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I read several questions, but I can't find the answer to these two

  1. How many non-trivial independent scalar quantities can we extract from a curved spacetime and why just $R$ appears in the action?
  2. Why not a generic function $f(R)$?

I thought about it a lot, but I don't find the reason why just $\mathscr{L}=R+2\Lambda$ and not a much more complicated expression

Qmechanic
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Rob Tan
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1 Answers1

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Although the "Lovelock gravity" reference in the comments probably takes care of this, it is worth pointing out here explicitly that the three quite natural requirements on the action functional $$ S[g]=\int_M L[g] $$of gravity, namely

  1. the Lagrangian must be equivariant with respect to diffeomorphisms;
  2. the metric is the only field that appears in the Lagrangian;
  3. the field equations are second order in the metric;

are quite restrictive. Lovelock's theorem gives the most general such Lagrangian, and it is a certain combination of curvature invariants, but in a four dimenional space, the only possiblity is $$ L[g]=(aR+b)\mu,\quad\mu=\sqrt{\mathfrak g}\mathrm d^4x. $$

For example the so-called Gauss-Bonnet (GB) term is also allowed, but in four dimensions its integral is a topological invariant, so it does not influence the Euler-Lagrange equations at all. In a spacetime whose dimension is larger than four, the GB term gives a genuine generalization of the Einstein-Hilbert action.

To answer one of OP's question directly, namely

Why not a generic function $f(R)$?

Such a generic function will give a fourth order field equation for gravity. Also linked in the comments are references to $f(R)$ gravity, which is a modification of GR that is actively researched, so people do consider such theories.

We generally however want to avoid higher order theories for two reasons. One is that Newtonian gravity is second order, and the other is that higher order theories tend to suffer from certain instabilities called Ostrogradski ghosts.

Neither problem is completely exclusionary, for example higher derivative parts tend to be suppressed at low energies, so they can still give acceptable Newtonian limits and Ostrogradski instabilities may be avoided if the Lagrangian is sufficiently degenerate, which is pretty much always the case for gravitation theories (the degeneracy, not the ghost avoidance). One must then do a case-by-case investigation via eg. the Dirac-Bergman formalism to determine whether the theory is still unstable.

Bence Racskó
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    +1. I think it might be interesting to add this comment that technically the action of GR is a little bit different, as one needs to remove a boundary term that occurs due to the Ricci scalar – Níckolas Alves Aug 14 '22 at 20:43
  • @NíckolasAlves That's true, but Lovelock's theorem is a classification of appropriate invariant Lagrangians, so it does not know anything about such boundary terms, hence why I thought to neglect to mention it. In fact Lovelock's theorem is a classification of divergenceless degree $2$ symmetric tensors that are constructed out of the metric $2$-jets, i.e. it classifies EL equations, rather than Lagrangians. The boundary term becomes relevant if one cares about boundary conditions but if the action principle is interpreted "formally", then the boundary term may be neglected anyways. – Bence Racskó Aug 15 '22 at 09:22
  • Perfectly fair. When mentioning the boundary term, I didn't mean to imply you should edit the answer nor was I correcting you, but rather I wanted to leave the comment for future readers. – Níckolas Alves Aug 15 '22 at 12:27
  • @NíckolasAlves Ah cool, I misread your comment and thought you suggested an edit. – Bence Racskó Aug 15 '22 at 19:40