Cosmology. April 13, 2015

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1 Cosmology April 3, 205 The cosmological principle Cosmology is based on the principle that on large scales, space (not spacetime) is homogeneous and isotropic that there is no preferred location or direction in the cosmos. This is strongly supported by the data. The principle means that we may write the spatial part of the spacetime metric as a time-dependent multiple of a constant-curvature space. We have found the curvature tensor for such maximally symmetric spaces. Combining this with a time-dependent factor, a (t), and the time part of the metric, the line element becomes ) ds 2 = dt 2 + a (δ 2 κ ij + κr 2 x ix j dx i dx j dt 2 + a 2 h ij dx i dx j where κ = ±, i, j =, 2, 3, and a = a (t) sets the cosmic distance scale at any given time and h ij is our constant curvature 3-metric. The connection is easily found, Γ 0ij = Γ i0j = Γ ij0 = aȧh ij Γ ijk = a 2 Γijk where Γ ijk is the connection arising purely from the maximally symmetric metric, h ij. Therefore, Γ i 0j = Γ i j0 = ȧ a δi j Γ 0 ij = aȧh ij Γ i jk = Γ i jk The curvature can also be written in terms of maximally symmetric parts, and parts depending on a (t). R i jkl = Γ i jl,k Γ i jk,l Γ i blγ b jk + Γ i bkγ b jl = R i jkl Γ i 0lΓ 0 jk + Γ i 0kΓ 0 jl = R i jkl + ȧ 2 ( δ i kh jl δ i lh jk ) and replacing R jkl i with the expression for the maximally symmetric curvature, Rjkl i = ( 2) ( δkh i jl δlh i ) jk Next, consider R 0 jkl = Γ 0 jl,k Γ 0 jk,l Γ 0 blγ b jk + Γ 0 bkγ b jl

2 This must be proportional to some rank-3 tensor in the maximally symmetric space, but there is none so we expect these components to vanish. Indeed, we find Rjkl 0 = aȧh jl,k aȧh jk,l aȧh ml Γm jk + aȧh ml Γm jk ) = aȧ (h jl,k h ml Γm jk h jm Γm lk h jk,l + h ml Γm jk + h jm Γm kl +h jm Γm lk h jm Γm kl = aȧ (h jl;k h jk;l ) = 0 where the derivatives of h ij in the penultimate step are with respect to the maximally symmetric connection. Since h ij is the metric compatible with this connection, the derivatives vanish. The final components are Collecting terms, we have and terms related to these by symmetry. The Ricci tensor follows immediately, R 0 j0l = Γ 0 jl,0 Γ 0 j0,l Γ 0 blγ b j0 + Γ 0 b0γ b jl = Γ 0 jl,0 Γ 0 blγ b j0 = ( aä + ȧ 2) h jl aȧh ml ȧ a δm j = aäh jl R 0 j0l = aäh jl R i jkl = ( 2) ( δ i kh jl δ i lh jk ) R 00 = R i 0i0 = a 2 hij R 0 j0i = a 2 hij aäh ji and = 3ä a R ij = Ri0j 0 + Rimj m = aäh ij + ( 2) (δmh m ij δi m h mj ) = ( aä + 2κ + 2ȧ 2) h ij and the Ricci scalar is R = g 00 R 00 + a 2 hij R ij = 3ä a + 3 a 2 ( aä + 2κ + 2ȧ = 6ä a + 6 a 2 ( 2

3 and Finally, we have the components of the Einstein tensor, G ij = R i 2 g ijr 2 The energy tensor G 00 = R 00 2 g 00R = 3ä a + 3ä a + 3 a 2 ( = 3 a 2 ( = ( aä + 2κ + 2ȧ 2) h ij 2 a2 h ij ( 6ä a + 6 a 2 ( ) = ( aä + 2κ + 2ȧ 2 ä 3 ( 2)) h ij = ( 2aä + 2) h ij Now consider possible energy tensors to serve as a source. At the current epoch, there is very little pressure and we may take the energy tensor for pressureless dust, ρ T µν = with its conservation equation, 0 = T µν ;ν = T µν,ν + T βν Γ µ βν + T µβ Γ ν βν The vanishing divergence is an identity for all but the time component, which becomes 0 = T 0ν,ν + T βν Γ 0 βν + T 0β Γ ν βν = T 00,0 + T 00 Γ T 00 Γ ν 0ν = ρ + T 00 Γ i 0i = ρ + ρ 3ȧ a Multiplying by a 3 we have 0 = a 3 ρ + 2 ȧρ = d ( ρa 3 ) dt so that ρa 3 remains constant. The meaning of this is easy to see. Consider a unit spatial volume, V 0 = l 3 0 =. The comoving lengths of the sides will expand with a factor of a, l (t) = a (t) so that the volume at any time is given by The mass contained in this volume is V (t) = a 3 m ρv = ρa 3 = constant 3

4 3 From the Einstein equation to the Friedmann equation We may now write the Einstein equation, including a possible cosmological constant, Λ, G αβ + Λg αβ = βt αβ where β = 8πG c. Each term in the equation is diagonal. 4 For the 00 component, while the spatial components give a second equation, G 00 + Λg 00 = βt 00 3 ( a 2 Λ = a 3 G ij + Λg ij = 0 ( 2aä + 2) h ij + Λa 2 h ij = 0 Notice that if we multiply the 00 equation by a 2 to get 3 ( 2) Λa 2 = a then differentiate with respect to time, 6ȧä aȧ = ȧ = a a 2 ( 3 ( Λa 2) ȧ where we substitute the original expression for a. Cancelling a common factor of ȧ and simplifying, so finally, 6ä 3Λa = 3 ( a 2aä Λa 2 = ( 2) 2aä Λa = 0 and we have derived the spatial equation from the 00 component. This is consistent with the requirements of the Bianchi identity. We have therefore reduced this cosmological model to the single, first-order Friedmann equation, ( 2) Λa 3 = 0 Solutions to this govern low energy cosmology. At early times, when the energies and pressures are substantial, this requires modification. 4 Curvature singularity The metric appears to be degenerate if a = 0 or if a diverges. The first of these is also a curvature singularity. We may confirm this by substituting the field equations into the expression for the scalar curvature, R = 6ä a + 6 a 2 ( 4

5 To evaluate this, start with the field equations, Solving the first for ä shows that 2aä + 2 Λa 2 = 0 3 ( a 2 Λ a 3 = 0 6ä a = 6 a ( 2 Λa 2a while the second immediately gives Combining in the expression for R, = 3 a 2 ( + 3Λ 3 ( a 2 = Λ + a 3 R = 6ä a + 6 a 2 ( = 3 a 2 ( + 3Λ + 6 a 2 ( = 3Λ + 3 a 2 ( = 4Λ + a 3 which diverges when a = 0. It is not hard to show that other curvature invariants also diverge at a = 0, and nowhere else. Properties of the Friedmann equation We now examine the Friedmann equation, ( 2) Λa 3 = 0 Solving for ȧ, Λa3 + 3κa ȧ = ± The rate of change of a is therefore divergent at a = 0, and when a diverges. The exact solution is given by integrating ˆa(t) da ± (t t 0 ) = Λa3 + 3κa a(t 0) but this is not necessary to see the qualitative properties of the possible solutions. Instead, for various signs of Λ and κ, we treat the problem as a single particle in a potential. We need to consider 4 cases, depending on the signs of Λ and κ. We may always take a > 0. 5

6 Case : Positive spatial curvature and positive cosmological constant When κ = and Λ > 0, If we write the Friedmann equation as ȧ = ± 3 Λa2 + = κ = ȧ 2 3 Λa2 we have conservation of the energy, κ. The effective potential is strongly attractive at all values of a: with a single maximum at ( (There are three roots: potential at this maximum is ) /3 ( ) /3, 2πi e V eff = 3 Λa2 0 = dv eff da = 2 Λa ( ) /3 a crit = 3, ( V eff (a crit ) = ( ) 2/3 3 Λ ( ) /3 3 = ( ) /3 ( ( ) ) Λ + 3 ( ( ) ) 3 /3 = 2 ( ) /3 = 4 β2 m 2 Λ ) /3 e 4πi 3 only one of which is real). The value of the The evolution of the system depends on how this value compares to the total energy, κ =. If 4 β2 m 2 Λ < then the energy exceeds the maximum of the potential and a increases or decreases monotonically to infinity or zero, depending on the sign of its initial condition, ȧ (t 0 ). On the other hand, if there is a turning point when V eff =, 4 β2 m 2 Λ > = 3 Λa2 turning turning 0 = 3 Λa3 turning a turning + 3 6

7 If the initial value, a (t 0 ), is below a crit, the universe expands to this turning point, then contracts to the singularity. If a (t 0 ) exceeds a crit, then the universe contracts to the turning point then expands forever. The complete solution is given by integrating, ˆt ± dt = t 0 ± (t t 0 ) = ˆa a(t 0) ˆa da 3 Λa2 + da Λa3 + a(t 0) Case 2: Positive spatial curvature and negative cosmological constant When κ = and Λ < 0, the energy equation takes the form and the effective potential is This has no extrema, = ȧ Λ a2 V eff = 3 Λ a2 since 0 = dv eff da = 2 3 Λ a + 2 a 3 = 2 Λ has only one negative and two complex roots. The potential becomes infinitely attractive for a small, and increases without bound for large a, with a single inflection point at 0 = d2 V eff da 2 = 2 2 Λ 3 3 ( ) /3 a = Λ The universe expands to a maximum at the single real positive root of then collapses to a = 0. 0 = 3 Λ a3 + a 3 Case 3: Negative spatial curvature and positive cosmological constant When κ = and the cosmological constant is positive, we have = ȧ 2 3 Λa2 7

8 which again has negative definite potential with a single maximum at ( ) /3 V eff (a crit ) = 4 β2 m 2 Λ Since the energy is now positive, ȧ 2 remains nonzero and there is no turning point. The universe continues to collapse or expand in accordance with the initial sign of ȧ. Case 4: Negative spatial curvature and negative cosmological constant When κ = and the cosmological constant is positive, so we again have the monotonic potential, = ȧ Λ a2 V eff = 3 Λ a2 This always eventually exceeds the energy, so we have a single turning point when 0 = 3 Λ a3 a 3 8