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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.

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  • $\begingroup$ Are you suggesting that one needs to include the $R/6$ for purposes of anomaly cancellation in a corresponding QFT? $\endgroup$
    – joshphysics
    Aug 16, 2013 at 1:53
  • $\begingroup$ I am not suggesting that. I do not know the origin of adding in the $R/6$. But what I've read is "In the framework of canonical quantum field theory in curved spacetime, the action must necessarily be of the form [mentioned above]". I do not understand why this should necessarily be the case. If there is some anomaly cancellation you are referring to, could you be more specific? $\endgroup$
    – Prahar
    Aug 16, 2013 at 1:58
  • $\begingroup$ Prahar when answering to someone's comment you should use @ (user name). e.g. @joshphysics $\endgroup$
    – user10001
    Aug 16, 2013 at 2:27
  • $\begingroup$ This may be helpful: arxiv.org/abs/gr-qc/0205066 'Only for ξ = 1/6 (so-called “conformal coupling”) does the flat space limit of the stress-energy tensor for the scalar field yield an expression with good renormalization properties.' $\endgroup$
    – user4552
    Aug 16, 2013 at 2:46
  • $\begingroup$ @Prahar I have no idea if this has anything to do with anomalies, but whenever I think of conformal invariance and I see the Ricci scalar sitting around, I think of the conformal anomaly...en.wikipedia.org/wiki/Conformal_anomaly $\endgroup$
    – joshphysics
    Aug 16, 2013 at 3:22

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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.

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    $\begingroup$ It should be noted that the above reasoning is classical. It has nothing to do with conformal anomaly, which is a quantum effect. Conformal invariance of the action implies that the trace of the energy-momentum tensor vanishes, but the vacuum expectation of the trace of the EMT maybe non-zero. This is the conformal anomaly, or trace anomaly. $\endgroup$
    – Aditya Bawane
    Aug 16, 2013 at 19:24
  • $\begingroup$ I see that. However, I am asking why is this conformal invariance imposed upon us by the requirement of renormalizability? $\endgroup$
    – Prahar
    Aug 16, 2013 at 19:35
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    $\begingroup$ @AdityaBawane: Welcome to physics.SE! If you like, you can edit your own answers rather than posting additional material as a separate comment. $\endgroup$
    – user4552
    Aug 16, 2013 at 21:21

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