## Wednesday, June 13, 2012

### static universe

would it be possible to construct a FLRW-cosmology with no time dependence? what would be needed for such a cosmology? what's a sensible starting point for constructing this cosmology? would the solution be stable?

bonus question: what's $|i!|$ with $i=\sqrt{-1}$?

1. hint for the bonus question: there's a generalisation of the factorial for real numbers, which can be extended to complex numbers.

1. Of course one can arrange the cosmological constant $\Lambda$ in a way to achieve a static universe corresponding to $\dot{a}$ = $\ddot{a}$ = 0.

If we consider Friedmann's equations

$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3} \rho -\frac{k}{a^2} +\frac{\Lambda}{3}$$
$$\frac{\ddot{a}}{a} = -\frac{4 \pi G}{3} \left(\rho + 3p\right) + \frac{\Lambda}{3}$$

In case of cold dark matter p=0 and negligible baryonic pressure we obtain the conditions:

$$\rho = \frac{3k}{4\pi G a^2}$$

$$\Lambda = 4\pi G \rho$$

The solution is of course not stable, because slight perturbations of the given choice of $\Lambda$ lead to completely different (non-static) results.

2. sorry, $$\rho = \frac{k}{4\pi G a^2}$$

3. I think you can also try with a properly chosen dynamical DE model
with $\rho_Q(a) = \rho_Q^{(0)} w(a)$ (at least mathematically):

$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3} \left(\rho+\rho_Q) -\frac{k}{a^2} = 0$$

$$\frac{ddot{a}}{a} = - \frac{4\pi G}{3} \left(\rho+\rho_Q (1+3w)\right) = 0$$

If I haven't messed up with this, this leads to the following constrains for $w$ and $\rho_Q$:

$$\rho_Q = -\rho + \frac{3k}{8\pi G a^2}$$

and

$$w(a) = \frac{\frac{k}{8 \pi G a^2}}{\frac{3k}{8\pi G a^2}-\rho}$$

2. can you do with one fluid less? :)

3. sorry for the two equations I messed up again

$$H^2(a) = \frac{8\pi G}{3} (\rho+\rho_Q) - \frac{k}{a^2} = 0$$

$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3} (\rho+\rho_Q (1+3w))= 0$$

4. I think it works like this:

$$\left|i!\right| = \left|\Gamma(i+1)\right| = \sqrt{\Gamma(i+1)\overline{\Gamma(i+1)}}$$

Then we make use of the relation

$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$$

for $z=i$. Thus

$$\Gamma(1-i) \Gamma(i) = \frac{\pi}{\sin(\pi i)} = \frac{\pi}{i \sinh(\pi)}$$

Using $\Gamma(i+1)= i \Gamma(i)$ we have

$$\Gamma(1-i) \Gamma(1+i) = \frac{i \pi}{i \sinh(\pi i)}$$

Thus

$$\left|i!\right| = \sqrt{\Gamma(i+1)\overline{\Gamma(i+1)}} = \sqrt{\frac{ \pi}{\sinh{\pi}}} \approx 0.52 ...$$