Homework 9 Solutions

Home , Big Crunch, False vacuum decay, Inflationary epoch

ASTR/PHYS 4080: Introduction to Cosmology Spring 2019

1. What upper limit is placed on Ω(tP ) by the requirement that the universe not −44 end in a Big Crunch between the Planck time, tP ∼ 5 × 10 s, and the start of the inflationary epoch at ti? Compute the maximum value of Ω(tP ), first −36 −26 assuming ti ∼ 10 s, then assuming ti ∼ 10 s. (Hint: prior to inflation, the Friedmann equation will be dominated by the radiation term and the curvature term.)

Solution

We start this problem with the Friedmann equation for radiation and curvature:

2 H(t) Ω0 1 − Ω0 2 = 4 + 2 (1) H0 a a then 2 a˙ Ω0 da p 2 2 = 2 + 1 − Ω0 ⇒ aa˙ = a = H0 Ω0 + (1 − Ω0)a (2) H0 a dt Z t2 Z a2 ada dt = (3) p 2 t1 a1 H0 Ω0 + (1 − Ω0)a

2 To integrate, we make the substitution x = (1−Ω0)a ⇒ dx = 2(1−Ω0)ada ⇒ 1/2 1/2 dx x1 x2 ada = , so then a1 = , a2 = , which results in 2(1−Ω0) 1−Ω0 1−Ω0 q q 1 2 2 t2 − t1 = Ω0 + (1 − Ω0)a2 − Ω0 + (1 − Ω0)a1 (4) H0(1 − Ω0)

and setting a1 = 0 when t1 = 0 results in " # 1 p 2 p t = Ω0 + (1 − Ω0)a − Ω0 (5) H0(1 − Ω0)

Solving for a(t), s Ω − [Ω1/2 − H t(Ω − 1)]2 a(t) = 0 0 0 0 . (6) Ω0 − 1

For a universe with a Big Crunch at ti, a(ti) → 0, which happens when Ω0 − 1/2 2 1/2 1/2 [Ω0 − H0ti(Ω0 − 1)] → 0 or Ω0 − H0ti(Ω0 − 1) = ±Ω0 . The positive solution is for t = 0, so we want the negative term solution:

1/2 2Ω0 H0ti = . (7) Ω0 − 1

1 ASTR/PHYS 4080: Introduction to Cosmology Spring 2019

The 0 subscripts simply refer to some reference time that we haven’t defined yet. Since we need to find Ω(tp) = Ωp, tp is a convenient reference time. The Planck time occurs very early, when radiation dominates over curvature, so we can use the solution for H in a radiation-only universe, Hp = 1/(2tp). Defining x ≡ ti/(4tp), we get a quadratic function in Ωp, 1 Ω2 − (2 + )Ω + 1 = 0 , (8) p x2 p with the solution (via the quadratic formula)

1 1 r 1 Ω = 1 + ± 4 + (9) p 2x2 x x2

−36 with the positive term as the physical solution (since Ωp > 1). If ti = 10 s, −7 −26 −17 this gives Ωp − 1 = 2 × 10 and for ti = 10 s, we get Ωp − 1 = 2 × 10 . There is not much room for a positively-curved universe during the Planck era the longer inﬂation takes to begin.

2. Current observational limits on the density of magnetic monopoles tells us that −6 their density parameter is currently ΩM,0 < 10 . If monopoles formed at the GUT time, with one monopole (mass mM = mGUT) per horizon, how many e- foldings of inﬂation would be required to drive the current density of monopoles −6 below the bound ΩM,0 < 10 ? Assume that inﬂation took place immediately after the formation of monopoles.

