Part F · Einstein · branch

Cosmology

Assume the universe looks the same everywhere and in every direction, and Einstein's equation collapses to one ordinary differential equation for one function of time — which, solved with measured numbers, says the universe is 13.8 billion years old.

needs 35, 22~3 h slides + 6 h problemsproblems: ./mr problems 39

Unit 35 ended by saying this unit drops two things at once: staticity and weakness. Unit 36 solves the field equations exactly by assuming spherical symmetry about one point. This unit solves them exactly by assuming far more symmetry — spatial homogeneity and isotropy — and the reward is correspondingly larger: the ten field equations become two ordinary differential equations, and one of them is the energy equation of unit 08.

This is a branch unit. Nothing later in the course depends on it, and it depends on nothing after unit 35. It also uses nothing from units 36 to 38: no Schwarzschild metric, no perturbation theory, no gravitational waves.

What is proved and what is assumed

The deck keeps three kinds of statement strictly apart. The cosmological principle is an assumption, and is labelled as one. The evidence for it is observational, and is stated separately, with its limits. Everything derived from it — the metric, the Friedmann equations, the age — is proved on the slides or in the notes, or labelled cited. Cosmology is a subject where these three are easily blurred, and blurring them is the most common way to believe more than has been shown.

How to read this deck

One example runs through every section: the flat universe with dust and a cosmological constant, fitted by the Planck satellite, $\Omega_{\mathrm m} = 0.315$, $\Omega_\Lambda = 0.685$, $H_0 = 67.4\ \mathrm{km\,s^{-1}Mpc^{-1}}$. Section 1 gives it its metric, section 2 its equations, section 3 reads them as a particle rolling in a potential, section 4 solves them for its age, section 5 reads its light, and section 6 asks what its early history cannot explain. Every number was computed, and the notes say from what.

Conventions

Those of unit 35, unchanged: signature $(-,+,+,+)$, $x^0 = ct$, unit 33's sign of the Riemann tensor, $R_{\sigma\nu} = R^\mu{}_{\sigma\mu\nu}$, and every factor of $c$ kept. Here also: the scale factor $a$ is dimensionless with $a = 1$ today; comoving coordinates carry units of length; $k$ has units of $\mathrm{m^{-2}}$ and only its sign is a qualitative fact. Dots are $\mathrm d/\mathrm dt$. A subscript $0$ means "today".

39 · overview

One function of time

The universe on large scales, idealised as uniform: how old is it? The route, in four steps.

steps 1–2 · the geometry and its equation

1. Assume space looks the same everywhere and in every direction (the cosmological principle). Symmetry forces the FLRW metric: every distance between galaxies scales with one function $a(t)$, $a = 1$ today.

2. Einstein's equation for it (unit 33's machinery: Christoffels, Ricci, Einstein) collapses to one ODE, the first of the Friedmann equations, $H^2 = \tfrac{8\pi G}{3}\rho - \tfrac{kc^2}{a^2} + \tfrac{\Lambda c^2}{3}$, with $H = \dot a/a$, $\rho$ the density, $k$ the curvature of space, $\Lambda$ unit 35's constant: unit 08's energy equation in disguise.

steps 3–4 · the numbers and the age

3. Energy conservation fixes how each density dilutes: matter as $a^{-3}$, $\Lambda$ not at all. With $\Omega_{\mathrm m}$, $\Omega_\Lambda$ their measured shares of today's density, flat space ($k = 0$) turns step 2 into $H(a) = H_0\sqrt{\Omega_{\mathrm m}a^{-3} + \Omega_\Lambda}$.

4. Since $\mathrm dt = \mathrm da/\dot a = \mathrm da/(aH)$, the age is one integral with a closed form: $$t_0 = \int_0^1\frac{\mathrm da}{aH(a)} = \frac{0.951}{H_0} = 13.8\ \text{Gyr}$$ for the Planck universe.

The plan. 1 the metric · 2 the Friedmann equations · 3 an energy equation · 4 solutions, the age · 5 light: redshift, horizons, $H_0$ · 6 the early universe.

A dotted underline is a glossary term: click it, G for the index, N for the notes and the long proofs. A solid underline like the running example shows it on hover.

The running example, stated once. A spatially flat universe containing pressureless matter (dust, unit 35) and a cosmological constant, with present density parameters $\Omega_{\mathrm m} = 0.315$ and $\Omega_\Lambda = 0.685$ and Hubble constant $H_0 = 67.4\ \mathrm{km\,s^{-1}Mpc^{-1}} = 2.184\times10^{-18}\ \mathrm{s^{-1}}$. These are the central values of the Planck 2018 cosmological parameters (Planck Collaboration, A&A 641, A6, 2020), rounded as they are usually quoted. Radiation, $\Omega_{\mathrm r} = 9.2\times10^{-5}$, is left out of the example and put back wherever it matters: it moves the age by $5.5$ Myr and the horizon at recombination by forty per cent.

The route, slide by slide. Step 1 is section 1: The cosmological principle states the assumption, and The FLRW metric, and why there is no other proves that it forces the metric. Step 2 is section 2: The Christoffel symbols of FLRW and Ricci, scalar curvature, Einstein tensor compute the curvature, and The Friedmann equations imposes Einstein's equation; section 3, from A Newtonian ball of dust on, reads the result as an energy equation. Step 3 is The fluid equation, twice and Equation of state, and how each component dilutes, with the $\Omega$s defined properly on Critical density and the density parameters. Step 4 is The age of the Planck universe, worked example 3, where the integral is done by one substitution.

The unit's arc, measured against it: by section 2 you can write its metric and its two equations; by section 4 you can compute its age to three figures with a pencil and a table of logarithms; by section 6 you can say, quantitatively, what it cannot explain.

section 1

The assumption, and the geometry it forces

One symmetry assumption; one metric, up to a sign and a function.

39 · assumption

The cosmological principle

the assumption

There is a family of fundamental observers — the mean motion of matter — and on large scales, at each instant of their common time, space is homogeneous (no point is special) and isotropic about each of them (no direction is special).

large scales only

It is false below roughly $100$ Mpc: galaxies, clusters, filaments and voids are not homogeneous. The claim is about averages over volumes larger than that, and the metric it produces describes the average, not the Milky Way.

the evidence, separately

The CMB: $T_0 = 2.7255$ K in every direction. Remove a dipole of $3.36$ mK — our own motion, $370\ \mathrm{km\,s^{-1}}$ — and what is left varies by parts in $10^5$.
Galaxy surveys (SDSS, WiggleZ): counts in spheres approach the homogeneous $N \propto R^3$ above about $100$ Mpc.
Radio sources and X-ray background: isotropic to about a per cent.

Isotropy is observed only from here. Homogeneity is inferred: isotropy about every point implies it, and the Copernican assumption that we are not special supplies "every point".

Why the assumption is stated apart from the evidence. Observations are made from one point, on our past light cone. Isotropy about us is measured; homogeneity is not, because we cannot visit another galaxy's present. The logical chain is: isotropy observed here, plus the assumption that we are not privileged, gives isotropy about every fundamental observer; and isotropy about every point implies homogeneity (given two points $p, q$, take the midpoint $m$ of a geodesic joining them; the rotation by $\pi$ about $m$ in a plane containing the geodesic is an isometry swapping $p$ and $q$ — this is Weinberg, Gravitation and Cosmology (1972) §13.1, cited). So the only genuinely unobservable input is "we are typical".

The numbers. $T_0 = 2.7255 \pm 0.0006$ K (Fixsen 2009). The dipole amplitude $3.3621$ mK is a Doppler shift (unit 26's formula to first order: $\Delta T/T = v/c$), giving $v = c \times 3.3621\times10^{-3}/2.7255 = 369.8\ \mathrm{km\,s^{-1}}$ for the Sun relative to the CMB rest frame. That frame is how "the fundamental observers" is made operational: they are the observers who see no dipole. The residual anisotropies are of order $10^{-5}$ in $\Delta T/T$ and are not noise: they are the seeds of structure, and their statistics are what Planck fitted to obtain the running example's parameters.

