Stau relic density at the Big-Bang nucleosynthesis era consistent with the abundance of the light element nuclei in the coannihilation scenario

0.32 MB·English

by  Toshifumi Jittoh

STUPP-09-205,TU-861, YITP-10-1,ICRR-Report-556
Stau relic density at the Big-Bang nucleosynthesis era consistent with the abundance
of the light element nuclei in the coannihilation scenario
Toshifumi Jittoh,1 Kazunori Kohri,2,3 Masafumi Koike,1 Joe Sato,1 Takashi Shimomura,4,5 and Masato Yamanaka6
1Department of Physics, Saitama University, Shimo-okubo, Sakura-ku, Saitama, 338-8570, Japan
2Physics Department, Lancaster University LA1 4YB, UK
3Department of Physics, Tohoku University, Sendai 980-8578, Japan
4Departament de F´isica Teo`rica and IFIC, Universitat de Vale`ncia-CSIC, E-46100 Burjassot, Vale`ncia, Spain
5Yukawa Institute for Theoretical Physics, Kyoto University, Kyoto 606-8502, Japan
6Institute for Cosmic Ray Research, University of Tokyo, Kashiwa 277-8582, Japan
0
1 WecalculatetherelicdensityofstauatthebeginningoftheBig-BangNucleosynthesis(BBN)era
0 inthecoannihilationscenarioofminimalsupersymmetricstandardmodel(MSSM).Inthisscenario,
2 stau can be long-lived and form bound states with nuclei. We put constraints on the parameter
space of MSSM by connecting the calculation of the relic density of stau to the observation of the
n
lightelementsabundance,whichstronglydependsontherelicdensityofstau. Consistencybetween
a
thetheoretical prediction andtheobservational result, both ofthedark matterabundanceand the
J
light elements abundance, requires the mass difference between the lighter stau and the lightest
1
neutralino to be around 100MeV, the stau mass to be 300 – 400 GeV, and themixing angle of the
1
left and right-handed staus to besinθτ =(0.65 – 1).

]
h
p I. INTRODUCTION lifetime of stau is longer than 1000 second for δm.100
- MeV. It is known[4–18] that the long-livedchargedpar-
p
e Cosmological observations have established the exis- ticlesaffecttherelicabundanceofthelightnucleiduring
h tence of the non-baryonic dark matter (DM) [1]. These or after the big-bang nucleosynthesis (BBN).
[ observationssuggestthattheDMisastableandweakly- There is a discrepancy, the so-called 7Li problem, on

interacting particle with a mass of (100) GeV. Many the primordial 7Li abundance between the prediction
2
O
v hypothetical candidates for the DM have been proposed from the standard BBN (SBBN) and the observations
7 in models of particle physics beyond the standardmodel [19,20]. Combinedwiththeup-to-datevaluesofbaryon-
1 (SM), and one of the most attractive candidates is the to-photonratio,η =(6.225 0.170) 10−10fromWilkin-
2 lightest neutralino, χ˜0, in supersymmetric extensions of son Microwave Anisotropy± Probe×(WMAP) [1], the
1 theSMwithRparityconservation. Neutralinoisalinear SBBN predicts the 7Li to proton ratio, (7Li/H) =
. SBBN
1 combination of the superpartners of U(1), SU(2) gauge 5.24+0.71 10−10, which is by about four-times larger
0 bosons and the two neutral Higgses, and is stable when than−i0t.s67o×bserved value in poor-metal halos [19, 21]. Be-
0
it is the lightest supersymmetric particle (LSP). Indeed cause there exists no general agreements about astro-
1
it accounts for the observed DM abundance when it is physical scenarios to reduce the 7Li abundance [22–25],
:
v degenerate in mass to the next lightest supersymmetric it is natural to consider nonstandard effects.
Xi particle (NLSP) and hence coannihilates with the NLSP The authors have investigated the BBN including the
[2]. We consider the setup that the LSP is a neutralino long-lived stau [14, 15]. The long-lived stau form a
r
a consistingofmainlybino,thesuperpartnerofU(1)gauge bound state with nuclei (τ˜N), and consequently con-
boson, and the NLSP is the lighter stau, the superpart- vert it into a nucleus with a smaller atomic number.
neroftaulepton. ThisisnaturallyrealizedintheMSSM Here N stands for a nucleus. In this scenario, the abun-
with the unification condition at the grand unified the- dance of 7Li is reduced through the conversion process,
oryscale. TheminimalsupersymmetricSM(MSSM)has (τ˜7Be) χ˜0+ν +7Liandthefurtherdestructionof7Li
τ
two eigenstates of stau as physical state. In absence of byeithe→racollisionwithabackgroundprotonoranother
inter-generational mixing the mass eigenstate of stau is conversion process, (τ˜7Li) χ˜0+ν +7He. Therefore,
τ
given by the linear combination of the left-handed stau the more these bound stat→es are formed, the more the
τ˜L and the right-handed stau τ˜R as 7Li abundance is reduced. The number density of the
τ˜=cosθ τ˜ +sinθ e−iγττ˜ , (1) boundstateisdeterminedbytherelicdensityofstau. In
τ L τ R
[14,15],weassumedY andδmtobe freeparameters
τ˜,FO
where θ is the left-right mixing angle and γ is the CP of the scenario, where Y is the yield value of stau
τ τ τ˜,FO
violating phase. at the time of decoupling from the thermal bath. The
Inascenarioofthe coannihilation,the NLSPstaucan full calculation of the light nucleus abundances showed
be long-lived if the mass difference, δm, between neu- a region in (δm,Y ) plane where the 7Li problem is
τ˜,FO
tralino and stau is small enough to forbid two-body de- solvedconsistently with the observationalconstraintson
caysof stau into neutralino. It was shownin [3] that the the other nuclei. The region points Y to be close to
τ˜,FO
2
dn
the yeildvalueofthe DMandδm tobe around0.1GeV, τ˜+ +3Hn =
dt τ˜+
respectively.
