Stau relic density at the Big-Bang nucleosynthesis era consistent with the abundance of the light element nuclei in the coannihilation scenario
Toshifumi Jittoh, Kazunori Kohri, Masafumi Koike, Joe Sato, Takashi Shimomura, Masato Yamanaka
aa r X i v : . [ h e p - ph ] J a n STUPP-09-205, TU-861, YITP-10-1, ICRR-Report-556
Stau relic density at the Big-Bang nucleosynthesis era consistent with the abundanceof the light element nuclei in the coannihilation scenario
Toshifumi Jittoh, Kazunori Kohri,
2, 3
Masafumi Koike, Joe Sato, Takashi Shimomura,
4, 5 and Masato Yamanaka Department of Physics, Saitama University, Shimo-okubo, Sakura-ku, Saitama, 338-8570, Japan Physics Department, Lancaster University LA1 4YB, UK Department of Physics, Tohoku University, Sendai 980-8578, Japan Departament de F ´ isica T e ` orica and IFIC, Universitat de V al ` encia -CSIC, E-46100 Burjassot, V al ` encia , Spain Yukawa Institute for Theoretical Physics, Kyoto University, Kyoto 606-8502, Japan Institute for Cosmic Ray Research, University of Tokyo, Kashiwa 277-8582, Japan
We calculate the relic density of stau at the beginning of the Big-Bang Nucleosynthesis (BBN) erain the coannihilation scenario of minimal supersymmetric standard model (MSSM). In this scenario,stau can be long-lived and form bound states with nuclei. We put constraints on the parameterspace of MSSM by connecting the calculation of the relic density of stau to the observation of thelight elements abundance, which strongly depends on the relic density of stau. Consistency betweenthe theoretical prediction and the observational result, both of the dark matter abundance and thelight elements abundance, requires the mass difference between the lighter stau and the lightestneutralino to be around 100MeV, the stau mass to be 300 – 400 GeV, and the mixing angle of theleft and right-handed staus to be sin θ τ = (0 .
65 – 1).
I. INTRODUCTION
Cosmological observations have established the exis-tence of the non-baryonic dark matter (DM) [1]. Theseobservations suggest that the DM is a stable and weakly-interacting particle with a mass of O (100) GeV. Manyhypothetical candidates for the DM have been proposedin models of particle physics beyond the standard model(SM), and one of the most attractive candidates is thelightest neutralino, ˜ χ , in supersymmetric extensions ofthe SM with R parity conservation. Neutralino is a linearcombination of the superpartners of U (1) , SU (2) gaugebosons and the two neutral Higgses, and is stable whenit is the lightest supersymmetric particle (LSP). Indeedit accounts for the observed DM abundance when it isdegenerate in mass to the next lightest supersymmetricparticle (NLSP) and hence coannihilates with the NLSP[2]. We consider the setup that the LSP is a neutralinoconsisting of mainly bino, the superpartner of U (1) gaugeboson, and the NLSP is the lighter stau, the superpart-ner of tau lepton. This is naturally realized in the MSSMwith the unification condition at the grand unified the-ory scale. The minimal supersymmetric SM (MSSM) hastwo eigenstates of stau as physical state. In absence ofinter-generational mixing the mass eigenstate of stau isgiven by the linear combination of the left-handed stau˜ τ L and the right-handed stau ˜ τ R as˜ τ = cos θ τ ˜ τ L + sin θ τ e − iγ τ ˜ τ R , (1)where θ τ is the left-right mixing angle and γ τ is the CPviolating phase.In a scenario of the coannihilation, the NLSP stau canbe long-lived if the mass difference, δm , between neu-tralino and stau is small enough to forbid two-body de-cays of stau into neutralino. It was shown in [3] that the lifetime of stau is longer than 1000 second for δm . Li problem, onthe primordial Li abundance between the predictionfrom the standard BBN (SBBN) and the observations[19, 20]. Combined with the up-to-date values of baryon-to-photon ratio, η = (6 . ± . × − from Wilkin-son Microwave Anisotropy Probe (WMAP) [1], theSBBN predicts the Li to proton ratio, ( Li / H) SBBN =5 . +0 . − . × − , which is by about four-times largerthan its observed value in poor-metal halos [19, 21]. Be-cause there exists no general agreements about astro-physical scenarios to reduce the Li abundance [22–25],it is natural to consider nonstandard effects.The authors have investigated the BBN including thelong-lived stau [14, 15]. The long-lived stau form abound state with nuclei (˜ τ N ), and consequently con-vert it into a nucleus with a smaller atomic number.Here N stands for a nucleus. In this scenario, the abun-dance of Li is reduced through the conversion process,(˜ τ Be) → ˜ χ + ν τ + Li and the further destruction of Liby either a collision with a background proton or anotherconversion process, (˜ τ Li) → ˜ χ + ν τ + He. Therefore,the more these bound states are formed, the more the Li abundance is reduced. The number density of thebound state is determined by the relic density of stau. In[14, 15], we assumed Y ˜ τ, FO and δm to be free parametersof the scenario, where Y ˜ τ, FO is the yield value of stauat the time of decoupling from the thermal bath. Thefull calculation of the light nucleus abundances showeda region in ( δm, Y ˜ τ, FO ) plane where the Li problem issolved consistently with the observational constraints onthe other nuclei. The region points Y ˜ τ, FO to be close tothe yeild value of the DM and δm to be around 0 . Y ˜ τ, FO with taking the relic abundance of DMinto account. The outline of this paper is as follows. Insection II, we review the formalism for the calculationof the relic density of stau, and derive the Boltzmannequations of stau and neutralino for the calculation. Insection III, we present the numerical results of the cal-culation, and prove the parameter space for solving the Li problem. Section IV is devoted to a summary anddiscussion.
