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This question already has an answer here:

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.

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marked as duplicate by ACuriousMind, Danu, Sebastian Riese, Daniel Griscom, John Rennie general-relativity Feb 19 '16 at 7:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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).

2) The role of the gauge field in GR elongated derivative plays the Christoffel symbol. The only difference is that GR transformation is explicit coordinate transformation which depends on the nature of field (vector, etc), while the EM gauge transformation is transformation in the isotopical space, for which the nature of the field isn't important. For example, we do not need to elongate the derivative of the scalar field in GR, but we have to elongate in in gauge invariant EM theory.

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    $\begingroup$ 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. $\endgroup$ – ACuriousMind Feb 16 '16 at 7:26

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