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BaryogenesisSeminar on Non-Accelerator Particle Physics
Christoph Nega
Physikalisches Institut Bonn
29.04.2016
Based on:E. Kolb and M. Turner: The Early Universe
1 / 27
Overview
Introduction
The Sakharov Conditions3 ConditionsSakharov Conditions in the Standard Model?
A Toy ModelThe Boltzmann EquationModel ContentSolutions
Further Models
Summary
Christoph Nega 2 / 27
Introduction
Christoph Nega 3 / 27
Motivation
Shortly after the Big Bang there was enough energy to create baryons (matter)and antibaryons (antimatter)
But also the annihilation process was possible
After a while the temperature drops
The baryon-antibaryon-production stops
Either the remainder baryons and antibaryons dissipate or maybe they freeze outand survive
Christoph Nega 4 / 27
Motivation
Shortly after the Big Bang there was enough energy to create baryons (matter)and antibaryons (antimatter)
But also the annihilation process was possible
After a while the temperature drops
The baryon-antibaryon-production stops
Either the remainder baryons and antibaryons dissipate or maybe they freeze outand survive
Why do we only have matter and no antimatter in the universe?
Christoph Nega 4 / 27
Cosmological Observations
From cosmological Observations we know thatNo antimatter on earthNo antimatter on the moonAlso nothing on larger scales like galaxy clusters
Christoph Nega 5 / 27
Cosmological Observations
From cosmological Observations we know thatNo antimatter on earthNo antimatter on the moonAlso nothing on larger scales like galaxy clusters
Maximal violation of matter-antimatter-symmetry!
But where does it come from?
Christoph Nega 5 / 27
Cosmological Observations
From cosmological Observations we know thatNo antimatter on earthNo antimatter on the moonAlso nothing on larger scales like galaxy clusters
Maximal violation of matter-antimatter-symmetry!
But where does it come from?1 Asymmetric initial conditions
Not satisfactory due to CPT invariance2 Dynamical generation of baryon asymmetry
BaryogenesisMaximal asymmetry nowadays caused by small difference
nq − nq
nq' 3 · 10−8
−→ today we have a baryon-to-entropy ratio of B := nB/s ' 10−10
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The Sakharov Conditions
Christoph Nega 6 / 27
The 3 Sakharov Conditions
1967 Andrei D. Sakharov formulated 3 necessary conditions togenerate a baryon asymmetry
(i) Baryon Number Violation:I Process with baryon number violation
(ii) C and CP Violation:I BNV alone produces baryons and antibaryons of the same rateI C and CP violation prefer particles or antiparticlesI Need really both!
(iii) Non-Equilibrium Condition:I CPT invariance is preserved, ensures mb = mbI Then thermal equilibrium arranges nb = nb
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A Simple Model
Take a particle X which has baryon number violating decays:
Decay BR B
X −→ qq r 2/3X −→ ql 1− r -1/3X −→ qq r -2/3X −→ ql 1− r 1/3
CPT: M(X −→ qq) =M(qq −→ X )C and CP violation: M(X −→ qq) 6=M(X −→ qq)
Christoph Nega 8 / 27
A Simple Model
Take a particle X which has baryon number violating decays:
Decay BR B
X −→ qq r 2/3X −→ ql 1− r -1/3X −→ qq r -2/3X −→ ql 1− r 1/3
CPT: M(X −→ qq) =M(qq −→ X )C and CP violation: M(X −→ qq) 6=M(X −→ qq)
BX = 2/3 · r − 1/3 · (1− r)BX = −2/3r + 1/3 · (1− r)
}ε = BX + BX = r − r
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C and CP ViolationConsider two heavy bosons X ,Y with tree level interactions:
�Xf2
f1
g1 �Xf4
f3
g2
�Yf1
f3
g3 �Yf2
f4
g4
εX ,Y =∑
k
Bf ·Γ(X ,Y → fk)− Γ(X , Y → fk)
ΓX ,Y
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C and CP ViolationConsider two heavy bosons X ,Y with tree level interactions:
�Xf2
f1
g1 ⇒ Γ(X → f1f2) = |g1|2 · ΠX
�Xf2
f1
g1 ⇒ Γ(X → f1 f2) = |g∗1 |2 · ΠX
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C and CP ViolationConsider two heavy bosons X ,Y with tree level interactions:
�Xf2
f1
g1 ⇒ Γ(X → f1f2) = |g1|2 · ΠX
�Xf2
f1
g1 ⇒ Γ(X → f1 f2) = |g∗1 |2 · ΠX
Since ΠX = ΠX ⇒ εX = 0 No C or CP violation!