n n
In this work, we improve our previous analysis by cal- σv n n neq neq X Y
culating Yτ˜,FO with taking the relic abundance of DM − i X,Yh iτ˜+i↔XY(cid:20) τ˜+ i− τ˜+ i (cid:18)neXqneYq(cid:19)(cid:21)
XX
into account. The outline of this paper is as follows. In
section II, we review the formalism for the calculation σ′v n n σ′v n n
τ˜+X→iY τ˜+ X iY→τ˜+X i Y
oeqfutahteiorneslicofdsetnasuitaynodfnsetuatur,alainnodfdoerritvheetchaelcuBloalttizomn.anInn −iX6=τ˜+XX,Y(h i (cid:20) (cid:21)−h i (cid:20) (cid:21))
(3)
section III, we present the numerical results of the cal-
culation, and prove the parameter space for solving the
7Li problem. Section IV is devoted to a summary and dn
χ˜
+3Hn =
discussion. χ˜
dt
n n
σv n n neqneq X Y
− h iχ˜i↔XY χ˜ i− χ˜ i neqneq
i X,Y (cid:20) (cid:18) X Y (cid:19)(cid:21)
II. FORMALISM FOR THE CALCULATION OF XX
RELIC DENSITY OF STAU AT THE BBN ERA
σ′v n n σ′v n n .
χ˜X→iY χ˜ X iY→χ˜X i Y
−i6=χ˜X,Y(h i (cid:20) (cid:21)−h i (cid:20) (cid:21))
Inthissection,wepreparetocalculatetherelicdensity XX
(4)
of stau at the BBN era. Firstly, in subsection IIA, we
briefly review the Boltzmann equations for the number Here n and neq representthe actual number density and
densityofstauandneutralinobasedonthe thermalrelic the equilibrium number density of each particle, and H
scenario. Then,insubsectionIIB,wediscussthenumber is the Hubble expansion rate. Index i denotes stau and
density evolution of stau and neutralino quantitatively. neutralino, and indices X and Y denote SM particles.
InsubsectionIIC,weinvestigatethesignificantprocesses Note that if relevant SM particles are in thermal equi-
for the calculation of the relic density of stau. Finally, librium, n = neq, n = neq, and (n n /neqneq) = 1
X X Y Y X Y X Y
we obtain the Boltzmann equations for the relic density then these equations are reduced into a familiar form.
of stau in a convenient form. σv and σ′v are the thermal averaged cross sections,
h i h i
which is defined by
A. Boltzmann equations for the number density σv g d3p1d3p2 f1f2 (σv)12→34
evolution of stau and neutralino h i12→34 ≡ 12R d3p1d3p2 f1f2 (5)
d3pR1d3p2 f1f2 (σv)12→34
=g ,
Inthissubsection,weshowtheBoltzmannequationsof 12 neqneq
R 1 2
stau andneutralino and briefly review their quantitative
structure based on [26] (see also a recent paper[27]). where f is the distribution function of a particle, v is
the relative velocity between initial state particles, and
We are interested in the relic density of stau in the
g =2(1) for same (different) particles 1 and 2. In this
coannihilation scenario. In this scenario, stau and neu- 12
work,weassumethatallofthesupersymmetricparticles
tralino are quasi-degenerate in mass and decouple from
except for stau and neutralino are heavy, and therefore
the thermal bath almost at the same time [2]. Thus the
do not involve them in the coannihilation processes.
relic density of stau is given by solving a coupled set of
the Boltzmann equations for stau and neutralino as si- The first line on the right-hand side of Eqs. (2), (3),
multaneous differential equation. For simplicity, we use and (4) accounts for the annihilation and the inverse
the Maxwell-Boltzmann statistics for all species instead annihilation processes of the supersymmetric particles
of the Fermi-Dirac for fermions and the Bose-Einstein (ij XY). Here index j denotes stau and neutralino.
↔
for bosons,and assume T invariance. With these simpli- As long as the R-parity is conserved, as shown later,
fications, the Boltzmann equations of them are given as the final number density of neutralino DM is controlled
follows only by these processes. The second line accounts for
the exchangeprocessesby scattering offthe cosmic ther-
dndτ˜t− +3Hnτ˜− = mstaaul bwaicthkgnreouutnrdali(niXoan↔dvjYice).veTrshae,seanpdrothceesrsmesaleizxechthaenmge.
