Neutron decay and interconversion
Particle processes are a lot like equations
•You can turn them around and they still work
•You can move particles to the other side by “subtracting them”
•This means replacing them with anti-particles
•The neutron (in isolation) is an unstable particle
•Decays to proton + electron + anti-neutrino
•Mean lifetime: 886 seconds
•Put the electron on the other side
•Put the neutrino on the other side
•All thee processes convert neutrons to
protons and vice versa
n0
n0
n0
p+
+
+
+
e-
e+
p+

p+
+
+
+


e-
Neutron/Proton Freezeout
•Weak interactions interconvert protons/neutrons
•These are slow processes, so they fall out of equilibrium fairly early
•At kBT = 0.71 MeV, the process stops
•What is ratio of protons to neutrons at this temperature? P  E   exp   E kBT 
2
2
P

exp

m
c
 n k BT 
•Non-relativistic, E = mc . n
•Ratio is:
Pp  exp   m p c 2 k BT 
2
exp

m
c
kBT
 mc 2 
nn
n
 1.294 MeV 

 exp  
  exp  
  0.162
2
n p exp  m p c k BT
 0.71 MeV 
 kBT 




nn
0.162

 0.139
nn  n p 1.162
2
2
2.42 s  MeV 
2.42 s  MeV 
 1.5 s
•This happens at about: t 





g eff  k BT 
10.75  0.71 MeV 
The Deuterium Bottleneck
•The next step in making more complex
+
+
0
p
p
n
+
elements is to make 2H, deuterium:
n0
•This releases about 2.24 MeV of energy
•Naively: this process will go ahead as soon as kBT drops below 2.24 MeV
•Actually, much lower temperature is required because of very low density of
nucleons
•Actual temperature is about factor of 20 lower: 0.1 MeV
2
2
•Age of universe at this time:
2.42
s
MeV




2.42 s MeV

•At this point, some neutrons t 

  130 s


3.36  0.1 MeV 
g eff  k BT 
are gone due to decay
 nn

 nn  n p

 nn


 nn  n p
132 s 

 130 
 exp  
  0.120
 886 
1.5 s
•Ratio depends weakly on density of protons/neutrons – more makes it happen sooner
Making Helium
•Once we make deuterium, we continue quickly to continue to helium:
p+
n0
0
n
p+ 0
n
+
+
n0
0
n
p+ 0
n
p+
0
n
p+ 0p+
n
p+ 0
n
p+ 0p+
n
+
+
p+
n0
p+ 0p+
n
0
n
p+ 0p+
n
•For every two neutrons, there will be two protons that combine to make 4He
•Mass fraction of 4He is twice that of neutron fraction
YP  

4
He
   total   2n
n
nB  2  0.12  0.24
•4He is extremely stable – once formed it won’t go back.
nB

•The sooner it happens, the more neutrons are left over
n
•Define  as the current ratio of baryons (protons + neutrons) to photons
•As  increases, YP increases weakly:
 

YP  0.248  0.011ln 
10 
6

10


Making Other elements
•When you run out of neutrons, 3He can still be turned into 4He via
p+ 0p+
n
+
p+
n0
p+
0
n
p+ 0p+
n
+
p+
+
p+
•The last few 2H, 3He, and 3H nuclei will have trouble finding partners
•There will be small amount of each of these isotopes left
•The more baryons there are, the easier it is to find a partner
•As  increases, 2H, 3He, and 3H all decrease
•There are other rare processes that produce a couple of other isotopes:
•7Li and 7Be are produced
0
n
•I don’t understand how they
p+ 0p+
+ p+
p
+
n
depend on 
n0
•Within a few hundred seconds, the
0
0
n
n
1
2
3
3
+
baryons are all in H, H, H, He,
p 0
p+ 0p+
+
n
4He, 7Be and 7Li
n
0
p0+ n
+p+
np
p+ n0
p0+ n0+
0p
nn
p+ n0
Anything we missed?
•Two of these isotopes are unstable:
•Add 3H to 3He and 7Be to 7Li
3
7
H  3He  e  e
Be  e  7 Li  e
•The process whereby stars make heavier elements do not work in the early universe
4
•Density is too low for unstable 8Be to find
He  4 He  8 Be*
another 4He to react with
4
He  8 Be*  12C
•In the end, we should be able to predict abundance (compared to hydrogen) of
2H, 3He, 4He, 7Li
•These have all been measured, mostly by studying light from quasars
•Back in the good old days (the 90s), this was how we estimated 
•Now we have an independent way of estimating it (later lecture)
•We should be able to compare the results with
nB
predictions
    6.2  0.2  1010
n
•A very strong test of Big Bang theory
The results
   6.2  0.2 1010
•Predictions for 4He, 2H and 3He all
work very well
•Prediction for 7Li seems to be off
•The Lithium problem
•Overall, success for the model
Summary of Events:
Event
Neutrinos Decouple
Neutron/Proton freezeout
Electron/Positron Annihilate
Primordial Nucleosynthesis
kBT or T
1 MeV
0.7 MeV
170 keV
80 keV
Time
0.4 s
1.5 s
30 s
200 s
Matter/Radiation Equality
Recombination
0.76 eV
0.26 eV
57 kyr
380 kyr
Structure formation
30 K
500 Myr
Now
2.725 K
13.75 Gyr
Lots of unsolved problems:
•What is the nature of dark matter?
•Why is the universe flat (or nearly so)?
•Where did all the structure come from?
•What is the nature of dark energy?
What we know and what we don’t:
•Up to now, everything we have discussed is based on pretty well understood physics
•And the experimental results match it well!
•As we move earlier, we reach higher temperatures/energies, and therefore things
become more uncertain
•For a while, we can assume we understand the physics and apply it, but we don’t
have any good tests at these scales
New particles appear as temperature rises:
•Muons, mass 105.7 MeV, at about kBT = 35 MeV (g = 4 fermions)
•Pions, mass 135-139 MeV, at about kBT = 45 MeV (g = 3 bosons)
•At a temperature of about kBT = 100 MeV, we have quark deconfinement
Quark Confinement
•There are a group of particles called baryons that have strong interactions
•Proton and neutron are examples
•There are also anti-baryons and other strong particles called mesons
•In all experiments we have done, the baryon number is conserved
•Baryon number = baryons minus anti-baryons
•All strongly interacting particle contain quarks or anti-quarks or both
•The quarks are held together by particles called “gluons”
u
u
g
•At
d low temperatures quarks are confined into these packets
•At high temperatures, these quarks become free (deconfined)
u
u
•Estimated kBT = 150 MeV
2
2
2.42 s  MeV 
2.42 s  MeV 
5

