What is the Partial Derivative of Metric in Gravitational Action?

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SUMMARY

The discussion focuses on the calculation of the partial derivative of the metric tensor \( g_{ab} \) in the context of gravitational action variations. The expression for the derivative is established as \( \frac{d g_{ab}}{d g_{cd}} \), which simplifies to the symmetrized Kronecker delta form \( \frac{1}{2}(\delta_{ac} \delta_{bd} + \delta_{ad} \delta_{bc}) \) when all components of \( g \) are independent. A key point raised is the concern regarding the factor of \( \frac{1}{2} \) when \( a \neq b \), which could lead to confusion in specific cases where \( a = c \) and \( b = d \).

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krishna mohan
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Could someone tell me what would be the partial derivative...

\frac{ d g_{ab} }{ d g_{cd}}

??
Such expressions occur when one does variations of gravitational action...

Note: For some reason, the d needed for derivative is not appearing in the post...although it was visible in the preview...
 
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Depends on exactly what you're doing, but if you're doing a variation in which all 10 components of g are independent, then your expression is equal to δac δbd, or more precisely the symmetrized version of that,
ac δbd + δad δbc)/2
 
Okay...

But if a is not equal to b, then won't the symmetrised version have an unwanted factor of (1/2)?

i.e if a=c and b=d, one would like to have 1 as the answer...but the symmetrised version would give 1/2...
 

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