Consequently, the number density ratio between them is
n n
− hσviτ˜−i↔XY nτ˜−ni−nτe˜q−neiq neXqneYq controlled by these processes. Instead, these processes
i X,Y (cid:20) (cid:18) X Y (cid:19)(cid:21) leave the total number density of the supersymmetric
XX
particles. Note that in general,althoughthere are terms
−iX6=τ˜−XX,Y(hσ′viτ˜−X→iY(cid:20)nτ˜−nX(cid:21)−hσ′viiY→τ˜−X(cid:20)n(i2n)Y(cid:21))twshthaeiumch(.τ˜aI↔tcciosχ˜ubXnetYcfa.o.u.r)sedinewcteahyearaBenodilnttzienmrveaesrntsenededqienucaastoyilovpnirnso,gcwetsehseeosm7Loitfi
3
problem by the long-lived stau, and the whole intention This is because the cross sections of the exchange pro-
of this work is to search parameters which can provide cesses are in the same order of magnitude as that of the
the solution for the 7Li problem. Hence we assume that annihilation and the inverse annihilation, but the num-
the stau is stable enough to survire until the BBN era, ber density of the SM particles is much larger than that
and focusing on the mass difference between stau and of the supersymmetric particles which is suppressed by
neutralino is small enough to make it possible. the Boltzmann factor. Thus even if the total number
density of stau and neutralino is frozen out at the tem-
perature T , each number density of them continue to
f
B. The evolution of the number density of stau evolve through the exchange processes.
and neutralino Thus, to calculate the relic density of stau, we have
to follow the two-step procedures. As a first step, we
Inthissubsection,wediscusstheevolutionofthenum- calculate the total relic density of the supersymmetric
ber density of each species. Firstly, we discuss the num- particles by solving the Eq. (7). We use the publicly
berdensityevolutionofneutralinoDM.Sincewehaveas- availableprogrammicrOMEGAs[28]tocalculateit. The
sumed R-parity conservation, all of the supersymmetric secondstepisthecalculationofthenumberdensityratio
particleseventuallydecayintotheLSPneutralino. Thus of stauand neutralino. The secondstep is significantfor
its final number density is simply described by the sum calculating the relic density of stau at the BBN era, and
ofthenumberdensityofallthesupersymmetricparticles hence we will discuss it in detail in the next subsection.
:
N = n . C. The exchange processes and Lagrangian for
i (6)
describing them
i
X
For N, that is the number density of the neutralino, we
Afterthefreeze-outofthetotalnumberdensityofstau
get the Boltzmann equation by summing up Eqs. (2),
and neutralino, each of them is exchanged through the
(3), and (4),
following processes
dN
+3HN = σv NN NeqNeq (7) τγ χτ
sum ←→ (10)
dt −h i −
(cid:20) (cid:21) χγ ττ.
←→
e e
Notice that although there are other exchange processes
e e
hσvisum ≡ hσviχ˜i↔XY . (8) via weak interaction (for example, τW ↔ χντ, τντ ↔
i=χ˜,τ˜X,Y χW, and so on), we can omit them. This is because
X X
the number density of W boson is not enough to work
e e e
Notice that the terms describing the exchange processes
these processes sufficiently due to the Boltzmann factor
e
ineachBoltzmannequationscanceleachotherout. Solv-
suppression,andthefinalstateWbosoniskinematically
ing the Eq. (7), we obtain N and find the freeze out
forbiddenwhenthethermalbathtemperatureislessthan
temperatureofthetotalnumberdensityofallthesuper-
T . These processes (Eq. (10)) are described by the
f
symmetric particles T by using the standard technique
f Lagrangian
[26]:
= τ˜∗χ˜0(g P +g P )τ
L L R R
m 0.038 g m m σv L (11)
Tfχ˜ =ln g∗1/2(mpχ˜l/Tfχ˜)h i ≃25 . (9) −ie(τ˜∗(∂µτ˜)−(∂µτ˜∗)τ˜)Aµ+h.c.,
whereeistheelectromagneticcouplingconstant,P and
L
Here, g and mχ˜ are the internal degrees of freedom and PR are the projection operators, and l e,µ . gL and
the mass of neutralino, respectively. The Planck mass g are the coupling constants, given by∈ { }
R
m = 1.22 1019 GeV, and g are the total number of
pl ∗
× g
the relativistic degrees of freedom. Consequently, we see g = sinθ cosθ ,
L W τ
that 4 GeV . Tf . 40 GeV for 100 GeV . mχ˜ . 1000 √2cosθW (12)
GeV. √2g
Now we will discuss the number density evolution of gR = sinθW sinθτeiγτ,
cosθ
W
stau. To obtain the relic density of stau, we solve a cou-
pled set of the Boltzmann equations, (2), (3), and (4), where g is the SU(2) gauge coupling constant, and θ
L W
as simultaneous differential equation. Each Boltzmann is the Weinberg angle.
equationcontains the contributions of the exchange pro- The evolution of the stau number density is governed
cesses. These processes exchange stau with neutralino onlybytheexchangeprocesses(Eq.(10))afterthefreeze-
and vice versa. At the temperature T , the interaction out of the total relic density of stau and neutralino.
f
rateoftheexchangeprocessesismuchlargerthanthatof When we calculate it, we should pay attention to two
the annihilation and the inverse annihilation processes. essential points relevant to the exchange processes.
