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宇宙論 · AIのノート · 院 · #cosmology

Complete Cosmology

A concise guide to the expanding universe, Friedmann equations, thermal history, perturbations, inflationary initial conditions, CMB, lensing, and structure.

· 17 分

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The Concordance Model of Cosmology

Core ideas

The concordance model, Λ\LambdaCDM, describes a nearly flat universe with radiation, baryons, cold dark matter, dark energy, and nearly scale-invariant primordial perturbations. It is not a list of facts: it is a parameterized model connecting expansion, nucleosynthesis, CMB anisotropies, and galaxy clustering.

For review, be able to state the components of Λ\LambdaCDM, define density parameters, explain flatness, and identify the key observational pillars. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

Ωi=ρiρcrit,ρcrit=3H028πG,∑iΩi≃1\Omega_i=\frac{\rho_i}{\rho_{\rm crit}},\qquad \rho_{\rm crit}=\frac{3H_0^2}{8\pi G},\qquad \sum_i\Omega_i\simeq1

Worked example

Consider a flat Λ\LambdaCDM universe with H0=70 km s−1 Mpc−1H_0 = 70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}, Ωm=0.3\Omega_m = 0.3, and ΩΛ=0.7\Omega_\Lambda = 0.7.

  1. Compute the critical density: ρcrit=3H028πG=3(70 km s−1 Mpc−1)28π(6.674×10−11 m3 kg−1 s−2)≈9.2×10−27 kg m−3.\rho_{\rm crit} = \frac{3H_0^2}{8\pi G} = \frac{3(70\ \mathrm{km\,s^{-1}\,Mpc^{-1}})^2}{8\pi (6.674\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}})} \approx 9.2\times10^{-27}\ \mathrm{kg\,m^{-3}}.
  2. The present matter density is ρm=Ωmρcrit≈0.3×9.2×10−27≈2.8×10−27 kg m−3\rho_m = \Omega_m \rho_{\rm crit} \approx 0.3 \times 9.2\times10^{-27} \approx 2.8\times10^{-27}\ \mathrm{kg\,m^{-3}}.
  3. The cosmological constant energy density is ρΛ=ΩΛρcrit≈6.4×10−27 kg m−3\rho_\Lambda = \Omega_\Lambda \rho_{\rm crit} \approx 6.4\times10^{-27}\ \mathrm{kg\,m^{-3}}, which corresponds to Λ=8πGρΛ≈1.2×10−52 m−2\Lambda = 8\pi G \rho_\Lambda \approx 1.2\times10^{-52}\ \mathrm{m^{-2}}.

Because Ωm+ΩΛ=1\Omega_m + \Omega_\Lambda = 1, the curvature parameter Ωk≃0\Omega_k \simeq 0, consistent with a spatially flat geometry.

Problems with Solutions

Problem 1. Calculate ρcrit\rho_{\rm crit} for H0=67.4 km s−1 Mpc−1H_0 = 67.4\ \mathrm{km\,s^{-1}\,Mpc^{-1}}.

Solution. Convert H0H_0 to SI: 67.4 km s−1 Mpc−1=2.184×10−18 s−167.4\ \mathrm{km\,s^{-1}\,Mpc^{-1}} = 2.184\times10^{-18}\ \mathrm{s^{-1}}. Then

ρcrit=3(2.184×10−18)28π(6.674×10−11)≈8.5×10−27 kg m−3.\rho_{\rm crit} = \frac{3(2.184\times10^{-18})^2}{8\pi(6.674\times10^{-11})} \approx 8.5\times10^{-27}\ \mathrm{kg\,m^{-3}}.

Problem 2. A model has Ωm=0.25\Omega_m = 0.25, ΩΛ=0.70\Omega_\Lambda = 0.70, and Ωk=0.05\Omega_k = 0.05. Verify that ∑iΩi=1\sum_i \Omega_i = 1 and state whether the universe is open or closed.

Solution. The sum is 0.25+0.70+0.05=1.000.25 + 0.70 + 0.05 = 1.00. Because Ωk>0\Omega_k > 0, the curvature term is negative in the Friedmann equation and the spatial geometry is open (negative curvature).

Problem 3. At redshift z=1z=1, by what factor is the matter density higher than today, assuming matter domination?

Solution. In a matter-dominated era, ρm∝a−3\rho_m \propto a^{-3} and a=(1+z)−1a = (1+z)^{-1}. At z=1z=1, a=1/2a=1/2, so ρm(z=1)=ρm,0(1+z)3=ρm,0 23=8ρm,0\rho_m(z=1) = \rho_{m,0}(1+z)^3 = \rho_{m,0}\, 2^3 = 8\rho_{m,0}. The density is 8 times larger.

Section summary. Modern cosmology is organized around the predictive Λ\LambdaCDM model.

The Expanding Universe

Core ideas

Expansion is encoded in the scale factor a(t)a(t). Cosmological redshift measures how wavelengths stretch with the universe. Distances are subtle because emission time, observation time, and geometry differ; comoving, angular-diameter, and luminosity distances answer different observational questions.

For review, be able to convert between scale factor and redshift, define Hubble rate, distinguish common cosmological distances, and interpret lookback time. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

1+z=a0a(t),H=a˙a,dL=(1+z)2dA1+z=\frac{a_0}{a(t)},\qquad H=\frac{\dot a}{a},\qquad d_L=(1+z)^2d_A

Worked example

A Type Ia supernova is observed at redshift z=0.5z = 0.5 in a flat universe. Its observed flux yields a luminosity distance dL=2.8 Gpcd_L = 2.8\ \mathrm{Gpc}.

  1. The angular-diameter distance is dA=dL(1+z)2=2.8 Gpc(1.5)2≈1.24 Gpc.d_A = \frac{d_L}{(1+z)^2} = \frac{2.8\ \mathrm{Gpc}}{(1.5)^2} \approx 1.24\ \mathrm{Gpc}.
  2. A physical diameter D=10 kpcD = 10\ \mathrm{kpc} at this redshift subtends an angle θ=DdA=10 kpc1.24×106 kpc≈8.1×10−6 rad≈1.7′′.\theta = \frac{D}{d_A} = \frac{10\ \mathrm{kpc}}{1.24\times10^6\ \mathrm{kpc}} \approx 8.1\times10^{-6}\ \mathrm{rad} \approx 1.7''.
  3. The scale factor at emission was a=(1+z)−1=2/3a = (1+z)^{-1} = 2/3, meaning the universe was 2/32/3 its present size when the light was emitted.

