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I know the definition of space-like and time-like coordinates in simple geometric, basically, we got: $$ds^2=dt^2-dx^2-dy^2-dz^2$$ so the coordinate with a positive contribution to the $ds^2$ is the timelike one, so far so good. but is it always the case that the metric has a $(+,-,-,-)$ form, so that the sign for one of them is different from the other one? what if I have a metric like $(+,+,-,-)$, then what would be the timelike one?

PS - I don't have a formal education in theoretical physics and I am self educating myself through various text books, any source or guidance is greatly appreciated.

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In general coordinates, the metric does not even have to be diagonal. However, for any metric you can always find a change of coordinates such that at a particular point in spacetime the metric is diagonal. It can be proven that the signs of the coefficients on the diagonal are the same (up to reordering, i.e. there will always be same number of positive and negative coefficients) for any choice of coordinates. Moreover, it can also be shown that the number of positive and negative coefficients does not depend on at which point the metric is diagonalized.

The number of positive and negative diagonal coefficients therefore is an intrinsic property of the metric. Typically this is denote something like $(-+++)$ or $(--++)$. In general relativity we are normally only interested in metrics with signature $(-+++)$ (or $(+---)$ if you are of that perverted persuasion).

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  • $\begingroup$ Is a metric with signature $(--++)$ Lorentz invariant? $\endgroup$ Commented Apr 8, 2020 at 12:31
  • $\begingroup$ @fogofmylife No. The Lorentz group (by definition) is the group of isometries of a diagonal metric with (-1,+1,+1,+1) on the diagonal. Note that almost all metrics with signature $(-+++)$, are in fact not Lorentz invariant. $\endgroup$
    – TimRias
    Commented Apr 8, 2020 at 12:35
  • $\begingroup$ @mmeent Thanks a lot. so if I understand correctly, in theory, there could be situation like (++--) but we either are not interested in them, meaning that they exclude most of the established and well known cases (rotating blackholes, ...), and if we find such metric we are not sure how to interpret the physical meaning of time and space out of it, yes? $\endgroup$
    – user81435
    Commented Apr 8, 2020 at 14:30

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