I'm new to conformal field theory and I expected this question to have a simple answer but I can't find any simple answers online. I'm starting to suspect that there's some subtlety, or that perhaps my question is not well defined.

More precisely, suppose I have a stress tensor $T^{ab}$ defined on a spacetime given by $g_{ab}$, and the theory is a conformal field theory. If I perform a conformal transformation $$ g_{ab} \rightarrow \tilde{g}_{ab} = \Omega^2 g_{ab}$$ what happens to the stress tensor? I would imagine it behaves like $$T^{ab} \rightarrow \tilde{T}^{ab} = \Omega^{-\Delta} T^{ab}$$ and for whatever reason I was under the impression that $\Delta$ was equal to the spacetime dimension $d$, which is odd for the case that I'm considering.

Is there more information needed to answer this question? Does the conformal transformation properties of the stress tensor depend on what kind of theory I'm considering? Is my question even well defined?


2 Answers 2


Properly speaking that isn't a conformal transformal, it's a Weyl transformation. But yes, under a Weyl transformation $T_{\mu \nu}$ transforms precisely like you mention, with $\Delta = d$, regardless of the theory you're considering. You can derive the fact that $\Delta = d$ in several ways, either starting from the definition of $T$ as $\sim \delta S/\delta g^{\mu \nu}$, or in the CFT language you can consider the two-point function $$\langle T_{\mu \nu}(x) T_{\rho \sigma}(y) \rangle$$ and prove that $\partial_{\mu} T_{\mu \nu} = 0 \leftrightarrow \Delta = d$.

  • $\begingroup$ Hi Marty, thank you for the quick response. I apologize, but I have a couple of follow up questions. 1) What's the difference between a Weyl transformation and a conformal transformation? One is a coordinate transformation and the other isn't? Which one is relevant in conformal field theory? 2) In section 4 (equation 4.6) of hep-th/0002230 it seems that the claim is that $\Delta$ should equal $d-2$ instead, how do I reconcile these? Thanks again for your help. $\endgroup$
    – pianyon
    Commented Nov 24, 2017 at 23:08
  • $\begingroup$ @pianyon tad late, but from what I understand, a conformal transformation is a coordinate transformation that induces a specific change in the metric, while a Weyl transformation is a direct transformation of the metric itself. In a Weyl transformation, you are physically changing the units of distance you use to measure things (by a scale factor), whereas for a coordinate transformation you are merely changing your perspective (which can be interpreted as a metric change taken from your previous coordinates). $\endgroup$
    – Zxv
    Commented Jun 6, 2023 at 11:37

Great questions. Well first of all I agree with Marty about the energy dimension of the Energy-Momentum tensor. Conserved currents of spin $l$ have scaling dimension $\Delta = d +l-2$. That makes $\Delta_T = d$.

As for the confusion between Weyl transformation and conformal transformation: a Weyl transformation physically changes the metric space that your theory lives in and that's why the metric changes. A conformal transformation however simply changes your coordinate system. The metric is the same, but it looks different in different coordinates. Just like when you rotate a vector - the object stays the same, but its components look different.

The clearest way I know to express this is the following. You have some physical space with points $P$. Some guy A writes labels on each point and calls them $x(P)$. Some other guy B also labels the space with different labels $\tilde{x}(P)$. Now given one of A's labels we are able to figure out which point it is and therefore we know what $\tilde{x}$ would correspond to it. This gives us a relation $\tilde{x}=f(x)$. If you now ask A and B what their metric at a specific point $P$ looks like they will give different answers. These answers will be related as follows. \begin{equation} g_{\mu \nu}(\tilde{x}(P)) = \frac{\partial x^\alpha}{\partial \tilde{x}^\mu} \frac{\partial x^\beta}{\partial \tilde{x}^\nu} g_{\alpha \beta}(x(P)) \end{equation} If this takes the specific form \begin{equation} g_{\mu \nu}(\tilde{x}(P)) = \Omega^2(x(P))g_{\mu \nu}(x(P)) \end{equation} we call that a conformal transformation.

Now let's move on to your actual question. As explained before your energy momentum tensor is physically the same as before, just the numbers describing it look different. It is almost true that it transforms the way you suggested. However a tensor also transforms under rotations. Aside from the fact that straight up rotations are indeed conformal transformation, you could also have local twisting going on in more general conformal transformations. Writing this down in all generality seems pretty nasty but I do know how to write it down infinitesimally. Taking the coordinate transform

\begin{equation} \tilde{x}^\mu = x^\mu + a^\mu +\omega^\mu _{\; \nu}x^\nu+\lambda x^\mu + 2\, x^\mu (x \cdot b) - b^\mu x^2 \end{equation}

leads to the following field transformation \begin{equation} \delta T = -i\left(a^\mu P_\mu +\lambda D + b^\mu K_\mu + \frac{1}{2}\omega^{\mu \nu} L_{\mu \nu}\right) T \end{equation}

with \begin{split} \mathrm{(Scaling) \qquad} \ D \ \ &= -i (\Delta - x^\mu \partial_\mu )\\ \mathrm{(Translation) \qquad} \ P_\mu \ &= -i \partial_\mu \\ \mathrm{(SCT) \qquad} K_\mu \ &= 2 x_\mu -i\Delta + i( x^2 \partial_\mu - 2x_\mu x^\nu \partial_\nu ) - 2 x^\nu S_{\mu \nu} \\ \mathrm{(Lorentz) \qquad} L_{\mu \nu} &= i(x_\mu \partial_\nu - x_\nu \partial_\mu) + S_{\mu \nu} \end{split}

I really hope I got the minus signs right. Anyway now you see why this is rarely written down explicitly. If you set for example $\omega=0=b$ this can actually be integrated to the finite version and will come out exactly like what you wrote.

  • $\begingroup$ Thank you AlexM for the clarification between Weyl transformations and conformal transformations. Would the answer be different if I performed a Weyl transformation, so the new metric is physically different from the old one? Is the stress tensor expected to behave nicely under the Weyl transformation? Does it behave differently than conformal transformations? $\endgroup$
    – pianyon
    Commented Nov 25, 2017 at 4:21
  • $\begingroup$ I am not 100% sure on this one but I believe that the EM-Tensor does not transform at all under a Weyl rescaling. In fact none of the fields transform. However if you now calculate correlation functions between points $x$ and $y$, these will change, because the physical distance between $x$ and $y$ has changed. $\endgroup$
    – AlexM
    Commented Nov 25, 2017 at 10:37
  • $\begingroup$ Alex, all fields certainly transform under a Weyl rescaling. This just means that correlation functions in different metrics are related via $< O \ldots >_{g} = \prod \Omega^{\Delta_O} < O \ldots >_{g'}$ if $g = \Omega^2 g$'. $\endgroup$
    – user159249
    Commented Nov 25, 2017 at 17:47
  • $\begingroup$ Thanks for the correction. In that case would you mind clarifying the transformation law of fields under Weyl transformations? $\endgroup$
    – AlexM
    Commented Nov 25, 2017 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.