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김관석
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2025-03-24 15:13:37, 조회수 : 70 |
4 Cosmological Inflation: Selected Problems
4.1 Oscillating scalar field
The action for a scalar field in a curved space time is
(4.1) 𝑆 = ∫ d4 𝑥√(-𝑔) [-1/2 𝑔𝜇𝜈∂𝜇𝜙∂𝜈𝜙 - 𝑉(𝜙)], where 𝑔 ≡ det 𝑔𝜇𝜈.
1. Evaluate the action for the homogeneous field 𝜙 = 𝜙(𝑡) in a flat FRW spacetime and determine the equation of the motion for the field.
[Solution] For the FRW metric,
(b) d𝑠2 = -d𝑡2 + 𝑎2(𝑡) d𝐱2.
the determinant of the matrix, 𝑔𝜇𝜈 = diag (-1, 𝑎2, 𝑎2, 𝑎2), is 𝑔 = -𝑎6(𝑡). The scalar field is constrained by the symmetries of the FRW spacetime to evolve over only in time 𝜙(𝑡, 𝐱) = 𝜙(𝑡). The Lagrangian in the example is then
(c) 𝐿 = 𝑎3[1/2 𝜙̇2 - 𝑉(𝜙)]
We use the Euler-Lagrange equation
(d) d/d𝑡 (∂𝐿/∂𝜙̇) - ∂𝐿/∂𝜙 = 0, ∂𝐿/∂𝜙̇ = 𝑎3𝜙̇, ∂𝐿/∂𝜙 = -𝑎3∂𝑉/∂𝜙.
(e) d/d𝑡 (𝑎3𝜙̇) = 3𝑎2 d𝑎/d𝑡 𝜙̇ + 𝑎3d𝜙̇/d𝑡
(f) 𝑎3(3𝑎̇/𝑎 𝜙̇ + d𝜙̇/d𝑡 + d𝑉/d𝜙) = 𝑎3(3𝐻𝜙̇ + d𝜙̇/d𝑡 + d𝑉/d𝜙) = 0.
Substituting this into (d), we get
(g) d𝜙̇/d𝑡 + 3𝐻𝜙̇ = -d𝑉/d𝜙, which is Klein-Gordon equation. ▮
2. Near minimum of the potential, we have 𝑉(𝜙) ≈ 1/2 𝑚2𝜙2. Making the ansatz 𝜙(𝑡) = 𝑎-3/2𝜒(𝑡), show that the equation of motion becomes
(4.1.2) d𝜒̇/d𝑡 + (𝑚2 - 3/2 Ḣ -9/4 𝐻2)𝜒 = 0.
Assuming that 𝑚2 ≫ 𝐻2 ~ Ḣ, find 𝜙(𝑡) and write the answer in terms of the maximum amplitude of the oscillations. What does this result imply for the evolution of the energy density during the oscillating phase after inflation?
[Solution] For 𝑚2 ≫ 𝐻2 ~ Ḣ the inflaton 𝜙 behave as harmonic oscillator, so that
(a) d𝜒̇/d𝑡 + 𝑚2𝜒 ≈ 0 ⇒ 𝜒(𝑡) = 𝛢 cos (𝑚𝑡 + 𝛣) = 𝑎3/2𝜙(𝑡).
so we get
(b) 𝜙(𝑡) = 𝜙mx𝑎-3/2 cos (𝑚𝑡), where 𝜙mx is the maximum amplitude of the oscillation.
This implies that 𝜙(𝑡) is an oscillating function whose amplitude decays as 𝑎-3/2.
Using (4.48) 𝜌𝜙 = 1/2 𝜙̇2 + 1/2 𝑚2𝜙2 with 𝑚 ≫ 𝐻, we find
(c) 𝜌𝜙 = 1/2 𝜙mx2𝑎-3[-𝑚 sin (𝑚𝑡) + 𝐻 cos (𝑚𝑡)]2 + 𝑚2𝜙mx2𝑎-3cos2 (𝑚𝑡) ≈ 1/2 𝑚2𝜙mx2𝑎-3 [sin2 (𝑚𝑡) + cos2 (𝑚𝑡)] = 1/2 𝑚2𝜙mx2𝑎-3.
