observational constraints of a matter-antimatter symmetric milne

Observational constraints of a matter-antimatter

symmetric Milne Universe

Aurélien Benoit-LévyRencontres de Moriond

March 2008

Rencontres de Moriond 2008

Introduction

The composition of the Universe according to concordance model is rather surprising: 95% is unknown!

Three components successively predominant and then completely negligible.

Before accepting such a strange Universe, necessity to consider the possibility of simpler alternatives

The Symmetric Milne Universe: a flat spacetime with no Dark Matter and no Dark Energy

Presentation and motivations

Confrontation to Type Ia SN, BBN, and CMB

2


Presentation of Milne Universe

The Symmetric Milne Universe

A coasting Universe, with linear scale factor, could be an interesting alternative to a strong deceleration at early times and then a phase of exponential expansion. This Universe has a slower evolution at high temperature. Spends about 108 times more time at QGP transition.

Universe very homogeneous above QGP transition. Separation of matter and antimatter during phase transition that creates domains is possible with such a long time available (cf Omnes 70’s)

Equal quantities of matter and antimatter. Antimatter has negative active gravitational mass

Without Dark Energy or Dark Matter

Ωb ≈ 0.3

(a

a

)2

= H20

(ΩM a−3 + ΩRa−4 + Ωka

−2 + ΩΛ)

(a

a

)2

= H20

[ΩM

(a0

a

)3+ ΩR

(a0

a

)4+ Ωk

(a0

a

)2+ ΩΛ

]

(a

a

)2

= H20

(a0

a

)2⇒ a(t) ∝ t

Ωr = 0

ΩM = 0

(= m)

t0 =1

H0= 13, 9× 109 H0 = 70

a(t) ∝ t

η =nb

nγ

p↔ n

× × × ×•Ci−1 Ci Ci+1m

!"d1 !" d2

Ci Ci+1 Ci Ci+1

M(Ci−1) = 0, M(Ci) = m`1− d1

dx

´, M(Ci+1) = m

`1− d2

dx

´

dx

Gravitationally empty Universe at large scales, no acceleration and no decelaration. Scale factor evolves linearly with time:

an in FRW metric implies flat space-time and open space. Compared to usual assumption of flat space.

Ωb ≈ 0.3

(a

a

)2

= H20


−2 + ΩΛ)

(a

a

)2

= H20

[ΩM

(a0

a

)3+ ΩR

(a0

a

)4+ Ωk

(a0

a

)2+ ΩΛ

]

(a

a

)2

= H20

(a0

a

)2⇒ a(t) ∝ t

Ωr = 0

ΩM = 0

(= m)

t0 =1

H0= 13, 9× 109 H0 = 70

t0 =1

H0= 13, 9× 109 years,with H0 = 70 km/s/Mpc

a(t) = t

k = −1

a(t) ∝ t

η =nb

nγ

p↔ n

× ×

× ×•

!" !"d1 d2

d3

d4

#

$

#$

Ci−1,j−1 Ci,j−1

Ci,jCi−1,j

Ωb ≈ 0.3

(a

a

)2

= H20


−2 + ΩΛ)

(a

a

)2

= H20

[ΩM

(a0

a

)3+ ΩR

(a0

a

)4+ Ωk

(a0

a

)2+ ΩΛ

]

(a

a

)2

= H20

(a0

a

)2⇒ a(t) ∝ t

Ωr = 0

ΩM = 0

(= m)

t0 =1

H0= 13, 9× 109 H0 = 70

t0 =1


a(t) = t

k = −1

a(t) ∝ t

η =nb

nγ

p↔ n

× ×

× ×•

!" !"d1 d2

d3

d4

#

$

#$


Ci,jCi−1,j

Milne Universe is the second “natural” universe

3


Motivations

MotivationsSymmetry in Kerr-Newmann geometry under space, mass and charge reversal. Two spaces connected by the ringElementary particles as “black holes”B. Carter 66&68, Arcos & Pereira 04

In the following, antimatter gravitational mass is taken negative. Gravitational repulsion between matter and antimatter.

