Field Theory – Exploring a Traceless Stress-Energy Tensor

Question

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.

Answer 1

"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.