Motivation Large cosmological observational findings: High orbital speeds of galaxies in clusters.( F.Zwicky, 1933) High orbital speeds of stars in spiral galaxies. ( Vera Rubin, at the end of 1960es ) Accelerated expansion of the Universe. ( 1998 )
Problem solving approaches There are two problem solving approaches: Dark matter and energy Modification of Einstein theory of gravity R µν 1 Rg µν = 8πGT µν Λg µν, c = 1 where T µν is stress-energy tensor, g µν are the elements of the metric tensor, R µν is Ricci tensor and R is scalar curvature of metric.
Dark matter and energy If Einstein theory of gravity can be applied to the whole Universe then the Universe contains about 4.9% of ordinary matter, 6.8% of dark matter and 68.3% of dark energy. It means that 95.1% of total matter, or energy, represents dark side of the Universe, which nature is unknown. Dark matter is responsible for orbital speeds in galaxies, and dark energy is responsible for accelerated expansion of the Universe.
Modification of Einstein theory of gravity Motivation for modification of Einstein theory of gravity The validity of General Relativity on cosmological scale is not confirmed. Dark matter and dark energy are not yet detected in the laboratory experiments. Another cosmological problem is related to the Big Bang singularity. Namely, under rather general conditions, general relativity yields cosmological solutions with zero size of the universe at its beginning, what means an infinite matter density. Note that when physical theory contains singularity, it is not valid in the vicinity of singularity and must be appropriately modified.
Approaches to modification of Einstein theory of gravity There are different approaches to modification of Einstein theory of gravity. Einstein General Theory of Relativity R From action S = ( 16πG L m Λ) gd 4 x using variational methods we get field equations R µν 1 Rg µν = 8πGT µν Λg µν, c = 1 where T µν is stress-energy tensor, g µν are the elements of the metric tensor, R µν is Ricci tensor and R is scalar curvature of metric. Currently there are mainly two approaches: f(r) Modified Gravity Gravity
Modified Gravity gravity is a modification of Einstein general relativity in such way that Einstein-Hilbert action contains a function f(, R). Our action is given by S = d 4 x ( R Λ g 16πG + Rp F( )R q) where = 1 g µ gg µν ν, F( ) = constant. f n n and C is a n=0
Also we assume that the variation of the metric coefficients and their first derivatives vanish on the border of the manyfold M, i.e. δg µν M = 0, δ λ g µν M = 0. Equations of motion Let us introduce the following axillary actions S 0 = S 1 = (R Λ) g d 4 x, R p F( )R q g d 4 x. Then the variation of the action S is δs = 1 16πG δs 0 + δs 1.
Variation of action S 0 Lemma On any manfold M we have g µν δr µν g d 4 x = 0. Lemma Variation of the action S 0 is δs 0 = G µν gδg µν d 4 x + Λ g µν gδg µν d 4 x,
Variation of the action S 1 Lemma For any scalar function h(r) the following equation holds hδr g d 4 x = (hr µν + g µν h µ ν h) δg µν g d 4 x. Proof. It can be shown that δr = R µν δg µν + g µν δg µν µ ν δg µν. Odavde sledi da za bilo koju skalarnu funkciju h vazhi Moreover, for any scalar function h(r) we have the following identity hδr g d 4 x = (hr µν δg µν + hg µν δg µν h µ ν δg µν ) g d 4 x. Last two terms can be transformed using the following two identities
hg µν δg µν g d 4 x = h µ ν δg µν g d 4 x = g µν hδg µν g d 4 x, µ ν h δg µν g d 4 x. To prove the first identity we use Stokes theorem in the following way hg µν δg µν g d 4 x = hg µν α α δg µν g d 4 x = α (hg µν ) α δg µν g d 4 x = α α (hg µν )δg µν g d 4 x = g µν α α h δg µν g d 4 x = g µν h δg µν g d 4 x.
To obtain the second identity, we introduce the following vector N µ = h ν δg µν ν hδg µν. From the above expression, one obtains µ N µ = µ (h ν δg µν ν hδg µν ) = µ h ν δg µν + h µ ν δg µν µ ν h δg µν ν h µ δg µν = h µ ν δg µν µ ν h δg µν. Integrating divergence µ N µ yields µ N µ g d 4 x = M N µ n µ d M = 0, where n µ is a unit normal. As N µ vanishes on the border M, we conclude that the last term is zero, which completes the proof.
