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

[jacdiam89]

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.

 

[George Jones]

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}}

 

[jacdiam89]

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]

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.