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Derivation of the Friedmann Equations• The universe is homogenous and isotropic • ds2 = -dt2 + a2(t) [ dr2/(1-kr2) + r2 (dθ2 + sinθ dɸ2)]
where k = 1, 0, -1• The energy momentum tensor is that of a perfect fluid and is
given by Tμν = (p + ρ ) Uμ Uν + p gμν , and becomes
Tνμ = diag (- ρ, p , p, p ) at the rest frame.
The conservation of energy; μ Tμ0 = 0 0ρ – 3(å/a)(ρ + p) = 0
For a perfect fluid, ρ and p are related by p = w ρ where w = 0 (matter ) dust , w = 1 radiation ,
w = -1 vacuum energy The energy conservation, [(dρ/dt)/ρ] = -3 (1+w) [(da/dt)/a]
The Einstein Equations, Rμν – (1/2) gμν R = 8π GN Tμν or Rμν = 8π GN ( Tμν - (1/2) gμν T ) results in
Friedmann Equations:
μν = 00 (ä/a) = -(4π GN /3)(ρ + 3p)
μν = ij (å/a)2 = (8π GN /3)ρ – (k/a2)
(å/a) : Hubble parameter
The observation of the acceleration of the expansion of
the universe implies (ä/a) > 0 , that is, most of the
dark energy is in the form of vacuum energy i.e.
cosmological constant
A symmetry for vanishing cosmological constant
Recai Erdem, 2006
The Theoretical Source of the Cosmological Constant
• Einstein field equations : Rμν – (1/2) gμν R = 8GNTμν - gμν Λ( i.e gravitational action: IG = (1/ 8GN)-g d4x (R - Λ) + -g d4x ℒ )
Λ: cosmological constant energy density with the energy-momentum tensor
Tμν = - gμνρ , ρ = Λ /(8GN)
(with positive energy density and negative pressure for Λ>0)ρ can be considered as a vacuum energy densityρ = <ℒ>.
Motivation for studying Cosmological Constant
• A conservative candidate for a dark energy (component) with a repulsive force to explain the accelerated expansion rate of the universe a positive Cosmological Constant
• Any contribution to the vacuum energy (e.g. vacuum polarization in quantum field theory) contributes to the cosmological constant
Cosmological Constant Problems• 1- Why so huge discrepancy between the observed and the
theoretical values of Λ?
(Λtheoretical /Λexperimetal ) ~ 1041 – 10118
Due to zero-point energies in chromodynamics ~ 1041 - Due to zero-point energies in
gravitation at Planck scalle ~ 10118 • 2- Why is Λ so small? ( in other words, is there a
mechanism which sets Λ to zero or almost to zero?) the subject of this talk
• 3- Why is Λ not exactly equal to zero?• 4- Why is the vacuum energy density so close to the
matter density today?• 5- Does Λ vary with time?
Basic schemes to explain “Why is the cosmological constant so small?”
• 1- Symmetries ( supersymmetry, supergravity, superstrings, conformal symmetry, Signature Reversal Symmetry the subject of this talk )
• 2- Anthropic Considerations• 3- Adjustment Mechanisms• 4- Changing Gravity• 5- Quantum Cosmology
A symmetry for vanishing cosmological constant ( Signature Reversal Symmetry)
The First Realization• Require action functional be invariant under
xA i xA , gAB gAB
( A = 0, 1, 2,.....D-1 where D is the dimension of the spacetime )
ds2 = gAB dxAdxB - ds2
( R.Erdem, “A symmetry for vanishing cosmological constant “, Phys.Lett. 621 (2005) 11-17, ArXiv hep-th/0410063
S. Nobbenhuis, “Categorizing different approaches to cosmological constant problems”, ArXiv gr-qc/0411093
G. ‘t Hooft, S. Nobbenhuis, “Invariance under complex transformations and its relevance to the cosmological constant problem”, Clas. Quant. Grav. 23 (2006) 3819-3832, ArXiv qr-qc/0602076 )
.
