Our Goal: take R(t) and physics (gravity) to calculate how R(t) varies with time. Then plug back...
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Transcript of Our Goal: take R(t) and physics (gravity) to calculate how R(t) varies with time. Then plug back...
Our Goal: take R(t) and physics (gravity) to calculate how R(t) varies with time.
Then plug back into (cdt)2 = R(t)2dr2/(1-kr2)
Get t versus R(t) and derive age of universe (t0) versus 0 and H0
Simple estimate of t0 = 1/H0
H0 = 50-70 km/sec-Mpc => 1/H0 has units of time = 19-14 billion years
Mpc = megaparsec = 3 million lt-years =
3 x 1024 cm
Want to show where the following come from:
• H0 = expansion rate for universe today
• c = critical density = 3H0/8G
• = /c <=> k relation
• q0 = de-acceleration parameter
• = cosmological constant <=> pressure and why positive (and causes an accelerating universe)
2
And, R(to)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(t0)/R(t) = 1+ z
t = the age of the universe when light left the object
t0 = age of the universe today by definition
cf. pages 374-376
Predicting the Future from the past:Also, R(t0)/R(t) = obem
=lambda(observed)/lambda(emitted).
the universe is expanding
R(t0) 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 aredshift!
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) mv2]
M is the self-gravitating mass of the universe
R is the scale factor of the universe.
= density() x volume[(4/3) x R3)]
How to get R(t), part 1, cont.
=> M = 4R3
density = ; volume = 4R3
v = R
Aside: A subscript 0 means “today” (R(t0) = R0 ) to keep from writing R(t) or R(t0).
mv2 = (1/2)mR2 and GMm/R = G4R3m/3R = GmR2/3
How to get R(t), part 1, cont.KE > PE, we get “escape”
KE < PE, the universe will collapse on itself.
(1/2)mv2 = 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”
R2 = G8R2/3 , now adding in the extra term
Yes! The k we used for our geometry and c is the speed of light.
R2 + kc2 = G8R2/3
R02 + kc2 = G8R0
2 /3 ; today
The KE, kc2, and PE connection
R0 = G8R0 /3 kc2
So, k = 1 means the KE is more than the PE, and we get escape, and vice versa
2 2
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 +kc2/(H0 R0 )= 0/c = ?
So, c as it is called is when k = 0 and we have c = 3R0/(8GR0), but R0/R0 = H0 ! (another old friend) = the expansion rate of the universe today 2
0
Or, c = 3H0/8G
Or kcR0)
We see the relationship between k and and the fate of the universe!
2
2 2
2 2
Aside on H0:
• How to use to get distances (good to 1+z of about 1.2)
• D = v/H0 where v = velocity of recession
• use km/sec along with H0 = 50 km/sec-Mpc for example
• D = v/H0 is the “Hubble Relation”
• Observation of this told us Universe is expanding
• For z << 1, z = v/c (approximately) z <=> v