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2024-11-08 17:16:07, 조회수 : 50 |
6.6 Summary
In this chapter, we developed the foundation of cosmological perturbation theory. By linearizing the Einstein Equations we derive the evolution equation for small fluctuations and solved them for various cases.
Focusing on scalar perturbations, the most general perturbation of spacetime metric is
(6.221) 𝑔𝜇𝜈 = 𝑎2⌈-(1+ 2𝛢) ∂𝑖𝛣 ⌉ * ipad view
⌊ ∂𝑖𝛣 (1 + 2𝐶)𝛿𝑖𝑗 + 2∂⟨𝑖∂𝑗⟩𝐸⌋,
(6.221) 𝑔𝜇𝜈 = 𝑎2⌈-(1+ 2𝛢) ∂𝑖𝛣 ⌉ * pc view
⌊ ∂𝑖𝛣 (1 + 2𝐶)𝛿𝑖𝑗 + 2∂⟨𝑖∂𝑗⟩𝐸⌋,
where ∂⟨𝑖∂𝑗⟩ = ∂𝑖∂𝑗 - 1/3 ∇2. The perturbed energy-momentum tensor is
(6.222) 𝛵00 ≡ -(𝜌̄ + 𝛿𝜌), 𝛵𝑖0 ≡ (𝜌̄ + 𝑃̄)∂𝑖𝑣, 𝛵0𝑖 ≡ (𝜌̄ + 𝑃̄)∂𝑖(𝑣 + 𝛣), 𝛵𝑖𝑗 ≡ (𝑃̄ + 𝛿𝑃)𝛿𝑖𝑗 + ∂⟨𝑖∂𝑗⟩𝛱,
and we often work in terms of the momentum density 𝑞 ≡ (𝜌̄ + 𝑃̄)𝑣 and the density contrast 𝛿 ≡ 𝛿𝜌/𝜌. Coordination transformation can change the values of these variables. The problem of unphyscial gauge mode can be addressed by working with combinations of perturbations that are invariant under coordinate transformation. Important gauge-invariant variables are the Bardeen potential and the comoving density contrast
(6.223-5) 𝛹 ≡ 𝛢 + 𝓗(𝛣 - 𝐸ʹ) + (𝛣 - 𝐸ʹ)ʹ, 𝛷 ≡ -𝐶 + 1/3 ∇2𝐸 - 𝓗(𝛣 - 𝐸ʹ), ∆ = 𝛿 + 𝜌̄ʹ/𝜌̄ (𝑣 + 𝛣).
as well as the two curvature perturbations
(6.226-7) 𝛇 = -𝐶 + 1/3 ∇2𝐸 + 𝓗 𝛿𝜌/𝜌̄ʹ, 𝓡 = -𝐶 + 1/3 ∇2𝐸 - 𝓗(𝑣 + 𝛣).
On superhorizon scales, 𝛇 and 𝓡 are equal and time independent if the matter perturbations are adiabatic. Perturbations are called adiabatic if
(6.228) 𝛿𝑃 = 𝑐𝑠2𝛿𝜌 = 𝑃̄ʹ/𝜌̄ʹ 𝛿𝜌,
and fluctuations in the densities of matter and radiation obey
(6.229) 𝛿𝑚 = 3/4 𝛿𝑟.
The initial conditions of our universe are well described by adiabatic perturbations.
An alternative to the gauge-invariant formalism is to work in a fixed gauge, track the evolution of all fluctuations and compute observables. Unphysical gauge modes must cancel in physical answers. Popular gauge are
• Newtonian: 𝛣 = 𝐸 = 0 • Spatially flat: 𝐶 = 𝐸 = 0 • Synchronous: 𝛢 = 𝛣 = 0
• Constant density: 𝛿𝜌 = 𝛣 = 0 • Comoving: 𝑣 = 𝛣 = 0
We derived the equations of motion for the perturbation in Newtonian gauge. The linearized Einstein equations are
(6.230) ∇2𝛷 - 3𝓗(𝛷ʹ + 𝓗𝛹) = 4π𝐺𝑎2𝛿𝜌,
(6.231) -(𝛷ʹ + 𝓗𝛹) = 4π𝐺𝑎2𝑞,
(6.232) ∂⟨𝑖∂𝑗⟩(𝛷 - 𝛹) = 8π𝐺𝑎2𝛱𝑖𝑗,
(6.233) 𝛷ʺ + 𝓗𝛹ʹ + 2𝓗𝛷ʹ + 1/3 ∇2(𝛹 - 𝛷) + (2𝓗ʹ + 𝓗2)𝛹 = 4π𝐺𝑎2𝛿𝑃.