Christoph Nega 9 / 27
C and CP ViolationNeed loop diagrams:
�Xf2
f1
g1 + �f4
f3
YX
f2
f1
⇒ Γ(X → f1f2) = g1g∗2 g3g∗4 ΠXY + c.c.
Get analogous expression for Γ(X → f1 f2)
⇒ Γ(X → f1f2)− Γ(X → f1 f2) = 4Im(g∗1 g2g∗3 g4)ImΠXY
⇒ εX = 4ΓXIm(g∗1 g2g∗3 g4)ImΠXY · {(Bf4 − Bf3 )− (Bf2 − Bf1 )}
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C and CP ViolationResults:
1 Only at loop level we get C and CP violation
Interference term
2 Need 2 massiv bosons with mX ,Y >∑
k mfk
ImΠXY 6= 0
3 Need a complex coupling constant
4 Attention εX 6= −εYmX 6= mY
Christoph Nega 9 / 27
Non-Equilibrium Condition
Consider:
At beginning of the universe: T � mX
Thermal equilibrium forces nX = nX ' nγ
At critical temperature T = mX we can have1 Interaction is effective compared to expansion of the universe2 Interaction is "too weak"
Forces overabundance of X and X
In the second case only at T � mX the bosons decay
Inverse decay and non-conserving scattering processes are suppressed
⇒ Can get net baryon number nBChristoph Nega 10 / 27
Non-Equilibrium Condition
Consider:
For a X decay we get a net baryon number of εAssume strong overabundance of nX = nX ' nγThe entropy density is s ' g∗nγ
B := nBs '
εnγg∗nγ
' ε
g∗
Out of Equilibrium Decay Scenario
Comparing with B ∼ 10−10 and g∗ of the order 100− 1000
Need only a very small C and CP violation ε = 10−8 − 10−7
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Sakharov Conditions in the Standard Model
Are the 3 Sakharov Conditions fulfilled in the SM?
1 In the electroweak theory we have baryon number violation
Chiral anomaly at loop level
∂µJµB =∑
f
∂µ (qf γµqf ) ∝ F · (∗F )
2 CP violation due to CKM matrix
LCC = − g√2
uLαγµVαβdLβW + + h.c→ BSM = nB
s ∼ 10−18
3 Electroweak phase transition as period of non-equilibrium
But was not strong enough
All 3 conditions are fulfilled but not strong enough!
There are other CP sources in SM but too small to measureGrand Unified Theories satisfy the 3 conditions naturally
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A Toy Model
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Decoupling
W. Buchmüller, Baryogenesis
Periods of non-equilibrium duringthe evolution of the Universe:
InflationDecoupling of WIMPSBaryogenesisNucleosynthesis
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Decoupling
W. Buchmüller, Baryogenesis
Periods of non-equilibrium duringthe evolution of the Universe:
InflationDecoupling of WIMPSBaryogenesisNucleosynthesis
How can we characterize decoupling?
