Simplest Dark Energy Model: The Cosmological Constant

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Simplest Dark Energy Model The Cosmological Constant Johar M. Ashfaque

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Einstein’s Equations: Variational Principles S=

1 16πG

Z

√ d4 x −g(R − 2Λ) + Sm

Then 0

= =

δS # " Z √ √ √ 1 µν µν µν 4 d x δ −g g Rµν − 2Λ + −gδg Rµν + −gg δRµν + δSm 16πG

Making use of the Palatini identity α δRµν = (δΓα µν );α − (δΓµα );ν

we see that g

µν

δRµν =

g

µν

δΓα µν

−g

µα

δΓνµν

;α

and therefore by Gauss’ theorem Z

√ d4 x −gg µν δRµν = 0.

Now we have 0

= δS # " Z √ √ 1 µν µν 4 = d x δ −g g Rµν − 2Λ + −gδg Rµν + δSm . 16πG

Using the fact that

δS

√ 1√ δ −g = − −ggµν δg µν 2

" # Z √ 1√ 1 4 µν µν µν d x − −ggµν δg (g Rµν − 2Λ) + −gδg Rµν + δSm = 16πG 2 √ 1 ⇒ −g Rµν − Rgµν + Λgµν δg µν + δSm 2

But δSm

1 =− 2

Z

√ d4 x −gTµν δg µν

which leads to the desired result 1 Rµν − Rgµν + Λgµν = 8πGTµν . 2

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2

The Perfect Fluid

The energy-momentum tensor of a perfect fluid takes the form T 00 T

ij

= ρ = pgij

or more precisely 

T µν

ρ 0  0 −p =  0 0 0 0

0 0 −p 0

 0 0  . 0  −p

The fully covariant expression for the components of the energy-momentum tensor of a perfect fluid in an arbitrary coordinate system reads p µν T = ρ + 2 uµ uν − pg µν . c Note. The energy-momentum tensor T µν is symmetric and is made up from two scalar fields ρ and p and the vector field u that characterise the perfect fluid.

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FRW Elements

The cosmological principle states that any particular time, the universe looks the same from all positions in space and all the directions in space at any point are equivalent.

3.1

Maximally Symmetric 3-Space

A maximally symmetric space is specified by just one number, the curvature κ which is independent of the coordinates. Such a constant curvature space is clearly homogeneous and isotropic. The curvature tensor Rijkl has six independent components defined as Rijkl = κ(gik gjl − gil gjk )

3.2

The FRW Metric

The Ricci tensor is Rjk

= g il Rijkl = −2κgjk

with the Ricci scalar −2κg jk gjk = −2κ · (3) = −6κ. Consider the line element dσ 2 = B(r)dr2 + r2 dθ2 + r2 sin2 θdφ2 where B(r) is an arbitrary function of r and gθθ = r2 .

grr = B(r), Then given

Rij = −2κgij and Rrr = −

1 dB(r) , rB(r) dr

Rθθ = 2

1 r dB(r) − −1 2 B(r) 2B (r) dr

we have

1 dB(r) 1 = 2κB(r) ⇒ B = rB(r) dr C − κr2

and 1+

r dB(r) 1 r − = 2κr2 ⇒ 1 + (2κr) − C = κr2 ⇒ C = 1, 2B 2 (r) dr B(r) 2

where by making use of the quotient rule we had 2κr dB(r) . = dr (C − κr2 )2 Hence dσ 2 = B(r)dr2 + r2 dθ2 + r2 sin2 θdφ2 ⇒ dσ 2 =

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1 dr2 + r2 dθ2 + r2 sin2 θdφ2 . 1 − κr2

What is Inflation?

Inflation is defined to be an era during which the rate of the scale factor accelerates corresponding the repulsive gravity Inflation ⇔ ä > 0. Inflation can also be described as a rapid expansion where it the meaning is ambiguous. There is an equivalent expression of the condition for inflation that gives it a more physical interpretation d H −1 < 0. dt a

Inflation ⇔ As

H −1 a is the co-moving Hubble length (the most important characteristic scale of the expanding Universe), the condition for inflation is that the co-moving Hubble length is decreasing with time. The condition for inflation can also be written as Inflation ⇔ −

Ḣ < 1. H2

On the usual assumption that H decreases with time, inflation can be seen as a era where H varies slowly on the Hubble time-scale. If this is a strong inequality |Ḣ| H 2 then H is practically constant over many Hubble times and we have almost exponential expansion α ∝ eHt . A universe with H exactly constant is called a de Sitter Universe. These definitions made no assumptions about the theory of gravity. Making contact with Einstein gravity, the condition for inflation can be written as a requirement on the pressure of the cosmic fluid Inflation ⇔ ρ + 3P < 0. As we always assume the energy density ρ to be positive it is necessary for the pressure density P to be negative to satisfy this condition. To have almost exponential inflation P ' −ρ.

