Invitation to a Contemporary Physics (2004)

to Planck’s law of radiation. The resulting distribution peaks at a wave-length λm = (2π/4.965)( c/kBT ). This relation is known as Wien’s displacement law. The photon energy at the peak is hc/λm = 4.965kBT .

The present universe has a temperature of T0 = 2.725 K. The corresponding number density is 414 photons per cubic centimeter. In comparison, the critical density ρc of the universe is about 5 nucleons per cubic meter, and the nucleon density is about 4% of the critical density, thus the nucleon number density is about 2 × 10−7 nucleons per cubic centimeter. This makes the ratio of nucleons to photons to be η 5 × 10−10, a very small number. This ratio is the same at all times when the nucleons remain non-relativistic, because both the photon number density and the nucleon number density vary with time like 1/a3(t).

10.5.3Baryogenesis

We might wonder why η is so small. Actually, the right question to ask is why is η so large!

In the beginning, when the temperature of the universe is high, there are as many protons as photons, because thermal energy is equally shared among the particles of all species. For that matter, there are also equal numbers of antiprotons, electrons, positrons, and so on.

As discussed before, when the temperature (kBT ) drops far below the rest energy E = mc2 of a particle, the particle may disappear through annihilation with its antiparticle. We therefore expect almost no protons, no neutrons, and no electrons left today, with an η many orders of magnitude smaller than the small number quoted above.

This means that somewhere along the line, a mechanism must have existed to create more protons than antiprotons, more neutrons than antineutrons, and more electrons than positrons. The present value of η reflects these extra nucleons that have been created. Such a mechanism is theoretically possible, though it must occur when the universe is not at thermal equilibrium. The force that creates them must be a special kind,6 and must violate nucleon number conservation. Although these conditions first pointed out by Sakharov are known, at what epoch it occured and the precise forces involved, are still a matter of debate. It is clearly a very important problem for it is at the very root of our existence. This is known as the baryogenesis problem.

10.5.4Relativistic Electrons and Neutrinos

The average energy and number densities of relativistic electrons are a bit smaller than those of photons, because electrons obey Pauli exclusion principle, forbidding them to get too close together. For the energy density, it is a factor of 7/8 down, and for the number density, it is a factor of 3/4 down. The same is true for positrons (anti-electrons).

For neutrinos, which also obey the Pauli exclusion principle, their total energy density is (6/2)×(7/8) = 21/8 times the photon energy density, and their total number density is (6/2)×(3/4) = 9/4 times the photon number density. The additional factor of 6/2 = 3 comes about because there are three kinds of neutrinos and three kinds of antineutrinos, but each has only one spin orientation rather than two.10

In the radiation-dominated era, the pressure is11 P = ρ/3, which as we saw implies that the energy density ρ be proportional to 1/a4, where a(t) is the scale factor describing the expansion. We have just seen that at thermal equilibrium ρ is also proportional to T 4 for all relativistic particles. Hence, a(t)T (t) remains the same at all times as long as the number of relativistic species does not change with time.

10.5.5Summary

Statistical mechanics predicts how matter and radiation are distributed at a given temperature. For example, it tells us that at the present temperature of 2.725 K, the universe should be about 400 photons per cubic centimeter, a factor of 2 × 109 larger than the number of nucleons present. The fact that there are any nucleons present at all is actually surprising; it points to some features of elementary particle physics which we have not yet understood.

10.6MeV Temperatures

Three important events take place when the temperature of the universe cools down from a few MeV to about 0.1 MeV. We shall discuss them separately below.

10We have assumed in this estimate three kinds of massless left-handed neutrinos. This cannot be strictly correct because neutrino oscillation experiments have shown that neutrinos have masses. In this chapter we will continue to use the massless estimate because presently we do not know what the neutrino masses are, except that they are all small.