The homogeneity scale. Hogg et al. (2005) with SDSS luminous red galaxies and Scrimgeour et al. (2012) with WiggleZ found the fractal dimension of the galaxy distribution reaching $3$ within a per cent at about $70\,h^{-1}$ Mpc, which is $104$ Mpc for $h = 0.674$. "About $100$ Mpc" is that number rounded.

History. Einstein assumed homogeneity in 1917 for his static model, and Milne named the principle in 1933. It was a simplifying assumption for forty years before the CMB (Penzias and Wilson, 1965) made it an observational statement.

39 · assumption

Maximally symmetric spaces

maximally symmetric space

A Riemannian $n$-manifold $(\Sigma, \gamma)$ whose Killing fields (unit 31) span the largest possible algebra, of dimension $n(n+1)/2$. Equivalently (cited): it is homogeneous, and at each $p$ every rotation of $T_p\Sigma$ is $\mathrm d\varphi_p$ for some isometry $\varphi$ fixing $p$.

Physics: the cosmological principle says each slice of constant time is one of these, with $n = 3$ and six Killing fields — three translations, three rotations.

theorem

A maximally symmetric space has constant sectional curvature: $$R_{ijkl} = K\,(\gamma_{ik}\gamma_{jl} - \gamma_{il}\gamma_{jk}),\quad K \text{ constant}.$$

proof

An isometry carries $R$ to $R$, since $R$ is built from $\gamma$ alone. At $p$, the isotropy isometries realise all of $SO(n)$ on $T_p\Sigma$, which acts transitively on 2-planes; so $K(\Pi)$ is the same for every plane at $p$, and unit 33's polarisation gives the displayed form. Homogeneity moves $p$ anywhere, so $K$ is constant. $\square$

Full proof. Three steps.

Isometries preserve curvature. If $\varphi$ is an isometry, $\varphi^*\gamma = \gamma$. The Levi-Civita connection is determined by $\gamma$ (unit 32's Koszul formula), so $\varphi$ carries it to itself, and hence carries $R$ to itself: $R_{\varphi(p)}(\mathrm d\varphi X, \mathrm d\varphi Y, \mathrm d\varphi Z, \mathrm d\varphi W) = R_p(X,Y,Z,W)$. Consequently sectional curvature satisfies $K(\mathrm d\varphi_p\Pi) = K(\Pi)$.

$SO(n)$ is transitive on 2-planes. Given planes $\Pi, \Pi'$ with orthonormal bases $(e_1, e_2)$ and $(e_1', e_2')$, extend each to an orthonormal basis of $T_p\Sigma$ and, if necessary, flip the sign of the last vector of one of them so both have the same orientation ($n \ge 3$; for $n = 2$ there is only one plane). The linear map sending one basis to the other is orthogonal with determinant $+1$ and carries $\Pi$ to $\Pi'$. So if every element of $SO(n)$ is realised by an isometry fixing $p$, all planes at $p$ have the same sectional curvature $K(p)$.

Polarisation. The tensor $K(p)(\gamma_{ik}\gamma_{jl} - \gamma_{il}\gamma_{jk})$ has all of unit 33's symmetries and sectional curvature $K(p)$ on every plane (unit 33's check). Two algebraic curvature tensors with the same sectional curvatures are equal (unit 33, cited there from do Carmo ch. 4, Lemma 3.3). So $R_p$ has the displayed form.

Constancy. Homogeneity gives, for any $p, q$, an isometry $\varphi$ with $\varphi(p) = q$, and then $K(q) = K(p)$ by the first step. $\square$

Cited, not proved. (i) The equivalence between "Killing algebra of dimension $n(n+1)/2$" and "homogeneous and isotropic": unit 31 cited the bound; the equivalence is Weinberg (1972) §13.1–13.2. (ii) The converse direction that matters for uniqueness: two Riemannian metrics of the same constant curvature $K$ and dimension are locally isometric (Riemann 1854; a proof via Cartan's theorem is in do Carmo, Riemannian Geometry, ch. 8, on space forms). So constant curvature fixes the geometry locally up to the one number $K$.

An aside worth having. For $n \ge 3$ homogeneity is not even needed: Schur's lemma (cited; it is among the exercises of do Carmo ch. 4) says that if $K(p)$ is independent of the plane at each point, the contracted Bianchi identity forces it to be independent of the point.

39 · assumption

Three model spaces

Up to local isometry there is one maximally symmetric 3-space for each $K = k$. With $\chi$ the geodesic distance from a chosen point, $$\gamma = \mathrm d\chi^2 + S_k(\chi)^2\,\mathrm d\Omega^2,$$ with $S_k$ from the table; or, with $r = S_k(\chi)$, the form used on the FLRW metric: $\gamma = \mathrm dr^2/(1 - kr^2) + r^2\,\mathrm d\Omega^2$.

check

Unit 32's Christoffels give $R_{ij} = 2k\,\gamma_{ij}$ (P4); in three dimensions Ricci determines Riemann (unit 33), so the curvature is $k$. As $k \to 0$, $S_k(\chi) \to \chi$: Euclidean polar coordinates ✓.

$k$	$S_k(\chi)$	model	circle of radius $\chi$
$> 0$	$\sin(\sqrt k\,\chi)/\sqrt k$	3-sphere, radius $1/\sqrt k$	$\lt 2\pi\chi$; finite volume
$0$	$\chi$	$\mathbb R^3$	$= 2\pi\chi$
$\lt 0$	$\sinh(\sqrt{\|k\|}\,\chi)/\sqrt{\|k\|}$	hyperbolic 3-space	$> 2\pi\chi$

Why $\mathrm d\chi^2 + S_k^2\,\mathrm d\Omega^2$. Choose a point and use geodesic polar coordinates about it: $\chi$ is geodesic distance, and the spheres $\chi = $ const are orbits of the rotations fixing the point, so by isotropy they are round, with some area $4\pi S(\chi)^2$. The metric is then $\mathrm d\chi^2 + S(\chi)^2\,\mathrm d\Omega^2$ (Gauss's lemma, that geodesic spheres are orthogonal to the radial geodesics, is cited from do Carmo ch. 3). The sectional curvature of the plane spanned by $\partial_\chi$ and $\partial_\theta$ is $-S''/S$: unit 32's formula gives $\Gamma^\chi_{\theta\theta} = -SS'$, $\Gamma^\theta_{\chi\theta} = S'/S$ and $\Gamma^\chi_{\chi\lambda} = 0$, so $R^\chi{}_{\theta\chi\theta} = \partial_\chi\Gamma^\chi_{\theta\theta} - \Gamma^\chi_{\theta\theta}\Gamma^\theta_{\chi\theta} = -(S'^2 + SS'') + S'^2 = -SS''$, and dividing by the Gram determinant $1\cdot S^2$ gives $K = -S''/S$. Constant curvature $k$ therefore means $S'' = -kS$ with $S(0) = 0$, $S'(0) = 1$ (so that the space is smooth at the centre). That is unit 06's constant- coefficient ODE, and its unique solution is the $S_k$ in the box.

The $r$ form. With $r = S_k(\chi)$, $\mathrm dr = S_k'(\chi)\,\mathrm d\chi$ and $S_k'^2 = 1 - kS_k^2$ in all three cases (for $k > 0$ this is $\cos^2 = 1 - \sin^2$). So $\mathrm d\chi^2 = \mathrm dr^2/(1 - kr^2)$. For $k > 0$ the $r$ coordinate covers only half the 3-sphere, since $r$ rises to $1/\sqrt k$ at the equator and falls again: the coordinate singularity at $kr^2 = 1$ is unit 05's lesson about spherical coordinates again, not a feature of the space.

The table's last column is drawn on the next slide.

