# Cosmological densities when universe was/will be a different size

Supposing a spatially flat, matter-dominated universe, the expression for the Hubble parameter is:

$$H^{2}=H_{0}^{2}\left(\frac{\Omega_{m}}{a^{3}}+\Omega_{\Lambda}\right)$$

The previous equation then allows us to interpret the evolution of matter density as $$\Omega_{m}(a)=\frac{\Omega_{m}}{a^{3}}$$. What I'm curious about is this - what was the value of $$\Omega_{m}$$ when the universe was one half its current size? What will its value be when it becomes two, ten, $$n$$ times its current size? Have I made an oversimplification or am I missing the point entirely?

You have to be very careful of the notation here, as different authors use different conventions. In the notation you use in your first equation, $$\Omega_m$$ is the matter density at $$a=1$$ relative to the critical density at $$a=1$$. To be explicit, $$\Omega_m=\frac{\rho_m}{\rho_{\rm crit}}=\frac{\rho_m}{\left(\frac{3H_0^2}{8\pi G}\right)}$$ Notice in particular that the critical density is a constant with time because it is defined in terms of $$H_0$$. To be clearer, I'm going to denote this $$\Omega_{m,0}$$ instead, analogous to $$H_0$$ being the value of $$H$$ at $$a=1$$. Then:
$$\Omega_m(a)=\frac{\Omega_{m,0}}{a^3}$$
Once the sloppy notation is cleaned up, it's easy to see that when the (linear) size of the universe doubles, the matter density drops by a factor of $$8$$, if it is a quarter the size then the density is $$4^3=64$$ times higher, etc.
• That's in line with what my initial thoughts were but I've got two "gripes" with this. The first one is that, after I tried thinking about this on my own, I came upon problem 7.4 in Andrew Liddle's book which asks to compute the value of $\Omega_{m}$ when the universe is five times its current size, supposing its value today is 0.3. His answer states that it's approximately 0.003 which isn't something you get dividing by $5^{3}$. (I'll continue in the following comment) – Kandrax Apr 30 at 13:07
• The other problem lies in trying to evaluate $z$ for larger or smaller sizes. Does a scaling of the universe to 10 times its size imply that $a=1/10$ or $a=10?$. I believe the answer is the former but then I still have to for the latter value when describing a universe one tenth of its current size which leads me to a negative value of $z$ - and I have no idea how to reconcile with this. – Kandrax Apr 30 at 13:10
• @Kandrax can't say for the book problem, as I don't have a copy and like I said, you have to be very careful about the notation with $\Omega$'s. As to $a$, when the universe is $10\times$ the size it is now, that is $a=10$, and this does lead to a negative redshift, which is perfectly well-defined. In the Universe we know, positive redshift is in the past when the Universe was smaller, negative redshift theoretically refers to the future when it is bigger (though you'd never observe a source in the future to actually measure a negative $z$). – Kyle Oman Apr 30 at 14:08