Gravitational waves, inflation and fate of the Universepaltani/Courses/Cosmo_gw.pdf ·...

Gravitational waves Problems of the Big Bang Inflation Future of the Universe

Gravitational waves, inflation and fate of theUniverse

Stéphane Paltani

[email protected]

Département d’astronomie, Université de Genève

Cosmologie I


Outline

Gravitational wavesGeneral relativity in the weak-field regimeWave equationObservations of gravitational waves

Problems of the Big BangThe baryon problemsThe flatness problemThe horizon problemThe magnetic monopole problem

Inflation

Future of the Universe


Weak-field metric (i)

In the weak-field regime, the metric gµν can be written:

gµν = ηµν + hµν

where ηµν is the Minkowski metric and |hµν | 1 is a (small)perturbation

We also request that first- and second-order derivatives arealso small. However, hµν can be (and will be) time variable

Note: such perturbation approach can be done around othermetrics than Minkowski’s; e.g., FLRW metric


Weak-field metric (ii)

The double contravariant metric gµν can be derived byrequiring that gµσgσν = δµν . We get (to 1st-order):

gµν = ηµν − hµν

We can also approximate the raising and lowering of indicesusing η instead of g:

hµν = gµσhσν = (ηµσ − hµσ)hσν = ηµσhσν

Note: h is a pseudo-tensor that does not follow rules ofchanges with all coordinate systems


Linearized Einstein equation (i)The Einstein equation:

Rµν −12

gµνR = −κTµν

The linearized Christoffel symbols are:

Γσµν =12ηρσ(∂νhρµ + ∂µhρν − ∂ρhµν) =

12

(∂νhσµ + ∂µhσν − ∂σhµν)

(where we define ∂µ ≡ ηµν∂ν). To first order again, we obtain:

Rσµνρ = ∂νΓσµρ − ∂ρΓσµν =

= 12(∂ν∂µhσρ + ∂ρ∂

σhµν − ∂ν∂σhµρ − ∂ρ∂µhσν )

The Ricci tensor is then:

Rµν =12

(∂ν∂µh + 2hµν − ∂ν∂ρhρσ − ∂ρ∂µhρν)


Linearized Einstein equation (ii)

We have introduced the D’Alembertian 2, which is (inCartesian coordinates):

2 =1c2

∂2

∂t2 −∂2

∂x2 −∂2

∂y2 −∂2

∂z2

where h = hµµ. The Ricci curvature is:

R = Rµµ = 2h − ∂ρ∂µhµρ

We can then substitute Rµν and R in the Einstein equation.Introducing hµν ≡ hµν − 1

2ηµνh, we obtain:

2hµν + ηµν∂ρ∂σhρσ − ∂ν∂ρhρµ − ∂µ∂ρhρν = −2κTµν


Lorentz gauge (i)

Let’s introduce an arbitrary, but very small, coordinatetransformation ξ:

x ′µ = xµ + ξν(x)

We can easily show that the linear perturbations of h′µνtransform as follows:

h′µν = hµν − ∂µξν − ∂νξµ

If hµν is a solution to the linearized Einstein equation, so is h′µν .This is equivalent to the Gauge invariance in electromagnetism.


Lorentz gauge (ii)

Therefore, we can choose ξν(x); in particular we can chooseξν(x) so that:

2ξµ = ∂ρhµρ

With this choice of the gauge, the Lorentz gauge, the Einsteinequation reduces to:

2hµν = −2κTµν

with the additional gauge condition:

∂µhµν = 0


Solutions in vacuumThe D’Alembertian equation has the form of a wave function. Astraightforward solution is:

hµν = Aµν exp(ikρxρ)

It can be easily seen that any solution must have: kσkσ = 0.This means that k is a null vector; therefore the wavepropagates on a null geodesics (at the speed of light)

The gauge condition is then: Aµνkν = 0

By construction, the system is linear, therefore the full solutionis the sum (integral) of the plane-wave solution over all k

Note however that in reality GR is NOT linear!


Energy loss through GW radiation

Without demonstration, the expression of GW emission for atwo-particle rotating body of size a and mass M is:

dEdt

= − G5c5 (128M2a4Ω6)

where Ω is the angular velocity

If the two bodies rotate with Keplerian velocities, we get:

dEdt

= −25

G4M5

a5

The energy decreases, which corresponds to a decrease in a


The binary pulsar of Hulse and Taylor (i)

PSR 1913+16


The binary pulsar of Hulse and Taylor (ii)

Hulse and Taylor, Nobel Prize in physics 1993


LIGO and VIRGO


LIGO principle


GW150914 (i)


GW150914 (ii)

Deformation of about 4 protons over 4km. Two ∼ 30 M blackholes. 3 M radiated away in 0.5 s, i.e. 10 times the radiatedpower of all galaxies in the observable universe


Outline



Inflation



Baryon asymmetry

The baryonic matter of the whole Universe is made of particles(and not anti-particles). Antimatter would be detectable throughannihilation with matter, so antimatter galaxies can beexcluded. However, the big bang should have created an equalamount of baryonic and antibaryonic matter

The baryon asymmetry problem could be resolved if thereexists a reaction that does not preserve the number of baryons.Such reactions are not observed. This would require anextension to the Standard Model of particle physics


Missing baryons

The latest Planck measurements of baryon density gives

ΩB = 0.049

Counting baryons in galaxies and clusters account only forabout 50–70 % of this fraction


SZ effect of stacked pairs of galaxies

SZ signal is observed in the stacked image. Some 30 % of thebaryons are located in the form of hot plasma in filamentslinking the large-scale structure (de Graaf et al. 2017)


The flatness problem

The standard cosmological model has ΩT ' 1. (it may beexactly 1). The Friedmann equation provides the evolution ofΩT :

ΩT (a)− 1 =ΩT − 1

1− ΩT + ΩΛa2 + ΩM a−1 + ΩRa−2

Obviously, if ΩT = 1 at any time, it will remain 1 at all times.However, this is an unstable equilibrium, and any deviation from1 gets strongly amplified


The flatness problem

Currently, ΩT ' 1 at a level of about 1 %. If ΩT ' 1 was notexactly 0 in the very early Universe, it would have to be 1 at alevel of 10−62, if the Universe expanded by a factor about 1060

since Planck time


The horizon problem

The Universe is very homogeneous. However, the horizon atthe time of recombination is of the order of 400 000 light years(comoving). Volumes more distant than this would have beencausally disconnected. Hence, any temperature differencewould have never been smoothed out


The magnetic monopole problem

The Maxwell equations are asymmetric, but they could bemade symmetric if there exist magnetic charges, the magneticmonopole:

∇ · ~E = ρEε0

∇ · ~B = 0 µ0ρB

∇× ~E = −∂B∂t −

(µ0JB + ∂B

∂t

)∇× ~B = µ0

(JE + ε0

∂E∂t

)Magnetic monopoles are predicted by great unification theory,but they have never been detected


Outline



Inflation



Exponentially increasing universe (i)If the fluid that dominates the Universe has a pressure thatcannot be neglected, the stress-energy tensor is:

Tµν =

ρc2 0 0 00 −p 0 00 0 −p 00 0 0 −p

The first and second Friedmann equation are :

2aa

+a2 + kc2

a2 =8πGc2 p

a2 + kc2

a2 =8πG

3ρ

which we can merge:

a = −8πG6

(ρ+3pc2 )a


Exponentially increasing universe (ii)

The basis of inflation is the condition:

a > 0

Using the equation of state p = wρ, this is formally satisfied if:

w < −13

The cosmological constant has w = −1. We get a solution:

a ∼ eHt

Formally, we are at the beginning of an inflation period!


Scalar fieldLet’s assume the Universe contains a scalar field φ with apotential V (φ). The stress-energy tensor must be:

Tµν = (∂µφ)(∂νφ)− gµν

[12

(∂σφ)∂σφ)− V (φ)

]But we also have:

Tµν = (ρ+ p)uµuν − pgµν

By equating the terms, we obtain:

ρφ = 12 φ

2 + V (φ) + 12

(~∇φ)2

pφ = 12 φ

2 − V (φ)− 16

(~∇φ)2

If the field is spatially and temporally constant, we get thecosmological constant and Λ = V (φ)


Continuity equation

If the field has no interaction except through gravity, thecontinuity equation ∇µTµν = 0 is:

ρ+ 3(ρ+ p)aa

= 0

Substituting ρφ and pφ, assuming no spatial variation, we get:

φ+ 3Hφ+dVdφ

= 0

This is the equation of motion in a field with a friction dependingon velocity


Inflationary epoch

The second Friedmann equation, neglecting curvature, is:

a2 =8πG

3ρa2

Therefore:

H2 ≡(

aa

)2

=8πG

3

[12φ2 + V (φ)

]The two differential equations determine H and φ

Inflation occurs if :φ2 < V (φ)


Slow-roll approximation

We assume φ2 V (φ), yielding the simplifications:

3Hφ = −dVdφ≡ −V ′ H2 =

8πG3

V (φ)

The inflation condition is equivalent to:(

V ′

V

) 1


Cosmological inflation


Inflaton field fluctuations

The inflaton scalar field is subject to microscopic quantumfluctuations

Inflation causes these fluctuations to become macroscopic,giving rise to the CMB fluctuations, and ultimately to thelarge-scale structure


Inflation and the flatness problem

Through inflation, causally connected regions expand waybeyond the observable horizon

About 60–70 e-folding of the Universe are needed


Inflation and the horizon problem

The Friedmann equation can be rewritten as:(Ω−1

T − 1)ρa2 = −3kc2

8πG

The right-hand side is constant. If a ∼ eλt and ρ ∼ const., then|Ω−1

T − 1| must decrease, which means ΩT → 1 at the end ofinflation

About 60–70 e-folding of the Universe are needed


Inflation and the magnetic monopole problem

Magnetic monopoles are presumably very heavy particles,which were created when the Universe was extremely hot

If magnetic monopoles were created before inflation, theexpansion of the Universe would have diluted the monopolesaway, so that extremely few remain in the observable Universe


Outline



Inflation



Evolution of the Universe


Big Crunch


Open Universes


Big Freeze

Age of the Universe (in logarithms of years)


Exponential expansion of the future Universe

• Under the effect of dark energy, the growth of the Universeis exponential if it overcomes ΩM

• If −1 ≤ w < −13 , galaxies that are not bound are separated

by the expansion of the Universe, and disappear from ourobservable Universe

• In 2000 billion years, the Local Group will be the onlyobservable structure of the Universe


Big Rip

• However, if w < −1, the energy density of dark energyincreases, and at some point dark energy will overcomeany other force; this is the “Big Rip”

• If H0 = 70 km s−1 Mpc−1, ΩM = 0, ΩDE = 0.7, butw = −1.5, the Big Rip will happen in 22 billion years

• 60 million yr before BR: stars leave the Galaxy• 3 months before BR: planets leave the Solar System• A few minutes before BR: Earth is torn apart• Just before BR: Nuclei are destroyed• At BR: a reaches infinity

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