39 · assumption

Circles in the three model spaces

Two-dimensional analogues, schematic. From the centre walk a geodesic distance $\chi$ in every direction: the circle traced has circumference $2\pi S_k(\chi)$. On the sphere the radial geodesics reconverge, so the circle is short; on the saddle they spread apart, and the circle has to ruffle to fit.

local, not global

Curvature fixes the geometry locally. A flat 3-torus has $k = 0$ and finite volume. "Flat means infinite" is not a theorem.

What to look for. The three pictures are 2-dimensional slices of the 3-dimensional model spaces, which cannot themselves be drawn. In each, the dot is the chosen point, the straight-looking line is a radial geodesic of length $\chi$, and the thick curve is the set of points at geodesic distance $\chi$. Only the sphere is drawn exactly. The saddle is a surface of negative curvature only near its centre, and its curvature is not constant; no surface in ordinary 3-space is a complete hyperbolic plane (Hilbert's theorem, cited from do Carmo, Differential Geometry of Curves and Surfaces, §5-11), which is why hyperbolic space is always drawn by some distorted model.

The circumference column. A circle at geodesic distance $\chi$ has circumference $2\pi S_k(\chi)$, and $\sin x \lt x \lt \sinh x$ for $x > 0$. Compare unit 30's Ehrenfest disc, which had circumference greater than $2\pi\chi$: negative spatial curvature, forced there by rotation. This is also how curvature would be measured from inside, with no embedding: count the galaxies, or measure the angular sizes, out to a given distance, and compare with the Euclidean $4\pi\chi^2$ and $2\pi\chi$.

Topology. Observations constrain only $k$ and only within the observable region. A non-trivial topology smaller than the horizon would repeat patterns in the CMB, and searches for matched circles have found none, so any identification scale exceeds roughly the distance to the last-scattering surface (Planck 2015 XVIII, cited).

39 · assumption

The FLRW metric, and why there is no other

theorem (FLRW metric)

If spacetime is foliated by spacelike slices orthogonal to the worldlines of fundamental observers, and each slice is homogeneous and isotropic about each of them, then there are coordinates in which $$\mathrm ds^2 = -c^2\,\mathrm dt^2 + a(t)^2\Bigl[\frac{\mathrm dr^2}{1-kr^2} + r^2\,\mathrm d\Omega^2\Bigr].$$

$t$ is cosmic time: proper time along every fundamental observer.
$(r, \theta, \phi)$ are comoving coordinates: each observer keeps fixed values.
$a(t)$ is the scale factor, dimensionless, $a(t_0) = 1$ today; $k$ is the curvature constant of the previous slides, in $\mathrm m^{-2}$; $\mathrm d\Omega^2 = \mathrm d\theta^2 + \sin^2\theta\,\mathrm d\phi^2$.
The bracket is the constant-curvature metric $\gamma$ of Three model spaces, in its $r$ form. The whole of the time dependence is the one factor $a(t)^2$.

The proof is drawn on the next slide.

Reading the metric. Along a fundamental observer ($r, \theta, \phi$ fixed), $\mathrm ds^2 = -c^2\mathrm dt^2$, so $t$ is that observer's proper time. Within a slice ($t$ fixed), $\mathrm ds^2 = a(t)^2\gamma$: the same shape at every time, uniformly rescaled. Those two sentences are the theorem's content; the next slide proves it.

Cited. The same theorem, with the step "the symmetry group of spacetime acts on the slices" treated more carefully, is Wald, General Relativity (1984) §5.1, and Weinberg (1972) §14.1–14.2. Friedmann (1922, 1924) wrote the metric by assumption; Robertson (1935) and Walker (1936) proved independently that the symmetries force it, which is why the four names are attached. Lemaître (1927), the L in "FLRW", found the expanding solution independently and was the first to connect it with observed redshifts.

What the theorem does not say. It says nothing about $a(t)$ or the sign of $k$. Those are what the Einstein equations and the matter content decide, in section 2. The geometry has been reduced from ten functions of four variables to one function of one variable and one number.

39 · assumption

Slices, observers and one scale factor

Schematic. Each observer keeps its comoving label from slice to slice; what changes is the spacing, by the same factor $a(t_2)/a(t_1)$ everywhere and in every direction.

proof, the three moves

1. Coordinates. Label each observer by where it crosses one slice, and each slice by the observers' proper time.

2. No cross terms. $g_{0i}$ would be a covector on the slice, and a rotation-invariant covector is zero: the worldlines meet the slices at right angles.

3. One factor. The slice metrics at two times differ by a map commuting with all rotations and translations: a constant factor, $a(t_2)^2/a(t_1)^2$.

Full proof in the notes.

What the figure shows. Two slices of constant cosmic time and nine fundamental observers, one dimension of space suppressed. The observers are drawn fanning out, which is how the separations look; in comoving coordinates the same worldlines are vertical, and that is move 1. Move 2 is that each worldline crosses each slice at a right angle, in the spacetime metric. Move 3 is the two brackets: one ratio, the same for every pair of neighbours and every direction.

Full proof. Let $u$ be the unit timelike field of fundamental observers, orthogonal to the slices $\Sigma$.

Step 1, coordinates. Pick one slice $\Sigma_*$ and coordinates $x^i$ on it; give every event the $x^i$ of the observer through it. Along one observer, proper time between two slices is the same for all observers, since an isometry of the slices (a symmetry of spacetime, by assumption) carries any observer to any other and preserves proper time; so label slices by that proper time $t$. Then $g_{00} = -1$: the observers' worldlines are the curves $x^i = $ const and $t$ is proper time along them.

Step 2, no cross terms. $g_{0i}\,\mathrm dx^i$ restricted to a slice is a covector field. Isotropy about a point $p$ means every rotation of $T_p\Sigma$ is induced by a symmetry fixing $p$ and $u$; such a symmetry preserves the covector, so $\omega_p$ is invariant under $SO(3)$. If $\omega_p \ne 0$, some rotation sends its dual vector $v$ to $-v$; invariance then gives $\omega_p = -\omega_p$. So $g_{0i} = 0$. (This is also simply "the observers are orthogonal to the slices", and problem P3 proves the lemma cleanly.)

Step 3, one scale factor. Now $g = -c^2\mathrm dt^2 + h_t$ with $h_t$ a metric on the comoving 3-space, maximally symmetric for each $t$. Compare $h_t$ and $h_{t'}$ at a point $p$: both are inner products on the same space $T_p$, both invariant under the same $SO(3)$ (the symmetries fix the comoving labels up to the isometry). The quadratic form $v \mapsto h_t(v,v)/ h_{t'}(v,v)$ is then constant on the $h_{t'}$-unit sphere, because $SO(3)$ is transitive on it; so $h_t = \lambda(t, t', p)\,h_{t'}$ by polarisation. Homogeneity makes $\lambda$ independent of $p$. Fix $t' = t_0$, write $\lambda = a(t)^2$ and $h_{t_0} = \gamma$, which by the previous slides is the constant-curvature metric with some $k$. $\square$

39 · assumption

Proper distance, Hubble's law, conformal time

proper distance

The length, in the slice at time $t$, of the radial geodesic to a comoving object at $\chi$: $d(t) = a(t)\,\chi$.

Hubble's law, exact

$\dot d = \dot a\,\chi = H(t)\,d$, with Hubble parameter $H = \dot a/a$.

One line: $\chi$ is constant. $[H] = \mathrm s^{-1}$; for the running example $H_0 = 67.4\ \mathrm{km\,s^{-1}Mpc^{-1}} = 2.184\times10^{-18}\ \mathrm s^{-1}$.

not a velocity

$\dot d$ is a rate of change of distance within slices. It exceeds $c$ beyond $c/H_0 = 4448$ Mpc $= 14.5$ Gly, and nothing is violated: unit 26's speed limit is local.

Computed for the running example. Comoving galaxies (grey) separate as $d = a(t)\chi$, each with slope $\dot d = Hd$. On the dashed Hubble sphere, $d = c/H$, that slope is $c$. Our past light cone is a teardrop: light reaching us from far away left its source when the source was near, never more than $5.85$ Gly off.

