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2024-10-21 22:17:39, 조회수 : 36 |
5.5 Summary
In this chapter, we have studied the growth of structure in Newtonian theory. The Newtonian approximation is valid for non-relativistic fluids-such as dark matter or baryons after decoupling-and on subhorizon scales. We began with the non-relativistic fluid equations
(5.134-6) Continuity equation: ∂𝜌/∂𝑡 + 𝛁 ⋅ (𝜌𝐮) = 0, Euler equation: ∂𝐮/∂𝑡 + (𝐮 ⋅ 𝛁)𝐮 = -𝛁𝑃/𝜌 - 𝛁𝛷,
Poisson equation: ∇2𝛷 = 4π𝐺𝜌,
and incorporated the expansion of the universe by writing the physical coordinates appearing in these equations as 𝐫 = 𝑎(𝑡)𝐱. We then expressed all quantities in terms of fluctuations around the homogeneous background solution
(5.137) 𝜌 = 𝜌̄[1 + 𝛿], 𝐮 = 𝐻𝑎𝐱 + 𝐯, 𝑃 = 𝑃̄ + 𝛿𝑃, 𝛷 = 𝛷̄ + 𝛿𝛷,
and expanded the equations of motion to linear order in these fluctuations:
(5.138-40) Continuity equation: 𝛿̇ = -1/𝑎 𝛁 ⋅ 𝐯, Euler equation: 𝐯̇ + 𝐻𝐯 = -1/𝑎𝜌̄𝛁𝛿𝑃 - 1/𝑎 𝛁𝛿𝛷,
Poisson equation: ∇2𝛷 = 4π𝐺𝑎2𝜌̄𝛿,
where ∇ is now a derivative with respect to 𝐱 and the overdot denotes a time derivative at fixed 𝐱. The linearized equations can be combined into a single equation for the evolution of the density contrast.
(5.141) 𝑑2𝛿/𝑑𝑡2 + 2𝐻𝑑𝛿/𝑑𝑡 - [𝑐𝑠2/𝑎2∇2 + 4π𝐺𝜌̄(𝑡)]𝛿 = 0,
where we have used that 𝛿𝑃 = 𝑐𝑠2𝛿𝜌 for barotropic fluid. On small scales, the pressure term dominates and we get oscillations with a frequency set by the speed of sound 𝑐𝑠. On large scales, the pressure is negligible and gravity dominates. The resulting growth of the fluctuations is the Jeans instability. We applied (5.141) to the evolution of the dark matter perturbations. During the radiation era, the rapid expansion of the space counteracts the gravitational instability and growth of the perturbations is only logarithmic, 𝛿 ~ log 𝑎. Once the universe becomes dominated by matter, the perturbations start to grow as 𝛿 ~ 𝑎.
Observations measure the spatial correlations in the large-scale structure of the universe. In linear theory, these correlations are easiest to predict in Fourier space. The simplest correlation function in Fourier space is the power spectrum
(5.142) ⟨𝛿(𝐤, 𝑡)𝛿*(𝐤', 𝑡)⟩ = (2π)3𝛿𝐷(𝐤 - 𝐤') 𝒫(𝑘, 𝑡),
where the Dirac delta function is a consequence of the homogeneity of the background. For a Gaussian random field. The late-time power spectrum can be written as
(5.143) 𝒫(𝑘, 𝑡) = 𝛵2(𝑘) 𝐷+2(𝑡)/𝐷+(𝑡𝑖) 𝒫(𝑘, 𝑡𝑖),
where 𝐷+(𝑡) is the linear growth factor and 𝛵(𝑘) is the transfer function. A nontrivial transfer function arises because of the suppressed growth of subhorizon modes during the radiation era. The asymptotic scaling of the transfer function are
(5.144) 𝛵(𝑘) ≈ { 1 𝑘 < 𝑘eq, (𝑘eq/𝑘)2 𝑘 > 𝑘eq,
where 𝑘eq is the wavenumber at matter-radiation equality. The primordial power spectrum is
(5.145) 𝒫(𝑘, 𝑡𝑖) = 𝛢𝑘𝑛,
where 𝑛 ≈ 1 for the Harrison-Zel'dovich spectrum. A key property of the spectrum is that gravitational potential is scale invariant. Inflation predicts percent-level deviations from a perfectly scale-invariant spectrum. This prediction of inflation was verified by observations of the CMB, with the measured value of the spectral index being 𝑛 = 0.967 ± 0.004.