SNIa observations reveals effective repulsive gravity which is unexplained

High level of fine-tuning in standard model

Removes need for inflation

4


Presentation of Milne Universe

The Symmetric Milne Universe

Age of the Milne Universe is almost exactly the same as the age of ΛCDM Universe

Ωb ≈ 0.3

(a

a

)2

= H20


−2 + ΩΛ)

(a

a

)2

= H20

[ΩM

(a0

a

)3+ ΩR

(a0

a

)4+ Ωk

(a0

a

)2+ ΩΛ

]

(a

a

)2

= H20

(a0

a

)2⇒ a(t) ∝ t

Ωr = 0

ΩM = 0

(= m)

t0 =1

H0= 13, 9× 109 H0 = 70

t0 =1


a(t) ∝ t

η =nb

nγ

p↔ n

× × × ×•Ci−1 Ci Ci+1m

!"d1 !" d2

Ci Ci+1 Ci Ci+1

M(Ci−1) = 0, M(Ci) = m`1− d1

dx

´, M(Ci+1) = m

`1− d2

dx

´

dx

Age of the Universe was a problem for a Einstein-de Sitter model, which was solved by ΛCDM, but is also solved by Milne Universe

χ(z) =∫ a0

a01+z

da

aaz→+∞−→ +∞

χ(z) z→+∞−→ +∞

µ = m∗ −M + α(s− 1)− βc

η ≈ 8× 10−9

m∗, s, c :

M,α, β :

c

µ = −5 + 5 log(

dL(z)1

)

Pk =⟨|ρk|2

⟩k′ , k′ =

√k2

x + k2y + k2

z = k

ρ(x, y, z)

ρ(kx, ky, kz)

(a

a

)2

= H20 a−2.

χ2 =∑ (m∗ −M + α(s− 1)− βc− µth)2

σ2(µ) + σ2int

χ2/dof = 7.29

σ2int :

(k = −1)

a(t) =a(t)a0

mi = mgp = mga = −m miB = mp

B = maB = −m, (m ≥ 0)

As radial coordinate of a z redshift object: , a Universe with linear scale factor has no horizon. There is no need for an inflation scenario.

oldest globular clusters(Chaboyer et al.,98)

5


Time scale of primordial Universe is extremely different !

Milne Universe spends much more time at high temperature than conventional Universe.

BBN duration:Standard BBN ≈ 200 secMilne BBN ≈ 30 years

Age of the Universe at recombinaison:14 Gy/1000 ≈ 14 My (compared to 0.38 My in ΛCDM)

BBN

CMB

4 x106

35

First noticed by Dev et al. 02

6


Big-Bang Nucleosynthesis

7



A. Coc et al., 2004

10-5

-3.5 -3 -2.5 -2 -1.5 -1 -0.5

WMAP( )

( )

[Si/H] or [O/H]

D/H

Upper(1!)

Mean(1!)

Q1009+2956 (Bur98)

PKS 1937-1009 (Tyt96)

HS 0105+1619 (OMe01)

Q2206-199 (Pet01)

Q0347-383 (DOd01)

Q1243+3047 (Kir03)

10-10

10-9

-3.5 -3 -2.5 -2 -1.5 -1 -0.5[Fe/H]

Li/H

Adopted (95% c.l.)(Ryan et al. 2000)

WMAPRyan et al. (1999)Ryan et al. (2000)

Thevenin et al. (2001)Bonifacio et al. (2002)

10-9

10-10

10-5

Predictions and observational status

16 Stellar Nucleosynthesis: 50 years after B2FH

0.22

0.23

0.24

0.25

0.26

WM

AP

!Bh

2M

ass

fracti

on

4He

10-2

10-6

10-5

10-4

3H

e/H

, D

/H

D

3He

10-10

10-9

1 10

7L

i/H

7Li

WM

AP

"!1010

Fig. 11. Abundances of 4He (mass fraction), D, 3He and 7Li (by number relative to H)

as a function of the baryon over photon ratio η (or Ωb·h2.) showing the effect of nuclearuncertainties.

during this period.Another question comes from recent observations of 6Li in the atmosphere of

A. Coc, 2007

8


Big-Bang Nucleosynthesis in Milne Universe

Production of adequate He-4 is possible in coasting BBN. It needs a greater baryon to photon ratio