Solution

In this problem we need to ﬁnd the number of e-foldings to reduce the current −6 −6 −3 density of monopoles to ΩM,0 < 10 . First, we know 10 × 4870 MeV m = −3 −3 2 4.87 × 10 MeV m . The rest energy of a magnetic monopole is mGUTc = 12 −36 10 TeV. One monopole is made within the horizon distance at tGUT ∼ 10 s, so the density of magnetic monopoles is given by Ryden Eq. 10.14:

1 monopole 82 −3 nM(tGUT) = 3 ∼ 10 m , (10) (2ctGUT) resulting from the Hubble volume being proportional to the horizon distance cubed. Monopoles will decrease in density ∝ a−3 like matter, and given that the temperature of radiation declines as the universe expands,

TCMB 2.73 K −28 a(tGUT) ≈ = 28 ≈ 3 × 10 (11) TGUT 10 K so 3 −82 82 −3 −3 nM,0 = a(tGUT) nGUT ∼ 10 × 10 m ∼ 1 m (12)

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Therefore, without inﬂation we expect to see 1 magnetic monopole per cubic meter. Assuming inﬂation is responsible for the reduced density, we have

−3 −3 εM,0 4.87 × 10 MeV m 3 −3 nM,0 = 2 = 2 = (ainflation) (1 m ) , (13) mGUTc mGUTc −N 2 12 18 where ainflation = e and mGUTc ∼ 10 TeV = 10 MeV. This number of e-foldings depends on the monopole energy at the time of GUT. Assuming this mass, we find e−N = (4.87 × 10−21)1/3 = 1.7 × 10−7 and N ≥ 7 ln(10) − ln(1.7) or N ≥ 15.6 . 3. What if the dark energy that dominates our universe today is just the manifes- tation of another round of inflation? It has been speculated that the present day acceleration of the universe is due to the existence of a false vacuum, much like the false vacuum that may be responsible for inflation between time ti and time tf in the early universe. Just as inflation is postulated to occur as a significant cooling phase during a period between a radiation dominated regime (or maybe magnetic monopole dominated?) and another radiation dominated regime, this model for the cosmological constant is postulated to eventually decay and re- lease energy back into the universe in the form of radiation. Suppose that the −3 energy density of this false vacuum today is εΛ = 0.69εc,0 = 3360 MeV m , −3 and the current matter density is εm,0 = 0.31εc,0 = 1510 MeV m . What will be the value of the Hubble parameter once the false vacuum becomes strongly dominant? Suppose that the false vacuum is fated to decay instantaneously to radiation at a time tf = 50t0. Assume that the radiation takes the form of blackbody photons. To what temperature will the universe be reheated at t = tf ? At what time will the universe again be dominated by matter?

Solution

When dark energy dominates, the universe evolves like a Λ-only universe, where the Hubble parameter is given by Ryden Eq. 5.72: 8πGε 1/2 H = Λ = Ω1/2H = 0.691/2H = 0.83H . (14) i 3c2 Λ 0 0 0 This lower value makes sense since the energy density of dark energy is constant and the matter density is falling rapidly ∝ a−3. At the time of false vacuum decay, the energy density of the vacuum is converted to radiation. Assuming blackbody photon radiation, we have ε = αT 4, where −3 −10 −3 −16 −3 −4 ε = εΛ = 3360 MeV m = 5.38×10 J m and α = 7.56×10 J m K . The temperature is then given by

5.38 × 10−10 J m−3 1/4 T = = 29 K . (15) 7.56 × 10−16 J m−3 K−4

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The re-emergence of matter-domination requires us to ﬁnd the scale factor at tf . The current age of the universe is t0 = 13.7 Gyr, and if tf = 50t0, then

16 19 Hi(tf −t0) (49·0.83·68·13.7·3.14×10 /3.086×10 ) 16 a(tf ) = a(t0)e = e = 5.5 × 10 . (16)

We also know the matter density has changed over time so:

−3 −3 16 −3 −48 −3 εm,f = εm,0a = 1510 MeV m (5.5 × 10 ) = 9 × 10 MeV m (17)

Now consider the moment of radiation matter equality:

εm,f εr,f εr,f 3 = 4 ⇒ arm = (18) a a εm,f

and using the relations for how a(t) evolves with time during the revived radiation era: t 1/2 a(t) = a(tf ) (19) tf we have: 2 2 εr,f 1 3360 70 trm = 50t0 = 50(13.7 Gyr) −48 16 ≈ 3×10 Gyr . εm,f a(tf ) 9 × 10 · 5.5 × 10 (20)