Calculation of Momentum Distribution Function of a Non-Thermal Fermionic Dark Matter, March 8, 2017. arxiv:1612.02793, with Anirban Biswas. Aritra Gupta
Why Non-Thermal? 1 / 31 The most widely studied dark matters are the thermal WIMPS relic satisfied by thermal freeze-out. Direct detection experiments however yielded null results upper bound on dark matter nucleon cross section. PandaX-II collaboration, arxiv:1607.07400 Non observation of dark matter may indicate that these particles are more feebly interacting than we think.
Thermal Freeze-Out : A Brief Recap 2 / 31 Particles which were in thermal equilibrium in early universe decouple or freezes out when the their rate of interactions can not keep up with the expansion of the universe. Boltzmann Equation : ( ) L f a = t H p a f = C a+b i+ j [ f a ] (1) p a where, C [ f a ] ( d 3 p b 2E b d 3 p i 2E i d 3 p j 2E j (2π) 4 δ 4 (p a +p b p i p j ) M 2 a+b i+ j f a f b (1± f i )(1± f j ) M 2 i+ j a+b f i f j (1± f a )(1± f b ) ) (2)
3 / 31 Eq.(2) can be simplified further using some approximations.... CP invariance is assumed i.e. M 2 i+ j a+b = M 2 a+b i+ j M 2 (say) d 3 p a d 3 p b d 3 p i d 3 p j C [ f ] (2π) 4 δ 4 (p a + p b p i p j ) 2E a 2E b 2E i 2E j ( ) M 2 f a f b (1 ± f i )(1 ± f j ) f i f j (1 ± f a )(1 ± f b )
4 / 31. Pauli blocking and stimulated emission terms are neglected. d 3 p a d 3 p b d 3 p i d 3 p j C [ f ] (2π) 4 δ 4 (p a + p b p i p j ) 2E a 2E b 2E i 2E j ( ) M 2 f a f b f i f j (3)
5 / 31. If particles i and j belong to the thermal soup, they are assumed to follow the classical Maxwell-Boltzmann distribution function i.e. f e E T f i f j = e E i +E j T = e E a+e b T = f eq a f eq b. Eq.(3) simplifies to : d 3 p a d 3 p b d 3 p i d 3 p j C [ f ] (2π) 4 δ 4 (p a + p b p i p j ) 2E a 2E b 2E i 2E j ( ) M 2 f a f b fa eq f eq b (4)
6 / 31 If a and b are both identical particles (say χ), then integrating Eq.(4) on both sides by 0 d 3 p, and remembering that n χ 0 f (p)d 3 p, we finally arrive at the conventional form : dn ( ) χ + 3Hn χ = σ v n 2 χ n 2 χ eq dt where, σ = σ χ χ all 1 σv = 8T M 4 χ K2 2 and, ( ( ) Mχ σ s 4M 2 4M 2 χ χ T ) ( sk1 s T ) ds Cosmic abundances of stable particles: Improved analysis, Gelmini & Gondolo, 1990.
7 / 31 If χ can decay then an extra term depletion term will be added to the "Rate Equation".. dn ( χ + 3Hn χ = σ v n 2 χ n 2 χ eq dt ) Γ χ ( n χ n eq ) χ where, Γ χ is the relevant width of χ, Γ χ = K 1(T ) K 2 (T ) Γ χ K 1 and K 2 are modified Bessel function of order 1 and 2. (5)
In terms of Y, ( ) dy χ dx = s σv Yχ 2 Yχ 2 eq H x 8 / 31 A formal solution of this equation is given by : The Early Universe, Kolb & Turner
Non-Thermal Dark Matter : Basic Concepts 9 / 31 Condition for thermalisation of a system of particles in the background of an expanding universe : where, Γ H 1 Γ = Interaction Rate = n eq σv, H = Expansion Rate. Freeze-out : Γ H 1 (T < T fo) Freeze-in : Γ H 1 (at all epochs)
10 / 31 Number density of non-thermal dark matter is very small initial abundance is almost negligible. Non-thermal dark matters hence needed to be produced comoving number density gradually rises and finally saturates to satisfy relic density. Usually the most dominant production channels are from decay of heavier particles. The decaying particle may be in thermal equilibrium or can itself be out-of-equilibrium.
If the decaying particle is in thermal equilibrium, then the usual form of rate equation in terms of n (or Y ) can be used. 11 / 31 For example, we can use : Γ = K 1(T ) Γ etc... K 2 (T ) But if the mother particle is not in equilibrium, then such an equation cannot be used. Now, Γ = Γ fdm (p)d 3 p fdm (p)d 3 p (from definition only...) Knowledge of the non-equilibrium distribution function, f mother is necessary to solve for f DM. Need for coupled set of Boltzmann equations...
12 / 31 The difference between the freeze-in and freeze-out scenario is best illustrated by the following plots : Y Ω X h 2 10 9 10 12! x 0.1 freeze-in freeze-in freeze-out freeze-out 10 15 1 10 100 x = m/t 10-12 "!"" λ, λ 1 Hall et.al., arxiv:0911.1120
A new U(1) B L model 13 / 31 Gauge Group : SU(2) L U(1) Y U(1) B L The Lagrangian : 2 i=1 L BL = iη L γ µ D µ η η L + iξ L γ µ D µ ξ ξ L + i 1 4 F µν Z BL F ZBL, µν + 2 i=1 2 i=1 χ i R γ µ D µ χ i χ i R (D µ ϕ i ϕ i ) (D ϕi µ ϕ i) ( ) y ξ i ξ L χ i R ϕ 2 + y η i η L χ i R ϕ 1 + h.c. V (H, ϕ 1, ϕ 2 ) + L Sudhanwa Patra et.al., arxiv:1607.04029
14 / 31 where, V (H,ϕ 1,ϕ 2 ) = µ 2 HH H + λ H (H H) 2 + µ 2 1ϕ 1 ϕ 1 + λ 1 (ϕ 1 ϕ 1) 2 + µ 2 2ϕ 2 ϕ 2 +λ 2 (ϕ 2 ϕ 2) 2 + ρ 1 (H H)(ϕ 1 ϕ 1) + ρ 2 (H H)(ϕ 2 ϕ 2) + λ 3 (ϕ 1 ϕ 1)(ϕ 2 ϕ 2) ) +µ (ϕ 2 ϕ 2 1 + ϕ 2 ϕ 2 1 The model is well motivated since :. It is anomaly free.. It can give rise to neutrino mass.. It can accommodate a thermal dark matter.
The charge assignment is given in the following table : Field SU(2) L U(1) Y U(1) B L VEV charge charge charge l L (ν L e L ) T 2-1 2-1 Q L (u L d L ) T 1 1 2 6 3 SM Fermions e R 1-1 -1 0 u R 1 2 3 1 3 d R 1-1 3 4 ξ L 1 0 3 1 BSM Fermions η L 1 0 3 0 χ 1 R 1 0 2 3 χ 2 R 1 0 2 3 1 H 2 2 0 v Scalars ϕ 1 1 0 1 v 1 ϕ 2 1 0 2 v 2 3 1 2 v 1 3 15 / 31
Points to highlight regarding the model : 16 / 31 The scalars acquire vev after symmetry breaking : H 0 = 1 (v + h) + i G, 2 2 ϕ 1 = 1 (v 1 + h 1 ) + i à 1, 2 2 ϕ 2 = 1 (v 2 + h 2 ) + i à 2, 2 2 The fields ( h 1, h 2, h 3 ) mix with each other 3 3 PMNS-like mixing matrix parametrised by θ 12, θ 13, θ 23. à 1 and à 2 also mix together to give rise to a massive physical pseudoscalar with mass M A.
For the Fermions : where, L fermion mass = ( ξ L η L ) Mfermion ( χ1 R χ 2 R M fermion = ( yξ 1 v 2 y ξ 2 v 2 y η 1 v 1 y η 1 v 1 The mass and gauge basis states are related by : ( ) ( ) ( ) ( ) ξl ψ2 = U L χ1 η L, R ψ2 = U R L ψ R. 1 L ψ 1 R where, χ 2 R ( ) cosθl,r sinθ U L,R = L,R sinθ L,R cosθ L,R The physical states are ψ 1 = ψ 1L + ψ 1R ψ 2 = ψ 2L + ψ 2R. ) and ) 17 / 31
18 / 31 The breaking of U(1) B L gauge group also gives the extra gauge boson its mass. M 2 Z BL = ( ) gbl v 2 2 (1 + 4β 2 ), where β = v 2 β v 1 The set of independent parameters relevant for our analysis are as follows : θ 12, θ 13, θ 23, θ L, θ R, M h2, M h3, M A, M ψ1, M ψ2, M ZBL, g BL and β.
19 / 31 ψ 1 is our dark matter candidate. The dominant production modes of this dark matter are : h 1 ψ 1 ψ 1, h 2 ψ 1 ψ 1, Z BL ψ 1 ψ 1. The decay widths strongly depend on g BL non-thermality g BL 10 10. and for Very small g BL Z BL is also not in equilibrium. Z BL Z BL production : h 2 Z BL Z BL. depletion : Z BL f f and Z BL ψ 1 ψ 1.
h 2 can however equilibrate through their mixing with SM Higgs (h 1 ). 20 / 31 We can not solve for Y ψ1 directly, rather we need to find f ZBL and consequently solve for f ψ1 coupled Boltzmann equations. ˆL f ZBL = C h 2 Z BL Z BL + C Z BL all ˆL f ψ1 = s=h 1,h 2 C S ψ 1ψ 1 + C Z BL ψ 1 ψ 1 SM Higgs (h 1 ) gets its mass after EWPT. So, when T > T EWPT, h 1 decay is not allowed.
Simplifying the Liouville Operator 21 / 31 For isotropic homogeneous universe, FRW metric gives : ( L = t H p ) p Using conservation of entropy : ( dt = HT 1 + T g s(t ) dt 3g s (T ) Change of coordinates : r = m 0 T, ( ξ p = g s (T 0 ) g s (T ) ) 1/3 p T, ) 1
The Liouville operator simplifies to : ( ) 1 ˆL = rh 1 + T g s 3g s r, 22 / 31 Chosen Benchmark (β = 10 3 ) : Input Parameters M ZBL M h2 M ψ1 g BL θ 12 θ 13 θ 23 θ L = θ R Corresponding values 1 TeV 5 TeV 10 GeV 1.75 10 11 0.1000 rad 0.01 rad 0.06 rad π/4 rad
Calculation of Collision Terms 23 / 31 Z BL (p) f(p ) f(q ) C Z BL f f [ f ZBL (p)] = 1 2E p g f d 3 p (2π) 3 2E p g f d 3 q (2π) 3 2E q (2π) 4 δ 4 ( p p q ) M 2 [ f ZBL (p)] = f ZBL (p) 1 2E p g f d 3 p (2π) 3 2E p g f d 3 q (2π) 3 2E q (2π) 4 δ 4 ( p p q ) M 2 In a general frame moving with momentum 'p' : Γ Z BL f f = 1 2E p g f d 3 p (2π) 3 2E p g f d 3 q (2π) 3 (2π) 2E 4 δ 4 ( p p q ) M 2 q ZBL f f
[ C Z BL f f [ f ZBL (ξ p )] = f ZBL (ξ p ) Γ rest Z BL f f M Z BL = f ZBL (ξ p ) Γ rest Z BL f f r ZBL ξp 2 B(r) 2 + r 2 ZBL E ZBL ] 24 / 31 Hence, C Z BL all = f ZBL (ξ p ) Γ ZBL all r ZBL ξ 2 p B(r) 2 + r 2 ZBL where, ( g s (T ) g s (T 0 ) ) 1/3 = ( ) 1/3 g s (M sc /r) B(r) g s (M sc /r 0 )
h 2 (k) Z BL (p)z BL (q ) C h 2 Z BL Z BL[ f ZBL (p)] = 2 1 2E p gh2 d 3 k g ZBL d 3 q (2π) 3 2E k (2π) 3 2E q 25 / 31 = 2 1 2E p Here, (2π) 4 δ 4 ( k p q ) M 2 h2 ZBL ZBL [ f h2 (1± f ZBL )(1± f ZBL ) f ZBL f ZBL (1± f h2 )] gh2 d 3 k g ZBL d 3 q (2π) 3 2E k (2π) 3 (2π) 2E 4 δ 4 ( k p q ) M 2 [ f h2 (k)]. q M 2 h2 ZBL ZBL= g 2 h 2 Z BL Z BL 2 9 ( 2+ (E p E q p. ) 2 q ) M 4 Z BL
26 / 31 g 2 C h 2 Z BL Z h 2 Z BL Z BL BL[ f ZBL (p)] = δ(e k E p E q (p,k,θ)) 2+ where, E q = 1 k 2 dk d(cosθ) 6(4π) 2 E p E k E q (p,k,cosθ) ( Ep E q (k, p,θ) + p 2 pk cosθ ) 2 M 4 Z BL k 2 + p 2 + M 2 Z BL 2pk cosθ [ f h2 (k)] In terms of the chosen coordinates, E q = T ξk 2 B(r)2 + ξp 2 B(r) 2 + rz 2 BL 2B(r) 2 ξ k ξ p cosθ
27 / 31 Finally, doing the cos θ integral and hence removing the delta function, we arrive at : C h 2 Z BL Z BL = r B 1 (r) 8πM sc ( ξ p ξpb(r) 2 2 MZBL r + g 2 h 2 Z BL Z BL 6 2+ (M 2 h 2 2M 2 Z BL ) 2 4M 4 Z BL M sc ) 2 (ξ min e k (ξp,r)) 2 ( ) Mh2 B(r) 2 r 2 + Msc (ξ k max (ξp,r)) 2 ( ) Mh2 B(r) 2 r 2 + Msc e
28 / 31 The distribution functions : ξ p 2 fzbl (ξ p, r) 10-4 10-6 10-8 10-10 10-12 r=0.02 r=0.05 r=0.2 r=2.0 r=700.0 β = 0.001 ξ p 2 fψ1 (ξ p, r) 10-6 10-8 10-10 10-12 10-14 β = 0.001 r=2.0 10-2 r=2.0 10-1 r=2.0 10 1 r=1.0 10 3 r=1.0 10 4 10-14 10-16 10-16 10-18 10-18 10-4 10-3 10-2 10-1 10 0 10 1 10 2 ξ p 10-20 10-4 10-3 10-2 10-1 10 0 10 1 10 2 10 3 10 4 10 5 ξ p
Knowing the distribution function, we are finally able to calculate the comoving number density Y n s. 29 / 31 Y ( n/s) 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14 10-15 10-16 M ψ1 = 10 GeV, M ZBL = 1 TeV, M h2 = 5 TeV, g BL = 1.75 10-11 ψ 1 production from Z BL decay Production of ψ 1 from scalars decay Y ZBL Y ψ1 10-1 10 0 10 1 10 2 10 3 10 4 10 5 10 6 10 7 r ( M h1 /T)
Comparing with the approximate solution 30 / 31 What if we had assumed that the system is close to equilibrium and used rate equations in terms of Y. Y ( n/s) 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14 10-15 10-16 M ψ1 = 10 GeV, M ZBL = 1 TeV, M h2 = 5 TeV, g BL = 1.75 10-11 Y ZBL < Γ ZBL all> NTh < Γ ZBL ψ 1 ψ 1 > NTh < Γ ZBL all> Th < Γ ZBL ψ 1 ψ 1 > Th Y ψ1 10-1 10 0 10 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 r ( M h1 /T)
31 / 31 THANK YOU