Redshift (Ryden 5.2), confusing step.

[Clever-Name]

Homework Statement

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

$$\frac{dz}{dt_{0}} = H_{0}(1+z) - H_{0}(1+z)^{3(1+w)/2}$$

Homework Equations

$$H_{0} = (\frac{\dot{a}}{a})|_{t = t_{0}} = \frac{2}{3(1+w)t_{0}}$$

$$(1+z) = \frac{a_{t_{0}}}{a_{t_e}} = \left( \frac{t_{0}}{t_{e}}\right )^{2/3(1+w)}$$

$$\frac{dz}{dt_{0}} = \frac{dz}{da_{0}}\frac{da_{0}}{dt_{0}} + \frac{dz}{da_{e}}\frac{da_{e}}{dt_{e}}\frac{dt_{e}}{dt_{0}}$$

$$t_{e} = \frac{t_0}{(1+z)^{3(1+w)/2}}$$

w is the component index (matter, radiation, lambda), not sure what its formal name is.

The Attempt at a Solution

Everything goes ok following my work below until the last step:

$$\frac{dz}{dt_{0}} = \frac{2}{3(1+w)t_{0}}\left( \frac{t_{0}}{t_{e}} \right)^{2/3(1+w)} - \frac{2}{3(1+w)t_{e}}\left( \frac{t_{0}}{t_{e}} \right)^{2/3(1+w)}\frac{dt_{e}}{dt_{0}}$$

$$= H_{0}(1+z) - \frac{2}{3(1+w)t_{e}}(1+z)\frac{dt_{e}}{dt_{0}}$$
Here is where I am stuck. Using the definition for $t_{e}$, as given in the chapter, we get:

$$\frac{dt_{e}}{dt_{0}} = (1+z)^{-3(1+w)/2} = \frac{t_e}{t_{0}}$$

But when subbing this in we end up with 0!

I have no idea how to proceed.

 

[mark726]

Please help!First of all, let's clarify some things. The component index w is usually called the equation of state parameter, and it represents the ratio of the pressure to the energy density of the component. For example, for matter w=0, for radiation w=1/3, and for the cosmological constant w=-1.

Now, let's look at your attempt at a solution. The first line is correct, but in the second line, you have made a mistake. The correct expression for dt_e/dt_0 is:

\frac{dt_e}{dt_0} = (1+z)^{-3(1+w)/2} = \frac{t_0}{t_e}

Here, you have used the relation t_e = t_0/(1+z)^{3(1+w)/2}, but you have forgotten to account for the fact that t_e is a function of z. So the correct expression should be:

\frac{dt_e}{dt_0} = \frac{t_0}{t_e}\frac{dt_e}{dz}\frac{dz}{dt_0}

Substituting this into your expression, we get:

\frac{dz}{dt_0} = H_0(1+z) - \frac{2}{3(1+w)t_e}\frac{t_0}{t_e}\frac{dt_e}{dz}\frac{dz}{dt_0}

Now, we can cancel out the dz/dt_0 terms, and we are left with:

\frac{dz}{dt_0} = H_0(1+z) - \frac{2}{3(1+w)t_e}\frac{t_0}{t_e}\frac{dt_e}{dz}

Finally, we can use the relation dt_e/dz = -t_e(1+z)^{-1}, and we get the final expression:

\frac{dz}{dt_0} = H_0(1+z) - \frac{2}{3(1+w)}\frac{t_0}{t_e}(1+z)^{-1}

Now, we can substitute the relation t_e = t_0/(1+z)^{3(1+w)/2}, and we get the desired expression:

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