Finally, we studied a simple toy model for the nonlinear growth of density perturbations. We showed that the evolution of spherically symmetric density perturbation in a flat Einstein-de Sitter universe is the same as that of a closed FRW universe and hence can be described analytically. Initially, the overdensity expands with the rest of the universe and the growth of the density contrast is consistent with the prediction of linear perturbation, 𝛿 ∝𝑎(𝑡). Like for a universe with positive curvature the expansion eventually stops and the matter recollapses. Small deviations from spherical symmetry get amplified during the collapse and the matter settles into a state of virial equilibrium. The predicted overdensity of the virialized halos is 𝛿vir -177. Virialization occurs when the linearly-evolved density contrast crosses the threshold 𝛿𝑐 = 1.69. Press and Schechter used these insights from the spherical collapse model to derive the halo mass function, i.e. the abundance of halos as a function of their mass.
5 Problems
Problem 5.1 Fluid equations
In Chapter 2, we showed that the energy-momentum tensor of a perfect fluid in a curved spacetime is
(P5.1) 𝛵𝜇𝜈 = (𝜌 + 𝑃)𝑈𝜇𝑈𝜈 + 𝑃𝑔𝜇𝜈, where the 4-velocity satisfies 𝑔𝜇𝜈𝑈𝜇𝑈𝜈 = -1.
1. Show that
(1) ∇𝛼𝑔𝜇𝜈 = 0, 𝑈𝜈∇𝜇𝑈𝜈 = 0, for arbitrary 𝑔𝜇𝜈.
[Solution] The covariant derivative of the metric is [RE (2.91)(2.28)]
(1.a) ∇𝛼𝑔𝜇𝜈 = ∂𝛼𝑔𝜇𝜈 - 𝛤𝜆𝛼𝜇𝑔𝜆𝜈 - 𝛤𝜆𝛼𝜈𝑔𝜆𝜇,
(1.b) 𝛤𝜆𝛼𝜇𝑔𝜆𝜈 = 1/2 𝑔𝜆𝜎[∂𝛼𝑔𝜎𝜈 + ∂𝜈𝑔𝜎𝛼 - ∂𝜎𝑔𝜈𝛼]𝑔𝜆𝜈 = 1/2 𝛿𝜎𝜈[∂𝛼𝑔𝜎𝜈 + ∂𝜈𝑔𝜎𝛼 - ∂𝜎𝑔𝜈𝛼] = 1/2 [∂𝛼𝑔𝜇𝜈 + ∂𝜈𝑔𝜇𝛼 - ∂𝜇𝑔𝜈𝛼]
The upper 𝜎 in 𝛿𝜎𝜈 changed the lower 𝜎 into 𝜇 in the combined components. Similarly,
(1.c) 𝛤𝜆𝛼𝜈𝑔𝜇𝜈 = 1/2 [∂𝛼𝑔𝜈𝜇 + ∂𝜇𝑔𝜈𝛼 - ∂𝜈𝑔𝜇𝛼].
(1.d) (f) + (g): 𝛤𝜆𝛼𝜇𝑔𝜆𝜈 + 𝛤𝜆𝛼𝜈𝑔𝜇𝜈 = ∂𝛼𝑔𝜇𝜈 Substituting (e), we get ∇𝛼𝑔𝜇𝜈 = 0.
