Cosmology

Idea. Everything in cosmology is measured in units of \(H_0\); this problem fixes those units once, so that the rest of the sheet can work in Hubble times and Hubble lengths.

1. \(H_0 = 6.74\times10^{4}/3.0857\times10^{22} = 2.184\times10^{-18}\ \mathrm s^{-1}\). Then \(1/H_0 = 4.578\times10^{17}\ \mathrm s = 14.51\) Gyr and \(c/H_0 = 1.373\times10^{26}\ \mathrm m = 4448\) Mpc (\(14.51\) billion light years, as it must be: a light-time of one Hubble time).

2. \[ \rho_{\mathrm c,0} = \frac{3(2.184\times10^{-18})^2}{8\pi(6.6743\times10^{-11})} = 8.53\times10^{-27}\ \mathrm{kg\,m^{-3}}, \] which is \(5.10\) protons per cubic metre. Matter: \(0.315\rho_{\mathrm c,0} = 2.69\times10^{-27}\ \mathrm{kg\,m^{-3}}\), \(1.61\) protons per \(\mathrm m^3\). \(\Lambda\): \(0.685\rho_{\mathrm c,0} = 5.85\times10^{-27}\ \mathrm{kg\,m^{-3}}\), \(3.49\) protons per \(\mathrm m^3\). Today \(\rho_\Lambda / \rho_{\mathrm m} = 2.17\).

3. \(\Lambda = 3\Omega_\Lambda H_0^2/c^2 = 3(0.685)(2.184\times10^{-18})^2/(2.998\times10^8)^2 = 1.091\times10^{-52}\ \mathrm m^{-2}\), and \(1/\sqrt\Lambda = 9.57\times10^{25}\ \mathrm m = 3103\) Mpc. Units: \([\Lambda c^2] = \mathrm{m^{-2}\cdot m^2s^{-2}} = \mathrm s^{-2} = [H^2]\) ✓.

4. \(300 \lt 0.01\times67.4\,d\) requires \(d \gt 445\) Mpc. At the deck's \(100\) Mpc the contamination is \(300/6740 = 4.5\%\).

Check. \(\rho_\Lambda = \Lambda c^2/8\pi G\) directly: \(1.091\times10^{-52}\times (2.998\times10^8)^2/(8\pi\times6.6743\times10^{-11}) = 5.85\times10^{-27}\) ✓, the same as \(\Omega_\Lambda\rho_{\mathrm c,0}\), as the definitions require.

Idea. One component makes the Friedmann equation separable, and the answer is a power law whose exponent is fixed by \(w\). The decay of peculiar momentum is Noether's theorem for the spatial translations, which remain symmetries of flat FLRW even though time translation is not.

1. The fluid equation gives \(\rho \propto a^{-3(1+w)}\), so \(H^2 = H_0^2a^{-3(1+w)}\) and \(\dot a = aH = H_0a^{1-\frac32(1+w)}\). Separate (unit 00) and integrate from \(a = 0\) at \(t = 0\): \[ \frac{a^{\frac32(1+w)}}{\tfrac32(1+w)} = H_0t,\qquad a(t) = \Bigl(\tfrac32(1+w)H_0t\Bigr)^{\frac{2}{3(1+w)}}. \] Then \(H = \dot a/a = \frac{2}{3(1+w)t}\), and \(a = 1\) at \(t_0 = \frac{2}{3(1+w)H_0}\). Matter, \(w = 0\): \(a \propto t^{2/3}\), \(t_0 = \tfrac23(14.51) = 9.67\) Gyr. Radiation, \(w = \tfrac13\): \(a \propto t^{1/2}\), \(t_0 = \tfrac12(14.51) = 7.25\) Gyr. For \(w = -1\), \(H = H_0\) is constant, \(\dot a = H_0a\), and \(a = e^{H_0(t-t_0)}\): no zero, no age.

2. With \(a = Ct^n\), \(n = \frac{2}{3(1+w)}\): \(\ddot aa/\dot a^2 = n(n-1)/n^2 = (n-1)/n\), so \(q = (1-n)/n = \frac{3(1+w)}{2} - 1 = \tfrac12(1+3w)\). For \(w = -1\), \(\ddot a a/\dot a^2 = 1\) and \(q = -1 = \tfrac12(1-3)\) ✓. Acceleration, \(q \lt 0\), exactly when \(w \lt -\tfrac13\) — the acceleration equation's \(\rho + 3p/c^2 \lt 0\).

3. Parametrise by proper time \(\tau\). Unit 32: geodesics are the stationary curves of \(\int L\,\mathrm d\tau\) with \(L = -c^2\dot t^2 + a(t)^2\delta_{ij}\dot x^i\dot x^j\) (dots here \(\mathrm d/\mathrm d\tau\)). The coordinates \(x^i\) do not appear, so by unit 12's cyclic-coordinate rule \(p_i = \partial L/\partial\dot x^i = 2a^2\dot x^i\) is constant. A comoving observer's orthonormal spatial basis is \(e_{\hat i} = a^{-1}\partial_i\), since \(g(\partial_i,\partial_i) = a^2\). The particle's four-velocity has spatial part \(\dot x^i\partial_i = (a\dot x^i)e_{\hat i}\), so the momentum the observer measures has components \(ma\dot x^i = m\,p_i/(2a)\), and its magnitude falls as \(1/a\).

4. At \(600\ \mathrm{km\,s^{-1}}\) the motion is non-relativistic, so the momentum is \(m\) times the velocity, which also falls as \(1/a\). From \(a = \tfrac12\) to \(a = 1\) it halves: \(300\ \mathrm{km\,s^{-1}}\).

Check. Dimensions: \(H_0t\) is dimensionless ✓. Limit: \(w \to -1^+\) sends \(t_0 \to \infty\), matching the exponential ✓. Part 3 for a photon: the same conservation law gives energy \(\propto 1/a\), which is the deck's cosmological redshift obtained a second way ✓. The same result from the Christoffels: \(\mathrm du^i/\mathrm d\tau + 2\Gamma^i_{0j}u^0u^j = 0\) with \(\Gamma^i_{0j} = (\dot a/ca)\delta^i_j\) and \(u^0 = c\,\mathrm dt/\mathrm d\tau\) gives \(\mathrm du^i/\mathrm dt = -2(\dot a/a)u^i\), so \(u^i \propto a^{-2}\) ✓.