II. FORMALISM FOR THE CALCULATION OFRELIC DENSITY OF STAU AT THE BBN ERA
In this section, we prepare to calculate the relic densityof stau at the BBN era. Firstly, in subsection II A, webriefly review the Boltzmann equations for the numberdensity of stau and neutralino based on the thermal relicscenario. Then, in subsection II B, we discuss the numberdensity evolution of stau and neutralino quantitatively.In subsection II C, we investigate the significant processesfor the calculation of the relic density of stau. Finally,we obtain the Boltzmann equations for the relic densityof stau in a convenient form.
A. Boltzmann equations for the number densityevolution of stau and neutralino
In this subsection, we show the Boltzmann equations ofstau and neutralino and briefly review their quantitativestructure based on [26] (see also a recent paper[27]).We are interested in the relic density of stau in thecoannihilation scenario. In this scenario, stau and neu-tralino are quasi-degenerate in mass and decouple fromthe thermal bath almost at the same time [2]. Thus therelic density of stau is given by solving a coupled set ofthe Boltzmann equations for stau and neutralino as si-multaneous differential equation. For simplicity, we usethe Maxwell-Boltzmann statistics for all species insteadof the Fermi-Dirac for fermions and the Bose-Einsteinfor bosons, and assume T invariance. With these simpli-fications, the Boltzmann equations of them are given asfollows dn ˜ τ − dt + 3 Hn ˜ τ − = − X i X X,Y h σv i ˜ τ − i ↔ XY (cid:20) n ˜ τ − n i − n eq ˜ τ − n eqi (cid:18) n X n Y n eqX n eqY (cid:19)(cid:21) − X i =˜ τ − X X,Y ( h σ ′ v i ˜ τ − X → iY (cid:20) n ˜ τ − n X (cid:21) − h σ ′ v i iY → ˜ τ − X (cid:20) n i n Y (cid:21)) (2) dn ˜ τ + dt + 3 Hn ˜ τ + = − X i X X,Y h σv i ˜ τ + i ↔ XY (cid:20) n ˜ τ + n i − n eq ˜ τ + n eqi (cid:18) n X n Y n eqX n eqY (cid:19)(cid:21) − X i =˜ τ + X X,Y ( h σ ′ v i ˜ τ + X → iY (cid:20) n ˜ τ + n X (cid:21) − h σ ′ v i iY → ˜ τ + X (cid:20) n i n Y (cid:21)) (3) dn ˜ χ dt + 3 Hn ˜ χ = − X i X X,Y h σv i ˜ χi ↔ XY (cid:20) n ˜ χ n i − n eq ˜ χ n eqi (cid:18) n X n Y n eqX n eqY (cid:19)(cid:21) − X i =˜ χ X X,Y ( h σ ′ v i ˜ χX → iY (cid:20) n ˜ χ n X (cid:21) − h σ ′ v i iY → ˜ χX (cid:20) n i n Y (cid:21)) . (4)Here n and n eq represent the actual number density andthe equilibrium number density of each particle, and H is the Hubble expansion rate. Index i denotes stau andneutralino, and indices X and Y denote SM particles.Note that if relevant SM particles are in thermal equi-librium, n X = n eqX , n Y = n eqY , and ( n X n Y /n eqX n eqY ) = 1then these equations are reduced into a familiar form. h σv i and h σ ′ v i are the thermal averaged cross sections,which is defined by h σv i → ≡ g R d p d p f f ( σv ) → R d p d p f f = g R d p d p f f ( σv ) → n eq n eq , (5)where f is the distribution function of a particle, v isthe relative velocity between initial state particles, and g = 2(1) for same (different) particles 1 and 2. In thiswork, we assume that all of the supersymmetric particlesexcept for stau and neutralino are heavy, and thereforedo not involve them in the coannihilation processes.The first line on the right-hand side of Eqs. (2), (3),and (4) accounts for the annihilation and the inverseannihilation processes of the supersymmetric particles( ij ↔ XY ). Here index j denotes stau and neutralino.As long as the R-parity is conserved, as shown later,the final number density of neutralino DM is controlledonly by these processes. The second line accounts forthe exchange processes by scattering off the cosmic ther-mal background ( iX ↔ jY ). These processes exchangestau with neutralino and vice versa, and thermalize them.Consequently, the number density ratio between them iscontrolled by these processes. Instead, these processesleave the total number density of the supersymmetricparticles. Note that in general, although there are termswhich account for decay and inverse decay processes ofstau (˜ τ ↔ ˜ χXY... ) in the Boltzmann equations, we omitthem. It is because we are interested in solving the Liproblem by the long-lived stau, and the whole intentionof this work is to search parameters which can providethe solution for the Li problem. Hence we assume thatthe stau is stable enough to survire until the BBN era,and focusing on the mass difference between stau andneutralino is small enough to make it possible.