Why $d = a\chi$. In the slice $t = $ const the metric is $a(t)^2\gamma$, and along a radial path $\gamma = \mathrm d\chi^2$. So the length is $\int_0^\chi a(t)\,\mathrm d\chi' = a(t)\chi$. It is the distance a chain of fundamental observers would measure by laying rulers end to end at one instant of cosmic time — a definition, not something anyone can do.

Dimensions and the unit of $H_0$. $\mathrm{km\,s^{-1}\,Mpc^{-1}}$ is a velocity per distance, so an inverse time: $67.4\times10^3\ \mathrm{m\,s^{-1}} / 3.0857\times10^{22}\ \mathrm m = 2.184\times10^{-18}\ \mathrm s^{-1}$. Its reciprocal, the Hubble time, is $4.578\times10^{17}$ s $= 14.51$ Gyr, and $c/H_0 = 4448$ Mpc $= 14.51$ Gly.

How the figure was made. Time runs up and proper distance across, as in unit 26's spacetime diagrams; we sit on the vertical axis. The galaxies are $d = a(t)\chi$ with $a(t)$ the closed form of worked example 3 (notes there), for $\chi = 5, 10, \dots, 25$ Gly. The Hubble sphere is $c/H(t)$; it is $14.5$ Gly today and tends to $c/(H_0\sqrt{\Omega_\Lambda}) = 17.5$ Gly. The light cone is $d = a(t)\,c\,[\eta(t_0) - \eta(t)]$, with the conformal time of the next slide. Its widest point, $5.85$ Gly at $z = 1.59$, $4.05$ Gyr after the Big Bang, is exactly where it crosses the Hubble sphere: there light moving towards us at $c$ through space that recedes at $c$ holds its distance.

Superluminal recession. Unit 26's statement is that no causal curve leaves the light cone at any event, and it is about vectors at one point. $\dot d$ compares a distance defined across a whole slice at two times, and there is no tangent vector at one event whose length it is. Section 5 returns to this, with numbers, when it says why redshift is not a Doppler shift.

39 · assumption

Conformal time straightens light rays

conformal time

$\mathrm d\eta = \mathrm dt/a(t)$. Then $$\mathrm ds^2 = a(\eta)^2\bigl[-c^2\,\mathrm d\eta^2 + \mathrm d\chi^2 + S_k(\chi)^2\mathrm d\Omega^2\bigr]$$ and radial light rays obey $\mathrm d\chi = \pm c\,\mathrm d\eta$: straight lines at $45°$, as in unit 26.

The factor $a^2$ in front changes lengths, not which curves are null. So the causal structure — who can signal whom — is that of the static bracket, and it can be read off with a ruler.

Our past light cone, computed for the running example with radiation, drawn twice on one grid. Against $\eta$ it is straight. Against $t$ it is bent: light now arriving from $46$ Gly covered $19$ Gly of the way in the first billion years, when $a$ was small.

Conformal time. The substitution is the whole content: $-c^2\mathrm dt^2 = -c^2a^2 \mathrm d\eta^2$. The causal structure of FLRW is that of the static metric in the bracket, because a conformal factor does not change which curves are null. That is why horizons in section 5 are computed as $\eta$-intervals, and why a Penrose-style diagram of FLRW is easy to draw. With $\mathrm d\eta = \mathrm dt/a$, $\eta$ has units of time; some books use $c\,\mathrm dt/a$ and give it units of length.

How the figure was made. A radial ray arriving at $\chi = 0$ today has $\chi = c[\eta_0 - \eta]$, and $\eta(a) = \int_0^a \mathrm da'/(a'^2H)$ was integrated numerically with $\Omega_{\mathrm r} = 9.21\times10^{-5}$, giving $\eta_0 = 46.1$ Gyr; $t(a) = \int_0^a\mathrm da'/(a'H)$ the same way. Both curves are the same set of events, plotted against two different time coordinates measured in the same unit. The straight one is the one to draw by hand; the bent one is what the same ray looks like in the time clocks keep.

The $19$ Gly. $\eta$ at $t = 1$ Gyr ($a = 0.150$) is $19.2$ Gyr, $42\%$ of $\eta_0$: in the early universe $a$ was small, so each second of cosmic time was many seconds of conformal time, and light crossed a great deal of comoving distance.

section 2

Curvature of an expanding space

Unit 33's machinery, run once from the metric to the Einstein tensor, then unit 35's equation.

39 · Friedmann

The Christoffel symbols of FLRW

worked example 1 · from the metric to the Einstein tensor, part 1

Write $g_{00} = -1$, $g_{ij} = a^2\gamma_{ij}$, $x^0 = ct$, so $\partial_0 = c^{-1}\partial_t$. Unit 32: $\Gamma^\lambda_{\mu\nu} = \tfrac12 g^{\lambda\sigma}(\partial_\mu g_{\sigma\nu} + \partial_\nu g_{\sigma\mu} - \partial_\sigma g_{\mu\nu})$.

$\lambda = 0$. $g^{00} = -1$ and $g_{0\nu}$ is constant, so only $-\partial_0 g_{\mu\nu}$ survives: $$\Gamma^0_{ij} = \tfrac12\partial_0(a^2\gamma_{ij}) = \frac{a\dot a}{c}\,\gamma_{ij},\quad \Gamma^0_{00} = \Gamma^0_{0i} = 0.$$

$\lambda = i$, one time index. $\Gamma^i_{0j} = \tfrac12 g^{ik}\partial_0 g_{kj} = \tfrac12 a^{-2}\gamma^{ik}\cdot\tfrac{2a\dot a}{c}\gamma_{kj}$: $$\Gamma^i_{0j} = \frac{\dot a}{ca}\,\delta^i_j,\qquad \Gamma^i_{00} = 0.$$

All spatial. The factor $a^2$ cancels between $g^{ik}$ and $\partial g$: $\Gamma^i_{jk} = {}^{(3)}\Gamma^i_{jk}$, the Christoffels of $\gamma$.

Check. $[\Gamma] = \mathrm m^{-1}$: $[\dot a/(ca)] = \mathrm s^{-1}/(\mathrm{m\,s^{-1}})$ ✓. For $a = $ const every time-bearing symbol vanishes and we are left with a static space ✓. Consequence used later: $\Gamma^\mu_{\mu 0} = 3\dot a/(ca)$.

The shortcut, and a cross-check. Unit 32 proved that geodesics are the stationary curves of $\int g_{\mu\nu}\dot x^\mu\dot x^\nu\,\mathrm d\lambda$, so the Christoffels can be read off the Euler-Lagrange equations of $L = -(\dot x^0)^2 + a(x^0)^2\gamma_{ij}\dot x^i\dot x^j$ (dots now $\mathrm d/\mathrm d\lambda$). The $x^0$ equation: $\frac{\mathrm d}{\mathrm d\lambda} (-2\dot x^0) - 2a\,\partial_0a\,\gamma_{ij}\dot x^i\dot x^j = 0$, i.e. $\ddot x^0 + a\,\partial_0a\, \gamma_{ij}\dot x^i\dot x^j = 0$. Comparing with $\ddot x^0 + \Gamma^0_{ij}\dot x^i\dot x^j = 0$ gives $\Gamma^0_{ij} = a\,\partial_0 a\,\gamma_{ij} = (a\dot a/c)\gamma_{ij}$ ✓. The $x^i$ equation: $\frac{\mathrm d}{\mathrm d\lambda}(2a^2\gamma_{ij}\dot x^j) - a^2\partial_i\gamma_{jk}\dot x^j\dot x^k = 0$; expanding, the term $4a\,\partial_0a\,\dot x^0\gamma_{ij}\dot x^j$ gives, after dividing by $2a^2$ and raising with $\gamma^{ki}$, $2(\partial_0 a/a)\dot x^0\dot x^k$, which is $2\Gamma^k_{0j} \dot x^0\dot x^j$ with $\Gamma^k_{0j} = (\partial_0 a/a)\delta^k_j$ ✓; the rest is the geodesic equation of $\gamma$ ✓.

