Cosmic Rays during BBN to Solve Lithium Problems Mingming Kang Collaborators : Yang Hu, Hongbo Hu, Shouhua Zhu College of physical science and technology, Sichuan University KIAA-WAP Ⅱ, Aug. 18, 2016

Contents Brief introduction to BBN The Lithium Problems 1 The Lithium Problems 2 Our method – MeV cosmic rays in BBN 3 Calculation and Results 4 Conclusion 5

A brief introduction to BBN The theory Big Bang Nucleosynthesis(BBN, for short) refers to the production of nuclei other than those of the lightest isotope of hydrogen(1H) during the early phases of the universe. BBN is believed to have taken place from tens of seconds to 20 minutes after the Big Bang. It is calculated to be responsible for the formation of most of the universe's 4He, along with small amounts of the deuterium(2H or D), 3He, and a very small amount of the lithium isotopes 7Li(decay from 7Be) and 6Li.

A brief introduction to BBN The main nuclear reaction chains in BBN Before the epoch of nucleosynthesis, there are electrons, positrons, neutrinos, neutrons and protons(heavier nuclei are almost absent) in the plasma. It was hot and dense enough for these particles to be in kinetic and chemical equilibrium. As the Universe expansion proceeds(the temprature was going down), neutrinos freeze out, the n/p density ratio departs from its equilibrium value and freezes out at the value n/p = 1/7. When the photon temperature become below the deuterium binding energy, deuterium formed via n + p → 2H + γ process. Once 2H starts forming, a whole nuclear process network sets in, leading to heavier nuclei production until BBN eventually stops.

The Lithium Problems Primordial nuclide abundance Our knowledge of the abundances of light elements in the early Universe has relied on measurements of the chemical content of old stars in the Milky Way’s halo. Theoretical predictions of the primordial abundances of light elements are based on the precise determination of the cosmic ratio of baryons to photons. The measured primordial amounts of hydrogen(H, D) and helium(4He, 3He(only upper limit)) match the predictions. But that of lithium does not. PDG 2014

The Lithium Problems The Spite plateau The chance to link primordial 7Li with the standard BBN abundance was first proposed by Spite & Spite (1982). They showed that the lithium abundance in the metal-poor dwarfs was independent of metallicity for [Fe/H] < -1.5. This constant lithium abundance is commonly called “the Spite plateau”. The Spite plateau suggests that this may be the lithium abundance in pre-Galactic gas provided by the standard BBN. Spite plateau

The Lithium Problems What are the “Lithium Problems”? The BBN predicted abundance of 7Li is higher than that of observations. The BBN predicted abundance of 6Li is lower than that of observations. Observational abundance 7Li/H = (1 ～2) × 10-10 6Li/H ～6 × 10-12 Theoretical abundance 7Li/H = (5.24+0.71-0.67) × 10-10 6Li/H ～ 10-14

The Lithium Problems Some methods solving the problems Modification to the BBN scenario: --Particle Decay, Evaporate, Annihilate, Catalyze --Variable constants (effective couplings; neutron lifetime …) The observed Lithium abundance is not primordial --Astrophysical origin Comments: Updated observations show that the second way are moving farther and farther away from the target. Some of the first solutions can solve the lithium-7 problem, but they will let the lithium-6 problem worse.

Our method – MeV cosmic rays in BBN Reasons for the existence of cosmic rays in BBN The absolute thermal equilibrium can’t be maintained because of the rapid cosmic expansion. Plasma turbulence acceleration? The field seeds born in the electroweak phase transition epoch evolve into BBN? Deviating from the thermal equilibrium because of the original perturbation? Anyway, we give the hypothesis that there are non-thermal energetic particles(BBNCR) in the BBN period !

e+, e- × quickly thermalized n × not easy to be accelerated p, d, t √ Our method – MeV cosmic rays in BBN Candidates for cosmic ray particles in BBN e+, e- × quickly thermalized ϒ × quickly thermalized n × not easy to be accelerated p, d, t √ 4He, 3He, etc. √ The nucleon energy loss due to Coulomb collisions; The electron energy loss due to Compton scattering and synchrotron.

Our method – MeV Cosmic rays in BBN BBN cosmic ray(BBNCR) spectrum The highest energy of energetic hydrogen isotopes is a model parameter, named Eupper. Assuming that electric-charged particles are accelerated by electromagnetic force, and that the gain energy is proportional to the electric charge, then Eupper(He) = 2Eupper . The proportion of energetic particles to the thermal ones is set as the other free paremeter, named ε, which is unchanging during the BBN period. The explicit expression of energy distribution with a “knee” and “GZK” cutoff：

Our method – MeV Cosmic rays in BBN The critical nuclear reactions For 7Li problem： 7Be + p -> 4He + 3He + p Eth = 1.814 MeV 2H + p -> n + 2p Eth = 3.337 MeV For 6Li problem： 4He + 3He -> p + 6Li Eth = 7.048 MeV 2H + 4He -> n + p + 4He Eth = 3.343 MeV We introduce energetic particles into BBN to open the endothermic reactions to account for the lithium problems. The endothermic reactions with the lowest threshold energy destroying 7Be and producing 6Li are 7Be(p, pα)3He and 3He(α, p)6Li, respectively. The reactions with the lowest threshold energy destroying D by proton and 4He are D(p, n)2H and 4He(d, α)np, respectively. The highest threshold energy of these 4 reactions is 7.048 MeV, if BBNCRs really work all reactions with a threshold below this energy will occur!. There are nearly 200 such reactions associated with light nuclides(isotopes of hydrogen, helium and lithium)! We have to pick out those important ones ignoring those that have no effect on the abundances of light elements.

