# Minimal vs. Non-minimal coupling

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What is the difference between Minimal vs. Non-minimal coupling in General Relativity? A brief introduction to Minimal Coupling in General Relativity could be useful too.

This post imported from StackExchange Physics at 2014-04-18 05:47 (UCT), posted by SE-user user38032
retagged Apr 19, 2014
I think its best to not get to caught up in this question in the specific context of GR. That is, look at the question in the more general context of gauge theories and effective field theory. For a discussion, see the reference: arxiv.org/abs/1305.0017

This post imported from StackExchange Physics at 2014-04-18 05:47 (UCT), posted by SE-user DJBunk

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I'll give as example the case of a scalar field. We assume the Einstein-Hilbert action: \begin{equation} S_{\text{grav}}=\int d^4x\ \sqrt{|g|}R \end{equation} Now, we would like to consider a quantum field in the spacetime: $$S=S_{\text{grav}}+S_{\text{matter}}+S_{\text{coupling}}$$ To first order in the curvature, the only scalar that couples gravity and the quantum field, which we can build out of a scalar field $\phi$ and curvature-related tensor objects, is $R \phi$. In general, there will be higher order terms if one considers higher energies. We will take the standard Lagrangian for a scalar field. The total action is now \begin{equation} S=\int d^4x \sqrt{|g|}\bigl(R+\frac{1}{2}g^{\mu\nu}\nabla_\mu\phi\nabla_\nu\phi+m^2\phi^2+\xi R\phi\bigr) \end{equation} Where $\xi$ is the coupling constant. Minimal coupling amounts to setting $\xi=0$. As you can imagine, this is the simplest (and perhaps most natural?) case. Another reasonably popular choice seems to be $\xi=\frac{1}{6}$. In this case, we say that the field is conformally coupled to gravity, because the action is now invariant under conformal transformations of the metric: $$g_{\mu\nu}\rightarrow \Omega^2(x)g_{\mu\nu}$$ Any possible $\xi\neq 0$ is a case of non-minimal coupling. Basically, minimal coupling means avoiding introducing any extra terms in the action.

This post imported from StackExchange Physics at 2014-04-18 05:47 (UCT), posted by SE-user Danu
answered Mar 17, 2014 by (505 points)
Danu, could please explain the difference between setting $\xi=0$ and $\xi\neq0$

This post imported from StackExchange Physics at 2014-04-18 05:47 (UCT), posted by SE-user user38032
Well, obviously the difference is whether the field directly couples to gravity or not.

This post imported from StackExchange Physics at 2014-04-18 05:47 (UCT), posted by SE-user Danu
Donu, What is the physical interpretation for a field to directly couple to gravity.

This post imported from StackExchange Physics at 2014-04-18 05:47 (UCT), posted by SE-user user38032

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