100: many quite rare
I. Big Bang Nucleosynthesis
1. Abundances by mass in our universe
a) 1H 75%
4He 25%
b) 6Li 7.75E-10
7Li 1.13e-8
9Be 3.13E-10
10B 5.22E-10
11B 2.30E-9
c) 12C 3.87E-3
14N 0.94E-3
16O 8.55E-3
20Ne 1.34E-3
24Mg 0.58E-3
28Si 0.75E-3
(these are products of stellar burning)
d) A 100: many quite rare
So how are the differences in abundances among these groups explained?
2. First question: Could 4He, 25% by mass, be generated by stellar burning during the lifetime of our galaxy?
Edington: 4p 
4He as an energy source for our sun
4p 4He + 2e
+ 2
mass difference for this reaction 
25.7 MeV
kinetic energy carried off by neutrinos 0.4 MeV
therefore about 25 MeV per four protons consumed
goes into generating sun's luminosity
solar constant 0.033 cal/sec/cm
distance to earth 
cm = r
therefore the power output is
but as 4 protons are consumed for every 25 MeV produced
4 
p/sec consumed
Now the sun has a mass equal to that of 
protons
So assume 5 billion years (b.y.) (the sun is estimated to be about
4.6 b.y. in age) of burning at the current power level. Then we
can estimate the number of protons consumed over that lifetime
So this is
Thus
ONLY 5% OF PROTONS CONVERTED IN 5BY
And many protons are not in stars. Thus the tentative conclusion is
that stellar burning contributes to, but cannot account for all,
of the 4He. In fact, this conclusion can be put on much firmer
ground by looking at the 4He abundance as a function of stellar
metallicity, which is kind of a ``clock" for the galaxy. Very
metal poor stars presumably were formed very early (before
supernovae and novae created many of the metals). The surfaces
of such stars should not know about the 4He synthesis in the
core, but rather be representative of the star at its birth.
So if the surface shows a large 4He abundance, one would conclude
that that 4He was primordial.
We will learn more about 4p 4He later. It was first
described by Bethe and Critchfield 1939; the theory was
further developed by Salpeter and by Burbidge, Burbidge, Fowler,
and Hoyle in the 1950s.
2. Big bang nucleosynthesis
Gamow in the 1940s: proposed a ``big bang" cosmology where the universe
began as a hot soup, then expanded and cooled. When cooled below
about kT 1 MeV, when e - e
annihilation would occur,
that soup would consists of the familiar stable particles like
p, n, e, and 
.
The basic idea of big bang nucleosynthesis: a nuclear reaction network
that begins with n + p d + and ends ....??
It is absolute clear this must happen as
So if there is no nucleosynthesis, there would be no neutrons now.
This is an important qualitative idea: neutrons exist in our
present day world only because they bind in nuclei. Free neutrons
have enough energy to decay to protons via beta decay. Bound
neutrons do not because their binding energy makes this decay
energetically impossible. So some kind of ``condensation" or
``freezeout" of nuclei from the hot big bang must have occurred.
Now we can calculate the n/p ratio. The mass abundance is 75%
H and 25% 4He. So this means for every 4He (4 mass units)
there must be about 12 protons (12 mass units). Thus there are
2 neutrons and 14 protons in the sum. Therefore
in our galaxy.
CHALLENGE: Can we understand this number?
Does the number tell us anything about the early universe?
Recall in a thermal gas a particle has a kinetic energy of
E 3kT/2, where k = Boltzman's constant = 0.862 
MeV/K. Here capital K stands for degrees K. So
1 MeV kT 
T = 1.16 
K
That is, an MeV is the typical thermal energy of a particle in
a heat bath of T = 10 billion degrees K.
**********
Before we proceed we need to have a few results about cosmology.
This will be very quick and without derivation, as it is the
subject of the next quarter (particle astrophysics etc) of this
two-quarter series. We start with the Robertson-Walker metric
(that is, measure of distance) for a homogeneous, isotropic
universe
where
k=+1 
curved space, finite
k= 0 flat space, infinite
k=-1 curved space, infinite
Note that R(t) sets the scale of the geometry, and is a function
of time. The coordinate r in the metric above is dimensionless:
it's not the usual r. The dimension of length is carried by
R(t). So if an observer sits at the comoving reference point
(r,
) (note t is the time such an at-rest observer
would measure), the length of the radius from r=0 to him/her is
Thus if R(t) increases with t, it would be like riding on the
surface of an expanding balloon.
