# Jacobi's matric formulation for tensors

Hi I am trying to derive the E field equation and am stuck using the Jacobi formula, is this correct: $$\delta \det g_{\mu \nu} = Tr(ADJ(g_{\mu\nu}\delta g_{\mu\nu})=\det(g_{\mu\nu})Tr(g^{\mu\nu}\delta g_{\mu \nu})$$ Then how can we remove the trace, or should it be: $$\delta \det g_{\mu \nu} =\det(g_{\mu\nu})Tr(g^{-1}\delta g)=\det(g_{\mu\nu})(g\delta g)^{\mu}_{\mu}=\det(g_{\mu\nu})(g\delta g)^{\mu}_{\mu}=\det(g_{\mu\nu})(g^{\mu\nu}\delta g_{\mu\nu})$$

-

You should be careful not to mix symbolic and index notation. $\text{Tr}(g^{\mu \nu}\delta g_{\mu \nu})$ does not make sense since $g^{\mu \nu}\delta g_{\mu \nu}$ is just a number. The correct symbolic notation would be:
$$\delta \text{det}(\mathbf{g})=\text{Tr}(\text{adj}(\mathbf{g})\delta \mathbf{g})=\text{det}(\mathbf{g})\text{Tr}(\mathbf{g}^{-T}\delta \mathbf{g}))$$ since $\text{adj}(\mathbf{g})=\text{det}(\mathbf{g})\mathbf{g}^{-T}$. Now you can transform to index notation and Einstein convention: $$\delta \text{det}(g_{\mu \nu})=\text{det}(g_{\mu \nu})(g^{\mu \nu} \delta g_{\mu \nu})$$
Thanks, could you if possible explain the transformation to index notation, I just don't see why you can effectively replace g with $g_{\mu\nu}$ –  user21119 Feb 25 '13 at 20:05
@user21119 $g_{\mu\nu}$ is a generic component of the metric $\mathrm{g}$, just as $A_{3,4}$ is the quantity in row 3 and column 4 of matrix $A$. Index notation is a clever way of writing scalar equations (sometimes many simultaneously) relating the components of tensors, without obscuring the tensorial nature of the equations. Unfortunately many people and textbooks abuse this notation, letting an indexed term stand for the entire object. –  Chris White Feb 26 '13 at 1:31
ok thx think i got it, so in including say $\mu=\nu=2$ is like saying this particular value, but them being there as $\mu$ and $\nu$ is basically a sum over all values, so just like g. –  user21119 Feb 26 '13 at 13:05