Comoving Distance in LCDM - Understanding an Approximation

AI Thread Summary
The discussion focuses on calculating the comoving distance in the ΛCDM model, specifically using the integral formula involving the Hubble parameter. The user encounters difficulties with manual integration and opts for an approximation method found in textbooks. They derive an expression for the scale factor's inverse, but notice discrepancies between their result and a reference from Barbara, particularly regarding the inclusion of the j0 term. The confusion arises from the treatment of higher-order terms in the expansion, leading to uncertainty about the correct approximation. The user seeks clarification on their mistake in the approximation process.
Arman777
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Homework Statement
Finding the Comoving Distance for LCDM universe.
Relevant Equations
$$\chi(t) = \int_{t_e}^{t_0}\frac{dt}{a(t)}$$
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 don't understand how this can be possible...I am doing a mistake somewhere but I don't know where.

Thanks
 
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a(t)^{-1}=\frac{1}{1+A}=1-A+A^2+..
where
A=H_0(t-t_0)-qH_0^2/2(t-t_0)^2+..
First order comes from A only . Second order comes from A and ##A^2## which is ##H_0^2(t-t_0)^2##. Similarly third order comes from A, A^2 and A^3.
 
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