Now the evolution of R(t) is determined by the Einstein equations
which, under certain simplifying assumptions, yield
where G is Newton's gravitational constant and 
(t) is the
energy density. One can show from energy conservation that
when the universe is dominated by relativistic particles
satisfies
(The assumption about relativistic particles is connected with
the need to know the equation of state to figure out how
changes when R(t) changes. For relativistic particles, the
pressure is just 1/3 of .)
So from this result we note
So we can plug in the expression for the Hubble constant and
integrate to get
where 
is an integration constant. Finally, one can write
still another expression for
by calculating the energy density, which in the early universe
is dominated by light, relativistic particles (the photon,
electron and positron, and the three neutrinos). This yields
where T is the temperature and N is the effective numbers of
degrees of freedom (=43/4 if all of the species above contribute).
This formula is relatively easy to understand. The density
of massive particles should go like 
because the
momentum states 
will fill up roughly to
(T,T,T). Then an extra power of T comes from the fact we are
calculating the energy density, and each particle has an
energy 
.
So examining these expressions yields
We can look up in Weinberg's book some numbers from more
careful calculations:
**********
So let's look at some of these epochs
A) t 
sec T 
K kT10 MeV
2m
c
Neutrinos and electrons/positrons are kept in chemical equilibrium by neutral- and charged-current interactions:
The neutral current reaction (Z
exchange) has the same
strength for all three neutrino flavors (
).
The charged-current reaction involves only electron-family
particles: the analogous reactions involve muon-family
particles does not occur because of the temperature. The
mass of two muons exceeds 200 MeV, and thus cannot be produced
at this time.
There are similar charge-changing reactions coupling
n p.
As we will be discussing weak interactions quite a bit (they
are involved in most of the stellar reactions of interest to
us), let's say a bit more at this time about the rules for
charged current reactions. The reactions p n
are
p + e
n +
n + e
p + 
p n + e
n p + e
We will treat the neutrinos as Dirac particles (that is, a
particle with a distinct antiparticle). The rules for the
reactions above are then simple. First, charge is conserved,
which means the sum of the charges going into a reaction are
equal to the sum of the outgoing charges. Second, the
lepton charge is conserved, which means the same thing for
a second charge, lepton number. The lepton numbers are
defined as follows
The n and p can be considered two possible states in which to
find the nucleon; these states have different masses,
m(n) = 939.566 MeV and m(p) = 938.272 MeV. A two state
system in thermal and chemical equilibrium - weak interactions
n p are the mechanism for maintaining chemical
equilibrium - has occupation probabilities
where 
is the number of states at energy 
(the two
spin states in this case). Thus
where 
m = 1.294 MeV. At T = 10
K, kT=8.62 MeV.
Thus at this temperature,
Although this temperature is far above the temperature of
nucleosynthesis, the n/p ratio has already begun to drop.
B) t 0.1 sec 
3 K 3 MeV
Somewhat before this temperature is reached the 
and
have fallen out of equilibrium. This occurs because
the rate for interactions with electrons is too slow to keep
up with the rate of expansion of the universe: we will do
an example below. Note that the muon and tauon flavor reactions
off electrons/positrons are only about 1/7 as strong as for
electron neutrinos: that is why these ``heavy flavors" decouple
first.
(Throughout our discussions we assume the mass can
be ignored; however experimentally this is not proved, as the
current limit is around 16 MeV.)
At around 3MeV the also decouple.
We will now explore this question of ``falling out of equilibrium"
in the context of the n p reactions to see
whether nucleons are still in equilibrium. To answer this
question we need to know:
What is the time scale for n p reactions?
What is the time scale describing the expansion of the
universe that forms the comparison scale?