Problems with Solutions

Problem 1. What is the scale factor corresponding to z=3z = 3?

Solution. a=(1+z)−1=1/4=0.25a = (1+z)^{-1} = 1/4 = 0.25.

Problem 2. A galaxy at z=2z = 2 has a physical separation of 100 kpc100\ \mathrm{kpc} from a quasar. What is their comoving separation?

Solution. Comoving distance is independent of expansion, defined as r=dphys/a=dphys(1+z)r = d_{\rm phys}/a = d_{\rm phys}(1+z). Thus r=100 kpc×3=300 kpcr = 100\ \mathrm{kpc} \times 3 = 300\ \mathrm{kpc}.

Problem 3. In a matter-dominated flat universe, the Hubble parameter scales as H(z)=H0(1+z)3/2H(z) = H_0(1+z)^{3/2}. Compute HH at z=1000z = 1000 for H0=70 km s−1 Mpc−1H_0 = 70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}.

Solution. H(1000)=70×(1001)3/2≈70×3.17×104≈2.2×106 km s−1 Mpc−1H(1000) = 70 \times (1001)^{3/2} \approx 70 \times 3.17\times10^4 \approx 2.2\times10^6\ \mathrm{km\,s^{-1}\,Mpc^{-1}}.

Section summary. Expansion turns cosmic time into observable redshift and distance relations.

The Fundamental Equations of Cosmology

Core ideas

The Friedmann equations follow from applying Einstein’s equation to a homogeneous and isotropic universe. Fluids dilute according to their equation of state: radiation as a−4a^{-4}, matter as a−3a^{-3}, and a cosmological constant as constant density.

For review, be able to derive density scaling, use Friedmann acceleration, compare radiation-, matter-, and dark-energy-dominated eras. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

H2=8πG3ρ−kc2a2+Λc23,ρ˙+3H(ρ+p/c2)=0H^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}+\frac{\Lambda c^2}{3},\qquad \dot\rho+3H(\rho+p/c^2)=0

Worked example

Consider a spatially flat (k=0k=0), matter-dominated universe with Λ=0\Lambda = 0. The Friedmann equation reduces to

H2=8πG3ρm.H^2 = \frac{8\pi G}{3}\rho_m.

Because ρm=ρm,0a−3\rho_m = \rho_{m,0} a^{-3}, we write H=a˙/aH = \dot a/a and obtain

a˙2=8πGρm,031a.\dot a^2 = \frac{8\pi G\rho_{m,0}}{3} \frac{1}{a}.

Let a∝t2/3a \propto t^{2/3}. Then a˙=23t−1/3a02/3\dot a = \frac{2}{3} t^{-1/3} a_0^{2/3} and

H=a˙a=23t.H = \frac{\dot a}{a} = \frac{2}{3t}.

The age of such a universe is t0=2/(3H0)t_0 = 2/(3H_0). For H0=70 km s−1 Mpc−1≈2.27×10−18 s−1H_0 = 70\ \mathrm{km\,s^{-1}\,Mpc^{-1}} \approx 2.27\times10^{-18}\ \mathrm{s^{-1}},

t0=23×2.27×10−18≈9.4×1017 s≈9.3 Gyr.t_0 = \frac{2}{3 \times 2.27\times10^{-18}} \approx 9.4\times10^{17}\ \mathrm{s} \approx 9.3\ \mathrm{Gyr}.

Problems with Solutions

Problem 1. Show that radiation density scales as ρr∝a−4\rho_r \propto a^{-4} using the continuity equation ρ˙+3H(ρ+p/c2)=0\dot\rho + 3H(\rho + p/c^2) = 0 with p=ρc2/3p = \rho c^2/3.

Solution. Substitute p=ρc2/3p = \rho c^2/3:

ρ˙+3H(ρ+ρ3)=ρ˙+4Hρ=0.\dot\rho + 3H\left(\rho + \frac{\rho}{3}\right) = \dot\rho + 4H\rho = 0.

Since H=a˙/aH = \dot a/a, this becomes dρ/ρ=−4 da/ad\rho/\rho = -4\, da/a, integrating to ρ∝a−4\rho \propto a^{-4}.

Problem 2. A flat universe has Ωm=0.3\Omega_m = 0.3, ΩΛ=0.7\Omega_\Lambda = 0.7. Write the Hubble parameter H(z)H(z) and evaluate it at z=1z = 1. Solution.

H(z)=H0Ωm(1+z)3+ΩΛ.H(z) = H_0\sqrt{\Omega_m(1+z)^3 + \Omega_\Lambda}.

At z=1z=1: H(1)=H00.3×8+0.7=H03.1≈1.76 H0≈123 km s−1 Mpc−1H(1) = H_0\sqrt{0.3 \times 8 + 0.7} = H_0\sqrt{3.1} \approx 1.76\,H_0 \approx 123\ \mathrm{km\,s^{-1}\,Mpc^{-1}}.

Problem 3. In a matter-dominated universe with small curvature, the Friedmann equation is H2=H02[Ωma−3+(1−Ωm)a−2]H^2 = H_0^2[\Omega_m a^{-3} + (1-\Omega_m)a^{-2}]. At what scale factor does the curvature term equal the matter term?

Solution. Set Ωma−3=(1−Ωm)a−2\Omega_m a^{-3} = (1-\Omega_m)a^{-2}, giving a=Ωm/(1−Ωm)a = \Omega_m/(1-\Omega_m). For Ωm=0.3\Omega_m = 0.3, a=0.3/0.7≈0.43a = 0.3/0.7 \approx 0.43.

Section summary. Friedmann dynamics connects cosmic contents to expansion.

The Origin of Species

Core ideas

The early universe was hot and dense, so particle abundances were set by equilibrium, freeze-out, decays, and nuclear reactions. Big-bang nucleosynthesis predicts light elements; recombination releases the CMB; baryogenesis and dark matter freeze-out are deeper origin questions.

For review, be able to explain thermal equilibrium, freeze-out, nucleosynthesis, recombination, and why relic abundances probe early physics. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

Γ(T)∼H(T)at freeze ⁣− ⁣out,T∝a−1\Gamma(T)\sim H(T)\quad\mathrm{at\ freeze\!-\!out},\qquad T\propto a^{-1}

Worked example

Estimate the temperature of the universe at photon decoupling (recombination). The ionization energy of hydrogen is 13.6 eV13.6\ \mathrm{eV}. Recombination occurs roughly when the typical photon energy is a few times below the ionization energy, i.e., Trec∼0.3 eV/kB≈3500 KT_{\rm rec} \sim 0.3\ \mathrm{eV}/k_B \approx 3500\ \mathrm{K}.

