Age vs. Red Shift
•In the recent past, the universe was dominated by matter
1 11 z 
•The age of the universe now is given by:
dx
11
tt0  H
H00 
•Recall: x = a/a0 = 1/(1+z)
22

x


x
 1  mm  
0
m

m

•Time-red-shift relation:
•For z > 1 or so, the m/x term dominates, ignore the others
17.8 Gyr
1 1 z 
1 1 z 
t
1
dx
 32
3/ 2
1
1  12
1  12
2
2
t  H0 
 H 0  m  x dx  3 H0 m 1  z 
1  z 
m x
0
0
•For small z, we neglected 
 1 

2 H 01 m
•12% error at z = 1, about 1% error at z = 4.
1
m
t
sinh 

3/
2
•For large z (z >1000) this formula in error
3 1  m 
 m 1  z  
because we neglected radiation
The Cosmic Microwave Background Radiation
•In 1964, Arno Allan Penzias and Robert Woodrow Wilson
discovered a uniform microwaves of extraterrestrial origin
•Since then, it has been studied in great detail
•It is thermal,
almost
perfectly so
•It is isotropic,
almost
T0  2.725 K
perfectly so
Why so
thermal?
Why so
isotropic?
Radiation in the evolving universe
•As the universe expands, the density of radiation energy changes
3
-3
•Number density of radiation changes proportional to a
 a0 
n  n0  
•However, the wavelength also expands proportional to a
 a 
•This implies frequency decreases proportional to a-1
 a0 
-1
E

E
•Each photon reduces its energy as a
0

a
 
•Total energy density proportional to a-4
4
r  a
-4
•Mass density proportional to a
Today it is thermal at temperature T0. What was it yesterday?
 E 
•Recall that for thermal distribution, probabilities satisfy P  E   exp  

k
T
 B 
•Number of photons in any given state doesn’t change
•But the energy of the states does change.
 E0 

 E 
a0
Ea 
P  E   P  E0  exp  
  exp  
  exp  
 if T  T0
a
 kBT 
 kBT0 
 kBT0 a0 
•A thermal distribution remains thermal at all times
•To explain why it’s thermal now, explain why it’s thermal in the past
Temperature as a label
•Another way to label times in the past (besides z) is in terms of temperature T
•Or even multiply by Boltzmann’s Constant to get energy kBT
a0
•The relation is simple:
T

T
T a0
0

 1 z
a
T0 a
•This works well back to very early times
•Imprecise at high z (109) due to electron-positron annihilation
•Breaks down completely at inflation (1027?)
•Matter dominated era time-temperature relation:
t
17.8 Gyr
1  z 
3/ 2
17.8 Gyr  kBT0 

 T0 
 17.8 Gyr    
3/ 2
T
 
 kBT 
3/ 2
17.8 Gyr   2.725 K  8.617 10
t
3/ 2
 kBT 
5

eV/K 
3/ 2
3/ 2
t
64.0 kyr
 k BT
eV 
1.5
Thermal or not Thermal?
•For something to be in thermal equilibrium, it must interact with something
•Let  by the rate at which something interacts, say by scattering
•To thermalize, it must make many scatterings in the age of the universe t
•Number of scatterings is t
•If t < 1, then no scattering
•Distribution of particles will be unmodified; universe is transparent
• If t >> 1, then lots of scattering
•Universe is opaque, particles may become “thermalized”
Cross Section and Rates
•The cross-section  is the area of the targets that are getting hit by
some sort of projectile
  n  v 
•Units of area: m2
•The rate will also be proportional to the number density of targets n
•Units of m-3
•It will also be proportional to the speed at which they come together
•Units m/s
What Thermalizes Photons?
  n  v 
•Photons can scatter off of any charged particle
•Most likely with light particles, like electrons
e•Cross section – comes from particle physics
•ke is Coulomb’s constant
2 2
8  ke e 
•m is mass of electron
T 
 2
•e is electron charge
3  mc 
29
2


