# Intuition for the trace-free energy-momentum tensor condition in CFTs

It is a textbook exercise to show that $$\begin{equation}T^{\mu}_{\,\,\,\mu}=0 \end{equation}$$ is a sufficient condition for there to be a conserved current associated with a dilation symmetry. This condition is very important in CFTs so my question is,

Question: is there a nice intuitive way to visualize the trace-free condition, or is there some physical example (like a fluid?) where one can better understand what the trace-free condition corresponds to physically?

As an example of the type of intuition I am looking for: the $$T^{00}$$ component is the energy density, whereas the diagonal components $$T^{ii}$$ have the interpretation of a pressure (the flux of $$p^i$$ momentum in the $$i^{th}$$ direction). So the trace-free condition is somehow equating the energy density with the pressure. Making this more precise or providing a physical example along these lines would be helpful.

• We have $$Tr[T] = \beta(g)$$ and CFTs are fixed points of RG flows so their traces vanish. Is this helpful? Nov 12, 2021 at 2:44
• Addendum: as pointed out by Peter in the comment it isn't an equals sign but a proportional sign. Nov 15, 2021 at 14:29

As mentioned the trace of the stress tensor in a QFT is related to the $$\beta$$ function by $$T_\mu^\mu \sim \beta(g).$$ The $$\beta$$ function indicates how couplings change with a change of scale but in a CFT there is no scale so the $$\beta$$ function vanishes; CFTs are fixed points of the RG flow.
There is more to the story. The stress tensor is usually defined at the quantum level by point-splitting: $$\lim_{\delta \rightarrow 0}\left[\partial_z \phi(z + \delta/2)\partial_z\phi(z - \delta/2) -\frac{1}{2 \delta^2}\right].$$ This definition fails to commute with conformal mappings because it introduces a scale $$1/2\delta^2$$ that is not mapped to $$1/2(|f'(z)|\delta)^2$$ in the image. An anomalous term appears in the transformation law because the regularization procedure does not respect the conformal symmetry: $$T(z) = f'(z)^2 T(f(z)) -\frac{c}{12}\{f,z\}$$ where $$c$$ is the famous central charge and $$\{f,z\}$$ is the Schwartzian derivative of $$f$$ with respect to $$z$$.
• I totally agree with what is written above. In a curved 1+1 dimension manyfold however, a supplementary term might arise due to curvature. This is the so called Weyl Anomaly one can rewrite as $T^\mu_\mu\propto\mathcal R$ Nov 23, 2021 at 0:07