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  1. One has that $ds^{2} = g_{ij}(x)dx^{i}dx^{j}$. I often see that the interval is re-expressed with a time "seperation" of the form: $$ ds^{2} = g_{00}(x)dt^{2} + \tilde{g}_{ab}dx^{a}dx^{b} \;\; a,b = 1,2,3 $$ When can this be done?

  2. Why can the proper time infinitesimal always be written in the form (according to Wiki "Proper time"): $ d\tau = \sqrt{g_{00}(x)}dt \; ? $

Thank you in advance for your answers

Thomas Hasler
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1 Answers1

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1)

I believe this can be done always, but I am not sure. You need to get rid of the cross terms $g_{0a}$ and you have 4 coordinate transformations at your disposal to get rid of the 3 metric functions, while keeping $g_{00}$ positive (assuming the signature is (+,-,-,-)). This looks like a problem that has a solution.

2)

It cannot. The proper time is always attached to some worldline. For any worldline whatsoever, the general formula is: $$d\tau=c^{-1}ds=c^{-1}\sqrt{g_{\mu\nu}dx^\mu dx^\nu}$$ This reduces to your formula only along worldlines that keep $x^1$,$x^2$ and $x^3$ constant.

Umaxo
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  • For (1), what about a non-static spacetime, e.g. the Kerr spacetime? – Nihar Karve Nov 18 '20 at 10:56
  • @NiharKarve interesting question. The thing is, in Kerr spacetime it is the timelike killing vector that is not hypersurface orthogonal. But we are not assuming $t$ being given by this vector. – Umaxo Nov 18 '20 at 11:03
  • @Umaxo Thanks! Sorry to ask again but if you look at https://en.wikipedia.org/wiki/Proper_time#In_general_relativity under general relativity they say that "In the same way that coordinates can be chosen such that x1, x2, x3 = const in special relativity, this can be done in general relativity too." Is this not always the case? – Thomas Hasler Nov 18 '20 at 11:25
  • @ThomasHasler I see. You can find such coordinates for any worldline P along which you integrate, that is true. I just assumed you look for global coordinates in which the formula is true for any worldline. – Umaxo Nov 18 '20 at 12:01