Let us next consider some key events in the evolutionary history of our universe [8]. The most well understood phase of the universe occurs when the temperature is less than about 10^{12} K. Above this temperature, thermal production of baryons and their strong interaction is significant and somewhat difficult to model. We can ignore such complications at lower temperatures and - as we shall see - several interesting physical phenomena did take place during the later epochs with T 10^{12}.
The first thing we need to do is to determine the composition of the universe when T 10^{12} K. We will certainly have, at this time, copious amount of photons and all species of neutrinos and antineutrinos. In addition, neutrons and protons must exist at this time since there is no way they could be produced later on. (This implies that phenomena which took place at higher temperatures should have left a small excess of baryons over anti baryons; we do not quite understand how this happened and will just take it as an initial condition.) Since the rest mass of electrons correspond to a much lower temperature (about 0.5 × 10^{10} K), there will be large number of electrons and positrons at this temperature but in order to maintain charge neutrality, we need to have a slight excess of electrons over positrons (by about 1 part in 10^{9}) with the net negative charge compensating the positive charge contributed by protons.
An elementary calculation using the known interaction rates show that all these particles are in thermal equilibrium at this epoch. Hence standard rules of statistical mechanics allows us to determine the number density (n), energy density () and the pressure (p) in terms of the distribution function f:
(10) |
(11) |
(12) |
Next, we can argue that the chemical potentials for electrons, positrons and neutrinos can be taken to be zero. For example, conservation of chemical potential in the reaction e^{+}e^{-} 2 implies that the chemical potentials of electrons and positrons must differ in a sign. But since the number densities of electrons and positrons, which are determined by the chemical potential, are very close to each other, the chemical potentials of electrons and positrons must be (very closely) equal to each other. Hence both must be (very close to) zero. Similar reasoning based on lepton number shows that neutrinos should also have zero chemical potential. Given this, one can evaluate the integrals for all the relativistic species and we obtain for the total energy density
(13) |
where
(14) |
The corresponding entropy density is given by
(15) |
As a simple application of the above result, let us consider the fate of neutrinos in the expanding universe. From the standard weak interaction theory, one can compute the reaction rate of the neutrinos with the rest of the species. When this reaction rate fall below the expansion rate H of the universe, the reactions cannot keep the neutrinos coupled to the rest of the matter. A simple calculation [8] shows that the relevant ratio is given by
(16) |
Thus, for T 1.6 × 10^{10} K, the neutrinos decouple from matter. At slightly lower temperature, the electrons and positrons annihilate increasing the number density of photons. Neutrinos do not get any share of this energy since they have already decoupled from the rest of the matter. As a result, the photon temperature goes up with respect to the neutrino temperature once the e^{+}e^{-} annihilation is complete. This increase in the temperature is easy to calculate. As far as the photons are concerned, the increase in the temperature is essentially due to the change in the degrees of freedom g and is given by:
(17) |
(In the numerator, one 2 is for electron; one 2 is for positron; the 7/8 factor arises because these are fermions. The final 2 is for photons. In the denominator, there are only photons to take care of.) Therefore
(18) |
The first equality is from Eq. (17); the second arises because the photons and neutrinos had the same temperature originally; the third equality is from the fact that for decoupled neutrinos aT_{} is a constant. This result leads to the prediction that, at present, the universe will contain a bath of neutrinos which has temperature that is (predictably) lower than that of CMBR. The future detection of such a cosmic neutrino background will allow us to probe the universe at its earliest epochs.
2.2. Primordial Nucleosynthesis
When the temperature of the universe is higher than the binding energy of the nuclei (~ MeV), none of the heavy elements (helium and the metals) could have existed in the universe. The binding energies of the first four light nuclei, ^{2}H, ^{3}H, ^{3}He and ^{4}He are 2.22 MeV, 6.92 MeV, 7.72 MeV and 28.3 MeV respectively. This would suggest that these nuclei could be formed when the temperature of the universe is in the range of (1 - 30)MeV. The actual synthesis takes place only at a much lower temperature, T_{nuc} = T_{n} 0.1MeV. The main reason for this delay is the `high entropy' of our universe, i.e., the high value for the photon-to-baryon ratio, ^{-1}. Numerically,
(19) |
To see this, let us assume, for a moment, that the nuclear (and other) reactions are fast enough to maintain thermal equilibrium between various species of particles and nuclei. In thermal equilibrium, the number density of a nuclear species ^{a} N_{Z} with atomic mass A and charge Z will be
(20) |
From this one can obtain the equation for the temperature T_{A} at which the mass fraction of a particular species-A will be of order unity (X_{A} 1). We find that
(21) |
where B_{A} is the binding energy of the species. This temperature will be fairly lower than B_{A} because of the large value of ^{-1}. For ^{2}H, ^{3}He and ^{4}He the value of T_{A} is 0.07 MeV, 0.11 MeV and 0.28 MeV respectively. Comparison with the binding energy of these nuclei shows that these values are lower than the corresponding binding energies B_{A} by a factor of about 10, at least.