(1.e) To show 𝑈𝜈∇𝜇𝑈𝜈 = 0, we take a partial derivative on 𝑔𝜇𝜈𝑈𝛼𝑈𝜈 = -1. Using ∇𝛼𝑔𝜇𝜈 = 0, we have
(1.f) 𝑔𝜇𝜈(∇𝛼𝑈𝜇)𝑈𝜈 + 𝑔𝜇𝜈∇𝛼𝑈𝜇∇𝛼𝑈𝜈 = 0. ⇒ 𝑈𝜇∇𝛼𝑈𝜇 = 0. ⇒ 𝑈𝜈∇𝜇𝑈𝜈 = 0. ▮
2. Show that the conservation of the energy-momentum tensor, ∇𝜇𝛵𝜇𝜈 = 0, implies
(2) Continuity: 𝑈𝜇∇𝜇𝜌 + (𝜌 + 𝑃)∇𝜇𝑈𝜇 = 0, Euler: (𝜌 + 𝑃)𝑈𝜇∇𝜇𝑈𝜇 = -(𝑔𝜇𝜈 + 𝑈𝜇𝑈𝜈)∇𝜇𝑃.
Do the fluid elements move on geodesics? If not, under what condition would they move on geodesics?
[Solution] Consider the energy-momentum tensor of a perfect fluid and the conservation equation implies
(2.a) 0 = ∇𝜇𝛵𝜇𝜈 = ∇𝜇[(𝜌 + 𝑃)𝑈𝜇𝑈𝜈 + 𝑃𝑔𝜇𝜈]
= ∇𝜇(𝜌 + 𝑃)𝑈𝜇𝑈𝜈 + (𝜌 + 𝑃)(∇𝜇𝑈𝜇)𝑈𝜈 + (𝜌 + 𝑃)𝑈𝜇∇𝜇𝑈𝜈 + (∇𝜇𝑃)𝑔𝜇𝜈,
where we used ∇𝜇𝑔𝜇𝜈 = 0. Now, multiplying 𝑈𝜈 at both sides and using equations 𝑈𝜈𝑈𝜈 = -1 and 𝑈𝜈∇𝜇𝑈𝜈 = 0, we find
(2.b) -𝑈𝜇∇𝜇(𝜌 + 𝑃) - (𝜌 + 𝑃)(∇𝜇𝑈𝜇) + 𝑈𝜇∇𝜇𝑃 = 0. ⇒ 𝑈𝜇∇𝜇𝜌 + (𝜌 + 𝑃)∇𝜇𝑈𝜇 = 0.
From (a) subtracting the result of multiplying 𝑈𝜈 at both sides of (b), then we find
(2.c) -(𝜌 + 𝑃)(∇𝜇𝑈𝜇)𝑈𝜈 + (∇𝜇𝑃)𝑈𝜇𝑈𝜈 + (𝜌 + 𝑃)(∇𝜇𝑈𝜇)𝑈𝜈 + (𝜌 + 𝑃)𝑈𝜇∇𝜇𝑈𝜈 + (∇𝜇𝑃)𝑔𝜇𝜈 = 0. ⇒ (𝜌 + 𝑃)𝑈𝜇∇𝜇𝑈𝜇 = -(𝑔𝜇𝜈 + 𝑈𝜇𝑈𝜈)∇𝜇𝑃.
Since in general 𝑈𝜇∇𝜇𝑈𝜇 ≠ 0 and the fluid elements do not move on geodesics. But if the the gradient of the pressure vanishes, then they would move on geodesics. ▮ [verification needed]
3. Let 𝑔𝜇𝜈 ≈ 𝜂𝜇𝜈, 𝑈𝜇 ≈ (1, 𝐮), with ∣𝐮∣ ≪ 1, and 𝑃 ≪ 𝜌. Show that the above equations become
(3) Continuity: 𝜌̇ + ∇ ⋅ (𝜌𝐮) = 0, Euler: 𝜌(𝐮̇ + 𝐮 ⋅ ∇𝐮) = -∇𝑃.
