Bsrgvty

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.
$xmapsto b x$ and $phi mapsto phi$. Show that the stress energy
tensor is traceless.

Writing the transformations as $xmapsto e^theta x$, $phimapsto e^omegathetaphi$, and $partial_muphimapsto e^(omega-1)thetapartial_muphi$, for $omega=0$.

I get the variations as $delta x_mu=theta x_mu$, $deltaphi=omegathetaphi=0$, and $partial_muphi=(omega-1)thetapartial_mu phi=-thetapartial_muphi$; and
I've tried to get some useful expression bassed on the variation of lagrangian:
$$
delta L=fracpartial Lpartialphidelta phi+fracpartial Lpartial(partial_muphi)delta(partial_muphi)=-thetafracpartial Lpartial(partial_muphi)partial_muphi.
$$
On the other hand, the trace of the tensor takes the form
$$
T_mu^mu=eta_munuT^munu=eta_munu(-eta^munuL+fracpartial Lpartial(partial_muphi)partial^nuphi)=-2L+fracpartial Lpartial(partial_muphi)partial_muphi
$$
Thus,
$$
T_mu^mu=-2L-fracdelta Ltheta.
$$
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 $eta_munu$),


• do not make a specific choice of the symmetry operation (why did you define $phi rightarrow e^omega thetaphi$? If $theta, omega in mathbbR$, 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:

$$ T_munu = frac-2sqrt-gfracdelta Sdelta g^munu,$$

where $S$ is the action.

Scaling transformations are a special case of conformal transformations where
$$ delta g^munu = epsilon g^munu,$$
in your specific example $epsilon = b^2$.

Inverting the formula to single out the variation of the action:
$$ delta S propto T_munudelta g^munu = epsilon T_munu g^munu = T^mu_mu,$$

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.
$$partial_muT^munu=0.$$
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,
$$partial^mu epsilon^nu + partial^nuepsilon^mu = frac2d(partial.epsilon)eta^munu.$$ 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 $j_mu = T_munuepsilon^nu$ which is conserved. Hence,
$$partial^mu j_mu = 0$$
$$beginalign
&implies (partial^mu T_munu)epsilon^nu + T_munupartial^muepsilon^nu=0\
&implies frac12 T_munu (partial^muepsilon^nu + partial^nuepsilon^mu)=0\
&implies frac12 T_munu frac2d(partial.epsilon)eta^munu=0\
&implies frac1d(partial.epsilon) T^mu_mu = 0.
endalign$$
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.