4
V) of tau lepton. When the SM particles X, Y, and
e
G Z are in the thermal equilibrium, (n n )/(neqneq) =
e ( 10 -10 tau inverse decay rate (n n ...)/(neqneq...) = 1, and henceY taZu lepYtonZs are
at X Y X Y
n r sufficiently produced through the inverse annihilation
o and/ortheinversedecayprocessesaslongastheseinter-
nsi 10 -20 Hubble expansion rate action rates are larger than the Hubble expansion rate.
a
p Therefore, whether tau leptons are in the thermal bath
x
e
d ornotcanbedistinguishedbycomparingtheHubbleex-
an 10 -30 tau inverse annihilation rate pansion rate H with the inverse annihilation rate of tau
e lepton σv neq, and the inverse decay rate of tau lepton
at h i X
on r hΓi. In other words, the inequality expression
acti 10 -40 0.18 0.14 0.1 0.06 0.02 σv neq > H and/or Γ > H (16)
er h i X h i
nt Temperature (GeV)
I
indicates that tau leptons are in the thermal bath. Fig.
FIG. 1: The inverse annihilation rate of tau leptons hΓineXq, 1 shows hΓineXq, hΓi, and H as a function of the thermal
the inverse decay rate of tau leptons hΓi, and the Hubble bathtemperature. As shownin Fig. 1, the inversedecay
expansionrateH asafunctionof thethermalbathtempera- rateoftauleptonismuchlargerthanthe Hubble expan-
ture. sionrate. Thus,wecanconcludethattauleptonsremain
in the thermal bath still at the beginning of the BBN.
One is the competition betweenthe interactionrateof
D. Calculation of the number density ratio of stau
the exchange processes and the Hubble expansion rate,
and neutralino
since when these interaction rates get smaller than the
Hubbleexpansionrate,therelicdensityofstauwouldbe
We arenowina positionto calculatethe numberden-
frozen out. The other is whether tau leptons are in the
sity ratio of stau and neutralino. In this subsection, we
thermalbathornot. Theinteractionrateoftheexchange
will show a set of relevant Boltzmann equations.
processesstronglydepends onthe number densityoftau
The right-hand side of the Boltzmann equations (Eqs.
leptons. When tau leptons are in the thermal bath, the
number density ratio between stau and neutralino are (2), (3), and (4)) depends only on temperature, and
hence it is convenient to use temperature T instead of
given by the thermal ratio,
time t as independent variable. To do this, we reformu-
nτ˜ e−mτ˜/T δm late the Boltzmann equations by using the ratio of the
=exp , (13)
nχ ≃ e−mχ˜/T − T number density to the entropy density s :
(cid:16) (cid:17)
n
through the exchange processes. On the contrary, once Y = i. (17)
i
tau leptons decouple from the thermal bath, the ratio s
cannot reach this value. To calculate the relic density
Consequently,weobtaintheBoltzmannequationsforthe
of stau, we have to comprehend the temperature of tau
number density evolution of stau and neutralino
lepton decoupling.
noTt,owseeecownhseidthererthtaeuBloelpttzomnasnanreeqinuatthieonthfeorrmitaslnbuamthboerr dYτ˜− = 3HTg∗(T) −1 3g∗(T)+Tdg∗(T) s
dT dT
density, nτ, h i (cid:20) (cid:21)
dnτ +3Hn = σv n n neqneq nYnZ ×(hσviτ˜−γ→χ˜τ−Yτ˜−Yγ −hσviχ˜τ−→τ˜−γYχ˜Yτ− (18)
dt τ −h i" τ X − τ X (cid:18)neYqneZq(cid:19)#
Γ n neq nXnY... , +hσviτ˜−τ+→χ˜γYτ˜−Yτ+ −hσviχ˜γ→τ˜−τ+Yχ˜Yγ)
−h i" τ − τ (cid:18)neXqneYq...(cid:19)#
(14)
dY −1 dg (T)
τ˜+ = 3HTg (T) 3g (T)+T ∗ s
∗ ∗
dT dT
σv = σv , Γ = Γ , h i (cid:20) (cid:21)
τX↔YZ τ↔XY...
h i h i h i h i
XX,Y,Z XX,Y,... (15) ×(hσviτ˜+γ→χ˜τ+Yτ˜+Yγ −hσviχ˜τ+→τ˜+γYχ˜Yτ+ (19)
awnhdereΓindriecpesresXen,tsY,thaendthZermdeanloatevetrhageedSMdepcaaryticrlaetse, +hσviτ˜+τ−→χ˜γYτ˜+Yτ− −hσviχ˜γ→τ˜+τ−Yχ˜Yγ)
h i
5
dY −1 dg (T)
χ˜ = 3HTg (T) 3g (T)+T ∗ s 0.15
∗ ∗
dT dT 2
h i (cid:20) (cid:21) hM 0.14 400GeV
D
σv χ˜τ−→τ˜−γYχ˜Yτ− σv τ˜−γ→χ˜τ−Yτ˜−Yγ (cid:10) 0.13
×(h i −h i e

(20) n 0.12 2(cid:27)
+hσviχ˜γ→τ˜−τ+Yχ˜Yγ −hσviτ˜−τ+→χ˜γYτ˜−Yτ+ da
+hσviχ˜τ+→τ˜+γYχ˜Yτ+ −hσviτ˜+γ→χ˜τ+Yτ˜+Yγ bun 0.11 WMAP
a 0.1 350GeV
+hσviχ˜γ→τ˜+τ−Yχ˜Yγ −hσviτ˜+τ−→χ˜γYτ˜+Yτ−) . eli
0.09 2(cid:27)
r
l
a 0.08
Here g (T) is the relativistic degrees of freedom, and we t
∗ o
use T 0.07 300GeV
0.4 0.5 0.6 0.7 0.8 0.9 1
2π2 T2
s= g (T)T3 , H =1.66g1/2 . (21)
∗ ∗
45 m
pl Mixing angle of left- and right-handed stau sin(cid:18)(cid:28)
We obtain the relic density of stau at the BBN era by
FIG.2: Totalabundanceofstausandneutralinos,whichcor-
integrating these equations from Tf to the temperature responds to the relic abundance of DM. Each line shows the
for beginning the BBN under the initial condition of the totalabundance,andattachedvaluerepresentsthestaumass.