B. The evolution of the number density of stauand neutralino
In this subsection, we discuss the evolution of the num-ber density of each species. Firstly, we discuss the num-ber density evolution of neutralino DM. Since we have as-sumed R-parity conservation, all of the supersymmetricparticles eventually decay into the LSP neutralino. Thusits final number density is simply described by the sumof the number density of all the supersymmetric particles: N = X i n i . (6)For N , that is the number density of the neutralino, weget the Boltzmann equation by summing up Eqs. (2),(3), and (4), dNdt + 3 HN = −h σv i sum (cid:20) N N − N eq N eq (cid:21) (7) h σv i sum ≡ X i =˜ χ, ˜ τ X X,Y h σv i ˜ χi ↔ XY . (8)Notice that the terms describing the exchange processesin each Boltzmann equations cancel each other out. Solv-ing the Eq. (7), we obtain N and find the freeze outtemperature of the total number density of all the super-symmetric particles T f by using the standard technique[26]: m ˜ χ T f = ln 0 . g m pl m ˜ χ h σv i g / ∗ ( m ˜ χ /T f ) ≃ . (9)Here, g and m ˜ χ are the internal degrees of freedom andthe mass of neutralino, respectively. The Planck mass m pl = 1 . × GeV, and g ∗ are the total number ofthe relativistic degrees of freedom. Consequently, we seethat 4 GeV . T f .
40 GeV for 100 GeV . m ˜ χ . T f , the interactionrate of the exchange processes is much larger than that ofthe annihilation and the inverse annihilation processes. This is because the cross sections of the exchange pro-cesses are in the same order of magnitude as that of theannihilation and the inverse annihilation, but the num-ber density of the SM particles is much larger than thatof the supersymmetric particles which is suppressed bythe Boltzmann factor. Thus even if the total numberdensity of stau and neutralino is frozen out at the tem-perature T f , each number density of them continue toevolve through the exchange processes.Thus, to calculate the relic density of stau, we haveto follow the two-step procedures. As a first step, wecalculate the total relic density of the supersymmetricparticles by solving the Eq. (7). We use the publiclyavailable program micrOMEGAs [28] to calculate it. Thesecond step is the calculation of the number density ratioof stau and neutralino. The second step is significant forcalculating the relic density of stau at the BBN era, andhence we will discuss it in detail in the next subsection. C. The exchange processes and Lagrangian fordescribing them
After the freeze-out of the total number density of stauand neutralino, each of them is exchanged through thefollowing processes e τ γ ←→ e χτ e χγ ←→ e τ τ. (10)Notice that although there are other exchange processesvia weak interaction (for example, e τ W ↔ e χν τ , e τ ν τ ↔ e χW , and so on), we can omit them. This is becausethe number density of W boson is not enough to workthese processes sufficiently due to the Boltzmann factorsuppression, and the final state W boson is kinematicallyforbidden when the thermal bath temperature is less than T f . These processes (Eq. (10)) are described by theLagrangian L = ˜ τ ∗ ˜ χ ( g L P L + g R P R ) τ − ie (˜ τ ∗ ( ∂ µ ˜ τ ) − ( ∂ µ ˜ τ ∗ )˜ τ ) A µ + h.c. , (11)where e is the electromagnetic coupling constant, P L and P R are the projection operators, and l ∈ { e, µ } . g L and g R are the coupling constants, given by g L = g √ θ W sin θ W cos θ τ ,g R = √ g cos θ W sin θ W sin θ τ e iγ τ , (12)where g is the SU (2) L gauge coupling constant, and θ W is the Weinberg angle.The evolution of the stau number density is governedonly by the exchange processes (Eq.(10)) after the freeze-out of the total relic density of stau and neutralino.When we calculate it, we should pay attention to twoessential points relevant to the exchange processes. -40 -30 -20 -10 I n t e r a c t i on r a t e and e x pan s i on r a t e ( G e V ) Temperature (GeV) tau inverse annihilation ratetau inverse decay rateHubble expansion rate
