# Could the gravitational field equations be formulated in term of the Riemann curvature tensor (as opposed to the Ricci curvature tensor)?

The most symmetries and identities in Riemannian geometry are in term of the Riemann curvature tensor. One may ask why the gravitational field equations are not in term of this main tensor of (pseudo)Riemannian geometry? i.e. without any contraction with the metric.

However contraction with the covariant derivative, torsion tensor (and the like) is OK.

• @Qmechanic I appreciate your kind edit. – user.3710634 Oct 2 '16 at 20:33

Of course Prahar's Answer that field equations completely constraining the Riemann Curvature tensor are not the Einstein field equations is completely correct, but here's a phenomenological / mathematical "motivation" as to why the field equations only constrain the contraction (i.e. the Ricci tensor): if the field equations alone completely constrained the curvature tensor, there would be no way for different boundary conditions to affect the curvature tensor, a system's geometry and therefore the system's gravitational physics.

To understand this, let's first recall how the field equations, together with boundary conditions, do fully set the Riemann tensor.

1. Directly, the field equations define the Ricci tensor (which defines how the local volume shrinks and swells at different positions in spacetime). This definition leaves the Weyl tensor unconstrained. The Weyl defines how the shape of spacetime test volumes is affected by gravitation;

2. However, from the Ricci tensor, one can form the Schouten tensor. The derivatives of the Schouten tensor do define the derivatives of the Weyl tensor. So we can now put in boundary conditions to complete the definition of the Weyl. Therefore, the derivatives of the Ricci together with appropriate boundary conditions on an appropriate hypersurface define the Weyl, and thus the Ricci-boundary condition condition set full define the Riemann tensor.

Think how weird the mathematical situation would be without the ability to set boundary conditions independently. Systems which had the same stress energy distribution within a control volume but which differed outside the volume would needfully have the same geometry within the control volume and there would be no way for the outside to "communicate" with the inside. This admission of boundary conditions through degrees of freedom, killed off by differentiation, in the Weyl allows for, amongst many other things, the existence of nontrivial vacuum solutions and gravitational waves.

• "Think how weird the mathematical situation would be without the ability to set boundary conditions independently" ... that's exactly how GR works in $1 + 1$ and $2 + 1$ spacetime dimensions, because the Weyl tensor vanishes trivially. And it is indeed weird. – tparker Oct 4 '16 at 2:28
• @tparker Indeed. I recall when I first met these low dimension solutions how it took me a great deal of reading to reassure myself that my understanding was right- I couldn't believe that things would be so different! – Selene Routley Oct 4 '16 at 4:51

Fine. But I mean the equations be formulated precisely in term of the Riemann curvature tensor without any contraction and the like.

Einstein's equations imposes a constraint on the Ricci tensor, which is a certain contraction of the Riemann tensor. However, it says nothing about the Weyl tensor, which the traceless part of the Riemann tensor.

Any set of equations which involves the full Riemann curvature tensor without any contraction would necessarily impose constraints on the Weyl tensor and are therefore not equivalent to the Einstein equations.

The Einstein field equations are in terms of the Riemann curvature tensor! It is just contracted with the metric once to give the Ricci curvature tensor. That then is contracted with the metric again to give the Ricci scalar. A combination of both tensor and scalar.

So you have the Ricci tensor (two indices) from the Riemann tensor (four indices): $R_{\nu\beta} = R^\alpha{}_{\nu\alpha\beta}$.

Then you form the Ricci scalar: $\mathcal R = g^{\nu\beta} R_{\nu\beta}$.

The Einstein tensor then is this: $G^{\mu\nu} = R^{\mu\nu} - \frac 12 \mathcal R$.

And finally the field equation is $G^{\mu\nu} = 8 \pi G T^{\mu\nu}$ where the left $G$ is the Einstein tensor and the right $G$ is the gravitational constant.

• Fine. But I mean the equations be formulated precisely in term of the Riemann curvature tensor without any contraction and the like. – user.3710634 Oct 2 '16 at 20:24
• There are coordinate free formulations of general relativity. One lecturer told me that in physics those are not that useful as for concrete problems one has to introduce coordinates anyway. I don't know how to write down the equations without coordinates. Perhaps you can reword your question as to ask for this formulation? – Martin Ueding Oct 2 '16 at 20:31
• @J.Pak, do you have in mind rank 4 tensor source term? – Alfred Centauri Oct 2 '16 at 20:33
• @Alfred Centauri, rank 3 tensor source actually. – user.3710634 Oct 2 '16 at 20:35
• @Martin Ueding, Actually not coordinate free formulations. – user.3710634 Oct 2 '16 at 20:39

@Martin Ueding gave a nice introduction to Einstein's field equations. I want to bring up some more points here related to the question.

