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Given these two expressions

1) $\sinh{d}=\frac{\sqrt{t^2−x^2}}{\sqrt{1−(t^2−x^2)}}$

2) $\sin{d}=\frac{\sqrt{t^2−x^2}}{\sqrt{1+(t^2−x^2)}}$

for distance $d$ from the origin $(0,0)$ to point $(x,t)$, which of these two options applies to de-Sitter space and which to anti de-Sitter space? For definiteness, assume $t$ is time and $(x,t)$ is time-like, that is $t^2-x^2$ is positive.

Does option (1) correspond to a space with the positive curvature and (2) to the negative one?

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Is there any reason in the former expression that you're implicitly constraining $t^2-x^2 < 1$? Beyond that, I'd recommend checking some extremal limits (eg. $t^2-x^2 \to \infty$, or whatever makes sense) and see which obtains the correct behavior. – Eugene Shvarts Jul 5 '12 at 1:27
In (1), the distance is defined only for $|t^2-x^2|<1$, which is what you get in a space with a hyperbolic distance measure. For instance, in the 2D hyperbolic space, $d$ has real values only for $|x^2+y^2|<1$. As for checking the limits, I don't see how that will help me decide which one is de-Sitter and which one is anti de-Sitter. – Andrey Sokolov Jul 5 '12 at 2:17
Sorry, I recall this being an easy way to tell between positive and negative curvature -- the correct application isn't coming to mind at the moment. – Eugene Shvarts Jul 5 '12 at 2:56
Hint: Anti De Sitter Space is hyperbolic, with negative curvature... – Alex R. Jul 5 '12 at 13:06
@Sam, Perhaps you could give me a hint as to what I should say to people claiming de Sitter has a hyperbolic distance measure, e.g. page 10 of – Andrey Sokolov Jul 5 '12 at 23:58

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