Coupled period: Γ & HDecoupled period: Γ . H
Need equation describing the evolution of decoupling
Christoph Nega 13 / 27
Non-Equilibrium Dynamics 1
Boltzmann equation:
Determine evolution of phase space density ρHolds in non-equilibriumValid if collision duration is small compared to inverse mean free path
L[ρ] = C [ρ]
Liouville operator:
Left side describes free evolution of ρCovariant form
L = pα ∂
∂xα − Γαβγpβpγ ∂
∂pα
Need model for the metricCosmological models use Robertson-Walker metric
gRW = dt2 − R(t)2(
11− κr2 dr2 + r2dθ2 + r2 sin2 θdϕ2
)Christoph Nega 14 / 27
Non-Equilibrium Dynamics 2
⇒ L = E ∂
∂t − Hp2 ∂
∂E , with H = RR
Use the number density n(t) = g∫ d3p
(2π)3 ρ(E , t)
n + 3Hn = g∫
d2p(2π)3
C [ρ]E
Boltzmann equation fornumber density in FRWmodel
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Non-Equilibrium Dynamics 2
⇒ L = E ∂
∂t − Hp2 ∂
∂E , with H = RR
Use the number density n(t) = g∫ d3p
(2π)3 ρ(E , t)
n + 3Hn = g∫
d2p(2π)3
C [ρ]E
Boltzmann equation fornumber density in FRWmodel
Collision operator:
Consider reaction ψ + a + · · · −→ i + · · ·
g∫
d2pψ(2π)3
C [ρ]Eψ
= −∫
dΠψdΠa · · · dΠi · · · (2π)4δ(4)(pψ + pa + · · · − pi − · · · )
×{|M(ψ + a + · · · → i + · · · )|2 ρψρa · · · − |M(i + · · · → ψ + a + · · · )|2 ρi · · ·
}Christoph Nega 15 / 27
Model Content
Ingredients:Self conjugate massive boson X , i.e. X = XInteractions violating B, C and CPParticles b and b with baryon number ±1/2Other particles having no baryon number (thermal bath), g∗ d.o.f.
I Interactions with thermal bath are in kinetic equilibriumI µb = µbI Can assume ρi (E ) = e−(E−µi )/T
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Model Content
Ingredients:Self conjugate massive boson X , i.e. X = XInteractions violating B, C and CPParticles b and b with baryon number ±1/2Other particles having no baryon number (thermal bath), g∗ d.o.f.
I Interactions with thermal bath are in kinetic equilibriumI µb = µbI Can assume ρi (E ) = e−(E−µi )/T
Form of the matrix elements
|M(X → bb)|2 =∣∣∣M(bb → X )
∣∣∣2 = 1/2(1 + ε) |M0|2∣∣∣M(X → bb)∣∣∣2 = |M(bb → X )|2 = 1/2(1− ε) |M0|2
Net baryon number εX = ε
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Boltzmann Equation for the Toy Model 1
Now, the Boltzmann equations determin the evolution of the number densities
1) X boson:
nx + 3Hnx =∫
dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)
×{−ρX (|MX ,bb |2 +
∣∣MX ,bb∣∣2) + ρbρb |Mbb,X |2 + ρbρb
∣∣Mbb,X∣∣2}
=∫
dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)
×{−ρX + e−(E1+E2)/T} · |M0|2 +O(ε, µ/T )
=−ΓD(nX − nEQX )
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Boltzmann Equation for the Toy Model 1
Now, the Boltzmann equations determin the evolution of the number densities
1) X boson:
nx + 3Hnx =∫
dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)
×{−ρX (|MX ,bb |2 +
∣∣MX ,bb∣∣2) + ρbρb |Mbb,X |2 + ρbρb
∣∣Mbb,X∣∣2}
=∫
dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)
×{−ρX + e−(E1+E2)/T} · |M0|2 +O(ε, µ/T )
=−ΓD(nX − nEQX )
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Boltzmann Equation for the Toy Model 1
Now, the Boltzmann equations determin the evolution of the number densities
1) X boson:
nx + 3Hnx =∫
dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)
×{−ρX (|MX ,bb |2 +
∣∣MX ,bb∣∣2) + ρbρb |Mbb,X |2 + ρbρb
