Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Based on the Friedmann equation for a universe with only cosmological constant, $$\left(\frac{\dot{a}}{a}\right)^2 \sim \Lambda$$

I would expect the scale factor $a(t) \sim e^{-it}$ if $\Lambda < 0$. Is this an anti de Sitter universe? How do we interpret the scale factor?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Because your solution is mathematically valid but unacceptable physically because $a(t)$ is required to be real, you have proved that the equation has no physical solutions. In other words, a negative-cosmological-constant space can't be sliced into flat spatial slices (you have assumed that the spatial curvature term $k$ is absent as well: with a negative spatial curvature, you could get solutions).

In particular, the AdS space can't be sliced into flat slices. This is also clear if you think about the isometry group. $AdS_4$ has the $SO(3,2)$ isometry group and the Euclidean affine symmetry of $R^3$, $SO(3)$ semidirect product with translations (which must commute with each other), is clearly not a subgroup of it. However, the $H^3$ negatively curved hyperbolic space has the $SO(3,1)$ isometry group (includes "translations" that don't commute) but it is a subgroup of $SO(3,2)$ so you can find solutions to the FRW equations with the negative $k$ added.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.