The time scale for the n p reactions can be
posed as that for a neutron in our thermal bath to convert to a
proton via n + 
p + e. That rate
is
were 
is the electron neutrino number density and v is
the relative velocity of the neutrino and neutron, which we
can take as c=1. Now has the dimensions of 1/cm
and v of cm/sec, so the product carries the dimensions of
flux, 1/cmsec. The cross section is roughly
What about ? This is our previous argument that the
number of states/unit volume for relativistic particles
should go like (kT). Thus we conclude
As discussed in class 
has the dimensions of 1/mass,
as it represents the exchange of a W boson: so there is a
coupling constant on each end and a propagator that goes
like 1/
(since all momenta of the scattering particles
are much, much less than 
). So we can look up the
numerical value in old papers that measured the strength of
decay of the neutron. A easy way to remember the
approximate result is
where 
is the nucleon mass. Thus
Note that this has the units of mass (we are free to insert
with each factor of ). But 
must have
the dimensions of 1/sec, as it is a rate. So setting
and inserting one factor of
to provide the needed units leads to
At 3 K this gives a rate of 17/sec. That is,
the typical lifetime of a neutron in such a thermal bath
is about 0.06 sec. Of course, we dropped all sorts of 
s
and 2's in this calculation. Had we done things more
carefully the result would be (see Weinberg's book)
So this would give a neutron lifetime at 
K
of about 0.011 sec.
Let's pause here for a remark. Even though we haven't yet
discussed to what we will compare this rate, it should be
clear that weak rates evolve VERY rapidly in the early
universe, dropping as 
. We don't have any other
quantity (t, R, etc) that changes so fast. So clearly at
some point we will reach a T where weak rates are so
slow that they can't keep up with the other changes in the
universe. Furthermore, the transition (range of T) over
which this ``freezeout" occurs should be rather narrow,
since the functional dependence on T is steep.
The first guess for a comparison timescale is the age of the
universe, which at this epoch is approximately
or about 0.12 sec. So this is 10 times longer than our neutron
lifetime: we conclude the weak rates are easily keeping the
chemical equilibrium between neutrons and protons at this
time.
Actually a better comparison rate is the Hubble parameter,
which clearly has dimensions of 1/sec and which clearly does
describe the rate of change of the universe at any given time.
It is given by
and
So
The value of G in MeV units is 6.7 
.
Thus, inserting the needed 
we have
We see this is close to our more naive guess based on
C) Epoch of decoupling
As we have these nice formulas, let's cut to the chase and find
the temperature characterizing the epoch of decoupling. This
should be when the neutron lifetime and the Hubble rate are
comparable. From what we have done above, this is easy:
which yields
So that's a nice round number to remember ( 1 MeV).
Now that we have the temperature at which the n p
system breaks out of chemical equilibrium, we can evaluate the
corresponding n/p ratio at this point:
Note this is not 1/7, but it is also not too far from it. It
is also ``on the right side": we expect it too continue to
drop. Why? First, we've just calculated the point where the
Hubble rate and weak rates are comparable. As the temperature
drops a bit, the 
-induced reactions continue to push things
toward the proton-rich side. Indeed, if we look in Weinberg's
book, in the next 10 seconds (1 MeV is about 1 second after the
big bang) the n/p ratio will drop to about .17 1/6.
And after that there continues to be a slow decrease in the
neutron percentage because of neutron decay - but this
has the timescale of 10 minutes.
So the next issue is to figure out when the nuclei form; and
the answer better be in less than 10 minutes, since we can't
tolerate much decay and still get 1/7 for the n/p ratio.
The nucleon gas in the early universe is relatively dilute,
with the consequence that nuclei must be made by two-body
reactions. The nucleon force has a range of only a few
fermis (1f = 10
cm), and the chance of getting three
or more fusing nuclei within this radius is negligible.
The only two-nucleon bound state is the deuteron, with a binding
energy of 2.24 MeV. But 2.24 MeV! We just found out that
n p chemical equilibrium persisted to about
1 MeV. With a 2.24 MeV binding energy, wouldn't all the
neutrons want to be in deuterium at or before that time?
(therefore freezing the n/p ratio at some value above 1/4)
The answer is no, for reasons having to do with the photon to
nucleon ratio. For reasons we don't fully understand, when
the very early universe cooled, it left over a net baryon
number. That is, we have neutrons and protons in our universe,
not antineutrons and antiprotons. Presumably at very early
times there were approximately equal numbers of quarks and
antiquarks, but not precisely equal numbers. As the universe
cools, quarks and antiquarks (nucleons and antinucleons)
annihilated each other. At the end - for reasons connected
with baryon number violation, CP violation, and nonequilibrium
physics - a small residual baryon number excess remained.