Using T∝a−1T \propto a^{-1} and T0=2.725 KT_0 = 2.725\ \mathrm{K} today, the corresponding redshift is

1+zrec=TrecT0≈35002.725≈1284,1+z_{\rm rec} = \frac{T_{\rm rec}}{T_0} \approx \frac{3500}{2.725} \approx 1284,

so zrec≈1100z_{\rm rec} \approx 1100, which matches detailed calculations. At this time, the photon mean free path became larger than the Hubble radius and the CMB was released.

Problems with Solutions

Problem 1. The CMB temperature today is T0=2.725 KT_0 = 2.725\ \mathrm{K}. What was the photon temperature at z=100z = 100?

Solution. T(z)=T0(1+z)=2.725×101≈275 KT(z) = T_0(1+z) = 2.725 \times 101 \approx 275\ \mathrm{K}.

Problem 2. The number density of CMB photons today is nγ≈410 cm−3n_\gamma \approx 410\ \mathrm{cm^{-3}}. What was it at z=1100z = 1100?

Solution. Photons dilute as a−3a^{-3} and are conserved (after decoupling), so nγ(z)=nγ,0(1+z)3n_\gamma(z) = n_{\gamma,0}(1+z)^3. Thus nγ(1100)≈410×(1101)3≈5.5×1011 cm−3n_\gamma(1100) \approx 410 \times (1101)^3 \approx 5.5\times10^{11}\ \mathrm{cm^{-3}}.

Problem 3. Big-bang nucleosynthesis occurs at T∼80 keVT \sim 80\ \mathrm{keV}. Convert this to kelvin and estimate the corresponding redshift.

Solution. T=80×103 eV/(8.617×10−5 eV K−1)≈9.3×108 KT = 80\times10^3\ \mathrm{eV}/(8.617\times10^{-5}\ \mathrm{eV\,K^{-1}}) \approx 9.3\times10^8\ \mathrm{K}. The redshift is z≈T/T0≈3.4×108z \approx T/T_0 \approx 3.4\times10^8.

Section summary. Cosmic species are fossils of early-universe reaction rates.

The Inhomogeneous Universe: Matter and Radiation

Core ideas

Small perturbations grow into structure. Density contrast, velocity, gravitational potential, and radiation perturbations evolve differently before and after horizon entry. Photon-baryon acoustic oscillations leave signatures in the CMB and matter power spectrum.

For review, be able to define density contrast, explain horizon entry, describe acoustic oscillations, and read a matter power spectrum qualitatively. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

δ(x,t)=ρ(x,t)−ρˉ(t)ρˉ(t),P(k)=⟨∣δk∣2⟩\delta(\bm x,t)=\frac{\rho(\bm x,t)-\bar\rho(t)}{\bar\rho(t)},\qquad P(k)=\langle|\delta_{\bm k}|^2\rangle

Worked example

In the linear regime, a matter overdensity has δ=0.01\delta = 0.01 inside a region of comoving radius R=10 MpcR = 10\ \mathrm{Mpc}. The mean matter density today is ρˉm≈Ωmρcrit≈2.5×10−27 kg m−3\bar\rho_m \approx \Omega_m \rho_{\rm crit} \approx 2.5\times10^{-27}\ \mathrm{kg\,m^{-3}}.

  1. The local density is ρ=ρˉ(1+δ)=1.01 ρˉ≈2.53×10−27 kg m−3\rho = \bar\rho(1+\delta) = 1.01\,\bar\rho \approx 2.53\times10^{-27}\ \mathrm{kg\,m^{-3}}.
  2. The excess mass in the region is ΔM=δ ρˉ×4π3R3=0.01×2.5×10−27 4π3(10×3.086×1022 m)3≈3.1×1044 kg≈1.5×1014 M⊙.\Delta M = \delta\,\bar\rho \times \frac{4\pi}{3}R^3 = 0.01 \times 2.5\times10^{-27}\ \frac{4\pi}{3} (10\times3.086\times10^{22}\ \mathrm{m})^3 \approx 3.1\times10^{44}\ \mathrm{kg} \approx 1.5\times10^{14}\ M_\odot.
  3. If the power spectrum at this scale is P(k)≈104 (Mpc/h)3P(k) \approx 10^4\ (\mathrm{Mpc}/h)^3, the variance of the Fourier mode is ⟨∣δk∣2⟩=P(k)≈104 (Mpc/h)3\langle|\delta_{\bm k}|^2\rangle = P(k) \approx 10^4\ (\mathrm{Mpc}/h)^3.

Problems with Solutions

Problem 1. Define the density contrast δ\delta and explain what δ=−0.5\delta = -0.5 means physically.

Solution. δ=(ρ−ρˉ)/ρˉ\delta = (\rho - \bar\rho)/\bar\rho. A value of −0.5-0.5 means the local density is half the cosmic mean; the region is an underdense void.

Problem 2. The matter power spectrum has dimensions of (length)3^3. Why?

Solution. In three dimensions, the Fourier transform of δ(x)\delta(\bm x) uses d3xd^3x, so ∣δk∣2|\delta_{\bm k}|^2 is dimensionless and P(k)=⟨∣δk∣2⟩P(k) = \langle|\delta_{\bm k}|^2\rangle must have dimensions of k−3k^{-3}, i.e., volume.

Problem 3. At what comoving scale does a mode enter the horizon during matter domination?

Solution. The comoving horizon is χH≈2c/H0Ωm(1+z)−1/2\chi_H \approx 2c/H_0\sqrt{\Omega_m}(1+z)^{-1/2}. At matter-radiation equality (zeq≈3400z_{\rm eq} \approx 3400), the horizon scale is roughly λeq≈100 Mpc\lambda_{\rm eq} \approx 100\ \mathrm{Mpc} comoving. Modes with k>2π/λeqk > 2\pi/\lambda_{\rm eq} entered earlier during radiation domination.

Section summary. Structure formation starts from small coupled matter-radiation perturbations.

The Inhomogeneous Universe: Gravity

Core ideas

Gravity turns perturbations into growing structure. In Newtonian gauge, potentials source motion and lensing; in the subhorizon matter era, density perturbations obey a simple growth equation. Relativistic gauge issues matter on large scales.