totalnumberdensityofstauandneutralino. Theseequa- Yellow band represents the allowed region from the WMAP
tions make it clear that if the tau number density is out observationatthe3σlevel,andtheregioninsidethehorizon-
of the equilibrium, the ratio between those of stau and tal dotted lines corresponds to allowed region at the 2σ level
neutralino does not satisfy the Eq.(13). [1]. In the left side of vertical lines, the LSP is left-handed
sneutrino. Three lines correspond to mτ˜ =400,350,300GeV
from left to right.
III. NUMERICAL RESULTS
In this section, we will first show the evolution of the sinθ 0.8. The increase of the abundance can be un-
τ
staunumberdensity,andthenstudytherelationbetween dersto≃odbythefactthattheannihilationcrosssectionof
the relic density of stau and the modification of nucle- τ˜+τ˜ τ+τ becomessmallerastheheavierstaumixes.
osynthesis. Finally, we study a solution of the 7Li prob- The in→crease is gradually compensated by two annihila-
lem with long-lived stau in the coannihilation scenario tionprocesses,τ˜+τ˜∗ W++W−andχ˜0+τ˜ W−+ν ,
based on Ref. [14–16]. astheleft-rightmixin→gbecomeslarge. T1hela→tterprocesτs
can not be ignored because left-handed sneutrino is de-
generatetostauandneutralinointhepresentparameter
A. Total abundance
set. These processes become significant for sinθ < 0.8.
τ
Another process which reduces the total abundance due
As a first step for the calculation of the relic number totheleft-rightmixingisτ˜+τ˜∗ t+t¯throughs-channel
densityofstau,basedonthediscussioninsectionIIB,we exchangeoftheheavyHiggses. T→hisannihilationprocess
calculatethetotalabundanceofstauandneutralinowith becomessignificantasthemixingreachestoπ/4andthe
micrOMEGAs [28]. Fig. 2 shows the total abundance, masses of two staus are split. In fig. 2, the mass differ-
which corresponds to the relic abundance of DM, as a ence between the lighter and the heavier stau is fixed to
function of sinθτ, where θτ is the mixing angle between be30GeVtomaximizetheDMabundance,butthesame
left and right-handed stau. Each curved line shows the resultcanbe obtainedby changingthe staumassforan-
totalabundanceforeachstaumass,andhorizontalband other values of the mass difference. As shown in Fig. 2,
represents the allowed region from the WMAP observa- thetotalabundancealsostronglydependsonm . Thisis
τ˜
tion at the 3σ level (0.0913 ΩDMh2 0.1285), and understoodas follows. In the non-relativistic limit, since
≤ ≤
the region inside the horizontal dotted lines corresponds therelicnumberdensityofrelicspeciesisproportionalto
0to.12a1ll3o)we[1d].regInionthaetlethftes2idσeleovfevle(r0t.i0c9a6l3lin≤esΩ, DthMehL2S≤P (tmo r1e/lmich2σvisu[3m0)]−,1thaendtohtσalvinsuummb(Eerq.de(n8s)i)tyisNproispoprrtoiopnoar-l
relic
is the left-handed sneutrino. Three lines correspond to tional to m ,
τ˜
m = 400,350,300GeV from left to right. Since the
τ˜
left-handed sneutrino DM has been ruled out by con- 1 1
N =m , (22)
straints from the direct detection experiments [29], we ∝ m σv ∝ 1/m τ˜
τ˜ sum τ˜
h i
focus on the right-side region. Here we took γ =0 and
τ
δm=100MeV. and the total abundance is given by Ω h2 m N.
DM τ˜
Thetotalabundanceincreasesfirstasthe heavierstau Thus the total abundance is proportinal to m2,∼and it is
τ˜
mixes to the lighter stau, then turns to decrease at consistent with the result in Fig. 2.
