Variation with scalar field coupled to gravity in 2D

[ismaili]

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!

 

[LightHero]

A:Let me give you some hints:The variation of $S_k$ must be a total derivative (Bianchi identity).Compute the variation of the Ricci scalar and the volume factor $\sqrt{h}$.Write down the Euler-Lagrange equations.Define the energy-momentum tensor as the variation of the action with respect to the metric.To answer your question, $\phi$ does not appear in the variation of $\sqrt{h}$, so it does not contribute to the energy-momentum tensor.

 

[sir-pinski]

The variation of the action with respect to the metric tensor h^{\alpha\beta} can be expressed as:

\delta S_k = \int d^2\sigma \left[\frac{\delta S_k}{\delta h^{\alpha\beta}}\delta h^{\alpha\beta}\right]

To find the energy-momentum tensor, we need to calculate the functional derivative of the action with respect to the metric tensor:

\frac{\delta S_k}{\delta h^{\alpha\beta}} = \sqrt{h}\left(\partial_\alpha\phi\partial_\beta\phi - \frac{i}{2}kR^{(2)}\phi\right) - \frac{1}{2}\sqrt{h}k\phi\frac{\delta R^{(2)}}{\delta h^{\alpha\beta}}

To find the second term, we can use the fact that the variation of the scalar curvature is given by:

\delta R^{(2)} = -\frac{1}{2}\partial_\alpha\partial_\beta h^{\alpha\beta}

Therefore, we can write the second term as:

-\frac{\delta R^{(2)}}{\delta h^{\alpha\beta}} = \frac{1}{4}\partial_\alpha\partial_\beta h^{\alpha\beta}

Substituting this into the expression for the functional derivative, we get:

\frac{\delta S_k}{\delta h^{\alpha\beta}} = \sqrt{h}\left(\partial_\alpha\phi\partial_\beta\phi - \frac{i}{2}kR^{(2)}\phi\right) + \frac{1}{4}\sqrt{h}k\phi\partial_\alpha\partial_\beta h^{\alpha\beta}

Now, we can use the fact that the variation of the volume factor \sqrt{h} is given by:

\delta\sqrt{h} = \frac{1}{2}\sqrt{h}h^{\alpha\beta}\delta h_{\alpha\beta}

Substituting this into the expression for the energy-momentum tensor, we get:

T_{\alpha\beta}^k = \frac{2}{\sqrt{h}}\frac{\delta S_k}{\delta h^{\alpha\beta}} = \partial_\alpha\phi\partial_\beta\phi -