η ≈ 3 10−10

η ≈ 6 10−10

η ≈ 7× 10−9

n

p= e−Q/T ≈ 1

6

n

p≈ 1

6n

p≈ 1

7

m∗, s, c :

M,α, β :

c

Yp =2(n/p)

1 + (n/p)≈ 0.25

p + n↔ d + γ

µ = m−M = −5 + 5 log(

dL(z)1pc

)

µ = m−M = −5 + 5 log(

f(z, cosmologie)1

)

m−M = −5 + 5 log(

dL(z)1

)

Pk =⟨|ρk|2

⟩k′ , k′ =

√k2

x + k2y + k2

z = k

ρ(x, y, z)

ρ(kx, ky, kz)

(a

a

)2

= H20 a−2.

χ2 =∑ (m∗ −M + α(s− 1)− βc− µth)2

σ2(µ) + σ2int

Studied in Lohiya et al. (astro-ph/9902022) and Kaplinghat et al. (astro-ph/9805114)

Due to slower evolution, weak reactions decouple around 80 keV, maintaining equilibrium between protons and neutrons. Slowly, deuterium is formed and burnt into helium.Neutrons are regenerated to restore equilibrium value


BBN lasts in Milne Universe lasts 30 years instead of 3 minutes

No deuterium left

9

In collaboration with A. Coc (CSNSM/Orsay)


Production of Deuterium and Lithium-6

Annihilation zone between domains of matter and antimatter is regulated by nucleon diffusion:

T ≥ 80 keV, massive annihilation by neutron diffusion

80 keV ≥ T ≥ 5 keV, no more neutrons, annihilation drops by a factor ≈ 104.

5 keV ≥ T ≥ 1 keV: Proton diffusion becomes efficient. Convection toward annihi lat ion zone . Nucleodisrupt ion and photodisintegration produce deuterium and lithium-6 by non-thermal reactions

1 keV > T: annihilation stops due to gravitational repulsion

Domain size: around 1015 m @ 1 keV (Mpc scale comoving)Deuterium production at the level of 10-5

BBN in presence of small-scale domains of antimatter studied in Jedamzik et al. 2001& Kurki Suonio et al. 2000


10


Type Ia Supernovae

11

Type Ia SN

SNLS Collaboration

(Astier et al. 06)


Einstein-de Sitter model seems totally ruled out

Milne Universe (blue) is very close to ΛCDM (green) as noted in Perlmutter et al. 99

Hubble diagram of Type Ia Supernovae

η ≈ 3 10−10

η ≈ 6 10−10

η ≈ 8× 10−9

n

p= e−Q/T ≈ 1

6n

p≈ 1

7

m∗, s, c :

M,α, β :

c

Yp =2(n/p)

1 + (n/p)≈ 0.25%

p + n↔ d + γ

µ = m−M = −5 + 5 log(

dL(z)1pc

)

µ = m−M = −5 + 5 log(

f(z, cosmologie)1

)

m−M = −5 + 5 log(

dL(z)1

)

Pk =⟨|ρk|2

⟩k′ , k′ =

√k2

x + k2y + k2

z = k

ρ(x, y, z)

ρ(kx, ky, kz)

(a

a

)2

= H20 a−2.

χ2 =∑ (m∗ −M + α(s− 1)− βc− µth)2

σ2(µ) + σ2int

χ2/dof = 7.29

Unknown parameter

12

Absolute magnitude parameter M unconstrained for Milne

Type Ia SN

Which one is Milne ? Which one is ΛCDM ?


Residuals of Hubble diagram for the two models

13

Res

idua

ls

Res

idua

lsredshift z redshift z

LCDM Best fit Milne - our analysis

Type Ia SN test most probably does not allow to exclude the Milne model !

Type Ia SN


Residues of Hubble diagram for the two modelsAbsolute magnitude parameter M unconstrained for Milne

14

Res

idua

ls

Res

idua



CMB

15


CMB

Position of the first acoustic peak in Milne Universe

In Milne Universe, spacetime is flat. Therefore space is hyperbolic, angular distance is drastically changed. An object in the sky will be seen with a much smaller angle than in standard cosmology

× ×

× ×•

!" !"d1 d2

d3

d4

#

$

#$


Ci,jCi−1,j

GM

c2R

ds2 = c2dt2 − a(t)2[

dr2

1− kr2+ r2

(dθ2 + sin2 θdφ2

)]

Gµν = Rµν −12Rgµν = 8πGTµν

cs =c√

3(1 + R), R =

3ρb

4ργ

∆θEdS

∆θMilne=

sinh(ln(1 + z))

2(1− 1√

1+z

) =z=1100

% 284

∆θ

∆θ=

2(1 = 1√

1+z

)

sinh(ln(1 + z))=

z=1100% 0.00352# 1.

