Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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: $$\delta \det g_{\mu \nu} =\det(g_{\mu\nu})Tr(g^{-1}\delta g)=\det(g_{\mu\nu})(g\delta g)^{\mu}_{\mu}=\det(g_{\mu\nu})(g\delta g)^{\mu}_{\mu}=\det(g_{\mu\nu})(g^{\mu\nu}\delta g_{\mu\nu}) $$

share|improve this question

1 Answer 1

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: $$\delta \text{det}(g_{\mu \nu})=\text{det}(g_{\mu \nu})(g^{\mu \nu} \delta g_{\mu \nu}) $$

share|improve this answer
    
Excellent thank you –  user21119 Feb 25 '13 at 19:39
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.