I Comoving distance and redshift relationship derivation

1. Jun 21, 2017

DoobleD

Hello PhysicsForum,

There is something I don't get at the end of this course notes PDF file. In the last section, titled "Comoving distance and redshift", which I have copied below, we have a short derivation of the comoving distance and redshift relation.

Almost all is well, the only thing that troubles me is : why is there no minus sign after da has been replaced by -a2dz ?

I have searched the web and found almost identical derivations in other courses or publications, but I never read the explanation for why the minus sign drops. I have found what seems to be the source material for most of those derivations : this paper from 93 (see section 6.3, "The General Redshift-Distance Relation" on 3rd page). It is referenced quite often by others when this comoving distance and redshift relationship shows up.

Maybe I am just missing some mathematical trick ? This is not super important of course, but it bugs me.

2. Jun 21, 2017

Bandersnatch

Could it be because the integration limits have been swapped? (consider what it means when the limits are $a_e -> a_0$, where e stands for emission, and $0 -> z$.)

3. Jun 21, 2017

kimbyd

Yes. The change of variables leads to two minus signs which cancel one another: $da = -a^2 dz$, and reversing the limits of integration.

4. Jun 21, 2017

DoobleD

Thank you !

Thay's what I thought, but then it means that the ae limit corresponds to z when you do the change of variable, and a0 to 0 redshift ? Sounds weird, shouldn't it be the other way around ? Since a0 = a(t0) is the expansion when we receive the redshifted signal.

5. Jun 21, 2017

kimbyd

This is probably easiest to see if you look at the equation for the scale factor in terms of the redshift:
$$a = {1 \over 1+z}$$

Here note that for $z=0$, $a=1$. That's the current scale factor and redshift. A far-away object, at, say, a redshift of $z=2$ is at a scale factor of $a=1/3$. The integral above over $da$ would integrate from $1/3$ to 1, while the integral over $dz$ integrates from 0 to 2.

6. Jun 21, 2017

Bandersnatch

Think of how far you need to look. Higher z is seen farther than lower z, while lower a is seen farther than high a.

7. Jun 22, 2017

DoobleD

Makes sense now. Thank you !