where the sources on the right-hand side include a sum over all components. In the absence of anisotropic stress, (6.232) implies that 𝛹 ≈ 𝛷, and (6.233) reduces to
(6.234) 𝛷ʺ + 3𝓗𝛷ʹ + (2𝓗ʹ + 𝓗2)𝛷 = 4π𝐺𝑎2𝛿𝑃.
(6.230) and (6.231) can be combined into
(6.235) ∇2𝛷 = 4π𝐺𝑎2𝜌̄∆,
which is of the same form as the Newtonian Poisson equation, but is now valid on all scales.
The conservation of the energy-momentum tensor gives the continuity and Euler equations
(6.236) 𝛿ʹ = -(1 + 𝜔)(𝛁 ⋅ 𝐯 - 3𝛷ʹ) - 3𝓗(𝑐𝑠2 - 𝜔)𝛿
(6.237) 𝑣𝑖ʹ = -[𝓗 - 3𝜔]𝐯 - 𝑐𝑠2/(1 + 𝜌) 𝛁𝛿 - 𝛁𝛷,
where 𝑐𝑠2 = 𝛿𝑃/𝛿𝜌 is the sound speed of the fluid and 𝜌 = 𝑃̄/𝜌̄ is its equation of state. These equations apply separately for every non-interacting fluid.
we discussed the solutions to the above equations in various special limits. Table 6.1 summarizes the results for 𝛷, ∆𝑟 and ∆𝑚. The density contrast 𝛿 is equal to ∆ on subhorizon scales, but differs on superhorizon scales, where it is a constant.
Before decoupling, photons and baryons can be treated as a single tightly-coupled fluid. The density contrast of photon-baryon fluid evolves as
(6.238) 𝛿𝑟ʺ + 𝑅ʹ/(1 + 𝑅) 𝛿𝑟ʹ - 1/3(1 + 𝑅) ∇2𝛿𝑟 = 4/3 ∇2𝛹 + 4𝛷ʺ + 4𝑅ʹ/(1 + 𝑅) 𝛷ʹ,
where 𝑅 ≡ 3/4 𝜌̄𝑏/𝜌̄𝛾 ∝ 𝑎(𝑡). This is the equation of a harmonic oscillator with a gravitational driving force, which plays an important role in he physics of the CMB anisotropies (see Chapter 7).
6 Problems
6.1 Problem Superhorizon evolution of the gravitational potential
Derive an analytic solution for the superhorizon evolution of the gravitational potential the remain valid in the transition from the radiation era to the matter era. The staring point is the following Einstein equation
(P6.1) ∇2𝛷 - 3𝓗(𝛷ʹ + 𝓗𝛹) = 4π𝐺𝑎2𝛿𝜌,
where 𝛿𝜌 = ∑𝑎𝛿𝜌𝑎. On superhorizon scales, we can drop ∇2𝛷 and interpret this as an evolution equation for 𝛷.
1. Assuming adiabatic perturbations, show that the superhorizon evolution of the potential in a universe with matter and radiation satisfies
(1) 𝑦 𝑑𝛷/𝑑𝑦 + 𝛷 = -(4 + 3𝑦)/6(1 + 𝑦) 𝛿𝑚,
where 𝑦 ≡ 𝑎/𝑎eq and 𝛿𝑚 is the density contrast of the matter perturbations. Use 𝛿𝑚ʹ = 3𝛷ʹ to write this as a closed equation for 𝛷.
[Solution] Consider matter density and radiation density for gravitational potential
(1.a) 𝓗(𝛷ʹ + 𝓗𝛷) = - 4π𝐺/3 𝑎2(𝜌̄𝛾𝛿𝛾 + 𝜌̄𝑚𝛿𝑚).
