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Redshift vs. Time:

$\displaystyle {\frac{dz}{dt}}$ = - H0(1 + z)2($\displaystyle \Omega_{0}^{}$z + 1)1/2

where z is the redshift, t the cosmic time, H0 is Hubble's constant, and $ \Omega_{0}^{}$ the current density of the universe, in units of

$\displaystyle {\frac{3 H_0^2}{8 \pi G}}$

Angular Size:

$\displaystyle \Delta$$\displaystyle \theta$ = $\displaystyle {\frac{d(1+z)}{D}}$

where $ \Delta$$ \theta$ is the angle subtended on Earth by an object of proper length d at redshift z. D is given by the expression

D = $\displaystyle {\frac{2c}{H_0\Omega_0^2(1+z)}}$$\displaystyle \left\{\vphantom{ \Omega_0 z +
\left( \Omega_0 - 2 \right) \left[ \left( \Omega_0 z + 1\right)^{1/2} - 1
\right] }\right.$$\displaystyle \Omega_{0}^{}$z + $\displaystyle \left(\vphantom{ \Omega_0 - 2 }\right.$$\displaystyle \Omega_{0}^{}$ - 2$\displaystyle \left.\vphantom{ \Omega_0 - 2 }\right)$$\displaystyle \left[\vphantom{ \left( \Omega_0 z + 1\right)^{1/2} - 1
}\right.$$\displaystyle \left(\vphantom{ \Omega_0 z + 1}\right.$$\displaystyle \Omega_{0}^{}$z + 1$\displaystyle \left.\vphantom{ \Omega_0 z + 1}\right)^{1/2}_{}$ - 1$\displaystyle \left.\vphantom{ \left( \Omega_0 z + 1\right)^{1/2} - 1
}\right]$ $\displaystyle \left.\vphantom{ \Omega_0 z +
\left( \Omega_0 - 2 \right) \left[ \left( \Omega_0 z + 1\right)^{1/2} - 1
\right] }\right\}$


S = $\displaystyle {\frac{L}{4 \pi D^2 (1+z)^2}}$

where S is the energy received per second per square metre on earth, and L is the luminosity emitted by the object at redshift z in a second (object frame).

Paul Francis