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 $R$?

  • 1
    $\begingroup$ 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. $\endgroup$
    – user44143
    Dec 15, 2020 at 16:28
  • $\begingroup$ 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. $\endgroup$ Dec 15, 2020 at 16:41
  • $\begingroup$ @MattF.: Thanks. Corrected. $\endgroup$
    – asv
    Dec 15, 2020 at 16:43
  • $\begingroup$ @AlexArvanitakis: I need more. I made it more clear now. $\endgroup$
    – asv
    Dec 15, 2020 at 16:43

2 Answers 2


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.

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

  • $\begingroup$ 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. $\endgroup$ Dec 15, 2020 at 17:10
  • $\begingroup$ @WillieWong: Yes, I meant "second order in metric". Corrected. $\endgroup$
    – asv
    Dec 15, 2020 at 17:17

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