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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?

Octavius
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  • 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. – Ghoster Sep 15 '22 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. – Ghoster Sep 15 '22 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. – Octavius Sep 15 '22 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. – Ghoster Sep 15 '22 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 $\xi=0$. – Ghoster Sep 15 '22 at 20:35
  • @Octavius This is an example of non-minimal coupling, and there isn't really any physical requirement to study this, it's just an alternative type of theory. From what we appear to observe, this type of coupling doesn't occur in nature, but it's interesting nonetheless. – Eletie Sep 15 '22 at 20:36
  • @Eletie What observations make you think it doesn’t occur? – Ghoster Sep 15 '22 at 20:39
  • @Ghoster Because to the best of my knowledge, no physical theories require it? (I.e. there is no observational evidence that seems to prefer non-minimal coupling). – Eletie Sep 15 '22 at 20:50
  • @Ghoster In hindsight my statement that it 'doesn't occur in nature' was too strong. I'd probably rephrase it to something along the lines of, we're not sure, but there currently isn't strong evidence for it. – Eletie Sep 15 '22 at 20:52
  • @Octavius The principle of “allow all possible terms” up to whatever energy scale you’re working is a hallmark of effective field theory. Regardless of what is going on at, say, the Planck energy, the low-energy approximation to that theory is probably going to have all possible low-energy terms. It’s another way in which the quantum aphorism “everything which is not forbidden is mandatory” applies. – Ghoster Sep 15 '22 at 21:03
  • @Ghoster I updated the question, please have a look at the "Edit" paragraph. – Octavius Sep 15 '22 at 21:28
  • The field equation must come from a Lorentz-invariant action. But there are lots of curvature invariants. The action could have terms with $\phi^2$ multiplied by $R^2$, $R^{;\mu}R_{;\mu}$, $R^{\mu\nu}R_{\mu\nu}$, etc. and higher powers up to infinity, but on dimensional grounds their coefficients would be dimensionfull, with negative powers of the Planck mass. At curvatures less than the Planck scale, such terms would be negligible compared to the $R$ term. There is only one curvature invariant involving no more than two derivatives of the metric, and that is $R$. – Ghoster Sep 15 '22 at 21:57
  • Note: Major edits to your question after getting an answer from Eletie are not appropriate. It’s better to ask a new question. – Ghoster Sep 15 '22 at 22:05

2 Answers2

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You ask whether there is any need to introduce this non-minimal coupling, and the answer depends what you mean by 'need'. If you mean, is there some reason it must be there, then I'd say no. You can look at this both from a mathematical consistency point of view.

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).

And, as far as I'm aware, there's no physical necessity from observational evidence. Depending on the specific context and type of data you're looking it, it may or may not be favoured over the standard minimally coupled models. But there is not hard evidence to indicate it's presence in nature. For a bit more about observational constraints, here's an example https://arxiv.org/abs/1511.08736.

Eletie
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  • Thank you for your answer. When the primary reason to include this term is because of generality considerations, then why are "other terms" not included? (see the updated question in the paragraph "Edit") – Octavius Sep 15 '22 at 21:30
  • 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. – Eletie Sep 16 '22 at 20:15
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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

Avantgarde
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