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I'm interested in finding the constraints on constants $p,\xi$ appearing on a scalar field lagrangian density conformally coupled to spacetime. The transformation is $$ \tilde{g}_{\alpha\beta} = a^2 g_{\alpha\beta},\quad \tilde{\phi} = a^{2p} \phi $$ and I should show that the lagrangian density $$ \tilde{\mathscr{L}} = \frac{1}{2} \tilde{g}^{\alpha\beta} \partial_\alpha \tilde{\phi} \partial_\beta \tilde{\phi} - \xi \tilde{R} \tilde{\phi}^2 $$ is equal to $\mathscr{L}$ before the transformation, up to a total divergence.

The problematic part is to express the transformed Ricci scalar $\tilde{R}$ in terms of $R$ before the transformation, and after a day and many mistakes I obtained (in a torsionless $n$-dimensional manifold) $$ \tilde{R} = a^{-2} R + a^{-3} ( 2 - n ) {\Gamma^\alpha}_{\beta\alpha} \partial^\beta a + 2 a^{-3} ( 1 - n ) \partial_\alpha \partial^\alpha a - a^{-4} ( n - 1 ) ( n - 4 ) \partial_\alpha a \partial^\alpha a $$ but first I'm not sure of the result, second is not at all easy to continue and finally demonstrate that $p$ should be equal to $(2-n)/2$ and $\xi$ to $(n-2)/(4(n-1))$ as stated for example in "Quantum Field Theory in Curved Spacetime" by Parker and Toms (pag. 44 in 2009 edition).

Is there any clear way to continue and demonstrate that $\tilde{\mathscr{L}}=\mathscr{L}$ with stated above conditions or is there an available source were the full calculation is shown?

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    $\begingroup$ Are you sure that $p=(2-n)/2$? Since for $n=4$, I know that scalar field transformation $\tilde{\phi} = a^{-1}\phi$ will preserve massless field equation $\endgroup$
    – KP99
    Commented Oct 27, 2022 at 14:45
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    $\begingroup$ Also, don't confuse this transformation with conformal transformation: Conformal transformation is a point wise transformation which induces conformal rescaling of metric. However, the transformation here is just a conformal rescaling metric where $a$ is any positive definite arbitrary function $\endgroup$
    – KP99
    Commented Oct 27, 2022 at 14:51
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    $\begingroup$ It's enough to consider infinitesimal Weyl transformations, and this will probably make the calculation easier. If we put $a=e^{\sigma}$, then the infinitesimal version of your transformation rules are $\delta g_{\alpha\beta} = 2\sigma g_{\alpha\beta}$ and $\delta \phi = 2p\sigma \phi$. For such a deformation of the metric, you can show that the Ricci scalar suffers a corresponding deformation of the form $\delta R = 2\sigma R -2(d-1)\Box \sigma$, where $d$ is the spacetime dimension and $\Box = \nabla^a\nabla_a$ is the covariant D'alembertian. $\endgroup$
    – NormalsNotFar
    Commented Oct 27, 2022 at 15:08
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    $\begingroup$ The brute force method is fine, but I don't think it should be brutal. You just have to plug in the pieces now, i.e. vary each part of the action and demand invariance under the Weyl rescaling. But the variation of $R$ that you gave is not manifestly generally covariant (if it is correct) and will make things alot harder. $\endgroup$
    – NormalsNotFar
    Commented Oct 27, 2022 at 15:26
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    $\begingroup$ To derive a nice expression for $\delta R$, start with the expression for a general metric variation, given e.g. in e.q. (2.47) of the book you quoted by $\delta R=-R^{\mu\nu}\delta g_{\mu\nu} + g^{\rho\sigma}g^{\mu\nu}(\nabla_{\mu}\nabla_{\nu}\delta g_{\rho\sigma}-\nabla_{\sigma}\nabla_{\nu}\delta g_{\rho\mu})$. (You may have seen a similar expression when deriving the Einstein-Hilbert equation). Now consider the special case when the deformation of the metric takes the form $\delta g_{\mu\nu} = 2\sigma g_{\mu\nu}$, i.e. a Weyl rescaling. You will get $\delta R = -2\sigma R +2(d-1)\Box R$. $\endgroup$
    – NormalsNotFar
    Commented Oct 27, 2022 at 15:42

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The way I did it is considering an infinitesimal transformation, so $a^2(x)=1+\lambda(x)$ with $|\lambda(x)|\ll1$, then

$$g_{\mu\nu}\to(1+\lambda)g_{\mu\nu}\\ \phi\to(1+p\lambda)\phi$$

In $n$ spacetime dimensions,

$$\sqrt{|g|}\to (1+\frac n2\lambda)\sqrt{|g|}\\ R_{\mu\nu}\to R_{\mu\nu}+\frac{1}{2}(n-2)\nabla_\mu(\partial_\nu \lambda)+\frac{1}{2}\nabla_\sigma(g_{\mu\nu}\partial^\sigma\lambda)\\ R\to(1-\lambda)R+(n-1)\square\lambda$$

You can check that if $p=(2-n)/4$ (it is wrong in the book, it's written as in Birrell and Davies, but their transformation is different) and $\xi=(n-2)/[4(n-1)]$, the lagrangian density changes

$$\mathcal{L}\to\mathcal{L}-\partial_\mu\left(\frac{n-2}{8}\phi^2\sqrt{|g|}\partial^\mu\lambda\right)$$

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  • $\begingroup$ This is invariant up to a surface term, so the equations of motion remain unchanged. $\endgroup$
    – Eletie
    Commented Oct 27, 2022 at 21:23
  • $\begingroup$ It's basically Parker and Toms method. Very effective, but I was asking myself if there was a way to consider the generic transformation and make conclusions from that; seemed just more powerful. Anyway, thank you again for your answer, I think I will have to use this method so thank you for explaining :) $\endgroup$
    – Rob Tan
    Commented Oct 28, 2022 at 8:32

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