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Is GR vacuum equation unique?

+ 7 like - 0 dislike
167 views

The title question would be too long if I tried to specify it clearly. So let me be more clear. Consider the class of theories having the following properties:

  1. The langrangian density is only dependent of scalars created from the curvature tensors (for example $R, R^{ab}R_{ab}, R^{abcd}R_{abcd}, C^{abcd}C_{abcd}, (R)^2, R^{ab}R_{bc}R^{cd}R_{da}$, etc.) + generic matter using the usual minimal coupling.
  2. The field theory is generated from the lagrangian by considering variations in the metric (ie. we will not consider Palatini or torsion theories)
  3. The field equation automatically provides a covariantly conserved stress energy tensor (this means like in GR, the covariant derivative of the Ricci terms are collectively zero just from geometric requirements and not an extra postulate. I've seen papers claim this follows automatically from 1, but I have not seen proof, so I'm specifying it just in case.)
  4. Let's stay with 3+1 dimensions.

My question is then: GR is the theory starting from using $R$ as the lagrangian density, and the vacuum equation in GR reduces to $R^{ab}=0$. Is there another function of curvature scalars in this class which would yield the same vacuum equations?

(For clarity: Yes, for this question I don't care if the field equations differ in non-vacuum. I'm just curious if they can possibly match in vacuum.)

One may be tempted to immediately say it is unique, but consider the theories from $(R)^2$, $R^{ab}R_{ab}$, and $R^{abcd}R_{abcd}$.(1) One might intuitively expect that $C^{abcd}C_{abcd}$ would then add yet another theory, but it turns out this theory can be written as a combination of the previous three. (2)(3) And it turns out that in 4-dimensions, there are actually only two linearly independent combinations of these four theories. (4)

Playing with it to try to find if there is a theory equivalent to GR in vacuum, I was looking for a theory which in vacuum after I contract the indices, I can show the only possible result is R=0 (For an example if the contraction gave a polynomial like $R + R^3 + R^{abcd}R_{abcd}R = 0$, then the only solution with real curvature is $R=0$). And then hoping that applying this to the field equations reduces everything to $R^{ab}=0$. This hunt and peck method is failing me, as I haven't found anything yet and clearly won't let me prove GR is unique if that is the answer.

Can someone give an example which yields the GR vacuum equation or prove mathematically no such theory exists and thus the GR vacuum equation is unique in this class?

This post imported from StackExchange Physics at 2014-12-15 16:08 (UTC), posted by SE-user Edward
asked Mar 20, 2011 in Theoretical Physics by Edward (95 points) [ no revision ]
In other words: "what Lagrangian densities reproduce given equations?". Very nice question! I'd be also interested in knowing the answer for simple matter fields. There it should be quite easy I suppose but I have no time to figure it out right now. Do you have an answer by chance?

This post imported from StackExchange Physics at 2014-12-15 16:08 (UTC), posted by SE-user Marek

All manners of sourceless gravitational wave solutions are also vacuum solutions?

2 Answers

+ 6 like - 0 dislike

You can add a Gauss-Bonnet term $R^2 - 4R^{\mu\nu}R_{\mu\nu} + R^{\mu\nu\rho\sigma}R_{\mu\nu\rho\sigma}$. It's purely topological.

This post imported from StackExchange Physics at 2014-12-15 16:08 (UTC), posted by SE-user QGR
answered Mar 20, 2011 by QGR (250 points) [ no revision ]
Hmm... so following your suggestion, if I take $R^{abcd}R_{abcd} - 4 R^{ab}R_{ab} + R^2$ (which according to en.wikipedia.org/wiki/Generalized_Gauss%E2%80%93Bonnet_theorem is a Euler character density in 4D) and add it to the lagrangian density, it would be equivalent to adding a constant to the total action and thus not have any effect. Is that the idea? Are there only a finite number of such "Gauss-Bonnet terms" in 4D? I don't know that much about them.

This post imported from StackExchange Physics at 2014-12-15 16:08 (UTC), posted by SE-user Edward
+ 3 like - 0 dislike

The vacuum equations are uniquely Einstein equations if

  1.  They are based on a covariant action principle of the geometry (right now, i am actually not sure if torsion is allowed or not, let's say it isn't)
  2. We are in dimension 3 or 4
  3. We only allow second-order derivatives of the metric

The action principle is however unique up to an infinite number of Lovelock terms which in the given dimension integrate out into topological (variation-conserved) terms and do not contribute to the equations of motion at all.

This trivially answers a part of your question, because if different than Lovelock terms are included, higher order derivatives appear and the equations cannot be Einstein. Additionally, new "longitudinal" degrees of freedom appear and the metric can be formally decomposed into Einstein equations with a peculiar scalar or vector source field (see this e.g. in $f(R)$). I haven't really dug into it but I am afraid that all these theories will have problems with causality and well-posedness of the Cauchy problem.

answered Jun 18, 2015 by Void (1,505 points) [ no revision ]

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