# Comoving distance in $\Lambda$CDM - Understanding an approximation

I am trying to find the comoving distance,

$$\chi = c\int_0^z \frac{dz}{H(z)}$$ for the $$\Lambda$$CDM model (spatially flat universe, containing only matter and $$\Lambda$$).

$$H^2 = H_0^2[\Omega_{m,0}(1+z)^3 + \Omega_{\Lambda, 0}]$$

When I put this into integral I am getting,

$$\chi = \frac{c}{H_0}\int_0^z \frac{dz}{\sqrt{\Omega_{m,0}(1+z)^3 + \Omega_{\Lambda, 0}}}$$

which seems that I cannot take the integral manually (only numeric integration is possible).

So I have decided to approximate the solution, as its done in many textbooks (including Weinberg and Barbara).

So I have written $$\chi(t) = \int_{t_e}^{t_0}\frac{dt}{a(t)}$$ and then I have expanded the $$a(t)$$ around $$t_0$$.

$$a(t) \simeq 1 + H_0(t-t_0) - \frac{q_0H_0^2}{2}(t-t_0)^2 + \frac{j_0H_0^3}{6}(t-t_0)^3$$

from here I need to find $$a(t)^{-1}$$ so that I can put that inside the integral. I have seen that $$1/x \simeq \sum_{n=0}^{\infty}(-1)^n(x-1)^n.$$

So by similar logic I should get

$$a(t)^{-1} \simeq 1 - H_0(t-t_0) + \frac{q_0H_0^2}{2}(t-t_0)^2 - \frac{j_0H_0^3}{6}(t-t_0)^3$$

However, it seems its not the case. In Barbara (the calculations do not include $$j_0$$) its written as,

$$a(t)^{-1} \simeq 1 - H_0(t-t_0) + (1+q_0/2)\frac{H_0^2}{2}(t-t_0)^2$$

I just dont understand how this can be possible...I am doing a mistake somewhere but I don't know where.

Write $$a=1+\delta a$$. Then (working to second order in $$H_0^2 (t-t_0)^2$$) $$\begin{equation} \delta a = H_0 (t-t_0) - \frac{q_0 H_0^2}{2} (t-t_0)^2 \end{equation}$$ So $$\begin{eqnarray} a^{-1} = \left(1+\delta a\right)^{-1} &=& 1 - \color{blue}{\delta a} + \color{red}{\delta a^2} + \cdots \\ &=& 1 - \color{blue}{\left[H_0 (t-t_0) - \frac{q_0 H_0^2}{2}(t-t_0)^2\right]} + \color{red}{\left[H_0^2 (t-t_0)^2 + \cdots \right]} + \cdots \\ &=& 1 - \color{blue}{H_0 (t-t_0)} + \left(\color{red}{1} + \color{blue}{\frac{q_0}{2}}\right)H_0^2(t-t_0)^2 + \cdots \end{eqnarray}$$ where $$+\cdots$$ refers to terms that are order $$H_0^3 (t-t_0)^3$$ or higher.
In other words, it appears you left out the $$\color{red}{\delta a^2}$$ term when you did the Taylor expansion.
• Okay thanks a lot. Do you think I should expand to $\delta a^3$ to collect third order terms from $a(t)$ expansion...it seems thats the case but I am not sure. May 1, 2022 at 20:53
• @Neptune If you want to expand $a(t)^{-1}$ to order $H_0^3 (t-t_0)^3$, then you need to expand the $\delta a^3$ term (and you'll also need to find the relevant term in $\delta a^2$, and include the cubic correction in $\delta a$ which I dropped). May 1, 2022 at 20:59
• Okay, thanks a lot. Btw, I have managed to approximate $\chi$ up to third order. Thanks a lot really. These approximation are really hard to find.. May 2, 2022 at 19:02