Our Goal: take R(t) and physics (gravity) to calculate how R(t) varies with time. Then plug back into (cdt) 2 = R(t) 2 dr 2 /(1-kr 2 ) Get t versus R(t) and derive age of universe (t 0 ) versus  0 and H 0 Simple estimate of t 0 = 1/H 0 H 0 = km/sec-Mpc => 1/H 0 has units of time = billion years Mpc = megaparsec = 3 million lt-years = 3 x cm

Want to show where the following come from: H 0 = expansion rate for universe today  c = critical density = 3H 0 /8  G   =   /  c k relation q 0 = de-acceleration parameter  = cosmological constant pressure and why positive  (and   causes an accelerating universe) 2

And, R(t o )r for the observed object translates into a distance to the object today, and our goal is to figure out how to calculate R and r The distance light travels on the surface is greatly affected by the value of k. k =  1 open k = 0 flat k = 1 closed

For the related figures, see page 217 (shows geometry), 283 (shows R changing in different ways), and 299 (shows R for k =  1, 0, +1)

For the math we will do, assume that there is no dark energy (cosmological constant) until further notice

Predicting the Future from the past: A primary goal of the cosmologist is to tell us what will happen to R as function of time, based on fitting models to the data

Predicting the Future from the past: Measure R(t) by looking back in time Measure how the geometry of the universe affects our measure of distance or apparent size. R(t 0 )/R(t) = 1+ z t = the age of the universe when light left the object t 0 = age of the universe today by definition cf. pages

Predicting the Future from the past: Also, R(t 0 )/R(t) = ob  em = lambda(observed)/lambda(emitted). the universe is expanding R(t 0 ) is always greater than R(t) (for us today) lambda(observed) must always be > lambda (emitted) longer lambda (now this means wavelength of light) means redder, we call this a redshift!

How to get R(t) We need to relate R(t) to some “force” The Universe affects itself. It has self gravity Self-gravity will slow down expansion

Equate potential energy (GMm/R) with kinetic energy [(1/2) mv 2 ] M is the self-gravitating mass of the universe R is the scale factor of the universe.  = density(  ) x volume[(4/3) x  R 3 )]

How to get R(t), part 1, cont. => M =  4  R 3 density =  ; volume = 4  R 3 v = R  Aside: A subscript 0 means “today” (R(t 0 ) = R 0 ) to keep from writing R(t) or R(t 0 ).  mv 2 = (1/2)mR 2 and GMm/R = G  4  R 3 m/3R = G  mR 2 /3 

How to get R(t), part 1, cont. KE > PE, we get “escape” KE < PE, the universe will collapse on itself. (1/2)mv 2 = GMm/R, KE = PE The little m’s cancel out. Put an energy term on the KE side to allow us to describe “to escape or not to escape”

R 2 = G8  R 2 /3, now adding in the extra term Yes! The k we used for our geometry and c is the speed of light.  R 2 + kc 2 = G8  R 2 /3  R kc 2 = G8   R 0 2 /3 ; today 

The KE, kc 2, and PE connection R 0 = G8   R 0 /3  kc 2  So, k =  1 means the KE is more than the PE, and we get escape, and vice versa 22

Critical density = when pull of gravity (PE) just balances the BB push (KE), i.e. the density when k = 0 !

How to get R(t), part 1, cont. Or, 1 +kc 2 /(H 0 R 0 )=  0 /  c = ? So,  c as it is called is when k = 0 and we have  c = 3R 0 /(8  GR 0 ), but R 0 /R 0 = H 0 ! (another old friend) = the expansion rate of the universe today 2 00   Or,  c = 3H 0 /8  G Or kc    R 0 )    We see the relationship between k and   and the fate of the universe! 

Aside on H 0 : How to use to get distances (good to 1+z of about 1.2) D = v/H 0 where v = velocity of recession use km/sec along with H 0 = 50 km/sec-Mpc for example D = v/H 0 is the “Hubble Relation” Observation of this told us Universe is expanding For z v