Idea. A rotation-invariant object can single out no direction. A covector singles out its dual vector, so it must vanish; a symmetric form singles out its eigenvectors, so it must have only one eigenvalue.

Transitivity. Given unit vectors \(u \ne \pm v\), the rotation about the axis \(u\times v\) through the angle between them carries \(u\) to \(v\); for \(v = -u\), rotate by \(\pi\) about any axis orthogonal to \(u\); for \(v = u\), the identity.

1. Let \(n = \omega^\sharp\), so that \(\omega(v) = \gamma(n, v)\). If \(n \ne 0\), take \(Q\) the rotation by \(\pi\) about an axis orthogonal to \(n\); then \(Qn = -n\) and \(Q^{-1} = Q\). Invariance gives \(\gamma(n,n) = \omega(n) = \omega(Qn) = -\gamma(n,n)\), so \(\gamma(n,n) = 0\) and \(n = 0\), a contradiction. So \(\omega = 0\). \(\square\)

2. Let \(q(v) = B(v,v)\), and fix a unit vector \(e\); put \(\lambda = q(e)\). For any unit \(u\), transitivity gives \(Q\) with \(u = Qe\), so \(q(u) = q(Qe) = q(e) = \lambda\). By homogeneity of degree two, \(q(v) = \lambda\gamma(v,v)\) for all \(v\). Polarisation, \(B(v,w) = \tfrac12[q(v+w) - q(v) - q(w)]\), then gives \(B = \lambda\gamma\). \(\square\)

3. At an event \(p\), the cosmological principle supplies, for each \(Q\in SO(3)\), an isometry of spacetime fixing \(p\) and the observer's velocity \(u\) and acting on the slice's tangent space \(T_p\Sigma\) by \(Q\). Such an isometry preserves \(g\), and preserves every tensor built from \(g\) alone — the Riemann, Ricci and Einstein tensors. Field equations whose left side is so invariant force the same on \(T\). Now apply the lemmas on \(T_p\Sigma\):

• \(v \mapsto g(u, v)\) is an invariant covector, so zero: the observers are orthogonal to the slices, \(g_{0i} = 0\). Likewise \(v\mapsto T(u,v)\) and \(v\mapsto G(u,v)\): \(T_{0i} = G_{0i} = 0\).
• \(T\) restricted to \(T_p\Sigma\) is an invariant symmetric form, so \(T_{ij} = p\,g_{ij}\) there, for some number \(p\) at \(p\).
• Homogeneity supplies isometries carrying any point of a slice to any other, so \(p\) and \(T_{00}\) are constant on each slice: functions of \(t\) alone.

4. With only the rotations about the \(z\)-axis, \(B = \mathrm{diag}(\lambda,\lambda,\mu)\) in an orthonormal basis with \(e_3\) along the axis is invariant for every \(\lambda, \mu\): each such rotation preserves the \(xy\)-plane with its inner product and fixes \(e_3\). So part 2 fails when \(\mu \ne \lambda\). A universe isotropic about one axis only can have different pressures along and across that axis, and a metric \(-c^2\mathrm dt^2 + a(t)^2(\mathrm dx^2 + \mathrm dy^2) + b(t)^2\mathrm dz^2\) with two scale factors, expanding at different rates in different directions. The CMB's isotropy to parts in \(10^5\) limits any such difference in the expansion rates severely.

Check. The proofs used orientation-preserving rotations only, so reflections were never needed ✓. Dimension three mattered only through transitivity, which holds for \(SO(n)\), \(n \ge 2\), on the unit sphere; so parts 1 and 2 hold in every dimension \(n \ge 2\) — the two-dimensional case being the rotation by \(\pi\) itself ✓.

Idea. Mechanical, and worth doing once: it is the spatial half of the deck's \(R_{ij}\), whose \(2kc^2\) comes entirely from here.

1. The \(r\) equation, expanded and multiplied by \((1-kr^2)/2\): \(\ddot r + \frac{kr}{1-kr^2}\dot r^2 - r(1-kr^2)\dot\theta^2 - r(1-kr^2)\sin^2\theta\,\dot\phi^2 = 0\). The \(\theta\) equation: \(\ddot\theta + \frac2r\dot r\dot\theta - \sin\theta\cos\theta\,\dot\phi^2 = 0\). The \(\phi\) equation: \(\ddot\phi + \frac2r\dot r\dot\phi + 2\cot\theta\,\dot\theta\dot\phi = 0\). Comparing with \(\ddot x^i + \Gamma^i_{jk}\dot x^j\dot x^k = 0\) (off-diagonal symbols appear twice):

\[ \Gamma^r_{rr} = \frac{kr}{1-kr^2},\ \ \Gamma^r_{\theta\theta} = -r(1-kr^2),\ \ \Gamma^r_{\phi\phi} = -r(1-kr^2)\sin^2\theta, \] \[ \Gamma^\theta_{r\theta} = \Gamma^\phi_{r\phi} = \frac1r,\ \ \Gamma^\theta_{\phi\phi} = -\sin\theta\cos\theta,\ \ \Gamma^\phi_{\theta\phi} = \cot\theta. \]

2. Write \(D = 1 - kr^2\). The four groups of \(R_{rr}\):

• \(\partial_\mu\Gamma^\mu_{rr} = \partial_r\frac{kr}{D} = \frac{k(1+kr^2)}{D^2}\).
• \(-\partial_r\Gamma^\mu_{\mu r} = -\partial_r\bigl(\frac{kr}{D} + \frac2r\bigr) = -\frac{k(1+kr^2)}{D^2} + \frac2{r^2}\).
• \(\Gamma^\mu_{\mu\lambda}\Gamma^\lambda_{rr} = \bigl(\frac{kr}{D} + \frac2r\bigr)\frac{kr}{D} = \frac{k^2r^2}{D^2} + \frac{2k}{D}\).
• \(-\Gamma^\mu_{r\lambda}\Gamma^\lambda_{\mu r} = -\bigl[(\Gamma^r_{rr})^2 + (\Gamma^\theta_{r\theta})^2 + (\Gamma^\phi_{r\phi})^2\bigr] = -\frac{k^2r^2}{D^2} - \frac2{r^2}\).

