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With regard to the density parameter derived from Friedmann Equations which is:

$$ Age = D_H\int_{z}^{\infty}\frac{1}{(1+z)\sqrt{\Omega_R(1+z)^4 + \Omega_M(1+z)^3 + \Omega_K(1+z)^2 + \Omega_L(1+z)^{(3(1+w))}}} dz $$

(setting $z$ to $0$ will provide the current age of the universe)

where $D_H=$ Hubble Distance, $z$ = Redshift, $\Omega_R =$ Radiation density, $\Omega_M =$ Matter density (incl. dark matter), $\Omega_K =$ Curvature, $\Omega_L =$ Dark energy density and $w=$ equation of state.

The above equation is used to compute age of the universe in almost any given density parameters except for when there is only matter-dominated scenario in which $\Omega_M > 1$. In such cases, because the integration is trying to compute the square root of negative numbers (negative redshifts; or scale factor ($a$) $>=1$ for future fate of the universe), we will obtain a complex answer.

What is the proper equation to use to compute for scenarios where $\Omega_M >1$ and with $a=1.5$ ?

$z= (1/a)-1 = -0.33$ for $a = 1.5$

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  • $\begingroup$ Everything looks positive...? $\endgroup$
    – Job Stancil
    Aug 11, 2018 at 10:49
  • $\begingroup$ $1+z$ is always positive. What's the problem? $\endgroup$
    – Indigo1729
    Aug 11, 2018 at 13:50
  • $\begingroup$ @GauthamAP Hi, I agree that as per the equation $1+z$ is always positive. However due to $K = 1 - (R+M+L)$ , $K$ will be negative (in this instance $-4$) when taking $M = 5$. Therefore we have the situation where we have to take the square root of a negative number! $\endgroup$
    – Vick
    Aug 11, 2018 at 14:39
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    $\begingroup$ You have assumed that $\Omega_M > 1$. Hence it has $k>0$. The constraint $\Omega_{tot}=1$ is only for a flat universe is not in general. Also since $\Omega_i 's$ represent densities they can never be negative. The formula is $\Omega_{tot} -1 = \frac{k}{a^{2} H^{2}}$ - From A.Liddle 'Modern Cosmology'. $\endgroup$
    – Indigo1729
    Aug 11, 2018 at 19:17
  • $\begingroup$ @GauthamAP If we use $\Omega_{tot} - 1$ for $\Omega_K$ instead, the results become very different when computing the age of the universe. For $\Omega_M = 5$ , the age in that case for $a = 1$ is $3.566 Gyr$. Whereas if we use the usual $1 - (\Omega_M + \Omega_R + \Omega_L)$ the age for same $a$ is $ 6.379 Gyr$ ! $\endgroup$
    – Vick
    Aug 12, 2018 at 5:57

1 Answer 1

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For example for the following density parameters: $\Omega_M = 5.0$ and $\Omega_L = 0$, the integration equation will compute till the point where the universe will stop expanding and start to contract due to a subsequent Big Crunch. This point for the above particular parameters is at scale factor $(a) = 1.25$. After this point the numbers will become negative and hence no computation for square root of negative numbers.

The above equation stands good therefore for any kind of parameters except when there is a reverse of the expansion (negative numbers for square root).

In Big Crunch scenarios, we just need to compute till the point where expansion ceases. After this, it is just a repeat of the data, albeit, in reverse order.Relative Expansion Rate

In order to complete the half circle we need to reverse the data gotten so far up to $a = 1.25$

In the end, with the above formula, we have been able to compute for $\Omega_M=5$ and for $a>1$ i.e: $a=1.25$ successfully as shown in the above picture.

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  • $\begingroup$ @Jim I used a specific scenario where $\Omega_M = 5$ and also corrected the scale factor for this particular scenario from the one in the question to the one I have to deal with, i.e instead of looking at $a=1.5$, I saw that only $a=1.25$ will yield anything useful. Also, using $\Omega_M = 5$ will yield $\Omega_K=-4$ which in effect becomes a closed curvature. This particular scenario will end in a Big Crunch. $\endgroup$
    – Vick
    Aug 17, 2018 at 18:31
  • $\begingroup$ @Jim I forgot to mention that the original question is based on a faulty assumption that states that for $\Omega_M > 1$ and for $ a>1$ the integration function will deal with the square root of negative numbers. This is not necessary as shown in the example in the answer: we integrated successfully $\Omega_M = 5$ and up to $ a=1.25$. $\endgroup$
    – Vick
    Aug 17, 2018 at 18:35
  • $\begingroup$ Okay, I now see that you were trying to indicate that the existing equation is valid in all cases but, by design, does not handle cases with a big crunch. It wasn't as noticeable as I had expected, but I see it there. I think you might have focused a little too much on one particular case, maybe add a final line summarizing your point at the end. You know, end on a high note $\endgroup$
    – Jim
    Aug 17, 2018 at 18:36

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