# Why must the Einstein tensor $G$ be linear in the Riemann curvature tensor?

In the classic book by Misner, Wheeler and Thorne, they justify the form of the Einstein tensor, $G$, by the fact that it is the unique tensor which satisfies

1. $G$ vanishes when spacetime is flat
2. $G$ is a function of the Riemann curvature tensor and metric only
3. $G$ is linear in the Riemann curvature tensor
4. $G$ is a symmetric and a second-rank tensor
5. $G$ has vanishing divergence.

Point 1 comes from the fact that if we equate gravitation to geodesic deviation by spacetime curvature, then the absence of spacetime curvature should mean no gravitation. Point 2 basically says gravitation is due to geodesic deviation only. Point 4 is basically because curvature is a two-form (or since we want the stress-energy tensor to be a source of spacetime curvature, which is a rank-two tensor) and point 5 is due to (local) energy-momentum conservation.

I do not understand why we need point 3 though. Possibly at the quantum level there might be corrections which are non-linear in Riemann curvature tensor, but why, at the classical level, do we demand linearity in the Riemann curvature tensor?

• Ah, I see. Would you accept "because $R_{\mu \nu}$ has dimensions of $m^{-2}$ and we don't have a fixed length scale in GR"? – Prof. Legolasov Apr 21 '17 at 4:28
• @SolenodonParadoxus. Interesting. Can you explain why that would lead to the linearity of it? Could it be that you'd want curvature to be linearly proportions to energy stress tensor (content)? – Bob Bee Apr 21 '17 at 4:46
• @BobBee my logic is the following. Imagine that $G$ includes both $\sim R^{\alpha}$ and $\sim R^{\beta}$ terms. Then we would have to compensate for $m^{\beta - \alpha}$ missing dimensions of lengths, but we don't have a dimensionfull constant in our disposal. – Prof. Legolasov Apr 22 '17 at 2:56
• @SolenodonParadoxus I tend to find dimensional analysis arguments to be very handwavy and not very rigorous. Why can't we just introduce more dimensionful constants? I was hoping there would be an argument based on symmetry. – thedoctar Apr 22 '17 at 8:03
• @thedoctar This is pure PDE! Check web.math.ucsb.edu/~grigoryan/124A.pdf for a discussion of causality of the plane wave. You should also check the discussions here physics.stackexchange.com/q/4102. The only specific example I can give now is the Abraham–Lorentz force, where a particle accelerates before it interacts. I will look for a more general reference. – CGH Apr 26 '17 at 13:11

Theories with more than two derivatives are always treated with great care. Causality and unitatiry might be broken. There could be ghosts (which actually means that there are less degrees of freedom than expected.) Perturbative degrees of freedom can differ from the Hamiltonian (non-perturbative) analysis. Among other issues.

The Abraham-Lorentz force is a canonical example of such behavior at the classical level. This force is proportional to the derivative of the acceleration, $F_{rad}=\frac{\mu_0 q^2}{6\pi c}\dddot x$. The problem with this force is that a particle accelerates before the force is applied.

When constructing gravity, you might want to couple it to matter. If you have a flat metric, you would expect to recover classical Electrodynamics. If, furthermore, you ask for causality, you don't expect third order derivatives in the equation of motion, since you want to avoid the Abraham-Lorentz force. If you linearize the Einstein EOM around a flat metric, you don't want this action to be the source of such forces. Thus, you require for the Einstein's equations to contain, at most, second order derivatives.

This is the reason why you require that "G is linear in the Riemann curvature tensor."

Of course, you don't avoid higher curvature corrections as if they were the plague. String theory corrections to the Einstein equation actually contain such terms. How to deal with them is still an active area of research.

For an old review on higher derivative theories, see this.

The Abraham-Lorentz force is well explained in wikipedia.