Since for adiabatic initial conditions 𝛿𝛾 = 4/3 𝛿𝑚, according to Friedmann equation and 𝓗 = 𝐻𝑎 weget
(1.b) 𝓗2 = 8π𝐺/3 𝑎2(𝜌̄𝛾,0𝑎-4 + 𝜌̄𝑚,0𝑎-3) = 8π𝐺/3 𝜌̄𝛾,0/𝑎2 (1 + 𝑎/𝑎eq) = 8π𝐺/3 𝜌̄𝛾,0/𝑎2(1 + 𝑦),
(1.c) 𝛷ʹ = d𝛷/d𝜂 = d𝑎/d𝜂 d𝛷/d𝑎 = 𝑎𝓗 d𝛷/d𝑎 = 𝓗𝑦 d𝛷/d𝑦.
(1.d) 𝓗(𝛷ʹ + 𝓗𝛷) = 𝓗2(𝑦 d𝛷/d𝑦 + 𝛷) = 8π𝐺/3 𝑎2(1 + 𝑦)(𝑦 d𝛷/d𝑦 + 𝛷).
The right-hand side of (1.a) becomes
(1.f) - 4π𝐺/3 𝑎2(𝜌̄𝛾𝛿𝛾 + 𝜌̄𝑚𝛿𝑚) = - 4π𝐺/3 𝜌̄𝛾,0/𝑎2 (4/3 + 𝑎/𝑎eq)𝛿𝑚 = - 4π𝐺/3 𝜌̄𝛾,0/𝑎2 (4/3 + 𝑦)𝛿𝑚
Combining (1.d) and (1.f), we get
(1.g) 𝑦 d𝛷/d𝑦 + 𝛷 = - (4 + 3𝑦)/6(1 + 𝑦) 𝛿𝑚.
Taking a derivative with respect to 𝑦, we have
(1.h) d/d𝑦 (𝑦 d𝛷/d𝑦 + 𝛷) = - d/d𝑦 [(4 + 3𝑦)/6(1 + 𝑦)] 𝛿𝑚 ⇒
𝑦 d2𝛷/d𝑦2 + 2 d𝛷/d𝑦 = 1/6(1 + 𝑦)2 - (4 + 3𝑦)/6(1 + 𝑦) d𝛿𝑚/d𝑦 = 1/(1 + 𝑦)(4 + 3𝑦)(𝑦 d𝛷/d𝑦 + 𝛷) - (4 + 3𝑦)/2(1 + 𝑦) d𝛷/d𝑦.
where we used 𝛿𝑚ʹ = 3𝛷ʹ. We can find a closed equation for the gravitational potential 𝛷,
(1.i) d2𝛷/d𝑦2 + [2/𝑦 + 1/(1 + 𝑦)(4 + 3𝑦) + (4 + 3𝑦)/2𝑦(1 + 𝑦)] d𝛷/d𝑦 + 𝛷/𝑦(1 + 𝑦)(4 + 3𝑦) = 0. ▮
6.2 Problem Superhorizon initial conditions
Let us derive the superhorizon initial conditions for pertirbations, accounting for the anisotropic stress due to neutrinos.
Cinsider the following Einstein equations during te radiation era:
(1) 𝑘2𝛷 + 3𝓗(𝛷ʹ + 𝓗𝛹) = -4π𝐺𝑎2(𝜌̄𝛾𝛿𝛾 + 𝜌̄𝜈𝛿𝜈),
(2) 𝑘2(𝛷 - 𝛹) = 8π𝐺𝑎2(𝜌̄𝜈 + 𝑃̄𝜈)𝛱𝜈,
where 𝛱𝜈 is the (rescaled) neutrino-induced anisotropic stress. In the tight-coupling approximation we have 𝛱𝜈ʹ ≈ - 2/5 𝑘𝜈𝜈 (see Section B.3.3).
1. Specializing to the case of adiabatic initial conditions, show that the superhorizon limit of (1) implies
(1.1) 𝜂𝛷ʺ + 3𝛷ʹ + 𝛹ʹ = 0.