θΛCDM

θMilne=

dMilneA (z)

dΛCDMA (z)

=z=1100

% 173

χ(z) =∫ a0

a01+z

da

aaz→+∞−→ +∞

dL = f(z, cosmologie)

χ(z) z→+∞−→ +∞

µ = m∗ −M + α(s− 1)− βc

rs =∫ trec

0csdη =

∫ trec

0cs

dt

a(t)

Angular scale of first peak corresponds to the angle under which is seen sound horizon at decoupling

rs =∫ ηrec

0csdη =

∫ trec

0cs

dt

a(t)

θ = rs/dA

θMilne ≈ 1.2

θMilne ≈ 1.2

rs =∫ trec

0cs

dt

a(t)

rs =∫ ηrec

η0

csdt

a(t)

rs =∫ trec

t170MeV

csdt

a(t)

θMilne

θΛCDM= 2.05

Ωb

Ωb ≈ 4 10−2

η ≈ 3 10−10

η ≈ 6 10−10

η ≈ 7× 10−9

n

p= e−Q/T ≈ 1

6

n

p≈ 1

6n

p≈ 1

7

m∗, s, c :

M,α, β :

c

Yp =2(n/p)

1 + (n/p)≈ 0.25

Sound generation during QGP transition, caused by matter-antimatter annihilation

rs =∫ ηrec

0csdη =

∫ trec

0cs

dt

a(t)

θ = rs/dA

θMilne ≈ 1.2

θMilne ≈ 1.2

rs =∫ trec

0cs

dt

a(t)

rs =∫ ηrec

η0

csdt

a(t)

rs =∫ trec

t170MeV

csdt

a(t)

θMilne

θΛCDM= 2.05

Ωb

Ωb ≈ 4 10−2

η ≈ 3 10−10

η ≈ 6 10−10

η ≈ 7× 10−9

n

p= e−Q/T ≈ 1

6

n

p≈ 1

6n

p≈ 1

7

m∗, s, c :

M,α, β :

c

Yp =2(n/p)

1 + (n/p)≈ 0.25

Finally, we obtain . One degree scale, just like the observed scale !

Angular distance

Sound horizon

16


Preliminary conclusions

A symmetric matter-antimatter Milne universe requires neither inflation, nor Dark Energy, nor Dark Matter and is in surprisingly good agreement with main cosmological tests

BBN: good agreement for helium-4, realistic mechanism for deuterium and lithium-6 production at the observed levelsType Ia SN: Milne Universe very hard to distinguish from ΛCDM with current data

CMB: degree scale for first peak for Milne Universe !

Thank you

Still, a number of questions remain: - angular spectrum of temperature fluctuations (CMB),- is it possible to hide so many baryons ? (molecular gas ?)- consistency with other cosmological probes

Work under the direction of G. ChardinIn collaboration with A. Coc, E. Keihanen, J-J. Aly

17


BACKUP SLIDES

18


Two characteristic scales in Milne Universe- Sound horizon- Size of domains

BAO

Future experiments will be decisive to conclude on BAO.

Eisenstein et al. 05

19

Residues of Hubble diagram for the two models in SNLS analysis

LCDM Best fit Milne - SNLS analysis

Type Ia SN

Rencontres de Moriond 2008 20

Res

idua

ls

Res

idua


LCDM Best fit Milne - our analysis

Type Ia SN test most probably does not allow to exclude the Milne model !

Type Ia SN


Residues of Hubble diagram for the two modelsAbsolute magnitude parameter M unconstrained for Milne

21

Res

idua

ls

Res

idua



Type Ia SN

Milne ΛCDM

Without low redshift SN, Milne fit is slightly better !

22

Res

idua

ls

Res

idua


observational constraints of a matter-antimatter symmetric milne

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