I'm trying to solve this exercise:

Suppose an arbitrary theory (Flat space-time?) with a single field (Is a scalar field?) invariant under dilations, i.e. $x\mapsto b x$ and $\phi \mapsto \phi$. Show that the stress energy tensor is traceless.

Writing the transformations as $x\mapsto e^\theta x$, $\phi\mapsto e^{\omega\theta}\phi$, and $\partial_\mu\phi\mapsto e^{(\omega-1)\theta}\partial_\mu\phi$, for $\omega=0$.

I get the variations as $\delta x_\mu=\theta x_\mu$, $\delta\phi=\omega\theta\phi=0$, and $\partial_\mu\phi=(\omega-1)\theta\partial_\mu \phi=-\theta\partial_\mu\phi$; and I've tried to get some useful expression bassed on the variation of lagrangian: $$ \delta L=\frac{\partial L}{\partial\phi}\delta \phi+\frac{\partial L}{\partial(\partial_\mu\phi)}\delta({\partial_\mu\phi})=-\theta\frac{\partial L}{\partial(\partial_\mu\phi)}\partial_\mu\phi. $$ On the other hand, the trace of the tensor takes the form $$ T_\mu^\mu=\eta_{\mu\nu}T^{\mu\nu}=\eta_{\mu\nu}(-\eta^{\mu\nu}L+\frac{\partial L}{\partial(\partial_\mu\phi)}\partial^\nu\phi)=-2L+\frac{\partial L}{\partial(\partial_\mu\phi)}\partial_\mu\phi $$ Thus, $$ T_\mu^\mu=-2L-\frac{\delta L}{\theta}. $$ Obviously, if the lagrangian is invariant then the Tensor isn't traceless. So I don't have idea how to proceed.


2 Answers 2


"Arbitrary theory" probably means

  • do not make a specific choice of metric (i.e. flat space time, your $\eta_{\mu\nu}$),
  • do not make a specific choice of the symmetry operation (why did you define $\phi \rightarrow e^{\omega \theta}\phi$? If $\theta, \omega \in \mathbb{R}$, then it does not have magnitude $1$. If one of them is an imaginary number, then you've selected a $U(1)$ symmetry.

So basically use equations that are general and apply to anything within the Lagrangian formalism.

For instance, the stress energy tensor can be generally written as:

$$ T_{\mu\nu} = \frac{-2}{\sqrt{-g}}\frac{\delta S}{\delta g^{\mu\nu}},$$

where $S$ is the action.

Scaling transformations are a special case of conformal transformations where $$ \delta g^{\mu\nu} = \epsilon g^{\mu\nu},$$ in your specific example $\epsilon = b^2$.

Inverting the formula to single out the variation of the action: $$ \delta S \propto T_{\mu\nu}\delta g^{\mu\nu} = \epsilon T_{\mu\nu} g^{\mu\nu} = T^\mu_\mu,$$

where the last step is the trace!

Since the action must be minimised, $\delta S =0$, you must have $T^\mu_\mu=0$, i.e. a traceless stress-energy tensor.


We know that the stress-energy tensor is divergenceless, i.e. $$\partial_{\mu}T^{\mu\nu}=0.$$ A proof of this follows from the definition of canonical stress-energy tensor considering the action is invariant under translation $x^\mu \rightarrow x^\mu + a^\mu$. Now the transformation that you mentioned is a conformal transformation. Infinitesimally, the conformal transformation can be written as, $$x^\mu \rightarrow x'^\mu = x^\mu + \epsilon^\mu(x),$$ where $\epsilon$ follows, $$\partial^\mu \epsilon^\nu + \partial^\nu\epsilon^\mu = \frac{2}{d}(\partial.\epsilon)\eta^{\mu\nu}.$$ This equation is known as confomal Killing equation which can be derived from the property about how the metric changes under a conformal transformation. Now, for a conformal symmetry, Noether's theorem suggests that there is a current $j_\mu = T_{\mu\nu}\epsilon^\nu$ which is conserved. Hence, $$\partial^\mu j_{\mu} = 0$$ $$\begin{align} &\implies (\partial^\mu T_{\mu\nu})\epsilon^\nu + T_{\mu\nu}\partial^\mu\epsilon^\nu=0\\ &\implies \frac{1}{2} T_{\mu\nu} (\partial^\mu\epsilon^\nu + \partial^\nu\epsilon^\mu)=0\\ &\implies \frac{1}{2} T_{\mu\nu} \frac{2}{d}(\partial.\epsilon)\eta^{\mu\nu}=0\\ &\implies \frac{1}{d}(\partial.\epsilon) T^{\mu}_\mu = 0. \end{align}$$ Since this is true for any $\epsilon(x)$, $T^\mu_\mu=0$. That is the stress-energy tensor is traceless. The $\epsilon(x)$ in your case is $(b-1)x$.

The answer given by SuperCiocia is perfectly correct and nice but uses the Einstein-Hilbert form of the stress-energy tensor. However, from the question, it seems that the OP is more familiar with the canonical stress-energy tensor. Hence, this is an alternative derivation of the tracelessness of the stress-energy tensor for a conformal transformation.

  • $\begingroup$ Wait can you write scaling $x\rightarrow bx$ as a translation $x\rightarrow x+a$? $\endgroup$
    – SuperCiocia
    Sep 30, 2019 at 6:32
  • $\begingroup$ Where does the condition $\phi\to\phi$ enter this argument (same goes for the answer above)? $\endgroup$
    – Winther
    Sep 30, 2019 at 12:27
  • $\begingroup$ @SuperCiocia You can write $x\rightarrow bx = x+(b-1)x=x+\epsilon(x)$ where $\epsilon(x) = (b-1)x$. Remember that in the derivation $\epsilon$ does not have to be a constant but can be a function of $x$. $\endgroup$
    – abhijit975
    Sep 30, 2019 at 14:27
  • $\begingroup$ There was a typo in the answer which I fixed. $\epsilon(x) = (b-1)x$ $\endgroup$
    – abhijit975
    Sep 30, 2019 at 14:27
  • $\begingroup$ Ah ok thank you $\endgroup$
    – SuperCiocia
    Sep 30, 2019 at 15:19

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