The resulting baryon/photon ratio is (deduced from nucleosynthesis
arguments we are about to describe)
So consider the two possible states of n+p
This reaction is electromagnetic and therefore fast, so we
can reasonably assume it is in equilibrium in the early
universe. The equilibrium condition is
This then yields
Roughly speaking, the logic of this is that the destruction
rate of deuterium is proportion to the number of deuterium
nuclei times the number of photons with the requisite energy,
etc.
Now 
. The time when deuterium
forms can be defined as the time when
, that is, when half of the neutrons are
in deuterium nuclei. It follows that the epoch of nucleosynthesis
is given by
and thus
That is, nucleosynthesis commences at about 100 seconds after
the big bang. If one were to have followed the n/p from
the weak freezeout ( 1 sec) to 100 sec by a numerical
integration one would find
That is, we understand the n/p ratio as a ``fingerprint" of the
thermal physics of the early universe.
Modern determinations of the primordial 
He abundance, which
is essentially the same as the n/p ratio, are quite precise.
Inspecting the above argument shows, therefore, that precise
calculation should be able to relate a fundamental and
unknown cosmological parameter to the primordial He
abundance! (That is where our value of 10
came from.)
Furthermore the large , the higher the temperature
T characterizing nucleosynthesis. Alternatively, one can state:
the smaller the baryon number of the universe
the later the epoch of nucleosynthesis
*******
Note that the discussion above shows that the temperature for
nucleosynthesis is about 100 keV, when the naive guess might
have placed it at about 2 MeV, the binding energy of deuterium.
The explanation is the very small value of .
The rest of the story involves the subsequent reaction network. One can look at the various steps:
n+p d + n/p 1/7
d(n,)H and d(d,p)H
d(p,)He and d(d,n)He
He(n,p)H (at long times H decays to He)
H(p,)He and H(d,n)He
He(n,)He and He(d,p)He and He(He,2p)He
Note: center of mass energies 100 kev
Coulomb suppression effects
trace amounts of 
Li,Be
He(H,)Li and He(He,)Be
(Be later decays to Li, but only when it becomes an atom
that is, after recombination: interesting story here)
the usual explanation for the termination of the reaction
network with the species mentioned above is that there are no
stable nuclei with A=5 and A=8. Thus the obvious potential
reactions for going further, He+He and He+p, are
ineffective. Actually, this common explanation is not quite true.
If one cheats and makes up stable isotopes at these mass numbers
with modest binding energies, the chain still largely terminates
as above. The reason is that the Coulomb barriers at these
low temperatures (100 keV) become increasingly hard to penetrate
as Z increases.
The repulsive Coulomb barrier near a nucleus is
So if 
and 
, this is about 3 MeV. So it
is very hard for charged particles with kinetic energies of
100 keV to tunnel through the Coulomb barrier to the
regions where the strong nuclear force can bind the reacting
particles to form a new nucleus.
Once the bottleneck to forming deuterium is broken,
the rest of the network described above proceeds quickly to
produce He.
**********
Now some comments about systematics:
1) The larger , we showed the larger 
,
and thus the larger n/p ratio. Therefore larger lead
to larger He abundance.
2) One can think of d, He as ``catalysts" in the network:
they are produced and then consumed, and thus reach an equilibrium
value that depends on the competition between production and
consumption. What happens is is increased?
The production channel is effectively n+p, where there is no
Coulomb barrier. Destruction channels include reactions like
d+d,d+p, etc, which are Coulomb inhibited. So increasing the T
effects the destruction channels more, since Coulomb barrier
penetration is exponential, enhancing the destruction. The conclusion is that higher
should produce lower d, He.
3) For low (and therefore low T) Li is made and
destroyed by:
The second reaction has the more effective Coulomb barriers
(effecting both initial and final states). Thus a lower T
will more effectively turn off the destruction of Li
than its production. Thus a lower T (with low ) means
more Li.
But it turns out that for high , the primary way to make
Li changes. High means more He, as we have noted,
so
becomes important. This production clearly benefits by high
T because of the Coulomb barrier. Furthermore there are very
few neutrons around to kill this isotope by (n,
).
Thus, Li produced as Be begins to turn up again at
high T and high .
This physics can be seen in the following results, taken from
Kolb and Turner.
(Add some comments about dependence on number of relativistic
species: adding another neutrino species increases
the energy density and therefore the Hubble rate. Therefore
weak interactions fall out of equilibrium earlier, when the
n/p ratio is higher. It follows that this forces He
production upward. And conversely...)