For review, be able to use Poisson’s equation for cosmological perturbations, solve qualitative growth, and distinguish physical perturbations from gauge artifacts. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

δ¨+2Hδ˙−4πGρˉmδ=0,∇2Φ=4πGa2ρˉmδ\ddot\delta+2H\dot\delta-4\pi G\bar\rho_m\delta=0,\qquad \nabla^2\Phi=4\pi Ga^2\bar\rho_m\delta

Worked example

In a matter-dominated Einstein—de Sitter universe (a∝t2/3a \propto t^{2/3}), the linear growth equation is

δ¨+2Hδ˙−4πGρˉmδ=0.\ddot\delta + 2H\dot\delta - 4\pi G\bar\rho_m\delta = 0.

Because H=2/(3t)H = 2/(3t) and 4πGρˉm=2/(3t2)4\pi G\bar\rho_m = 2/(3t^2) (from the Friedmann equation), the equation becomes

δ¨+43tδ˙−23t2δ=0.\ddot\delta + \frac{4}{3t}\dot\delta - \frac{2}{3t^2}\delta = 0.

Trying a power law δ∝tn\delta \propto t^n gives n(n−1)+43n−23=0n(n-1) + \frac{4}{3}n - \frac{2}{3} = 0, or 3n2+n−2=03n^2 + n - 2 = 0, with solutions n=2/3n = 2/3 (growing mode) and n=−1n = -1 (decaying mode). Thus the growing mode is δ+(t)∝a(t)\delta_+(t) \propto a(t), and a perturbation that was δ=10−5\delta = 10^{-5} at recombination (z≈1100z \approx 1100) grows to δ0=10−5(1+z)−1≈10−2\delta_0 = 10^{-5}(1+z)^{-1} \approx 10^{-2} today.

Problems with Solutions

Problem 1. Show that in a matter-dominated universe, δ+∝a\delta_+ \propto a is a solution of the growth equation.

Solution. With a∝t2/3a \propto t^{2/3}, δ=a\delta = a, δ˙=a˙=Ha\dot\delta = \dot a = Ha, δ¨=H˙a+Ha˙=(H˙+H2)a\ddot\delta = \dot H a + H\dot a = (\dot H + H^2)a. Using H˙=−4πGρˉm/3=−H2/2\dot H = -4\pi G\bar\rho_m/3 = -H^2/2 (EdS) gives δ¨=H2a/2\ddot\delta = H^2 a/2. Substituting into the growth equation:

H2a2+2H(Ha)−4πGρˉma=H2a2+2H2a−3H2a2=0.\frac{H^2 a}{2} + 2H(Ha) - 4\pi G\bar\rho_m a = \frac{H^2 a}{2} + 2H^2 a - \frac{3H^2 a}{2} = 0.

Problem 2. A density contrast δ=0.02\delta = 0.02 exists on a scale of 5 Mpc/h5\ \mathrm{Mpc}/h. Use Poisson’s equation to estimate the gravitational potential Φ\Phi.

Solution. ∇2Φ=4πGa2ρˉmδ\nabla^2\Phi = 4\pi G a^2 \bar\rho_m \delta. Approximating ∇2∼−k2\nabla^2 \sim -k^2 with k∼2π/(5 Mpc/h)≈1.26 h Mpc−1k \sim 2\pi/(5\ \mathrm{Mpc}/h) \approx 1.26\,h\ \mathrm{Mpc^{-1}}, and using 4πGa2ρˉm≈(3/2)ΩmH02a−14\pi G a^2\bar\rho_m \approx (3/2)\Omega_m H_0^2 a^{-1} today (a=1a=1), we get ∣Φ∣∼(3/2)ΩmH02δ/k2∼10−5|\Phi| \sim (3/2)\Omega_m H_0^2 \delta/k^2 \sim 10^{-5}, consistent with CMB potential wells.

Problem 3. What is the physical difference between the growing and decaying modes?

Solution. The growing mode (δ∝a\delta \propto a) becomes more important with time and seeds structure. The decaying mode (δ∝t−1\delta \propto t^{-1} or a−3/2a^{-3/2} in EdS) redshifts away and is negligible after a short period unless artificially excited by initial conditions.

Section summary. Gravity amplifies primordial perturbations into cosmic structure.

Initial Conditions

Core ideas

Inflation explains why initial perturbations are nearly adiabatic, Gaussian, and scale invariant. The curvature perturbation is the central variable because it remains conserved on superhorizon scales for adiabatic modes. Its power spectrum seeds both CMB anisotropy and structure.

For review, be able to define adiabatic perturbations, read the scalar power spectrum, explain spectral tilt, and connect inflation to horizon and flatness problems. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

PR(k)=As(kk∗)ns−1,ns≈1P_{\mathcal R}(k)=A_s\left(\frac{k}{k_*}\right)^{n_s-1},\qquad n_s\approx1

Worked example

Planck 2018 reports As≈2.1×10−9A_s \approx 2.1\times10^{-9} and ns≈0.965n_s \approx 0.965 at k∗=0.05 Mpc−1k_* = 0.05\ \mathrm{Mpc^{-1}}. Evaluate the primordial curvature power spectrum at k=0.002 Mpc−1k = 0.002\ \mathrm{Mpc^{-1}}.

Using

PR(k)=As(kk∗)ns−1,P_{\mathcal R}(k) = A_s\left(\frac{k}{k_*}\right)^{n_s-1},

we find

PR(0.002)=2.1×10−9(0.0020.05)−0.035=2.1×10−9×(0.04)−0.035.P_{\mathcal R}(0.002) = 2.1\times10^{-9}\left(\frac{0.002}{0.05}\right)^{-0.035} = 2.1\times10^{-9} \times (0.04)^{-0.035}.

Since ln⁡(0.04)≈−3.22\ln(0.04) \approx -3.22, (0.04)−0.035=e0.113≈1.12(0.04)^{-0.035} = e^{0.113} \approx 1.12. Thus PR(0.002)≈2.35×10−9P_{\mathcal R}(0.002) \approx 2.35\times10^{-9}. The slight scale dependence means there is about 12%12\% more power on larger scales (smaller kk) than at the pivot.

Problems with Solutions

Problem 1. Evaluate PRP_{\mathcal R} at k=0.1 Mpc−1k = 0.1\ \mathrm{Mpc^{-1}} for As=2.1×10−9A_s = 2.1\times10^{-9}, ns=0.965n_s = 0.965, k∗=0.05 Mpc−1k_* = 0.05\ \mathrm{Mpc^{-1}}.