First I have to admit that I do not have an definite answer to it: I do not know if there is a formulation of GR in terms of the plain Riemann curvature tensor but I can give some points why even if one could construct one it will be inferior to the "standard formulation" over the Ricci tensor and the courvature scalar.

"The most symmetries and identities in Riemannian geometry are in term of the Riemann curvature tensor." that might be true in a sense that the Riemann tensor has Skew symmetry, Interchange symmetry and two Bianchi identities but it also is a tensor of rank 4. If one uses all those symmetries one can reduce the number of independent components to 20 (in case of 4 dimensions). The Ricci tensor has only one symmetry but is only of rank 2; it therefore has only 10 independent components (in case of 4 dimensions). The Riemann tensor has more symmetries but still twice the amount of independent components compared to the Ricci tensor.

That being said: the Ricci tensor is the contraction of the Riemann tensor (@Martin Ueding pointed that out already). But it is in fact the only meaning full contraction of the Riemann tensor: all others vanish or are proportional to it. The reason for that are the symmetries of the Riemann tensor.

In GR we consider the curvature of four dimensional space time: which gives 10 independent components of the metric (metric potentials). Einstein's field equations give us 10 equations for those 10 metric potentials. Plus 4 additional constrain equations based on one very important property of the Einstein tensor: the covariant divergence of the Einstein tensor vanishes: $$G^{\alpha\mu}_{~~~~~;\mu}=0.$$

The professor who told me GR mentioned once that Einstein struggled quite a bit to get this right: that is the reason why one needs the curvature scalar in the Einstein tensor. From the field equations the vanishing of the covariant divergence of $G^{\alpha\beta}$ implies the vanishing of the divergence of the energy momentum tensor. This is the GR equivalent to the classical Euler equation: the energy momentum conservation of GR. This is a very important feature of the theory form a physical point of view. From a tensor calculus point of view the vanishing of the covariant divergence of $R^{\alpha\beta}-\frac12g^{\alpha\beta}R$ is however a direct consequence of the second Bianchi identity.

So Einstein's field equations give us $10+4$ equations for our 10 metric potentials, using the only non vanishing contraction of the Riemann tensor (and again its contraction). The symmetries of the Riemann tensor and its Bianchi identities are actually an integral part of the Einstein equations. They are the reason why the Einstein tensor looks like it does and behaves like it does.

I think just looking at the degrees of freedom it would make no sense to even try to formulate field equations over the Riemann tensor. Apart from that one would need to get the source terms (Energy momentum tensor) to a rank 4 tensor while guaranteeing energy momentum conversation, which might be impossible or only possible with a lot of additional constrains. @Prahar actually made a point in that direction about the Weyl tensor and that an equation for the Riemann tensor it self can not be equivalent to Einstein's field equations without additional constrains on its contraction.

Maybe it is possible to blow the field equations up to rank 4 equations but doing so would make the problem much more complex and bring in degrees of freedom that are just not there. To get something "physical" out, meaning some equations capable of describing the effect of GR, one would need additional constrains on the equations of the rank 4 tensors. If it where possible those field equations on the Riemann tensor would encode the much simpler problem of Einstein's field equations of rank 2.

Sure they can. The answer comes from the Ricci decomposition and the Einstein equations.

Let $T$ be the trace of the stress-energy tensor, let $S_{ab}$ be the traceless part of the stress-energy tensor, and let $C_{abcd}$ be the Weyl tensor, let $\kappa$ be the proportionality constant for the Einstein equations $8\pi G/c^4$, and the result is

$$R_{abcd} = -\frac{\kappa T}{6} g_{a[c} g_{b]d} + C_{abcd} + \kappa g_{a[c} S_{b]d} - g_{b[c} S_{d]a}$$

The Weyl tensor $C_{abcd}$ holds all the gravitational radiation degrees of freedom. All the other terms stem directly from the stress-energy tensor.

In other words, the Einstein equations eliminate the gravitational radiation degrees of freedom to directly relate stress energy to curvature. Writing the equations in terms of the Riemann tensor merely makes those gravitational radiation terms explicit again.

• Thank you, but some parts of the Weyl tensor are written in term of the Ricci tensor and Ricci scalar (?). – user.3710634 Oct 4 '16 at 3:16
• @J.Pak And through the Einstein equations, you can exchange the Ricci tensor and Ricci scalar for terms involving the stress-energy tensor and its trace, which is exactly what I've done here. – Muphrid Oct 4 '16 at 3:19