# Should Anti de Sitter Universe collapse exponentially? (n0)

The de Sitter Universe is described by the Friedmann equation with $k=0$, $\rho=0$ and a positive cosmological constant $\Lambda$:

$$\Big(\frac{\dot{a}}{a}\Big)^2=\frac{\Lambda c^2}{3}$$

The solution for the scale factor $a(t)$ is given by:

$$a(t) \sim e^{\pm\sqrt{\Lambda c^2/3}\ t}$$

Thus a Universe with a positive cosmological constant $\Lambda$ expands exponentially. I suppose the solution with a negative square root is unphysical.

I understand that it is possible to have an anti-de Sitter Universe with a negative cosmological constant $-\Lambda$.

I have heard people say that such a Universe would shrink exponentially but surely it should oscillate?

$$a(t) \sim e^{\pm i\sqrt{\Lambda c^2/3}\ t}$$

Correction

Andrew pointed out that the metric must be real so that $a(t) \sim e^{iwt}$ is not a legitimate solution.

Instead of the first Friedmann equation I can take the acceleration equation for a de Sitter Universe with positive cosmological constant $\Lambda$ given by:

$$\frac{\ddot{a}}{a}=\frac{\Lambda c^2}{3}$$

The solution is given by:

$$a(t) \sim \sinh{\sqrt{\Lambda c^2/3}\ t}$$

Let us consider an anti-de Sitter Universe, with a negative cosmological constant $-\Lambda$. The acceleration equation is given by:

$$\frac{\ddot{a}}{a}=-\frac{\Lambda c^2}{3}$$

The real solution to this equation is given by:

$$a(t) \sim \sin{\sqrt{\Lambda c^2/3}\ t}$$

Therefore the anti-de Sitter Universe collapses as expected - not exponentially but like a sine function.

By substituting into the first Friedmann equation one finds that for both anti-de Sitter and de Sitter Universes we have negative spatial curvature $k \sim -\Lambda c^2/3$ (again as pointed out by Andrew).

• The oscillatory behavior is present in the final formula for the scale factor(just remember the way you used in electrodynamics the exponential numbers) – Nikey Mike Mar 4 '16 at 21:39
• Here the deceleration can be calculated in eq. 32. You can also solve easily the Friedman eqs. numerically using a Runge Kutta. – Nikey Mike Mar 4 '16 at 21:42
• It is called anti-de Sitter space – Slereah Mar 4 '16 at 21:52

The metric should be real, so $a(t)\sim e^{i \omega t}$ is not a legitimate solution of Einstein's equations [unless you want to talk about instanton solutions, in which case you are interested in complex geometries, but that is a whole different kettle of fish]. Also note that the Friedman equation is non-linear so you can't just take linear combinations of the positive and negative frequency solutions to make $a$ real.

What you are discovering is that there are no spatially flat slicings of Anti-de Sitter space. Or if you prefer, when $\Lambda<0$ we must also have spatial curvature (in FLRW-type coordinates).

Even if there was no matter, and just a negative CC, the most general form of the Friedman equation following from Einstein's equations is not what you wrote but instead $$H^2 = \frac{8\pi G}{3}\left(-\frac{k}{a^2} + \frac{1}{3}\Lambda\right)$$ where $H=\dot{a}/a$ and $k=+1,-1$, or $0$ tells you if the spatial curvature is positive, negative, or zero.

If we want real solutions of this (which we do) when $\Lambda<0$, then we need $k<0$ which corresponds negative spatial curvature.