baryogenesis - seminar on non-accelerator particle · pdf fileandantibaryons(antimatter)...

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

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

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?

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?

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

The Sakharov Conditions

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

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)

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

C and CP ViolationConsider two heavy bosons X ,Y with tree level interactions:

�Xf2

g1 �Xf4

�Yf1

g3 �Yf2

εX ,Y =∑

Bf ·Γ(X ,Y → fk)− Γ(X , Y → fk)

ΓX ,Y

�Xf2

g1 ⇒ Γ(X → f1f2) = |g1|2 · ΠX

�Xf2

g1 ⇒ Γ(X → f1 f2) = |g∗1 |2 · ΠX

�Xf2

g1 ⇒ Γ(X → f1f2) = |g1|2 · ΠX

�Xf2

g1 ⇒ Γ(X → f1 f2) = |g∗1 |2 · ΠX

Since ΠX = ΠX ⇒ εX = 0 No C or CP violation!

C and CP ViolationNeed loop diagrams:

�Xf2

g1 + �f4

⇒ 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 )}

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 >∑

ImΠXY 6= 0

3 Need a complex coupling constant

4 Attention εX 6= −εYmX 6= mY

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

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

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 =∑

∂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

A Toy Model

Decoupling

W. Buchmüller, Baryogenesis

Periods of non-equilibrium duringthe evolution of the Universe:

InflationDecoupling of WIMPSBaryogenesisNucleosynthesis

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

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

⇒ 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

⇒ 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 + · · ·

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

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

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 )

1) X boson:

nx + 3Hnx =∫

dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)

×{−ρX (|MX ,bb |2 +

∣∣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 )

1) X boson:

nx + 3Hnx =∫

dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)

×{−ρX (|MX ,bb |2 +

∣∣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 )

1) X boson:

nx + 3Hnx =∫

dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)

×{−ρX (|MX ,bb |2 +

∣∣Mbb,X∣∣2}

dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)

×{−ρX + e−(E1+E2)/T} · |M0|2 +O(ε, µ/T )

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

1) X boson:

nx + 3Hnx =∫

dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)

×{−ρX (|MX ,bb |2 +

∣∣Mbb,X∣∣2}

dΠX dΠ1dΠ2(2π)4δ(4)(px − p1 − p2)

×{−ρX + e−(E1+E2)/T} · |M0|2 +O(ε, µ/T )

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

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

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

The Generell Solution

One can find a general solution

∆(z) = ∆0 exp{−K

0z ′γD(z ′)dz ′

}−∫ z

0X ′EQ(z ′) · exp

z′z ′′γD(z ′′)dz ′′

}dz ′

B(z) = B0 exp{−K

0z ′γB(z ′)dz ′

}+ εK

0∆(z ′)z ′γD(z ′) · exp

z′z ′′γB(z ′′)dz ′′

}dz ′

Can consider two limits K � 1 and K � 1

We will assume ∆0 = B0 = 0 for a moment

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

Out of Equilibrium Decay

Produced baryon asymmetry: Bf := B(∞) = εX (0) = εg∗

Limit 1: K � 1

0 20 40 60 80 100

Baryon

B/ε g ∗

Time t

Baryon asymmetry in Out of Equilibrium Decay scenario

K = 0.5K = 0.1

K = 0.01

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

Bf ∝ε

a · exp{−zf − K

∫ ∞zf

γBz dz}

with zf from 1 = KγB(zf )zfand a := −∂z (KγBz)|zf

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

Bf ∝ε

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

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

Limit 2: K � 1

102 103 104 105

Baryon

B f/ε g ∗10−

Asymptotic baryon asymmetry for 1 . K . KC

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

Limit 2: K � 1

105 106 107 108 109

Baryon

B f/ε g ∗10−

Asymptotic baryon asymmetry for KC & K

Numerical Integration

Kolb/Turner, The Early Universe

Further Models

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

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

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

baryogenesis - seminar on non-accelerator particle · pdf fileandantibaryons(antimatter)...

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