6
Eq. (13), n /n exp( δm/T), since they follow the
Y~(cid:28) DM abundance Boltzmanndτ˜istrχ˜ib∼utionb−eforetheirfreeze-out. Thus,the
u
a WMAP relic density of stau strongly depends on δm.
st 10(cid:0)12 Here, we comment onthe dependence of the stau relic
he density nτ˜− on other parameters such as mτ˜, θτ, and
t γτ. The number density of the negatively charged stau
of is expressed in terms of the total relic density N by
10 MeV
y
t 10 MeV(thermal) N
nsi 50 MeV nτ˜− = 2(1+eδm/Tf(ratio)) . (23)
e 50 MeV(thermal)
d 100 MeV
Here,thefreeze-outtemperatureT hardlydepends

(cid:0)13 f(ratio)
eli 10 100 MeV(thermal) on these pararameters. This is because the cross sec-
R 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 depends logarithmically on σv
0.1 0.07 0.04 0.01 f(ratio) h i
as shown in Eq. (9). On the other hand, the total relic
Temperature (GeV)
density N is proportional to m as in Eq. (22). The
τ˜
FIG. 3: The evolution of the number density of negative value of N is also affected by the left-right mixing θτ
charged stau. Each line attached [δm] shows the actual evo- as seen in Fig. 2 since the annihilation cross section de-
lution ofthenumberdensityofstau,while theoneatattched pends on this parameter. In contrast, γ scarcely affects
τ
[δm(thermal)] shows its evolution under the equilibrium de- the relic density, since this parameter appears in the an-
termined given by Eq. (13) and the total relic abundance. nihilation section through the cross terms of the contri-
Yellow band represents the allowed region from the WMAP butions from the left-handed stau and the right-handed
observation at the2σ level [1].
one, and such terms always accompany the suppression
factor of m /m compared to the leading contribution.
τ τ˜
Thus the relic number density of stau n strongly de-
τ˜
pends on m and θ while scarecely depends on γ .
τ˜ τ τ
B. Stau relic density at the BBN era
Wecommentonthegeneralityofourmethodtocalcu-
late the density of exotic heavy particles that coannihi-
Next, we solve the Boltzmann equations (18), (19), late with other (quasi)stable particles: we calculate the
and (20) numerically, and obtain the ratio of the stau total number density of these particles and then calcu-
number density to the total number density of stau and late the ratio among them by evaluating the exchange
neutralino. Fig. 3 shows the evolution of the number processes such as Eq. (10). This method of calculation
density of stau as a function of the universe tempera- can be found versatile in various scenarios including the
ture. Here we took mτ˜ = 350 GeV, sinθτ = 0.8, and γτ catalyzedBBNandtheexoticcosmologicalstructurefor-
= 0 and chose δm = 10 MeV, 50 MeV, and 100 MeV as mation [31–33].
sample points. Each line attached [δm] shows the actual
evolution of the number density of stau, while the one
atattched [δm(thermal)] shows its evolution under the C. Long-lived stau and BBN
equilibrium determined by Eq. (13) and the total relic
abundance. Horizontal dotted line represents the relic After the number density of stau freezes out, stau de-
density of DM, which is the total abundance calculated cays according to its lifetime [3], or forms a bound state
above. We took itas a initial conditionoftotal value for with a nuclei in the BBN era. Their formation rate has
the calulation of the number density ratio. Yellow band beenstudiedinliteratures[14,16,17]. Theboundstates
represents the allowed region from the WMAP observa- modify the predictionsofSBBN,andmakeit possibleto
tion at the 2σ level [1]. solve the 7Li problem via internal conversion processes
The number density evolution of stau is qualitatively in the bound state [14–16].
understood as follows. As shown in Fig. 3, the freeze- In Fig. 4 we show parameter regions that are consis-
out temperature of stau almost does not depend on δm. tentwiththe observedabundancesofthe DM andofthe
It is determined by the exchange processes Eq. (10), light elements. We calculate the relic density of stau by
whose magnitude hσviYτ˜Yγ is governed by the factor varying the value of δm with the values of mτ˜ = 350
e−(mτ−δm)/T, where mτ represents the tau lepton mass. GeV, sinθτ = 0.8, and γτ = 0. With these parameters,
Thefreeze-outtemperatureofthestaudensityT is the allowed region is shown inside the dotted oval. We
f(ratio)
gcrivoesnssbeyct(imonτo−fδtmhe)/eTxcfh(raantigo)e≃pr2o5ceasssiins oEfqt.h(e9)s,asminecmetahge- sfoeretδhmat.th1e3re0aMreeVallotowesdolrveegitohnes7aLtiYpτ˜r−ob∼le1m0−a1t33–σ1.0−O1n2
nitude as weak processes. Thus T hardly depends the other hand, it is found that the observational 6Li to
f(ratio)
on δm. In contrast, the ratio of the number density be- 7Li ratio excludes Yτ˜− &10−15 and δm. 100 MeV. We
tween stau and neutralino depends on δm according to will explain this feature as follows.
7
10-11 with σv 6Li the thermal average of the cross section
h i
3σ CL times the relative velocity for this process [38], and nD
Upper Bound from Bonifacio et al. the number density of deuterium. By using (24), we
Upper Bound from Melendez-Ramilez Overproduction of Dark Matter see that the additional 6Li production is constrained to
Yu~τ10-12 Lower Bound from Bonifacio et al. bσev∆6YLi6nLDi /<HO(10−(2110)−.6N) uatmTerica1l0vkaelVue. Tofhheσnvfir6oLmi g(i2v7es)
a h i ∼ O ∼
St Allowed it is found that the upper bound on the abundance of
e Lower Bound from Melendez-Ramilez stau should be Yτ˜− . 10−15. Because the bound state
ic Density of th 1100--1134 Overproduction of 6Li mstau =Ca l3cu5lat0ed RGelic SetauV Density cttb(toihe4ofommHenu6eseLnprt,τ˜daeir−wartisaisi)hotntamiuefcthodrheuσre.cmvfchodTiosr6ehrmLcrτaiiresnτo˜tes−rDiapseTs/o&eteHnhs.ff,de1e1sar0cr0anet4tpiadovkwsiedeotδiVlhntmjyhui,swsdtτt.hnehτ˜yaoc−isr1fnwte0tbepaeh0resrecieofMrnasocmgnreeamasVetsslhsa.tpteitishNimrosoeostnadtatctrueuoeoocfsnt(tlmht2iigfoha7leinyec)t-
l
e
R 10-15 sinθτ = 0.8 at around 10 keV.
