The most general spacetime metric is given by $$ds^2=g_{\mu\nu}dx^\mu dx^\nu=c^2dt^2+g_{0i}dtdx^i+g_{ij}dx^i dx^j.$$

Why is the second term said to violate isotropy of 3-space?

It is true that spatial displacements $dx^i$ and $-dx^i$ contribute differently to the metric on the surface of a sphere, which is isotopic. Therefore, we cannot have terms like $dxdy$ etc. in the metric.

Can we make the same argument here as well because we are talking about isotropy of 3-space, not of spacetime? So how can we draw a conclusion about the spacetime metric from the isotropy of 3-space?

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    $\begingroup$ Consider a rotation of $\pi$ around some axis inducing the transformation $x_i \to -x_i$. Then $g_{0i} dt dx^i \to - g_{0i} dt dx^i $. $\endgroup$ – Slereah Jul 9 '17 at 21:59

The sub/superscripts $i$, $j$ represent the different spatial dimensions to be summed over in the metric. Because the second term contains a $0$ and only $i$s and is thus direction dependent, making it anisotropic.


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