6.65

10
m
•This assumes the electrons are free
T
•If bound in atoms, cross-section is much lower
What is current density of electrons in the universe?
H 02  B
-3



0.256
u

m
•Density of ordinary matter, worked out in homework B 0
8
3 G
•Each proton or neutron weighs about 1 u:
nB 0  0.254 m -3
•Most of them are protons
•Neutral universe  Comparable # electrons
ne 0  0.222 m-3
• The density in the past was higher
3
-3
ne  0.222 m 1  z 


  n  v 
Thermal or Not?
•Relative velocity is c:
v  c
•Age of universe at red shift z is:
•Number of collisions is:
t  n  v  t



 0.222 m -3 1  z  6.65 1029 m 2
3
t

17.8 Gyr
1  z 
3/ 2
 T  6.65 1029 m 2

ne  0.222 m
 1  z 
16
3.156

10
s
17.8
Gyr
8

2.998 10 m/s
3/ 2
Gyr
1  z 

 0.00249 1  z 
•At present, (z = 0), universe is transparent
•In the past (z  53) universe is opaque if electrons were free
•Electrons are free now because of colliding galaxies, hot stars, etc.
•Before z = 10.5, most electrons were bound
•Universe was too cold to free electrons from atoms at this time:
•Temperature: T  T0 1  z   150 K
•Corresponding energy: kBT  0.013 eV
•Compare hydrogen binding energy Eb  13.6 eV
3/ 2
-3
3
Announcements
ASSIGNMENTS
Day
Homework Read
Today
Hwk. R
Time/Temperature Relations
Friday
Hwk. S
Particle Physics
Monday
Hwk. T
Primordial Nucleosynthesis
Error on
2 2
8  ke e 
Previous  T 
 2
3
 mc 
Lecture
 T  6.65 1029 m 2
4/7
Recombination (1)
•The universe has thin gas of atoms, mostly hydrogen nuclei
•There are a matching number of electrons
•At high temperatures, the electrons prefer to be free
•As temperature drops, they are attracted to the nuclei
•The process where they come together is called recombination
•Before recombination, the universe is opaque
•And therefore in thermal equilibrium
e
•After it, universe is transparent
•The microwave background shows us
universe at recombination
•Throughout recombination, universe in thermal
e+
p
equilibrium
•We want to know when this happens.
•First step: Find number density of hydrogen nuclei:
nB 0  0.254 m -3
3
-3
nH  0.191 m 1  z 
-3
nH 0  0.75nB 0  0.191 m


p+
ep+
e-
p+
The Saha Equation
We want to know what fraction x of electrons are free (same as free protons)
•For thermal distributions
P  E   exp   E kBT 
nb  xnH exp  Eb kBT 
Bound electrons in hydrogen:
3
•Binding energy Eb = -13.6 eV
-3
nH  0.191 m T T0 
•Density proportional to available free protons:
Naïve conclusion: Electrons bound when kBT drops below 13.6 eV
•But density of hydrogen is very low
•There are many free states
•Need to bind 99.5% of electrons to make Universe
Transparent
universe transparent


3
2
 13.6 eV 
1 x
 me k BT 
 nH 
exp 

2
2 
x
 2

 k BT 
 13.6 eV 
1  x  k BT eV 

exp 

2
17
x
2.05 10
k
T
 B

32
Universe
Opaque

Recombination (2)
•Universe becomes transparent at about kBT* = 0.256 eV
•This corresponds to z* = 1091 or T = 2976 K
•Age of universe can be approximated as:
•This is inaccurate because we
17.8 Gyr
are too close to radiation era t* 
3/ 2  493,000 yr
1 z 

z*  1091
T*  2976 K
k BT*  0.256 eV
t*  380,000 yr
•The cosmic microwave background shows us the
temperature of the universe at this time
•The dipole moment is an effect of our peculiar velocity
•There is a galactic foreground, which we can subtract
•The small fluctuations are interesting – discuss later