maverick280857
- 1,774
- 5
[SOLVED] Diffeomorphism invariance of the Polyakov action
Hi,
I'm struggling with something that is quite elementary. I know that the Polyakov action is diffeomorphism invariant and Weyl invariant. Denoting the world-sheet coordinates \sigma^0 = \sigma and \sigma^1 = t and the independent world-sheet metric by h_{\alpha\beta}(\sigma), the Polyakov action can be written as
S_{P} = -\frac{T}{2}\int d^{2}\sigma \, \sqrt{-h}h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu
where T is the string tension and \alpha, \beta run over 0 and 1 (labeling \sigma and t).
Now, if I define
\delta_{E}h_{\alpha\beta}(\sigma) = h_{\alpha\beta}'(\sigma) - h_{\alpha\beta}(\sigma)
as the Einstein variation of the independent world-sheet metric, where the primed coordinates denote the transformed world-sheet coordinates under a diffeomorphism defined by
{\sigma'}^{\gamma} = \sigma^\gamma - \xi^\gamma(\sigma)
then I would like to show that
\delta_E h_{\alpha\beta} = \partial_\alpha \xi^\gamma h_{\gamma\beta} + \partial_\beta \xi^\gamma h_{\alpha\gamma} + \xi^\gamma \partial_\gamma h_{\alpha\beta}
This is supposed to follow from chain rule, but I do not quite get all the terms.
My working
h'_{\gamma\delta}(\sigma') = h_{\alpha\beta}(\sigma)\frac{d\sigma^\alpha}{d{\sigma'}^\gamma}\frac{d{\sigma}^\beta}{d{\sigma'}^\delta}
Now by definition,
\sigma^\alpha = {\sigma'}^\alpha + \xi^\alpha(\sigma)
So,
\frac{d\sigma^\alpha}{d{\sigma'}^\beta} = \delta^\alpha_\beta + \frac{d\sigma^\delta}{d{\sigma'}^\beta}\frac{d{\xi}^\alpha(\sigma)}{d {\sigma}^{\delta}}
But this seems to give only the first two terms of \delta_E h and not the third one, which is the usual derivative (change along gradient) term that I'd expect to get.
Any suggestions will be greatly appreciated.
Thanks!
Hi,
I'm struggling with something that is quite elementary. I know that the Polyakov action is diffeomorphism invariant and Weyl invariant. Denoting the world-sheet coordinates \sigma^0 = \sigma and \sigma^1 = t and the independent world-sheet metric by h_{\alpha\beta}(\sigma), the Polyakov action can be written as
S_{P} = -\frac{T}{2}\int d^{2}\sigma \, \sqrt{-h}h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu
where T is the string tension and \alpha, \beta run over 0 and 1 (labeling \sigma and t).
Now, if I define
\delta_{E}h_{\alpha\beta}(\sigma) = h_{\alpha\beta}'(\sigma) - h_{\alpha\beta}(\sigma)
as the Einstein variation of the independent world-sheet metric, where the primed coordinates denote the transformed world-sheet coordinates under a diffeomorphism defined by
{\sigma'}^{\gamma} = \sigma^\gamma - \xi^\gamma(\sigma)
then I would like to show that
\delta_E h_{\alpha\beta} = \partial_\alpha \xi^\gamma h_{\gamma\beta} + \partial_\beta \xi^\gamma h_{\alpha\gamma} + \xi^\gamma \partial_\gamma h_{\alpha\beta}
This is supposed to follow from chain rule, but I do not quite get all the terms.
My working
h'_{\gamma\delta}(\sigma') = h_{\alpha\beta}(\sigma)\frac{d\sigma^\alpha}{d{\sigma'}^\gamma}\frac{d{\sigma}^\beta}{d{\sigma'}^\delta}
Now by definition,
\sigma^\alpha = {\sigma'}^\alpha + \xi^\alpha(\sigma)
So,
\frac{d\sigma^\alpha}{d{\sigma'}^\beta} = \delta^\alpha_\beta + \frac{d\sigma^\delta}{d{\sigma'}^\beta}\frac{d{\xi}^\alpha(\sigma)}{d {\sigma}^{\delta}}
But this seems to give only the first two terms of \delta_E h and not the third one, which is the usual derivative (change along gradient) term that I'd expect to get.
Any suggestions will be greatly appreciated.
Thanks!
Last edited: