Phys 231B Winter 2017Assignment 6 Solutions

1. (a) The Einstein equation is Rab + (Λ− 12R)gab = 8πTab and its trace gives 4Λ−R =

8πT . For dust, Tab = ρuaub and T = −ρ. This reduces the Einstein equation to

Rab = (Λ + 4πρ)gab + 8πρuaub . (1)

Contracting with uaub gives

Rabuaub = −(Λ + 4πρ) + 8πρ = 4πρ− Λ . (2)

Combining this with the result from class, Rabuaub = −3(H + H2), we conclude

that

H +H2 =a

a= −4πρ

3+

Λ

3. (3)

To get the equation for ρ, we use conservation of the stress tensor:

0 = −ub∇aTab

= −ub[ub(ua∇aρ) + ρub(∇au

a) + ρ(ua∇aub)]

= ρ + 3Hρ + 0ρ

ρ= −3

a

a. (4)

Integrating this result then gives

4π

3ρa3 = b , (5)

where b is a constant (the factor of 4π/3 is for convenience). Using (5), we findthat (3) becomes

a = − b

a2+

Λa

3. (6)

If we multiply this by a, the result says dε/dt = 0 where ε is

ε =a2

2− b

a− Λa2

6. (7)

To get the equation for 3R we use the result given in class:

3R = 2Gabuaub − 6H2

= 2(8πTab − Λgab)uaub − 6H2

= 16πρ+ 2Λ− 6H2

=12b

a3+ 2Λ− 6

a2

a2

= −12ε

a2(8)

where we used (5) and (7) in the last steps.

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(b) For a static solution aS we have aS = aS = 0. In this case, (6) implies

Λ =3b

a3S

= 4πρ > 0 . (9)

Using this, (7) gives

ε = −(b

aS+

Λa2S

6

)= −

(4πρ

3+

Λ

6

)a2S

= −2πρa2S (10)

which is the general relationship between aS and ρ for a static solution, and sincein this case ε < 0 we have 3R > 0 from (8). Using our standard normalization,3R = 6/a2, (8) shows that ε = −1/2 and we get

ρ =1

4πa2S

.

(c) We can write (7) and (3) as, respectively:

1

2a2 + V (a) = ε (11)

a = −V ′(a) (12)

where

V (a) = −(b

a+

Λa2

6

)= −b

(1

a+

a2

2a3S

). (13)

This describes a “particle” with coordinate a in a concave-down potential V (a).It’s easy to see that

V ′(aS) = b

(1

a2− a

a3S

)∣∣∣∣aS

= 0 (14)

V ′′(aS) = b

(− 2

a3− 1

a3S

)∣∣∣∣aS

= − 3b

a3S

= −Λ < 0 . (15)

Since V ′(aS) = 0 and V ′′(aS) < 0 we see that a = aS is a local maximum of V ,i.e. an unstable equilibrium point. Hence, the Einstein static universe is unstable.Alternatively, if we consider a perturbation δ(t) and write a(t) = aS + δ(t), then(12) becomes, to first order in δ,

δ = − b

a2S

(1− 2δ

aS+ . . .

)+

b

a2S

(1 +

δ

aS

)=

3b

a3S

δ = Λδ . (16)

The general solution for this perturbation is

δ(t) = Ae√

Λt +Be−√

Λt , (17)

which admits a mode that exponentially increases, indicating an instability.

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2. The solutions of interest are

ds2 = −dt2 + a(t)2[dχ2 + S2

k(χ)(dθ2 + sin2 θ dφ2)]. (18)

The function Sk(χ) depends on the spatial geometry, but the exact form won’t beneeded here. We define the conformal time η by dt/dη = a. For a null (ds2 = 0) andradial (dθ = dφ = 0) geodesic, eq. (18) becomes

dψ = ±dτa

= ±dη , (19)

∆ψ = ±∫ τ2

τ1

dτ

a= ±

∫ η2

η1

dη . (20)

Note: the ± sign must be chosen appropriately below, taking into account that ψ isa polar angle with 0 ≤ ψ ≤ π for the 3-sphere geometry (k = 1).

The dust-filled universe with k = 1 has

a(η) =1

2C(1− cos η) , τ(η) =

1

2C(η − sin η) . (21)

We have a = 0 at a big bang and big crunch. The light ray’s path from big bang(η = 0, τ = 0) to big crunch (η = 2π, τ = πC) then has

∆ψ =

∫ πC/2

0

dτ

a−∫ πC

πC/2

dτ

a=

∫ π

0

dη −∫ 2π

π

dη = π − π = 0 . (22)

This corresponds to a full trip around the universe (the analog of traveling all the wayaround the surface of a globe along a longitude line).

3. (a) The metric is homogeneous since translations in xi are symmetries (the metricdoesn’t depend on xi). The metric is not isotropic since it’s not spherically sym-metric about any point (unless p1 = p2 = p3). To see that curves of constantxi are geodesics, note that we can take ua = (1, 0, 0, 0). Using the Γabc that areworked out in (c) below,

ua∇aub = ua

(∂au

b + Γbacuc)

= ua(0 + Γbacu

c)

= Γbtt = 0 , (23)

so the curve is an affinely parameterized geodesic. Alternatively, note that (ξi)aua =

0 where (ξi)a = (∂/∂xi)

a = δai are the three spatial Killing vectors, and hence

0 = ub∇b[(ξi)aua] = (ξi)a[u

b∇bua] + uaub∇b(ξi)a . (24)

The last term vanishes by the Killing equation ∇(b(ξi)a) = 0, so ub∇bua is orthog-

onal to each of ξi, and therefore ub∇bua ∝ ua, which is the geodesic equation.

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(b) For the Killing vectors (ξi)a we have ka(ξi)

a= constant along the null geodesic withtangent ka, so the initial values ka(ξ2)a = ka(ξ3)a = 0 don’t change and the lightray stays in the x1 direction. The energy seen by an observer with 4-velocity ua

is E = −kaua where ua ∝ (1, 0, 0, 0). Since ka is null, at any point its projectiononto ua must have the same magnitude and opposite sign as its projection ontothe x1 direction. With ξa ≡ (ξ1)a this means:

kaua = −ka

ξa

(ξbξb)1/2. (25)

Since kaξa= constant on the geodesic, for a signal emitted at t1 = 1 and received

at t,

E(t)

E(t1)=

[kaξa/(ξbξ

b)1/2](t)

[kcξc/(ξdξd)1/2](t1)=

[(ξdξd)1/2](t1)

[(ξcξc)1/2](t)=tp11

tp1, =⇒ E(t) =

E0

tp1. (26)

(c) First find Γabc by comparing the geodesic equation xa+Γabcxbxc = 0 with the Euler

equations,

d

dλ

∂L

∂xa− ∂L

∂xa= 0 where L = −t2 +

∑i

t2pix2i . (27)

The Euler equation for t and xi (respectively) are

t+∑i

pit2pi−1x2

i = 0 =⇒ Γtii = pit2pi−1 (28)

xi + 2pit−1xit = 0 =⇒ Γiit = pit

−1 . (29)

Using the Christoffel symbols to evaluate the Ricci tensor, we find that the nonzerocomponents are (we omit the algebra)

Rtt = t−2∑i

(pi − p2i ) (30)

Rii = pit2pi−2

(∑j

pj − 1

). (31)

This means that the vacuum Einstein equations Rab = 0 will only be satisfied ifeither p1 = p2 = p3 = 0 (Minkowski space), or if∑

i

pi =∑i

p2i = 1 . (32)

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