For the $r$ form of $\gamma$, the spatial symbols are, for reference (problem P4 computes them): $^{(3)}\Gamma^r_{rr} = kr/(1-kr^2)$, $^{(3)}\Gamma^r_{\theta\theta} = -r(1-kr^2)$, $^{(3)}\Gamma^r_{\phi\phi} = -r(1-kr^2)\sin^2\theta$, $^{(3)}\Gamma^\theta_{r\theta} = {}^{(3)}\Gamma^\phi_{r\phi} = 1/r$, $^{(3)}\Gamma^\theta_{\phi\phi} = -\sin\theta\cos\theta$, $^{(3)}\Gamma^\phi_{\theta\phi} = \cot\theta$. The full four-dimensional list was also checked by machine against the formula; every entry agrees.

Why the physics is already visible. $\Gamma^i_{0j} = H\delta^i_j/c$ says that the spatial basis vectors are "stretched" at the rate $H$ as one moves forward in time; $\Gamma^0_{ij} \ne 0$ says that a freely moving particle with spatial momentum loses it (problem P2's last part and the redshift slide are this fact for massive and massless particles).

39 · Friedmann

Ricci, scalar curvature, Einstein tensor

worked example 1, continued · part 2

Unit 33: $R_{\sigma\nu} = \partial_\mu\Gamma^\mu_{\nu\sigma} - \partial_\nu\Gamma^\mu_{\mu\sigma} + \Gamma^\mu_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\mu_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}$.

$R_{00}$. $\Gamma^\mu_{00} = 0$ kills terms one and three. Term two: $-\partial_0(3\dot a/ca) = -\frac{3}{c^2}\bigl(\frac{\ddot a}{a} - \frac{\dot a^2}{a^2}\bigr)$. Term four: $-\Gamma^i_{0j}\Gamma^j_{i0} = -\frac{3\dot a^2}{c^2a^2}$. Sum: $$R_{00} = -\frac{3\ddot a}{c^2 a}.$$

$R_{ij}$ (notes): time terms give $(a\ddot a + 2\dot a^2)\gamma_{ij}/c^2$; spatial terms give $^{(3)}R_{ij} = 2k\gamma_{ij}$: $$R_{ij} = \frac{a\ddot a + 2\dot a^2 + 2kc^2}{c^2}\,\gamma_{ij},\quad R_{0i} = 0.$$

Scalar. $R = -R_{00} + a^{-2}\gamma^{ij}R_{ij}$: $$R = \frac{6}{c^2}\Bigl(\frac{\ddot a}{a} + \frac{\dot a^2}{a^2} + \frac{kc^2}{a^2}\Bigr).$$

Einstein, $G_{\mu\nu} = R_{\mu\nu} - \tfrac12Rg_{\mu\nu}$: $$G_{00} = \frac{3}{c^2}\Bigl(\frac{\dot a^2}{a^2} + \frac{kc^2}{a^2}\Bigr),\quad G_{ij} = -\frac{2a\ddot a + \dot a^2 + kc^2}{c^2}\gamma_{ij}.$$

Checks. $a = 1, k = 0$: $G = 0$, Minkowski ✓. $k \lt 0$, $a = c\sqrt{-k}\,t$: $\dot a^2 = -kc^2$, $\ddot a = 0$, so $G = 0$ — the empty Milne universe is flat spacetime in expanding coordinates ✓. $[G] = \mathrm m^{-2}$ ✓.

Full computation of $R_{ij}$. Split each sum into terms whose summed indices are time or space. Only $\Gamma^0_{ij}$, $\Gamma^i_{0j}$ and the spatial $^{(3)}\Gamma$ are nonzero.

$\partial_\mu\Gamma^\mu_{ji}$: $\mu = 0$ gives $\partial_0\Gamma^0_{ij} = \frac1c\partial_t \bigl(\frac{a\dot a}{c}\bigr)\gamma_{ij} = \frac{\dot a^2 + a\ddot a}{c^2}\gamma_{ij}$; $\mu = k$ gives the spatial term $\partial_k{}^{(3)}\Gamma^k_{ji}$.
$-\partial_j\Gamma^\mu_{\mu i}$: $\Gamma^0_{0i} = 0$, so this is purely spatial.
$\Gamma^\mu_{\mu\lambda}\Gamma^\lambda_{ji}$: $\lambda = 0$ gives $\Gamma^\mu_{\mu 0} \Gamma^0_{ij} = \frac{3\dot a}{ca}\cdot\frac{a\dot a}{c}\gamma_{ij} = \frac{3\dot a^2}{c^2} \gamma_{ij}$; $\lambda = k$ gives $\Gamma^\mu_{\mu k}\Gamma^k_{ji}$, and since $\Gamma^0_{0k} = 0$ this is spatial.
$-\Gamma^\mu_{j\lambda}\Gamma^\lambda_{\mu i}$: $(\mu,\lambda) = (0,k)$ gives $-\Gamma^0_{jk} \Gamma^k_{0i} = -\frac{a\dot a}{c}\gamma_{jk}\frac{\dot a}{ca}\delta^k_i = -\frac{\dot a^2}{c^2} \gamma_{ij}$; $(k, 0)$ gives $-\Gamma^k_{j0}\Gamma^0_{ki}$, the same; $(0,0)$ vanishes; $(k,l)$ is spatial.

Time terms: $(\dot a^2 + a\ddot a + 3\dot a^2 - 2\dot a^2)/c^2 = (a\ddot a + 2\dot a^2)/c^2$, times $\gamma_{ij}$. Spatial terms: exactly the Ricci tensor of $\gamma$, which is $R_{ij} = \gamma^{kl}R_{kilj} = k\gamma^{kl}(\gamma_{kl}\gamma_{ij} - \gamma_{kj}\gamma_{il}) = k(3 - 1) \gamma_{ij} = 2k\gamma_{ij}$ by the constant-curvature form. $R_{0i}$: every term needs an odd number of time indices on the Christoffels and there is no nonzero symbol with one time index and a free spatial index left over after contraction; directly, each term is either zero or proportional to $^{(3)}\Gamma^j_{ji} - {}^{(3)}\Gamma^j_{ij} = 0$. Isotropy says the same thing faster: $R_{0i}$ would be an invariant covector.

Scalar. $g^{00}R_{00} = +3\ddot a/(c^2a)$ and $g^{ij}R_{ij} = a^{-2}\gamma^{ij}\gamma_{ij} (a\ddot a + 2\dot a^2 + 2kc^2)/c^2 = 3(a\ddot a + 2\dot a^2 + 2kc^2)/(c^2a^2)$. Adding: $\frac{3}{c^2}\bigl(\frac{2\ddot a}{a} + \frac{2\dot a^2}{a^2} + \frac{2kc^2}{a^2}\bigr)$ ✓.

Einstein. $G_{00} = R_{00} + \tfrac12R = -\frac{3\ddot a}{c^2a} + \frac{3}{c^2}\bigl( \frac{\ddot a}{a} + \frac{\dot a^2 + kc^2}{a^2}\bigr)$, the $\ddot a$ cancelling. $G_{ij} = R_{ij} - \tfrac12Ra^2\gamma_{ij} = \bigl[(a\ddot a + 2\dot a^2 + 2kc^2) - 3(a\ddot a + \dot a^2 + kc^2) \bigr]\gamma_{ij}/c^2$ ✓.

The cancellation in $G_{00}$ is structural. $G_{00}$ contains no second time derivative. That is the general fact that $G^0{}_\mu$ is a constraint (unit 35's count $10 - 4 - 4$): the $00$ equation restricts initial data rather than evolving it. Here it becomes the first-order Friedmann equation.

Verification. All components were also computed symbolically from the metric in the $r$ form with the unit 33 sign convention; they agree term by term.

