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Say we have the transform such that $x^i \rightarrow (x')^{i'}=M^{i'}_i(x)$ where $M$ is an orthogonal rotation matrix

I've been asked to show that a general metric $g_{ij}(x)$ invariant under the transformation using the fact that $M^{i'}_iM^{j'}_j \delta_{i'j'} = \delta_{ij} $[1]

Here is my thinking,

The line element $ds^2 = g_{ij}(x)dx^i dx^j$ becomes $\bar{g}_{i'j'}(x')d(M^{i'}_i x^i)d(M^{j'}_j x^j)$, but when I use the fact that $M$ is constant so they can be taken out of the derivatives and use relation [1] I get

$ds^2 = \bar{g}_{i'j'}(x')dx^i dx^j \delta_{ij}$

which doesn't make sense as the i' and j' indices are not contracted

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  • $\begingroup$ I just realised that I implicitly assumed the some properties of the metric in my previous answer. Are you sure the metric is general? Are you sure you are not only concerned with its spatial elements? I cannot see why this should be true for a general, curved geometry. $\endgroup$ – Ollie113 May 8 at 13:56
  • $\begingroup$ Perhaps consider a LIF and prove it there? $\endgroup$ – Ollie113 May 8 at 14:04
  • $\begingroup$ @Ollie113 It also specifies that the matrix is a rotation matrix, so I think it means on the spatial coordinates. I'll correct the question with that. $\endgroup$ – Nonsematter May 8 at 15:24
  • $\begingroup$ I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. $\endgroup$ – Ben Crowell May 8 at 15:29

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