Sum: \(R_{rr} = \frac{2k}{1-kr^2} = 2k\gamma_{rr}\) ✓. The four groups of \(R_{\theta\theta}\):

• \(\partial_r\Gamma^r_{\theta\theta} = \partial_r(-r + kr^3) = -1 + 3kr^2\).
• \(-\partial_\theta\Gamma^\mu_{\mu\theta} = -\partial_\theta\cot\theta = \frac1{\sin^2\theta}\).
• \(\Gamma^\mu_{\mu r}\Gamma^r_{\theta\theta} = \bigl(\frac{kr}{D} + \frac2r\bigr)(-rD) = -kr^2 - 2D = kr^2 - 2\).
• \(-\Gamma^\mu_{\theta\lambda}\Gamma^\lambda_{\mu\theta}\): the pairs \((r,\theta)\) and \((\theta,r)\) each give \(-r D\cdot\frac1r = -D\), and \((\phi,\phi)\) gives \(\cot^2\theta\); with the sign, \(2D - \cot^2\theta = 2 - 2kr^2 - \cot^2\theta\).

Sum: \(-1 + 3kr^2 + kr^2 - 2 + 2 - 2kr^2 + \bigl(\frac1{\sin^2\theta} - \cot^2\theta\bigr) = 2kr^2 = 2k\gamma_{\theta\theta}\) ✓. \(R_{\phi\phi} = 2k\gamma_{\phi\phi}\) follows by the same computation, or by isotropy (P3 part 2), which also gives the off-diagonal components zero.

3. \({}^{(3)}R = \gamma^{ij}R_{ij} = 2k\cdot3 = 6k\). From the constant-curvature form, \(R_{jl} = \gamma^{ik}R_{ijkl} = k(3\gamma_{jl} - \gamma_{jl}) = 2k\gamma_{jl}\) ✓, and in three dimensions unit 33's decomposition makes Ricci determine Riemann, so the space has constant sectional curvature \(k\).

4. \(S_k(\chi) = \sin(\sqrt k\chi)/\sqrt k = \chi - k\chi^3/6 + O(\chi^5)\), so \[ 4\pi S_k^2 = 4\pi\chi^2\Bigl(1 - \frac{k\chi^2}{3} + O(\chi^4)\Bigr). \] Spheres are smaller than Euclidean ones by the fraction \(k\chi^2/3\): the deficit measures positive curvature, and the same expansion with \(\sinh\) shows an excess for \(k \lt 0\).

Check. Dimensions: \([k] = \mathrm m^{-2}\) and \([R_{rr}] = [k/(1-kr^2)] = \mathrm m^{-2}\) ✓. Limit \(k = 0\): every \(k\)-term vanishes and the list is that of flat spherical coordinates, with \(R_{ij} = 0\) ✓.

Idea. A second derivation in different coordinates, which cannot be reproduced from memory of the first. Conformal time makes light simple and matter-plus-radiation solvable, and it trades \(\ddot a\) for \(\mathcal H'\).

1. \(g_{\mu\nu} = a^2\eta_{\mu\nu}\), \(g^{\mu\nu} = a^{-2}\eta^{\mu\nu}\), and only \(\partial_0 g_{\mu\nu} = 2aa'\eta_{\mu\nu}\) is nonzero. So \(\Gamma^\lambda_{\mu\nu} = \mathcal H\,(\delta^\lambda_\mu\delta^0_\nu + \delta^\lambda_\nu\delta^0_\mu - \eta^{\lambda0}\eta_{\mu\nu})\), that is

\[ \Gamma^0_{00} = \mathcal H,\qquad \Gamma^0_{ij} = \mathcal H\,\delta_{ij},\qquad \Gamma^i_{0j} = \mathcal H\,\delta^i_j, \]

and all others vanish (\(\Gamma^0_{0i}\), \(\Gamma^i_{00}\), \(\Gamma^i_{jk}\)). Contractions: \(\Gamma^\mu_{\mu0} = 4\mathcal H\), \(\Gamma^\mu_{\mu i} = 0\).

2. \(R_{00}\): \(\partial_\mu\Gamma^\mu_{00} = \mathcal H'\); \(-\partial_0\Gamma^\mu_{\mu0} = -4\mathcal H'\); \(\Gamma^\mu_{\mu\lambda}\Gamma^\lambda_{00} = 4\mathcal H\cdot\mathcal H\); \(-\Gamma^\mu_{0\lambda}\Gamma^\lambda_{\mu0} = -(\Gamma^0_{00})^2 - \Gamma^i_{0j}\Gamma^j_{0i} = -\mathcal H^2 - 3\mathcal H^2\). Sum: \(R_{00} = -3\mathcal H'\).

\(R_{ij}\): \(\partial_0\Gamma^0_{ij} = \mathcal H'\delta_{ij}\); \(-\partial_j\Gamma^\mu_{\mu i} = 0\); \(\Gamma^\mu_{\mu0}\Gamma^0_{ij} = 4\mathcal H^2\delta_{ij}\); \(-\Gamma^\mu_{j\lambda}\Gamma^\lambda_{\mu i}\) has two nonzero groups, \(\Gamma^0_{jk}\Gamma^k_{0i}\) and \(\Gamma^k_{j0}\Gamma^0_{ki}\), each \(\mathcal H^2\delta_{ij}\), giving \(-2\mathcal H^2\delta_{ij}\). Sum: \(R_{ij} = (\mathcal H' + 2\mathcal H^2)\delta_{ij}\). \(R_{0i} = 0\) by P3.

Scalar: \(R = -a^{-2}R_{00} + a^{-2}\delta^{ij}R_{ij} = a^{-2}\bigl[3\mathcal H' + 3\mathcal H' + 6\mathcal H^2\bigr] = 6(\mathcal H' + \mathcal H^2)/a^2\), and \(\mathcal H' + \mathcal H^2 = a''/a\), so \(R = 6a''/a^3\).

3. \(G_{00} = R_{00} - \tfrac12R(-a^2) = -3\mathcal H' + 3(\mathcal H' + \mathcal H^2) = 3\mathcal H^2\) (the second derivative cancels, as in the deck). \(G_{ij} = (\mathcal H' + 2\mathcal H^2)\delta_{ij} - 3(\mathcal H' + \mathcal H^2)\delta_{ij} = -(2\mathcal H' + \mathcal H^2)\delta_{ij}\).

