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In a new article, Cosmological constraints on slow-roll inflation, the authors write: "Notice that this theory is said to be “minimal coupled to gravity” because there is not a direct coupling between the inflaton field and the metric tensor in the action ".

How to understand this? That the inflaton field does not interact with the gravitational field in any way, or is this interaction so small that it can be neglected? After all, if the inflaton field has energy-momentum, it must interact gravitationally.

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Minimal coupling means that couplings of the matter part of the action with curvature invariants or non-minimal derivative couplings are abscent, so matter "'feels" gravity only through the volume element $d^4x\sqrt{-g}$ in the action. In the action of the paper $(1)$, there exists only the kinetric term for the scalar field $g_{ab}\partial^{a}\phi\partial^{b}\phi$ and also the potential $V(\phi)$. There exists no term that directly couples $\phi$ to curvature such as $R\phi^2, R^{ab}R_{ab}\phi^2$ or couplings of the scalar field with the Gauss-Bonnet invariant for example see Gauss-Bonnet Inflation. For combined non-minimal and derivative couplings check Successful Higgs inflation from combined nonminimal and derivative couplings, where in their action the kinetric term of the scalar field is modified $G(\phi)G_{ab}\partial^{a}\phi\partial^{b}\phi$ and also a non-minimal coupling term exists $Rf(\phi)$, see their action $(30)$.

For a very interesting discussion about what exactly is the "minimal coupling" i suggest these notes http://www.blau.itp.unibe.ch/newlecturesGR.pdf pages 167-181!

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    $\begingroup$ Thanks for the answer. I guess I misunderstood the quote from the article. The gravitational field and the inflaton field interact. $\endgroup$ Oct 7, 2021 at 9:40
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Suppose a field $\phi$ has a Langrangian $\mathcal{L}_{\phi}$ without any coupling to gravity specified in this Lagrangian. When you write down the action of $\phi$ in a general space-time you use $$S_{\phi}=\int d^4x\sqrt{-g}\mathcal{L}_{\phi},$$ where $g$ is the determinant of the metric tensor. This pops up so that the measure $d^4x\sqrt{-g}$ in the integral is Lorentz-invariant. It turns out that this way to write the action couples the field $\phi$ minimally to gravity, since if you now want to find the equations of motion, you vary the action with respect to the field and say that the action must be stationary when the field fulfills the equation of motion: $$\frac{\delta S_{\phi}[\phi(x)]}{\delta \phi(y)}\overset{!}{=}0$$(This is a functional derivative.) But since the metric is also, in general, space and time dependent, the equations of motion of the field will depend on the metric (because when calculating the functional derivative, derivatives of the metric will pop up. For example in cosmology, they will depend on the Hubble factor). This means, the field is sensitive to the evolution of the metric, BUT the metric is not sensitive to the evolution of the field. (The dynamics of the metric has to be already known or found elsewhere but does not depend on the equation of motion for $\phi$.) This can be done if the energy density in the field is negligible, or if you account for the energy density of the field when finding the evolution of the metric tensor (what one does in cosmology when finding the evolution of the scale factor by looking at the energy density of the universe's constitutents).

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  • $\begingroup$ Thanks for the answer. I guess I misunderstood the quote from the article. The gravitational field and the inflaton field interact. $\endgroup$ Oct 7, 2021 at 9:06
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"Minimal coupling" means that there's no term of the form $\phi^2 R$ in the Lagrangian density. That's all.

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  • $\begingroup$ That is, there is an interaction between the inflaton field and the gravitational field? $\endgroup$ Oct 7, 2021 at 8:51

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