11This equation of state comes about in the following way. Imagine a large semi-infinite tube with cross-sectional area A immersed in the relativistic fluid. The tube is capped at one end, and we wish to compute the pressure exerted on the end cap by the relativistic fluid. First, concentrate on all the relativistic particles moving towards the end cap at an angle θ to its normal. These particles all move with a speed c, so their velocity component normal to the

end cap is c cos θ. After a time T , all particles within a length L = c(cos θ)T of the end cap would have hit the end cap. The total amount of energy these particles carry is then Etot = ρAL = ρc(cos θ)AT . Each such particle carries an energy E and a momentum of magnitude p = E/c. The normal component of the momentum is (E/c) cos θ. The total amount of momentum hitting the end cap in a normal direction is then (Etot/c) cos θ = ρ(cos2 θ)AT . The particles hitting the end cap bounce backward, so the total amount of change of momentum at the end cap is 2ρ(cos2 θ)AT . The pressure Pθ the end cap endures is the force divided by A, and the force is the total change of momentum divided by T . Hence, Pθ = 2ρ cos2 θ. This formula is true when 0 ≤ θ ≤ π/2. For π/2 < θ ≤ π, the relativistic particles are moving away from the end cap, so the corresponding Pθ = 0. The final pressure is obtained by averaging Pθ over all angles, with a volume factor proportional to (sin θ)dθ (in a spherical coordinate system). Hence, P = 0π Pθ(sin θ)dθ/ 0π(sin θ)dθ = ρ/3. For non-relativistic particles, their momenta are much less than mc, or their total energy divided by c. The pressure P is only a small fraction of ρ, hence it is a good approximation to set P = 0.

10.6.1Neutrino Decoupling

At a temperature well below 100 MeV, the only relativistic particles present are electrons, positrons, neutrinos and photons. To be sure, there are also neutrons and protons around, but their rest energy is close to 1000 MeV, so they are nonrelativistic. Besides, there are very few of them left by that time.

Thermal equilibrium is established by sharing energy through collisions. The rate of collision increases with temperature, and also with the strength of interaction of the colliding particles. Neutrinos interact weakly. By the time the temperature drops to a few MeV, neutrinos would have fallen out of thermal equilibrium with the rest of the universe. This means that they will no longer share their energies through collisions. Nevertheless, the temperature of the universe T is still proportional to12 1/a. As we shall see in the next subsection, this decoupling makes the presentday temperature of the cosmic neutrinos lower than the temperature of the cosmic photons.

10.6.2Electron–Positron Annihilation

The electron has a rest energy of 0.5 MeV, which is the same for the positron. At a temperature well below 1 MeV, most of the positrons and electrons have disappeared through the annihilation process into photons.

The annihilation increases the photon temperature by an amount13 equal to (11/4)1/3 = 1.4, but it does not increase the neutrino temperature because neutrinos are no longer in thermal equilibrium with the rest of the universe. The present photon temperature is T0 = 2.725 K; this means that the present neutrino temperature is T0/1.4 = 1.95 K.

10.6.3Temperature and Time

In the radiation era, time and temperature are related by a simple formula:

12This is because the expansion of the universe stretches all the wavelengths. From Wien’s displacement law and Planck’s black-body radiation law, this means a decrease of temperature proportional to 1/a.

13In an isotropic universe there is no exchange of heat between di erent comoving volumes. This fact has already been used in Eq. (10.2). The change of energy inside a comoving sphere of

volume V a3 is −d(ρV ), and the work done during the expansion against pressure is PdV . The di erence of these two is the heat flowing out of the volume, which by definition is −Td(sV ), with s being the entropy density. We therefore have dρ = (sT − ρ − P )(dV/V ) + Tds. Since ρ depends only on T but not on V , we must have s = (ρ + P )/T = 4ρ/3T , which is proportional to gT 3, where g is the e ective number of relativistic species present. With no heat flow S = sV must be the same at all times, which means g(aT )3 must be constant. Discounting the neutrinos because they have already decoupled, we have g = 1 + 2 × 7/8 = 11/4 before annihilation, and g = 1 after. Since annihilation takes place quite suddenly, a remains essentially constant during that period, hence (11/4)Tbefore3 = Tafter3 , or Tafter = (11/4)1/3 Tbefore.

The number 1.71/√g is equal to 1.32 after e+e− annihilation and 0.74 before,14 up to a temperature of about 100 MeV.

10.6.4Big Bang Nucleosynthesis

The chemical element helium is so named because it was first discovered in the sun’s atmosphere (by J. N. Lockyer in 1868; helios is Greek for sun). It is a latecomer in the annals of the discovery of the elements, but it is the second element synthesized in the universe, and also the second most abundant one.