This implies that inflaton behaves like particles which disperse as 𝑎-3. ▮
4.2 Quadratic inflation
The equation of motion of the inflation is
(4.2) d𝜙̇/d𝑡 + 3𝐻𝜙̇ + 𝑉,𝜙 = 0, with 3𝑀2𝑃𝑙𝐻2 = 1/2 𝜙̇2 + 𝑉.
1. For the potential 𝑉(𝜙) = 1/2 𝑚2𝜙2, use the slow-roll approximation to obtain the inflationary solutions
(4.2.1a) 𝜙(𝑡) = 𝜙𝑖 - √(2/3) 𝑚𝑀𝑃𝑙𝑡,
(4.2.1b) 𝑎(𝑡) = 𝑎𝑖 exp[{𝜙𝑖2 - 𝜙2(𝑡)}/4𝑀2𝑃𝑙],
where 𝜙𝑖 > 0 is the field value at the tart of inflation (𝑡𝑖 ≡ 0).
[Solution] In the slow-roll approximation we have
(a) 3𝐻𝜙̇ ≈ -𝑚2𝜙, and 3𝑀2𝑃𝑙𝐻2 ≈ 1/2 𝑚2𝜙2. ⇒
(b) 𝐻 ≈ (𝑚𝜙)/(√6𝑀𝑃𝑙),
(c) 𝜙̇ ≈ -√(2/3)𝑚𝑀𝑃𝑙.
Integrating (c) we get
(d) 𝜙(𝑡) ≈ 𝜙𝑖 - √(2/3)𝑚𝑀𝑃𝑙𝑡, where 𝜙𝑖 is the 𝜙 at the beginning if inflation.
From (b) and (c) we find
(e) (d ln a)/d𝜙 = (d ln a)/d𝑡 d𝑡/d𝜙 = 𝐻/𝜙̇ ≈ -1/2 𝜙/𝑀2𝑃𝑙
(f) d ln a = -1/2 𝜙/𝑀2𝑃𝑙 d𝜙
Integrating (f) from 𝑡𝑖 to 𝑡 we get
(g) ∫𝑡𝑡𝑖 d ln a = ln a(𝑡)/a𝑖 = {𝜙𝑖2 - 𝜙2(𝑡)}/4𝑀2𝑃𝑙, ⇒
(h) a(𝑡) = a𝑖 exp[{𝜙𝑖2 - 𝜙2(𝑡)}/4𝑀2𝑃𝑙]. ▮
2. What is the value of 𝜙 where inflation ends? Find an expression for the number of 𝑒-folds. If 𝑉(𝜙𝑖) ~ 𝑀4𝑃𝑙, estimate the total number of 𝑒-folds of inflation.
[Solution] According to (4.65) the potential slow-roll parameter is
(a) 𝜀𝑉 ≡ 𝑀2𝑃𝑙/2 (𝑉,𝜙/𝑉)2 = 2 (𝑀𝑃𝑙/𝜙)22.
If we define 𝜙𝑒 as inflaton at the end of inflation, where 𝜀𝑉(𝜙𝑒) ≡ 1, we get
(b) 𝜙𝑒 = √2𝑀𝑃𝑙.
Since the total number of 𝑒-folds of inflation is definded
(c) 𝛮tot = ln(𝑎𝑒/𝑎𝑖).
Using (g) and (h) in the previous problem, we find
(d) 𝛮tot = {𝜙𝑖2 - 𝜙𝑒2}/4𝑀2𝑃𝑙 = 𝜙𝑖2/4𝑀2𝑃𝑙 - 1/2.
