Proof Scalar action is conformally invariant

[LAHLH]

Hi,

So if we have the Lagrange density for a massless scalar field: $$L=\sqrt{-g}\left(-\frac{1}{2}g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi-\frac{(n-2)}{4(n-1)} R\phi^2\right)$$

Then under a conformal transformation $$g_{\mu\nu}=\omega^{-2}\tilde{g_{\mu\nu}}$$, then the Ricci sclar goes to $$R=\omega^2\tilde{R}+2(n-1)\tilde{g^{\mu\nu}}\omega\left(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega\right)-n(n-1)\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)$$. We also have $$g^{\mu\nu}=\omega^2\tilde{g}^{\mu\nu}$$ and $$\tilde{\nabla}_{\mu}\phi=\nabla_{\mu}\phi$$

Using all these suggests that,

$$\tilde{L}=\omega^{-1}\sqrt{-\tilde{g}}\left(-\frac{1}{2}\omega^2 \tilde{g}^{\mu\nu}\tilde{\nabla}_{\mu}\phi\tilde{\nabla}_{\nu}\phi-\frac{(n-2)}{4(n-1)} \left[\omega^2\tilde{R}+2(n-1)\tilde{g^{\mu\nu}}\omega\left(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega\right)-n(n-1)\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\right]\phi^2\right)$$

Which can be written as,


$$\tilde{L}=\sqrt{-\tilde{g}}\left(-\frac{1}{2}\omega \tilde{g}^{\mu\nu}\tilde{\nabla}_{\mu}\phi\tilde{\nabla}_{\nu}\phi-\frac{(n-2)}{4(n-1)} \omega\tilde{R}\phi^2-\frac{1}{2}(n-2)\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega\right)\phi^2+\frac{n}{4}(n-2)\omega^{-1}\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\phi^2\right)$$

The statement is that this conformal transformation leaves the scalar field theory invariant, so I was expecting to obtain the same form of the Lagrange density after applying the CT as I had before, but I can't see how the above reduces to this?

 

[LAHLH]

Seems that I forgot that the field should transform too, but never the less I still can't seem to show that this is conformally invariant, anyone know any books that show this?

 

[Bill_K]

You can get rid of the term containing the second derivative of ω by subtracting a divergence. Then you need to rescale φ by some power of ω, chosen to cancel the terms with first derivatives of ω.

 

[LAHLH]

Thanks for the reply, I have tried this and seem to still not be getting the result. Firstly my determinant was off by a power of n so I correct this, and also use $$\phi=\omega^{(n-2)/2} \tilde{\phi}$$ (Birrell and Davies)

$$\tilde{L}=\omega^{-n}\sqrt{-\tilde{g}}\left(-\frac{1}{2}\omega^2 \tilde{g}^{\mu\nu}\tilde{\nabla}_{\mu}(\omega^{(n-2)/2} \tilde{\phi})\tilde{\nabla}_{\nu}(\omega^{(n-2)/2} \tilde{\phi})-\frac{(n-2)}{4(n-1)} \left[\omega^2\tilde{R}+2(n-1)\tilde{g^{\mu\nu}}\omega\left(\tilde{\nabla}_{\m u}\tilde{\nabla}_{\nu}\omega\right)-n(n-1)\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\right](\omega^{(n-2)/2} \tilde{\phi})^2\right)$$

$$\tilde{L}=\omega^{-n}\sqrt{-\tilde{g}}\Big(-\frac{1}{2}\omega^2 \tilde{g}^{\mu\nu}\left[\frac{(n-2)}{2}\omega^{(n-4)/2}\tilde{\nabla}_{\mu}(\omega)\tilde{\phi}+\omega^{(n-2)/2}\tilde{\nabla}_{\mu}( \tilde{\phi})\right]\left[\frac{(n-2)}{2}\omega^{(n-4)/2}\tilde{\nabla}_{\nu}(\omega)\tilde{\phi}+\omega^{(n-2)/2}\tilde{\nabla}_{\nu}( \tilde{\phi})\right]$$ $$-\frac{(n-2)}{4(n-1)} \left[\omega^2\tilde{R}+2(n-1)\tilde{g^{\mu\nu}}\omega\left(\tilde{\nabla}_{\m u}\tilde{\nabla}_{\nu}\omega\right)-n(n-1)\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\right](\omega^{(n-2)/2} \tilde{\phi})^2\right) \Big)$$

I will have to finish this shortly as I need to go somewhere, but then I essentially expand out, and use product rule to get replace the double derivative on omega...