[Solution] We can drop the 𝑘2𝛷 on large scales, so
(1.a) 3𝓗(𝛷ʹ + 𝓗𝛹) = -4π𝐺𝑎2(𝜌̄𝛾𝛿𝛾 + 𝜌̄𝜈𝛿𝜈) = - 3/2 𝓗2𝛿𝛾
where we used 𝛿𝛾 = 𝛿𝜈 for adiabatic initial conditions and as for (6.130), 𝓗2 = 8/3 π𝐺𝑎2(𝜌̄𝛾 + 𝜌̄𝜈) during the radiation era. Substituting 𝓗 = 1/𝜂, we then get
(1.b) 𝜂𝛷ʹ + 𝛹 = -1/2 𝛿𝜈.
Taking a derivative respect to 𝜂 and using 𝛿𝜈ʹ = 4𝛷ʹ, we get
(1.c) (𝜂𝛷ʹ + 𝛹)ʹ = -1/2 𝛿𝜈ʹ ⇒ 𝜂𝛷ʺ + 3𝛷ʹ + 𝛹ʹ = 0. ▮
2. Taking multiple time derivative of (2) shows that 𝛷 satisfies
(2.1) 𝜂3𝛷ʺʺ + 12 𝜂2𝛷ʹʺ + 4(9 + 2/5 𝑓𝜈)𝜂𝛷ʺ + 8(3 + 2/5 𝑓𝜈)𝛷ʹ = 0,
where 𝑓𝜈 ≡ 𝜌̄𝜈/(𝜌̄𝛾 + 𝜌̄𝜈). Determine the four solutions of this equation.
[Solution] At this time we don't set 𝛷 ≈ 𝛹, using radiation era 𝑃̄/𝜌̄ = 1/3 and 𝓗 = 1/𝜂, we get
(2.a) 𝑘2(𝛷 - 𝛹) = 8π𝐺𝑎2(𝜌̄𝜈 + 𝑃̄𝜈)𝛱𝜈 ≈ 32/3 π𝐺𝑎2𝜌̄𝜈 𝛱𝜈 = 4[8π𝐺𝑎2/3 (𝜌̄𝜈 + 𝜌̄𝛾)]𝑓𝜈𝛱𝜈 = 4𝓗2𝑓𝜈𝛱𝜈. ⇒ 𝑘2𝜂2(𝛷 - 𝛹) = 4𝑓𝜈𝛱𝜈.
Taking a time derivative of the last equation gives
(2.b) 𝑘2𝜂2(𝛷ʹ - 𝛹ʹ) + 2𝑘2𝜂(𝛷 - 𝛹) = -8/5 𝑘𝑓𝜈𝑣𝜈,
Taking a time derivative again and similar to (6.89) 𝑣𝜈ʹ = -𝑘(𝛿𝜈/4 + 𝛹)
(2.c) 𝜂2(𝛷ʺ - 𝛹ʺ) + 4𝜂(𝛷ʹ - 𝛹ʹ) + 2(𝛷 - 𝛹) = 8/5 𝑓𝜈(𝛿𝜈/4 + 𝛹).
Another time derivative lead to
(2.c) 𝜂2(𝛷ʹʹʹ - 𝛹ʹʹʹ) + 6𝜂(𝛷ʺ - 𝛹ʺ) + 6(𝛷ʹ - 𝛹ʹ) = 8/5 𝑓𝜈(𝛷ʹ + 𝛹ʹ),
where we used 𝛿𝜈ʹ = 4𝛷ʹ. Substituting 𝛹ʹ = -𝜂𝛷ʺ - 3𝛷ʹ from (1.1), we get
(2.d) 𝜂2(𝜂𝛷ʹʹʹʹ + 6𝛷ʹʹʹ) + 6𝜂(𝜂𝛷ʹʹʹ + 5𝛷ʹʹ) + 6(𝜂𝛷ʹʹ + 4𝛷ʹ) = - 8/5 𝑓𝜈(𝜂𝛷ʹʹ + 2𝛷ʹ), ⇒
(2.e) 𝜂3𝛷ʹʹʹʹ + 12𝜂2𝛷ʹʹʹ + 4(9 + 2/5 𝑓𝜈)𝜂𝛷ʹʹ + 8(3 + 2/5 𝑓𝜈)𝛷ʹ = 0.
