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  Akin to gauge field, why GR's lagrangian is not $R_{abcd}R^{abcd}$? What's the mathematical or physical meaning of $R_{abcd}R^{abcd}$?

+ 2 like - 0 dislike
For gauge field theory, the Lagrangian of the gauge field is $$\mathcal{L}=-\frac{1}{4}\mathrm{tr}(\mathcal{F}_{\mu\nu}\mathcal{F}^{\mu\nu})=-\frac{1}{8}F_{a\ \mu\nu}F^{a \ \mu\nu}$$ The field strength $F^a_{\phantom\a \mu\nu}$ where $\mu\nu$ is the coordinate index and $a$ is the fiber index. So analogous to the gauge field, $R_{abcd}$ where $ab$ is the fiber index and $cd$ is the coordinate index. And akin to the Lagrangian of gauge field, the Lagrangian of gravity should be $R_{abcd}R^{abcd}$. While it is actually the Einstein-Hilbert action $R$. My questions are: 1. What's the mathematical or physical meaning of $R_{abcd}R^{abcd}$? 2. Why gravity's Lagrangian is not $R_{abcd}R^{abcd}$? If some field's Lagrangian is $R_{abcd}R^{abcd}$, what's the physical properties of this field?
asked Dec 15, 2014 in Theoretical Physics by Alienware (185 points) [ revision history ]
recategorized Dec 15, 2014 by Alienware

The term arises upon renormalization of quantum gravity, as one of the simpler ones of infinitely many counterterms needed for this nonrenormalizable theory. 

2 Answers

+ 6 like - 0 dislike

1. The first question, namely, the physical meaning of \(R_{abcd}R^{abcd}\) was correctly pointed out in the other answer here, it appears in Gauss-Bonnet gravity.

2. But I cannot agree that gravity is not a gauge theory. Actually, there exists many approaches to GR as a gauge theory of the gravitational field. It started already with Utiyama in 1956, and then followed a crowd of people trying to formulate GR as a gauge theory, like Trautman, Kibble, Sciama, Feynman, Weinberg and Thirring. A complete list of references can be found here: https://arxiv.org/pdf/1210.3775v2.pdf

3. One cannot confuses a gauge theory as being necessarily a Yang-Mills theory. A Lagrangian of the form \(\mathcal L = \frac{1}{2}tr(D\omega\wedge \star D\omega)\) is typical of a Yang-Mills theory, but a gauge theory is a much more general framework. Y-M is a kind of gauge theory. To be a gauge theory, what one needs is a p-bundle where the potential \(\omega\) of the theory is a 1-form that takes its values on the p-bundle's Lie group, the field strength is the exterior covariant derivative \(\Omega = D \omega\) of that 1-form, namely, the associated curvature 2-form, and the Lagrangian of the theory being any combination of \(\omega\), \(\Omega\), \(\wedge\) and \(\star\), i.e., a 4-form \(\mathcal L = \mathcal L(\omega,D\omega)\in \bigwedge^4\mathcal M\), it is necessarily going to be gauge invariant, since nowhere we have introduced a local trivialization!

So how can we describe Einstein's GR in terms of a gauge theory? We choose our base manifold as spacetime \(\mathcal M\), as gauge group the group of general linear transformations \(GL(4,\mathbb R)\), acting on the frame bundle of \(\mathcal M\). The p-bundle of GR is therefore the bundle of frames \(F(\mathcal M)\) (the construction of the frame bundle and the action of \(GL\) is performed in all standard references, e.g., Choquet-Bruhat or Nakahara). Let \((e_a)\in \bigwedge^1\mathcal M\) be a tetrad in the spacetime (an orthonormal coframe), and \(R_{\mu \nu}\) be the curvature 2-form. Substituting Cartan's structure equation \(R_{\mu \nu} = d\omega_{\mu \nu}+\omega_\mu ^{.\alpha}\wedge \omega_{\alpha \nu}\) in the Einstein-Hilbert Lagrangian \(\mathcal L = \frac{1}{2} R\star1\), one shows (take this as an execise!) that the action of GR can be written like \(S=\int R_{ab}\wedge \star (e^a\wedge e^b)\). This is just the gauge theoretical reformulation of GR as the gauge theory with group \(GL\).