Source: the fluid is at rest, \(u^\mu = \mathrm dx^\mu/\mathrm d\tau = (c/a, 0,0,0)\) since \(\mathrm d\tau = a\,\mathrm d\eta\); so \(u_0 = -ac\) and \(T_{00} = (\rho + p/c^2)a^2c^2 - pa^2 = \rho c^2a^2\), \(T_{ij} = pa^2\delta_{ij}\). Einstein's equation with \(\kappa = 8\pi G/c^4\):

\[ 3\mathcal H^2 = \frac{8\pi G}{c^2}\rho a^2,\qquad -(2\mathcal H' + \mathcal H^2) = \frac{8\pi G}{c^4}pa^2. \]

Convert: \(\mathcal H = \frac1{ca}\frac{\mathrm da}{\mathrm d\eta} = \frac{a\dot a}{ca} = \frac{\dot a}{c}\), and \(\mathcal H' = \frac1c\frac{\mathrm d}{\mathrm d\eta}\frac{\dot a}c = \frac{a\ddot a}{c^2}\). The first becomes \(\dot a^2 = \frac{8\pi G}{3}\rho a^2\), i.e. \(H^2 = \frac{8\pi G}3\rho\) ✓. The second becomes \(2a\ddot a + \dot a^2 = -8\pi Gpa^2/c^2\); subtract the first and divide by \(2a^2\): \(\ddot a/a = -\frac{4\pi G}{3}(\rho + 3p/c^2)\) ✓.

4. \(\mathrm da/\mathrm d\eta = a\dot a = a^2H\), so \((\mathrm da/\mathrm d\eta)^2 = a^4H_0^2(\Omega_{\mathrm r}a^{-4} + \Omega_{\mathrm m}a^{-3}) = H_0^2(\Omega_{\mathrm r} + \Omega_{\mathrm m}a)\). Differentiating: \(a_{\eta\eta} = \tfrac12H_0^2\Omega_{\mathrm m}\), constant. With \(a(0) = 0\) and \(a_\eta(0) = H_0\sqrt{\Omega_{\mathrm r}}\) (the ODE at \(a = 0\)):

\[ a(\eta) = \tfrac14\Omega_{\mathrm m}H_0^2\eta^2 + \sqrt{\Omega_{\mathrm r}}\,H_0\eta,\qquad H_0\eta(a) = \frac{2}{\Omega_{\mathrm m}}\Bigl(\sqrt{\Omega_{\mathrm m}a + \Omega_{\mathrm r}} - \sqrt{\Omega_{\mathrm r}}\Bigr). \]

Cosmic time is \(t = \int a\,\mathrm d\eta = \frac{\Omega_{\mathrm m}H_0^2\eta^3}{12} + \frac{\sqrt{\Omega_{\mathrm r}}H_0\eta^2}2\). At equality \(a_{\mathrm{eq}} = \Omega_{\mathrm r}/\Omega_{\mathrm m} = 2.924\times10^{-4}\):

\[ H_0\eta_{\mathrm{eq}} = \frac{2(\sqrt2 - 1)\sqrt{\Omega_{\mathrm r}}}{\Omega_{\mathrm m}} = \frac{2(0.41421)(0.0095969)}{0.315} = 0.025239, \]

so the comoving horizon is \(c\eta_{\mathrm{eq}} = 0.025239\times4448 = 112\) Mpc, and \(H_0t_{\mathrm{eq}} = 4.220\times10^{-7} + 3.057\times10^{-6} = 3.479\times10^{-6}\), i.e. \(t_{\mathrm{eq}} = 3.479\times10^{-6} \times14.51\ \mathrm{Gyr} = 50\,500\) years — the deck's figure, now in closed form.

Check. \(\Omega_{\mathrm r} = 0\): \(a \propto \eta^2\), \(t \propto \eta^3\), so \(a \propto t^{2/3}\) ✓. \(\Omega_{\mathrm m} = 0\): \(a = \sqrt{\Omega_{\mathrm r}}H_0\eta\), \(t = \sqrt{\Omega_{\mathrm r}}H_0\eta^2/2\), so \(a = (2\sqrt{\Omega_{\mathrm r}}H_0t)^{1/2}\), which is P2's radiation law at \(\Omega_{\mathrm r} = 1\) ✓. Dimensions: \([\mathcal H] = \mathrm m^{-1}\) and \([G\rho a^2/c^2] = \mathrm{s^{-2}/(m^2s^{-2})}\) ✓.

Idea. Unit 08's picture: a particle rolls in \(V(a)\) with energy \(\Omega_k/2\) (in units of \(H_0^2\)). It turns round where its kinetic energy vanishes, which is where \(f = 0\). A positive \(k\) makes the energy negative, but \(\Lambda\) pulls the potential down without limit, so negative energy is not enough to be trapped.

1. The first Friedmann equation divided by \(H_0^2\) and multiplied by \(a^2\) is \(\dot a^2/H_0^2 = f(a)\). Also \(f = \Omega_k - 2V/H_0^2\) with \(V/H_0^2 = -\Omega_{\mathrm m}/2a - \Omega_\Lambda a^2/2\), so \(f = 0\) is "energy equals potential". Today \(f(1) = 1 \gt 0\). If \(f \gt 0\) on all of \([1,\infty)\), then \(\dot a \gt 0\) there and \(a\) never stops growing. If \(f\) is negative somewhere beyond \(1\), it first vanishes at some \(a_1 \gt 1\) with \(f'(a_1) \lt 0\) — the zero is simple, because \(af(a) = \Omega_\Lambda a^3 + \Omega_ka + \Omega_{\mathrm m}\) is a cubic with at most two positive roots and each is a simple root unless the two coincide. Then \(\ddot a = -V'(a_1) = \tfrac12H_0^2f'(a_1) \lt 0\) at a moment when \(\dot a = 0\), and the universe turns round. Conversely, if \(f \ge 0\) on \([1,\infty)\) the universe never recollapses: the only way to touch zero without crossing is a double zero, approached in infinite time (part 4).

2. \(\Omega_k = 1 - 0.4 - 0.7 = -0.1 \lt 0\), so \(k \gt 0\): closed. But \(f' = -\Omega_{\mathrm m}/a^2 + 2\Omega_\Lambda a\) vanishes only at \(a_{\mathrm{top}} = (\Omega_{\mathrm m}/2\Omega_\Lambda)^{1/3} = 0.6586\), and \(f\) increases on \([a_{\mathrm{top}},\infty)\); since \(a_{\mathrm{top}} \lt 1\) and \(f(1) = 1\), \(f\) never vanishes for \(a \gt 1\). It expands for ever. Its minimum is \(f(a_{\mathrm{top}}) = 0.6073 - 0.1 + 0.3037 = 0.811 \gt 0\), so \(\dot a\) never vanished in the past either: it began at \(a = 0\). In the deck's language the energy \(-0.05\) is far above the potential's top, \(V_{\max}/H_0^2 = -\tfrac34\Omega_{\mathrm m}/a_{\mathrm{top}} = -0.455\).

3. \(\Omega_k = -1 - \Omega_\Lambda\). Tangency at \(a_c\): \(f'(a_c) = 0\) gives \(\Omega_\Lambda a_c^3 = 1\); then \(f(a_c) = 2/a_c - (1+\Omega_\Lambda) + 1/a_c = 0\), so \(a_c = 3/(1+\Omega_\Lambda)\). Substituting into the first,

\[ 27\,\Omega_\Lambda = (1 + \Omega_\Lambda)^3. \]

Iterate \(x \mapsto (1+x)^3/27\) from \(x = 0\): \(0.03704,\ 0.04131,\ 0.04182,\ 0.04188,\ 0.04189\). The map's derivative there is \((1+x)^2/9 \approx 0.12\), so it contracts (unit 01's fixed-point theorem) and converges. Answer: \(\Omega_\Lambda = 0.0419\), touching at \(a_c = 3/1.0419 = 2.88 \gt 1\) ✓. Below it, \(f\) dips below zero past \(a = 1\) and the universe recollapses; above it, \(f \gt 0\) throughout.

4. At the threshold the universe arrives at \(a_c\) with \(\dot a \to 0\) and \(\ddot a \to 0\): it approaches the Einstein static universe (P7) asymptotically, taking infinite time, since near a double zero \(\dot a \propto (a_c - a)\) and \(\int\mathrm da/(a_c - a)\) diverges. The cubic's other positive root, \(\Omega_\Lambda = 3.596\), has \(a_c = 0.653 \lt 1\): that is the threshold in the past, above which there was no Big Bang but a bounce.

Check. \(\Omega_\Lambda = 0\), \(\Omega_{\mathrm m} = 2\): \(f = 2/a - 1\) vanishes at \(a = 2\), recollapse, as unit 08's bound orbit and P8 say ✓. Everything here is dimensionless ✓.

Idea. Einstein's 1917 universe sits at the top of the potential hill of the deck's effective-potential slide. A top is an equilibrium, and an unstable one.

1. With \(p = 0\) and \(\ddot a = 0\): \(\frac{4\pi G}{3}\rho_E = \frac{\Lambda c^2}3\), so \(\rho_E = \Lambda c^2/4\pi G\). With \(\dot a = 0\): \(0 = \frac{8\pi G}3\rho_E - \frac{kc^2}{a_E^2} + \frac{\Lambda c^2}3 = \frac{2\Lambda c^2}3 + \frac{\Lambda c^2}3 - \frac{kc^2}{a_E^2}\), so \(k/a_E^2 = \Lambda \gt 0\): closed, a 3-sphere of radius \(a_E/\sqrt k = 1/\sqrt\Lambda\).

2. With \(\rho a^3 = \rho_Ea_E^3\) fixed, the acceleration equation is \(\ddot a = -A/a^2 + \Lambda c^2a/3 = -V'(a)\), \(A = \frac{4\pi G}3\rho_Ea_E^3\). Linearising about \(a_E\), where \(V'(a_E) = 0\): \(\ddot\delta = -V''(a_E)\delta\), and

\[ -V''(a_E) = \frac{2A}{a_E^3} + \frac{\Lambda c^2}{3} = \frac{8\pi G}{3}\rho_E + \frac{\Lambda c^2}3 = \frac{2\Lambda c^2}3 + \frac{\Lambda c^2}3 = \Lambda c^2. \]

So \(\delta = \alpha e^{c\sqrt\Lambda t} + \beta e^{-c\sqrt\Lambda t}\) (unit 06), and any perturbation with \(\alpha \ne 0\) — every perturbation but a fine-tuned one — grows. A slight excess of \(a\) runs away into expansion; a slight deficit into collapse.

3. \(\rho_E = 1.091\times10^{-52}(2.998\times10^8)^2/(4\pi\times6.6743\times10^{-11}) = 1.17\times10^{-26}\ \mathrm{kg\,m^{-3}}\), seven protons per cubic metre — twice \(\rho_\Lambda\), as part 1 says. Curvature radius \(1/\sqrt\Lambda = 9.57\times10^{25}\ \mathrm m = 3103\) Mpc \(= 10.1\) Gly. E-folding time \(1/(c\sqrt\Lambda) = 3.19\times10^{17}\ \mathrm s = 10.1\) Gyr.

Check. Dimensions: \([c\sqrt\Lambda] = \mathrm{m\,s^{-1}\,m^{-1}} = \mathrm s^{-1}\) ✓. Limit \(\Lambda \to 0\): \(\rho_E \to 0\), the radius \(\to\infty\) and the growth rate \(\to 0\) — flat, empty, static Minkowski space, which is neutrally stable ✓. The equality of the radius in light years and the e-folding time in years is not a coincidence: both are \(1/\sqrt\Lambda\), once in metres and once divided by \(c\).

Idea. In conformal time a closed dust universe is a harmonic oscillator about a displaced centre, which is why its history is a cycloid: the curve traced by a point on a rolling wheel.

1. \((\mathrm da/\mathrm d\eta)^2 = a^4H^2 = H_0^2(\Omega_{\mathrm m}a + \Omega_ka^2)\). Differentiate: \(a_{\eta\eta} + \beta^2a = \tfrac12H_0^2\Omega_{\mathrm m}\) with \(\beta^2 = -\Omega_kH_0^2 = (\Omega_{\mathrm m} - 1)H_0^2 \gt 0\). Unit 06: \(a = A + C\cos\beta\eta + D\sin\beta\eta\) with \(A = H_0^2\Omega_{\mathrm m}/(2\beta^2)\). The condition \(a(0) = 0\) gives \(C = -A\); the undifferentiated ODE at \(a = 0\) gives \(a_\eta(0) = 0\), so \(D = 0\). Hence

\[ a = A(1 - \cos\theta),\quad \theta = \beta\eta,\quad A = \frac{\Omega_{\mathrm m}}{2(\Omega_{\mathrm m} - 1)},\quad \beta = H_0\sqrt{\Omega_{\mathrm m} - 1}, \]

and \(t = \int_0^\eta a\,\mathrm d\eta' = \frac A\beta(\theta - \sin\theta)\).