There are four solutions
(2.e) 𝛷 = const, 𝛷 ∝ 𝜂-5/2±1/2√ (1-32𝑓𝜈/5), 𝛷 ∝ 𝜂-1. ▮
3. Focusing on the growing mode 𝛷 = 𝛷𝑖. show that
(3.1) 𝛹 = 𝛷𝑖(1 + 2/5 𝑓𝜈)-1 ≈ 0.86 𝛷𝑖 and 𝛿𝛾 = -2𝛹 ≈ -1.72𝛷𝑖.
[Solution] Substituting 𝛿𝛾 = 4𝛷𝑖 + 𝐶𝛾 in (1.b) and (2.c) we get
(3.a) 2𝛷𝑖 + 𝛹 = -𝐶𝛾/2,
(3.b) 𝛷𝑖 - 𝛹 = 4/5 𝑓𝜈(𝛹 + 𝛷𝑖 + 𝐶𝛾/4).
Eliminating 𝐶𝛾, we get
(3.c) 𝛹 = 𝛷𝑖(1 + 2/5 𝑓𝜈)-1 ≈ 0.86𝛷𝑖,
and using (1.b) we get 𝛿𝛾 = -2𝛹 ≈ -1.72𝛷𝑖. ▮
4. Repeating the analysis for the nutrino density isocurvature mode (see Section 6.2.3) show that
(4.1) 𝛹 = -2𝛷𝑖, 𝑆𝜈 ≡ 𝛿𝜈 - 𝛿𝛾 = (15 + 4𝑓𝜈)/𝑓𝜈(1 - 𝑓𝜈) 𝛷𝑖 ≈ 68.8 𝛷𝑖, and 𝛿𝛾 ≈ -24.2𝛷𝑖.
[Solution] We can start with (1.a)
(4.a) 3𝓗(𝛷ʹ + 𝓗𝛹) = -4π𝐺𝑎2(𝜌̄𝛾𝛿𝛾 + 𝜌̄𝜈𝛿𝜈) = - 3/2 𝓗2𝛿𝛾 = - 3/2 𝓗2(𝑓𝛾𝛿𝛾 + 𝑓𝜈𝛿𝜈),
where 𝑓𝜈 ≡ 𝜌̄𝜈/(𝜌̄𝛾 + 𝜌̄𝜈). Referring to isocurvature perturbation at early times 𝛿𝛾 = 4𝛷 - 𝑓𝜈𝑆𝜈 = 𝛿𝜈 - 𝑆𝜈, 𝑓𝛾 ≈ 1 - 𝑓𝜈 and 𝓗 = 1/𝜂, so we can calculate
(4.a) 𝑓𝛾𝛿𝛾 + 𝑓𝜈𝛿𝜈 = (1 - 𝑓𝜈)𝛿𝛾 + 𝑓𝜈(𝛿𝛾 + 𝑆𝜈) = 𝛿𝛾 + 𝑓𝜈𝑆𝜈 = 4𝛷. ⇒ 3𝓗(𝛷ʹ + 𝓗𝛹) = - 3/2 𝓗2 × 4𝛷.
(4.b) 3(1/𝜂)[𝛷ʹ + (1/𝜂)𝛹] = - 6 (1/𝜂)2𝛷. ⇒ 𝜂𝛷ʹ + 2𝛷 + 𝛹 = 0.
Taking a time derivative we get
(4.c) 𝛷ʺ + 3𝛷ʹ + 𝛹ʹ = 0.
which is the same as (1.1). Because (2.c) still hold, so we have a constant mode solution 𝛷 = 𝛷𝑖. (4.b) and (4.c) give
(4.d) 2𝛷𝑖 + 𝛹 = 0, 𝛷𝑖 - 𝛹 = 4/5 𝑓𝜈(𝛿𝜈/4 + 𝛹) = 4/5 𝑓𝜈[𝛹 + 𝛷𝑖 + (1 - 𝑓𝜈)/4 𝑆𝜈],
Since 𝑓𝜈 ≈ 0.41 [RE Exercise 6.9]
(4.e) 𝛹 = -2𝛷𝑖, 𝑆𝜈 = 𝛿𝜈 - 𝛿𝛾 = (15 + 4𝑓𝜈)/𝑓𝜈(1 - 𝑓𝜈) 𝛷𝑖 ≈ 68.8 𝛷𝑖
(4.f) 𝛿𝛾 = 4𝛷 - 𝑓𝜈𝑆𝜈 ≈ -24.2𝛷𝑖. ▮
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