# A traceless stress energy tensor?

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.

"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.

• Wait can you write scaling $x\rightarrow bx$ as a translation $x\rightarrow x+a$? Sep 30 '19 at 6:32
• Where does the condition $\phi\to\phi$ enter this argument (same goes for the answer above)? Sep 30 '19 at 12:27
• @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 $x$. Sep 30 '19 at 14:27
• There was a typo in the answer which I fixed. $\epsilon(x) = (b-1)x$ Sep 30 '19 at 14:27
• Ah ok thank you Sep 30 '19 at 15:19