If 𝑉(𝜙𝑖) ~ 𝑀4𝑃𝑙, using (4.69) where 𝑚 is the inflaton mass,, we can estimate 𝛮tot so that
(e) 𝑉(𝜙𝑖) = 1/2 𝑚2𝜙𝑖2 ~ 𝑀4𝑃𝑙, ⇒ 𝛮tot ~ 𝜙𝑖2/4𝑀2𝑃𝑙 ~ (𝑀𝑃𝑙/𝑚)2. ▮
4.3 Power law inflation
Consider the potential [41]
(4.3) 𝑉(𝜙) = 𝑉0 exp(-𝑐 𝜙/𝑀𝑃𝑙),
where 𝑐 and 𝑉0 are constants, Find the solution for the field and the scale factor without making the slow-roll approximation. For what values of 𝑐 does this solution correspond to inflation? What is the problem with this model of inflation?
Hints: Try an ansatz of the form 𝜙(𝑡) = 𝛼 ln(𝛽𝑡) and 𝑎(𝑡) = (𝑡/𝑡0)𝛾.
[Solution] Since the dynamics during inflation is determined by a combination of the Friedmann and Klein-Gordon equations, we start with 𝜙(𝑡) = 𝛼 ln(𝛽𝑡) and 𝑎(𝑡) = (𝑡/𝑡0)𝛾
(a) 𝜙̇ = 𝛼/𝑡, d𝜙̇/d𝑡 = -𝛼/𝑡2, 𝐻 = 𝑎̇/𝑎 = 𝛾/𝑡, d𝑉/d𝜙 = -𝑐𝑉0/𝑀𝑃𝑙 exp(-𝑐 𝜙/𝑀𝑃𝑙) = -𝑐𝑉0/𝑀𝑃𝑙 (𝛽𝑡)-𝑐𝛼/𝑀𝑃𝑙.
From the Klein-Gordon equation, d𝜙̇/d𝑡 + 3𝐻𝜙̇ + d𝑉/d𝜙 = 0, we get
(b) -𝛼/𝑡2 + 3 𝛾/𝑡 𝛼/𝑡 - 𝑐𝑉0/𝑀𝑃𝑙 (𝛽𝑡)-𝑐𝛼/𝑀𝑃𝑙 = 0.
If we set 𝛼 = 2𝑀𝑃𝑙/𝑐 the equation, all the three terms have the common elements 𝑡-2, it becomes
(c) -2𝑀𝑃𝑙/𝑐 + 6𝑀𝑃𝑙/𝑐 𝛾 - 𝑐/𝑀𝑃𝑙 𝑉0/𝛽2 = 0.
From e Friedmann equation 𝐻2 = 1/3𝑀2𝑃𝑙 (1/2 𝜙̇2 + 𝑉), with 𝛼 = 2𝑀𝑃𝑙/𝑐, we get
(d) 𝐻2 = (𝛾/𝑡)2 = 1/3𝑀2𝑃𝑙 [1/2 (𝛼/𝑡)2 + 𝑉0(𝛽𝑡)-𝑐𝛼/𝑀𝑃𝑙] ⇒ 3𝑀2𝑃𝑙𝛾2 = 2𝑀2𝑃𝑙/𝑐2 + 𝑉0/𝛽2 ⇒ 𝑉0/𝛽2 = 3𝑀2𝑃𝑙𝛾2 - 2𝑀2𝑃𝑙/𝑐2.
Substituting the result into (c), we get
(e) -2𝑀𝑃𝑙/𝑐 + 6𝑀𝑃𝑙/𝑐 𝛾 - 𝑐/𝑀𝑃𝑙 (3𝑀2𝑃𝑙𝛾2 - 2𝑀2𝑃𝑙/𝑐2) = 0 ⇒ 𝑐2/2 𝛾2 - 𝛾 = 0 ⇒ 𝛾 = 0, or 𝛾 = 2/𝑐2.
In order to get 𝛽, we substitute 𝛾 = 2/𝑐2 into (d), so that
(f) 𝑉0/𝛽2 = 3𝑀2𝑃𝑙 2/𝑐2 - 2𝑀2𝑃𝑙/𝑐2 ⇒ 𝛽 = 𝑐2/√(12 - 2𝑐2) √𝑉0/𝑀𝑃𝑙.
