- #1

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- 605

At first he says that under rotations, Cartesian coordinates change by ##x^\mu \rightarrow x^\mu+\xi^\mu ## where ## \xi^0=0 \ , \ \xi^i=\epsilon^{ij}x_j ## and ## \epsilon^{ij}=-\epsilon^{ji} ## is a set of arbitrary infinitesimal constants. The change in functional form of the metric is given by ## \delta g^{\mu\nu}=-\xi^\sigma \partial_\sigma g^{\mu\nu}+g^{\mu \sigma}\partial_\sigma \xi^\nu+g^{\nu\sigma}\partial_\sigma \xi^\mu ## so if we set ## \delta g_{\mu\nu}=0 ##(yeah, he sets the covariant thing to zero!), we'll get ## \xi^\sigma \partial_\sigma g^{\mu\nu}=g^{\mu \sigma}\partial_\sigma \xi^\nu+g^{\nu\sigma}\partial_\sigma \xi^\mu##. Then by considering the several different values of ## \mu ## and ## \nu ##, he gets several equations which constrain the form of the metric.

When he sets ## \mu=\nu=0##, he gets ## (\partial_i g_{00})x_j=(\partial_j g_{00})x_i ##. Which I'm not sure how he gets it.

By setting ## \mu=0##, he gets ## g_{0i}\epsilon^i_{\ j}+(\partial g_{0j})\epsilon^{ik}x_k=0 ##. He says this indicates that ## g_{0i}dx^idt=a_1(r,t)x^idx_idt=f(r,t)dr dt ##(where ##a_1## and f are arbitrary functions. But I neither understand how he gets the equation nor how he gets that form of the metric from the equation.

Then he assumes ## \mu,\nu \neq 0## which gives him ## g_{ik}\epsilon^k_j+g_{kj}\epsilon^k_i+(\partial_k g_{ij})\epsilon^{kl}x_l=0## . Then he says it shows that ##g_{ij}dx^idx^j=[a_2(r,t)\delta^{ij}+a_3(r,t)x^ix^j]dx_idx_j ## which again I don't get the whole thing!

Any help is much appreciated!

Thanks