Solutions Exam FY3452 Gravitation and Cosmology Fall 2016

Solutions Exam FY3452 Gravitation and Cosmology Fall 2016

Lecturer: Professor Jens O. Andersen Department of Physics, NTNU

Phone: 73593131 Monday December 12 2016

09.00-13.00

Permitted examination support material:

Approved calculator

Rottmann: Matematisk Formelsamling Rottmann: Matematische Formelsammlung Barnett & Cronin: Mathematical Formulae

Angell og Lian: Fysiske størrelser og enheter: navn og symboler.

Problem 1

a) Taking the differentials, we obtain dt⁰ = γ(dt− v

c²dx)

= γdt

1−vV_x c²

, (1)

dV_x⁰ = dV_x 1− ^vV_c2^x

+ V_x−v (1− ^vV_c2^x)²

v

c²dV_x . (2)

Dividing Eq. (2) by Eq. (1), we obtain a⁰_x = 1

γ a_x

(1− ^vV_c2^x)² + 1 γ

V_x−v (1− ^vV_c2^x)³

v

c²a_x . (3)

1

If S⁰ is the instantaneous rest frame, we havev =V_x and Eq. (3) reduces to

a⁰_x = γ³a_x , (4)

where we have used that 1− ^vV_c2^x = 1− ^V_c^x2² = _γ¹2.

b) Since a⁰_x =g, Eq. (4) can be written as dV_x

dt = g 1−V_x² c²

!³₂

. (5)

or

dV_x

1− ^V_c^x2²

³

2

= gdt . (6)

Changing variables, V_x =csinu, we obtain cdu

cos²u = gdt . (7)

Integrating yields

ctanu = gt+C , (8)

where C is an integration constant.

V_x

q

1− ^V_c^x2²

= gt+C . (9)

Solving with respect to V_x, this finally yields V_x(t) = gt+C

q1+(gt+C)² c²

. (10)

C = 0 since V_x(0) = 0. Thus

V_x(t) = gt

q

1 + ^g_c²2^t2

. (11)

The limiting velocity is V_lim =cas seen from Eq. (11).

c) We have

dτ

dt = 1 γ

= 1

q

1− ^V_c^x2²

= 1

q

1 + ^g_c²2^t2

(12)

Changing variables t= ^c_gsinhu, we can write dτ = c

gdu . (13)

Integration yields

τ = c g

Z u 0

du+C

= c gu+C

= c

g sinh⁻¹(^g_ct) +K , (14) where K is an integration constant. K = 0 since τ(0) = 0. This yields

t(τ) = c

g sinh(^g_cτ). (15)

d) Integrating Eq. (11), we find x(t) = c²

g



 s

1 + g²t² c² −1



 , (16)

where we have used that x(τ = 0) = x(t = 0) = 0. Substituting Eq. (15) into Eq. (16), we finally obtain

x(τ) = c² g

hcosh(^g_cτ)−1ⁱ, (17)

e) Taking the differentials oft and x yields

1 gt^0! c x^0! gt^0!g

Inserting these expressions into the line element and using dy = dy⁰ and dz =dz⁰, we find

ds² = −c²dt²+dx²+dy²+dz²

= −c²dt⁰² 1 + gx⁰ c²

!2

+dx⁰²+dy⁰²+dz⁰² , (20)

f ) Since the line element is independent of time, the vectorξ = (1,0,0,0) is a Killing vector. The quantityξ·p is a conserved quantity along a geodesic.

g) A stationary observer with spatial coordinates (h,0,0) has four-velocity vector

u =



 1 + gx⁰ c²

!−1

,0,0,0





= 1 + gx⁰ c²

!−1

ξ . (21)

The energy of a photon with four-momentum p and frequency ω is ¯hω =

−p·u_obs. This yields

¯

hω = − 1 + gx⁰ c²

!−1

ξ·p. (22)

or

¯

hω 1 + gx⁰ c²

!

= −ξ·p. (23)

The energy of a photon emitted at x⁰ =h is denoted by ¯hω_h and the energy of the same photon absorbed at x⁰ = h is denoted by ¯hω0. Eq. (23) then gives

ω₀ = ω_h 1 + gh c²

!

, (24)

since ξ·p is constant along the photon’s geodesic.

According to the equivalence principle acceleration is equivalent to a gravitional field. The blueshift of the photon is an example of this principle.

Problem 2

a) Subtracting one-third of the first Friedman equation from the second Friedman equation gives

¨

a = −4π

3 aρ_m+1

3aΛ. (25)

where we have used that the pressure p vanishes.

b) For a time-independent solution, we have ˙a = ¨a = 0. Equation (25), then yields

ρ^c_m = Λ

4π . (26)

For a static solution the first Friedman equation reduces to 31

a²_c = 8πρ^c_m+ Λ, (27)

or

a_c = 1

√Λ . (28)

c) We write a = a_c+δa. Note that ˙a = _dt^dδa and ¨a = _dt^d22δa since a_c is constant in time. For p= 0, the second Friedman equation can be rewritten as

2¨aa+ ˙a²+ 1 = Λa² . (29) To first order in the perturbation, Eq. (29) reads

2ad²

dt²δa+ 1 = Λ(a²_c+ 2acδa). (30) Using the result for a_c, we find

d²

dt²δa = Λδa , (31)

which corresponds to B = Λ. This is a second-order differential equation for δa, whose solution is

δa = A₁e

√

Λt+A₂e⁻

√

Λt, (32)

whereA andA are constants. The perturbation is growing and so the static