We are so used to the richness of elements around us that sometimes we do not realize most of what we see on earth is made up of extremely rare materials indeed. The universe is predominantly made up of hydrogen, then helium, with about one helium atom for every twelve hydrogen atoms. The rest of the elements occur only in trace amounts. The reason why hydrogen and helium seldom occur on earth in a natural form is because they are too light to be retained by earth’s relatively weak gravitational field.

In the beginning, right after the Big Bang, the universe is very hot. The intense heat prevents any compound nucleus from being formed, so the primordial soup consists of single protons and neutrons, as well as other particles. The only chemical element (or rather, the nucleus of a chemical element) present is hydrogen, which is just the proton. As the universe expands and cools down, at some point neutrons and protons can combine to form other nuclei, such as helium and a (very) few others. This formation is called primordial nucleosynthesis, or Big Bang nucleosynthesis (BBN). Much later when stars appear, hydrogen and other fuels are burned in their interiors to produce thermonuclear energy. This also produces heavier elements as a by-product. The larger the star is, the heavier the elements that can be produced. But no matter how large the star is, elements beyond iron can only be produced in supernova explosions (see Chapter 8 for more details).

Nuclei are formed from two-body collisions. The probability of three or more particles colliding is so small that they can be discounted. Two protons cannot bind into a nucleus because of their electrostatic repulsion, and two neutrons cannot bind into a nucleus because of their tendency to β-decay.6 A proton and a neutron, however, can form a deuteron (D), which is the nucleus of deuterium, or heavy hydrogen. The binding energy is 2.225 MeV, meaning that it takes that amount of energy to tear them apart. This binding energy is su ciently large to prevent the neutron inside from undergoing a β-decay, n → p + e− + ν¯e, for the following reason. The neutron (n) has a rest energy 1.3 MeV above that of the proton (p), and the electron (e−) itself has a rest energy of 0.5 MeV. The electron antineutrino

14The total radiation density is ρ = gργ, where g = 1 + (21/8)(4/11)4/3 = 1.68 after e+e−

annihilation and g = 1 + (2 + 3) × (7/8) = 5.375 before, up to about 100 MeV. In the radiation

√

era, a(t) is proportional to t , so [(da/dt)/a]2 = 1/4t2. Moreover, according to Eq. (10.1), for a critical universe this is also equal to 8πGρ/3c2 = 8πGgT 4/3c2. Equating these two, we obtain the equation above.

(¯νe) may have a mass, but it is too small to be presently detectable. If the neutron decays, it would disintegrate the deuteron, and that requires 2.225 MeV of energy, of which only 0.8 MeV can be supplied by the β-decay process. For that reason, the neutron cannot decay and the deuteron is stable.

However, a binding energy of 2.225 MeV is very small compared to the typical binding energies of other nuclei, so the deuteron is fragile. When two deuterons come together, they can actually form a much more stable nucleus, helium (4He), which contains two protons and two neutrons. The binding energy of a helium nucleus is 28.31 MeV, a large number. In fact, helium is so stable that when radioactive nuclei decay, sometimes the whole helium nucleus is ejected. In that form it is called an α particle and the decay is called α-decay.

Deuteron is formed from the process p + n → D + γ. At high temperatures, the inverse process γ + D → p + n also takes place to tear the deuteron apart. With a binding energy of E0 = 2.225 MeV, one might think that the photo-disintegration of deuteron would stop at a temperature approximately equal to E0. This is not the case because η−1 2 ×109 is a very large number, with two billion photons present for each nucleon. It is true that the probability of any photon having an energy E0 is given by the Boltzmann factor exp(−E0/kBT ). This factor is small when kBT E0, but if it is not smaller than one billionth, there would still be one of the two billion photons that can attain the energy E0 to disintegrate the deuteron. For that reason the deuteron would not become stable until the temperature is such that exp(−E0/kBT ) η, which works out to be kBT 0.1 MeV. According to Eq. (10.6), that occurs about 130 seconds after the Big Bang.

Neutrons and protons have a mass (meaning rest energy) di erence of 1.3 MeV. If they were in thermal equilibrium until deuterium formation at 0.1 MeV, there would be too few neutrons left to account for the approximately 25% of 4He (by weight) left over from the early universe. What happens is that proton and neutron numbers are frozen at their values at kBT 0.7 MeV, so their ratio is given by the Boltzmann factor to be exp(−1.3/0.7) 1/7. This means one 4He (2 protons and 2 neutrons) to every 12 protons, or a weight ratio between 4He and H of 1/3. Consequently, some 25% of the nucleonic mass is locked up in the helium nucleus.

