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

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jacdiam89
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Homework Statement


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

Homework Equations


Friedmann equation

[itex](\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}][/itex]

The Attempt at a Solution


With this model of universe Friedmann equation becomes:

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

so

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

I should use the equation:

[itex]X=\int^{t 0}_{t em}\frac{cdt}{a\dot{a}}[/itex]

[itex]X=[/itex] comobile distance

..but i don't know how to put the scale factor into it.
 
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Since you know [itex]z[/itex], and since [itex]z[/itex] is easily expressible in terms of the scale factor [itex]a[/itex], maybe you should use [itex]\dot{a} = da/dt[/itex] to eliminate [itex]dt[/itex] in
jacdiam89 said:
[itex]X=\int^{t 0}_{t em}\frac{cdt}{a\dot{a}}[/itex]
 
First of all, thank you,

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

...is it true?

I guess it is wrong, i think i didn't understand why the scale factor must be normalized..
 
I have had a closer look at this, and I think I have worked it out.
jacdiam89 said:
I should use the equation:

[itex]X=\int^{t 0}_{t em}\frac{cdt}{a\dot{a}}[/itex]

Did you mean
[tex]\chi = \int^{a_0}_{a_{em}}\frac{cda}{a\dot{a}} ?[/tex]
jacdiam89 said:
[itex]X=[/itex] comobile distance

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

I don't think so. I think that [itex]\chi[/itex] is the comoving coordinate and that the the comoving distance (what you are looking for) is [itex]a_0 \chi[/itex]
jacdiam89 said:
First of all, thank you,

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

...is it true?

I guess it is wrong

I think
[tex]z = \frac{a_0}{a_{em}} - 1[/tex]
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.