Big Bang Nucleosynthesis with Bound States of Long-lived Charged Particles

0.13 MB·English

by  Manoj Kaplinghat

Big Bang Nucleosynthesis with Bound States of Long-lived Charged Particles
Manoj Kaplinghat and Arvind Rajaraman
Department of Physics and Astronomy
University of California, Irvine, California 92697, USA
(Dated: February 5, 2008)
Charged particles (X) decaying after primordial nucleosynthesis are constrained by the require-
ment that their decay products should not change the light element abundances drastically. If the
−
decayingparticleis negatively charged (X ) thenit will bindtothenuclei. Weconsider theeffects
ofthedecayofXwhenboundtoHelium-4andshowthatthiswillmodifytheLithiumabundances.
INTRODUCTION binding processstarts belowabout0.5 keVandcouldbe
importantforlifetimeslargerthanaboutayear. Another
7
interestingprocesswehavenotconsideredhereisthatof
0 Many models of physics beyond the standard model
0 have long-lived charged particles in their spectrum. For forming stable Be8-X bound states [28]. A stable Be8-X
2 bound state has important implications for the produc-
example, in the class of supergravity models where the
n gravitino is the lightest supersymmetric particle (LSP), tion of Carbon in the early universe. We hope to return
a to these issues in future work.
thenextlightest(NLSP)canbealong-livedchargedslep-
J Whilethisworkwasbeingcompleted,tworelatedarti-
ton[1]. SimilarphenomenologyarisesinUniversalExtra
8 clesappearedonthearXiv. Thethreearticlesallconsider
Dimension models [2]. Another example is the class of
1
different processes and hence complement each other.
SUSY models with axino dark matter where again the
The first by Pospelov [32] pointed out that the nuclear
3 NLSP could be long-lived [3, 4, 5]. A third example
v is a supersymmetric model with almost degenerate LSP processes with a photon in the final state will be mod-
9 ified by the presence of the strong electric field of the
andNLSP suchthat the dominantdecaychannelfor the
0 X particle. Pospelov showed that the enhanced D on
NLSP is kinematically suppressed [6, 7].
2 X-4He cross-section could produce orders of magnitude
6 Here we consider a new aspect of the early universe
larger 6Li abundance. However, it is important to note
0 cosmologyoftheseparticlesthataffectsthelightelement
6 abundances. Othereffectsofthesechargedparticleswere that if the decay lifetime is about 106 seconds or larger,
0 then 6Li will also be destroyed due to the non-thermal
previously considered in [8, 9, 10, 11, 12, 13, 14, 15, 16,
/ photons from the decay.
h 17, 18, 19, 20, 21]
p ThesecondpaperbyKohriandTakayama[33]worked
If the particle is negatively charged, it will eventually
- out the changes in the light element abundances due to
o form a bound state with the positively charged nuclei. −
thechangeincoulombbarrierinthepresenceoftheX ,
r When the particle decays, the decay products will elec-
t and due to the changes in the kinematics and energy
s tromagnetically interact with the nearby nucleus. If the
a balance for the reactions. They found that the Lithium
nucleus is 4He, it may break up, creating 3He, T, D, p,
: abundances can be significantly modified from the stan-
v n. These 3He, T and D nuclei will find other 4He nuclei
Xi and interact to form 6Li and 7Li . dard BBN predictions.
r In addition to the production mechanism described
a
above, charged particle decays result in a large non- RATES
thermalphotonbackgroundthatcandestroythecreated
6Li and 7Li nuclei. When both the production and de-
Tocalculatethemagnitudeoftheeffectswementioned
struction processes are taken into account, we find that
intheprevioussection,weneedtocalculatethefollowing
the overall 6Li production could be in the same range
quantities.
as the Lithium isotopic abundance measured in few low
metallicity popII stars [22, 23, 24, 25, 26, 27]. 1. f , the fraction of Helium-4 nuclei bound to the
b
Other effects can come from the elastic scattering of charged particles.
the charged particle produced in the decay from the
4He nucleus, and also from the change in the motion of 2. fP, the fraction of the bound 4He nuclei which are
the 4He in the bound state. We find that these effects destroyed by photodisintegration.
produce insignificant changes to standard BBN abun- 3. f3,theaveragefractionsof3HeandTproducedby
dances. the photodisintegration of a 4He nucleus.
In this work we only consider singly charged bound
state of 4He nuclei [28, 29, 30, 31]. We ignore the pro- 4. f , the fraction of 3He and T that formed in the
L
duction of neutral atoms of 4He bound to two X. We photo-disintegrationthat find another 4He nucleus
also do not consider the bound state of protons. This and form 6Li .
2
10-1 In the above equation, R is the recombination rate and
β is the ionization rate. The recombination rate may be
calculatedinamanneranalogoustotherecombinationof
n HydrogentoyieldR=5(α/m) (BE/T)ln(BE/T). The
o
ti10-2 ionization cross-section and hepnce β can then be found
c
a from the usual analysis of the equilibrium case.
r
f

4
-
m
u10-3 PHOTO-DISINTEGRATION OF HELIUM-4
i
l
e
H
Thephoto-disintegrationoccursthroughaone-photon

d
n -4 exchange diagram, where the charged decay product
ou10 emitsaphotonofinvariantmomentumq2 =−Q2,which
B
hits the Helium nucleus and breaks it up.
Photo-disintegration can occur as long as the energy
-5 transferred to the hadronic system, is larger than the
10
-6 -4 -2 0 binding energy, which is here 28.8 MeV. Unfortunately,
10 10 10 10
Temperature (keV) the cross-sectionfor the process γ+He4 →any has not
beenmeasuredforalltheenergiesthatweneed. Forlarge
Q2,onecanusetheformulaefordeepinelasticscattering,
FIG. 1: Solid curve is the fraction of helium nuclei that are
but these are only valid when Q2 is greater than Q2 ∼
bound to a long-lived massive charged particle. The number 0
density of these charged particles is set by the requirement 1GeV2.
that the massive neutral particle it decays into be all the Toproceed,wewillrestrictouranalysistophotonmo-
dark matter in the universe and the curve here is for a dark mentasatisfyingQ2 >Q2. Thiswillresultinaconserva-
0
matter particle mass of 200 GeV. We have taken the decay tive estimate of the cross-section; the true cross-section
lifetimetobemuchlargerthantheageoftheuniverseatthe
will certainly receive additional contributions from pho-
temperatures shown above.
tons below this cutoff.
For Q2 >Q2, the cross section is given by
0
In what follows we will concentrate on the dσ 2πα2 1
D
6Li abundance. Some 7Li is also produced through = dx
4He(3He,γ)7Be and 4He(T,γ)7Li. However, these reac- dQ2 Q4 Z0
Q2
tions have an electromagnetic component and hence are × f (x)Q2[1+(1− )2]θ(xs−Q2).(2)
slower than the 6Li production reactions. Xf f f xs
The total fraction of 6Li to Hydrogen (by number) is
then fbfPf3fLYP/4/(1 − YP/2) where YP is the total Hereff(x)isthe partondistributionfunctionofflavor
4He mass fraction (bound or not) in the absence of the f at longitudinal fraction x. Qf is the corresponding
charge. The final factor above is a kinematic constraint.
decay. The effects on the abundances of the other nuclei
Unlike the usual situation in deep inelastic scattering,
can also be similarly determined.
hereweshouldintegrateovertheallowedQ2. Forafixed
x the range is Q2 <Q2 < xs. The allowed range of x is
0
CHARGED PARTICLE RECOMBINATION then between Q20 and 1.
s
We then have to perform an integration over x. This
is hampered by our lack of knowledge of f (x). Quali-
The bulk of 4He forms at T ≡ T ≃ 70keV. If the f
α,f tatively, f (x) increases at smaller x, so this integral is
decayhappenssignificantlybeforethistemperatureT , f
α,f expected to be dominated by the smaller values of x.
then the effects we are interested in here are not impor- We will assume that ǫ = Q20 ≪ 0.1, and restrict
tant. For the rest of the discussion we will assume that s
the range of the integral between x = ǫ and x = 0.1.
the decay lifetime is larger than 1/H(T ).
α,f
We alsoassumethatthe predominantcomponentofX Over this range, we can take approximately F2(x) =
xf (x)Q2 ∼1 [34].
is negatively charged. Thus we are guaranteed that X f f
With these approximations,we find
will bind to the positively charged nuclei.
The equation for the evolution of the fraction of 2πα2
σ ∼− ln(10Q2/s). (3)
charged X particles that are not bound (xf) is given by Q2 0
0
1 Using 1GeV−1 ∼ 0.2fm, the cross section evaluates
x˙f =(cid:20)β(1−xf)−RxfnB4Yp(t)(1−fb)(cid:21) . (1) to about 2π × 10−4 × 0.04 × 10−26cm2ln(10Q20/s) ∼
3
0.3×10−3ln(10Q2/s)mbinalmostexactagreementwith pansion rate. As a consequence one may write the prob-
0
a calculation from GEANT. ability of 6Li to form as [8]
To derive a probability from this quantity, we must
multiply by the flux, and for this we must examine the K0 Y dK −1
P
geometry. The helium nucleus orbits around X at a fL =Z dKσ3+4(K)v 4 nB(T)(cid:18)dT (cid:19) (5)
distance given by d ≃ m−1/Zα = 3.6 fm where we Eth
have taken m to be the mass of the helium nucleus. where v is the relative speed, K is the kinetic energy
The size of the helium nucleus may be taken to be of the nucleon, dK/dT is the total energy loss rate and
r = 1.44A1/3 fm = 2.3 fm. The solid angle subtended σ3+4 is the cross-section for 3He on 4He reaction. Eth is
bythenucleusatthe positionofXis2π(1−cosφ)where the minimum energy required for the reaction to occur.
cos2φ = 1−r2/d2 and the probability of disassociation The upper limit to the kinetic energy K0 is the energy
is with which the 3He nucleus is born. As it turns out, the
integral is insensitive to the upper limit as long as it is
σD 2(1−cosφ) ≃0.012σ fm−2. (4) largerthanabout100MeV.Thisisduetothefastenergy
4πd2sin2φcos2φ D
degradation rate.
The probability of an interaction between the emitted Thisisnottheendofthestory. Whenthedecayoccurs
particle and the nucleus is thus 0.012×3×10−5 =3.6× the charged daughter particle carries off a large amount
10−7 for a cross-sectionof 3×10−4mb. ofenergy. All ofthis gets convertedto non-thermalpho-
tons [36]. In the time it takes for these photons to ther-
malize,theycanfinda 6Li(andothernuclei)andphoto-
NUCLEI FROM THE PHOTO-DISINTEGRATION dissociateitiftheenergyissufficienttocrossthe thresh-
old.
The photo-disintegration of 4He described in the pre- We can calculate the non-thermal distribution of the
vioussectiontypicallybreaksupthe nucleus. The nuclei photonsfollowing[8,15,36]. Theinitialenergyinjection
will split into lighter elements 3He, T, D, p and n. The is rapidly brought to a quasi-static equilibrium state by
exclusive cross-sectionsfor these processes are unknown. pair productionoff ofthe background(thermal) photons
We will assume following [35] that the helium nucleus and inverse Compton scattering. On a longer timescale
is always destroyed, and produces 3He and T with the (but still shorter than the age) processes like photon-
cross-sections σ3 =σD = 49σP. In other words, the pho- photon scattering and Compton scattering redistribute
todisintegration produces a 3He or T nucleus in about the energy and lead to thermalization. We assume that
half the interactions, and D in about half the interac- the thermalization is mainly due to Compton scattering
tions. We will assume that 3He and T are produced in andcalculatethedifferentialnon-thermalphotonnumber
equal amounts. density following [15].
Withtheexpressionforthenon-thermalphotondistri-
butionathand,onemaywritetheratefor6Liproduction
PRODUCTION OF LITHIUM-6
as
The3Heinthephoto-dissociationof4Heareextremely d Y6(T) = −nB(T)fL(T)f3fPfb(T)
relativistic. They can then scatter off background dln(T) τH(T)
4He nuclei, and produce Lithium. The rate for this is Γ6,γ
given by Γα =3.6×10−15YPTM3eV(σ/40mb)(η10/6)MeV + Y6(T)H(T) (6)
whereσ istheappropriatenuclearcross-section. Incom-
parison, the expansion rate of the universe is H(T) = where subscript “6” stands for 6Li and Γ6,γ is the total
2.5×10−22T2 MeV. Comparing the two rates, we see photo-dissociation rate for γ+6Li → anything. Y de-
MeV
that the nuclear reactionrate will be larger than the ex- notes the number density relative to the number density
pansion rate of the universe until T ∼ 0.3eV(40mb/ < of baryons.
σv >). Similarly, the interaction of 3He and T with the back-
The nuclear reaction rate that we have quoted above groundprotonsandphotons candegradethe nuclearen-
assumes that the 3He are energetic enough to overcome ergy. We have checked that these are negligible effects
theCoulombbarrier. Thisiscertainlytrueforthenuclei forthetemperaturesofaround10keVandlowerthatwe
at the moment of decay. However, these nuclei will lose are interested in.
energy due to a variety of processes [9, 36], the most The non-thermal photons may also dissociate the
importantofwhicharetheenergylossesduetoCoulomb 3He before they cancollide with 4He nuclei. To estimate
and Compton scattering. this, we note that the minimum threshold for 3He de-
The time scale for these energy loss processes is much structionis about 5.5 MeV. In orderto produce photons
shorter than the nuclear reaction rate as well as the ex- with this energy, one would have to be at a temperature
4
10-11 amounts of 7Li and 7Be .
The 4He nuclei could also receive a large kick due to
elasticscatteringoffofthedecayproducedchargedparti-
cle. Our calculations show that for small energy releases
oforder10GeV,itispossibletoproduce7Liinquantities
comparabletothestandardBBNyield. However,incon-
H
/ cretecasesliketheSuper-WIMPscenario[1,16],thelife-
6 time for such small energy releases is large which means
m-10-12 that the destruction rate due to non-thermal photons is
u
hi large. The destruction effect dominates and significant
it amounts of 7Li are not produced.
L
We also note that the break-up of 4He produces en-
ergetic neutrons and protons that in turn can break up
the background 4He nuclei. However, we have seen that
only about 1 in a million bound 4He is broken up and
-13 hence the destructiondue to theseenergetic protonsand
10
3 4 5 6 7 neutrons is negligible.
10 10 10 10 10
Lifetime (seconds)
DISCUSSION
FIG. 2: Curve shows the ratio of number density of 6Li to
that of H produced due to the process described in the text.
The model assumed to make this plot is a gravitino Super- Wehaveconcentratedhereonthecollisionof3Hewith
WIMP with mass of 100 GeV and a heavier charged slepton background 4He nuclei that would produce 6Li nuclei.
likestau. Westart with an initial6Li/H ratioof 1.4×10−13. Standard BBN does not produce much 6Li and thus the
expectation is that this could be an important probe.
The SBBN yield (without decays) is about 10−13 com-
lowerthanabout2keV.Thus,thiscouldonlybeaprob- pared to the number density of H. This should be com-
lem for long lifetimes. Also, the spectrum at these ener- paredtothemeasuredlithiumisotopicratiosinsomelow
gies is falling sharply [15, 36]. We estimate that this ef- metallicity popII stars which (conservatively) range be-
fectcannotbesignificantunlessdecaylifetimesarelarger tween 0.01 and 0.1. A reasonable constraint on 6Li then
than106 secondsandenergyreleasedislargerthanTeV. is that its abundance should not be much larger than
The bottom-line from the above calculation is that it that of 7Li nuclei (about 10−10 relative to the H number
is possible to produce 6Li of order 10−12 (relative to the density). It is of course possible that the 6Li primordial
numberdensityofH)inlargeregionsofdecayingcharged abundance is larger, but then we are left with the ques-
particle model parameter space. This is clear from the tion of why 6Li is depleted in much larger amounts than
Figure 2 where we have assumed that a slepton is de- 7Li by astrophysicalmechanisms.
caying to a gravitinoandlepton, with the gravitino hav- We found that in some regions of parameter space it
ing a mass of 100 GeV. As the mass of the gravitino is is possible to produce 6Li in quantities that match the
lowered, the amount of 6Li produced decreases. Note observed“plateau”abundance. Thisdoesnotimplythat
thatthemasseswhereareasonableamountof6Liispro- the decaying charged particle scenario can explain the
ducedcorrespondstoastrophysicallyinterestinglifetimes 6Li abundance. A full calculation including all the light
of about a month [37, 38]. We note in closing that the elements would be necessary to ascertain that. We also
dissociationby non-thermalphotons willbe insignificant note that there are astrophysical scenarios wherein the
for models where the mass of the decayingand daughter observedabundanceof6Lifindsanexplanation(e.g.,[39,
particles is fine-tuned to be small [6]. 40, 41, 42, 43, 44, 45]).
The processes we have described here are clearly rele-
vantfordeterminingthelightelementabundances. Much
PRODUCTION OF LITHIUM-7 AND OTHER more work needs to be done before these processes can
REACTIONS
becombinedwithpreviousanalysestoputboundsonthe
model parameter space. Throughout this work we have
After the decay, all the 4He nuclei have the orbital ki- neglected the bound states of other light elements. The
netic energy ∼ 0.16 MeV. The kinetic energy is still not fraction of bound states of D, T, 3He , 6Li and 7Li are
enoughtoinitiateanyendothermicnuclearreactions. We smallowingtothemuchsmallerabundanceoftherespec-
therefore consider the following processes 3He(α,γ)7Be tivelightelements. However,they couldstillhavesignif-
and 3H(α,γ)7Li. Proceeding as in the 6Li production icant effects because of enhanced nuclear cross-sections.
case, we find that this process produces insignificant The correct treatment of this problem including all the
5
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