2. \(\Omega_{\mathrm m} = 2\): \(A = 1\), \(\beta = H_0\). Maximum at \(\theta = \pi\): \(a_{\max} = 2\), at \(t = \pi/H_0 = 45.6\) Gyr. Crunch at \(\theta = 2\pi\): \(2\pi/H_0 = 91.2\) Gyr. Today \(1 - \cos\theta_0 = 1\), so \(\theta_0 = \pi/2\) and \(t_0 = (\pi/2 - 1)/H_0 = 0.5708\times14.51 = 8.28\) Gyr, younger than Einstein-de Sitter's \(9.67\) Gyr: more matter decelerates harder, so the same \(H_0\) today implies a later Big Bang.

3. The comoving space is a 3-sphere of radius \(1/\sqrt k\), so geodesic distance and the angle \(\psi\) subtended at its centre are related by \(\chi = \psi/\sqrt k\). Light covers \(\mathrm d\chi = c\,\mathrm d\eta\), so since \(t = 0\) it has covered \(\psi = \sqrt k\,c\eta = \beta\eta = \theta\), using \(\beta^2 = kc^2\). Today \(\theta_0 = \pi/2\): a quarter of the way round a great circle, so we see half-way to the antipode and nothing beyond. Light from the antipode arrives at \(\theta = \pi\), at maximum expansion. By the crunch, \(\theta = 2\pi\): exactly once round, so light emitted here at the Big Bang returns here just as the crunch arrives, and no light ever gets further.

Check. Small \(\theta\): \(a \approx A\theta^2/2\), \(t \approx A\theta^3/(6\beta)\), so \(a \propto t^{2/3}\), matter-dominated as it must be near the Big Bang ✓. Dimensions: \([A] = 1\), \([\beta] = \mathrm s^{-1}\), \([\eta] = \mathrm s\) ✓. The age by direct quadrature, \(H_0^{-1}\int_0^1\mathrm da/\sqrt{2/a - 1}\), is also \(8.28\) Gyr ✓. Limit \(\Omega_{\mathrm m} \to 1^+\): \(A \to \infty\) and the maximum recedes to infinity, becoming Einstein-de Sitter ✓.

Idea. The age is \(1/H_0\) times a dimensionless function of the \(\Omega\)s. Raising \(H_0\) alone shortens it in proportion; but the CMB ties \(\Omega_{\mathrm m}\) to \(H_0\), and less matter means less deceleration and a longer history, which claws much of the difference back.

1. \(\mathrm dt = \mathrm da/(aH)\) and \(H = H_0\sqrt{\Omega_{\mathrm m}a^{-3} + \Omega_\Lambda}\), so \(t_0 = H_0^{-1}\int_0^1\sqrt a\,\mathrm da/\sqrt{\Omega_{\mathrm m} + \Omega_\Lambda a^3}\). With \(u = a^{3/2}\), \(\mathrm du = \tfrac32\sqrt a\,\mathrm da\):

\[ t_0 = \frac{2}{3H_0}\int_0^1\frac{\mathrm du}{\sqrt{\Omega_{\mathrm m} + \Omega_\Lambda u^2}} = \frac{2}{3H_0\sqrt{\Omega_\Lambda}}\operatorname{arsinh}\sqrt{\frac{\Omega_\Lambda}{\Omega_{\mathrm m}}}, \]

using \(\frac{\mathrm d}{\mathrm dx}\operatorname{arsinh}x = (1+x^2)^{-1/2}\). Planck: \(\sqrt{0.685/0.315} = 1.4747\), \(\operatorname{arsinh}1.4747 = 1.1806\), \(2/(3\sqrt{0.685}) = 0.8055\), product \(0.9510\); \(t_0 = 0.9510\times14.51 = \mathbf{13.80}\) Gyr. (Radiation, by quadrature, lowers it by \(5.5\) Myr.)

2. Same \(\Omega\)s, so the same \(0.9510\); \(1/H_0 = 13.39\) Gyr at \(73.0\); \(t_0 = 12.74\) Gyr, lower by \(1.06\) Gyr: \(7.7\%\), since \(t_0 \propto 1/H_0\) and \(67.4/73.0 = 0.923\).

3. \(\Omega_{\mathrm m}h^2 = 0.315\times0.674^2 = 0.1431\). At \(h = 0.730\): \(\Omega_{\mathrm m} = 0.1431/0.5329 = 0.2685\), \(\Omega_\Lambda = 0.7315\). Then \(\sqrt{0.7315/0.2685} = 1.6505\), \(\operatorname{arsinh}1.6505 = 1.2754\), \(2/(3\sqrt{0.7315}) = 0.7795\), product \(0.9942\), and \(t_0 = 0.9942\times13.39 = \mathbf{13.32}\) Gyr.

4. Only \(0.48\) Gyr of the \(1.06\) survives: the age falls by \(3.5\%\) instead of \(7.7\%\). The dimensionless factor rose from \(0.951\) to \(0.994\): with less matter the expansion decelerated less, so extrapolating back from the same present rate reaches \(a = 0\) earlier in the past. Near the Planck point, \(\mathrm d\ln t_0/\mathrm d\ln H_0 \approx -0.44\) at fixed \(\Omega_{\mathrm m}h^2\), against \(-1\) at fixed \(\Omega_{\mathrm m}\). Neither age is within \(0.3\) Gyr of \(13.8\), which is why the checkpoint names its parameters.

Check. \(\Omega_\Lambda \to 0\) at \(\Omega_{\mathrm m} = 1\): \(\operatorname{arsinh}x \approx x\) and \(t_0 \to 2/(3H_0) = 9.67\) Gyr, Einstein-de Sitter ✓. \([t_0] = [1/H_0]\) ✓. The closed form agrees with direct numerical quadrature of the first integral to five figures.

Idea. The patch of the last-scattering surface that one point there could have heard from has comoving radius \(\chi_{\mathrm p}(a_*)\); we see it from comoving distance \(\chi_{\mathrm{ls}}\); in flat space the angle is the ratio. The whole problem is getting two integrals right.

