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Variation with scalar field coupled to gravity in 2D

  1. Sep 14, 2008 #1
    The two dimensional action is:
    [tex]S_k = \int d^2\sigma\sqrt{h}\left(\partial_\alpha\phi\partial^\alpha\phi - \frac{i}{2}kR^{(2)}\phi\right)[/tex]
    where [tex]k[/tex] is a constant, [tex]R^{(2)}[/tex] is the two dimensional scalar curvature. I'm trying to derive the following energy momentum tensor:
    [tex]T_{\alpha\beta}^k = \partial_\alpha\phi\partial_\beta\phi - \frac{1}{4}ik\partial_\alpha\partial_\beta\phi[/tex]
    The first term can be understood as variation from [tex]\partial_\alpha\phi\partial_\beta\phi h^{\alpha\beta}[/tex], but I can't understand how the variation of [tex]\frac{i}{2}kR^{(2)}\phi[/tex] w.r.t. [tex]h^{\alpha\beta}[/tex] yields [tex]\frac{1}{4}\partial_\alpha\partial_\beta\phi[/tex]. There is one more thing that I'm confused, we didn't consider the variation of the volume factor [tex]\sqrt{h}[/tex]?! 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 [tex]R^{(2)} = \frac{1}{2}\partial_\alpha\partial_\beta h^{\alpha\beta}[/tex], 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!
  2. jcsd
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