Suppose we have a theory with conformal invariance that has been extended to a diffeomorphism invariant theory in a way that the resulting energy-momentum tensor is traceless on-shell (which can always be done, as e.g. explained in my answer in Relation of conformal symmetry and traceless energy momentum tensor). In general we have that $\xi^\mu T_{\mu\nu}$ is the conserved current associated to any infinitesimal isometry $x\mapsto x+\xi$. For a traceless energy momentum tensor, this current is also conserved whenever $\xi$ induces an infinitesimal conformal transformation as well. However, this doesn't make it clear to me that this current gives the charges that generate conformal transformations. In particular, I don't know how to obtain this current from Noether's theorem. More physically, the current $\xi^\mu T_{\mu\nu}$ only seems to know about the change in coordinates. Why would it know about the scaling of the fields $\phi(x)\mapsto\Omega^{-\Delta}\tilde{\phi}(\tilde{x})$ in a conformal transformation?
1 Answers
Here is how to understand the association between a conformal transformation generated by a conformal Killing vector field $\xi^\mu$ and the conserved current $J_\nu=\xi^\mu T_{\mu\nu}$ using Noether's theorem.
First, replace $\xi\mapsto \xi'=\epsilon\xi$ for an $\epsilon$ vanishing at the boundaries of spacetime. In this replacing, keep the same conformal factor for the fields $\Omega^2=1+2\omega$ and let $\tilde{\phi}(\tilde{x})$ be the field transformed by the diffeomorphism $\epsilon\xi$. Consider then the transformation $$\begin{pmatrix} x \\ \phi(x) \\ g_{\mu\nu}(x) \end{pmatrix}\mapsto\begin{pmatrix} \tilde{x} \\ \tilde{\Omega}^{-\Delta}(\tilde{x})\tilde{\phi}(\tilde{x}) \\ g_{\mu\nu}(\tilde{x}) \end{pmatrix},$$ which reduces to a conformal transformation for $\epsilon=1$. This transformation can be broken down into the following composition $$\begin{pmatrix} x \\ \phi(x) \\ g_{\mu\nu}(x) \end{pmatrix}\mapsto\begin{pmatrix} x \\ \Omega^{-\Delta}(x)\phi(x) \\ \Omega^{2}(x)g_{\mu\nu}(x) \end{pmatrix}\mapsto\begin{pmatrix} \tilde{x} \\ \tilde{\Omega}^{-\Delta}(\tilde{x})\tilde{\phi}(\tilde{x}) \\ \tilde{\Omega}^{2}(\tilde{x})\tilde{g}_{\mu\nu}(\tilde{x}) \end{pmatrix}\mapsto\begin{pmatrix} \tilde{x} \\ \tilde{\Omega}^{-\Delta}(\tilde{x})\tilde{\phi}(\tilde{x}) \\ g_{\mu\nu}(\tilde{x}) \end{pmatrix}$$
The first is a Weyl transformation while the second is a diffeomorphism. Therefore, in a Weyl and diffeomorphism invariant theory, the variation of the action is only due to the last transformation $$\begin{pmatrix} x \\ \phi(x) \\ g_{\mu\nu}(x) \end{pmatrix}\mapsto\begin{pmatrix} x \\ \phi(x) \\ \Omega^{-2}(x)\bar{g}_{\mu\nu}(x) \end{pmatrix}$$ where $x\mapsto\bar{x}$ is the inverse transformation to $x\mapsto\tilde{x}$. In the latter we have $$\delta g_{\mu\nu}=-2\omega g_{\mu\nu}+\nabla_\mu\xi'_\nu+\nabla_\nu\xi'_\mu$$ and the variation of the action is $$\delta S=\int\text{d}^D{x}\frac{\delta S}{\delta g_{\mu\nu}}\delta g_{\mu\nu}\propto\int\text{d}^D{x}(-\omega T+T^{\mu\nu}\nabla_\mu\xi'_\nu).$$ In a Weyl invariant theory $T\equiv T^\mu_\mu=0$. Thus, on-shell, using the fact that $\epsilon$ vanishes at the boundaries to integrate by parts, and using the definition of a conformal Killing vector field, we have $$0=\int\text{d}^Dx (J^\mu\nabla_\mu\epsilon+\epsilon T^{\mu\nu}\nabla_\mu\xi_\nu)=\int\text{d}^Dx (J^\mu\nabla_\mu\epsilon+\frac{1}{D}\epsilon T\nabla_\alpha\xi^\alpha)=-\int\text{d}^Dx \nabla_\mu J^\mu\epsilon.$$ Thus $J^\mu$ is conserved.
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