinear Bianchi equations and can be seen in the mentioned article, eq. 125-128.

I imagine that in looking for possible metrics on a manifold, the above equation is not exactly helpful, although perhaps imposing some fall-off behaviour functions of $$C_{abcd}$$ in the vacuum ($$\nabla^{d}C_{abcd}=0$$) would be an example of its potential use. Just rambling at this point.