# Proper distance integral limits seem wrong

1. Dec 24, 2015

### deneve

I've seen in some lecture notes that the proper distance dp(t) can be written as
$\int_{t_e}^{t_0} c dt/a = \int_0^z c dz /H(z)$

I can perform this integral ok using
$H =\dot a/a$ and the fact that $1 + z = 1/a(t_e)$ but it requires associating the limits of the integration as te transforming to z and t0 to z= 0 - there is a minus sign which creeps in when you find dz/dt because da = - a2dz so the limits have to be switched. Thus they don't match as it appears when you read the integral.

I don't see how to interpret this because I feel it should be the other way round. That is t 0 should be associated with red shift z (that's what we measure today) and at time t = te the red shift should be 0 . Why am I wrong here?

2. Dec 24, 2015

### Chalnoth

If the photon is emitted at t_0, then its observed redshift will be zero. If it's emitted at t_e, then its observed redshift will be z.

3. Dec 25, 2015

### deneve

Thank you Chalnoth that makes so much sense. I just couldn't get it right in my mind but this has settled the matter.

Kindest regards