Graduate Differences between Actions in Curved Spacetime

[Haorong Wu]

TL;DR

I have found two different actions of scalar field in curved spacetime. I am not sure their differences.

First, in Anastopoulos C, Hu B L. A master equation for gravitational decoherence: probing the textures of spacetime[J]. Classical and Quantum Gravity, 2013, 30(16): 165007. , the Einstein-Hilbert action is used to analysis a quantum matter field interacting with the gravitational field, $$S=\frac 1 \kappa \int d^4 x \sqrt{-g} R + \int d^4 x \sqrt {-g} (-\frac 1 2 g^{\mu \nu} \nabla_{\mu} \phi \nabla_{\nu} \phi-\frac 1 2 m^2 \phi^2) .$$

Then, in Spacetime and geometry by Sean M. Carroll, section 9.4, the quantum field theory in curved spacetime consider the following Lagrange density $$L=\sqrt {-g} (-\frac 1 2 g^{\mu \nu} \nabla_{\mu} \phi \nabla_{\nu} \phi-\frac 1 2 m^2 \phi^2 -\xi R \phi^2) .$$

It appears that in the first paper, the curvature scalar $R$ does not couple to the scalar field, while the one in the second case does.

Are the two actions/Lagrangians describe different situations?

 

[samalkhaiat]

Summary:: I have found two different actions of scalar field in curved spacetime. I am not sure their differences.
S=\frac 1 \kappa \int d^4 x \sqrt{-g} R + \int d^4 x \sqrt {-g} (-\frac 1 2 g^{\mu \nu} \nabla_{\mu} \phi \nabla_{\nu} \phi-\frac 1 2 m^2 \phi^2) .L=\sqrt {-g} (-\frac 1 2 g^{\mu \nu} \nabla_{\mu} \phi \nabla_{\nu} \phi-\frac 1 2 m^2 \phi^2 -\xi R \phi^2).

In 4 dimensions and for m = 0, the first action is not conformal invariant while the second one is invariant for the specific value \xi = \frac{1}{6}. Also, the case \xi = 0 is called minimally coupled.

Are the two actions/Lagrangians describe different situations?

Yes they are, for you can convince yourself by deriving the equations of motions and the energy-momentum tensors from both actions.