t

1.4

10
s




61.75  150 MeV 
g eff  k BT 
Electroweak Phase Transition
•There are three forces that particle physicist understand:
•Strong, electromagnetic, and weak
•Electromagnetic and weak forces affected by a field called the Higgs field
•The shape of the Higgs potential is interesting:
•Sometimes called a Mexican Hat potential
•At low temperatures (us), one direction is easy to
move (EM forces) and one is very hard (weak forces)
•At high temperatures, (early universe) you naturally
move to the middle of the potential
•All directions are created equal
•Electroweak unification becomes apparent at
perhaps kBT = 50 GeV
2
2.42 s  MeV 
2.42 s 
MeV

10
t


10
s




g eff  k BT 
100  50, 000 MeV 
2
The Standard Model
•Above the electroweak phase
transition, all known particles of the
standard model should exist with
thermal densities
geff  28  78  90  106.75
g
4
2
12
12
4
2
12
12
4
2
12
12
mc2 (GeV)
0.0005
~0
~0.005
~0.010
0.1057
~0
1.27
~0.10
1.777
~0
173
4.7
gggggggg
W
Z
1
1
1
1
2
16
6
3
0
0
80.4
91.2
H
0
1
115–285
Particle
symbols spin
Electron
e
½
Electron neutrino e
½
Up quark
uuu
½
Down quark
ddd
½
Muon

½
Muon neutrino 
½
Charm quark
ccc
½
Strange quark
sss
½
Tau

½
Tau neutrino

½
Top quark
ttt
½
Bottom quark
bbb
½
•From here on, we will be
speculating on the physics
•Cosmology sometimes indicates we
are guessing right
•Goal: Learn physics from
Photon
cosmology
Gluon
W-boson
Z-boson
Higgs

Supersymmetry
•In conventional particle physics, fermions and bosons are fundamentally different
•And never the twain shall meet
•In a hypothesis called supersymmetry, fermions and bosons are interrelated
•There must be a superpartner for every particle:
•Supersymmetry also helps solve a
problem called the hierarchy problem
•But only if it doesn’t happen at
too high an energy
•If supersymmetry is right, then scale of
supersymmetry breaking probably around kBT = 500 GeV or so.
•If this is right, the LHC should discover it
2
2
•In most versions of
2.42 s  MeV  2.42 s 
MeV

12
t


10
s

supersymmetry, the lightest




5
g eff  k BT 
100  5  10 MeV 
super partner (LSP) should be
absolutely stable Could this be dark matter?
Grand Unification Theories (GUT’s)
•In the standard model, there are three fundamental forces, and three corresponding
coupling constants
•These have rather different values
•But their strength changes as you change the energy of the experiment, theortically
•How much they change depends on whether supersymmetry is right or not
•If supersymmetry is right, then at an energy
of about 1016 GeV, the three forces are
equal in strength
•At kBT = 1016 GeV, there will be another
phase transition – the Grand Unification
transition
2
2.42 s  MeV 
39
t

10
s
 19

210  10 MeV 
Baryogenesis might occur at this scale
No
Supersymmtery
With
Supersymmtery
Scale could be right for inflation
Summary of Events:
Event
Grand Unification
Supersymmetry Scale
Electroweak Scale
Quark Confinement
kBT or T
1016 GeV
500 GeV
50 GeV
150 MeV
Time
10-39 s
10-12 s
10-10 s
1.410-5 s
Neutrinos Decouple
Neutron/Proton freezeout
Electron/Positron Annihilate
Primordial Nucleosynthesis
1 MeV
0.7 MeV
170 keV
80 keV
0.4 s
1.5 s
30 s
200 s
Matter/Radiation Equality
Recombination
0.76 eV
0.26 eV
57 kyr
380 kyr
Structure formation
Now
30 K
2.725 K
500 Myr
13.75 Gyr