Assuming that the density of the universe is dominated by matter, so that $\rho = \rho_m$, how can the deceleration parameter today be shown to be $q_0 = \frac{\Omega_m}{2} - \Omega_{\Lambda}$

The acceleration equation; $\frac{\ddot{a}}{a}=-\frac{4\pi G}{3}(\rho + \frac{3p}{c^2}) + \frac{\Lambda}{3}$,

the continuity/fluid equation; $\dot{\rho}=-3H(\rho+\frac{p}{c^2})$,

and the Friedmann equation $H^2=(\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\rho - \frac{kc^2}{a^2} + \frac{\Lambda c^2}{3}$ are all needed as far as I can see, as well as $q=\frac{-\ddot{a}a}{\dot{a}^2}$.

$\rho_\Lambda = \frac{\Lambda}{8\pi G}$, and $p=-\rho_\Lambda c^2$ as matter does not contribute to $p$. I just can't manage it.



1 Answer 1


Start the definition of the deceleration parameter:

$$\begin{align} q_o &= -\frac{\ddot{a}a}{\dot{a}^2} \\ &= -\frac{\ddot{a}}{a}\left(\frac{a}{\dot{a}}\right)^2 \\ &= -\frac{\ddot{a}}{a}\frac{1}{H^2} \\ &= \left( \frac{4\pi G}{3}(\rho + \frac{3p}{c^2}) - \frac{\Lambda}{3} \right) \frac{1}{H^2}\\ \end{align}$$

We are assuming that only matter is present, so $\rho=\rho_m$, and we assume the matter is pressureless, $p=0$. Then we get:

$$ q_0 = \frac{4\pi G}{3H^2}\rho_m - \frac{\Lambda}{3H^2}$$

And now it's just a matter of using the definition of $\Omega$:

$$ \Omega_m = \frac{8\pi G}{3H^2}\rho_m $$

$$ \Omega_\Lambda = \frac{\Lambda}{3H^2}$$

  • $\begingroup$ Would $\Lambda$ not contribute to the pressure, where $p_\Lambda = -\rho_\Lambda c^2$? $\endgroup$
    – Noah P
    Commented Jul 14, 2016 at 16:09
  • $\begingroup$ Further to this, where do the definitions of $\Omega_m$ and $\Omega_\Lambda$ come from? $\endgroup$
    – Noah P
    Commented Jul 14, 2016 at 16:12
  • $\begingroup$ For the definition of the density parameter see this Wikipedia article. Note that $\Omega_\Lambda$ doesn't get the factor of $8\pi G$ because it's normally put on the left side of the Einstein equation. $\endgroup$ Commented Jul 14, 2016 at 16:20

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