Our method – MeV Cosmic rays in BBN Reactions added into BBN code D(p, n)2H Eth = 3.337 destroy D D(d, n)3He ----- destroy D D(d, p)T ----- destroy D T(d, n)4He ----- destroy D D(p, g)3He ----- destroy D 4He(d, np) 4He Eth = 3.343 destroy D 7Be(p, pα)3He Eth = 1.814 destroy 7Be 7Be(α, 2α)3He Eth = 2.492 destroy 7Be 7Be(α, p)10B Eth = 1.8 destroy 7Be, produce 10B 7Li(p, n)7Be Eth = 1.880 destroy 7Li 7Li(p, α)4He ----- destroy 7Li 4He(t, g)7Li ----- produce 7Li 4He(t, n)6Li Eth = 8.388 produce 6Li 3He(α, p)6Li Eth = 7.048 produce 6Li 7Li(d, t)6Li Eth = 1.279 produce 6Li 7Be(d, 3He)6Li Eth = 0.146 produce 6Li 4He(d, g)6Li ----- produce 6Li D(α, g)6Li ----- produce 6Li 6Li(α, p)9Be Eth = 3.540 destroy 6Li, produce 9Be 6Li(α, d)24He Eth = 2.454 destroy 6Li 6Li(p, α)3He ----- destroy 6Li

Calculation and Results The set of differential equations ruling the BBN Definition of Hubble parameter H Baryon number conservtion Entropy conservtion Boltzmann equations for Nnuc Universe charge neutrality Boltzmann equations for neutrino species

Calculation and Results Modification in the BBN code – calculation of new processing rates Each new reaction will contribute a processing rate to the Boltzmann equation to change the abundance of related nuclide.

Calculation and Results The production and destruction rate of 7Be (7Li) Processing rates of the reactions of Destroying and producing 7Be(or 7Li). The reactions with BBNCR work later than the thermal ones The reactions with BBNCR will delay BBN.

Calculation and Results The production and destruction rate of 6Li Processing rates of the reactions of Destroying and producing 6Li.

Calculation and Results The evolution of light element abundance as a function of the Universe’s temperature ε = 2x10-5 Eupper = 3.7MeV 7Be 6Li Element abundances as a function of the Universe’s temperature. Solid lines Denote the BBNCR results and dashed lines those for standard BBN.

Calculation and Results The parameter space 7Li/H = (1.3 ～1.9) × 10-10 6Li/ 7Li < 0.05 and 6Li /H > 10-13 7Li/H = (1.3 ～1.9) × 10-10 6Li/ 7Li < 0.05 and 6Li /H > 10-12 Because only the upper bound of 6Li is measured, we give two results corresponding to two lower bounds. The red area indicate the BBNCR predicted abundances fall into the observationally allowed range.

Calculation and Results Abundance variation range 7Li/H = (1.3 ～1.9) × 10-10 6Li/ 7Li < 0.05 and 6Li /H > 10-12 D, 7Be, and 6Li abundances as a function of the Universe’s temperature with the observationally allowed ε and Eupper range.

Calculation and Results Model test On cross-section uncertainties -- trial cross-sections (7Be(p, pα)3He and 3He(α, p)6Li) time 1/10 -- the cross-sections are constant, using the value once the energy exceeds the threshold. (0.22 mb and 5.2 mb, respectivly) On energy spectrum uncertainty -- uniform distribution all through Our hypothesis is insensitive to the amplitude or resonance peak of the uncertain cross-sections and also insensitive to the explicit shape of the BBNCR energy spectrum.

Calculation and Results Model test The parameter space of different observations, cross-section test, and energy spectrum test.

Calculation and Results Predictions BBNCRs may promote 10B via 7Be(α, p)10B to 10-12 approaching the upper limit of observations, and also produce more 9Be, isotopes of C, N, O.

Conclusion The Lithium Problem is still a puzzle after 30 years, to answer this problem requires new thoughts. We take the concept of cosmic rays into BBN for the first time and find it effective on both 7Li and 6Li problems without excessive depletion of deuterium. The BBN cosmic ray(BBNCR) flux in the model is small enough (~10-5 relative to the background) to keep the success of standard BBN(SBBN) and the BBNCR need to be accelerated to MeV. Our hypothesis is insensitive to the amplitude or resonance peak of the uncertain cross-sections and also insensitive to the explicit shape of the BBNCR energy spectrum. We predict more primordial 10B and more CNO which would affect the first generation of stars(or Population Ⅲ), precise observations may or may not support our hypothesis. The acceleration mechanism in BBN and possible observational consequences of BBNCR are interesting subjects to explore.

Thanks

Recent observation data

The spectrum of BBNCR

On the reaction cross sections 1， D（p，n）2H 3.337 √ 2， T（a，n）Li6 11.131 √ 3， （He4（t，n）Li6 8.388） shift the data of reaction 2 4， Be7（a，p）B10 1.8 √ 5， Be7（a，aa）He3 2.492 √ 6， He4（He3，p）Li6 7.048 √ 7， He4（d，np）He4 3.343 shift the data of reaction 1 and times 5 8， Li6（a，p）Be9 3.540 √ 9， Li6（a，d）Be8 2.607 √

Calculation and Results The 7Be abundance with different ε

preliminary preliminary

Calculation and Results The evolution of light nuclide abundance (with high energy H isotopes only) ε = 1.6x10-5 Eupper = 3.5 MeV 7Be 6Li