To understand freezing, we must first understand how neutrons and protons turn into each other at thermal equilibrium. That proceeds mainly through the weak interaction processes νe + n ↔ p + e− and e+ + n ↔ p + ν¯e. As the temperature decreases, the reaction rate goes down. The disappearance of e± through annihilation around a temperature of 1 MeV slows down these processes even further. At some point these reactions e ectively stopped freezing out the number of neutrons and protons. Calculations show that this happens at about 0.7 MeV.

As mentioned before, deuterons are fragile so they tend to combine to form the more stable 4He. If we increase η, i.e., the number of nucleons, then D = 2H would be formed at a higher temperature, at which time more neutrons would be present. With more neutrons around, there would be more deuterons formed. At a

The abundance of these elements all depends on the parameter η, or the nucleon density, as shown in Fig. 10.3. By comparing observation with calculations, one can determine η to be around 5 × 10−9. This is equivalent to a nucleonic density equal to 4% of the critical density, as quoted before.

It should also be pointed out that it is a di cult task to observe the abundance of these primordial elements, not only because they occur in small quantities, but also because they must be isolated from the elements formed much later in stars.

10.6.5Summary

The energy density of relativistic particles is proportional to T 4, and the number density is proportional to T 3, where T is the temperature of the universe.

Neutrinos are decoupled from the rest of the universe at a temperature of a few MeV. The permanent annihilation of electron–positron pairs at about 1 MeV reheats the photons, but not the already decoupled neutrinos. As a result, the present neutrino temperature is only 1.95 K, although the photon temperature is 2.725 K.

Deuteron is the first nucleus to be created, at a temperature of about 0.1 MeV. It is fragile; most of it is converted into the much more stable 4He. Besides 3He and 7Li formed in trace amounts, practically no other nuclei are formed at this stage of the universe. By comparing calculation with observation, one finds η 5 × 10−9, which translates into a nucleon density equal to about 4% of the critical density.

Below 1 MeV, the temperature of the universe is inversely proportional to the scale factor a(t). In the radiation era, temperature and time are related by Eq. (10.6).

10.7The Cosmic Microwave Background (CMB)

10.7.1Observation

The age of communication satellites brought us an unexpected gift from heaven. It led to the accidental discovery of the cosmic microwave background radiation (CMB) by Arno Penzias and Robert Wilson in 1965.

Penzias and Wilson worked for the Bell Telephone Laboratories in New Jersey. They used a radio horn, originally designed to link up with communication satellites, to make astronomical observations at a microwave wavelength of 7.35 cm. After eliminating all known sources of background and interference, an annoying noise remained. It could not be removed no matter what they tried, and in what direction they pointed the horn. That turned out to be one of the most important discoveries in astronomy, and the unwanted noise became a new and powerful tool for astroarchaeology.

412 Cosmology

Later observations show that this electromagnetic noise has a black-body distribution with a temperature T0 = 2.725 K.

This was not the kind of accuracy obtained initially. The peak wavelength of this temperature occurs at 1 mm. Electromagnetic waves in that vicinity are absorbed by the water vapor in the atmosphere, so this kind of accuracy can be obtained only when water vapor is greatly reduced or eliminated. This can be achieved by building ground-based observatories at high altitude and/or high latitude, by putting the detectors in a balloon, and better still, on a satellite.

Unknown to Penzias and Wilson, the noise was actually predicted by George Gamow in the late 1940’s to be the reverberating radiation left over from the big bang. He even estimated the temperature of this radiation to be around five or six degrees Kelvin; not a bad estimate at such an early stage!

The CMB was discovered at the beginning to be isotropic to an accuracy of 10−3 (one part in a thousand). Soon afterwards, using a borrowed U2 spy plane flying high above most of the atmosphere, Luis Alvarez and his group were able to improve the accuracy to 10−4. With that accuracy, they found a ‘dipole’ deviation from isotropy, namely, one direction of the sky seems to be a bit hotter than the opposite direction. This is attributed to the Doppler shift arising from the motion of the solar system,15 with a velocity of 371 km/s. The COBE satellite, now retired, then found a complicated temperature fluctuation to set in at the level of 10−5. This fluctuation turns out to yield very rich information on the state of the universe, stretching all the way back to the inflationary era. We will discuss the physics of that in the next subsection.

