Primary Nucleosynthesis

6.7.1

Problem 6.7.1

Find the ratio of neutrons to protons number densities in the case of thermal equilibrium between them.

solution

Though total number of baryons is conserved, they can transform to each other in reactions of the type $n{\nu _e} \leftrightarrow p{e^-}, n{e^+} \leftrightarrow p{\bar \nu}_e.$ If their rates are sufficiently high to support the thermal equilibrium in the expanding Universe, then $$ \frac{n_n}{n_p} = \left(\frac{m_n}{m_p}\right)^{3/2}e^{-\left(m_n - m_p\right)/T} \approx e^{-\Delta m/T}, $$ where $m_p = 938.272\,MeV,\;m_n = 939.565\,MeV;\Delta m = m_n - m_p = 1.293\,MeV.$

6.7.2

Problem 6.7.2

Up to what temperature can the reaction $n\nu_e\leftrightarrow pe^-$ support thermal equilibrium between protons and neutrons in the expanding Universe

solution

In order to know whether the thermal equilibrium is supported one needs to compare the rate of Universe expansion with that of the reactions, required to provide the equilibrium. The reaction rate (per neutron) is given by the expression \[\Gamma = n\left\langle \sigma v\right\rangle,\] where $n$ is the number density of the target particles (neutrinos or electrons in the case under consideration), $$ n_{\nu ,e} = \frac{3} {4}\frac{\zeta (3)}{\pi ^2}g_{\nu ,e}T^3, $$ and the cross-section $\sigma$ can be found in frames of the Standard Model. To estimate the thermal mean $\left\langle \sigma v\right\rangle $ one can use the expression \[ \left\langle {\sigma v} \right\rangle \approx G_F^2{T^2},\] where $G_F \approx 1.166 \times 10^{ - 5} \,GeV^{-2}$ is the Fermi constant of weak interaction. Omitting all the factors of order of unity one obtains $ \Gamma \approx G_F^2T^5.$ The expression for the Hubble parameter in radiation-dominated Universe was obtained in problem of the present Chapter: $$ H = 1.66\sqrt {g^*} \frac{T^2}{M_{Pl} }. $$ The point, where $\Gamma = H$, determines the temperature ${T_f}$ of freeze-out (fixation) for the ratio ${n_n}/{n_p}.$ Equating the expressions for $H$ and $\Gamma,$ one obtains $$ G_F^2T^5 = 1.66\sqrt {g^*} \frac{T^2}{M_{Pl} }, $$ and then $$ T_f = \left(\frac{1.66} {G_F^2M_{Pl} } \right)^{1/3}(g^*)^{1/6}. $$ Using that ${g^*} = 10.75,$ one can obtain that $T_f\simeq1.2\,MeV$. More accurate calculations give the value ${T_f} \approx 0.7\,MeV.$

6.7.3

Problem 6.7.3

Determine the ratio $n_n/n_p$ at the temperature of freeze-out.

solution

$$\frac{n_n}{n_p} = e^{ - \left( m_n - m_p \right)/T_f} \approx e^{ - 1.3/0.7} \approx 0.16. $$

6.7.4

Problem 6.7.4

Determine the age of Universe when it reached the temperature of freeze-out.

solution

In the region $T \sim 1\,MeV,\;{g^*} = 10.75\;$ the relation between time and temperature can be presented in the following form (see problem and problem) $tT^2 \approx 0.75\ \mbox{s}\cdot\,MeV^2$, and it follows for $T \approx 0.7\,\,MeV$ that $t \approx 1.5\,s$.

6.7.5

Problem 6.7.5

At what temperature and at what time did efficient deuterium synthesis start?

solution

6.7.6

Problem 6.7.6

Determine the time period during which the synthesis of light elements took place.

solution

6.7.7

Problem 6.7.7

Determine the ratio of neutrons to protons number densities at temperature interval from the freeze-out to the creation of deuterium.

