# Metric in dilatation transformation of massless scalar field

The lagrangian density of the massless real scalar field is \begin{align} L = \frac{1}{2}\eta^{\mu\nu}\partial_\mu\Phi\partial_\nu\Phi = \frac{1}{2}\partial_\mu\Phi\partial^\mu\Phi. \end{align} I want to check if the action is invariant under dilatation transformation \begin{align} x'^\mu=e^\alpha x^\mu \\ \Phi'(x')=e^{-\alpha}\Phi(x). \end{align} Since $$d^4x'=e^{4\alpha}d^4x$$ the lagrangian density must thus transform as $$L' = e^{-4\alpha}L$$. My question here is, how the metric $$\eta^{\mu\nu}$$ transforms. If I have a look at the line element \begin{align} ds^2=\eta_{\mu\nu}dx^\mu dx^\nu = e^{-2\alpha}\eta_{\mu\nu}dx'^\mu dx'^\nu . \end{align} I find, that the transformed metric is $$g_{\mu\nu}=e^{-2\alpha}\eta_{\mu\nu}$$. When I use this to transform the Lagrangian I get \begin{align} L' = \frac{1}{2}g^{\mu\nu}\partial'_\mu\Phi'(x')\partial'_\nu\Phi'(x') = e^{-6\alpha} L. \end{align} So this differs from the correct result by a factor of $$e^{-2\alpha}$$. I think my mistake here is, that I also transform the metric, but I dont get why the metric should stay the same since it obviously transforms under this coordinate transformation.

• What about the transformation of $\sqrt{|g|}$? Commented Nov 29, 2022 at 14:51

We are in flat spacetime, so the action is

$$S=\int d^4x \, \frac{1}{2}(\partial \phi)^2$$

The Lagrangian transforms as

$$\mathcal{L}'=\frac{1}{2}(\partial \phi')^2=\frac{1}{2}\partial'_{\mu}\Phi'\partial'^{\mu}\Phi'=e^{-4\alpha}\mathcal{L}$$

which cancels with

$$d^4x'=e^{4\alpha}d^4x$$.

Note: this is not a Lorentz transformation, so $$ds^2$$ does not remain invariant. It changes as

$$ds'^2=\eta_{\mu\nu}dx'^{\mu}dx'^{\nu}=e^{2\alpha}ds^2$$

• Why do you integrate over $(\partial_\mu\Phi)^2$ and not over $\partial_\mu\Phi\partial^\mu\Phi$? These two differ by a sign on the 0th component Commented Nov 30, 2022 at 10:08
• When I wrote $(\partial \Phi)^2$ I meant $\partial_{\mu}\Phi\partial^{\mu}\Phi$, it is just another way to write the same thing. Commented Nov 30, 2022 at 12:19