- #1
LAHLH
- 409
- 1
Hi,
So if we have the Lagrange density for a massless scalar field: [tex] 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) [/tex]
Then under a conformal transformation [tex] g_{\mu\nu}=\omega^{-2}\tilde{g_{\mu\nu}} [/tex], then the Ricci sclar goes to [tex] 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) [/tex]. We also have [tex] g^{\mu\nu}=\omega^2\tilde{g}^{\mu\nu} [/tex] and [tex] \tilde{\nabla}_{\mu}\phi=\nabla_{\mu}\phi [/tex]
Using all these suggests that,
[tex] \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) [/tex]
Which can be written as,
[tex] \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) [/tex]
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?
So if we have the Lagrange density for a massless scalar field: [tex] 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) [/tex]
Then under a conformal transformation [tex] g_{\mu\nu}=\omega^{-2}\tilde{g_{\mu\nu}} [/tex], then the Ricci sclar goes to [tex] 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) [/tex]. We also have [tex] g^{\mu\nu}=\omega^2\tilde{g}^{\mu\nu} [/tex] and [tex] \tilde{\nabla}_{\mu}\phi=\nabla_{\mu}\phi [/tex]
Using all these suggests that,
[tex] \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) [/tex]
Which can be written as,
[tex] \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) [/tex]
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?