 

[LAHLH]

So expanding:

$$\tilde{L}=\omega^{-n}\sqrt{-\tilde{g}}\Big(-\frac{1}{2}\omega^2 \tilde{g}^{\mu\nu}\left[\frac{(n-2)^2}{4}\omega^{(n-4)}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\omega)\tilde{\phi}^2+\omega^ {(n-2)}\tilde{\nabla}_{\mu}( \tilde{\phi})\tilde{\nabla}_{\nu}( \tilde{\phi})+(n-2)\omega^{(n-3)}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\tilde{\phi})\tilde{\phi}\right]$$
$$+\left[ -\frac{(n-2)}{4(n-1)}\omega^2\tilde{R}-\frac{1}{2}(n-2)\tilde{g^{\mu\nu}}\omega\left(\tilde{\nabla}_{\m u}\tilde{\nabla}_{\nu}\omega\right)+\frac{n}{4}(n-2)\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\right]\omega^{(n-2)} \tilde{\phi}^2 \Big)<br />$$


Finally,

$$\tilde{L}=\sqrt{-\tilde{g}}\Big(-\tilde{g}^{\mu\nu}\frac{(n-2)^2}{8}\omega^{-2}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\omega)\tilde{\phi}^2-\frac{1}{2}\tilde{g}^{\mu\nu}\tilde{\nabla}_{\mu}( \tilde{\phi})\tilde{\nabla}_{\nu}( \tilde{\phi})-\frac{(n-2)}{2}\tilde{g}^{\mu\nu}\omega\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\tilde{\phi})\tilde{\phi}$$
$$-\frac{(n-2)}{4(n-1)}\omega^2\tilde{\phi}^2\tilde{R}-\frac{1}{2}(n-2)\tilde{g^{\mu\nu}}\omega\tilde{\phi}^2\left(\tilde{\nabla}_{\m u}\tilde{\nabla}_{\nu}\omega\right)+\frac{n}{4}(n-2)\tilde{\phi}^2\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right) \Big)<br />$$


Now we can use product rule e.g. $$\tilde{\phi}^2\tilde{\nabla}_{\m u}\tilde{\nabla}_{\nu}\omega\right) =\tilde{\nabla}_{\m u}\left(\tilde{\phi}^2}\tilde{\nabla}_{\nu}\omega\right)\right)-2\tilde{\phi}\tilde{\nabla}_{\m u}\left(\tilde{\phi}\right)\left(\tilde{\nabla}_{\nu}\omega\right)$$

But now what?

 

[LAHLH]

I made a mistake the last line should be:

$$\tilde{L}=\sqrt{-\tilde{g}}\Big(-\tilde{g}^{\mu\nu}\frac{(n-2)^2}{8}\omega^{-2}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu} (\omega)\tilde{\phi}^2-\frac{1}{2}\tilde{g}^{\mu\nu}\tilde{\nabla}_{\mu}(\tilde{\phi})\tilde{\nabla}_{\nu}( \tilde{\phi})-\frac{(n-2)}{2}\tilde{g}^{\mu\nu}\omega^{-1}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\tilde{\phi})\tilde{\phi}$$
$$-\frac{(n-2)}{4(n-1)}\tilde{\phi}^2\tilde{R}-\frac{1}{2}(n-2)\tilde{g^{\mu\nu}}\omega^{-1}\tilde{\phi}^2\left(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega\right)+\frac{n}{4}(n-2)\tilde{\phi}^2\tilde{g^{\mu\nu}}\omega^{-2}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right) \Big)$$

