# Comoving Distance in LCDM - Understanding an Approximation

• Arman777
In summary, the conversation discusses the process of finding the comoving distance in the ##\Lambda##CDM model using an integral. The integral involves the Hubble parameter and the density parameters for matter and dark energy. The speaker then outlines their approach of approximating the solution by expanding the scale factor around a certain time, and finding the inverse of the scale factor. The discrepancy between their result and that in the textbook is also mentioned.
Arman777
Gold Member
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

$$a(t)^{-1}=\frac{1}{1+A}=1-A+A^2+..$$
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.

Last edited:
Arman777

## 1. What is comoving distance in LCDM?

Comoving distance in LCDM refers to the distance between two points in the universe that is adjusted for the expansion of the universe. It takes into account the expansion of space and is therefore a better measure of distance than physical distance.

## 2. Why is comoving distance important in LCDM?

Comoving distance is important in LCDM because it allows us to accurately measure the distances between objects in the expanding universe. It also helps us understand the effects of cosmic expansion on the distribution of matter and the evolution of the universe.

## 3. How is comoving distance calculated in LCDM?

Comoving distance in LCDM is calculated using the Hubble parameter, which describes the rate of expansion of the universe, and the redshift of an object. The formula for calculating comoving distance is d = c/H0 * z, where d is the comoving distance, c is the speed of light, H0 is the Hubble parameter, and z is the redshift.

## 4. What is the difference between comoving distance and physical distance?

The main difference between comoving distance and physical distance is that comoving distance takes into account the expansion of the universe, while physical distance does not. Physical distance is the actual distance between two points in space, while comoving distance is a measure of the distance that would exist if the universe were not expanding.

## 5. What are the limitations of using comoving distance in LCDM?

One limitation of using comoving distance in LCDM is that it assumes a homogeneous and isotropic universe, which may not be accurate at small scales. Additionally, it does not take into account the effects of gravity on the expansion of the universe. It is also important to note that comoving distance is an approximation and may not be completely accurate in all cases.

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