Under this symmetry transformation
• RAB -RAB , R = gABRAB -R , g g , dDx (i)DdDx
• So IG = ∫ g dDx R remains invariant only if
• D = 2(2n+1) , i.e. D = 2, 6, 10, 14, ....
while
IG = ∫ g dDx Λ is forbidden.
If the symmetry is imposed on the Lagrangian then
• We require ℒ - ℒthen through the kinetic terms we obtain
Φ +/- Φ , Φμ +/- Φμ for scalar and gauge
fields
Fermion kinetic term is not allowed on the bulk it may be allowed only on a lower dimenion with D=2n+1, n=1,2,..
In D=2n+1
Ψ eiΨ constant
So the n-point functions < Φ1 Φ2..... Φn> are invariant • and• D=2n+1 for fermions for any n, and for D=4n+2 for scalars and
gauge fields (if one takes + sign) and for even number of n if one adopts – sign
• Moreover two-point functions (propagators), in any case, respect the symmetry
• So this symmetry is more promising for extension into quantum field theory unlike the usual scale symmetry.
• The straightforward application of the symmetry in its present realization only forbids bulk cosmological constant. In order to forbid brane cosmological constant one must impose an additional symmetry.
• The straightforward application of the symmetry in its this realization only forbids bulk cosmological constant in order to forbid brane cosmological constant (that may be induced by the extra dimensional part of the curvature scalar) one must impose an additional symmetry.
• In order to make the brane cosmological constant vanish one may let, in D=6,
• 1- abondon the singnature reversal so that the extra dimensional piece of the metric is invariant under the symmetry
gabdxadxb gabdxadxb
• 2- let gab - gab as x4(5) x5(4) • Then g44 = - g55 which makes the extra dimensional piece of the
curvature scalar zero for metrics with Poincaré invariance
• In the next realization of the symmetry we make both the bulk and the brane cosmological constants zero by signature reversal symmetry alone.
Second Realization• Require the action be invariant under
gAB - gAB , xA xA that is, ds2 = gAB dxadxB - ds2 ( as in the first realization )
(R. Erdem, “A symmetry for vanishing cosmological constant: Another realization”, Phys. Lett. B ( to be published)M.J. Duff and J. Kalkkinen, “Signature reversal invariance”, ArXiv, hep-th/0605273M.J. Duff and J. Kalkkinen, “Metric and coupling reversal in string theory”, ArXiv, hep-th/0605274 )
thenRAB RAB , R = gABRAB -R , g dDx (-1)D/2 g dDx
• So IG = ∫ g dDx R remains invariant if
D = 2(2n+1) , i.e. D = 2, 6, 10, 14, ....
• This realization of the symmetry as well forbids a bulk cosmological constant provided one assumes that the Einstein-Hilbert action is non-zero.
• This symmetry may be used to forbid a possible contribution of the extra dimensional curvature scalar to the 4-dimensional cosmological constant
by taking the 4-dimensional space-time be in the intersection of a 2(2n+1) and a 2(2m+1) dimensional space ( n,m = 1,2,3,.........)
• The 4-Dimensional Space as the Intersection of two Spaces
2(2n+1)dimensionalspace
2(2m+1) dimensional space
The usual 4-dimensional space
• In this case the symmetry transformation becomes
gAB - gAB ; A,B = 0,1,2,3,4’,5’....D’-1; D’ = 2(2n+1)
gCD - gCD ; C,D = 0,1,2,3,4’’,5’’....D’’-1; D’’ = 2(2m+1)
• Under this transformation the metric (with 4-dimensional Poincaré invariance), the curvature scalars, and the volume element transform as
ds2 = gμν dxμdxν + gab dxadxb gμν dxμdxν - gab dxadxb
R4 R4 , Re -Re , g dDx g dDx
where R4, Re are the 4-dimensional, and the extra dimensional pieces
of the curvature scalar.