Solution. PR(0.1)=2.1×10−9(0.1/0.05)−0.035=2.1×10−9×2−0.035≈2.1×10−9×0.976≈2.05×10−9P_{\mathcal R}(0.1) = 2.1\times10^{-9}(0.1/0.05)^{-0.035} = 2.1\times10^{-9} \times 2^{-0.035} \approx 2.1\times10^{-9} \times 0.976 \approx 2.05\times10^{-9}.

Problem 2. A model has ns=1n_s = 1 exactly. What does this imply about the power spectrum?

Solution. PR(k)=AsP_{\mathcal R}(k) = A_s, independent of scale. This is the Harrison—Zel’dovich scale-invariant spectrum.

Problem 3. Inflation solves the horizon problem because a mode that is inside the horizon today was inside the inflationary horizon before inflation ended. Estimate the number of e-folds needed if the current horizon is 14 Gpc14\ \mathrm{Gpc} and the Hubble radius at the end of inflation was 10−28 m10^{-28}\ \mathrm{m}.

Solution. The ratio of physical scales is ∼14 Gpc/10−28 m≈1.4×1052\sim 14\ \mathrm{Gpc}/10^{-28}\ \mathrm{m} \approx 1.4\times10^{52}. The required e-folds are N∼ln⁡(1.4×1052)≈120N \sim \ln(1.4\times10^{52}) \approx 120, but because expansion after inflation also stretches scales, only ∼50\sim 50—6060 e-folds of inflation are typically needed.

Section summary. Initial conditions are encoded in the primordial curvature power spectrum.

Growth of Structure: Linear Theory

Core ideas

Linear theory describes perturbations while ∣δ∣≪1|\delta|\ll1. Matter perturbations grow after matter-radiation equality, baryon acoustic features survive statistically, and transfer functions encode the processing of primordial fluctuations.

For review, be able to use growth factors, transfer functions, equality scale, and linear bias at a conceptual and formula level. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

Pm(k,z)=D2(z)T2(k)Pprim(k),f=dln⁡Ddln⁡aP_m(k,z)=D^2(z)T^2(k)P_{\rm prim}(k),\qquad f=\frac{d\ln D}{d\ln a}

Worked example

Suppose the primordial power spectrum is Pprim(k)=AknsP_{\rm prim}(k) = A k^{n_s} with A=2×10−9A = 2\times10^{-9} at a reference scale, and the transfer function at k=0.1 h Mpc−1k = 0.1\ h\ \mathrm{Mpc^{-1}} is T(k)≈0.9T(k) \approx 0.9. The linear growth factor today is normalized to D(0)=1D(0)=1, and at z=1z=1 it is D(1)≈0.55D(1) \approx 0.55 for Λ\LambdaCDM.

The matter power spectrum at z=1z=1 is

Pm(k,z=1)=D2(1) T2(k) Pprim(k)=(0.55)2×(0.9)2×Pprim(k)≈0.245 Pprim(k).P_m(k,z=1) = D^2(1)\,T^2(k)\,P_{\rm prim}(k) = (0.55)^2 \times (0.9)^2 \times P_{\rm prim}(k) \approx 0.245\,P_{\rm prim}(k).

If Pprim(0.1)≈2×103 (Mpc/h)3P_{\rm prim}(0.1) \approx 2\times10^3\ (\mathrm{Mpc}/h)^3 (after normalization to match observed large-scale amplitude), then Pm≈490 (Mpc/h)3P_m \approx 490\ (\mathrm{Mpc}/h)^3. The growth rate is

f=dln⁡Ddln⁡a≈Ωm(z)0.55≈0.89f = \frac{d\ln D}{d\ln a} \approx \Omega_m(z)^{0.55} \approx 0.89

at z=1z=1 for Ωm=0.3\Omega_m = 0.3.

Problems with Solutions

Problem 1. If D(z=0)=1D(z=0)=1 and D(z=1)=0.55D(z=1)=0.55, by what factor have perturbations grown between z=1z=1 and today?

Solution. The ratio is D(0)/D(1)=1/0.55≈1.82D(0)/D(1) = 1/0.55 \approx 1.82. Perturbations grew by about 82%82\%.

Problem 2. A galaxy survey measures δg=0.1\delta_g = 0.1 in a region. If linear bias is b=1.5b = 1.5, what is the underlying matter overdensity?

Solution. Using δg=b δm\delta_g = b\,\delta_m, we get δm=δg/b=0.1/1.5≈0.067\delta_m = \delta_g/b = 0.1/1.5 \approx 0.067.

Problem 3. The comoving wavenumber entering the horizon at matter-radiation equality is keq≈0.01 (Ωmh2) Mpc−1k_{\rm eq} \approx 0.01\ (\Omega_m h^2)\ \mathrm{Mpc^{-1}}. Evaluate keqk_{\rm eq} for Ωmh2=0.14\Omega_m h^2 = 0.14.

Solution. keq≈0.01×0.14 Mpc−1=1.4×10−3 Mpc−1k_{\rm eq} \approx 0.01 \times 0.14\ \mathrm{Mpc^{-1}} = 1.4\times10^{-3}\ \mathrm{Mpc^{-1}}, corresponding to a comoving scale λeq≈2π/keq≈4500 Mpc\lambda_{\rm eq} \approx 2\pi/k_{\rm eq} \approx 4500\ \mathrm{Mpc}. (More precise calculations give keq∼0.015 h Mpc−1k_{\rm eq} \sim 0.015\,h\ \mathrm{Mpc^{-1}}, i.e., ∼400 Mpc/h\sim 400\ \mathrm{Mpc}/h, but this illustrates the scaling.)

Section summary. Linear theory links primordial fluctuations to large-scale structure.

The Cosmic Microwave Background

Core ideas

The CMB is radiation from last scattering. Its temperature anisotropies come from acoustic oscillations, gravitational redshift, Doppler motion, diffusion damping, and projection effects. Peak positions measure geometry; peak heights measure contents.