So we can find the solution of the field
(g) 𝑎(𝑡) = (𝑡/𝑡0)2/𝑐2, and 𝜙(𝑡) = 2𝑀𝑃𝑙/𝑐 ln[𝑐2/√(12 - 2𝑐2) √𝑉0/𝑀𝑃𝑙 𝑡].
The problem with this model of inflation is that since 𝑐 in (g) is constant, the inflation would not stop forever. ▮
4.4 Natural inflation
An influential idea in inflationary model-building is that the inflaton could be a pseudoscalar axion. At the perturbative level, axions enjoy a continuous shift symmetry, but nonperturbatively this is broken to a discrete symmetry, leading to a potential of the form
(4.4) 𝑉(𝜙) = 𝑉0/2 [1 -cos(𝜙/𝑓)],
where 𝑓 is the axion decay constant. At what value 𝜙𝑖 close to 𝜙 = π𝑓 does the field have to start in order for the evolution to give more than 50 𝑒-folds of inflation? The mode is called natural inflation [42].
[Solution] From the slow-roll parameter (4.63), we can estimate the the total number of 𝑒-foldings of inflation (4.68), so that
(a) 𝜀𝑉 ≈ 𝑀2𝑃𝑙/2 (𝑉,𝜙/𝑉)2 = 1/2 𝑀2𝑃𝑙/𝑓2 sin(𝜙/𝑓)2/[1 -cos(𝜙/𝑓)]2.
The the total number of 𝑒-foldings between 𝜙𝑖 ≡ 𝜃𝑖𝑓 and 𝜙𝑒 ≡ 𝜃𝑒𝑓 or 𝜃𝑖 ≈ π and 𝜃𝑒𝑓 ≈ 0,
(b) 𝑁tot ≈ ∫𝜙𝑒𝜙𝑖 1/√(2𝜀𝑉) ∣d𝜙∣/𝑀𝑃𝑙 = 𝑓2/𝑀2𝑃𝑙 ∫𝜃𝑖𝜃𝑒 [1 - cos(𝜃)]/sin(𝜃) d𝜃 = - 2𝑓2/𝑀2𝑃𝑙 ln[cos(𝜃𝑖/2)/cos(𝜃𝑒/2)] ≈ - 2𝑓2/𝑀2𝑃𝑙 ln[cos(𝜃𝑖/2)].
For more than 50 𝑒-folds, we need
(c) 𝑁tot ≈ - 2𝑓2/𝑀2𝑃𝑙 ln[cos(𝜃𝑖/2)] > 50 ⇒ 𝜃𝑖 > 2 cos-1(𝑒-25𝑀2𝑃𝑙/𝑓2).
Hence
(d) 𝜙𝑖 > 2𝑓 cos-1(𝑒-25𝑀2𝑃𝑙/𝑓2). ▮
4.5 Kinetic inflation
Consider single-field inflation with a more general term [43]
(4.5) 𝑆 = ∫ d4𝑥 √(-𝑔) 𝑃(𝑋, 𝜙),
where 𝑃(𝑋, 𝜙) is an arbitrary function of 𝑋 ≡ -1/2 𝑔𝜇𝜈∂𝜇𝜙∂𝜈𝜙 and 𝜙.
1. By varying the action with respect to the metric, show that this corresponds to a perfect fluid with pressure and energy density
(4.5.1) 𝜌 = 2𝑋 𝑃,𝑋 - 𝑃, where 𝑃,𝑋 ≡ d𝑃/d𝑋.
We may easily check that this gives the expected result for the case of slow-roll inflation, 𝑃 = 𝑋 - 𝑉(𝜙), namely 𝜌 = 𝑋 + 𝑉.
[Solution] Varying the action with respect to the metric, with 𝛿√(-𝑔) = - 1/2 √(-𝑔) 𝑔𝜇𝜈𝛿𝑔𝜇𝜈 from Exercise 4.3, gives
(a) 𝛿𝑆 = ∫ d4𝑥 (𝛿√(-𝑔) 𝑃 + √(-𝑔) 𝛿𝑃) = ∫ d4𝑥 √(-𝑔) (-1/2 𝑔𝜇𝜈𝛿𝑔𝜇𝜈 𝑃 - 𝑃,𝑋 1/2 ∂𝜇𝜙∂𝜈𝜙𝛿𝑔𝜇𝜈)
= 1/2 ∫ d4𝑥 √(-𝑔) (-𝑔𝜇𝜈𝑃 - 𝑃,𝑋 ∂𝜇𝜙∂𝜈𝜙) 𝛿𝑔𝜇𝜈.