4.1

The Inflaton Potential & The Klein-Gordon Equation

The Friedmann equation reads H2 =

8πG ρ 3

H2 =

ρ 3MP2 l

which simplifies to

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after making use of the reduced Planck mass r MP l =

~c 8πG

instead of the Newton’s constant G. Combining Friedmann’s equation and the fluid equation (continuity equation) H2

=

ρ̇ + 3H(ρ + P )

=

ρ 3MP2 l 0

we have the second Friedmann equation 1 (ρ + 3P ) 6MP2 l H2 3P = − 1+ . 2 ρ

Ḣ + H 2

= −

For ρ

=

P

=

1 2 φ̇ + V (φ) 2 1 2 φ̇ − V (φ) 2

Substituting ρ into the Friedmann equation yields 1 1 2 H = φ̇ + V (φ) 3MP2 l 2 2

and taking the time derivative gives 2ḢH =

1 0 φ̇ φ̈ + V φ̇ . 3MP2 l

Substituting ρ and P into the second Friedmann equation, we obtain Ḣ =

−(φ̇2 ) −(ρ + P ) = . 2 2MP l 2MP2 l

Ḣ =

−(ρ + P ) −(φ̇2 ) = . 2 2MP l 2MP2 l

By substitution of

into 2ḢH =

1 0 φ̇ φ̈ + V φ̇ . 3MP2 l

gives us the Klein-Gordon equation φ̈ + 3H φ̇ + V 0 (φ) = 0.

4.2

The Slow-Roll Regime

The potential must dominate over kinetic energy 1 2 φ̇ < V (φ). 2 In the limit of vanishing kinetic energy the acceleration is exponential and never ending. 4

4.2.1

Derivations

We consider

1 2 φ̇ V (φ). 2 Moreover, if the damping the KG equation is large, the acceleration term can be neglected when compared to the friction term Ď†Ěˆ 3H φ̇. This enforces the slow-roll regime where 2

=

8πG 1 2 φ̇ + V (φ) 3 2

Ď†Ěˆ + 3H φ̇ + V 0 (φ)

=

0

H

simplify to H2

8Ď€G V (φ) 3 −V 0 (φ).

'

3H φ̇ '

We now wish to recast these equations. We begin by replacing φ̇ ' −

1 0 V (φ) 3H

in the condition 12 φ̇2 V (φ) to obtain 2 1 V 0 (φ) V 2 3H

1 V 0 (φ)2 V 2 9( 8πG 3 V (φ)) 2 MP2 l V 0 (φ) 1. 6 V (φ)

�⇒ �⇒

Similarly, the condition on acceleration Ď†Ěˆ 3H φ̇ is rewritten as

Ď†Ěˆ = ∂t (φ̇) ' ∂t

V 0 (φ) 3H

=

V 00 (φ) φ̇ 3H

yielding V 00 (φ) φ̇ 3H φ̇ 3H

1 V 00 (φ) 1 9H 2 1 V 00 (φ) 1 3(8πG)V (φ) MP2 l V 00 (φ) 1 3 V (φ)

�⇒ �⇒ �⇒

Defining the potential slow-roll parameters M2 V ≥ P l 6 and ΡV ≥

V 0 (φ) V (φ)

2

MP2 l V 00 (φ) 3 V (φ) 5

the conditions determining the range of validity of the slow-roll regime are simply V 1,

|ΡV | 1.

The parameters constrain the potential to be almost flat such that the inflaton field slowly rolls down the potential. As long as the these parameters are small the space-time inflates (technically speaking the universe will inflate if the condition V 1 is met) and is approximately de Sitter a(t) âˆź eHt .

4.3 4.3.1

The Conditions of Slow-Roll Inflation Summarized The First Condition

Motion of the inflaton is overdamped so that the force V 0 (φ) in Ď†Ěˆ + 3H φ̇ + V 0 (φ) = 0,

Fluid Equation

balances the friction term 3H φ̇ such that φ̇ ' −

1 0 V (φ). 3H

This condition is referred to as the attractor solution. 4.3.2

The Second Condition M2 V ≥ P l 6

V 0 (φ) V (φ)

2 1

which means that φ̇ < V (φ) is well satisfied and H2 '

4.3.3

κ2 V (φ). 3

The Third Condition |ΡV | 1

for ΡV ≥

5

MP2 l V 00 (φ) . 3 V (φ)

The Simplest Dark Energy Model

The simplest dark energy models is the so called cosmological constant invented by Einstein. When the equation of state reads ω=

P = −1 ⇒ P = âˆ’Ď Ď

then Ď Ě‡ + 3H(Ď + (âˆ’Ď )) = 0 that is to say that Ď is a constant. Since H is constant in the flat universe (Îş = 0) the scale factor evolves exponentially a âˆ? exp(Ht).

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The cosmological constant however can not be held responsible for inflation in the early universe because otherwise the accelerated expansion would not end. However, it is possible that the cosmological constant is responsible for dark energy because the current cosmic acceleration might indeed continue without end. Consider some non-interacting quantum field Φ which can be thought of as infinitely many harmonic oscillators. The relativistic energy-momentum relation is p ωk = k2 + m2 and in order to obtain the vacuum contribution the sum over all energies ω is needed. In order to get a result that is finite some ultra-violet cut-off Λ is required.

Z ρvac

Λ

d3 k 1 2 ωk + k2 + m2 3 (2π) 2ωk 2

Λ

d3 k 1 2 2ω (2π)3 2ωk 2 k

Λ

d3 k ω2 (2π)3 2ωk k

Λ

d3 k 1 ωk (2π)3 2

= 0

Z = 0

Z = 0

Z = 0

≈ = =

Λ

d3 k 1 |k| 3 0 (2π) 2 Z Λ 1 dkk 3 4π 2 0 1 Λ4 16π 2

Z

The estimate for ρvac is 120 orders of magnitude higher than the energy density of the our universe. Given that there is supersymmetry at 103 GeV, before the Planck scale MP l this mismatch is reduced to 60 orders of magnitude. To sum up, the cosmological constant • matches observations well • is simple • does not solve either of the problems – “Coincidence” Problem – “Why-Now” Problem

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