GR Text: Zee's Formulas & Where I'm Going Wrong

  • #1
363
4
It seems like Zee lost a trace in his new GR text, but I am sure it is me confusing things.

First, he establishes:

[itex]
log\: det\: M = tr\: log\: M[/itex]

Then, differentiating:

[itex]
(det\: M)^{-1} \: \partial (det\: M)=\partial (tr\: log\: M)=tr(\partial\: log\: M)=tr(M^{-1}\partial M)[/itex]

Then he applies to the metric, giving:

[itex]
\frac{1}{\sqrt{-g}}\partial _{\nu }\sqrt{-g}=\frac{1}{2g}\partial _{\nu }g=\frac{1}{2}\partial _{\nu }log\: g[/itex]

where g is the determinant of the metric.

Sure seems like he lost a trace on the very last term. Where I am going wrong here?
 
  • #2
In Zee's equation, ##g## means the determinant of the metric tensor. In more explicit notation, the steps are

##\frac{1}{\sqrt{-\det g}} \partial_\nu \sqrt{-\det g} = \frac{1}{2 \det g} \partial_\nu (\det g) = \frac{1}{2} \partial_\nu ({\rm tr} \log g) = \frac{1}{2} \partial_\nu (\log \det g)##

In what I've written here, ##g## is always a matrix and the determinant is kept explicit.
 

Suggested for: GR Text: Zee's Formulas & Where I'm Going Wrong

Replies
1
Views
861
Replies
3
Views
667
Replies
5
Views
788
Replies
5
Views
627
Replies
13
Views
1K
Replies
5
Views
913
Replies
4
Views
666
Replies
27
Views
1K
Replies
4
Views
856
Back
Top