- #1

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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

$$\chi = c\int_0^z \frac{dz}{H(z)}$$ 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