When coupling a scalar field to gravity, one sometimes introduces an additional term into the action: $S=\int d^4x \sqrt{-g}(L-\frac{1}{2}H_0 R\phi_0^2)$ where $R$ is the Ricci scalar, $L$ is the matter lagrangian, and $H_0$ is a parameter tuned to make the kinetic energy conformally invariant. This so-called "non-minimal" term is zero in flat space, but it's derivative with respect to the metric gives a modification to the canonical energy momentum tensor, defining the "Improved Energy Momentum Tensor".

In the paper https://doi.org/10.1103/PhysRevD.14.1965 (apologies for the paywall), Collins writes that

"One might say the minimal way to go from flat to curved space is not for the kinetic energy term to be $\frac{1}{2}(\partial\phi)^2$, but for it to be conformally invariant".

If a theory exhibits conformal invariance, great. But I don't understand why such a thing should be imposed, especially for a massive theory. Is there an intuitive reason the kinetic term in a lagrangian should be conformally invariant?

  • $\begingroup$ The minimally coupled theory has an improved and unimproved stress tensor. The conformally coupled theory also has an improved and unimproved stress tensor. All four are different. Improvement means adding terms that vanish on-shell to make the symmetry more manifest. It has nothing to do with changing the theory. $\endgroup$ Jul 22, 2021 at 0:41
  • $\begingroup$ Maybe "conformally covariant" would be more appropriate? $\endgroup$
    – TLDR
    Jul 22, 2021 at 0:41
  • $\begingroup$ If it is a massive theory, the $R\phi^2$ term will not make the theory conformal. $\endgroup$
    – fewfew4
    Jul 22, 2021 at 2:22

1 Answer 1


The massless scalar in flat space is classically conformal, so it is natural to do whatever is necessary to maintain that symmetry when lifting to a curved background.

The massive scalar is not conformal, and cannot be conformal.


The above is in some sense correct, but it requires clarification. A conformal transformation is a special case of a general coordinate transformation, and so any theory on a curved background is invariant under conformal transformations! When we say a theory is conformal, the nontrivial statement is that it is conformal on flat space. In order for a theory to be conformal on flat space, the theory on curved space must be both diffeomorphism invariant and Weyl invariant.

A Weyl transformation is not a coordinate transformation, it is a local rescaling of the metric $g'_{\mu\nu}(x)=\Omega^2(x)g_{\mu\nu}(x)$ (note that a conformal transformation is coordinate transformation which results in the metric transforming the same way). Now, in order for the theory on curved space to correspond to the one on flat space, the symmetries in the curved space need to correspond to the ones on flat space. So, the massless scalar which is classically conformal on flat space needs to have Weyl invariance on curved space.

Just so we have a concrete theory, the action is $$S=\frac{1}{2}\int d^dx\sqrt{g}\Big(g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi+\frac{d-2}{4(d-1)}R\phi^2-V(\phi)\Big)$$

The point is if you don't include the $R\phi^2$ term, and take the flat space limit, the theory actually won't be conformal.

  • $\begingroup$ But wouldn’t curvature introduce a lenghtscale? I imagine this should spoil the conformal invariance. $\endgroup$
    – Andrea
    Jul 22, 2021 at 5:42
  • $\begingroup$ @Andrea I think I had a misconception when I wrote this answer. A conformal transformation is a diffeomorphism, and so any theory on a curved background is conformal. The extra symmetry on a curved background which allows the flat theory to be conformal is Weyl symmetry. I will edit my answer in a bit. $\endgroup$
    – fewfew4
    Jul 22, 2021 at 7:32
  • $\begingroup$ Cool! I don’t know much myself about conformal transformations, that’s why I was asking $\endgroup$
    – Andrea
    Jul 22, 2021 at 15:39
  • $\begingroup$ I think there two different uses of the expression “conformal transformation”: either a rescaling of the coordinates or a rescaling of the coordinates and the fields $\endgroup$
    – Andrea
    Jul 22, 2021 at 16:03
  • 1
    $\begingroup$ Ok I think I see my confusion. The conformal symmetry of the kinetic term for a free scalar field in flat space is a consequence of the theory, its not something imposed as I initially thought. Then in generalizing to curved space, Weyl and Diffeomorphism invariance should be preserved which motivates the improvement term in the action that vanishes in flat space. Is this more on target? $\endgroup$
    – Adots005
    Jul 22, 2021 at 20:47

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