So we find the energy-momentum tensor 𝛵𝜇𝜈 = 𝑃,𝑋 ∂𝜇𝜙∂𝜈𝜙) 𝛿𝑔𝜇𝜈 + 𝑔𝜇𝜈𝑃.
(b) 𝛵𝜇𝜈 = 𝑃,𝑋 ∂𝜇𝜙∂𝜈𝜙) 𝛿𝑔𝜇𝜈 + 𝑔𝜇𝜈𝑃. ⇒ 𝛵𝜇𝜈 = 𝑃,𝑋 𝑔𝜇𝜆∂𝜆𝜙∂𝜈𝜙) 𝛿𝑔𝜇𝜈 + 𝛿𝜇𝜈𝑃.
Since we are interested in the evolution of a homogeneous field configuration, 𝜙 = 𝜙(𝑡), we get
(c) 𝜌𝜙 = -𝛵00 = 2𝑃,𝑋 𝑋 - 𝑃. 𝑃𝜙 = 1/3 𝛵𝑖𝑖 = 𝑃, where 𝑋 = 1/2 𝜙̇2, as in (4.44)
which correspond to a perfect fluid with pressure and energy density. ▮
2. By varying the action with respect to 𝜙, show that the equation of motion for the inflation is
(4.5.2) - d/d𝑡 (𝑎3𝑃,𝑋 𝜙̇) + 𝑎3𝑃,𝜙 = 0.
For slow-roll inflation, this gives the Klein-Gordon equation.
[Solution] Varying the action with respect to 𝜙 gives
(a) 𝛿𝑆 = ∫ d4𝑥 √(-𝑔) 𝛿𝑃 = ∫ d4𝑥 √(-𝑔) (𝑃,𝑋 𝛿𝑋 + 𝑃,𝜙𝛿𝜙) = ∫ d4𝑥 𝑎3(𝑃,𝑋 𝜙̇𝛿𝜙̇ + 𝑃,𝜙𝛿𝜙) = ∫ d4𝑥 [-d/d𝑡 (𝑎3𝑃,𝑋 𝜙̇) + 𝑎3𝑃,𝜙]𝛿𝜙,
where we used 𝑋 = 1/2 𝜙̇2 for the evolution of a homogeneous field configuration. Hence we find
(b) -d/d𝑡 (𝑎3𝑃,𝑋 𝜙̇) + 𝑎3𝑃,𝜙 = 0.
For 𝑃 = 𝑋 - 𝑉(𝜙), this gives Klein-Gordon equation, d/d𝑡 𝜙̇ + 3𝐻𝜙̇ + d𝑉/d𝜙 = 0. ▮
3. Show that the slow-roll parameter is
(4.5.3) 𝜀 = - Ḣ/𝐻2 = 3𝑋𝑃,𝑋/(2𝑋𝑃,𝑋 - 𝑃).
For suitable 𝑃(𝑋), this may lead to inflation even without a flat potential.
[Solution] Using Friedmann equation with (3.2) and the continuity equation (2.106), we can find
(a) 𝐻2 = 𝜌/3𝑀2𝑃𝑙 ⇒ 2𝐻Ḣ = 𝜌̇/3𝑀2𝑃𝑙 = 1/3𝑀2𝑃𝑙 (-3𝐻)(𝜌 + 𝑃) ⇒ Ḣ = -1/2𝑀2𝑃𝑙 (𝜌 + 𝑃).
So the slow-roll parameter is by using the result in Problem 4.5.1
(b) 𝜀 = - Ḣ/𝐻2 = 3(𝜌 + 𝑃)/2𝜌 = 3𝑋𝑃,𝑋/(2𝑋𝑃,𝑋 - 𝑃). ▮
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