Gauge Theory of Gravity – Understanding Lagrangian for Gauge Theory of Gravity

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There are a number of questions here discussing gravity as a gauge theory of the Lorentz group. I am trying to find the Lagrangian this gauge produces, and the other discussions stop just short of providing that. For example consider those questions:

https://physics.stackexchange.com/a/127587/747

where the answer states:

Gravity can be seen as a gauge theory of the Lorentz group (which acts on the tangent space). These was pointed out by Kibble and Sciama during the 50s and 60s.

Another relevant question is:

https://physics.stackexchange.com/a/46367/747

Where the author specifically shows the correspondence between a general linear gauge and the Christoffel symbols.


I am just trying to get a bit more complete with these answers and finish the answer to something useable. I assume these theories are suppose to produce a Lagrangian, but all of these questions and answers seem to stop just short of doing that.

Knowing that:

$$
\psi'=g\psi g^{-1} \tag{1}
$$

and

$$
D_\mu = \partial_\mu \psi – [ig A_\mu , \psi] \tag{2}
$$

and

$$
R_{\mu\nu} = [D_\mu,D_\nu] \tag{3}
$$

Can I construct a Lagrangian from this?

Inspired by QED, I previously suggested (as a draft).

$$
\mathcal{S}=\int \bar{\psi} (i\hbar c \gamma^\mu D_\mu – m c^2)\psi-\frac{1}{4} R_{\mu\nu}R^{\mu\nu} \tag{4}
$$

I don't actually necessarily think (4) is correct, but I propose it only to stimulate creativity and to set fix the expectations.

Is there a prescription to construct a Lagrangian from 1,2 and 3? Using the "Yang-Mills method" it produces 4, but it appears this is not what people has in mind when describing gravity as a gauge theory (as I was told in another question). How would you produce the Lagrangian from 1,2 and 3 that satisfy the people who claim it can be understood as a gauge theory of the Lorentz group?

Best Answer

You can write the action in terms of differential forms as \begin{equation} S = \int \epsilon_{abcd} R^{ab} \wedge e^c \wedge e^d \end{equation} Up to a numerical factor, this is equivalent to the Einstein-Hilbert action \begin{equation} S = \int {\rm d}^4 x \sqrt{-g} R = \int {\rm d}^4 x |\det e| R \end{equation} The fact that the Einstein-Hilbert action is linear, not quadratic, in the curvature, is one of the differences between gravity and a Yang-Mills gauge theory.

Incidentally, in differential form language, the cosmological constant contribution to the action is \begin{equation} S = \int \epsilon_{abcd} e^a \wedge e^b \wedge e^c \wedge e^d \end{equation}

Going the other direction, you can try to "increase the number of $R$" in the wedge product, for example \begin{equation} S = \int \epsilon_{abcd} R^{ab} \wedge R^{cd} \end{equation} However, this is a total derivative in 4 dimensions (the Gauss-Bonnet term). In higher dimensions, this kind of pattern of $R$'s wedged with $e$'s give you the Lovelock interactions. In addition to the fact that these terms are trivial in four dimensions, by power-counting you would expect them to be parametrically suppressed relative to the Einstein-Hilbert and cosmological constant terms, below the Planck scale.

There are other higher order interactions you can write down that are invariant under diffeomorphisms and local Lorentz transformations other than the Lovelock terms, that one expects to be generated by loop corrections. Of course, the equations of motion will be higher than second order, since by definition the Lovelock terms are the terms that give you second-order equations of motion. But these terms are expected to be Planck-suppressed and so irrelevant (although people do search for them in observational data).

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