How is the variation of the determinant of the metric computed?

[christodouloum]

in varying an action like Polyakov's action with respect to the metric on the world sheet we have to consider the variation of the square root of the determinant. I have not found how to express the variation of the determinant of the metric. From reverse engineering I found that

$$\delta(h)=2 h h_{\alpha\beta}\delta(h^{\alpha\beta})$$

can someone give a hint on how this is computed?

 

[dextercioby]

in varying an action like Polyakov's action with respect to the metric on the world sheet we have to consider the variation of the square root of the determinant. I have not found how to express the variation of the determinant of the metric. From reverse engineering I found that

$$\delta(h)=2 h h_{\alpha\beta}\delta(h^{\alpha\beta})$$

can someone give a hint on how this is computed?

That's inaccurate

$$\delta(h^2)=2h \delta h = 2 h^2 h^{\alpha\beta}\delta h_{\alpha\beta}$$

The indexed part in the last term of the muliple equality comes simply by considering the way one computes the inverse of a square matrix. The argument can be found in most GR books when discussing the HE action and, of course, in maths books when discussing integration on arbitrary manifolds.

 

[christodouloum]

Ok I derived it for the 2d case by writing out the components. Still can anyone provide a general proof or a hint for it?

 

[Avodyne]

Start from det = exp Tr log.