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 $i=\sqrt{-1}$?

7 comments:

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

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    Replies
    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.

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

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    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

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

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

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  3. sorry for the two equations I messed up again

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

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

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  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

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

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