[Solution] 𝑈𝜇 ≈ (1, 𝐮) = (1, 𝑢𝑖), The continuity equation (2.b) becomes
(3.a) 𝑈0∇0𝜌 + 𝑈𝑖∇𝑖𝜌 + (𝜌 + 𝑃)[∇0𝑈0 + ∇𝑖𝑈𝑖]= 0, ∂0𝜌 + 𝑈𝑖∇𝑖𝑈𝑖∇𝑖𝜌 + 𝜌∇0𝑈0 + ∇𝑖𝑈𝑖 = 0, with 𝑃 ≪ 𝜌
Using that ∇0𝑈0 = ∇0(1) = 0. With 𝑔𝜇𝜈 ≈ 𝜂𝜇𝜈, ∂0𝜌 = 𝜌̇, ∇𝑖 = ∂𝑖 so we find
(3.b) 𝜌̇ + ∂𝑖(𝜌𝑢𝑖) or 𝜌̇ + ∇ ⋅ (𝜌𝐮) = 0
Similarly, the Euler equation (2.c) becomes
(3.c) 𝜌(𝑈0∇0𝑈𝜈 +𝑈𝑖∇𝑖𝑈𝜈) = -(𝜂𝜇𝜈 + 𝑈𝜇𝑈𝜈)∇𝜇𝑃
𝜌(∇0𝑈𝜈 +𝑈𝑖∇𝑖𝑈𝜈) = -𝜂𝜇𝜈∇𝜇𝑃
𝜌(∂0𝑢𝑖 +𝑢𝑗∂𝑖𝑢𝑖) = -∂𝑖𝑃 or 𝜌(𝐮̇ + 𝐮 ⋅ ∇𝐮) = -∇𝑃. ▮
Problem 5.2 Growth of matter perturbations
Above the Jean scale, but below the Hubble radius, the evolution of matter perturbations is governed by
(P5.2) 𝑑𝛿̇𝑚/𝑑𝑡 + 2𝐻𝛿̇𝑚 = 4π𝐺𝜌̄𝑚𝛿𝑚, [𝛿̇𝑚 ≡ 𝑑𝛿𝑚/𝑑𝑡]
where 𝜌̄𝑚 = 𝜌̄𝑚,0 𝑎-3. We have ignored the fluctuations in the radiation and the dark energy on the right-hand side, but still include their homogeneous densities as a source for the Hubble rate.
1. Show that the evolution equation can be written as
(1) 𝑑/𝑑𝑎 (𝑎3𝐻 𝑑𝛿𝑚/𝑑𝑎) = 4π𝐺𝜌̄𝑚,0 𝛿𝑚/𝐻𝑎2.
[Solution] with 𝜌̄𝑚 = 𝜌̄𝑚,0 𝑎-3 for the right-hand side
(1.a) 𝛿̇𝑚 = 𝑑𝑎/𝑑𝑡 𝑑𝛿𝑚/𝑑𝑎 = 𝑎𝐻 𝑑𝛿𝑚/𝑑𝑎, 𝑑𝛿̇𝑚/𝑑𝑡 = 𝑎𝐻 𝑑/𝑑𝑎 (𝑎𝐻 𝑑𝛿𝑚/𝑑𝑎)
(1.b) 𝑑𝛿̇𝑚/𝑑𝑡 + 2𝐻 𝛿̇𝑚 = 𝑎𝐻 𝑑/𝑑𝑎 (𝑎𝐻 𝑑𝛿𝑚/𝑑𝑎) + 2𝑎𝐻2𝛿̇ = 𝑑/𝑑𝑎 (𝑎2𝐻2𝛿̇)
(1.c) 𝑑/𝑑𝑎 (𝑎2𝐻2𝛿̇) = 4π𝐺𝜌̄𝑚,0𝛿𝑚/𝑎3 ⇒ 𝐻/𝑎 𝑑/𝑑𝑎 (𝑎3𝐻 𝑑𝛿𝑚/𝑑𝑎) = 𝐻/𝑎 4π𝐺𝜌̄𝑚,0𝛿𝑚/𝐻𝑎2. ▮
2. Assuming that the universe contains a combination of matter, curvature and a cosmological constant, show that the solutions to (1) are
(2) 𝛿𝑚 ∝ { 𝐻, 𝐻 ∫ 𝑑𝑎/(𝑎𝐻)3.
What is the scaling of the growing mode solution when the expansion of the universe is by matter, by curvature, or by a cosmological constant?
[Solution] Let us denote derivative with respect to 𝑎 by primes and assume 𝑢 = 𝛿𝑚/𝐻 and 𝑑𝛿𝑚/𝑑𝑎 = 𝑑(𝐻𝑢)/𝑑𝑎 = 𝐻𝑢' + 𝐻'𝑢.
