Assume a flat, matter-dominated universe. (a) Use the Friedmann equation to deri

Question

Assume a flat, matter-dominated universe. (a) Use the Friedmann equation to derive the time evolution of the scale factor. You may assume the scale factor has a power-law form, a(t) t^q. (b) Consider a galaxy at redshift z. Using the scaling derived in part a, derive a relationship between 1+z, the time light was emitted from the galaxy, t_e, and time at which the light is observed, t_0. (b) Show that a(t) = (t/t_0)^q, where t_0 is the age of the universe. Derive an equation for t_0 in terms of the present energy density, elementof_0. Re-write in terms of H_0, noting the relationship between H(t) and epsilon (t) listed on the equation page. (c) Derive an equation for the proper distance to a galaxy at redshift z, at the time of observation, t_0. Your answer should be written in terms of c, H_0, and z. (d) The LMC is about 50 kpc from the Milky Way. What angle would this correspond to if we saw a similar pair of galaxies at z = 1 in an ohm_m = 1 universe? Express your answer in terms of arcseconds. You may assume that the separation is transverse to the line of sight.

Explanation / Answer

(a)

Taking Hubble rate as H, H = a’ / a

H’ = -H2 + a” / a = - H2 [1- (a” / H2a) = -H2 (1+q)

Where q is the deceleration parameter = q = a” / H2 a

Considering Non-relativistic matter-dominated Universe is modeled by dust approximation: P = 0.

a”/a + (4 pi G / 3) *p = 0 , and H would be –H2q + (4 pi G / 3) *p = 0

Therefore p = (3H2 / 4 pi G) *q

This results as (a’/a)2 –(8 pi G/3)*p = k / a2 and H2 – 2H2q = k/a2

So –k = a2 H2(1-2q).

Since both a 6= 0 and H 6= 0, for flat Universe (k = 0), q = 1/2 (q > 1/2 for k = 1 and q < 1/2 for k = 1), the density needed to yield flat universe is as follows:

Pcr = 3H02 / 8 pi G = {3 [h/0.98*1010 years)2 * [1 year/ 3600*24*365 sec]2} / 8 pi (6.67-10-8cm3g-1s-2)

= 1.87*10-29 h2 g/cm3 = 10-29 g/ cm3

Considerinf h = 0.72+- 0.02, the quantity q provides the relationship between the density of the Universe and the critical density cr

Second Friedman’s equation for matter dominated universe:

p’ + 3p (a’/a) = 0

a3p’ + 3pa’a2 = 0 , therefore leading to a3p = a30p0 = constant.

With P = 1/3p, we have flat universe k=0, q0 = ½ , P = 0, a3 p = constant

A’/a2 = (8 pi G / 3)*p0 (a0/a)3 = square root of (8 piGp0a03 /3)*t

At the Big Bang, t = 0, a = 0, so K = 0. Upon adopting convention a0 = 1, and the fact that the Universe is flat 0 = cr, we finally have

A=(6 pi G p0)13t2/3 = (6 pi Gpcr)1/3t2/3 = [6 pi G (3H20 / 8 pi G)]1/3 * t2/3 = (9H20 / 4) * t2/3 = (3H0/2)2/3t2/3

Universe t0, which corresponds to the Hubble rate H0 and the scale factor a = a0 = 1 to be:

T0 = 2/3H0, taking H0 = h / 0.98*1010 years and h = 72 we have

T0 = 2*0.98*1010 years / 3*0.72 = 9.1*109 years = 9.1 Acon.

This making Friedman’s equation as a’2 / a2 = (8 pi G /3 )*p0 (a0/a)4 = square root of (8 pi Gp0a04 / 3) *t

At big bang t = 0, a = 0, so K = 0, and a0=1. Also 0 = cr resulting:

A = (2/3 *pi G p0)1/4 , t1/2 = (2/3 pi G pcr)1/4 , t1/2 = [2/3 pi G (3H20/8 pi G)] , t1/2 =(H0 / 2)1/2 t1/2.

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