In general relativity we have [itex]c^2ds^2 = c^2dt^2 - dx^2 -dy^2-dz^2[/itex]. From this we can derive the not commonly used equation:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

(c\frac{ds}{dt})^2 + (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2 =c^2

\qquad\qquad\qquad (1)

[/tex] I think this is the "velocity" squared of an object relative to an observer who's own proper time is [itex]t[/itex]. I find this equation slightly remarkable, because of the constancy of this relative velocity which forces a tradeoff between speed through space and speed through time [itex]ds/dt[/itex]: the slower you move the faster you get older ;).

Now I wonder what the same would be for general relativity. The respective equation for [itex]ds[/itex] is [itex]ds^2 = g_{\mu\nu} dx^\mu dx^\nu[/itex]. With [itex]i,j\in\{1,2,3\}[/itex] and [itex]t:=x^0[/itex] I can expand the sum on the right a bit more explicit to get:

[tex]

ds^2= g_{ij} dx^i dx^j + 2 g_{0j} dt dx^j + g_{00}dt^2

[/tex] There I have the [itex]dt^2[/itex] again and could just divide by it, where it not for the mixed term. Is it possible to "simplify" this equation more into the direction of (1) such that all the ##dt## are in the denominator? Under which circumstances would the ##g_{0j}## be zero such the cross terms with ##dt## disappear gracefully?

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# 4-velocity of an observed object in GRT

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