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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?

Qmechanic
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Anon21
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    Your Lagrangian isn’t a scalar density so it can’t be integrated over spacetime. In Yang-Mills theory, we use a fixed spacetime metric (usually Minkowski) to form a scalar density out of it (this is just multiplication by $d^4 x$ in Galilean coordinates). In gravity there is no background metric because gravity is itself described by a dynamical metric. For example, the Einstein-Hilbert Lagrangian contains the $\sqrt{\det g}$ multiplier which makes it a scalar density. Your theory doesn’t have a value that is a scalar density unless you add the tetrad field too – Prof. Legolasov Nov 04 '21 at 21:15
  • @Prof.Legolasov you are talking about this part $\frac{1}{4} R_{\mu\nu}R^{\mu\nu}$... its not a scalar? So I need to multiply it with a tetrad $| \det e | e_a^{\mu} a_b^{\nu} R_{\mu\nu}$ instead to make it scalar? What about the kinetic energy part... is that fine or do I need to remove that completely? – Anon21 Nov 04 '21 at 21:21
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    @Anon21 it is a scalar which is exactly the issue. You can’t integrate scalars over a manifold, only scalar densities. There is nothing in your field content that transforms as a density and doesn’t lead to a trivial or pathological theory (one term that does transform as a density is $\varepsilon_{\mu \nu \sigma \tau} \text{tr} F^{\mu \nu} F^{\sigma \tau}$, note that $\varepsilon$ is not a tensor which is precisely what makes the term a density; but it leads to a trivial theory because this term is in fact a total derivative). $\det e$ is how GR gets a density term – Prof. Legolasov Nov 04 '21 at 21:33
  • @Prof.Legolasov I am a bit surprised that it works with GL(4,R), because I intuitively expected translations to also matter? For example, why do I not need the poincare group, or the affine group T(4)xGL(4,R) to derive EFE from the gauge approach? Do I miss something important by not adding translation... it's seems I don't because it is claimed that I get the EFE... but my god is it counter-intuitive to me. Can you shed some light on this please? – Anon21 Dec 04 '21 at 13:59
  • it isn't true that symmetries of SR are "gauged" to become symmetries of GR. The relationship between SR and GR is much more elaborate than the relationship between the global phase shift symmetry and gauged phase shift symmetry with electromagnetism. From the discussions we had earlier I get the impression that you have a good understanding of SR-compatible field theories, but almost no understanding of GR. Perhaps you could try reading up on GR? I couldn't possibly fit the answer to your question in a comment, but I perhaps could in an answer. Ask another question and ping me maybe? – Prof. Legolasov Dec 04 '21 at 14:45

1 Answers1

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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).

Andrew
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  • So it just gives the Einstein-Hilbert action. That makes sense. I am not sure why I was expecting something different... Thanks for the response. Much appreciated! – Anon21 Nov 05 '21 at 13:36
  • I am a bit surprised that it works with GL(4,R), because I intuitively expected translations to also matter? For example, why do I not need the poincare group, or the affine group T(4)xGL(4,R) to derive EFE from the gauge approach? Do I miss something important by not adding translation... it's seems I don't because it is claimed that I get the EFE... but my god is it counter-intuitive to me. Can you shed some light on this please? – Anon21 Dec 04 '21 at 14:03
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    @Anon21 The vielbiens $e^a$ are the "gauge fields" associated with translations, and the spin connections $\omega^{ab}$ are the fields associated with Lorentz transformations. The "curvature" of the vielbiens $d e^a + \omega^{ab}\wedge e^b$ is the torsion (which is zero in ordinary GR without spinors) and of the spin connection $d \omega^{ab} + \omega^{ac} \wedge \omega^{cb}$ is the Riemann curvature. – Andrew Dec 04 '21 at 16:22
  • What is the interpretation of the gauge approach in this context... Say I start with a manifold M, I assign a frame field to each point of M, then demand gauge invariance with respect to a general linear transformation of said frame field - What happens physically? Does applying a general linear transformation, or a sequence of, to the frame field transform the manifold into a new manifold... ? Or is it just a change of coordinate... thus simply a statement on what the observer experiences whilst leaving the manifold intact? Does anything in there transforms the manifold? – Anon21 Dec 06 '21 at 04:17
  • Is gauge invariance with respect to general linear transformations of the frame field simply a fancy way to say "invariant with respect to change of coordinates"? – Anon21 Dec 06 '21 at 21:20
  • What is the difference between gauge the general linear group versus the Lorentz group. My understanding is that they both produce the EFE. But is one version more general than the other, or do they reduce somehow to the same EFE? – Anon21 Dec 16 '21 at 18:28