bogusław broda and michał szanecki- dark energy from quantum vacuum fluctuations

8/3/2019 Bogusaw Broda and Micha Szanecki- Dark Energy from Quantum Vacuum Fluctuations

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Dark Energy from Quantum Vacuum Fluctuations

Dark Energy from Quantum Vacuum

Fluctuations

Bogusaw Broda and Micha Szanecki

Department of Theoretical Physics, University of dz

Warszawa, 7.IV.2009

Broda, B., Bronowski, P., Ostrowski, M., & Szanecki, M., Vacuum Driven Accelerated Expansion, Ann. Phys.

(Berlin), 17, 855863, 2008; [arXiv: 0708.0530].

Broda, B. & Szanecki, M., Quantum Vacuum and Accelerated Expansion, CRAL-IPNL conferenceDark

Energy and Dark Matter", Lyon 2008; [arXiv: 0812.4892].

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Introduction

3 problems in modern physics and cosmology

accelerated expansion of the Universe

dark energy

cosmological constant

quantum vacuum energy density

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Introduction

Dozens of candidates for the solution of the problem of the

accelerated expansion.

One of the possibility: quantum vacuum energy. But it does

not work well. Traditionally calculated, Casimir-like value ofquantum vacuum energy density is 2 orders of orders too big

than required!

One should solve the quantum vacuum energy problem

anyway.

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Introduction

Potential resolutions:

lowering the ultraviolet cutoff scale UV using

supersymmetry arguments

quantum vacuum energy does not influence gravity

or zero-point energy does not gravitate in vacuum

Do not work!

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Introduction

But

Possible to reasonably estimate the value of quantum

vacuum energy obtaining an experimentally acceptable

result

Approach does not appeal to any clever, arbitrary or exotic

assumption

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Quantum vacuum energy

Standard estimation of the quantum vacuum energy densityvac for a bosonic mode

vac =1

2

UV

0

4

(2)3

c(mc)2 + k2 k2dk. (1)

For a large UV cutoff UV

vac 1

(4)2UV

4

3c . (2)

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D k E f Q V Fl i


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Setting UV = P (Planck momentum)

P =

c3

G 6.5 kgm/s, (3)

Pvac c5(4)2G2 3.4 1094 kg/m3. (4)

Experimentally estimated value

crit = 3H02/8G( 1026 kg/m3).

More than 120 orders less!

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D k E f Q t V Fl t ti


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Lowering UV to SUSY 1TeV/c yields

SUSYvac 1.5 1030 kg/m3.

Idea that gravitational field is insensitive to quantum vacuum

fluctuations yields

0vac = 0.

Incorrect results. Obvious also from theoretical point of view.

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Casimir-like calculation should not give any contribution to

gravitational (or any other) field by construction: No

contributions coming from closed loops without external lines,

Above, no classical external lines.

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One should consider loops with classical external lines.

Symbolically

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Implementation of the estimation

Full quantum contribution

Seff =

2

log det

D, (5)

+ bosonic statistics

fermionic statistics

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Formally

Seff = 2Tr logD =

2

0

ds

sTr esD. (6)

UV regularized

Seff =

2

ds

sTr esD

2

ds

sTD(s). (7)

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SeeleyDeWitt expansion

TD(s)

t(s; x)

gd4x , (8)

t(s; x) =1

(4s)2

n=0

an(x)sn. (9)

Other an(x) (for n > 0) contain powers and derivatives of

curvature.

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gy


Example:a1(x) =

16 R induces classical HilbertEinstein gravity

Sind = 2

1

1

(4)2 1

6R

gd4x = 112

c3

16GR

gd4x,

(10)

where = LP2 = G

c3.

Induced coupling constant is 12 times less than the

standard classical value!

a2(x) and further an(x) yield quantum corrections.

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gy


Cosmological constant or dark energy induced by a0,

SCas = 2

1

221

(4)2

gd4x

=

4

1

LP4

1

(4)2gd4x

= 14

c6

(4)2G2

gd4x.

(11)

Extract part corresponding to gravitational field!

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Simplification.

Take the flat FRWL metric

g =

1 0 0 0

0 a2(t) 0 0

0 0 a2(t) 00 0 0 a2(t)

. (12)

t = 0 present moment

Normalize

a(0) = 1. (13)

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Expanding around t = 0

a(t) = a(0) + a(0)t + a(0)t2 + O(t3)

= 1 + H0t 12

q0H02t2 + O(t3),

(14)

where

H0 a(0)a(0)

= a(0),

q0 deceleration parameter

q0 H02a1(0)a(0) = H02a(0).

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Hence

g = a

2(t)32

= 1 + 2H0t + (1 q0)H02t2 + O(t3)

32

. (15)

Now discard term linear in t gauge symmetry.

[I. L. Shapiro, Effective action of vacuum: semiclassical approach, Class. Quant.

Grav. 25, 103001 (2008); Arxiv: 0801.0216 [gr-qc], Chapt. 3.1.]

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Proof: Infinitesimal gauge transformations

g = + , (16)

= (

0

, i) gauge parameters.

1st Eq.

g00 = 2 = 0, (17)

g00 = 1 should be unchanged.

General solution: = (x).

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2nd Eq.

g0i gi0 = i + i = 0, (18)

g0i = 0 should be unchanged.

Hence i = i(x), and consequently

i = t i(x) + i(x). (19)

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For spatial indices

gij =ij + ji

= 2t ij(x) + ij(x) + ji(x).(20)

Put

gij =

0, for i= jf(t), for i = j.

(21)

Linear in t function can be gauged away.

Solution

(x) =1

2

3i,j=1

ijxixj, (x) = 0. (22)

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Constant matrix ij, diagonal

ij =12 H0ij.

Solution

=

1

4H0x

2,12

H0txi

. (23)

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Now

g = 1 +3

2 (1 q0)H02

t2

+ O(t3

). (24)

Divide by the spatial volumed3x.

Dividing by time is time averaging.

Analysis perturbative in t

the longer t the smaller the reliability.

Shortest time t = TP (the Planck time).

Averaging t around present moment (t = 0)

t limTTP

1

T

T0

dt ( ). (25)

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I l t ti f th ti ti


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Estimated

= 14

c5

(4)2G2lim

TTP

1

T

T

0

dt (

g 1)

14

c5

(4)2G21

2(1 q0)H02TP2.

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No ad hoc subtraction!

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Concl sions


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Conclusions

TP2

= G/c5

,

H02 =

8

3Gcrit,

1

48

(1

q0)crit. (27)

exp 0.72 crit,

q0 0.7

0.01exp,

per mode.

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Conclusions


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Conclusions

Thanks for your attention

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bogusław broda and michał szanecki- dark energy from quantum vacuum fluctuations

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