# Variation with scalar field coupled to gravity in 2D

1. Sep 14, 2008

### ismaili

The two dimensional action is:
$$S_k = \int d^2\sigma\sqrt{h}\left(\partial_\alpha\phi\partial^\alpha\phi - \frac{i}{2}kR^{(2)}\phi\right)$$
where $$k$$ is a constant, $$R^{(2)}$$ is the two dimensional scalar curvature. I'm trying to derive the following energy momentum tensor:
$$T_{\alpha\beta}^k = \partial_\alpha\phi\partial_\beta\phi - \frac{1}{4}ik\partial_\alpha\partial_\beta\phi$$
The first term can be understood as variation from $$\partial_\alpha\phi\partial_\beta\phi h^{\alpha\beta}$$, but I can't understand how the variation of $$\frac{i}{2}kR^{(2)}\phi$$ w.r.t. $$h^{\alpha\beta}$$ yields $$\frac{1}{4}\partial_\alpha\partial_\beta\phi$$. There is one more thing that I'm confused, we didn't consider the variation of the volume factor $$\sqrt{h}$$?! And, the variation of the scalar curvature yields the surface term in the usual four dimensional gravity, but here, it seems that we have to write $$R^{(2)} = \frac{1}{2}\partial_\alpha\partial_\beta h^{\alpha\beta}$$, however, I didn't see the reason...
Is there anybody can solve this for me or can give me some hint of how to do it, many thanks in advance!