I recommend you to give a look in





answered Dec 4, 2016 by Igor Mol (420 points) [ no revision ]

General relativity is a gauge theory in the vague sense that its space of dynamical variables has a natural quotient description (eg, in the metric formulation, space of riemannian metrics divided by the action of the group of diffeomorphisms, to compare to the "usual gauge theory": space of connections on a principal bundle divided by the action of the group of gauge transformations) but it is not a gauge theory in the precise technical sense of a theory formulated in terms of a connection on a principal bundle and of other fields which are sections of associated vector bundles. The tetrad formulation of general relativity indeed contains as variable a connection on a principal bundle, which is certaintly of "gauge-theoretic nature", but also (as correctly written in the answer) a tetrad which can not be written as a section of an associated vector bundle because of the invertibility condition it has to satisfy.

I am ready to agree that my remark is more about terminology than physics. Something physical: at a tree-level quantum perturbative expansion around a given background, a "usual gauge theory" describes spin 1 massless particles whereas general relativity describes spin 2 massless particles. The two situations are certainly different.

In the first link I added above, you can see a paper by Wallner where he tries to make the gravitational potentials very much alike a connection on a p-bundle with the following "trick." You let \((b_a)\) be any basis for \(\mathbb R^4\) and define \(\mathcal G = e^a\otimes b_a\in \bigwedge^1M\otimes\mathbb R^4\). Then he wrote the Einstein-Hilbert Lagrangian like \(\mathcal L = \frac{1}{2} tr(\mathcal G\wedge \star\mathcal G)+...\) so that \(\mathcal G\) can be regarded as a 1-form connection on the p-bundle whose Lie group is that of the affine translations, namely, \(T(4,\mathbb R)\). It looks artificial indeed, but now GR is formulated as a gauge theory in the trivial p-bundle \(\mathcal M\times T(4,\mathbb R)\) instead of the frame bundle, but at least formally, I think it is a gauge theory as the connection now can be seen as a section on the associated vector bundle. Of course, these are all classical considerations...

I would say the characteristics of a general gauge theory is that its phase space is not a cotangent space since the description by an action principle contains redundant degrees of freedom. This is the view taken in Thirring's treatise of mathematical physics (see, e.g., Vol. 2, remark 4.2.2): ''theories that are invariant under position-dependent transformations are called gauge theories, opf which gravitational theory is an example'' - It is also the view taken in https://en.wikipedia.org/wiki/Gauge_theory.

+ 2 like - 0 dislike

Because General Relativity is not a gauge theory. The Lagrangian of General Relativity is the Ricci Scalar R, which is the trace of \(R_{abcd}\), or \(g^{ab}g^{cd}R_{abcd}\).

As for the presence of terms like \(R^{abcd}R_{abcd}\), the term can be found in Gauss-Bonnet gravity, see e.g. the question Is the GR vacuum equation unique? Note that the term is purely topological in 4 spacetime dimensions. As to the properties of a field whose Lagrangian is just \(R^{abcd}R_{abcd}\), that sounds like an interesting calculation to do. I'll try it an update the answer (or write another if it's interesting enough).

Edit: Okay, before trying out anything, a quick search revealed this as the variation of an action whose Lagrangian is \(R_{abcd}R^{abcd}\):

\[- \frac12 \sqrt{g} g_{\mu\nu}R_{abcd} R^{abcd} +\sqrt{g} R_{( \mu}{}^{bcd} R_{\nu ) bcd} + \ldots\]

See this PhysicsForums thread for the source. I haven't verified the equation yet, but this yields a first-order approximation of this form. 

\[R_{( \mu}{}^{bcd} R_{\nu ) bcd} - \frac12 g_{\mu\nu}R_{abcd} R^{abcd} \approx 0\]

answered Dec 15, 2014 by dimension10 (1,955 points) [ revision history ]
edited Dec 15, 2014 by dimension10

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