∣∣Mbb,X∣∣2}
=∫
dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)
×{−ρX + e−(E1+E2)/T} · |M0|2 +O(ε, µ/T )
=− ΓD(nX − nEQX )
Christoph Nega 17 / 27
Boltzmann Equation for the Toy Model 1
Now, the Boltzmann equations determin the evolution of the number densities
1) X boson:
nx + 3Hnx =∫
dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)
×{−ρX (|MX ,bb |2 +
∣∣MX ,bb∣∣2) + ρbρb |Mbb,X |2 + ρbρb
∣∣Mbb,X∣∣2}
=∫
dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)
×{−ρX + e−(E1+E2)/T} · |M0|2 +O(ε, µ/T )
=− ΓD(nX − nEQX )
nx + 3Hnx = −ΓD(nX − nEQX )
∆′= −X ′EQ − KγDz∆
∆ = X − XEQ
K = ΓD(z = 1)2H(M)
γD = ΓD(z)ΓD(z = 1)
Want evolution of number density independent of expansion of the universe
X := nX/s and use dimensionless coordinates z := mX/T ∝√
tChristoph Nega 17 / 27
Boltzmann Equation for the Toy Model 1
Now, the Boltzmann equations determin the evolution of the number densities
1) X boson:
nx + 3Hnx =∫
dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)
×{−ρX (|MX ,bb |2 +
∣∣MX ,bb∣∣2) + ρbρb |Mbb,X |2 + ρbρb
∣∣Mbb,X∣∣2}
=∫
dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)
×{−ρX + e−(E1+E2)/T} · |M0|2 +O(ε, µ/T )
=− ΓD(nX − nEQX )
nx + 3Hnx = −ΓD(nX − nEQX )
∆′ = −X ′EQ − KγDz∆
∆ = X − XEQ
K = ΓD(z = 1)2H(M)
γD = ΓD(z)ΓD(z = 1)
Want evolution of number density independent of expansion of the universe
X := nX/s and use dimensionless coordinates z := mX/T ∝√
tChristoph Nega 17 / 27
Boltzmann Equation for the Toy Model 2
For the b-particles we can derive analogously
∆′ = X ′EQ − KzγD∆B′ = εKzγD∆− KzγBB
with B = nb−nb2s and γB = ΓB(z)
ΓD(z=1)
Have now two coupled integral-differential equations
Christoph Nega 18 / 27
Boltzmann Equation for the Toy Model 2
For the b-particles we can derive analogously
∆′ = X ′EQ − KzγD∆B′ = εKzγD∆− KzγBB
with B = nb−nb2s and γB = ΓB(z)
ΓD(z=1)
Have now two coupled integral-differential equations
Physics:
∆ is driven by derivative of XEQ
For vanishing X ′EQ exponential decay of ∆
We need expanding universe X ′EQ 6= 0Baryon asymmetry B is driven by the departure from equilibrium ∆γB damps the asymmetry
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The Generell Solution
One can find a general solution
∆(z) = ∆0 exp{−K
∫ z
0z ′γD(z ′)dz ′
}−∫ z
0X ′EQ(z ′) · exp
{−K
∫ z
z′z ′′γD(z ′′)dz ′′
}dz ′
B(z) = B0 exp{−K
∫ z
0z ′γB(z ′)dz ′
}+ εK
∫ z
0∆(z ′)z ′γD(z ′) · exp
{−K
∫ z
z′z ′′γB(z ′′)dz ′′
}dz ′
Can consider two limits K � 1 and K � 1
We will assume ∆0 = B0 = 0 for a moment
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Limit 1: K � 1
This corresponds to H � ΓD(T = mx )
Then we can assume: exp{−K
∫ zz′ z′′γ(z ′′)dz ′′
}≈ 1
For T � mX we have a baryon symmetry
At T = mX the “expansion rate of the universe is faster then decay rate”
At lower temperatures the X boson decays following
∆′ ≈ −Kz∆
B ≈ εK∫ z
0z ′∆ dz ′
X = X0 · e−K2 z2
B = εX0
(1− e− K
2 z2)
Out of Equilibrium Decay
Produced baryon asymmetry: Bf := B(∞) = εX (0) = εg∗
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Limit 1: K � 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 20 40 60 80 100
Baryon
asym
metry
B/ε g ∗
Time t
Baryon asymmetry in Out of Equilibrium Decay scenario
K = 0.5K = 0.1
K = 0.01
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Limit 2: K � 1
So what happens for mX � T if ΓD(T = mx )� H?