γτ = 0 On the other hand, the rates of 7Be and 7Li destruc-
η = 6.225 10–10 tion through the internal conversion [14–16] could be
nearly equal to the formation rates of the bound sate
10-16 (7Beτ˜−) and (7Liτ˜−), respectively. This is because the
10 100 timescale ofthe destruction throughthe internalconver-
Stau-Neutralino Mass Differenceδm/[MeV] sionismuchfasterthanthatofanyothernuclearreaction
ratesandtheHubbleexpansionrate. Thenthedestroyed
amount of 7Be (or 7Li after its electron capture) is ap-
FIG. 4: Parameter regions that are consistent with the ob-
proximately represented by
served abundances of the DM and the light elements. Con-
straintsfromtheobserved7LiabundancesareduetoBonifa- σv n
cioet al.[36]andMelendezandRamilez[37]. Thecalculated ∆Y7Be ∼ h ibnHd,7 7BeYτ˜−, (28)
relic number density of stau with the indicated parameter is
alsoshown. Thetopleftregion isexcludedfrom 6Lioverpro- where σv 10−2GeV−2(T/30keV)−1/2(Z/4)2
duction for Yτ˜− &10−15 and δm.100 MeV. (A/7)−h3/2(iEbnbd7B,7e/∼1350keV) is the thermally-average×d
cross section times the relative velocity of the bound-
state formation for (7Beτ˜−) [16, 17]. We request ∆Y
7Be
to become (10−20) to reduce the abundance of 7Be
Wehaveadoptedfollowingobservatinalabundancesof ∼ O
to fit the observational data. Then the abundance of
6Li and 7Li. Throughout this subsection, n denotes the
i τ˜− should be the order of ∆Y ( σv n /H)−1
numberdensityofaparticle“i”,andobservationalerrors (10−12) with σv n /7HBe h10i−b8nda,7t T7B=e 30 keV∼.
are given at 1σ. For the n6Li to n7Li ratio, we use the OBecause σv h nibnd/,7H7Bdeecrea∼ses as the cosmic tem-
upper bound [34], h ibnd,7 7Be
perature decreases ( T1/2), the destruction is more ef-
∝
(n /n ) <0.046 0.022+0.106, (24) fective justafter the formationofThe boundstate. This
6Li 7Li p ± validates that we have estimated (28) at 30 keV. There-
with a conservative systematic error (+0.106) [35]. For fore the parameter region at around Yτ˜− ∼ 10−12 and
the 7Li abundance,we adopt two observationalvalues of δm.130MeVis allowedby the observational7Li. Here
the n7Li to nH ratio. Recentlyit hasbeenreportedto be δm . 130 MeV corresponds to ττ˜ & 103 s. The case for
the destruction of (7Liτ˜−) through the internal conver-
log (n /n ) = 9.90 0.09, (25) sion is also similar to that of (7Beτ˜−) [14, 15].
10 7Li H p − ±
Further constraints come from the relic density of the
by Ref. [36], and on the other hand, a milder one was DM,whichcanbestatedintermsofthestaurelicdensity.
also given by Ref. [37], It is calculated as shown in Fig. 4 for the present values
ofparameters. Applyingalltheconstraints,weareledto
log (n /n ) = 9.63 0.06. (26)
10 7Li H p − ± theallowedintervalshownbythethicklineinthefigure.
Inthecurrentscenario6Licanbeoverproducedbythe
scatteringoftheboundstate(4Heτ˜−)offthebackground
D. Constraint on parameter space of stau
deuterium through (4Heτ˜−)+D 6 Li+τ˜− [4]. The
→
abundance of the nonthermally-produced 6Li through
Finally,weshowinFig.5theparameterspaceinwhich
this process is approximately represented by
the calculated abundances of the DM and that of the
σv n light elements are consistent with their values from the
∆Y6Li ∼ h iH6Li DYτ˜−, (27) observations. Here, based on the discussion in previous
8
550 of m is determined.
τ˜
δm = 100 MeV
500
]
V
e
G
[450 IV. SUMMARY AND DISCUSSION
s / 3σ
s
a400
m We have studied the evolution of stau number density
tau 350 2σ in the MSSM coannihilation scenario, in which the LSP
S andthe NLSP are the lightest neutralino and the lighter
stau,respectively,andhaveasmallmassdifferenceδm.
300
(1GeV). In this case, stau can survive until the BBN
0.4 0.5 0.6 0.7 0.8 0.9 1.0 O
era, and provide additional nucleosynthesis processes. It
Left-Right mixing of stau, sinθτ is therefore necessary to see how large the relic density
of stau is at the BBN era.