For review, be able to interpret the CMB angular power spectrum, name the Sachs-Wolfe and acoustic effects, and connect peaks to parameters. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

ΔTT(n^)=∑ℓmaℓmYℓm(n^),Cℓ=⟨∣aℓm∣2⟩\frac{\Delta T}{T}(\hat{\bm n})=\sum_{\ell m}a_{\ell m}Y_{\ell m}(\hat{\bm n}),\qquad C_\ell=\langle|a_{\ell m}|^2\rangle

Worked example

The first acoustic peak of the CMB temperature power spectrum is observed at ℓ1≈220\ell_1 \approx 220. The angular diameter distance to last scattering in a flat Λ\LambdaCDM model is dA≈14 Gpcd_A \approx 14\ \mathrm{Gpc}. The physical sound horizon at recombination is rs≈150 Mpcr_s \approx 150\ \mathrm{Mpc}.

The angular scale of the sound horizon is

θs=rsdA≈15014×103≈1.07×10−2 rad≈0.61∘.\theta_s = \frac{r_s}{d_A} \approx \frac{150}{14\times10^3} \approx 1.07\times10^{-2}\ \mathrm{rad} \approx 0.61^\circ.

The corresponding multipole is

ℓ≈πθs≈π1.07×10−2≈293.\ell \approx \frac{\pi}{\theta_s} \approx \frac{\pi}{1.07\times10^{-2}} \approx 293.

A more accurate treatment including projection effects, early integrated Sachs—Wolfe, and the precise sound speed shifts this to ℓ1≈220\ell_1 \approx 220, but the estimate captures the basic geometry.

Problems with Solutions

Problem 1. The temperature anisotropy ΔT/T∼10−5\Delta T/T \sim 10^{-5} corresponds to what temperature fluctuation in μK\mu\mathrm{K} at T0=2.725 KT_0 = 2.725\ \mathrm{K}?

Solution. ΔT=10−5×2.725 K=2.725×10−5 K=27.25 μK\Delta T = 10^{-5} \times 2.725\ \mathrm{K} = 2.725\times10^{-5}\ \mathrm{K} = 27.25\ \mu\mathrm{K}.

Problem 2. Estimate the angular separation on the sky corresponding to ℓ=1000\ell = 1000.

Solution. θ≈π/ℓ≈π/1000≈3.14×10−3 rad≈0.18∘≈10.8′\theta \approx \pi/\ell \approx \pi/1000 \approx 3.14\times10^{-3}\ \mathrm{rad} \approx 0.18^\circ \approx 10.8'.

Problem 3. The aℓma_{\ell m} coefficients for ℓ=2\ell = 2 are measured as a20=1.0×10−5a_{20}=1.0\times10^{-5}, a21=0.5×10−5a_{21}=0.5\times10^{-5}, a2,−1=−0.3×10−5a_{2,-1}=-0.3\times10^{-5}, etc. What is the ensemble average ⟨∣a2m∣2⟩\langle|a_{2m}|^2\rangle if all five modes have the same variance?

Solution. For a given ℓ\ell, ⟨∣aℓm∣2⟩=Cℓ\langle|a_{\ell m}|^2\rangle = C_\ell. If the variance is the same for all mm, then C2=⟨∣a2m∣2⟩C_2 = \langle|a_{2m}|^2\rangle. If the measured values are one realization, the estimator is C^2=(2ℓ+1)−1∑m∣a2m∣2\hat C_2 = (2\ell+1)^{-1}\sum_m |a_{2m}|^2. With hypothetical equal variance ∣a2m∣2=10−10|a_{2m}|^2 = 10^{-10} for all five modes, C^2=10−10\hat C_2 = 10^{-10}.

Section summary. The CMB is a precision image of early-universe perturbations.

The Polarized CMB

Core ideas

CMB polarization is produced by Thomson scattering of radiation with a quadrupole anisotropy. Scalar perturbations generate E-modes; tensor perturbations can generate primordial B-modes. Lensing converts some E-mode power into B-modes.

For review, be able to distinguish E and B polarization, explain Thomson scattering origin, and state why B-modes are important. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

scalar⇒E,tensor/lensing⇒B,CℓEE,CℓBB\mathrm{scalar}\Rightarrow E,\qquad \mathrm{tensor/lensing}\Rightarrow B,\qquad C_\ell^{EE},C_\ell^{BB}

Worked example

Thomson scattering of CMB photons off free electrons produces linear polarization with a characteristic EE-mode pattern. The polarization amplitude is roughly 10%10\% of the temperature quadrupole anisotropy at last scattering. If the temperature quadrupole is ΔTℓ∼20 μK\Delta T_\ell \sim 20\ \mu\mathrm{K} at the relevant angular scales, the polarization amplitude is

P∼0.1×20 μK2.725 K∼7×10−7.P \sim 0.1 \times \frac{20\ \mu\mathrm{K}}{2.725\ \mathrm{K}} \sim 7\times10^{-7}.

In terms of CℓC_\ell, the first peak of the EEEE power spectrum is CℓEE≈0.1 CℓTTC_\ell^{EE} \approx 0.1\, C_\ell^{TT} near ℓ∼300\ell \sim 300. For scalar perturbations, the polarization pattern is purely EE-mode; primordial BB-modes from tensor perturbations are predicted at a much lower level, with r=CℓBB/CℓTT∼0.01r = C_\ell^{BB}/C_\ell^{TT} \sim 0.01 or less.

Problems with Solutions

Problem 1. Why does Thomson scattering produce polarization only when the radiation has a quadrupole anisotropy?

Solution. In the electron rest frame, Thomson scattering is symmetric for unpolarized radiation only if the incoming intensity is isotropic. A dipole anisotropy produces no net polarization when averaged over all scattering directions. A quadrupole intensity pattern creates a linear polarization because scattering perpendicular to the hot and cold directions differs, leaving a net linear polarization in the scattered radiation.

Problem 2. If the tensor-to-scalar ratio is r=0.05r = 0.05 and CℓTT≈5×10−10C_\ell^{TT} \approx 5\times10^{-10} at ℓ=80\ell = 80, estimate CℓBBC_\ell^{BB} from primordial tensors at that scale.

Solution. Roughly, CℓBB≈(r/10) CℓTTC_\ell^{BB} \approx (r/10)\,C_\ell^{TT} near the reionization bump. For r=0.05r=0.05, CℓBB≈0.005×5×10−10=2.5×10−12C_\ell^{BB} \approx 0.005 \times 5\times10^{-10} = 2.5\times10^{-12}. This is well below current detection limits (r≲0.03r \lesssim 0.03 from BICEP/Keck).

Problem 3. Lensing converts EE-modes to BB-modes. If the lensing potential has power Cℓϕϕ≈10−7C_\ell^{\phi\phi} \approx 10^{-7} at ℓ∼100\ell \sim 100, why is the lensing BB-mode considered a foreground for primordial tensor searches?