1. \(\Lambda\) is negligible: at \(a_*\), \(\Omega_\Lambda/(\Omega_{\mathrm m}a_*^{-3}) \sim 2\times10^{-9}\). P5:

\[ \chi_{\mathrm p}(a_*) = \frac{2c}{\Omega_{\mathrm m}H_0}\Bigl(\sqrt{\Omega_{\mathrm m}a_* + \Omega_{\mathrm r}} - \sqrt{\Omega_{\mathrm r}}\Bigr) = \frac{2}{0.315}(0.019515 - 0.009597)\frac c{H_0} = 0.06297\,\frac c{H_0}, \]

that is \(280\) Mpc comoving (at the time, \(280/1091 = 0.26\) Mpc proper).

2. Without radiation, \(\chi_{\mathrm p}(1) = \frac c{H_0}\int_0^1\frac{\mathrm da}{\sqrt{\Omega_{\mathrm m}a + \Omega_\Lambda a^4}}\); with \(a = u^2\), \(\mathrm da = 2u\,\mathrm du\) and the root is \(u\sqrt{\Omega_{\mathrm m} + \Omega_\Lambda u^6}\), so the integrand is \(f(u) = 2/\sqrt{0.315 + 0.685u^6}\). Values at \(u = 0, \tfrac14, \tfrac12, \tfrac34, 1\): \(3.5635,\ 3.5625,\ 3.5044,\ 3.0257,\ 2.0000\). Simpson with \(h = \tfrac14\):

\[ \tfrac1{12}\bigl[3.5635 + 4(3.5625) + 2(3.5044) + 4(3.0257) + 2.0000\bigr] = \tfrac{38.9255}{12} = 3.2438. \]

Radiation correction: for \(a \gg a_{\mathrm{eq}}\), P5 gives \(H_0\eta = \frac{2}{\Omega_{\mathrm m}} \bigl(\sqrt{\Omega_{\mathrm m}a}\sqrt{1 + \Omega_{\mathrm r}/\Omega_{\mathrm m}a} - \sqrt{\Omega_{\mathrm r}}\bigr) = \frac{2\sqrt a}{\sqrt{\Omega_{\mathrm m}}} - \frac{2\sqrt{\Omega_{\mathrm r}}}{\Omega_{\mathrm m}} + O\bigl(\Omega_{\mathrm r}a^{-1/2}\bigr)\). The first term is the radiation-free horizon; so by the time radiation is negligible, it has left a constant deficit \(2\sqrt{\Omega_{\mathrm r}}/\Omega_{\mathrm m} = 0.0609\), and afterwards the two integrands agree. Hence \(\chi_{\mathrm p}(1) = (3.2438 - 0.0609)\,c/H_0 = 3.1829\,c/H_0\).

3. \(\chi_{\mathrm{ls}} = 3.1829 - 0.0630 = 3.1199\,c/H_0 = 13\,880\) Mpc. Then

\[ \theta_{\mathrm H} = \frac{0.06297}{3.1199} = 0.02018\ \text{rad} = 1.16°. \]

A cap of angular radius \(\theta_{\mathrm H}\) has solid angle \(\approx \pi\theta_{\mathrm H}^2\), so the sky holds \(4\pi/\pi\theta_{\mathrm H}^2 = 4/\theta_{\mathrm H}^2 \approx 9800\) of them. (Numerical quadrature gives \(3.2402\) for the radiation-free integral, so Simpson's rule is high by \(0.0036\); with radiation it gives \(\chi_{\mathrm{ls}} = 3.1170\) and \(\theta_{\mathrm H} = 0.02020\) rad. Four Simpson steps were enough for three figures in the angle.)

4. Two points on the last-scattering surface in opposite directions are \(2\chi_{\mathrm{ls}}\) apart in comoving distance. Their past light cones intersect — so that some event could have influenced both — only if that separation is below \(2\chi_{\mathrm p}(a_*)\). The ratio is \(\chi_{\mathrm{ls}}/\chi_{\mathrm p}(a_*) = 49.5\). So no signal, and hence no physical process, has ever connected them, nor any two points more than about \(2.3°\) apart; yet their temperatures agree to one part in \(10^5\). The model does not contradict this. It must assume it, as an initial condition imposed separately on some ten thousand causally disconnected regions. The problem is that the observed uniformity is put in by hand, not explained.

Check. \(H_0\) cancels from the angle, as it must ✓. Leaving out radiation altogether gives \(\chi_{\mathrm p}(a_*) = 2\sqrt{a_*/\Omega_{\mathrm m}} = 0.1079\) and \(\theta_{\mathrm H} = 1.97°\): radiation matters at the forty per cent level, because recombination is only \(3.1\) times later in \(a\) than equality ✓. The deck's Einstein-de Sitter estimate, \(1.79°\), differs in having \(\Omega_{\mathrm m} = 1\) and neither radiation nor \(\Lambda\); all three versions agree that the angle is of order a degree, which is all the argument of part 4 needs.

Idea. In an expanding universe "the distance" and "the velocity" each split into several honest quantities. Computing them for one galaxy shows that none coincides with the special-relativistic reading of its redshift.

1. Integrand \(g(u) = 2/\sqrt{0.315 + 0.685u^6}\) at the five nodes: \(3.1598,\ 2.9184,\ 2.6264,\ 2.3109,\ 2.0000\). Simpson:

\[ \frac{0.073223}{3}\bigl[3.1598 + 4(2.9184) + 2(2.6264) + 4(2.3109) + 2.0000\bigr] = 0.024408\times31.3295 = 0.7647. \]

So \(\chi = 0.7647\times4448 = 3401\) Mpc \(= 11.09\) billion light years. (Quadrature: \(0.76468\).)

2. Now: \(\dot d = H_0\chi = 0.765c\). At emission: \(a = \tfrac12\), \(E(\tfrac12) = \sqrt{2.52 + 0.685} = 1.7903\), and \(\dot d = aH\chi = 0.5\times1.7903\times0.7647\,c = 0.684c\). Unit 26's Doppler formula, \(1 + z = \sqrt{(1+\beta)/(1-\beta)}\), gives \(\beta = \frac{(1+z)^2 - 1}{(1+z)^2 + 1} = \tfrac35 = 0.600\). Three different numbers, and no reason for any to agree: the redshift records only \(a(t_0)/ a(t_e) = 2\).