FIG.5: Parameterspaceinwhichthecalculatedabundances
We have shown the Boltzmann equations for the cal-
of the DM and that of the light elements are consistent with
culation of the relic number density of stau, and have
their values from the observations. Parameter region sur-
foundthatthenumberdensityofstaucontinuestoevolve
rounded by black solid (blue dashed) line shows the allowed
through the exchange processes Eq.(10), even after the
region from the WMAP observation at the 3σ (2σ) level [1].
relic abundance of the DM is frozen out. Thus, we need
Redcrisscrosspointsshowtheparameterswhichisconsistent
with observational abundances for the light elements includ- to calculate the stau relic density by a two-step proce-
ing 7Li at 3σ level. dure. In the first step, we calculate the total abundance
of stau and neutralino, which corresponds to the relic
abundanceofthe DM.Thetotalabundanceiscontrolled
only by pair annihilation processes of the supersymmet-
subsection, we took δm = 100MeV. Parameter region ric particles. In the second step, we calculate the ratio
surrounded by black solid (blue dashed) line is allowed of the stau number density to the total number density
by the relic abundance of the DM from the WMAP ob- ofstauandneutralino,whichisgovernedonlybythe ex-
servation at the 3σ (2σ) level [1], which corresponds to change processes. We have calculated the relic density
0.0913 Ω h2 0.1285 (0.0963 Ω h2 0.1213). of stau at the BBN era by solving the Boltzmann equa-
DM DM
≤ ≤ ≤ ≤
Redcrisscrosspointsshowtheparameterswhicharecon- tions numerically. The freeze-out temperature T
f(ratio)
sistentwiththeobservationalabundancesforthelightel- is determined by (m δm)/T 25 and the relic
τ f(ratio)
− ≃
ements including 7Li, where the observational 7Li abun- density of stau are given by Eq. (23). Thus it becomes
dance is yielded by literature [37]. largerasthemassdifferencebetweenstauandneutralino
Theabundanceofthelightelementsconstrainsthepa- gets smaller. Our method of calculation is generally ap-
rameterspaceduetothefollowingreasons. First,there- plicable to obtain the density of exotic heavy particles
gionwherethestaumassislessthan300GeVisexcluded that coannihilate with other (quasi)stable particles: we
since the relic density becomes too small to destruct 7Li calculate the total number density of these particles and
sufficiently. Next, the top-left region of the figure is ex- then calculate the ratio among them by evaluating the
cluded since the lifetime of stau becomes too long and exchange processes such as Eq. (10). This method of
hence overproduces 6Li through the catalyzed fusion [4]. calculationcanbefoundversatileinvariousscenariosin-
Indeed, the lifetime of stau gets longer as its mass gets cluding the catalyzed BBN and the exotic cosmological
heavier due to the small phase space of the final state structure formation.
[3]. On the other hand, its lifetime gets shorter as the AttheBBNera,thelong-livedstauformboundstates
left-rightmixingangleincreasesinthepresentparameter withnuclei,andprovideexoticnucleosynthesisprocesses.
space [3]. As a result, the final allowed region becomes One of them is the internal conversion process, which
as shown by the red crisscrosses in Fig. 5. offers a possible solution to the 7Li problem. Apply-
The black solid curve and the blue dashed curve are ing the calculated relic density of stau, we have calcu-
the constraints from the relic abundance of the DM lated the primordial abundance of light elements includ-
as discussed in subsection IIIA. Note that the relic ing these exotic processes. We have found the param-
abundance is insensitive to the mass difference δm for eter space consistent with both of the calculational re-
δm<<m . Combinationoftheconstraintsontheabun- sults and the observations for the relic abundance of the
χ˜
dance of the light elements and of the DM strongly re- DMandthe lightelementsabundanceincluding 7Li. We
stricts the allowed region and leads to δm 100MeV, haveshowna predictionfor the values ofthe parameters
≃
m = (300 – 400)GeV, and sinθ = (0.65 – 1). In relevant to stau and neutralino, which is shown in Fig.
τ˜ τ
Fig. 4, these parameter values correspond to the white 5. Consistency between the theoretical prediction and
triangular region below the allowed region (thick line). the observationalresult, both of the DM abundance and
Our model can thus provide a handle to the mixing an- the light elements abundance requires δm 100MeV,
≃
gle, which has few experimental signals, once the value m =(300 – 400)GeV, and sinθ =(0.65 – 1).
τ˜ τ
9
Acknowledgments was supported in part by the Grant-in-Aid for the Min-
istry of Education, Culture, Sports, Science, and Tech-
The work of K. K. was supported in part by PPARC nology, Government of Japan (No. 20025001, 20039001,
Grant No. PP/D000394/1, EU Grant No. MRTN-CT- and 20540251). The work of T. S. was supported in
2006-035863, the European Union through the Marie part by MEC and FEDER (EC) Grants No. FPA2005-
Curie Research and Training Network “UniverseNet,” 01678. The work of M. Y. was supported in part by
MRTN-CT-2006-035863,and Grant-in-Aid for Scientific theGrant-in-Aidforthe MinistryofEducation,Culture,
researchfromtheMinistryofEducation,Science,Sports, Sports, Science, and Technology, Government of Japan
and Culture, Japan, No. 18071001. The work of J. S. (No. 20007555).
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