Estimate of time of the universe

[bananabandana]

Homework Statement

Show that the time of matter-radiation equality, t_{eq} can be written:
$$ t_{eq} =\frac{a_{eq}^{\frac{3}{2}}}{H_{0}\sqrt{\Omega_{m}}} \int_{0}^{1} \frac{x}{\sqrt{x+1}} dx $$

Homework Equations

$$ t = \int_{0}^{t} dt = \int_{0}^{a} \frac{1}{H(a)} \frac{da}{a} $$ [Given]
$$ H^{2}(a) \approx H^{2}_{0} \bigg( \frac{\Omega_{m}}{a^{3}} + \frac{\Omega_{r}}{a^{4}}\bigg)$$

The Attempt at a Solution

[/B]
I won't write it out here - it's just a lot of algebra - but substitute the definition of $H(a)$ into the equation for the time and rearrange is clearly what you need to do.

I got stuck however, because apparently you are meant to say: [this is in the solution set provided by my lecturer]
$$ \bigg( \frac{\Omega_{m}}{\Omega_{r}} \bigg) = \frac{1}{1+z_{eq}} = a_{eq}$$
This makes no sense to me at all! At matter-radiation equality, we could expect, by definition:
$$ \frac{\Omega_{m}}{\Omega_{r}}=1 \implies z_{eq} = 0$$
i.e matter-radiation equilibrium is occurring right now, which is obviously nonsense. [and would conflict completely with the result we are trying to show]

Have I misunderstood something?

Thanks!

 

[bananabandana]

Problem solved!

The point is that at matter radiation equality, we must have that:

$$ \frac{\rho_{M}}{\rho_{R}} = 1 $$

This does NOT mean that:

$$ \frac{\Omega_{M}}{\Omega_{R}} = 1 $$ Since :

$$ \frac{\rho_{M}}{\rho_{R}} = \frac{\frac{\Omega_{M}}{a^{3}}}{\frac{\Omega_{R}}{a^{4}}} = \frac{1}{a}\frac{\Omega_{M}}{\Omega_{R}} $$

So at radiation matter equality, we have:

$$ \frac{1}{a_{eq}}\frac{\Omega_{M}}{\Omega_{R}} = 1 \implies \frac{\Omega_{M}}{\Omega_{R}} = \frac{1}{1+z_{eq}} $$ as req'd.

 

[Vector1962]

Opps, I posted an incorrect statement and don't know how to delete this post... sorry.

Correction:
I don't think the 3rd equation is correct? is it?