The COBE satellite had two antennae separated by some 7◦ to measure the temperature di erence ∆T in these two directions. Subsequent balloon and groundbased experiments are able to measure down to a much smaller angular di erence θ. The present available data of these measurements are shown in Figs. 10.4 and 10.5. The vertical axis can be thought of as the average of (∆T/T )2, and the horizontal axis is related to θ roughly by 2π/θ. Figure 10.4 shows the data from the WMAP data released in February, 2003. The curve is a fit to a popular model of cosmology. Figure 10.5 summarizes the earlier data, together with the fitted curve taken from Fig. 10.4. Some of the parameters determined from the fits are, the Hubble constant h0 = 0.72 ± 0.05, the total matter content Ωm = 0.29 ± 0.07, and the baryonic (ordinary) matter content Ωb = 0.047 ± 0.006. The latter number translates into a baryon number density nb = 0.27 ± 0.01/m3, and a baryon-to- photon ratio η = (6.5±0.4)×10−10. The age of the universe is t0 = 13.4±0.3 billion years, and decoupling (the time when CMB was emitted — see next subsection) occurs at a red-shift of z = 1088 + 1/− 2, and a time 372 000 ± 14 000 years after the Big Bang. The matter-dominated era took over from the radiation-dominated era at a red-shift of z = 3454.

15Doppler shift changes the peak and other wavelengths of the black-body radiation, hence the e ective temperature.

The universe is flat, with a total Ω = 1.02 ± 0.02. The equation of state for the dark energy component must have a w < −0.78.

It is remarkable that the WMAP parameters, determined from the physics at a red-shift z of about 1000 and at a time more than 300 000 years after the Big Bang (see the next subsection), are essentially the same as those determined in a completely di erent way, and at a much di erent epoch of the universe. For example, h0 determined directly from the Hubble plot involves objects with z < 2, and the baryonic content determined from BBN concerns the universe at about 3 minutes old.

The European Planck satellite to be launched in a few years will have an angular resolution of 5 , which enables it to obtain very precise results at large .

10.7.2The Physics of CMB

The energy required to tear the electron away from a hydrogen atom is E0 = 13.6 eV. Above that temperature, the universe consists of a plasma of charged electrons and nuclei, with photons trapped in between because they are being continuously absorbed and emitted by the plasma.

Below 13.6 eV, an electron and a proton can combine into a hydrogen atom H. In the process, a photon γ is emitted to carry away the energy E0. Just like the photodisintegration of deuteron considered earlier, the large number of photons around makes it possible for the inverse reaction γ + H → n + p to take place all the way down to a temperature determined by E0 · ln(η−1) 1, viz., kBT 0.6 eV. A more careful calculation, taking distribution and reaction rates into account, shows that the neutralization of plasma into hydrogen atoms takes place at about (1/4) eV. This epoch, known as recombination, or decoupling, will be indicated by an asterisk subscript. It is called decoupling because after the recombination, photons in this neutral environment are hardly absorbed or scattered, so they e ectively decouple from the rest of the environment. With kBT (1/4) eV, the corresponding temperature is T 3000 K. Since the photon temperature is inversely proportional to the scale factor a, which in turn is inversely proportional to the red-shift factor z + 1, we get z 1000. This is so because at the present time t0, the parameters are T0 3 K, z(t0) = 0, and a(t0) = 1.

The wiggly structure in Figs. 10.4 and 10.5 is caused by the acoustic waves set up at the beginning of the universe. In the inflationary era when the universe is very small, quantum fluctuations are important. These fluctuations set up an acoustic wave in the cosmic fluid, in much the same way a snap of your finger generates a sound wave in the air. The amplitude of a sound wave in the air, with wave number k = 2π/λ and generated with no initial velocity, has a time dependence

cos(ωt) = cos(kcst), where ω is the angular frequency and cs is the speed of sound.

√

With the equation of state P = ρ/3, the speed of sound is cs = c/ 3 . The amplitude of an acoustic wave in the cosmic fluid is similar, but with one crucial di erence.