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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.

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  • $\begingroup$ What about the transformation of $\sqrt{|g|}$? $\endgroup$
    – paul230_x
    Commented Nov 29, 2022 at 14:51

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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

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

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  • $\begingroup$ 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 $\endgroup$
    – Aralian
    Commented Nov 30, 2022 at 10:08
  • $\begingroup$ When I wrote $(\partial \Phi)^2$ I meant $\partial_{\mu}\Phi\partial^{\mu}\Phi$, it is just another way to write the same thing. $\endgroup$ Commented Nov 30, 2022 at 12:19

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