FIG. 1: The inverse annihilation rate of tau leptons h Γ i n eqX ,the inverse decay rate of tau leptons h Γ i , and the Hubbleexpansion rate H as a function of the thermal bath tempera-ture. One is the competition between the interaction rate ofthe exchange processes and the Hubble expansion rate,since when these interaction rates get smaller than theHubble expansion rate, the relic density of stau would befrozen out. The other is whether tau leptons are in thethermal bath or not. The interaction rate of the exchangeprocesses strongly depends on the number density of tauleptons. When tau leptons are in the thermal bath, thenumber density ratio between stau and neutralino aregiven by the thermal ratio, n ˜ τ n χ ≃ e − m ˜ τ/T e − m ˜ χ/T = exp (cid:16) − δmT (cid:17) , (13)through the exchange processes. On the contrary, oncetau leptons decouple from the thermal bath, the ratiocannot reach this value. To calculate the relic densityof stau, we have to comprehend the temperature of taulepton decoupling.To see whether tau leptons are in the thermal bath ornot, we consider the Boltzmann equation for its numberdensity, n τ , dn τ dt + 3 Hn τ = − h σv i " n τ n X − n eqτ n eqX (cid:18) n Y n Z n eqY n eqZ (cid:19) − h Γ i " n τ − n eqτ (cid:18) n X n Y ...n eqX n eqY ... (cid:19) , (14) h σv i = X X,Y,Z h σv i τX ↔ Y Z , h Γ i = X X,Y,... h Γ i τ ↔ XY... , (15)where indices X , Y , and Z denote the SM particles,and h Γ i represents the thermal averaged decay rate of tau lepton. When the SM particles X , Y , and Z are in the thermal equilibrium, ( n Y n Z ) / ( n eqY n eqZ ) =( n X n Y ... ) / ( n eqX n eqY ... ) = 1, and hence tau leptons aresufficiently produced through the inverse annihilationand/or the inverse decay processes as long as these inter-action rates are larger than the Hubble expansion rate.Therefore, whether tau leptons are in the thermal bathor not can be distinguished by comparing the Hubble ex-pansion rate H with the inverse annihilation rate of taulepton h σv i n eqX , and the inverse decay rate of tau lepton h Γ i . In other words, the inequality expression h σv i n eqX > H and/or h Γ i > H (16)indicates that tau leptons are in the thermal bath. Fig.1 shows h Γ i n eqX , h Γ i , and H as a function of the thermalbath temperature. As shown in Fig. 1, the inverse decayrate of tau lepton is much larger than the Hubble expan-sion rate. Thus, we can conclude that tau leptons remainin the thermal bath still at the beginning of the BBN. D. Calculation of the number density ratio of stauand neutralino
We are now in a position to calculate the number den-sity ratio of stau and neutralino. In this subsection, wewill show a set of relevant Boltzmann equations.The right-hand side of the Boltzmann equations (Eqs.(2), (3), and (4)) depends only on temperature, andhence it is convenient to use temperature T instead oftime t as independent variable. To do this, we reformu-late the Boltzmann equations by using the ratio of thenumber density to the entropy density s : Y i = n i s . (17)Consequently, we obtain the Boltzmann equations for thenumber density evolution of stau and neutralino dY ˜ τ − dT = h HT g ∗ ( T ) i − (cid:20) g ∗ ( T ) + T dg ∗ ( T ) dT (cid:21) s × ( h σv i ˜ τ − γ → ˜ χτ − Y ˜ τ − Y γ − h σv i ˜ χτ − → ˜ τ − γ Y ˜ χ Y τ − + h σv i ˜ τ − τ + → ˜ χγ Y ˜ τ − Y τ + − h σv i ˜ χγ → ˜ τ − τ + Y ˜ χ Y γ ) (18) dY ˜ τ + dT = h HT g ∗ ( T ) i − (cid:20) g ∗ ( T ) + T dg ∗ ( T ) dT (cid:21) s × ( h σv i ˜ τ + γ → ˜ χτ + Y ˜ τ + Y γ − h σv i ˜ χτ + → ˜ τ + γ Y ˜ χ Y τ + + h σv i ˜ τ + τ − → ˜ χγ Y ˜ τ + Y τ − − h σv i ˜ χγ → ˜ τ + τ − Y ˜ χ Y γ ) (19) dY ˜ χ dT = h HT g ∗ ( T ) i − (cid:20) g ∗ ( T ) + T dg ∗ ( T ) dT (cid:21) s × ( h σv i ˜ χτ − → ˜ τ − γ Y ˜ χ Y τ − − h σv i ˜ τ − γ → ˜ χτ − Y ˜ τ − Y γ + h σv i ˜ χγ → ˜ τ − τ + Y ˜ χ Y γ − h σv i ˜ τ − τ + → ˜ χγ Y ˜ τ − Y τ + + h σv i ˜ χτ + → ˜ τ + γ Y ˜ χ Y τ + − h σv i ˜ τ + γ → ˜ χτ + Y ˜ τ + Y γ + h σv i ˜ χγ → ˜ τ + τ − Y ˜ χ Y γ − h σv i ˜ τ + τ − → ˜ χγ Y ˜ τ + Y τ − ) . (20)Here g ∗ ( T ) is the relativistic degrees of freedom, and weuse s = 2 π g ∗ ( T ) T , H = 1 . g / ∗ T m pl . (21)We obtain the relic density of stau at the BBN era byintegrating these equations from T f to the temperaturefor beginning the BBN under the initial condition of thetotal number density of stau and neutralino. These equa-tions make it clear that if the tau number density is outof the equilibrium, the ratio between those of stau andneutralino does not satisfy the Eq.(13). III. NUMERICAL RESULTS
In this section, we will first show the evolution of thestau number density, and then study the relation betweenthe relic density of stau and the modification of nucle-osynthesis. Finally, we study a solution of the Li prob-lem with long-lived stau in the coannihilation scenariobased on Ref. [14–16].
A. Total abundance
As a first step for the calculation of the relic numberdensity of stau, based on the discussion in section II B, wecalculate the total abundance of stau and neutralino withmicrOMEGAs [28]. Fig. 2 shows the total abundance,which corresponds to the relic abundance of DM, as afunction of sin θ τ , where θ τ is the mixing angle betweenleft and right-handed stau. Each curved line shows thetotal abundance for each stau mass, and horizontal bandrepresents the allowed region from the WMAP observa-tion at the 3 σ level (0 . ≤ Ω DM h ≤ . σ level (0 . ≤ Ω DM h ≤ . m ˜ τ = 400 , ,
300 GeV from left to right. Since theleft-handed sneutrino DM has been ruled out by con-straints from the direct detection experiments [29], wefocus on the right-side region. Here we took γ τ = 0 and δm = 100 MeV.The total abundance increases first as the heavier staumixes to the lighter stau, then turns to decrease at Mixing angle of left- and right-handed stau sin (cid:18)(cid:28) T o t a l r e li a bund a n e (cid:10) D M h WMAP400GeV350GeV300GeV 2(cid:27)2(cid:27)
FIG. 2: Total abundance of staus and neutralinos, which cor-responds to the relic abundance of DM. Each line shows thetotal abundance, and attached value represents the stau mass.Yellow band represents the allowed region from the WMAPobservation at the 3 σ level , and the region inside the horizon-tal dotted lines corresponds to allowed region at the 2 σ level[1]. In the left side of vertical lines, the LSP is left-handedsneutrino. Three lines correspond to m ˜ τ = 400 , ,
300 GeVfrom left to right. sin θ τ ≃ .
8. The increase of the abundance can be un-derstood by the fact that the annihilation cross section of˜ τ + ˜ τ → τ + τ becomes smaller as the heavier stau mixes.The increase is gradually compensated by two annihila-tion processes, ˜ τ +˜ τ ∗ → W + + W − and ˜ χ +˜ τ → W − + ν τ ,as the left-right mixing becomes large. The latter processcan not be ignored because left-handed sneutrino is de-generate to stau and neutralino in the present parameterset. These processes become significant for sin θ τ < . τ + ˜ τ ∗ → t +¯ t through s-channelexchange of the heavy Higgses. This annihilation processbecomes significant as the mixing reaches to π/ m ˜ τ . This isunderstood as follows. In the non-relativistic limit, sincethe relic number density of relic species is proportional to( m relic h σv i sum ) − and h σv i sum (Eq. (8)) is proportionalto 1 /m relic [30], the total number density N is propor-tional to m ˜ τ , N ∝ m ˜ τ h σv i sum ∝ /m ˜ τ = m ˜ τ , (22)and the total abundance is given by Ω DM h ∼ m ˜ τ N .Thus the total abundance is proportinal to m τ , and it isconsistent with the result in Fig. 2.