39 · Friedmann

The source: a perfect fluid at rest in the slices

Unit 35's perfect fluid, with unit 22's stress $p\,\delta_{ij}$ as its spatial part: $$T_{\mu\nu} = \Bigl(\rho + \frac{p}{c^2}\Bigr)u_\mu u_\nu + p\,g_{\mu\nu}.$$ The fluid moves with the fundamental observers: $u^\mu = (c, 0, 0, 0)$, $u_0 = -c$. Hence $$T_{00} = \rho c^2,\quad T_{0i} = 0,\quad T_{ij} = p\,a^2\gamma_{ij},$$ $$T^\mu{}_\nu = \mathrm{diag}(-\rho c^2,\ p,\ p,\ p).$$

the form is forced

Any symmetric $T$ respecting the symmetries of FLRW has this form, with $\rho, p$ functions of $t$ only.

Why: $T_{0i}$ would be an invariant covector, so $0$. $T_{ij}$ restricted to $T_p\Sigma$ is an $SO(3)$-invariant symmetric form, hence $\lambda\gamma_{ij}$. Homogeneity makes $\lambda$ and $T_{00}$ constant on slices. (P3.)

dimensions of $\rho$

$\rho$ is a mass density, $\mathrm{kg\,m^{-3}}$, and $\rho c^2$ is the energy density; $p$ is in $\mathrm{Pa} = \mathrm{J\,m^{-3}}$. Books with $c = 1$ write $T^\mu{}_\nu = \mathrm{diag}(-\rho, p, p, p)$.

The isotropy lemma, in full. Let $B$ be a symmetric bilinear form on a 3-dimensional inner-product space $(V, \gamma)$ with $B(Qv, Qw) = B(v,w)$ for all $Q \in SO(3)$. Then the quadratic form $q(v) = B(v,v)$ takes one value $\lambda$ on the unit sphere, because $SO(3)$ acts transitively on it (any unit vector is carried to any other by a rotation about their common normal). So $q(v) = \lambda\gamma(v,v)$ for all $v$, and polarisation, $B(v,w) = \tfrac12[q(v+w) - q(v) - q(w)]$, gives $B = \lambda\gamma$. For a covector $\omega$: pick $Q$ the rotation by $\pi$ about an axis orthogonal to $\omega^\sharp$; then $Q\omega^\sharp = -\omega^\sharp$ and invariance gives $\omega = -\omega$. $\square$

Connection with unit 22. Unit 22 flagged that the triple (energy density, momentum density, momentum flux) would become $T^{\mu\nu}$. In the comoving frame the momentum density $T^{0i}/c$ is zero — nothing flows, on average — and the momentum flux is Cauchy's stress, which is isotropic, so it is a pressure: unit 22's "pressure as the isotropic part", which here is the only part.

The index positions. $T^\mu{}_\nu = g^{\mu\lambda}T_{\lambda\nu}$: $T^0{}_0 = g^{00} T_{00} = -\rho c^2$ and $T^i{}_j = a^{-2}\gamma^{ik}pa^2\gamma_{kj} = p\,\delta^i_j$. The mixed form is the one to remember because it has no factors of $a$.

What the "fluid" is. Galaxies are the molecules. For them $p$ is utterly negligible (random velocities of a few hundred $\mathrm{km\,s^{-1}}$ give $p/\rho c^2 \sim 10^{-6}$), so they are dust. Radiation has $p = \rho c^2/3$, and a cosmological constant behaves as $p = -\rho c^2$; both are proved on the equation-of-state slide and in its notes. Several components add: $\rho = \sum\rho_i$, $p = \sum p_i$.

39 · Friedmann

The Friedmann equations

Unit 35's $G_{\mu\nu} + \Lambda g_{\mu\nu} = \kappa T_{\mu\nu}$, $\kappa = 8\pi G/c^4$. The $00$ component: $\frac{3}{c^2}\frac{\dot a^2 + kc^2}{a^2} - \Lambda = \frac{8\pi G}{c^2}\rho$. The $ij$ component, divided by $a^2\gamma_{ij}$: $-\frac{2a\ddot a + \dot a^2 + kc^2}{c^2a^2} + \Lambda = \frac{8\pi G}{c^4}p$. Rearranging the first, and subtracting it from the second:

the Friedmann equations $$H^2 = \frac{8\pi G}{3}\rho - \frac{kc^2}{a^2} + \frac{\Lambda c^2}{3}$$ $$\frac{\ddot a}{a} = -\frac{4\pi G}{3}\Bigl(\rho + \frac{3p}{c^2}\Bigr) + \frac{\Lambda c^2}{3}$$

The second is the acceleration equation.

dimensions

$[H^2] = \mathrm s^{-2}$. $[G\rho] = \mathrm{m^3kg^{-1}s^{-2}\cdot kg\,m^{-3}} = \mathrm s^{-2}$ ✓. $[kc^2/a^2] = \mathrm{m^{-2}\,m^2s^{-2}}$ ✓. $[\Lambda c^2]$ ✓. $[Gp/c^2] = [G\rho]$ since $[p] = [\rho c^2]$ ✓.

limits

Empty, $\Lambda = 0$: $H^2 = -kc^2/a^2$, the Milne universe ✓. $\Lambda$ only, $k = 0$: $H = c\sqrt{\Lambda/3}$ constant, $a \propto e^{Ht}$, de Sitter (1917) ✓. Static, $\rho = p = \Lambda = k = 0$: Minkowski ✓.

The rearrangement, written out. First equation times $c^2/3$: $\dot a^2/a^2 + kc^2/a^2 - \Lambda c^2/3 = 8\pi G\rho/3$, which is the first box. Second: multiply by $c^2$ to get $2\ddot a/a + \dot a^2/a^2 + kc^2/a^2 = \Lambda c^2 - 8\pi Gp/c^2$. Replace $\dot a^2/a^2 + kc^2/a^2$ using the first: $2\ddot a/a = \Lambda c^2 - 8\pi Gp/c^2 - 8\pi G\rho/3 - \Lambda c^2/3$, so $\ddot a/a = \Lambda c^2/3 - (4\pi G/3)(\rho + 3p/c^2)$ ✓.

Only two equations. Ten components, but $G_{0i} = T_{0i} = 0$ identically and the three diagonal $ij$ equations coincide. Symmetry has done the work that makes a PDE system into ODEs.

Pressure gravitates. The combination $\rho + 3p/c^2$ in the acceleration equation is the trace-reversed source of unit 35's weak-field computation, $T_{00} - \tfrac12 Tg_{00}$ per unit $c^2$, now with pressure kept. A positive pressure decelerates expansion even though it pushes outward, because in general relativity pressure is a source of gravity, not a force on the boundary — there is no boundary. Negative pressure below $-\rho c^2/3$ accelerates.

The running example. For the Planck universe, $k = 0$, $p = 0$ for the matter, and the first equation today reads $H_0^2 = 8\pi G\rho_{\mathrm m,0}/3 + \Lambda c^2/3$. Section 3 turns this into the density parameters.

History. Friedmann found both equations in 1922 (for $k > 0$) and 1924 ($k \lt 0$); Einstein first published a note claiming an error, then retracted it. Lemaître rederived them independently in 1927 and added the connection to galaxy redshifts.

39 · Friedmann

The fluid equation, twice

route 1 · $\nabla_\mu T^\mu{}_0 = 0$

Unit 32's divergence of a (1,1)-tensor: $\nabla_\mu T^\mu{}_\nu = \partial_\mu T^\mu{}_\nu + \Gamma^\mu_{\mu\lambda}T^\lambda{}_\nu - \Gamma^\lambda_{\mu\nu}T^\mu{}_\lambda$. For $\nu = 0$:

$\partial_0(-\rho c^2) = -c\dot\rho$; $\ \Gamma^\mu_{\mu0}T^0{}_0 = \frac{3\dot a}{ca}(-\rho c^2)$; $\ -\Gamma^i_{j0}T^j{}_i = -\frac{\dot a}{ca}\cdot3p$.