The contribution of Re vanishes after integration
so the contribution of Re to the cosmological constant is zero
Realization of the symmetry through reflections
For example one may take
gμν = Ω4ημν(x) , gAB = Ω1eηAB(x) , gCD = Ω2eηCD(x)
where Ω4= Ω1(xA) Ω2(xC) , Ωe1= Ω1(xA) , Ωe2 = Ω2(xC)
Ω1(xA) = cos k1x5’ ; Ω2(xC) = cos k2x6’
Then the reflections in extra dimensions given by
k1x5’ π - k1x5’ ; k2x6’ π - k2x6’
induce the symmetry transformation
gAB - gAB ; A,B = 0,1,2,3,4’,5’....D’-1; D’ = 2(2n+1)
gCD - gCD ; C,D = 0,1,2,3,4’’,5’’....D’’-1; D’’ = 2(2m+1)
If the symmetry is identified as a reflection in extra dimensions then the forbiddance of a term materializes as the vanishing of that term after integration. For example
considerds2 = Ω4gμν(x) dxμdxν + Ωe1gAB dxAdxB + Ωe2gCD dxCdxD
where Ω4= Ω1(xA) Ω2(xC) , Ωe1= Ω1(xA) , Ωe2 = Ω2(xC)
and the metric transforms in accordance with the symmetryThe corresponding volume element and the curvature scalar are
g dDx = Ω12(2n+1) Ω2
2(2m+1) g’ dDx , where g’FG = Ω1-1 Ω2
-1 g FG
R = Ω1-1 Ω2
-1 {R’4 + R’e – (D-1)[g’4’(d2 ln Ω1/(dx4’)2)
+ g’4’’(d2 ln Ω’/(dx4’’)2)] – [(D-1)(D-2)/4][g’4’(d ln Ω1/dx4’)2
+ g’4’’(d ln Ω’/dx4’’)2]
Then only the term, R’4 survives in the action. If we simply consider the signature reversal we say that the other terms are forbidden. However if one identifies it as a reflection then it manifest itself as; the other terms vanish after integration.
Transformation rules for fields
• The invariance of the action under the symmetry requires
• ℒ - ℒThen, as in the first realization,
Φ +/- Φ for scalars
while the gauge fields are allowed in D = 4n
Like the first realization fermions live in
D=2n+1. However an ambiguity is introduced by
the veilbeins. A veilbein may roughly be considered as
the square root of the corresponding metric tensor. So
+/- ambiguity due to the signature reversal
transformation is introduced. To circumvent this
problem we double the dimension of the spinor space.
So if we take gAB = cos u ηAB
then the corresponding vielbein may be taken as
EAK’ = [cos (u/2) τ3 + i sin (u/2) τ1] IA
K’
Where τ3(1) denote sigma matrices and denotes tensor product.
Some similarities with and differences from Linde’s model
• The action for Linde’s model is• S = Sψ - Sφ where
Sψ = N ∫ g(x) g’(y) d4x d4y [(Mpl /16π)2 R(x) + ℒ(ψ(x)]
Sφ = N ∫ g(x) g’(y) d4x d4y [(Mpl /16π)2 R(y) + ℒ(φ(y)]• Under the symmetry
gμν (x) g’μν (y) , ψ(x) φ(y)
If one imposes
S - S
then this symmetry forbids a cosmological constant
This conclusion is somewhat similar to what we
have found in the sense that S - S corresponds to
a vanishing cosmological constant. On the other
hand our scheme does not contain ghosts.
In SummaryBoth realizations of the symmetry are quite similar:• Both impose zero bulk cosmological constant• In the first realization one needs an extra symmetry to
make the possible contribution of the curvature scalar to cosmological constant zero while in the second realization this may be accomplished through the same symmetry provided we put the usual 4-dimensional space to the intersection of two spaces.
• Scalars live in D = 2(2n+1) and fermions in D = 2n+1 for both realizations while the gauge fields live in D =2(2n+1), D = 4n for the first and the second realizations, respectively. The n-point functions are invariant in both.
• The symmetry seems to be promising.