10 MeVWMAP10 MeV(thermal)50 MeV50 MeV(thermal) 100 MeV100 MeV(thermal)DM abundance
Temperature (GeV) R e li d e n s i t y o f t h e s t a u Y ~ (cid:28) FIG. 3: The evolution of the number density of negativecharged stau. Each line attached [ δm ] shows the actual evo-lution of the number density of stau, while the one atattched[ δm (thermal)] shows its evolution under the equilibrium de-termined given by Eq. (13) and the total relic abundance.Yellow band represents the allowed region from the WMAPobservation at the 2 σ level [1]. B. Stau relic density at the BBN era
Next, we solve the Boltzmann equations (18), (19),and (20) numerically, and obtain the ratio of the staunumber density to the total number density of stau andneutralino. Fig. 3 shows the evolution of the numberdensity of stau as a function of the universe tempera-ture. Here we took m ˜ τ = 350 GeV, sin θ τ = 0.8, and γ τ = 0 and chose δm = 10 MeV, 50 MeV, and 100 MeV assample points. Each line attached [ δm ] shows the actualevolution of the number density of stau, while the oneatattched [ δm (thermal)] shows its evolution under theequilibrium determined by Eq. (13) and the total relicabundance. Horizontal dotted line represents the relicdensity of DM, which is the total abundance calculatedabove. We took it as a initial condition of total value forthe calulation of the number density ratio. Yellow bandrepresents the allowed region from the WMAP observa-tion at the 2 σ level [1].The number density evolution of stau is qualitativelyunderstood as follows. As shown in Fig. 3, the freeze-out temperature of stau almost does not depend on δm .It is determined by the exchange processes Eq. (10),whose magnitude h σv i Y ˜ τ Y γ is governed by the factor e − ( m τ − δm ) /T , where m τ represents the tau lepton mass.The freeze-out temperature of the stau density T f (ratio) isgiven by ( m τ − δm ) /T f (ratio) ≃
25 as in Eq. (9), since thecross section of the exchange process is of the same mag-nitude as weak processes. Thus T f (ratio) hardly dependson δm . In contrast, the ratio of the number density be-tween stau and neutralino depends on δm according to Eq. (13), n ˜ τ /n ˜ χ ∼ exp( − δm/T ), since they follow theBoltzmann distribution before their freeze-out. Thus, therelic density of stau strongly depends on δm .Here, we comment on the dependence of the stau relicdensity n ˜ τ − on other parameters such as m ˜ τ , θ τ , and γ τ . The number density of the negatively charged stauis expressed in terms of the total relic density N by n ˜ τ − = N e δm/T f (ratio) ) . (23)Here, the freeze-out temperature T f (ratio) hardly dependson these pararameters. This is because the cross sec-tion of the exchange processes are changed by these pa-rameters at most by factors but not by orders of magni-tudes, and the T f (ratio) depends logarithmically on h σv i as shown in Eq. (9). On the other hand, the total relicdensity N is proportional to m ˜ τ as in Eq. (22). Thevalue of N is also affected by the left-right mixing θ τ as seen in Fig. 2 since the annihilation cross section de-pends on this parameter. In contrast, γ τ scarcely affectsthe relic density, since this parameter appears in the an-nihilation section through the cross terms of the contri-butions from the left-handed stau and the right-handedone, and such terms always accompany the suppressionfactor of m τ /m ˜ τ compared to the leading contribution.Thus the relic number density of stau n ˜ τ strongly de-pends on m ˜ τ and θ τ while scarecely depends on γ τ .We comment on the generality of our method to calcu-late the density of exotic heavy particles that coannihi-late with other (quasi)stable particles: we calculate thetotal number density of these particles and then calcu-late the ratio among them by evaluating the exchangeprocesses such as Eq. (10). This method of calculationcan be found versatile in various scenarios including thecatalyzed BBN and the exotic cosmological structure for-mation [31–33]. C. Long-lived stau and BBN
After the number density of stau freezes out, stau de-cays according to its lifetime [3], or forms a bound statewith a nuclei in the BBN era. Their formation rate hasbeen studied in literatures [14, 16, 17]. The bound statesmodify the predictions of SBBN, and make it possible tosolve the Li problem via internal conversion processesin the bound state [14–16].In Fig. 4 we show parameter regions that are consis-tent with the observed abundances of the DM and of thelight elements. We calculate the relic density of stau byvarying the value of δm with the values of m ˜ τ = 350GeV, sin θ τ = 0.8, and γ τ = 0. With these parameters,the allowed region is shown inside the dotted oval. Wesee that there are allowed regions at Y ˜ τ − ∼ − – 10 − for δm .
130 MeV to solve the Li problem at 3 σ . Onthe other hand, it is found that the observational Li to Li ratio excludes Y ˜ τ − & − and δm .