(2.a) 𝐻2 = 𝐻02(𝛺𝑚𝑎-3 + 𝛺𝑘𝑎-2 + 𝛺𝛬).
(2.b) 𝐻𝐻' = 𝐻02(-3/2 𝛺𝑚𝑎-4 + 𝛺𝑘𝑎-3)
(2.c) (𝐻𝐻')' = 𝐻02(6𝛺𝑚𝑎-5 + 3𝛺𝑘𝑎-4)
(2.d) 𝑑/𝑑𝑎 (𝑎3𝐻 𝑑𝛿𝑚/𝑑𝑎) = [𝑎3𝐻(𝐻𝑢' + 𝐻'𝑢)]' = 3𝑎2𝐻2𝑢' + 𝑎32𝐻𝐻'𝑢' + 𝑎3𝐻2𝑢" + 3𝑎2𝐻𝐻'𝑢 + 𝑎3(𝐻𝐻')'𝑢 + 𝑎3𝐻𝐻'𝑢'
= 𝑎3[𝐻2𝑢" + 3𝐻𝐻'𝑢' + (𝐻𝐻')'𝑢] + 3𝑎2(𝐻2𝑢' + 𝐻𝐻'𝑢) = 𝑎3[𝐻2𝑢" + 3(𝐻𝐻' + 𝑎-1𝐻2)𝑢' + {(𝐻𝐻')' + 3𝑎-1𝐻𝐻'}𝑢].
Using (b) and (c) we can calculate the last term with 𝑢 and 𝐻2 = 𝐻02𝛺𝑚𝑎-3 = 8/3 π𝐺𝜌̄𝑚,0 𝑎-3
(2.e) 𝑎3{(𝐻𝐻')' + 3𝑎-1𝐻𝐻'}𝑢 = 𝑎3𝛿𝑚/𝐻 𝐻02[6𝛺𝑚𝑎-5 + 3𝛺𝑘𝑎-4 + 3(-3/2 𝛺𝑚𝑎-5 + 𝛺𝑘𝑎-4)] = 3/2 𝛿𝑚/𝐻 𝐻02𝛺𝑚𝑎-2 = 4π𝐺𝜌̄𝑚,0𝛿𝑚/𝐻𝑎2
Since (e) is the same as 𝑑/𝑑𝑎 (𝑎3𝐻 𝑑𝛿𝑚/𝑑𝑎), we have
(2.f) 𝐻2𝑢" + 3(𝐻𝐻' + 𝑎-1𝐻2)𝑢' = 0 ⇒ 𝑢" + 3[𝐻'/𝐻 + 𝑎-1]𝑢' = 𝑢" + 3[𝑑 ln (𝑎𝐻)]/𝑑𝑎 𝑢' = 0.
We can find two solutions
(2.g) 𝑢' = 0 ⇒ 𝛿𝑚 ∝ 𝐻,
(2.h) 𝑢"/𝑢' = (ln 𝑢')/𝑑𝑎 = -3[𝑑 ln (𝑎𝐻)]/𝑑𝑎 ⇒ 𝑢' ∝ (𝑎𝐻)-3 ⇒ 𝛿𝑚 ∝ 𝐻 ∫ 𝑑𝑎/(𝑎𝐻)3.
The scaling of the growing mode solution is when the expansion of the universe dominated by matter, by curvature or a cosmological constant in case of (g) and (h):
(2.i) 𝛿𝑚 ∝ 𝐻 ∝ 𝑎-3/2, 𝑎-1 and 𝑎0 respectively according to Friedmann equation (2.144).