Variation of the action S 1 Lemma Let θ(r) and ψ(r) scalar functions such that δψ M = 0. Then, it follows θδ ψ g d 4 x = 1 g αβ α θ β ψg µν δg µν g d 4 x µ θ ν ψδg µν g d 4 x + θ δψ g d 4 x + 1 g µν θ ψδg µν g d 4 x.
Variation of the action S 1 Proof. θδ ψ g d 4 x = θ α δ( gg αβ β ψ) d 4 x ( ) 1 + θδ α ( gg αβ β ψ) g d 4 x g = α (θδ( gg αβ β ψ)) d 4 x α θ δ( gg αβ β ψ) d 4 x + 1 θg µν ψδg µν g d 4 x. After a short calculation one obtains α (θδ( gg αβ β ψ)) d 4 x = 0.
Then, it follows θδ ψ g d 4 x = g αβ α θ β ψδ( g) d 4 x α θ β ψδg αβ g d 4 x g αβ g α θ β δψ d 4 x + 1 θg µν ψδg µν g d 4 x = 1 g αβ α θ β ψg µν δg µν g d 4 x µ θ ν ψδg µν g d 4 x β (g αβ g α θ δψ) d 4 x + β (g αβ g α θ) δψ d 4 x + 1 g µν θ ψδg µν g d 4 x = 1 g αβ α θ β ψg µν δg µν g d 4 x µ θ ν ψδg µν g d 4 x + θ δψ g d 4 x + 1 g µν θ ψδg µν g d 4 x.
At the end we obtain θδ ψ g d 4 x = 1 g αβ α θ β ψg µν δg µν g d 4 x µ θ ν ψδg µν g d 4 x + θ δψ g d 4 x + 1 g µν θ ψδg µν g d 4 x.
Variation of the action S 1 At the end we can prove the following lemma Lemma Variation of the naction S 1 is given by δs 1 = 1 g µν R p F( )R q δg µν g d 4 x + (R µν W K µν W ) δg µν g d 4 x + 1 f n n 1 n=1 l=0 ( ( g µν α l R p α n 1 l R q + l R p n l R q) µ l R p ν n 1 l R q) δg µν g d 4 x, Where, K µν = µ ν g µν, W = pr p 1 F( )R q + qr q 1 F( )R p, and denotes derivation over R.
Variation of the action S and EOM Theorem Variation of the action S vanishes iff G µν + Λg µν + 1 16πG g µνr p F( )R q + (R µν W K µν W ) + 1 n 1 ( f n gµν g αβ α l R p β n 1 l R q n=1 l=0 µ l R p ν n 1 l R q + g µν l R p n l R q) = 0, (1)
Trace and 00-component of EOM Theorem Assume that we use FRW metric. Then there are two linearly independent equations. The most convinent choice is: 4Λ R ( 16πG + C R P F( )R q + (RW + 3 W ) n 1 ( + f n µ l R p µ n 1 l R q + l R p n l R q)) = 0, n=1 l=0 G 00 + Λg ( 00 + C 1 16πG g 00R p F( )R q + (R 00 W K 00 W ) + 1 n 1 ( f n g00 g αβ α l R p β n 1 l R q n=1 l=0 0 l R p 0 n 1 l R q + g 00 l R p n l R q)) = 0.
FLRW metric In case of Friedmann-Lemaître-Robertson-Walker (FLRW) metric ds = dt + a (t) ( dr 1 kr + r dθ + r sin θdφ ), k { 1, 0, 1}. there are two linearly independent equations 4Λ R 16πG Rp F( )R q + (RW + 3 W ) n 1 ( + f n µ l R p µ n 1 l R q + l R p n l R q) = 0, n=1 l=0 G 00 + Λg 00 1 16πG g 00R p F( )R q + (R 00 W K 00 W ) + 1 n 1 ( f n g00 g αβ α l R p β n 1 l R q n=1 l=0 0 l R p 0 n 1 l R q + g 00 l R p n l R q) = 0.