Thus, even when the thermal equilibrium is maintained, significant synthesis of nuclei can occur only at T 0.3 MeV and not at higher temperatures. If such is the case, then we would expect significant production (X_{A} 1) of nuclear species-A at temperatures T T_{A}. It turns out, however, that the rate of nuclear reactions is not high enough to maintain thermal equilibrium between various species. We have to determine the temperatures up to which thermal equilibrium can be maintained and redo the calculations to find non-equilibrium mass fractions. The general procedure for studying non equilibrium abundances in an expanding universe is based on rate equations. Since we will require this formalism again in Section 2.3 (for the study of recombination), we will develop it in a somewhat general context.
Consider a reaction in which two particles 1 and 2 interact to form two other particles 3 and 4. For example, n + _{e} p+e constitutes one such reaction which converts neutrons into protons in the forward direction and protons into neutrons in the reverse direction; another example we will come across in the next section is p+e H + where the forward reaction describes recombination of electron and proton forming a neutral hydrogen atom (with the emission of a photon), while the reverse reaction is the photoionisation of a hydrogen atom. In general, we are interested in how the number density n_{1} of particle species 1, say, changes due to a reaction of the form 1 + 2 3 + 4.
We first note that even if there is no reaction, the number density will change as n_{1} a^{-3} due to the expansion of the universe; so what we are really after is the change in n_{1} a^{3}. Further, the forward reaction will be proportional to the product of the number densities n_{1} n_{2} while the reverse reaction will be proportional to n_{3} n_{4}. Hence we can write an equation for the rate of change of particle species n_{1} as
(22) |
The left hand side is the relevant rate of change over and above that due to the expansion of the universe; on the right hand side, the two proportionality constants have been written as µ and (Aµ), both of which, of course, will be functions of time. (The quantity µ has the dimensions of cm^{3}s^{-1}, so that nµ has the dimensions of s^{-1}; usually µ v where is the cross-section for the relevant process and v is the relative velocity.) The left hand side has to vanish when the system is in thermal equilibrium with n_{i} = n_{i}^{eq}, where the superscript `eq' denotes the equilibrium densities for the different species labeled by i=1 - 4. This condition allows us to rewrite A as A = n_{1}^{eq} n_{2}^{eq} / (n_{3}^{eq} n_{4}^{eq}). Hence the rate equation becomes
(23) |
In the left hand side, one can write (d / dt) = H a (d / da) which shows that the relevant time scale governing the process is H^{-1}. Clearly, when H / nµ >> 1 the right hand side becomes ineffective because of the (µ / H) factor and the number of particles of species 1 does not change. We see that when the expansion rate of the universe is large compared to the reaction rate, the given reaction is ineffective in changing the number of particles. This certainly does not mean that the reactions have reached thermal equilibrium and n_{i} = n_{i}^{eq}; in fact, it means exactly the opposite: The reactions are not fast enough to drive the number densities towards equilibrium densities and the number densities "freeze out" at non equilibrium values. Of course, the right hand side will also vanish when n_{i} = n_{i}^{eq} which is the other extreme limit of thermal equilibrium.
Having taken care of the general formalism, let us now apply it to the process of nucleosynthesis which requires protons and neutrons combining together to form bound nuclei of heavier elements like deuterium, helium etc.. The abundance of these elements are going to be determined by the relative abundance of neutrons and protons in the universe. Therefore, we need to first worry about the maintenance of thermal equilibrium between protons and the neutrons in the early universe. As long as the inter-conversion between n and p through the weak interaction processes ( + n p+e), ( + n p + ) and the `decay' (n p + e + ), is rapid (compared to the expansion rate of the universe), thermal equilibrium will be maintained. Then the equilibrium (n / p) ratio will be
(24) |
where Q = m_{n} - m_{p} = 1.293 MeV. At high (T >> Q) temperatures, there will be equal number of neutrons and protons but as the temperature drops below about 1.3 MeV, the neutron fraction will start dropping exponentially provided thermal equilibrium is still maintained. To check whether thermal equilibrium is indeed maintained, we need to compare the expansion rate with the reaction rate. The expansion rate is given by H = (8 G / 3)^{1/2} where = (^{2} / 30) g T^{4} with g 10.75 representing the effective relativistic degrees of freedom present at these temperatures. At T = Q, this gives H 1.1 s^{-1}. The reaction rate needs to be computed from weak interaction theory. The neutron to proton conversion rate, for example, is well approximated by
(25) |
At T=Q, this gives 5 s^{-1}, slightly more rapid than the expansion rate. But as T drops below Q, this decreases rapidly and the reaction ceases to be fast enough to maintain thermal equilibrium. Hence we need to work out the neutron abundance by using Eq.(23).