3. \(t(a) = \frac{2}{3H_0\sqrt{\Omega_\Lambda}}\operatorname{arsinh}\bigl(\sqrt{\Omega_\Lambda/ \Omega_{\mathrm m}}\,a^{3/2}\bigr)\). At \(a = \tfrac12\): argument \(1.4747\times0.35355 = 0.52137\), \(\operatorname{arsinh} = 0.50024\), \(t = 0.8055\times0.50024\times14.51 = 5.85\) Gyr. Lookback \(13.80 - 5.85 = 7.95\) Gyr. The light travelled \(7.95\) billion light years of path, but the galaxy is now \(11.09\) billion light years away: the space behind the light kept stretching.

4. \(d_L = 2\times3401 = 6803\) Mpc, against \(cz/H_0 = 4448\) Mpc. Flux goes as \(d^{-2}\), so it is \((6803/4448)^2 = 2.34\) times fainter than the naive estimate. Comparisons of this kind, between \(d_L(z)\) and a decelerating model's prediction, are how acceleration was found in 1998.

Check. Small \(z\): \(\chi \approx cz/H_0\), \(d_L \approx \chi\), and all three speeds tend to \(cz\) ✓. Dimensions: \(\chi\) in units of \(c/H_0\) is dimensionless ✓. Radiation at \(a = \tfrac12\): \(\Omega_{\mathrm r}a^{-4} = 1.5\times10^{-3}\) against \(E^2 = 3.2\), negligible ✓.

Idea. \(\Omega = 1\) is an equilibrium, and in a decelerating universe a repelling one; so the nearer to \(1\) we find it now, the nearer it must have started.

1. Divide \(H^2 = \frac{8\pi G}3\rho - \frac{kc^2}{a^2}\) (with \(\rho_\Lambda\) in \(\rho\)) by \(H^2\): \(1 = \Omega - kc^2/(a^2H^2)\) ✓. And \(kc^2 = -\Omega_kH_0^2\), so \(\Omega - 1 = -\Omega_k/(a^2E^2)\).

2. At \(a_* = 1/1091\) the three terms of \(E^2\) are: radiation \(9.21\times10^{-5}\times 1091^4 = 9.21\times10^{-5}\times1.4168\times10^{12} = 1.305\times10^8\); matter \(0.315\times1091^3 = 4.091\times10^8\); \(\Lambda\), \(0.685\). So \(E^2 = 5.395\times10^8\) — radiation is still a quarter of it, the trap in this part — and with \(a_*^{-2} = 1.190\times10^6\), \(|\Omega - 1| = 0.002\times1.190\times10^6/5.395\times10^8 = 4.4\times10^{-6}\). At nucleosynthesis radiation dominates completely: \(E^2 \approx \Omega_{\mathrm r}a^{-4}\), so \(|\Omega - 1| \approx |\Omega_k|a^2/\Omega_{\mathrm r} = 0.002\times(2.35\times10^{-10})^2/9.21\times10^{-5} = 1.2\times10^{-18}\).

3. With \(H\) constant, \(a \propto e^{Ht}\) and \(a^2H^2 \propto e^{2Ht}\); after \(N\) e-folds, \(a\) has grown by \(e^N\), so \(|\Omega - 1| = |k|c^2/(a^2H^2)\) has shrunk by \(e^{-2N}\). For \(N = 60\): \(e^{-120} = 7.7\times10^{-53}\). From the end of inflation to today the universe is radiation-dominated almost throughout the growth of \(|\Omega - 1|\), so part 1's formula with \(E^2 \approx \Omega_{\mathrm r}a^{-4}\) gives the value today's bound demands at \(a_{\mathrm{end}}\): \(0.002\,a_{\mathrm{end}}^2/\Omega_{\mathrm r} = 0.002\times(2.35\times10^{-28})^2/9.21\times10^{-5} = 1.2\times10^{-54}\). Starting from order one, \(e^{-2N} \le 1.2\times10^{-54}\) needs \(N \ge \tfrac12\ln(8.3\times10^{53}) = 62\). The deck's horizon estimate for the same reheating temperature was \(59\): the two problems ask for the same number of e-folds, "about sixty", which is part of why one mechanism is offered for both.

Check. Dimensions: \([kc^2/(a^2H^2)] = \mathrm{m^{-2}m^2s^{-2}/s^{-2}} = 1\) ✓. Special case \(k = 0\): \(\Omega = 1\) at all times ✓. Single-component scaling: in the radiation era \(a^{-2}E^{-2} \propto a^2\), and in the matter era \(\propto a\), which is the deck's \(|\Omega - 1| \propto a^{1+3w}\) ✓.

Idea. Helium is where the neutrons end up, so predicting \(Y\) is predicting the neutron fraction at about three minutes: a Boltzmann factor at freeze-out, then a radioactive decay.

1. With \(N_n\) neutrons and \(N_p\) protons, \(N_n/2\) helium nuclei form, of mass \(4\times N_n/2 = 2N_n\) nucleon masses, out of \(N_n + N_p\). So \(Y = 2N_n/(N_n + N_p) = 2x/(1+x)\), neglecting the neutron–proton mass difference and the binding energy, both under one per cent. \(\square\)

2. \(x_f = e^{-1.724} = 0.1784\), \(X_n = 0.1784/1.1784 = 0.1514\). Decay: \(e^{-200/878} = 0.7963\), so \(X_n = 0.1205\) and

\[ Y = 2X_n = 0.241. \]

(The textbook \(0.25\) rounds \(x\) to \(1/7\); here \(x = 0.1205/0.8795 = 0.137 = 1/7.3\).)

3. \(kT_f = 0.70\): \(x_f = 0.1577\), \(X_n = 0.1362\to0.1085\), \(Y = 0.217\). \(kT_f = 0.80\): \(x_f = 0.1986\), \(X_n = 0.1657\to0.1320\), \(Y = 0.264\). A \(7\%\) change in the freeze-out temperature moves \(Y\) by \(10\%\), because it sits in an exponent. The measurement, at about \(1\%\), therefore pins the freeze-out temperature, and with it the expansion rate at one second — which is how nucleosynthesis limits extra relativistic species. The full network calculation gives \(0.247\).

Check. Limits: \(x = 0\) gives \(Y = 0\), no neutrons, no helium; \(x = 1\) gives \(Y = 1\), all helium ✓. The exponents \(\Delta mc^2/kT\) and \(t/\tau_n\) are dimensionless ✓. A longer lifetime, \(\tau_n = 888\) s, raises \(Y\) only to \(0.2417\): the decay stage is the less sensitive one.