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I read the following statement in one of Penrose's paper

zero rest-mass field equations can, with suitable interpretations, be regarded as being conformally invariant.

I take this to imply that if I would like to describe massless scalar fields (for example) in curved spacetimes I should couple them conformally. More precisely, the curved space generalization of the action is $$ - \phi \partial^2 \phi \to - \phi \left( \nabla^2 + \frac{1}{6} R \right) \phi ~~~(d=4) $$ instead of the naive $\partial^2 \to \nabla^2$. Why is this the case?

EDIT: I believe this holds for massive scalar fields as well, though we no longer have conformal invariance.

Danu
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Prahar
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1 Answers1

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With the choice of $R/6$ term in four dimensions, the action is invariant under conformal transformations of the metric $g_{\mu\nu} \rightarrow \Omega(x)g_{\mu\nu}$. Such actions are of interest, because if the spacetime itself is conformally flat (for example, spatially flat FRW metric), then the action is equivalent to that of a field in flat spacetime! This is of course a huge simplification and implies that the scalar field is decoupled from gravity.

A detailed derivation may be found in chapter 3 of the book by Birrell and Davies. In $n$ dimensions, the corresponding factor in front of $R$ is $(1/4)(n-2)/(n-1)$.

One could also not put the factor and the study the consequent theory. Conformal invariance action is of course not mandatory but for reasons explained above, nice nonetheless.

Aditya Bawane
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