Solution. Gravitational lensing by large-scale structure recombines EE-mode polarization into BB-modes, producing a roughly scale-invariant lensing BB-mode power spectrum with amplitude CℓBB,lens∼10−5 μK2C_\ell^{BB,\rm lens} \sim 10^{-5}\,\mu\mathrm{K}^2 at ℓ∼1000\ell \sim 1000. This is larger than the primordial tensor BB-mode for r≲0.01r \lesssim 0.01, making it a confusion signal that must be modeled and subtracted.

Section summary. CMB polarization adds geometry and tensor information beyond temperature.

Probes of Structure I: Tracers

Core ideas

Galaxies, quasars, clusters, and the Lyman-alpha forest trace matter imperfectly. Bias, redshift-space distortions, selection functions, and shot noise must be modeled to infer the underlying density field.

For review, be able to define bias, correlation functions, redshift-space distortions, and the relation between tracers and matter. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

δg=b δm,ξ(r)=⟨δ(x)δ(x+r)⟩\delta_g=b\,\delta_m,\qquad \xi(r)=\langle\delta(\bm x)\delta(\bm x+\bm r)\rangle

Worked example

A galaxy survey measures the two-point correlation function ξgg(r)\xi_{gg}(r). At large scales, the galaxy overdensity is δg=b δm\delta_g = b\,\delta_m with bias b=1.8b = 1.8. The matter power spectrum at k=0.1 h Mpc−1k = 0.1\ h\ \mathrm{Mpc^{-1}} is Pm(k)≈2×104 (h−1Mpc)3P_m(k) \approx 2\times10^4\ (h^{-1}\mathrm{Mpc})^3.

  1. The galaxy power spectrum is Pg(k)=b2Pm(k)=(1.8)2×2×104≈6.5×104 (h−1Mpc)3P_g(k) = b^2 P_m(k) = (1.8)^2 \times 2\times10^4 \approx 6.5\times10^4\ (h^{-1}\mathrm{Mpc})^3.
  2. The real-space correlation function at separation r=10 h−1Mpcr = 10\ h^{-1}\mathrm{Mpc} can be estimated by Fourier transforming Pg(k)P_g(k), but a rough estimate is ξ(r)∼(b σ8)2(r/8 h−1Mpc)−1.8\xi(r) \sim (b\,\sigma_8)^2 (r/8\ h^{-1}\mathrm{Mpc})^{-1.8}, giving ξ(10)∼0.5\xi(10) \sim 0.5 for typical values.
  3. Shot noise adds a constant 1/nˉ1/\bar n to Pg(k)P_g(k). If the number density is nˉ=3×10−4 (h−1Mpc)−3\bar n = 3\times10^{-4}\ (h^{-1}\mathrm{Mpc})^{-3}, the shot noise is 1/nˉ≈3.3×103 (h−1Mpc)31/\bar n \approx 3.3\times10^3\ (h^{-1}\mathrm{Mpc})^3, non-negligible on small scales.

Problems with Solutions

Problem 1. If δm=0.02\delta_m = 0.02 in a region and linear bias is b=1.2b=1.2, what is the expected galaxy overdensity?

Solution. δg=b δm=1.2×0.02=0.024\delta_g = b\,\delta_m = 1.2 \times 0.02 = 0.024.

Problem 2. Explain why redshift-space distortions cause the clustering of galaxies to appear anisotropic.

Solution. Along the line of sight, peculiar velocities add to the Hubble flow, compressing structures on small scales (fingers of God) and producing a characteristic quadrupole anisotropy on large scales (Kaiser effect). Transverse distances are unaffected, breaking isotropy.

Problem 3. The correlation function is ξ(r)=⟨δ(x)δ(x+r)⟩\xi(r) = \langle\delta(\bm x)\delta(\bm x+\bm r)\rangle. What is ξ(0)\xi(0) if the field has variance σ2\sigma^2?

Solution. ξ(0)=⟨δ2(x)⟩=σ2\xi(0) = \langle\delta^2(\bm x)\rangle = \sigma^2, the variance of the density field.

Section summary. Observed tracers are biased maps of the matter distribution.

Probes of Structure II: Gravitational Lensing

Core ideas

Lensing maps projected mass by measuring deflection, shear, convergence, and magnification. Weak lensing statistically measures large-scale structure; strong lensing probes dense systems; CMB lensing maps matter to high redshift.

For review, be able to define convergence and shear, connect lensing to projected potential, and distinguish weak and strong regimes. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

κ=ΣΣcrit,Σcrit=c24πGDsDlDls\kappa=\frac{\Sigma}{\Sigma_{\rm crit}},\qquad \Sigma_{\rm crit}=\frac{c^2}{4\pi G}\frac{D_s}{D_lD_{ls}}

Worked example

A foreground cluster at zl=0.3z_l = 0.3 lenses a background galaxy at zs=1.0z_s = 1.0. In a flat Λ\LambdaCDM model with H0=70 km s−1 Mpc−1H_0 = 70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}, the angular diameter distances are approximately Dl≈1.2 GpcD_l \approx 1.2\ \mathrm{Gpc}, Ds≈3.3 GpcD_s \approx 3.3\ \mathrm{Gpc}, and Dls≈2.3 GpcD_{ls} \approx 2.3\ \mathrm{Gpc}.

The critical surface mass density is

Σcrit=c24πGDsDlDls=(3×108)24π(6.674×10−11)3.31.2×2.3×1027≈1.7×1015 M⊙ Mpc−2≈2 g cm−2.\Sigma_{\rm crit} = \frac{c^2}{4\pi G}\frac{D_s}{D_l D_{ls}} = \frac{(3\times10^8)^2}{4\pi(6.674\times10^{-11})} \frac{3.3}{1.2\times2.3\times10^{27}} \approx 1.7\times10^{15}\ \mathrm{M_\odot\,Mpc^{-2}} \approx 2\ \mathrm{g\,cm^{-2}}.

If the cluster has a projected mass density Σ=3 g cm−2\Sigma = 3\ \mathrm{g\,cm^{-2}} at a certain radius, the convergence is κ=Σ/Σcrit=1.5\kappa = \Sigma/\Sigma_{\rm crit} = 1.5, indicating strong lensing (multiple images possible).

