How does the conformal scaling behavior of the stress-energy tensor depend on the spacetime dimension?

Question

I'm new to Weyl transformations and am struggling to find online the answer to what should be a simple question. Consider an $n=D+1$ dimensional spacetime with metric $g_{ab}$ and some stress-energy tensor $T_{ab}$ (e.g., from the variation of some action, say of a perfect fluid). If I perform a Weyl transformation $$g_{ab}\mapsto\bar{g}_{ab}=\Omega^2 g_{ab}$$ then what happens to the stress-energy tensor? In particular, what is its scaling behavior, $\Delta$?

$$T_{ab}\mapsto\bar{T}_{ab}=\Omega^{-\Delta} T_{ab}$$

Specifically, how does this depend on the spacetime's dimensions $n=D+1$? I have found the answer online in the $n=4$ case in several books (edit: For instance, with raised indices, we have $T^{ab}$ having $\Delta=6$ in Eq.(2.14) of [ref]). I have found some discussion of the issue in a generic dimension here but the discussion there is inconsistent with itself. I know that the scaling behavior is different if the indices are raised or lowered, e.g., $T_{ab}$ vs $T_{a}{}^b$ vs $T^{ab}$ But I don't think the discussion I found distinguishes these properly.

The hyper-specific question which I am looking for an answer to is as follow: What is the scaling behavior of the rest mass density distribution, $\rho$, which would go into the stress-energy of a perfect fluid, $$T_{ab}=(P+\rho) u_a u_b + g_{ab} P$$ I am particularly interested in the $n=1+1$ case generalizing Eq.(2.21) of [ref]. Would rest charge density transform any differently?

Note: I am interested in this question outside of the context of CFT. Please do not assume that the theory that I am working with is conformally invariant or that the stress-energy is traceless.

Edit: I have replaced "conformal weight" with "scaling behavior" everywhere. The way that both I and this post are defining $\Delta$ it does depend on how the indices are raised or lowered.

[ref] Beyond Einstein Gravity: A Survey of Gravitational Theories for Cosmology by Valerio Faraoni and Salvatore Capozziello.

Answer 1

Under Weyl transformations, the partition function of a CFT transforms as $$ Z[\Omega^2g] = e^{-S_{an}[\Omega,g]} Z[g] $$ where $S_{an}[\Omega,g]$ is the anomaly action. This is non-vanishing only in even dimensions.

The stress tensor in a CFT is defined as (up to some factor of $i$) $$ \langle T_{\mu\nu}(x) \rangle_g \equiv \frac{2}{\sqrt{-\det g(x)}} \frac{\delta }{\delta g^{\mu\nu}(x)} \ln Z[g] $$ It then follows that \begin{align} \langle T_{\mu\nu}(x) \rangle_{\Omega^2 g} &= \frac{2}{\sqrt{-\det [ \Omega^2(x) g(x) ] }} \frac{\delta }{\delta [ \Omega(x)^{-2} g^{\mu\nu}(x) ] } \ln Z[\Omega^2 g] \\ &= \Omega(x)^{-n+2} \frac{2}{\sqrt{-\det g(x)}} \frac{\delta }{\delta g^{\mu\nu}(x) } \ln \left( e^{-S_{an}[\Omega,g]} Z[g] \right) \\ &= \Omega(x)^{-n+2} \langle T_{\mu\nu}(x) \rangle_g - \Omega(x)^{-n+2} \frac{2}{\sqrt{-\det g(x)}} \frac{\delta S_{an}[\Omega,g] }{\delta g^{\mu\nu}(x) } \end{align} We now evaluate this in flat spacetime, in which case, the anomaly term vanishes. Setting $g_{\mu\nu} = \eta_{\mu\nu}$, we find \begin{align} \langle T_{\mu\nu}(x) \rangle_{\Omega^2 \eta} &= \Omega(x)^{-n+2} \langle T_{\mu\nu}(x) \rangle_\eta \tag{1} \end{align} Noting that the stress tensor has spin $s=2$, we can read off its scaling dimension as $\Delta = n$.

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PS - For a primary operator of scaling dimension $\Delta$ and spin $s$, we have $$ \langle O(x) \cdots \rangle_{\Omega^2 g} = \Omega(x)^{-\Delta+s} \langle O(x) \cdots \rangle_{g} $$