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  • $\begingroup$ Thanks for your answer. I don't need all of the CFT trappings so maybe I can simplify your reply. Ignoring your $S_{an}[\Omega,g]$ and replacing $Z[g]$ with the matter Lagrangian, $\sqrt{-g} \ \mathcal{L}_m[g]$, we have the usual definition of $T_{\mu\nu}$, correct? Now in these terms, I think that your computation goes through so long as we have that $\sqrt{-g} \ \mathcal{L}_m[g]$ doesn't change under the Weyl transformation. Is this a usual condition for Lagrangians to obey outside of CFT? Is it obeyed for a perfect fluid? If so then $T_{\mu\nu}$ has this scaling behavior very generally. $\endgroup$
    – Daniel Grimmer
    Commented Apr 16 at 17:40
  • $\begingroup$ Yes, by definition, classical Weyl invariance is equivalent to the statement that $\sqrt{-g} {\cal L}[g]$ is invariant under $ g \to \Omega^2 g$. $\endgroup$
    – Prahar
    Commented Apr 16 at 18:51
  • $\begingroup$ I don't think you have understood the end of my comment. Is this condition often obeyed outside of CFT? For instance, see Eq.(2.14) of the [ref] from my post. They come to the same result as we have but without any CFT assumption (as far as I can tell). They then go on to apply this $T_{\mu\nu}$ scaling to a perfect fluid in order to determine the scaling behavior of the mass density, $\rho$. But I don't think the action of a perfect fluid is Weyl invariant... All I want to know is the scaling behavior of the mass (and charge) densities for a perfect fluid outside of a CFT context. $\endgroup$
    – Daniel Grimmer
    Commented Apr 16 at 19:15
  • $\begingroup$ I did misunderstand, sorry. Though it’s weird you’re talking about Weyl transformations and then you say, let’s NOT talk about CFTs since QFTs with Weyl symmetries, by definition, are CFTs. You probably meant to say scale invariance and NOT Weyl invariance. $\endgroup$
    – Prahar
    Commented Apr 17 at 9:18
  • $\begingroup$ If you only want to talk about scale invariance, then the other answer on this post answers it for you, because you only need to look at the engineering dimension of various quantities. Strictly speaking, $\Delta$ simply has no meaning whatsoever outside of CFTs. $\endgroup$
    – Prahar
    Commented Apr 17 at 9:20