100 MeV. Wewill explain this feature as follows.
Allowed -11 -12 -13 -14 -16 -15
10 100 R e li c D e n s i t y o f t h e S t a u Y τ ~ Stau-Neutralino Mass Difference δ m / [MeV] σ CL m stau = 350 GeVsin θ τ = 0.8 γ τ = 0 η = 6.225 10 –10 L o w e r B o u n d f r o m M e l e n d e z -Ra m i l e z U pp e r B o u n d f r o m M e l e n d e z - R a m i l e z U p p e r B o u n d f r o m B o n if a cio e t a l . Overproduction of Dark MatterOverproduction of Li L o w e r B o u nd f r o m B o n i f a c i o e t a l . C a l c u l a t e d R e l i c S t a u D e n s i t y FIG. 4: Parameter regions that are consistent with the ob-served abundances of the DM and the light elements. Con-straints from the observed Li abundances are due to Bonifa-cio et al. [36] and Melendez and Ramilez [37]. The calculatedrelic number density of stau with the indicated parameter isalso shown. The top left region is excluded from Li overpro-duction for Y ˜ τ − & − and δm .
100 MeV.
We have adopted following observatinal abundances of Li and Li. Throughout this subsection, n i denotes thenumber density of a particle “ i ”, and observational errorsare given at 1 σ . For the n Li to n Li ratio, we use theupper bound [34],( n Li /n Li ) p < . ± .
022 + 0 . , (24)with a conservative systematic error (+0.106) [35]. Forthe Li abundance, we adopt two observational values ofthe n Li to n H ratio. Recently it has been reported to belog ( n Li /n H ) p = − . ± . , (25)by Ref. [36], and on the other hand, a milder one wasalso given by Ref. [37],log ( n Li /n H ) p = − . ± . . (26)In the current scenario Li can be overproduced by thescattering of the bound state ( He˜ τ − ) off the backgrounddeuterium through ( He˜ τ − ) + D → Li + ˜ τ − [4]. Theabundance of the nonthermally-produced Li throughthis process is approximately represented by∆ Y Li ∼ h σv i Li n D H Y ˜ τ − , (27) with h σv i Li the thermal average of the cross sectiontimes the relative velocity for this process [38], and n D the number density of deuterium. By using (24), wesee that the additional Li production is constrained tobe ∆ Y Li < O (10 − ). Numerical value of h σv i Li gives h σv i Li n D /H ∼ O (10 − ) at T ∼
10 keV. Then from (27)it is found that the upper bound on the abundance ofstau should be Y ˜ τ − . − . Because the bound state( He˜ τ − ) forms at T .
10 keV, this process is stronglyconstrained for τ ˜ τ − & with τ ˜ τ − being the stau life-time, which corresponds to δm .
100 MeV. Note thatthe ratio h σv i Li n D /H rapidly decreases as the cosmictemperature decreases, and this nonthermal productionof Li is much more effective just after formation of thebound state. This is the reason why we can estimate (27)at around 10 keV.On the other hand, the rates of Be and Li destruc-tion through the internal conversion [14–16] could benearly equal to the formation rates of the bound sate( Be˜ τ − ) and ( Li˜ τ − ), respectively. This is because thetimescale of the destruction through the internal conver-sion is much faster than that of any other nuclear reactionrates and the Hubble expansion rate. Then the destroyedamount of Be (or Li after its electron capture) is ap-proximately represented by∆ Y Be ∼ h σv i bnd , n Be H Y ˜ τ − , (28)where h σv i bnd , ∼ − GeV − ( T / − / ( Z/ × ( A/ − / ( E b Be / Be˜ τ − ) [16, 17]. We request ∆ Y Be to become ∼ O (10 − ) to reduce the abundance of Beto fit the observational data. Then the abundance of˜ τ − should be the order of ∆ Y Be ( h σv i bnd , n Be /H ) − ∼O (10 − ) with h σv i bnd , n Be /H ∼ − at T = 30 keV.Because h σv i bnd , n Be /H decreases as the cosmic tem-perature decreases ( ∝ T / ), the destruction is more ef-fective just after the formation of The bound state. Thisvalidates that we have estimated (28) at 30 keV. There-fore the parameter region at around Y ˜ τ − ∼ − and δm .