Using n_{1} = n_{n}, n_{3} = n_{p} and n_{2}, n_{4} = n_{l} where the subscript l stands for the leptons, Eq.(23) becomes
(26) |
We now use Eq.(24), write (n_{l}^{eq}µ) = _{np} which is the rate for neutron to proton conversion and introduce the fractional abundance X_{n} = n_{n} / (n_{n} + n_{p}). Simple manipulation then leads to the equation
(27) |
Converting from the variable t to the variable s = (Q / T) and using (d / dt) = - HT(d / dT), the equations we need to solve reduce to
(28) |
It is now straightforward to integrate these equations numerically and determine how the neutron abundance changes with time. The neutron fraction fall out of equilibrium when temperatures drop below 1 MeV and it freezes to about 0.15 at temperatures below 0.5 MeV.
As the temperature decreases further, the neutron decay with a half life of _{n} 886.7 sec (which is not included in the above analysis) becomes important and starts depleting the neutron number density. The only way neutrons can survive is through the synthesis of light elements. As the temperature falls further to T = T_{He} 0.28 MeV, significant amount of He could have been produced if the nuclear reaction rates were high enough. The possible reactions which produces ^{4}He are [D(D,n) ^{3}He(D,p) ^{4}He, D(D,p) ^{3}H(D,n) ^{4}He, D(D, ) ^{4}He]. These are all based on D, ^{3}He and ^{3}H and do not occur rapidly enough because the mass fraction of D, ^{3}He and ^{3}H are still quite small [10^{-12}, 10^{-19} and 5× 10^{-19} respectively] at T 0.3 MeV. The reactions n + p d+ will lead to an equilibrium abundance ratio of deuterium given by
(29) |
The equilibrium deuterium abundance passes through unity (for _{B} h^{2} = 0.02) at the temperature of about 0.07 MeV which is when the nucleosynthesis can really begin.
So we need to determine the neutron fraction at T = 0.07 MeV given that it was about 0.15 at 0.5 MeV. During this epoch, the time-temperature relationship is given by t = 130 sec (T / 0.1 MeV)^{-2}. The neutron decay factor is exp(-t/_{n}) 0.74 for T = 0.07 MeV. This decreases the neutron fraction to 0.15 × 0.74 = 0.11 at the time of nucleosynthesis. When the temperature becomes T 0.07 MeV, the abundance of D and ^{3}H builds up and these elements further react to form ^{4}He. A good fraction of D and ^{3}H is converted to ^{4}He (See Fig. 1 which shows the growth of deuterium and its subsequent fall when helium is built up). The resultant abundance of ^{4}He can be easily calculated by assuming that almost all neutrons end up in ^{4}He. Since each ^{4}He nucleus has two neutrons, (n_{n}/2) helium nuclei can be formed (per unit volume) if the number density of neutrons is n_{n}. Thus the mass fraction of ^{4}He will be
(30) |
where x_{c} = n / (n + p) is the neutron abundance at the time of production of deuterium. For _{B} h^{2} = 0.02, x_{c} 0.11 giving Y 0.22. Increasing baryon density to _{B} h^{2} = 1 will make Y 0.25. An accurate fitting formula for the dependence of helium abundance on various parameters is given by
(31) |
where _{10} measures the baryon-photon ratio today via Eq.(19) and g_{*} is the effective number of relativistic degrees of freedom contributing to the energy density and _{1/2} (n) is the neutron half life. The results (of a more exact treatment) are shown in Fig. 1.
