# Klein-Gordon equation coupled to scalar curvature

Consider the Klein-Gordon equation of the form

$$\square_g \psi - m^2 \psi - \xi R \psi \enspace = \enspace 0 \quad .$$

This equation describes the relativistic propagation of a scalar field with mass $$m$$ on a curved spacetime with metric $$g$$. My question now concerns the interpretation of the term which couples the field to the scalar curvature $$R$$. Since the d'Alembertian operator $$\square_g$$ already includes the properties of the curved space on which the wave propagates, why is there any need to couple the equation to the scalar curvature? The d'Alembertian operator should already "cover" the influence of the curvature on the wave, shouldn't it?

E D I T :

Thank you for your answers so far. GHOSTER's answer of "allowing all possible terms" makes of course sense in how to fundamentally justify the occurence of this term in the Klein-Gordon equations most general form. What still confuses me - and this question may only arise due to a lack of basic knowledge in this sector - is the question: what makes the scalar curvature term "possible" in contrast to whatever other term, say e.g. the Kretschmann scalar $$R^{abcd} \, R_{abcd}$$? In other words, what condition does $$R$$ satisfy such that it is considered "possible"? Some condition must be satisfied otherwise all scalar fields on the spacetime would be "possible" and therefore by the principle "allow all possible terms" be added to the equation, wouldn't they?

• They are two different ways that spacetime curvature can influence the scalar field. Neither “includes” the influence of the other. One advantage of the $\xi$ term is that for $\xi=1/6$ the equation has conformal symmetry. Commented Sep 15, 2022 at 20:18
• Note that adding higher-order curvature scalars would require introducing a new scale into the equation. With just $R$, the coefficient is dimensionless. Commented Sep 15, 2022 at 20:22
• Thank you for your quick answer. I understand that $\xi = 1/6$ ensures conformal invariance. However, this being the "only" explanation seems unsatisfactory, since it suggests an approach along the lines of "artificially manipulation the equation until one gets the desired results" (I hope it is somewhat clear what I mean). $\quad$ May I ask you to elaborate on the two different ways that spacetime curvature influences the scalar field. Commented Sep 15, 2022 at 20:27
• I don’t understand what elaboration you are asking for. It seems obvious that in any metric with $R\ne 0$ the solution to the equation with the $\xi$ term is going to be different from the solution without it. Commented Sep 15, 2022 at 20:30
• Why would you not allow the $\xi$ term? Ultimately Nature will tell us whether it should be included or not. Until then, I see no reason to assume that $\xi=0$. Commented Sep 15, 2022 at 20:35

You can put $$\xi=0$$ and you'll get a minimally-coupled massive scalar in curved spacetime. Nothing prevents you from doing that. The reason you usually see the $$\xi R \psi$$ term in the equations of motion is because it is that, beyond minimal coupling, this is the simplest way in which you can couple the scalar to gravity (it comes from a quadratic in $$\psi$$ term in the action $$\sim \xi R \psi^2$$). You are free to add higher dimensional/derivative operators to the action to get a more general (effective field) theory, like $$\frac{1}{\Lambda^2}\xi \psi^2 R^2$$ or $$\frac{1}{\Lambda^8} \xi \psi^2 R^5$$ or $$\frac{1}{\Lambda^4} \xi \psi^2 R \square R$$ or $$\frac{1}{\Lambda^2} \xi \psi^2 R^{\alpha \beta \gamma \beta} R_{\alpha \beta \gamma \beta}$$, etc. Note the high energy mass scale $$\Lambda$$ which must accompany these terms with appropriate powers so that the action remains dimensionless. The point is that these higher order terms are not relevant at low energies ($$< \Lambda$$). Moreover, these negative dimensional couplings make the theory non-renormalizable in general (which is not necessarily a problem, though). See my answer here: https://physics.stackexchange.com/a/467869/133418
There is no theoretical necessity, because as you said, the field is already coupled to geometry through $$\nabla$$ and the action is covariant. There is motivation in various different contexts (e.g., from string theory and theories of quantum gravity or EFT considerations), but some of these arguments are speculative and they don't constitute any real reason that non-minimal coupling is incorrect. If you want to read more, there's a tonne of literature on minimal coupling in different areas (especially inflation).
• I think that edit constitutes a new question, but in short, the reason is that the 'next order' in the EFT treatment only has this $\psi R$ term. Commented Sep 16, 2022 at 20:15