Setting the sum to zero and dividing by $-c$: $$\dot\rho + 3H\Bigl(\rho + \frac{p}{c^2}\Bigr) = 0.$$

route 2 · from the Friedmann pair

Differentiate $\dot a^2 = \frac{8\pi G}{3}\rho a^2 - kc^2 + \frac{\Lambda c^2}{3}a^2$: $\ 2\dot a\ddot a = \frac{8\pi G}{3}(\dot\rho a^2 + 2\rho a\dot a) + \frac{2\Lambda c^2}{3}a\dot a$. Substitute $\ddot a$ from the acceleration equation; the $\Lambda$ and $k$ terms cancel and what remains is the same equation (notes).

why twice is not a coincidence

Unit 33's contracted Bianchi identity makes $\nabla^\mu G_{\mu\nu} = 0$ identically, so the field equations imply $\nabla_\mu T^{\mu\nu} = 0$. It is the first law: $\mathrm d(\rho c^2 a^3) = -p\,\mathrm d(a^3)$.

Route 1 in full. $\partial_\mu T^\mu{}_0$: only $\mu = 0$, since $T^i{}_0 = 0$ and nothing depends on $x^i$; it is $\frac1c\partial_t(-\rho c^2) = -c\dot\rho$. Second term: $\lambda$ must be $0$ because $T^i{}_0 = 0$, and $\Gamma^\mu_{\mu0} = \Gamma^0_{00} + \Gamma^i_{i0} = 0 + 3\dot a/(ca)$. Third term: $\Gamma^\lambda_{\mu0}$ is nonzero only for $\lambda = i$, $\mu = j$, where it is $(\dot a/ca)\delta^i_j$; then $T^j{}_i = p\delta^j_i$ and the sum over $i, j$ gives $3p\dot a/(ca)$. Total: $-c\dot\rho - 3\dot a\rho c/a - 3p\dot a/(ca) = 0$; divide by $-c$ ✓. The $\nu = i$ equations are $\partial_i p = 0$, true by homogeneity.

Route 2 in full. The acceleration equation times $2a\dot a$: $2\dot a\ddot a = -\frac{8\pi G}{3}(\rho + 3p/c^2)a\dot a + \frac{2\Lambda c^2}{3}a\dot a$. Equate with the derivative of the first equation; the $\Lambda$ terms are identical and cancel; $k$ disappeared on differentiating. Divide by $\frac{8\pi G}{3}a$: $\dot\rho a + 2\rho\dot a = -(\rho + 3p/c^2)\dot a$, i.e. $\dot\rho a = -3(\rho + p/c^2)\dot a$ ✓. This needs $\dot a \ne 0$; at a turning point route 1 still holds.

The first law. $\frac{\mathrm d}{\mathrm dt}(\rho c^2a^3) = c^2a^3(\dot\rho + 3H\rho) = -3Hpa^3 = -p\frac{\mathrm d}{\mathrm dt}a^3$. A comoving volume $a^3$ of fluid does work $p\,\mathrm dV$ as it expands, and its energy falls by exactly that. For radiation this is why photons lose energy; for $\Lambda$, with $p = -\rho c^2$, the energy in a comoving volume grows as $a^3$, consistently with $\rho_\Lambda$ being constant. Where the energy "comes from" has no answer, because unit 35 showed there is no global energy conservation law in a spacetime without a timelike Killing field, and FLRW with $\dot a \ne 0$ has none.

Structure. So only two of the three equations (Friedmann, acceleration, fluid) are independent. In practice one uses the first Friedmann equation and the fluid equation, which are both first order.

39 · Friedmann

Equation of state, and how each component dilutes

equation of state

$p = w\rho c^2$ with $w$ constant. The fluid equation becomes $\dot\rho/\rho = -3(1+w)\dot a/a$, separable (unit 00): $$\rho = \rho_0\,a^{-3(1+w)}.$$

component	$w$	$\rho \propto$	why
matter (dust)	$0$	$a^{-3}$	number conserved, volume $\propto a^3$
radiation	$\tfrac13$	$a^{-4}$	number $a^{-3}$, each quantum redshifts $a^{-1}$
$\Lambda$	$-1$	$a^{0}$	a property of spacetime, not of stuff

$\Lambda$ as a fluid

Move $\Lambda g_{\mu\nu}$ to the right: $T^{(\Lambda)}_{\mu\nu} = -\frac{\Lambda c^4}{8\pi G}g_{\mu\nu}$. Compare with the perfect fluid: $\rho_\Lambda = \frac{\Lambda c^2}{8\pi G}$, $p_\Lambda = -\rho_\Lambda c^2$.

For the running example: $\Lambda = 1.091\times10^{-52}\ \mathrm m^{-2}$, $\rho_\Lambda = 5.85\times10^{-27}\ \mathrm{kg\,m^{-3}}$ — about three and a half hydrogen atoms per cubic metre, everywhere, always.

The three powers are drawn on the next slide.

Solving. $\int\mathrm d\rho/\rho = -3(1+w)\int\mathrm da/a$ gives $\ln\rho = -3(1+w)\ln a + $ const, and $a(t_0) = 1$ fixes the constant as $\ln\rho_0$. Limit: $w = 0$ recovers the dust result $\rho a^3 = $ const, which is also what unit 22's continuity equation gives for a fluid in uniform expansion $\vec v = H\vec x$: $\partial_t\rho + \rho\,\mathrm{div}\,\vec v = \dot\rho + 3H\rho = 0$ ✓.

Radiation's $w$, proved. Take a gas of quanta moving at $c$ in directions distributed isotropically, $n$ per unit volume, each with energy $\varepsilon$ and hence, by unit 27's $p\cdot p = 0$ for a massless particle, momentum of magnitude $\varepsilon/c$. The pressure is the flux of $x$-momentum across a plane $x = $ const (unit 22's momentum flux): a quantum moving at angle $\theta$ to the $x$-axis carries $x$-momentum $(\varepsilon/c)\cos\theta$ across at rate $c\cos\theta$, so $p = n\varepsilon\langle\cos^2\theta\rangle$. By isotropy $\langle\cos^2\rangle$ is the same for the three axes, and the three add to $1$, so it is $\tfrac13$: $p = \tfrac13 n\varepsilon = \tfrac13\rho c^2$. Equivalently the trace $-\rho c^2 + 3p$ vanishes; unit 35 cited from unit 29 that the electromagnetic $T_{\mu\nu}$ is trace-free, which is the same fact for a field. It holds for any gas of particles moving at $c$, including neutrinos while they were relativistic.

The $\Lambda$ numbers. $\Lambda = 3\Omega_\Lambda H_0^2/c^2 = 3\times0.685\times (2.184\times10^{-18})^2/(2.998\times10^8)^2 = 1.091\times10^{-52}\ \mathrm m^{-2}$, consistent with the $1.09\times10^{-52}$ of unit 35. $\rho_\Lambda = \Lambda c^2/8\pi G = 5.845\times10^{-27}\ \mathrm{kg\,m^{-3}}$; dividing by the proton mass, $1.673\times10^{-27}$ kg, gives $3.5$.

What $\Lambda$ is. Unit 35 said that geometry allows the term and does not fix its value. Whether it is a constant in the field equations or the energy of the vacuum (which Lorentz invariance forces to have $T_{\mu\nu} \propto g_{\mu\nu}$, hence $w = -1$ exactly) cannot be distinguished by any measurement of the expansion. The name "dark energy" includes the possibility that $w \ne -1$ or varies; current data are consistent with $w = -1$ at the level of several per cent, with some recent survey results suggesting mild evolution that is not yet settled.

39 · Friedmann

Three slopes, two crossings

On log–log axes $\rho = \rho_0a^{-3(1+w)}$ is a straight line of slope $-3(1+w)$: $-4$, $-3$, $0$.

