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.

  • 1
    $\begingroup$ 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 $\endgroup$ – DJBunk Apr 6 '14 at 18:21
  • $\begingroup$ There are many reasons as why should consider non-minimal coupling. they are discussed in this paper $\endgroup$ – Nikey Mike Feb 19 '16 at 19:20

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 $R \phi^2$. 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+\xi R)\phi^2\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.

  • $\begingroup$ Danu, could please explain the difference between setting $\xi=0$ and $\xi\neq0$ $\endgroup$ – user38032 Mar 17 '14 at 17:13
  • $\begingroup$ Well, obviously the difference is whether the field directly couples to gravity or not. $\endgroup$ – Danu Mar 17 '14 at 18:43
  • $\begingroup$ Donu, What is the physical interpretation for a field to directly couple to gravity. $\endgroup$ – user38032 Mar 18 '14 at 1:50
  • $\begingroup$ Danu, the coupling that one usually finds in these non-minimally coupled inflation models is $R\phi^2$ and not $R\phi$ (see e.g. Wikipedia). Is that what you meant? And if yes, why does one not consider the linear field coupling to the Ricci scalar, can it be absorbed somehow? $\endgroup$ – physicus Apr 3 '15 at 21:51
  • $\begingroup$ @physicus You're absolutely right! I think a linear coupling would probably yield a pathological theory because the classical Hamiltonian seems to be unbounded from below. It is also wholly possible that a linear term such as this can be made to disappear through a field redefinition somehow, although I don't immediately see how. $\endgroup$ – Danu Apr 3 '15 at 22:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.