Variation of determinant

  • #1

Main Question or Discussion Point

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
12,985
540
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
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
87
Start from det = exp Tr log.
 

Related Threads on Variation of determinant

Replies
1
Views
414
  • Last Post
Replies
1
Views
1K
  • Last Post
2
Replies
30
Views
1K
  • Last Post
Replies
7
Views
2K
Replies
6
Views
3K
  • Last Post
Replies
3
Views
1K
Replies
7
Views
3K
Replies
4
Views
2K
  • Last Post
Replies
3
Views
564
Top