Figure 1. The evolution of mass fraction of different species during nucleosynthesis |
As the reactions converting D and ^{3}H to ^{4}He proceed, the number density of D and ^{3}H is depleted and the reaction rates - which are proportional to X_{A}( n_{}) < v> - become small. These reactions soon freeze-out leaving a residual fraction of D and ^{3}H (a fraction of about 10^{-5} to 10^{-4}). Since it is clear that the fraction of (D, ^{3}H) left unreacted will decrease with . In contrast, the ^{4}He synthesis - which is not limited by any reaction rate - is fairly independent of and depends only on the (n / p) ratio at T 0.1 MeV. The best fits, with typical errors, to deuterium abundance calculated from the theory, for the range = (10^{-10} - 10^{-9}) is given by
(32) |
The production of still heavier elements - even those like ^{16}C, ^{16}O which have higher binding energies than ^{4}He - is suppressed in the early universe. Two factors are responsible for this suppression: (1) For nuclear reactions to proceed, the participating nuclei must overcome their Coulomb repulsion. The probability to tunnel through the Coulomb barrier is governed by the factor F = exp[-2 A^{1/3} (Z_{1} Z_{2})^{2/3}(T / 1 MeV)^{-1/3}] where A^{-1} = A_{1}^{-1} + A_{2}^{-1}. For heavier nuclei (with larger Z), this factor suppresses the reaction rate. (2) Reaction between helium and proton would have led to an element with atomic mass 5 while the reaction of two helium nuclei would have led to an element with atomic mass 8. However, there are no stable elements in the periodic table with the atomic mass of 5 or 8! The ^{8}Be, for example, has a half life of only 10^{-16} seconds. One can combine ^{4}He with ^{8}Be to produce ^{12}C but this can occur at significant rate only if it is a resonance reaction. That is, there should exist an excited state ^{12}C nuclei which has an energy close to the interaction energy of ^{4}He + ^{8}Be. Stars, incidentally, use this route to synthesize heavier elements. It is this triple-alpha reaction which allows the synthesis of heavier elements in stars but it is not fast enough in the early universe. (You must thank your stars that there is no such resonance in ^{16}O or in ^{20}Ne - which is equally important for the survival of carbon and oxygen.)
The current observations indicate, with reasonable certainty that: (i) (D / H) 1 × 10^{-5}. (ii) [(D + ^{3}He) / H] (1-8)× 10^{-5} and (iii) 0.236 < (^{4}He / H) <0.254. These observations are consistent with the predictions if 10.3 min 10.7 min, and = (3 - 10) × 10^{-10}. Using = 2.68 × 10^{-8} _{B} h^{2}, this leads to the important conclusion: 0.011 _{B} h^{2} 0.037. When combined with the broad bounds on h, 0.6 h 0.8, say, we can constrain the baryonic density of the universe to be: 0.01 _{B} 0.06. These are the typical bounds on _{B} available today. It shows that, if _{total} 1 then most of the matter in the universe must be non baryonic.
Since the ^{4}He production depends on g, the observed value of ^{4}He restricts the total energy density present at the time of nucleosynthesis. In particular, it constrains the number (N_{}) of light neutrinos (that is, neutrinos with m_{} 1 MeV which would have been relativistic at T 1 MeV). The observed abundance is best explained by N_{} = 3, is barely consistent with N_{} = 4 and rules out N_{} > 4. The laboratory bound on the total number of particles including neutrinos, which couples to the Z^{0} boson is determined by measuring the decay width of the particle Z^{0}; each particle with mass less than (m_{z} / 2) 46 GeV contributes about 180 MeV to this decay width. This bound is N_{} = 2.79 ± 0.63 which is consistent with the cosmological observations.
2.3. Decoupling of matter and radiation
In the early hot phase, the radiation will be in thermal equilibrium with matter; as the universe cools below k_{B} T (_{a} / 10) where _{a} is the binding energy of atoms, the electrons and ions will combine to form neutral atoms and radiation will decouple from matter. This occurs at T_{dec} 3 × 10^{3} K. As the universe expands further, these photons will continue to exist without any further interaction. It will retain thermal spectrum since the redshift of the frequency a^{-1} is equivalent to changing the temperature in the spectrum by the scaling T (1 / a). It turns out that the major component of the extra-galactic background light (EBL) which exists today is in the microwave band and can be fitted very accurately by a thermal spectrum at a temperature of about 2.73 K. It seems reasonable to interpret this radiation as a relic arising from the early, hot, phase of the evolving universe. This relic radiation, called cosmic microwave background radiation, turns out to be a gold mine of cosmological information and is extensively investigated in recent times. We shall now discuss some details related to the formation of neutral atoms and the decoupling of photons. Cosmic Microwave Background Radiation
The relevant reaction is, of course, e + p H + and if the rate of this reaction is faster than the expansion rate, then one can calculate the neutral fraction using Saha's equation. Introducing the fractional ionisation, X_{i}, for each of the particle species and using the facts n_{p} = n_{e} and n_{p} + n_{H} = n_{B}, it follows that X_{p} = X_{e} and X_{H} = (n_{H} / n_{B}) = 1 - X_{e}. Saha's equation now gives
(33) |
where = 2.68 × 10^{-8}(_{B} h^{2}) is the baryon-to-photon ratio. We may define T_{atom} as the temperature at which 90 percent of the electrons, say, have combined with protons: i.e. when X_{e} = 0.1. This leads to the condition:
(34) |
where = (T / 1 eV). For a given value of (_{B} h^{2}), this equation can be easily solved by iteration. Taking logarithms and iterating once we find ^{-1} 3.084 - 0.0735ln(_{B} h^{2}) with the corresponding redshift (1 + z) = (T / T_{0}) given by
(35) |
For _{B} h^{2} = 1, 0.1,0.01 we get T_{atom} 0.324 eV, 0.307 eV, 0.292 eV respectively. These values correspond to the redshifts of 1367, 1296 and 1232.
