# Gauge transformations in gravity [duplicate]

The Maxwell equations are invariant under the transformation

$$A_{\mu} \rightarrow A_{\mu} - \dfrac{1}{e}\partial_{\mu}\alpha(x)$$

where $\alpha(x)$ is a phase transformation varying from point to point. The Maxwell Lagrangian can be coupled to a scalar field Lagrangian by stipulating that the scalar field remains invariant under the local phase transformation, and therefore a covariant derivative needs to be defined in order to properly effect the transformation instead of the ordinary derivative given by

$$D_{\mu} \phi(x) = \partial_{\mu} \phi(x) + ieA_{\mu}(x)\phi(x)$$ with the $A_{\mu}$ transforming again as

$$A_{\mu} \rightarrow A_{\mu} - \dfrac{1}{e}\partial_{\mu}\alpha(x)$$

Thus the combined Lagrangian can be written as a gauge invariant function $$L = L_{Maxwell} + L_{scalar}$$. A similar procedure can be done for the general class of Yang-Mills theories.

1. Suppose if we take only the Einstein - Hilbert Lagrangian, are there gauge transformations on $g_{\mu\nu}$ which leave the Lagrangian invariant in the similar sense as above? (Also I've heard that coordinate invariance constitutes some sort of gauge invariance, I'm not sure how, and whether this sort of gauge transformation is the answer to my problem)

2. Can I construct and couple gauge invariant scalar fields like in the above example to the Einstein-Hilbert action? How do I do so?

EDIT : Later I found this http://web.mit.edu/edbert/GR/gr5.pdf to be particularly useful to understand diffeomorphism invariance.

• – ACuriousMind Feb 16 '16 at 7:26
• @ACuriousMind TY for suggestions to both questions.. :) – Bruce Lee Feb 16 '16 at 7:28
• In a way the GR analogue of gauge transformations are general coordinate transformations (which leave the metric intact, due to it being a tensor). – Sebastian Riese Feb 18 '16 at 22:01

1) Of course, there are such transformations, called isometrical transformatooms: $$g_{\mu\nu}\to g_{\mu\nu} + D_{\mu}\epsilon_{\nu} + D_{\nu}\epsilon_{\mu}$$ In some sense they are gauge (unphysical) transformations too. The metric plays the same role for curvature (Riemann tensor) as the 4-potential plays for EM strength tensor. In fact, due to such gauge invariance, the metric tensor has only two independent "physical" components, as well as 4-potential. This is related to the fact that for linearized theory the metric represents massless states with helicity two (in fact, the true tensor which represents these states is the Weyl tensor, but here this is not important).
• 1. The transformation you write is an isometry only if $\epsilon$ is a Killing vector. 2. To say the metric is to the curvature as the four-potential is to the field strength is wrong, and you even say the correct thing later on: The Christoffels are the gauge fields/potentials of GR, and they only become proper gauge fields if you consider them as independent of the metric as in the Palatini formalism. 3. The diffeomorphism invariance of GR is not a full gauge invariance since there are gauge transformations of the tangent bundle that are not induced by diffeomorphisms. – ACuriousMind Feb 16 '16 at 7:26