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The Lagrangian of general relativity is linear (first order) in Riemann curvature. The cosmological term, if included, is zero order in Riemann curvature.

From an effective field theory point of view, all symmetry-permitting action terms should be included. Nothing prevents us from adding terms with 2 or more Riemann curvatures.

These high order terms are usually suppressed at low energies so that they are not relevant under normal circumstances. That being said, they could play an important role at the early stages of our universe involving high energy effects.

Are these high order terms rightfully taken into account for the prevalent cosmological models (especially the inflationary model which is supposedly applicable to early universe)?

And for that matter, one can even entertain terms with multiple torsion tensors to account for the spin current's effect (in addition to the energy-momentum current) on space-time for early universe.

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    $\begingroup$ related: Could the full theory of quantum gravity just be a nonrenormalizable quantum field theory?. $\endgroup$
    – AccidentalFourierTransform
    Commented Jun 20, 2018 at 21:03
  • $\begingroup$ It sounds like you're describing f(R) gravity. See, e.g., Sotiriou, arxiv.org/abs/0710.4438 $\endgroup$
    – user4552
    Commented Jun 21, 2018 at 1:53
  • $\begingroup$ f(R) gravity only concerns functions of the scalar R, with Riemann tensor's Lorentz indices contracted with those of the metric tensor. Whereas broader high order terms can contract Lorentz indices between Riemann tensors. And for that matter, one can even entertain terms with multiple torsion tensors. $\endgroup$
    – MadMax
    Commented Jun 21, 2018 at 4:21
  • $\begingroup$ Sure, there are infinitely many curvature scalars that you can construct. You can construct curvature polynomials in the Riemann tensor, and you can also build expressions involving covariant derivatives. But what is the physical motivation for including these in the action, and how would you propose to determine all the dimensionful constants that would appear? From an effective field theory point of view, all symmetry-permitting action terms should be included. Don't you mean "could" rather than "should?" There are infinitely many actions we can write down for any system. $\endgroup$
    – user4552
    Commented Jun 21, 2018 at 12:11
  • $\begingroup$ arxiv.org/abs/1507.08194 $\endgroup$
    – Count Iblis
    Commented Jun 21, 2018 at 16:48

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Classically, if you add any more terms like $R^2$, $R_{\mu \nu} \, R^{\mu \nu}$, $R^3$, etc, to the lagrangian density of general relativity, you'll get field equations that are of higher order than 2 in the metric derivatives. That would imply problems with causality.

The Einstein-Hilbert action (including the $\Lambda$ term) is the only one that give second order field equations for the metric components.

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  • $\begingroup$ Wow, I didn't know this. Why isn't this considered a showstopper for f(R) gravity? $\endgroup$
    – user4552
    Commented Jun 22, 2018 at 20:24
  • $\begingroup$ Because many researchers are looking for a path to quantum gravity, which allows extra terms (non-classical field equations). I'm not a specialist of $f(R)$ theories, but I think they are using some (artificial?) tricks to "stabilize" the theory and/or to prevent the causality problems. That would make an interesting question : how do $f(R)$ theories prevent the causality problems from the higher metric derivatives? $\endgroup$
    – Cham
    Commented Jun 23, 2018 at 1:59
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    $\begingroup$ @BenCrowell: $f(R)$ gravity has a fourth-order equation for the metric, but it turns out to be entirely equivalent to a scalar-tensor theory with two second-order equations of motion. It turns out to be a equivalent to a weird sort of Brans-Dicke theory, which does have some tight experimental constraints, but isn't mathematically pathological. Certain forms for $f(R)$ also give rise to a potential for the scalar that is unbounded below. See Chiba, "$1/R$ gravity and scalar-tensor gravity", Phys. Lett. B 575, 1–3 (2003). $\endgroup$
    – Michael Seifert
    Commented Jun 25, 2018 at 15:14
  • $\begingroup$ @Cham --As a layperson, I'm just curious about what you think of Nikodem J. Poplawsi's "cosmology with torsion" as a description of the extremely early universe. I remembering noticing on some other postings that you're familiar with the Einstein-Cartan Theory that's its basis. I hear its math is bloody murder, but his English-language abstracts and conclusions, in all the papers about it that he's preprinted on Arxiv since 2009, make it sound mechanically simple and plausible. (I'm a total procrastinator, so it's the lack of any need for a beginning to his cosmos that fascinates me.) $\endgroup$
    – Edouard
    Commented Jun 11, 2020 at 4:52
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    $\begingroup$ @Edouard, the ECKS (Einstein-Cartan-Kibble-Sciama) theory is a viable one, and physically sound. It's the natural extension of GR, but it is mathematically very complicated. Still, it's worth the study. $\endgroup$
    – Cham
    Commented Jun 11, 2020 at 14:21
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To date there have not been any cases where general relativity failed to meet the data. There are one or two things people are looking at very carefully with still some puzzled expressions. But, so far as solid data, GR is still champion.

So such considerations would be pretty speculative. As far as we can tell, we don't need any other gravity theory. It can be pretty discouraging.

Not to say it couldn't be interesting. It could be very much so. But you have some mighty big shoes to fill with existing data. Those higher order terms need to be really seriously suppressed for things like solar system data.

A book you might get on the subject is Clifford Will's book about GR confronting the data.

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