Because the preceding analysis was based on equilibrium densities, it is important to check that the rate of the reactions p + e H + is fast enough to maintain equilibrium. For _{B} h^{2} 0.02, the equilibrium condition is only marginally satisfied, making this analysis suspect. More importantly, the direct recombination to the ground state of the hydrogen atom - which was used in deriving the Saha's equation - is not very effective in producing neutral hydrogen in the early universe. The problem is that each such recombination releases a photon of energy 13.6 eV which will end up ionizing another neutral hydrogen atom which has been formed earlier. As a result, the direct recombination to the ground state does not change the neutral hydrogen fraction at the lowest order. Recombination through the excited states of hydrogen is more effective since such a recombination ends up emitting more than one photon each of which has an energy less than 13.6 eV. Given these facts, it is necessary to once again use the rate equation developed in the previous section to track the evolution of ionisation fraction.
A simple procedure for doing this, which captures the essential physics, is as follows: We again begin with Eq. (23) and repeating the analysis done in the last section, now with n_{1} = n_{e}, n_{2} = n_{p}, n_{3} = n_{H} and n_{4} = n_{}, and defining X_{e} = n_{e} / (n_{e} + n_{H}) = n_{p}/ n_{H} one can easily derive the rate equation for this case:
(36) |
This equation is analogous to Eq. (27); the first term gives the photoionisation rate which produces the free electrons and the second term is the recombination rate which converts free electrons into hydrogen atom and we have used the fact n_{e} = n_{b} X_{e} etc.. Since we know that direct recombination to the ground state is not effective, the recombination rate is the rate for capture of electron by a proton forming an excited state of hydrogen. To a good approximation, this rate is given by
(37) |
where r_{0} = e^{2} / m_{e} c^{2} is the classical electron radius. To integrate Eq. (36) we also need to know / . This is easy because in thermal equilibrium the right hand side of Eq. (36) should vanish and Saha's equation tells us the value of X_{e} in thermal equilibrium. On using Eq. (33), this gives
(38) |
We can now integrate Eq. (36) using the variable B / T just as we used the variable Q / T in solving Eq. (27). The result shows that the actual recombination proceeds more slowly compared to that predicted by the Saha's equation. The actual fractional ionisation is higher than the value predicted by Saha's equation at temperatures below about 1300. For example, at z = 1300, these values differ by a factor 3; at z 900, they differ by a factor of 200. The value of T_{atom}, however, does not change significantly. A more rigorous analysis shows that, in the redshift range of 800 < z < 1200, the fractional ionisation varies rapidly and is given (approximately) by the formula,
(39) |
This is obtained by fitting a curve to the numerical solution.
The formation of neutral atoms makes the photons decouple from the matter. The redshift for decoupling can be determined as the epoch at which the optical depth for photons is unity. Using Eq. (39), we can compute the optical depth for photons to be
(40) |
where we have used the relation H_{0} dt -_{NR}^{-1/2} z^{-5/2} dz which is valid for z >> 1. This optical depth is unity at z_{dec}=1072. From the optical depth, we can also compute the probability that the photon was last scattered in the interval (z, z + dz). This is given by (exp -) (d / dz) which can be expressed as
(41) |
This P(z) has a sharp maximum at z 1067 and a width of about z 80. It is therefore reasonable to assume that decoupling occurred at z 1070 in an interval of about z 80. We shall see later that the finite thickness of the surface of last scattering has important observational consequences.