Brief Recalls on Homogeneous Cosmology
1. FLRW Metric and Friedmann Equations ¶ 1.1 The FLRW Metric ¶ For a homogeneous and isotropic universe:
d s 2 = d t 2 − a ( t ) 2 d x 2 ds^2 = dt^2 - a(t)^2 d\mathbf{x}^2 d s 2 = d t 2 − a ( t ) 2 d x 2 where a ( t ) a(t) a ( t )
1.2 The Friedmann Equation ¶ From Einstein’s equations for the FLRW metric:
H 2 = 8 π G 3 ρ t o t = 8 π 3 M p l 2 ρ t o t H^2 = \dfrac{8 \pi G}{3} \rho_{\rm tot} = \dfrac{8 \pi }{3 M_{pl}^2} \rho_{\rm tot} H 2 = 3 8 π G ρ tot = 3 M pl 2 8 π ρ tot where H = a ˙ / a H = \dot{a}/a H = a ˙ / a ρ t o t \rho_{\rm tot} ρ tot
2. Energy Constituents of the Universe ¶ 2.1 Matter, Radiation, and Cosmological Constant ¶ The total density comprises:
Matter (cold dark matter and baryons): ρ m = ρ c + ρ b ∝ a − 3 \rho_m = \rho_c + \rho_b \propto a^{-3} ρ m = ρ c + ρ b ∝ a − 3
Radiation (photon and neutrino): ρ r = ρ γ + ρ ν ∝ a − 4 \rho_r = \rho_\gamma + \rho_\nu \propto a^{-4} ρ r = ρ γ + ρ ν ∝ a − 4
Cosmological constant : ρ Λ ∝ a 0 \rho_\Lambda \propto a^0 ρ Λ ∝ a 0
2.2 Density Parameters and Their Evolution ¶ Define the critical density today:
ρ c , 0 = 3 H 0 2 8 π G \rho_{c,0} = \dfrac{3H_0^2}{8\pi G} ρ c , 0 = 8 π G 3 H 0 2 The density parameter for component i i i
Ω i = ρ i ρ c \Omega_i = \dfrac{\rho_i}{\rho_c} Ω i = ρ c ρ i Today, we have:
Ω i , 0 = ρ i , 0 ρ c , 0 \Omega_{i,0} = \dfrac{\rho_{i,0}}{\rho_{c,0}} Ω i , 0 = ρ c , 0 ρ i , 0 For a flat universe (as supported by observations):
∑ i Ω i , 0 = 1 \sum_i \Omega_{i,0} = 1 i ∑ Ω i , 0 = 1 The Friedmann equation becomes:
H 2 = H 0 2 [ Ω m , 0 a − 3 + Ω r , 0 a − 4 + Ω Λ , 0 ] H^2 = H_0^2 \left[ \Omega_{m,0} a^{-3} + \Omega_{r,0} a^{-4} + \Omega_{\Lambda,0} \right] H 2 = H 0 2 [ Ω m , 0 a − 3 + Ω r , 0 a − 4 + Ω Λ , 0 ] where we set a 0 = 1 a_0 = 1 a 0 = 1
Exercise 1 (Derivation of
∑ Ω i = 1 \sum \Omega_i = 1 ∑ Ω i = 1 for flat universe)
For k = 0 k=0 k = 0
H 2 = 8 π G 3 ρ t o t H^2 = \dfrac{8\pi G}{3} \rho_{\rm tot} H 2 = 3 8 π G ρ tot At present time t 0 t_0 t 0
H 0 2 = 8 π G 3 ρ t o t , 0 H_0^2 = \dfrac{8\pi G}{3} \rho_{\rm tot,0} H 0 2 = 3 8 π G ρ tot , 0 Dividing the general equation by H 0 2 H_0^2 H 0 2
( H H 0 ) 2 = ρ t o t ρ t o t , 0 ρ t o t , 0 ρ c , 0 = ρ t o t ρ t o t , 0 Ω t o t , 0 \left(\dfrac{H}{H_0}\right)^2 = \dfrac{\rho_{\rm tot}}{\rho_{\rm tot,0}} \dfrac{\rho_{\rm tot,0}}{\rho_{c,0}} = \dfrac{\rho_{\rm tot}}{\rho_{\rm tot,0}} \Omega_{\rm tot,0} ( H 0 H ) 2 = ρ tot , 0 ρ tot ρ c , 0 ρ tot , 0 = ρ tot , 0 ρ tot Ω tot , 0 But ρ t o t / ρ t o t , 0 \rho_{\rm tot}/\rho_{\rm tot,0} ρ tot / ρ tot , 0