From which one has the conformally the Lagrangian in the conformal frame, plus a bit I want to get ride of:

$$\tilde{L}=L_0+\sqrt{-\tilde{g}}\Big(-\tilde{g}^{\mu\nu}\frac{(n-2)^2}{8}\omega^{-2}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu} (\omega)\tilde{\phi}^2-\frac{(n-2)}{2}\tilde{g}^{\mu\nu}\omega^{-1}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\tilde{\phi})\tilde{\phi}$$
$$-\frac{1}{2}(n-2)\tilde{g^{\mu\nu}}\omega^{-1}\tilde{\phi}^2\left(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega\right)+\frac{n}{4}(n-2)\tilde{\phi}^2\tilde{g^{\mu\nu}}\omega^{-2}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right) \Big)$$

Which looks prettier as:

$$\tilde{L}=L_0+\sqrt{-\tilde{g}}\Big(\tilde{g^{\mu\nu}}\omega^{-2}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\tilde{\phi}^2\frac{(n^2-4)}{8} -\frac{(n-2)}{2}\tilde{g}^{\mu\nu}\omega^{-1}\left(\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\tilde{\phi})\tilde{\phi} <br /> +\left(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega\right)\tilde{\phi}^2\Big)$$

 

[LAHLH]

I think I have it now:

Take $$\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\tilde{\phi}^2= \tilde{\nabla}_{\mu}\left[\omega(\tilde{\nabla}_{\nu}\omega)\tilde{\phi}^2\right]-\omega(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega)\tilde{\phi}^2-\omega(\tilde{\nabla}_{\nu}\omega)2\tilde{\phi}(\tilde{\nabla}_{\mu}\tilde{\phi})$$

If we include $$\tilde{g}^{\mu\nu}$$ the first term is a total divergence so vanishes. So we are left with:

$$\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\tilde{\phi}^2= -\omega(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega)\tilde{\phi}^2-\omega(\tilde{\nabla}_{\nu}\omega)2\tilde{\phi}(\tilde{\nabla}_{\mu}\tilde{\phi}) (1)$$

Now using the product rule again on the first term:

$$(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega)\tilde{\phi}^2=\tilde{\nabla}_{\mu}(\tilde{\phi}^2(\tilde{\nabla}_{\nu}\omega)) -(\tilde{\nabla}_{\nu}\omega)2\tilde{\phi}(\tilde{\nabla}_{\mu}\tilde{\phi})$$

again if we bring in the external metric the first term here is a total divergence so vanishes leaving:

$$(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega)\tilde{\phi}^2= -(\tilde{\nabla}_{\nu}\omega)2\tilde{\phi}(\tilde{\nabla}_{\mu}\tilde{\phi})$$

Plugging this into (1)

$$\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\tilde{\phi}^2= +\omega(\tilde{\nabla}_{\nu}\omega)2\tilde{\phi}(\tilde{\nabla}_{\mu}\tilde{\phi})-\omega(\tilde{\nabla}_{\nu}\omega)2\tilde{\phi}(\tilde{\nabla}_{\mu}\tilde{\phi})=0$$

So that takes care of the first term in the undesired Lagrangian density. We also have:

$$\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla} _{\nu}(\tilde{\phi}) \tilde{\phi}$$

$$\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\tilde{\phi}\right)\tilde{\phi}= -\omega(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\tilde{\phi})\tilde{\phi}-\omega(\tilde{\nabla}_{\nu}\tilde{\phi})(\tilde{\nabla}_{\mu}\tilde{\phi})$$
Use product rule again on the first term:

$$(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\tilde{\phi})\tilde{\phi}=\tilde{\nabla}_{\mu}\left[\tilde{\phi}(\tilde{\nabla}_{\nu}\tilde{\phi})\right]-(\tilde{\nabla}_{\mu}\tilde{\phi})(\tilde{\nabla}_{\nu} \tilde{\phi}) =-(\tilde{\nabla}_{\mu}\tilde{\phi})(\tilde{\nabla}_{\nu} \tilde{\phi})$$

So we get:


$$\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\tilde{\phi}\right)\tilde{\phi}= -\omega\left[ -(\tilde{\nabla}_{\mu}\tilde{\phi})(\tilde{\nabla}_{\nu} \tilde{\phi})\right]-\omega(\tilde{\nabla}_{\nu}\tilde{\phi})(\tilde{\nabla}_{\mu}\tilde{\phi})=0$$

The only other term not yet considered is:

This everything undesired vanishes.

I have used the symmetry of the metric and the divergence theorem to convert total divergences into surface terms that one can set to zero.