solution

Even after the fixation (freezing-out) of the neutron concentration at $T_f \approx 0.7\, \, MeV $ (see problem of the present Chapter) the neutrons can decay up to the deuterium creation, which starts near $T = 0.085\, \,MeV$. In order to determine the corresponding moment of time use the relation (see problem) \[ H \approx 1.66\sqrt{g^*}\frac{T^2}{M_{Pl} }. \] Calculation effective number of degrees of freedom $g^* $ at the temperature corresponding to the deuteron creation threshold should be performed with due care. Firstly, at the temperature under consideration only photons and neutrinos remain relativistic, which is trivially taken into account. Secondly one should remember that in the considered energy range the temperature of neutrino background differs from that of microwave. Therefore in order to calculate the effective number of internal degrees of freedom the following expression is to be used (see problem) \[ g^* = \sum\limits_{i = bosons} g_i \left(\frac{T_i}{T}\right)^4 + \frac{7}{8}\sum\limits_{j = fermions}g_i \left(\frac{T_j}{T}\right)^4 . \] For $T \gg m_e \approx 0.5\,\, MeV$ the reaction $e^ + e^ - \leftrightarrow \gamma \gamma $ proceeds with the same rates in both directions. As long as temperature falls down below the electron mass, the photons are already unable to support the reaction $\gamma \gamma \to e^ + e^ - $. The fact that the reaction $e^ + e^ - \leftrightarrow \gamma \gamma $ presents an additional source of photons, leads to result that the photon temperature appears higher than that of neutrino. Using solely thermodynamic considerations, one can shove that \[ \frac{T_\nu}{T_\gamma} = \left(\frac{4}{11}\right)^{1/3} \approx 0.714,\] and therefore \[ g_* = g_\nu + \frac{7}{8}2 \times 3\frac{T_\nu ^4}{T^4} =2 + \frac{7}{8}6\left(\frac{4}{11}\right)^{4/3} \approx 3.36. \] In the considered energy range (see problem and problem) \[ tT^2 \approx \frac{ {0.301} } { {\sqrt {g^* } } }M_{Pl} = 1.32\ \,s\cdot \,MeV^2. \] and it follows that $t \approx 180\,s$. As a rough estimate one can take that the neutrons can decay during that time and they are afterwards absorbed due to creation of deuterium, and later helium. Therefore from the point of neutron number density fixation to starting point of deuterium creation one obtains \[ \frac{n_n}{n_p} = e^{-\left(m_n - m_p \right)/T_f }e^{ - t/\tau _n }. \] Here $\tau _n \approx 885.7\, s$ is the mean lifetime of neutron. At $t \approx 180\,s$ \[ \frac{n_p}{n_n} \approx 0.13. \] Recall that at the beginning of the latter process the considered ratio was equal to $0.16$ (see problem).

6.7.8

Problem 6.7.8

Determine the relative abundance of ${}^4He$ in the Universe

solution

6.7.9

Problem 6.7.9

How many helium atoms are there for each hydrogen atom?

solution

6.7.10

Problem 6.7.10

What changes in relative ${}^4 He$ abundance would be caused by

a) decreasing of average neutron lifetime $\tau_n$;

b) decreasing or increasing of the temperature of freeze-out $T_f$?

solution

6.7.11

Problem 6.7.11

What nuclear reactions provided the ${}^4 He$ synthesis in the early Universe?

solution

Synthesis of deuterium presents a critically important step to synthesis of $^4He$, because the direct synthesis of helium from two protons and two neutrons is highly improbable event. After creation of deuterium the helium synthesis proceeds along the following reactions: $$ \begin{gathered} d + p \to ^3He + \gamma ;\\ d + d \to ^3He + n; \\ d + d \to t + p; \\ ^3He + d \to ^4He + p;\\ t + d \to ^4He + n.\\ \end{gathered} $$

6.7.12

Problem 6.7.12

Why is synthesis of elements heavier than ${}^7 Li$ suppressed in the early Universe?

solution

There are no natural stable nuclei with $A=5$, therefore one should consider only fusion of $^4He$ with tritium and $^3He$: $$ \begin{gathered} t + ^4He \to ^7Li + \gamma ; \\ ^3He + ^4He \to ^7Be + \gamma \to ^7Be + e^ - \to ^7Li + \nu _e. \\ \end{gathered} $$ Synthesis of two nuclei $^4He$ leads to unstable nucleus $^8Be.$ The reaction $$ ^8Be + ^4He \to ^{12}C + \gamma $$ is inefficient: low density of the reactants leads to the fact that the mean time between the collisions of the nuclei considerable exceeds the lifetime of the unstable nucleus $^8Be.$ This reaction becomes important in stars, but it does not make importance in early Universe. Thus the absence of stable elements with $A=5$ and $A=8$ makes it impossible to proceed beyond the $^7Li$ in primary nucleosynthesis.

6.7.13

Problem 6.7.13

In our Universe the neutron half-value period (the life-time) approximately equals to 600 seconds. What would the relative helium abundance be if the neutron life-time decreased down to 100 seconds?

solution

6.7.14

Problem 6.7.14

At what temperature in Universe did the synthesis reactions stop?

solution

At $t\sim 1000\, s$ the temperature falls down to $T \approx 0.03\,MeV$. After that moment the kinetic energy of nuclei is insufficient to overcome the Coulomb barrier and the synthesis processes stop.