# In what sense exactly are the Einstein metrics distinguished?

EDIT: In general relativity given a manifold $$M$$ one can consider a functional on (pseudo-) Riemannian metrics $$g$$ $$\int_M R\,\, dvol_g,$$ where $$R$$ is the scalar curvature and $$vol_g$$ is the (pseudo-) Riemannian measure. The extremal metrics for this functional (solutions of the Euler-Lagrange equation) satisfy the so called Einstein equation.

As far as I heard the distinguished property of the above functional is that $$R$$ involves second derivatives of the metric, but the Euler-Lagrange equation is a differential equation not of the fourth order in the metric, but only of the second order in the metric.

I am looking for a precise statement within what class of functionals on metrics the above property distinguishes the scalar curvature $$R$$. Say assume I have a generally covariant expression involving at most second order derivatives of the metric such that the Euler-Lagrange equation is also of second order in the metric. Does it imply that this expression is proportional to $$R$$?

• I'm not sure what you mean by "this statement". It would help to indent the claim that you're asking about, and leave the commentary and unproblematic claims unindented.
– user44143
Dec 15, 2020 at 16:28
• Reading between the lines the question seems to be about the apparent mismatch in the number of derivatives in the Einstein-Hilbert action and the number of derivatives in the field equation...? That mismatch is resolved because the "extra" derivatives in the EH action come in a total derivative term. Dec 15, 2020 at 16:41
• @MattF.: Thanks. Corrected.
– asv
Dec 15, 2020 at 16:43
• @AlexArvanitakis: I need more. I made it more clear now.
– asv
Dec 15, 2020 at 16:43

If I understood your question correctly, the answer indeed is due to Lovelock. I think it's important to state all the hypotheses clearly, because they are not always reported accurately.

Theorem. (Lovelock, 1971) Given a metric $$g_{ab}$$ and a covariantly constructed symmetric 2-tensor $$T^{ab}(g,\partial g, \partial^2 g)$$ such that $$\nabla_a T^{ab} = 0$$, then necessarily the tensor $$T^{ab} dvol_g = \frac{\delta L}{\delta g_{ab}}$$ is the Euler-Lagrange derivative of the Lagrangian density $$L = \sum_k \alpha_k R_{[a_1 a_2}{}^{a_1 a_2} \cdots R_{a_{2k-1} a_{2k}]}{}^{a_{2k-1} a_{2k}} dvol_g$$, where $$R_{abcd}$$ is the Riemann curvature tensor, $$[{\cdots}]$$ denotes full antisymmetrization, and $$\alpha_k$$ are arbitrary constants (their values determine $$T^{ab}$$).

The number of terms in $$L$$ is finite in any given dimensions, because antisymmetrizing over more indices than the dimension always gives zero. In 4 dimensions, the only possibilities are $$L = (\alpha_0 + \alpha_1 R + \alpha_2 R_{abcd} R^{abcd}) dvol_g$$.

If $$T^{ab} dvol_g = \delta L/\delta g_{ab}$$, for any covariantly constructed Lagrangian $$L$$, then $$T^{ab}$$ is obviously symmetric and $$\nabla_a T^{ab}=0$$. So Lovelock's theorem actually classifies all Lagrangians that depend depend covariantly on the metric and Riemann curvature (including its derivatives) that have second order Euler-Lagrange equations, up to null Lagrangians, those whose Euler-Lagrange equations are identically zero. In fact, in 4 dimensions, a particular choice of the constants with $$\alpha_2\ne 0$$ gives such a null Lagrangian, the so-called Gauss-Bonnet term. So, among 4-dimensional Lagrangians, by adding a null Lagrangian, one can always reduce the desired variational principle to the Einstein-Hilbert form, $$L= (\alpha_0 + \alpha_1 R) dvol_g$$.

Lovelock, D., The Einstein tensor and its generalizations, J. Math. Phys. 12, 498-501 (1971). ZBL0213.48801.

• In the definition of $L$ did you really mean that both upper and lower indices are the same- $a_i$?
– asv
Dec 15, 2020 at 17:48
• @makt Yes, it means that they are contracted with the metric, by the Einstein summation convention. Dec 15, 2020 at 17:58

OK I think I understand the question now.

You want to look at Lovelock gravities.

These are the most general theories (not bothering with a list of qualifiers here, see wikipedia) that produce 2nd order EOMS.

For dimensions d=3 and d=4 the only Lovelock functional is indeed the Einstein-Hilbert one.

• Doesn't the general Lovelock EOM involve second order derivatives of the curvatures (and so 4th order on the metric)? I guess it is not clear to me when the OP wrote that the ELE "is also of second order" whether "second order on the metric" is meant. Dec 15, 2020 at 17:10
• @WillieWong: Yes, I meant "second order in metric". Corrected.
– asv
Dec 15, 2020 at 17:17