Physical Foundations of Cosmology

100 The hot universe the cross-section is reduced by the factor 1 −ne= [ 1 +exp(−e/T) ]− 1 to account for the Pauli exclusion ...

3.5 Nucleosynthesis 101 Before electron–positron annihilation the temperatures of the electrons and neu- trinos are equal, that ...

102 The hot universe is the equilibrium concentration of neutrons.To obtain the second equality in (3.109) we used the relations ...

3.5 Nucleosynthesis 103 us to see how the freeze-out concentration depends on the number of relativistic species present at the ...

104 The hot universe is the sole remaining cause for a change in the number of neutrons. As a result, the neutron concentration ...

3.5 Nucleosynthesis 105 is the binding energy of the deuterium. We have parameterized the baryon-to-photon ratio by η 10 ≡ 1010 ...

106 The hot universe become efficient. Within the relevant temperature interval, 0.06 MeV to 0.09 MeV, the experimentally measur ...

3.5 Nucleosynthesis 107 Problem 3.18Verify that after electron–positron annihilation, the value ofκin (3.85) becomes κ 1. 11 + ...

108 The hot universe pn pn D DD1 DD2 Dγ Dp TD (^3) Hen (^3) He D (^3) He (^4) He (^4) He T (^7) Lip (^7) Ben (^7) Be (^7) Li (^4 ...

3.5 Nucleosynthesis 109 1/60 1 51560 10 0 10 −^4 10 −^9 10 −^14 time [minutes] temperature mass fraction p n^4 He D T (^3) He 10 ...

110 The hot universe deuterium to produce^4 He: TD→^4 Hen. (3.136) In this sequence, two of the three neutrons end up in the new ...

3.5 Nucleosynthesis 111 10 −^2 10 −^11 10 −^10 10 −^9 10 −^8 10 −^7 10 −^6 10 −^5 10 −^4 10 −^3 10 −^2 10 −^1 10 0 10 −^1 10 1 1 ...

112 The hot universe see later thatη 10 can be determined with high precision from data on deuterium abundance and CMB fluctuati ...

3.5 Nucleosynthesis 113 If the universe were not expanding, then nearly all free neutrons would end up in^4 He nuclei and there ...

114 The hot universe It is convenient to rewrite these equations using a temperature variable instead of a time variable (see (3 ...

3.5 Nucleosynthesis 115 and henceXn 0 .12. Neglecting the temperature dependence ofα,we then find the approximate solution of ( ...

116 The hot universe to be a constant. Equation (3.160) is then readily integrated: ( 1 + 2 R 2 XD(T) ) = ( 1 + 2 R 2 XD(T∗) ) e ...

3.5 Nucleosynthesis 117 3.5.5 The other light elements Now we can calculate the final abundances of the other light elements by ...

118 The hot universe Assuming that tritium freeze-out occurs at about the same time as for deuterium, and substitutingXnX^2 D/R ...

3.5 Nucleosynthesis 119 deuterium reaches its final abundance, and we can substitute into (3.173) the values ofX 3 fHeandXTfobta ...