# Trace of stress tensor vanishes ==> Weyl invariant

You often see in textbooks the statement that ${T^\mu}_\mu = 0$ implies Weyl invariance or conformal invariance. The proof goes like

$\delta S \sim \int \sqrt{g} T^{\mu\nu} \delta g_{\mu\nu} \sim \int \sqrt{g} {T^\mu}_\mu$

where I have varied the action with respect to the metric and assumed $\delta g_{\mu\nu} \propto g_{\mu\nu}$ (i.e. a Weyl transformation).

This does not seem to be completely general because I can imagine a Lagrangian containing matter fields with non-trivial conformal weights. Then the full variation under Weyl tranformation contains a term proprotional to the matter equation of motion.

So I would conclude that the correct statement is more like

${T^\mu}_\mu = 0, \quad \& \quad \frac{\delta S}{\delta \phi} = 0\implies \textrm{Weyl invariant}$

Is it true that Weyl invariance only holds when the matter fields are on-shell or am I missing something?

Thanks.

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The stress tensor $T_{\mu \nu}$ is defined as the total variation of $S$ as you vary the metric, and as such contains the contributions of all matter fields, regardless of their conformal weights (in $d \neq 2$, the term is conformal dimension, by the way). I hope this clears things up. – Vibert Feb 4 '13 at 7:44
To clarify, I don't believe you need the matter to be on shell. – Vibert Feb 4 '13 at 7:55
The stress tensor is defined as the variation of $S$ wrt the metric, not wrt the metric and all matter fields. – user11881 Feb 4 '13 at 14:57
Dear user11881, we're miscommunicating. Your definition of the stress tensor is 100% correct, but since the action itself contains a bunch of matter fields, they all contribute when you vary the metric. All details are given in sec. 19.5 of Peskin-Schroeder, formula 19.150 for example gives the (symmetrized) stress tensor for QED. If you want to I can look up similar results for $\phi^4$ or QCD, but I don't have the references at hand. – Vibert Feb 4 '13 at 19:21