H 2 = H 0 2 ∑ i Ω i , 0 a − 3 ( 1 + w i ) H^2 = H_0^2 \sum_i \Omega_{i,0} a^{-3(1+w_i)} H 2 = H 0 2 i ∑ Ω i , 0 a − 3 ( 1 + w i ) where w i w_i w i t 0 t_0 t 0 a = 1 a=1 a = 1
H 0 2 = H 0 2 ∑ i Ω i , 0 ⇒ ∑ i Ω i , 0 = 1 H_0^2 = H_0^2 \sum_i \Omega_{i,0} \quad \Rightarrow \quad \sum_i \Omega_{i,0} = 1 H 0 2 = H 0 2 i ∑ Ω i , 0 ⇒ i ∑ Ω i , 0 = 1 3. Radiation in the Early Universe ¶ 3.1 Photon Energy Density ¶ For a relativistic boson with g g g
ρ = g ( 2 π ) 3 ∫ d 3 p E ( p ) f ( p ) \rho = \dfrac{g}{(2\pi)^3} \int d^3p \, E(p) f(p) ρ = ( 2 π ) 3 g ∫ d 3 p E ( p ) f ( p ) with Bose-Einstein distribution f ( p ) = ( e E / T − 1 ) − 1 f(p) = (e^{E/T} - 1)^{-1} f ( p ) = ( e E / T − 1 ) − 1 E = p E = p E = p
For photons (g γ = 2 g_\gamma = 2 g γ = 2
ρ γ = π 2 15 T 4 \rho_\gamma = \dfrac{\pi^2}{15} T^4 ρ γ = 15 π 2 T 4 Exercise 2 (Detailed derivation of
ρ γ \rho_\gamma ρ γ )
We have:
ρ γ = g γ ( 2 π ) 3 ∫ d 3 p p 1 e p / T − 1 \rho_\gamma = \dfrac{g_\gamma}{(2\pi)^3} \int d^3p \, p \dfrac{1}{e^{p/T} - 1} ρ γ = ( 2 π ) 3 g γ ∫ d 3 p p e p / T − 1 1 In spherical coordinates:
ρ γ = g γ 2 π 2 ∫ 0 ∞ d p p 3 e p / T − 1 \rho_\gamma = \dfrac{g_\gamma}{2\pi^2} \int_0^\infty dp \dfrac{p^3}{e^{p/T} - 1} ρ γ = 2 π 2 g γ ∫ 0 ∞ d p e p / T − 1 p 3 Set ξ = p / T \xi = p/T ξ = p / T
ρ γ = g γ 2 π 2 T 4 ∫ 0 ∞ d ξ ξ 3 e ξ − 1 \rho_\gamma = \dfrac{g_\gamma}{2\pi^2} T^4 \int_0^\infty d\xi \dfrac{\xi^3}{e^\xi - 1} ρ γ = 2 π 2 g γ T 4 ∫ 0 ∞ d ξ e ξ − 1 ξ 3 The integral is Γ ( 4 ) ζ ( 4 ) = 3 ! ⋅ π 4 90 = π 4 15 \Gamma(4)\zeta(4) = 3! \cdot \dfrac{\pi^4}{90} = \dfrac{\pi^4}{15} Γ ( 4 ) ζ ( 4 ) = 3 ! ⋅ 90 π 4 = 15 π 4
ρ γ = g γ 2 π 2 T 4 ⋅ π 4 15 = π 2 30 g γ T 4 \rho_\gamma = \dfrac{g_\gamma}{2\pi^2} T^4 \cdot \dfrac{\pi^4}{15} = \dfrac{\pi^2}{30} g_\gamma T^4 ρ γ = 2 π 2 g γ T 4 ⋅ 15 π 4 = 30 π 2 g γ T 4 For g γ = 2 g_\gamma = 2 g γ = 2 ρ γ = π 2 15 T 4 \rho_\gamma = \dfrac{\pi^2}{15} T^4 ρ γ = 15 π 2 T 4
3.2 Neutrino Energy Density and Temperature ¶ Neutrinos are fermions with Fermi-Dirac distribution:
f ( p ) = 1 e E / T + 1 f(p) = \dfrac{1}{e^{E/T} + 1} f ( p ) = e E / T + 1 1 After electron-positron annihilation, the photon temperature increases relative to neutrinos.