Problems with Solutions

Problem 1. Evaluate Σcrit\Sigma_{\rm crit} for a lens at zl=0.5z_l = 0.5 and source at zs=2.0z_s = 2.0 using Dl≈1.7 GpcD_l \approx 1.7\ \mathrm{Gpc}, Ds≈5.3 GpcD_s \approx 5.3\ \mathrm{Gpc}, Dls≈3.9 GpcD_{ls} \approx 3.9\ \mathrm{Gpc}.

Solution. Σcrit=c24πGDsDlDls≈9×10164π(6.674×10−11)5.31.7×3.9×1027≈8.5×1015 kg m−2≈0.85 g cm−2\Sigma_{\rm crit} = \frac{c^2}{4\pi G}\frac{D_s}{D_l D_{ls}} \approx \frac{9\times10^{16}}{4\pi(6.674\times10^{-11})}\frac{5.3}{1.7\times3.9\times10^{27}} \approx 8.5\times10^{15}\ \mathrm{kg\,m^{-2}} \approx 0.85\ \mathrm{g\,cm^{-2}}.

Problem 2. A circular source has intrinsic radius Rsrc=1′′R_{\rm src} = 1''. If the magnification is μ=4\mu = 4, what is the observed angular area?

Solution. Area magnification is ∣μ∣|\mu|. The intrinsic area is π(1′′)2\pi (1'')^2. The observed area is 4π(1′′)2≈12.6 arcsec24\pi (1'')^2 \approx 12.6\ \mathrm{arcsec^2}. The observed radius is 2′′2''.

Problem 3. Weak lensing measures shear γ∼0.02\gamma \sim 0.02 for a population of galaxies. If the mean ellipticity dispersion is σϵ≈0.3\sigma_\epsilon \approx 0.3, how many galaxies NN are needed to measure the shear with signal-to-noise S/N=5S/N = 5?

Solution. The uncertainty on the mean shear is σϵ/N\sigma_\epsilon/\sqrt{N}. Setting 0.02/(0.3/N)=50.02/(0.3/\sqrt{N}) = 5 gives N=5×0.3/0.02=75\sqrt{N} = 5 \times 0.3 / 0.02 = 75, so N≈5625N \approx 5625 galaxies.

Section summary. Lensing observes mass through its effect on light paths.

Probes of Structure III: Nonlinear Growth

Core ideas

On small scales, density contrasts become nonlinear. Spherical collapse, halo formation, virialization, N-body simulations, halo mass functions, and baryonic feedback describe the transition from smooth perturbations to galaxies and clusters.

For review, be able to explain nonlinear collapse, virial equilibrium, halo profiles, and why simulations are required. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

2K+U=0,ρNFW(r)=ρs(r/rs)(1+r/rs)22K+U=0,\qquad \rho_{\rm NFW}(r)=\frac{\rho_s}{(r/r_s)(1+r/r_s)^2}

Worked example

A galaxy cluster of mass M=1015 M⊙M = 10^{15}\ M_\odot virializes at z=0z=0 with virial radius Rvir≈2 MpcR_{\rm vir} \approx 2\ \mathrm{Mpc}. The circular velocity is

vc2=GMRvir=(6.674×10−11 m3 kg−1 s−2)(2×1045 kg)2×3.086×1022 m≈2.2×1012 m2 s−2,v_c^2 = \frac{GM}{R_{\rm vir}} = \frac{(6.674\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}})(2\times10^{45}\ \mathrm{kg})}{2\times3.086\times10^{22}\ \mathrm{m}} \approx 2.2\times10^{12}\ \mathrm{m^2\,s^{-2}},

so vc≈1500 km s−1v_c \approx 1500\ \mathrm{km\,s^{-1}}. The virial temperature for an ideal monatomic gas is

Tvir=μmpvc22kB≈0.6×1.67×10−27 kg×2.2×1012 m2 s−22×1.38×10−23 J K−1≈8×107 K≈7 keV.T_{\rm vir} = \frac{\mu m_p v_c^2}{2k_B} \approx \frac{0.6 \times 1.67\times10^{-27}\ \mathrm{kg} \times 2.2\times10^{12}\ \mathrm{m^2\,s^{-2}}}{2 \times 1.38\times10^{-23}\ \mathrm{J\,K^{-1}}} \approx 8\times10^7\ \mathrm{K} \approx 7\ \mathrm{keV}.

This temperature matches the X-ray emission observed from hot intracluster gas. The virial theorem 2K+U=02K + U = 0 ensures the cluster is in dynamical equilibrium.

Problems with Solutions

Problem 1. For a self-gravitating system in virial equilibrium, show that the total energy is E=−KE = -K.

Solution. The virial theorem states 2K+U=02K + U = 0, so U=−2KU = -2K. The total energy is E=K+U=K−2K=−KE = K + U = K - 2K = -K.

Problem 2. An NFW halo has scale density ρs=107 M⊙ kpc−3\rho_s = 10^7\ M_\odot\ \mathrm{kpc^{-3}} and scale radius rs=200 kpcr_s = 200\ \mathrm{kpc}. Compute the density at r=2rsr = 2r_s.

Solution. Using ρNFW(r)=ρs/[(r/rs)(1+r/rs)2]\rho_{\rm NFW}(r) = \rho_s / [(r/r_s)(1+r/r_s)^2], at r=2rsr=2r_s:

ρ=1072×(1+2)2=10718≈5.6×105 M⊙ kpc−3.\rho = \frac{10^7}{2 \times (1+2)^2} = \frac{10^7}{18} \approx 5.6\times10^5\ M_\odot\ \mathrm{kpc^{-3}}.

Problem 3. In an Einstein—de Sitter universe, spherical collapse turnaround occurs at redshift ztaz_{\rm ta}. What is the collapse redshift zcollz_{\rm coll}?

Solution. In EdS, parametric time gives tcoll=2ttat_{\rm coll} = 2t_{\rm ta}. Since t∝(1+z)−3/2t \propto (1+z)^{-3/2} during matter domination,

(1+zcoll)−3/2=2(1+zta)−3/2⇒1+zcoll=1+zta22/3≈1+zta1.587.(1+z_{\rm coll})^{-3/2} = 2(1+z_{\rm ta})^{-3/2} \quad\Rightarrow\quad 1+z_{\rm coll} = \frac{1+z_{\rm ta}}{2^{2/3}} \approx \frac{1+z_{\rm ta}}{1.587}.

For zta=1z_{\rm ta}=1, zcoll≈0.26z_{\rm coll} \approx 0.26.

Section summary. Nonlinear structure requires collapse physics and simulations.