2

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?

Qmechanic
  • 201,751
thedoctar
  • 516
  • 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 04:28
  • @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 04:46
  • @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 02:56
  • @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 08:03
  • @thedoctar I totally agree, but then again you should expect something handwavy to (apriori) support a claim like this. I am not aware of any symmetry-based arguments that support this claim. – Prof. Legolasov Apr 22 '17 at 08:17
  • @SolenodonParadoxus I'm not sure why I should expect a handwavy argument. There might be a physical justification. Or there might not be. Maybe it's something to do with gravity being a gauge theory? – thedoctar Apr 22 '17 at 09:08
  • Lovelace has shown that it is the only divergence free two rank tensor that depends on derivatives of the metric tensor no higher than second. It is of course also linear in second derivatives of the metric tensor, leading to linearity in the curvature. Einstein-Cartan, with torsion, I'm not sure if also linear (should be easy to find), but it and pure Einstein are the only possibilities if one uses the Einstein Lagrangian (and with Cartan allowing torsion). Still doesn't answer the question, but 3 basically also then leads to 10 pseudo linear equation equations for 4 dimensions – Bob Bee Apr 22 '17 at 20:15
  • @BobBee Would I be correct in saying that if we were to remove condition 3 then our Einstein tensor would be linear in Riemann curvature tensor terms? – thedoctar Apr 23 '17 at 07:25
  • sounds like it, if Lovelace is right. I've not seen the proof, it'd be good to find some reference in some other book. I saw that in Wiki on the Ricci tensor. – Bob Bee Apr 24 '17 at 00:15
  • Although the dimensional analysis argument is very interesting, the reason why G has to be linear in R comes from classical mechanics. In classical mechanics we learn that the dynamical equations cannot have more than two derivatives with respect to time if we require causality. Since R already contains such terms, higher power of R would lead to acasual dynamics. – CGH Apr 26 '17 at 01:10
  • @CGH Could you provide a reference? Is the context of your statement purely field-theoretic? I don't see how an equation like a=v^3 is acausal. – thedoctar Apr 26 '17 at 12:46
  • 2
    @thedoctar This is pure PDE! Check http://web.math.ucsb.edu/~grigoryan/124A.pdf for a discussion of causality of the plane wave. You should also check the discussions here https://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
  • @CGH Please do not use comments to answer questions, but write an answer instead. – ACuriousMind May 06 '17 at 10:59
  • @CGH Do you have a proof or example of how higher time derivatives break causality? – thedoctar May 08 '17 at 00:27
  • @thedoctar I don't have a proof, but examples. See my answer for a link discussing such issues. – CGH Jun 13 '17 at 16:53

1 Answers1

1

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.

CGH
  • 191