Exercise 3 (Derivation of
T ν / T γ = ( 4 / 11 ) 1 / 3 T_\nu/T_\gamma = (4/11)^{1/3} T ν / T γ = ( 4/11 ) 1/3 )
Before annihilation (T ≳ 1 T \gtrsim 1 T ≳ 1 T γ = T ν = T e T_\gamma = T_\nu = T_e T γ = T ν = T e
The entropy density is:
s = ρ + p T = 2 π 2 45 g ∗ s T 3 s = \dfrac{\rho + p}{T} = \dfrac{2\pi^2}{45} g_{*s} T^3 s = T ρ + p = 45 2 π 2 g ∗ s T 3 where g ∗ s g_{*s} g ∗ s
Before annihilation: g ∗ s = g γ + 7 8 ( g e + g e ˉ + g ν + g ν ˉ ) g_{*s} = g_\gamma + \frac{7}{8}(g_e + g_{\bar{e}} + g_\nu + g_{\bar{\nu}}) g ∗ s = g γ + 8 7 ( g e + g e ˉ + g ν + g ν ˉ )
Photons: g γ = 2 g_\gamma = 2 g γ = 2
Electrons and positrons: g e = g e ˉ = 2 g_e = g_{\bar{e}} = 2 g e = g e ˉ = 2
Neutrinos: 3 flavors × 2 spin states = 6 (only left-handed)
Antineutrinos: also 6
So:
g ∗ s before = 2 + 7 8 ( 2 + 2 + 6 + 6 ) = 2 + 7 8 × 16 = 2 + 14 = 16 g_{*s}^{\text{before}} = 2 + \frac{7}{8}(2+2+6+6) = 2 + \frac{7}{8} \times 16 = 2 + 14 = 16 g ∗ s before = 2 + 8 7 ( 2 + 2 + 6 + 6 ) = 2 + 8 7 × 16 = 2 + 14 = 16 After annihilation (T ≲ 0.5 T \lesssim 0.5 T ≲ 0.5
But neutrinos have decoupled and don’t receive the entropy from e + e − e^+e^- e + e − e + e − e^+e^- e + e −
g ∗ s before T 3 a 3 = g ∗ s after T ′ 3 a 3 g_{*s}^{\text{before}} T^3 a^3 = g_{*s}^{\text{after}} T'^3 a^3 g ∗ s before T 3 a 3 = g ∗ s after T ′3 a 3 where T ′ T' T ′
g ∗ s after, photons = 2 g_{*s}^{\text{after, photons}} = 2 g ∗ s after, photons = 2 For neutrinos, which decoupled earlier:
g ∗ s after, neutrinos = 7 8 × 6 = 42 8 = 5.25 g_{*s}^{\text{after, neutrinos}} = \frac{7}{8} \times 6 = \frac{42}{8} = 5.25 g ∗ s after, neutrinos = 8 7 × 6 = 8 42 = 5.25 But we consider the photon temperature change. Before annihilation, the entropy in photons+e + e − e^+e^- e + e −
s γ + e ± = 2 π 2 45 ( 2 + 7 8 × 4 ) T 3 = 2 π 2 45 × 11 2 T 3 s_{\gamma+e^\pm} = \frac{2\pi^2}{45} \left(2 + \frac{7}{8} \times 4\right) T^3 = \frac{2\pi^2}{45} \times \frac{11}{2} T^3 s γ + e ± = 45 2 π 2 ( 2 + 8 7 × 4 ) T 3 = 45 2 π 2 × 2 11 T 3 After annihilation, only photons:
s γ ′ = 2 π 2 45 × 2 × T ′ 3 s_{\gamma}' = \frac{2\pi^2}{45} \times 2 \times T'^3 s γ ′ = 45 2 π 2 × 2 × T ′3 Entropy conservation (for the coupled plasma) gives:
11 2 T 3 a 3 = 2 T ′ 3 a 3 ⇒ T ′ T = ( 11 4 ) 1 / 3 \frac{11}{2} T^3 a^3 = 2 T'^3 a^3 \quad \Rightarrow \quad \frac{T'}{T} = \left(\frac{11}{4}\right)^{1/3} 2 11 T 3 a 3 = 2 T ′3 a 3 ⇒ T T ′ = ( 4 11 ) 1/3 Thus after annihilation:
T ν = T , T γ = T ′ = ( 11 4 ) 1 / 3 T T_\nu = T, \quad T_\gamma = T' = \left(\frac{11}{4}\right)^{1/3} T T ν = T , T γ = T ′ = ( 4 11 ) 1/3 T So:
T ν T γ = ( 4 11 ) 1 / 3 ≈ 0.714 \frac{T_\nu}{T_\gamma} = \left(\frac{4}{11}\right)^{1/3} \approx 0.714 T γ T ν = ( 11 4 ) 1/3 ≈ 0.714 For neutrinos (g ν = 6 g_\nu = 6 g ν = 6
ρ ν = 7 8 ( π 2 30 g ν T ν 4 ) = 3 7 8 ( T ν T γ ) 4 ρ γ \rho_\nu = \dfrac{7}{8} \left( \dfrac{\pi^2}{30} g_\nu T_\nu^4 \right) = 3 \dfrac{7}{8} \left( \dfrac{T_\nu}{T_\gamma} \right)^{4} \rho_\gamma ρ ν = 8 7 ( 30 π 2 g ν T ν 4 ) = 3 8 7 ( T γ T ν ) 4 ρ γ Using ( T ν / T γ ) 4 = ( 4 / 11 ) 4 / 3 (T_\nu/T_\gamma)^4 = (4/11)^{4/3} ( T ν / T γ ) 4 = ( 4/11 ) 4/3
ρ ν = 3 7 8 ( 4 11 ) 4 / 3 ρ γ \rho_\nu = 3 \dfrac{7}{8} \left( \dfrac{4}{11} \right)^{4/3} \rho_\gamma ρ ν = 3 8 7 ( 11 4 ) 4/3 ρ γ 4. The Epoch of Decoupling ¶ 4.1 Recombination and Photon Decoupling ¶ Decoupling occurs at z ≃ 1100 z \simeq 1100 z ≃ 1100 t ∼ 3.8 × 1 0 5 t \sim 3.8 \times 10^{5} t ∼ 3.8 × 1 0 5 T ∼ 3000 T \sim 3000 T ∼ 3000
e − + p ↔ H + γ e^- + p \leftrightarrow H + \gamma e − + p ↔ H + γ The reduction of free electrons drops the Thomson scattering rate below the expansion rate.
4.2 The Saha Equation ¶ For the reaction e − + p ↔ H + γ e^- + p \leftrightarrow H + \gamma e − + p ↔ H + γ
n e n p n H = ( m e T 2 π ) 3 / 2 e − B / T \dfrac{n_e n_p}{n_H} = \left( \dfrac{m_e T}{2\pi} \right)^{3/2} e^{-B/T} n H n e n p = ( 2 π m e T ) 3/2 e − B / T where B = 13.6 B = 13.6 B = 13.6 X e = n e / ( n e + n H ) X_e = n_e/(n_e + n_H) X e = n e / ( n e + n H ) n b = n p + n H n_b = n_p + n_H n b = n p + n H
X e 2 1 − X e = 1 n b ( m e T 2 π ) 3 / 2 e − B / T \dfrac{X_e^2}{1 - X_e} = \dfrac{1}{n_b} \left( \dfrac{m_e T}{2\pi} \right)^{3/2} e^{-B/T} 1 − X e X e 2 = n b 1 ( 2 π m e T ) 3/2 e − B / T At z ≃ 1100 z \simeq 1100 z ≃ 1100 X e X_e X e ∼ 1 0 − 3 \sim 10^{-3} ∼ 1 0 − 3
5. Cosmological Distances and Times ¶ 5.1 Hubble Radius ¶ R H ≡ c H = 3000 h − 1 M p c R_H \equiv \dfrac{c}{H} = 3000 h^{-1} \, \rm Mpc R H ≡ H c = 3000 h − 1 Mpc is the radius of curvature of spatial sections.
