# Invariance of Liouville action under rescaling

I was studying the Liouville action

$$S=\frac{1}{8\pi} \int d^2 x\ \left[ \partial_\mu \phi \partial^\mu \phi + e^{\beta\phi} \right] \tag{1}$$

under the following general form of transformation:

$$\delta \phi = \xi^\mu \partial_\mu \phi + \frac{h}{d} \left(\partial_\mu \xi^\mu \right), \tag{2}$$

with $$d=2$$, as already specified in the measure of $$(1)$$. I want to show that the action is invariant under rescaling, i.e. $$\xi_\nu = \alpha x_\nu$$, which implies

$$\partial_\mu \xi_\nu = \alpha \eta_{\mu\nu}. \tag{3}$$

Using the general form of the transformation given by $$(2)$$ as well as the equation of motion ($$\partial_\mu \partial^\mu \phi = \frac{\beta}{2} e^{\beta \phi}$$), I was able to derive the following expression for the variation of the action:

$$\delta S = \frac{1}{4 \pi} \int d^2x\ \left( \partial_\mu \xi_\nu \right) \left( \partial^\mu \phi \partial^\nu \phi - \frac{1}{2} \eta^{\mu\nu} \partial_\alpha \phi \partial^\alpha \phi - \frac{1}{2} \eta^{\mu\nu} e^{\beta\phi} \right). \tag{4}$$

Then my expectation would be that inserting $$(3)$$ in $$(4)$$ should result in the desired $$\delta S = 0$$. But here is what happens:

\begin{align} \delta S & = \frac{\alpha}{4 \pi} \int d^2x\ \eta_{\mu\nu} \left( \partial^\mu \phi \partial^\nu \phi - \frac{1}{2} \eta^{\mu\nu} \partial_\alpha \phi \partial^\alpha \phi - \frac{1}{2} \eta^{\mu\nu} e^{\beta\phi} \right) \\ & =\frac{\alpha}{4 \pi} \int d^2x\ \left( \partial^\mu \phi \partial_\mu \phi - \partial_\alpha \phi \partial^\alpha \phi - e^{\beta\phi} \right) \\ & = - \frac{\alpha}{4 \pi} \int d^2x\ e^{\beta\phi}. \tag{5} \end{align}

As far as I can see, $$(5)$$ is not vanishing. Of course, inserting the equation of motion would make the action vanish, but I don't think that I am allowed to do that, since the equation of motion was precisely derived by setting $$\delta S = 0$$ (then the action would be invariant under any transformation!). Any idea of what I am missing?

## Edit:

I meant to also mention that in order to obtain $$(4)$$, one has to set $$h / d = 1/ \beta$$, otherwise there are terms which cannot take the form $$\left( \partial_\mu \xi_\nu \right) \left( ... \right).$$

• What steps can you show for deriving (4). Did you vary the potential function?
– user196418
May 10 '19 at 21:45
• @ggcg Yes I varied the potential, and I forgot to mention one crucial step, which I now added as an edit to the question. In the derivation, the potential term can be rewritten in the form $(\partial_\mu \xi_\nu)(...)$ only if we set $h/d=1/\beta$. This is because two terms coming from the kinetic part and from the potential term cancel, leaving us with a total derivative (that we can throw away) and the term $(\partial_\mu \xi_\nu) e^{\beta \phi}$ that appears in $(4)$. Is that what you were thinking of?
– Jxx
May 10 '19 at 22:38
• @ggcg (I can show all the steps if you think that it is useful)
– Jxx
May 10 '19 at 22:38
• I'm pretty sure that in order for the variation to be identically zero you need to assume that $\xi$ is a conformal Killing vector. May 10 '19 at 22:39
• @Jxx, thanks...
– user196418
May 10 '19 at 23:07

The variation done above is correct and corresponds to the canonical procedure of deriving the stress-energy tensor. However it is a rigid requirement for conformal invariance that the trace of the stress-energy tensor vanishes or $$\eta_{\mu\nu}T^{\mu\nu}=0$$. However it is easy to see that the canonical stress energy tensor doesn't satisfy this relation. It is possible however, that this tensor can be improved a la Belinfante , so that it becomes traceless.

The correct stress energy tensor is given by

$$\hat{T}_{\mu\nu}=T_{\mu\nu}+\frac{2}{\beta}(\eta_{\mu\nu}\square-\partial_\mu\partial_\nu)\phi$$

which, as can be checked, is traceless, using the equations of motion. Note in passing that using the equations of motion is necessary for the stress energy tensor to be conserved here, the conservation law is on shell, just as it is in classical mechanics (a particle conserves energy only if the equations of motion are obeyed).

However, this seems to have come out of thin air, and indeed, if you would want to derive the improved stress-energy tensor from first principles, you would need to couple the theory to non-minimal gravity with the following Lagrangian density:

$$\mathcal{L}=\sqrt{-g}(g_{\mu\nu}\partial_\mu\phi\partial_\nu \phi+QR\phi+ \lambda e^{\beta\phi})$$ then vary with respect to the metric, the Christoffel symbols (that are implicit here in the Ricci scalar $$R$$) and the scalar field, using the Palatini variation or the flatness of the metric connection (either should work) to obtain that in fact the conserved energy momentum tensor with respect to the covariant derivative is $$\hat{T}_{\mu\nu}$$ and it's trace vanishes when the metric is flat.

[here page 25 for the Palatini action]

and of course the original paper from Belinfante (1940) which is really at the heart of matters.

Happy reading and hope this helps!

EDIT: Say that you we have an action that is invariant under the whole set of Poincare and conformal transformations. Then from poincare invariance, we can find a tensor such that under the transformation $$x\rightarrow x+\epsilon(x)$$

$$\delta S=\int(\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu)T_{\mu\nu}$$

Then if we decompose the tensor into a traceless part and it's trace as $$T_{\mu\nu}=t_{\mu\nu}+\frac{1}{d}\eta_{\mu\nu}T$$ and use the conformal Killing equation $$\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu=\frac{2}{d}\eta_{\mu\nu}\partial_ \rho\epsilon^\rho$$ we can easily see that the only way that we can make the variation of the action for infinitesimal nonlocal translation that satisfies the conformal Killing equation be proportional to it is to set $$T=0$$. This concludes why the stress energy tensor has to be traceless if conformal invariance is to be obeyed.

• Thanks for the answer, that is indeed helpful and very nicely presented. I have already constructed the same traceless energy-momentum tensor that you gave, in an exercise. But when you say that the trace of the stress-energy tensor should vanish for conformal invariance, why should that be the case? I thought that all $T_{\mu\nu}+K_{\mu\nu}$ were equivalent, as long as $\partial_\mu K^{\mu\nu} = 0$, and that this freedom could be used for building a tensor with more pleasant properties, such as symmetry in the indices of tracelessness.
– Jxx
May 11 '19 at 21:57
• Edited to answer part of your question. And the answer to the rest is, no, we don't modify the stress tensor as we wish, it is the product of a very standard procedure. It's just that some physicists like hiding the big calculations under carpet statements like "we have the recipe, so we modify like that and we forget about the origin of the entire thing". The gravitational theory allows through the diffeomorphism invariance the ability to explore statements about the stress energy tensor that you cannot derive otherwise. Varying the metric is what defines the stress tensor after all :) May 12 '19 at 2:08
• Alright, that sounds convincing :) I have accepted your answer and will take a look at the references you mentioned!
– Jxx
May 12 '19 at 8:08