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

Question

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.

Answer 1

I will elaborate the answer I gave in a comment.

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 $z$.