Variation of determinant

  • #1
christodouloum
35
0
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

[tex]\delta(h)=2 h h_{\alpha\beta}\delta(h^{\alpha\beta})[/tex]

can someone give a hint on how this is computed?
 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
Insights Author
13,238
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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

[tex]\delta(h)=2 h h_{\alpha\beta}\delta(h^{\alpha\beta})[/tex]

can someone give a hint on how this is computed?

That's inaccurate

[tex] \delta(h^2)=2h \delta h = 2 h^2 h^{\alpha\beta}\delta h_{\alpha\beta}[/tex]

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.
 
  • #3
christodouloum
35
0
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?
 
  • #4
Avodyne
Science Advisor
1,396
90
Start from det = exp Tr log.
 

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