# [cosmology] Find the comobile distance of a galaxy given redshift and H0.

## Homework Statement

Calculate the comobile distance of a galaxy with z=7.3, H$_{0}$=72 km/s/Mpc, universe with $\Omega_{0}=\Omega_{0,m}=1$
Calculate the scale factor when the galaxy emitted the light we receive today.

## Homework Equations

Friedmann equation

$(\frac{\dot{a}}{a})^{2}=(H_{0})^{2}[ \Omega_{0,r}(\frac{a}{a_{0}})^{-4}+\Omega_{0,m}(\frac{a}{a_{0}})^{-3}+(1-\Omega_{0})(\frac{a}{a_{0}})^{2}+\Omega_{\Lambda}]$

## The Attempt at a Solution

With this model of universe Friedmann equation becomes:

$(\frac{\dot{a}}{a})^{2}=(H_{0})^{2}[\Omega_{0,m}(\frac{a}{a_{0}})^{-3}]$

so

$(\frac{\dot{a}}{a})=(H_{0})(\frac{a}{a_{0}})^{-3/2}$

I should use the equation:

$X=\int^{t 0}_{t em}\frac{cdt}{a\dot{a}}$

$X=$ comobile distance

..but i don't know how to put the scale factor into it.

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George Jones
Staff Emeritus
Gold Member
Welcome to Physics Forums!

Since you know $z$, and since $z$ is easily expressible in terms of the scale factor $a$, maybe you should use $\dot{a} = da/dt$ to eliminate $dt$ in
$X=\int^{t 0}_{t em}\frac{cdt}{a\dot{a}}$

First of all, thank you,

i did the substitution, now the integral is in $da$. I have problems using $a_{0}$. If i use the scale factor without it, the conclusions should be the same....i think! I mean, if
$z_{mis}=\frac{1}{a_{em}}-1$
then
$a_{em}$ is $\frac{a}{a_{0}}$

.....is it true?

I guess it is wrong, i think i didn't understand why the scale factor must be normalized..

George Jones
Staff Emeritus
Gold Member
I have had a closer look at this, and I think I have worked it out.
I should use the equation:

$X=\int^{t 0}_{t em}\frac{cdt}{a\dot{a}}$
Did you mean
$$\chi = \int^{a_0}_{a_{em}}\frac{cda}{a\dot{a}} ?$$
$X=$ comobile distance

..but i don't know how to put the scale factor into it.
I don't think so. I think that $\chi$ is the comoving coordinate and that the the comoving distance (what you are looking for) is $a_0 \chi$
First of all, thank you,

i did the substitution, now the integral is in $da$. I have problems using $a_{0}$. If i use the scale factor without it, the conclusions should be the same....i think! I mean, if
$z_{mis}=\frac{1}{a_{em}}-1$
then
$a_{em}$ is $\frac{a}{a_{0}}$

.....is it true?

I guess it is wrong
I think
$$z = \frac{a_0}{a_{em}} - 1$$
that is, you don't need to worry about the normalization.

I wrote the above in a hurry (my wife is pulling me out the door for a social engagement), so it might have some mistakes.