(2.j) 𝛿𝑚 ∝ 𝐻 ∫ 𝑑𝑎/(𝑎𝐻)3 ∝ 𝑎, 𝑎0 and 𝑎-2 respectively. [𝐻 = { 𝑎-3/2, 𝑎-1, 𝑎0 ⇒ 𝐻 ∫ 𝑑𝑎/(𝑎𝐻)3 = { 2𝑎/5 + C, 𝑎0 + C, -1/2𝑎2 + C.] ▮
3. At early times the universe was dominated by radiation and pressreless matter and we may ignore baryons.
Let 𝑦 ≡ 𝑎/𝑎eq, where 𝑎eq is the scale factor at matter-radiation equality. Show that the Hubble parameter can be written as
(3.1) 𝐻(𝑦) = 𝛢/𝑦2√(1 + 𝑦),
where the constant 𝛢 should be determined in terms of the energy density at equality, 𝜌eq. Show that (1) is equivalent to
(3.2) 𝑑2𝛿𝑚/𝑑𝑦2 + (2 + 3𝑦)/2𝑦(1 + 𝑦) 𝑑𝛿𝑚/𝑑𝑦 = 3/2 𝛿𝑚/𝑦(1 + 𝑦).
Verify that the solutions of this equation are
(3.3) 𝛿𝑚 ∝ { 1 + 3/2 𝑦 ;(1 + 3/2 𝑦) ln [√(1 + 𝑦) + 1]/[√(1 - 𝑦) + 1] - 3√(1 + 𝑦).
What are the growing and decaying modes at early late times?
[Solution] According to Friedmann equations in matter and radiation era with 𝑦 ≡ 𝑎/𝑎eq, we have
(3.a) 𝐻2 = 8π𝐺/3 (𝜌̄𝑟,0𝑎-4 + 𝜌̄𝑚,0𝑎-3) = 8π𝐺𝜌̄𝑟,0/3𝑎4 (1 + 𝑎/𝑎eq) = 8π𝐺𝜌eq/3 (1 + 𝑦)/𝑦4, 𝑎eq ≡ 𝜌̄𝑟,0/𝜌̄𝑚,0 and 𝜌eq ≡ 𝜌̄𝑟,0𝑎eq-4. ⇒
(3.b) 𝐻 = 𝛢/𝑦2 √(1 + 𝑦), with 𝛢 ≡ √(8π𝐺𝜌eq/3)
(3.c) 4π𝐺 𝜌̄𝑚,0 = 4π𝐺 𝜌eq𝑎3eq = 3/2 𝛢2𝑎3eq,
Using this we can write (1) as
(3.d) 𝑑/𝑑𝑎 (𝑎3𝐻 𝑑𝛿𝑚/𝑑𝑎) = 4π𝐺𝜌̄𝑚,0 𝛿𝑚/𝐻𝑎2. ⇒ 𝑑/𝑑𝑦 (𝑦3𝐻 𝑑𝛿𝑚/𝑑𝑦) = 3/2 𝛢2 𝛿𝑚/𝐻𝑦2.
Substituting (3.b)
(3.e) 𝑑2𝛿𝑚/𝑑𝑦2 + (2 + 3𝑦)/2𝑦(1 + 𝑦) 𝑑𝛿𝑚/𝑑𝑦 = 3/2 𝛿𝑚/𝑦(1 + 𝑦).
To find a solution, multiplying by 2𝑦(1 + 𝑦), we get
(3.f) 2𝑦(1 + 𝑦) 𝑑2𝛿𝑚/𝑑𝑦2 + (2 + 3𝑦) 𝑑𝛿𝑚/𝑑𝑦 - 3𝛿𝑚 = 0.
(3.g) 𝛿𝑚 = 𝐶1(1 + 3/2 𝑦) + 𝐶2[(1 + 3/2 𝑦) ln {√(1 + 𝑦) + 1}/{√(1 - 𝑦) + 1} - 3√(1 + 𝑦)] ⇒
(3.h) 𝛿𝑚 ∝ { 1 + 3/2 𝑦 ;(1 + 3/2 𝑦) ln [√(1 + 𝑦) + 1]/[√(1 - 𝑦) + 1] - 3√(1 + 𝑦).
At early times (𝑦 ≪ 1) the growing mode scales logarithmically as
(3.i) 𝛿𝑚 { 2/3 ; 3/2 ln 𝑦 by expanding and linearizing.
At late times (𝑦 ≫ 1) the growing mode scales linearly as
(3.i) 𝛿𝑚 { 𝑦 ; -5/2 𝑦-3/2 by expanding and linearizing. ▮
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