- #1

Clever-Name

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## Homework Statement

A light source in a flat, single-component universe has a redshift z when observed at a time [itex] t_{0} [/itex]. Show that the observed redshift changes at a rate

[tex]

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

[/tex]

## Homework Equations

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

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

[tex] \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}} [/tex]

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

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:

[tex]\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}}[/tex]

[tex] = H_{0}(1+z) - \frac{2}{3(1+w)t_{e}}(1+z)\frac{dt_{e}}{dt_{0}} [/tex]

Here is where I am stuck. Using the definition for [itex] t_{e} [/itex], as given in the chapter, we get:

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

But when subbing this in we end up with 0!

I have no idea how to proceed.

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