Why does Weyl invariance imply a traceless energy-momentum tensor?

Question

I've begun to self-study String Theory from Polchinski and Becker, Becker and Schwarz. I don't see why the fact that the Polyakov action is invariant under Weyl transformations is related to the tracelessness of the energy-momentum tensor. I can follow the argument in BBS with gauge freedom just fine, but then they mention that this is related to Weyl invariance. On the other hand, Polchinski simply says

The invariance of $S_\text{P}$ under arbitrary Weyl transformations further implies that $$\gamma_{ab}\frac{\delta}{\delta\gamma_{ab}}S_\text{P}=0\implies T_a^{\;a}=0$$

(Here $\gamma_{ab}$ is the worldsheet metric.)

How does this follow from Weyl invariance?

Answer 1

The (Belinfante-Rosenfeld) stress energy momentum tensor is defined as

$$T^{\mu\nu}\propto \frac{1}{\sqrt{-g}} \frac{\delta S}{\delta g_{\mu\nu}}$$

where the worldsheet metric is $g_{\mu\nu}$. By definition of the functional derivative, for any variation $\delta g_{\mu\nu}$, we have

$$\delta S = \int \frac{\delta S}{\delta g_{\mu\nu}} \delta g_{\mu\nu}$$

Consider now the case where $\delta g_{\mu\nu}$ is an infinitesimal Weyl invariance, or

$\delta g _{\mu\nu} = \Omega^2 g_{\mu\nu}$, where $\Omega$ is any function.

Weyl invariance of $S$ means that $\delta S$ has to vanish for all $\delta g_{\mu\nu}$ of this form, or

$$0=\int \frac{\delta S}{\delta g_{\mu\nu}}\Omega^2 g_{\mu\nu}$$

Then, the fundamental lemma of the calculus of variations implies tracelessness of the functional derivative $\frac{\delta S}{\delta g_{\mu\nu}}$, which is equal to the stress energy tensor up to various proportionality factors.

Incidentally, this line of reasoning also gives you things like Bianchi identities (try this for the Einstein action).