Cosmology Question that I really don't have a clue on.

[ppyadof]

Homework Statement

a) Using the Robertson-Walker metric, write down an expression for the proper distance in terms of the coordinate r. For objects moving with the expansion of the universe, show that the proper distance increases with time in the manner described by the Hubble Law.

b) Use the Friedmann equations for a matter-dominated universe Universe with $\Lambda = 0$ to show that:
[tex]H(z) = H_0 (1+z)(1 + \Omega_0 z)^{1/2}[/itex]<br /> <br /> c) Define the angular diameter distance $d_A$ of an object and relate this to the coordinate r appearing in the Robertson-Walker metric. Show that in a matter dominated universe with $\Omega_0 = 1$,<br /> $$d_A = \frac{2c}{H}\lbrace(1+z)^{-1} - (1+z)^{-3/2}\rbrace$$<br /> <br /> <br /> <h2>Homework Equations</h2><br /> $$ds^2 = c^2 dt^2 - a^2(t)\lbrace \frac{dr^2}{1-kr^2} \rbrace + r^2(d\theta^2 + sin^2\theta d\phi)$$<br /> <br /> $$\lbrace\frac{da}{dt}\rbrace^2 + kc^2 = \frac{8 \pi G \rho^2}{3} + \frac{\Lambda c^2 a^2}{3}$$<br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> a) If ds=0, then let $l_p = cdt = adr$ and so by considering<br /> $$\frac{dr}{dt} = Hr$$<br /> and<br /> $$H = \frac{ \dot{a}}{a}$$<br /> it means that:<br /> $$\dot{l_p} = \dot{a}r$$<br /> <br /> I don't even have the slightest idea where to start for parts b and c. Any help on those would be greatly appreciated.[/tex]

 

[vacuum]

b) To show that the proper distance increases with time in the manner described by the Hubble Law, we can use the Friedmann equations for a matter-dominated universe with \Lambda = 0:

[tex]H^2 = \frac{8 \pi G \rho}{3} <br /> <br /> and<br /> <br /> [tex]\frac{\ddot{a}}{a} = - \frac{4 \pi G \rho}{3} <br /> <br /> From the first Friedmann equation, we can substitute in the definition of Hubble's constant (H_0 = H(t=0)) and the critical density (\rho_c = \frac{3H_0^2}{8 \pi G}) to get:<br /> <br /> [tex]H^2 = \frac{8 \pi G \rho}{3} = H_0^2 \frac{\rho}{\rho_c} <br /> <br /> Then, using the second Friedmann equation and substituting in the definition of the critical density, we get:<br /> <br /> [tex]\frac{\ddot{a}}{a} = - \frac{4 \pi G \rho}{3} = - \frac{4 \pi G}{3} H_0^2 \frac{\rho}{\rho_c} = - \frac{4 \pi G}{3} H_0^2 \frac{1}{a^3} <br /> <br /> We can then solve for \rho and substitute it into the first Friedmann equation to get:<br /> <br /> [tex]H^2 = \frac{8 \pi G \rho}{3} = \frac{8 \pi G}{3} H_0^2 \frac{1}{a^3} = H_0^2 \left(\frac{\ddot{a}}{a}\right)^2 <br /> <br /> This can be simplified to:<br /> <br /> [tex]\frac{\dot{a}}{a} = H_0 \left(\frac{\ddot{a}}{a}\right)^{1/2} <br /> <br /> Using the definition of Hubble's constant and the second Friedmann equation, we can substitute in for \frac{\ddot{a}}{a} to get:<br /> <br /> $$\frac{\dot{a}}{a} = H_0 \left(- \frac{4 \pi G}{3} H_0^2 \frac{1}{a^3}\right)^{1/2} =$$[/tex][/tex][/tex][/tex][/tex][/tex]