Since the decay rate is fast enough ∆′ = 0 and thus
∆ '−X ′EQKγDz '
XEQKz
∆ ' XEQKz
Bf ∝ε
g∗
√z3
a · exp{−zf − K
∫ ∞zf
γBz dz}
with zf from 1 = KγB(zf )zfand a := −∂z (KγBz)|zf
Christoph Nega 21 / 27
Limit 2: K � 1
So what happens for mX � T if ΓD(T = mx )� H?
Since the decay rate is fast enough ∆′ = 0 and thus
∆ '−X ′EQKγDz '
XEQKz
∆ ' XEQKz
Bf ∝ε
g∗
√z3
a · exp{−zf − K
∫ ∞zf
γBz dz}
with zf from 1 = KγB(zf )zfand a := −∂z (KγBz)|zf
Need specific form of γB
γB ' z 32 e−z + 1
z5 resulting from Boltzmann
distribution and bb scattering
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Limit 2: K � 1
(i) Damping dominated by inverse decay
1 ' Kz5/2f e−zf ⇒ zf ' 4.2 log0.6 K
Bf ' 0.3 εg∗· 1
K log0.6 K
Christoph Nega 21 / 27
Limit 2: K � 1
0
0.2
0.4
0.6
0.8
1.01
1.2
102 103 104 105
Baryon
asym
metry
B f/ε g ∗10−
3
K
Asymptotic baryon asymmetry for 1 . K . KC
Christoph Nega 21 / 27
Limit 2: K � 1
(i) Damping dominated by inverse decay
1 ' Kz5/2f e−zf ⇒ zf ' 4.2 log0.6 K
Bf ' 0.3 εg∗· 1
K log0.6 K
(ii) Damping dominated by bb scattering
zf ' z1/4
Bf 'ε
g∗·√
K e− 43 K 1/4
Christoph Nega 21 / 27
Limit 2: K � 1
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
105 106 107 108 109
Baryon
asym
metry
B f/ε g ∗10−
8
K
Asymptotic baryon asymmetry for KC & K
Christoph Nega 21 / 27
Numerical Integration
Kolb/Turner, The Early Universe
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Further Models
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Further Models 1
Pre-Existing Asymmetries:I Assume a preexisting asymmetry: Bi = nB
s∣∣i , for T � mX
Bf = Bi · exp{−K
∫ ∞0
γBz ′dz ′}
+ εfasym(K )
Exponential suppression of pre-existing baryon asymmetryI There exist so called non-thermalized modes without suppression
Lepton number violations:I So far we have considered only a baryon asymmetryI Perhaps there might be also an asymmetry in the lepton number?I Is there a relation between both asymmetries?
Yes! Relation through so called Sphaleron processes
Christoph Nega 24 / 27
Further Models 2
Alternative Theories:I Variations containing the 3 Sakharov conditionsI Containing primordial black holes
Spontaneous Baryogenesis:I We need completely different conditions:
temporal, dynamical CPT violationI Consider Lagrangian
Lspon = 1f ∂µφJµB
Get baryon asymmetry Bspon ∼ − φg∗fT
Christoph Nega 25 / 27
Summary
Christoph Nega 26 / 27
Summary
As yet there have been no cosmological observations which verify anybaryogenesis theory!
The Sakharov conditions are the most popular ingredients to generate abaryon asymmetry
The Standard Model alone can not explain the asymmetry although all 3conditions are fulfilled
Gand Unified Theories?
The Boltzmann equation determine the evolution of the asymmetry
Christoph Nega 27 / 27