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

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Zee's formulation in the GR text appears to have a missing trace in the final differentiation step involving the metric determinant. The key equations establish the relationship between the determinant of a matrix and its trace through logarithmic differentiation. Specifically, the expression for the metric determinant leads to a transformation that simplifies to a derivative of the logarithm of the determinant. The confusion arises in the interpretation of the determinant and its trace, particularly in the last term of the derivation. Clarification is needed on how the trace is applied in this context to resolve the apparent discrepancy.
dm4b
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It seems like Zee lost a trace in his new GR text, but I am sure it is me confusing things.

First, he establishes:

<br /> log\: det\: M = tr\: log\: M

Then, differentiating:

<br /> (det\: M)^{-1} \: \partial (det\: M)=\partial (tr\: log\: M)=tr(\partial\: log\: M)=tr(M^{-1}\partial M)

Then he applies to the metric, giving:

<br /> \frac{1}{\sqrt{-g}}\partial _{\nu }\sqrt{-g}=\frac{1}{2g}\partial _{\nu }g=\frac{1}{2}\partial _{\nu }log\: g

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?
 
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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.
 
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