How Does Redshift Change Over Observational Time?

[fliptomato]

Greetings--this is from Ryden's Introduction to Cosmology, question # 5.2. Let me restate the question:

A light source in a flat, single-component universe has a redshift z when observed at a time $$t_0$$. Show that the observed redshift changes at a rate

$$\frac{dz}{dt_0} = H_0(1+z) - H_0(1+z)^{\frac{3(1+w)}{2}}$$

Here w is the equation of state parameter for whatever the universe is made of (e.g. w = 0 for matter, 1/3 for radiation).

Here is my best attempt so far:
Simplify notation by introducing $$y=\frac{2}{3(1+w)}$$

Then (eq. 5.51 in Ryder):
$$1+z = \left(\frac{t_0}{t_e}\right)^y$$
Where $$t_e$$ is the time at which the light was emitted.

Also (eq. 5.48 in Ryder):
$$t_0 = \frac{y}{H_0}$$

So differentiating the first equation w/rt $$t_0$$, we get:

$$\frac{dz}{dt_0} = yt_0^{y-1}t_e^{-y} - y t_e^{-y-1}t_0^y \frac{dt_e}{t0}$$

$$\frac{dz}{dt_0} = H_0(1+z) - y t_e^{-1} (1+z) \frac{dt_e}{t0}$$

But $$\frac{dt_e}{t0} = (1+z)^{-1/y}$$ from our equation (5.51) above. Equation (5.52) in Ryder also tell us:

$$t_e = \frac{y}{H_0}(1+z)^{-1/y}$$

So plugging these two in:
$$\frac{dz}{dt_0} = H_0(1+z) - y\left(\frac{H_0}{y}(1+z)^{1/y}\right)(1+z)(1+z)^{-1/y}$$

Which becomes identically zero! Where am I making my mistake?

Thanks,
Flip

 

[CarlB]

Is this one of those derivatives like you get in thermodynamics where the answer depends on whether you are holding $$t_e$$ constant? And you end up using something like

$$\frac{dz}{dt_0}\frac{dt_0}{dt_e}\frac{dt_e}{dz} = -1 ?$$

These always confuse me.

Carl