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Almost all is well, the only thing that troubles me is : why is there no minus sign after da has been replaced by -a2dz ?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