5.2 Age of the Universe ¶ Define d η = d t / a ( t ) d\eta = dt/a(t) d η = d t / a ( t )
d s 2 = a 2 ( η ) ( d η 2 − d x 2 ) ds^2 = a^2(\eta) (d\eta^2 - d\mathbf{x}^2) d s 2 = a 2 ( η ) ( d η 2 − d x 2 ) Conformal time elapsed:
η = ∫ d t a = ∫ d a a 2 H = ∫ d z H ( z ) \eta = \int \dfrac{dt}{a} = \int \dfrac{da}{a^2 H} = \int \dfrac{dz}{H(z)} η = ∫ a d t = ∫ a 2 H d a = ∫ H ( z ) d z Setting η B B = 0 \eta_{\rm BB} = 0 η BB = 0
η 0 = ∫ 0 ∞ d z H ( z ) \eta_0 = \int_0^\infty \dfrac{dz}{H(z)} η 0 = ∫ 0 ∞ H ( z ) d z 5.4 Causal Horizons ¶ Exercise 4 (Radiation-dominated universe)
For a flat, radiation-dominated universe (Ω r , 0 = 1 \Omega_{r,0}=1 Ω r , 0 = 1
H 2 = H 0 2 a − 4 H^2 = H_0^2 a^{-4} H 2 = H 0 2 a − 4 The observable distance (setting c = 1 c=1 c = 1
d o b s = ∫ 0 1 d a a 2 H 0 a − 2 = 1 H 0 ∫ 0 1 d a = R H 0 d_{\rm obs} = \int_0^1 \dfrac{da}{a^2 H_0 a^{-2}} = \dfrac{1}{H_0} \int_0^1 da = R_{H_0} d obs = ∫ 0 1 a 2 H 0 a − 2 d a = H 0 1 ∫ 0 1 d a = R H 0 where R H 0 = c / H 0 R_{H_0} = c/H_0 R H 0 = c / H 0 c c c d o b s = c / H 0 d_{\rm obs} = c/H_0 d obs = c / H 0
Exercise 5 (Matter-dominated (Einstein-de Sitter) universe)
For a flat, matter-dominated universe (Ω m , 0 = 1 \Omega_{m,0}=1 Ω m , 0 = 1
H 2 = H 0 2 a − 3 H^2 = H_0^2 a^{-3} H 2 = H 0 2 a − 3 Then:
d o b s = 1 H 0 ∫ 0 1 d a a 2 a − 3 / 2 = 1 H 0 ∫ 0 1 d a a − 1 / 2 = 2 H 0 [ a 1 / 2 ] 0 1 = 2 R H 0 d_{\rm obs} = \dfrac{1}{H_0} \int_0^1 \dfrac{da}{a^2 a^{-3/2}} = \dfrac{1}{H_0} \int_0^1 da \, a^{-1/2} = \dfrac{2}{H_0} \left[ a^{1/2} \right]_0^1 = 2 R_{H_0} d obs = H 0 1 ∫ 0 1 a 2 a − 3/2 d a = H 0 1 ∫ 0 1 d a a − 1/2 = H 0 2 [ a 1/2 ] 0 1 = 2 R H 0 6. Fourier Decomposition and Horizon Crossing ¶ We decompose perturbations into comoving Fourier modes with wavenumber k k k physical wavelength evolves as:
λ ( t ) = a ( t ) λ c o m = a ( t ) 2 π k \lambda(t) = a(t) \lambda_{\rm com} = a(t) \dfrac{2\pi}{k} λ ( t ) = a ( t ) λ com = a ( t ) k 2 π where λ c o m = 2 π / k \lambda_{\rm com} = 2\pi/k λ com = 2 π / k
The physical wavenumber is k p h y s = k / a k_{\rm phys} = k/a k phys = k / a
Horizon crossing occurs when:
λ ( t ) ≃ H − 1 ( t ) ⇔ k ≃ a H \lambda(t) \simeq H^{-1}(t) \quad \Leftrightarrow \quad k \simeq aH λ ( t ) ≃ H − 1 ( t ) ⇔ k ≃ a H The comoving Hubble radius is ( a H ) − 1 = ( a ˙ ) − 1 (aH)^{-1} = (\dot{a})^{-1} ( a H ) − 1 = ( a ˙ ) − 1