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The scale factor satisfies the following equation - $$\frac{\dot a^{2}}{a^{2}} + \frac{k}{a^{2}} = \frac{8\pi G}{3} \rho(t)$$ where $k$ is constant proportional to total energy of the dynamical system. Now numerical value of $k$ can be absorbed in the definition of $a(t)$ by rescaling it hence $k$ can be treated having one of the three values $(0,\pm1)$. From this equation how can we show that cosmological scale factor $a(t)$ will have maximum value followed by a contracting phase to the universe if $k=1$ and $\Omega >1$, where $\Omega = \frac{\rho}{\rho_{c}}$.

I tried following - if $k=1$ and $\Omega >1$, we get - $$\frac{\dot a^{2}}{a^{2}} + \frac{1}{a^{2}} = \frac{8\pi G}{3} \rho_{c}\cdot(\Omega>1) >\frac{8\pi G}{3} \rho_{c}\cdot\Omega$$ $\implies \dot a^{2} + 1 > \frac{8\pi G}{3} \rho_{c}\cdot\Omega \cdot a^{2} \implies a < \bigg[ \frac{3(\dot a^{2} + 1)}{8\pi G\rho_{c}\cdot\Omega}\bigg]^\frac{1}{2} \implies a_{max} = \bigg[ \frac{3(\dot a^{2} + 1)}{8\pi G\rho_{c}\cdot\Omega}\bigg]^\frac{1}{2}$

But, from here how will I prove that this maximum value will be followed by a contracting phase to the universe?

To find the form of $a(t)$ we need to know $\Omega(t)$, Could somebody please tell what kind of models are available to determine $\Omega(t)$?

Ref: Theoretical Astrophysics, T.Padmanabhan, Vol 3, pg - 4

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  • $\begingroup$ I guess, you have to solve the differential equation for $a(t)$. $\endgroup$ – Frederic Thomas May 23 '18 at 15:36
  • $\begingroup$ @FredericThomas But we need $\Omega(t)$ for that. $\endgroup$ – luv_phy May 25 '18 at 5:06
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You don't know $\Omega(t)$. How $\Omega$ varies with time depends on the specific nature of $\Omega$, e.g. $\Omega_r \propto a^{-4}$, but $\Omega_m \propto a^{-3}$. A universe with different amounts of radiation and matter will have different behaviors for $\Omega(t)$.

However you do know that after this maximum value there will be a contracting phase. That comes from the definition of a maximum point - all nearby points have lower values than the maximum. So both approaching the maximum and going away from it, $a(t)$ must decrease. If it starts increasing again, there must be a minimum, which does not exist (from the differential equation you solved). Hence it keeps decreasing, until the assumptions that went into the calculation break down.

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  • $\begingroup$ Thanks for the answer. I understand your point, so what leads to the expansion of universe to a maximum and then to a contraction phase? $\endgroup$ – luv_phy May 25 '18 at 10:54
  • $\begingroup$ @luv_phy as in, why is the universe expanding? $\endgroup$ – Allure May 25 '18 at 11:07

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