# Lookback Time & Age of the Universe Calculations

I try to calculate the age of the universe with the FLRW model: $$H(a) = H_0 \sqrt{\Omega_{\mathrm{R},0} \left(\frac{a_0}{a}\right)^4 + \Omega_{\mathrm{M},0} \left(\frac{a_0}{a}\right)^3 + (1-\Omega_{\mathrm{T},0}) \left(\frac{a_0}{a}\right)^2 + \Omega_{\Lambda,0}}.$$

I set $\Omega_{\mathrm{M},0} = 0.317$ (matter density) and $\Omega_{\Lambda,0} = 0.683$ (dark energy), as delivered by Planck 2013; $\Omega_{\mathrm{T},0} = 1.02$ (space curvature), according to this site; and $\Omega_{\mathrm{R},0} = 4.8\times10^{-5}$ (radiation density), according to this document.

For the time $t(a)$ I take the scale factor $a$ and divide it through the integrated recessional velocity $$t(a) = \frac{a}{\int_0^a{H(a')a'\ \mathrm{d}a'}/(a-0)}$$ and finally simplify to $$t(a) = \frac{a^2}{\int_0^a{H(a')a'\ \mathrm{d}a'}}.$$

But the problem is, I then get about $8\times10^9$ years for the age of the universe, but it should be around $12\times10^9$ years (which I get when I set $\Omega_{\mathrm{R},0}$ to zero):

$\Omega_{\mathrm{R},0} = 4.8\times10^{-5}$:

$\Omega_{\mathrm{R},0} = 0 \to 0.00001$:

Do I have to use some other models than FLRW/ΛCDM, or is one of my parameters outdated?

-
Hi Симон Тыран, and welcome to Physics Stackexchange. A couple of us did some major editing to typeset your formulas - you should make sure we didn't inadvertently change the meaning of your question. For future questions, we encourage the use of Latex-style formatting in order to make questions easier to read, and to help them garner more upvotes ;) – Chris White May 7 '13 at 4:49

The total energy density is by definition $$\Omega_{T,0} = \Omega_{R,0} + \Omega_{M,0} + \Omega_{\Lambda,0},$$ so with the values you cite ($\Omega_{R,0}=4.8\times 10^{-5}$, $\Omega_{M,0}=0.317$, $\Omega_{\Lambda,0}=0.683$), we get $\Omega_{T,0} = 1$, or in a more common notation $\Omega_{K,0}=1-\Omega_{T,0}=0$, i.e. a space with zero curvature.
It is also common to define the present-day value of the scale-factor as $a_0=1$, so that $$H(a) = H_0\sqrt{\Omega_{R,0}a^{-4} + \Omega_{M,0}a^{-3} + \Omega_{K,0}a^{-2} + \Omega_{\Lambda,0}}.$$
The age of the universe can then be derived as follows: from $$\frac{\text{d}a}{\text{d}t} = \dot{a},$$ we get \begin{align} \text{d}t &= \frac{\text{d}a}{\dot{a}} = \frac{\text{d}a}{aH(a)} = \frac{a\,\text{d}a}{a^2H(a)}\\ &= \frac{1}{H_0}\frac{a\,\text{d}a}{a^2\sqrt{\Omega_{R,0}a^{-4} + \Omega_{M,0}a^{-3} + \Omega_{K,0}a^{-2} + \Omega_{\Lambda,0}}}\\ &= \frac{1}{H_0}\frac{a\,\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}a + \Omega_{K,0}a^2 + \Omega_{\Lambda,0}a^4}}. \end{align} Integrating yields the difference between the time that a signal is emitted and the time it is observed: $$t_{\text{ob}} - t_{\text{em}} = \frac{1}{H_0}\int_{a_{\text{em}}}^{a_{\text{ob}}} \frac{a\,\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}a + \Omega_{K,0}a^2 + \Omega_{\Lambda,0}a^4}},$$ and the age of the universe is $$t_0 = \frac{1}{H_0}\int_0^1 \frac{a\,\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}a + \Omega_{K,0}a^2 + \Omega_{\Lambda,0}a^4}}.$$ This should give you the correct age.
Yes, notice that I've written an extra $a$ in the numerator, so that $dt = a\,da/(a^2\,H(a))$, and the $a^2$ in the denominator can be written under the sqrt. It's also interesting to note that the peak in the integrand corresponds with the time at which the expansion of the universe started to accelerate. – Pulsar May 7 '13 at 20:12
I've added two extra steps in my answer, I hope things are clear now. It really is the correct formula, you'll find it in every cosmology course (for instance, on page 14 of casa.colorado.edu/~ajsh/phys5770_08/frw.pdf). I don't know which value of $H_0$ you used, but with $H_0=67.3\;\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}$, you should get an age of 13.79 billion years. – Pulsar May 9 '13 at 16:09