130 MeV is allowed by the observational Li. Here δm .
130 MeV corresponds to τ ˜ τ & s. The case forthe destruction of ( Li˜ τ − ) through the internal conver-sion is also similar to that of ( Be˜ τ − ) [14, 15].Further constraints come from the relic density of theDM, which can be stated in terms of the stau relic density.It is calculated as shown in Fig. 4 for the present valuesof parameters. Applying all the constraints, we are led tothe allowed interval shown by the thick line in the figure. D. Constraint on parameter space of stau
Finally, we show in Fig. 5 the parameter space in whichthe calculated abundances of the DM and that of thelight elements are consistent with their values from theobservations. Here, based on the discussion in previous
Left-Right mixing of stau, sin θ τ S t a u m a ss / [ G e V ] δ m = 100 MeV σ σ FIG. 5: Parameter space in which the calculated abundancesof the DM and that of the light elements are consistent withtheir values from the observations. Parameter region sur-rounded by black solid (blue dashed) line shows the allowedregion from the WMAP observation at the 3 σ (2 σ ) level [1].Red crisscross points show the parameters which is consistentwith observational abundances for the light elements includ-ing Li at 3 σ level. subsection, we took δm = 100 MeV. Parameter regionsurrounded by black solid (blue dashed) line is allowedby the relic abundance of the DM from the WMAP ob-servation at the 3 σ (2 σ ) level [1], which corresponds to0 . ≤ Ω DM h ≤ . . ≤ Ω DM h ≤ . Li, where the observational Li abun-dance is yielded by literature [37].The abundance of the light elements constrains the pa-rameter space due to the following reasons. First, the re-gion where the stau mass is less than 300 GeV is excludedsince the relic density becomes too small to destruct Lisufficiently. Next, the top-left region of the figure is ex-cluded since the lifetime of stau becomes too long andhence overproduces Li through the catalyzed fusion [4].Indeed, the lifetime of stau gets longer as its mass getsheavier due to the small phase space of the final state[3]. On the other hand, its lifetime gets shorter as theleft-right mixing angle increases in the present parameterspace [3]. As a result, the final allowed region becomesas shown by the red crisscrosses in Fig. 5.The black solid curve and the blue dashed curve arethe constraints from the relic abundance of the DMas discussed in subsection III A. Note that the relicabundance is insensitive to the mass difference δm for δm << m ˜ χ . Combination of the constraints on the abun-dance of the light elements and of the DM strongly re-stricts the allowed region and leads to δm ≃
100 MeV, m ˜ τ = (300 – 400) GeV, and sin θ τ = (0 .
65 – 1). InFig. 4, these parameter values correspond to the whitetriangular region below the allowed region (thick line).Our model can thus provide a handle to the mixing an-gle, which has few experimental signals, once the value of m ˜ τ is determined. IV. SUMMARY AND DISCUSSION
We have studied the evolution of stau number densityin the MSSM coannihilation scenario, in which the LSPand the NLSP are the lightest neutralino and the lighterstau, respectively, and have a small mass difference δm . O (1GeV). In this case, stau can survive until the BBNera, and provide additional nucleosynthesis processes. Itis therefore necessary to see how large the relic densityof stau is at the BBN era.We have shown the Boltzmann equations for the cal-culation of the relic number density of stau, and havefound that the number density of stau continues to evolvethrough the exchange processes Eq.(10), even after therelic abundance of the DM is frozen out. Thus, we needto calculate the stau relic density by a two-step proce-dure. In the first step, we calculate the total abundanceof stau and neutralino, which corresponds to the relicabundance of the DM. The total abundance is controlledonly by pair annihilation processes of the supersymmet-ric particles. In the second step, we calculate the ratioof the stau number density to the total number densityof stau and neutralino, which is governed only by the ex-change processes. We have calculated the relic densityof stau at the BBN era by solving the Boltzmann equa-tions numerically. The freeze-out temperature T f (ratio) is determined by ( m τ − δm ) /T f (ratio) ≃
25 and the relicdensity of stau are given by Eq. (23). Thus it becomeslarger as the mass difference between stau and neutralinogets smaller. Our method of calculation is generally ap-plicable to obtain the density of exotic heavy particlesthat coannihilate with other (quasi)stable particles: wecalculate the total number density of these particles andthen calculate the ratio among them by evaluating theexchange processes such as Eq. (10). This method ofcalculation can be found versatile in various scenarios in-cluding the catalyzed BBN and the exotic cosmologicalstructure formation.At the BBN era, the long-lived stau form bound stateswith nuclei, and provide exotic nucleosynthesis processes.One of them is the internal conversion process, whichoffers a possible solution to the Li problem. Apply-ing the calculated relic density of stau, we have calcu-lated the primordial abundance of light elements includ-ing these exotic processes. We have found the param-eter space consistent with both of the calculational re-sults and the observations for the relic abundance of theDM and the light elements abundance including Li. Wehave shown a prediction for the values of the parametersrelevant to stau and neutralino, which is shown in Fig.5. Consistency between the theoretical prediction andthe observational result, both of the DM abundance andthe light elements abundance requires δm ≃
100 MeV, m ˜ τ = (300 – 400) GeV, and sin θ τ = (0 .
65 – 1).
Acknowledgments
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