Steeper lines win in the past. So, going back in time: $\Lambda$ dominates now and for ever after, matter before that, radiation earliest. The eras are set by the powers alone; the crossings by today's densities.

For the running example, where today's densities are the fractions $0.315$, $0.685$ and $9.2\times10^{-5}$ of the total, the crossings fall at $a = 2.9\times10^{-4}$ and $a = 0.77$.

Redshift $z$, used in the labels, is $1/a - 1$; section 5 explains the name.

Computed: each component's density in units of today's total, against $a$. The CMB was released at $a = 1/1091$, in the matter era but only $3.1$ times later in $a$ than the crossing with radiation.

How the figure was made. $\log_{10}(\rho_i/\rho_{\mathrm c,0}) = \log_{10}\Omega_i - 3(1+w_i)\log_{10}a$ with $\Omega_{\mathrm r} = 9.21\times10^{-5}$, $\Omega_{\mathrm m} = 0.315$, $\Omega_\Lambda = 0.685$; the $\Omega$s, and the critical density $\rho_{\mathrm c}$ they are fractions of, are defined properly in section 3 (Critical density and the density parameters). Radiation and matter cross where $\Omega_{\mathrm r}a^{-4} = \Omega_{\mathrm m}a^{-3}$, $a = 2.92\times10^{-4}$, $z = 3420$; matter and $\Lambda$ where $\Omega_{\mathrm m} a^{-3} = \Omega_\Lambda$, $a = 0.772$, $z = 0.296$. Section 4 dates both.

What the picture hides. The transitions are not sharp: near each crossing two components matter, and the solutions of section 4 interpolate between the pure power laws. On a log axis that blur spans less than a decade either side of each crossing.

section 3

Friedmann as an energy equation

Unit 08's effective potential, one dimension, and what $\Lambda$ does to it.

39 · energy

A Newtonian ball of dust

A uniform ball of dust expanding as $R(t) = a(t)\,r_c$. By unit 05's shell theorem, a particle on its surface feels only the mass inside, $M = \frac{4\pi}{3}\rho a^3r_c^3$, constant as it expands. Unit 00's energy integral: $$\tfrac12\dot R^2 - \frac{GM}{R} = E.$$ Divide by $r_c^2$: $$\tfrac12\dot a^2 - \frac{4\pi G}{3}\rho a^2 = \frac{E}{r_c^2}.$$

Schematic. Around any fundamental observer, a sphere of dust of radius $R = a\,r_c$; the shells outside pull equally every way and drop out, leaving the mass $M$ inside to decelerate the surface.

39 · energy

The ball and Friedmann: one equation

the same equation

Multiply the first Friedmann equation by $a^2/2$ (with $\Lambda = 0$): $\tfrac12\dot a^2 - \frac{4\pi G}{3}\rho a^2 = -\tfrac12 kc^2$. So $-kc^2/2$ plays the part of the energy per unit mass per comoving length squared: bound if $k > 0$, marginal if $k = 0$, unbound if $k \lt 0$.

motivation, not derivation

The Newtonian argument cannot produce the $3p/c^2$ in the acceleration equation, needs a centre and an edge that the universe lacks, and gets $k$'s geometric meaning as a coincidence. The derivation is the previous section.

Where it breaks. An infinite uniform medium has no well-defined Newtonian potential: the integral for $\Phi$ diverges, and the shell theorem's "outside exerts no force" relies on spherical symmetry about a centre that an infinite homogeneous medium does not single out. In general relativity the corresponding statement is Birkhoff's theorem, proved in unit 36, which makes the argument honest in retrospect; this unit does not need it. Second, pressure: a Newtonian pressure is uniform and so exerts no net force, whereas the relativistic equation has $\rho + 3p/c^2$ as the active density. For radiation that doubles the deceleration.

Dimensions of the energy. $E$ is energy per unit mass, $\mathrm{m^2s^{-2}}$; $E/r_c^2$ is $\mathrm s^{-2}$; and $kc^2$ is $\mathrm{m^{-2}\,m^2s^{-2}}$ ✓.

39 · energy

The effective potential

$V(a)/H_0^2 = -\Omega_{\mathrm m}/2a - \Omega_\Lambda a^2/2$ for the running example, $a$ from $0$ to $2$, where $\Omega_{\mathrm m} = 8\pi G\rho_{\mathrm m,0}/3H_0^2 = 0.315$ and $\Omega_\Lambda = \Lambda c^2/3H_0^2 = 0.685$ (named on Critical density and the density parameters). Dashed grey: matter alone.

With $\rho_{\mathrm m} = \rho_{\mathrm m,0}a^{-3}$ and $\Lambda$, the first Friedmann equation times $a^2/2$ is $$\tfrac12\dot a^2 + V(a) = -\tfrac12kc^2,$$ $$V(a) = -\frac{4\pi G\rho_{\mathrm m,0}}{3a} - \frac{\Lambda c^2}{6}a^2.$$

A particle of unit mass in one dimension: unit 08's picture with $a$ for $r$. $\Lambda$ adds an inverted harmonic well — a repulsion growing with distance.

The acceleration equation is $\ddot a = -V'(a)$ ✓.

The potential, derived. $H^2 = \frac{8\pi G}{3}\rho_{\mathrm m,0}a^{-3} - kc^2/a^2 + \Lambda c^2/3$; multiply by $a^2/2$: $\tfrac12\dot a^2 - \frac{4\pi G\rho_{\mathrm m,0}}{3a} - \frac{\Lambda c^2}{6}a^2 = -\tfrac12kc^2$. Dividing by $H_0^2$ and using the density parameters of Critical density and the density parameters, $V/H_0^2 = -\Omega_{\mathrm m}/(2a) - \Omega_\Lambda a^2/2$ and the energy is $\Omega_k/2$ with $\Omega_k = -kc^2/H_0^2$.

$\ddot a = -V'$. $-V'(a) = -\frac{4\pi G\rho_{\mathrm m,0}}{3a^2} + \frac{\Lambda c^2}{3}a$, and the acceleration equation with $p = 0$ gives $\ddot a = -\frac{4\pi G}{3}\rho_{\mathrm m,0} a^{-2} + \frac{\Lambda c^2}{3}a$ ✓. The first Friedmann equation is the first integral of the second — unit 00's "multiply by $\dot x$ and integrate", with $-kc^2/2$ as the constant of integration. That is the cleanest way to see why $k$ is constant: it is an energy.

The figure's numbers. $V$ has its maximum where $V' = 0$: $\Omega_{\mathrm m}/(2a^2) = \Omega_\Lambda a$, so $a_{\max} = (\Omega_{\mathrm m}/2\Omega_\Lambda)^{1/3} = 0.613$, i.e. $z = 0.632$, and $V_{\max}/H_0^2 = -3\Omega_{\mathrm m}/(4a_{\max}) = -0.386$. Today $V(1)/H_0^2 = -\tfrac12$ and $\tfrac12\dot a^2/H_0^2 = \tfrac12$, summing to $0$ ✓ (flat). The ball passed over the top at $a = 0.61$; since then $V' \lt 0$ and it has been accelerating.

Comparison with unit 08. There, $V_{\mathrm{eff}} = V(r) + L^2/2\mu r^2$ and the centrifugal term was a repulsive barrier at small $r$. Here the repulsive term is at large $a$, and its shape — $-a^2$ — is that of an inverted harmonic oscillator, unit 02's saddle: the equilibrium at the top is unstable, and motion away from it is exponential.

39 · energy

What the potential says

$\Lambda = 0$. $V = -A/a$ rises monotonically to $0$. Energy $\ge 0$ ($k \le 0$): expand for ever. Energy $\lt 0$ ($k > 0$): reach a turning point and recollapse. Unit 08's unbound and bound orbits, exactly.
$\Lambda > 0$. $V$ has a single maximum. Energy above it: expand for ever, eventually exponentially. Below it: either recollapse, or (starting large) a bounce with no Big Bang.
At the top. A universe balanced there never moves.

The potential of the running example, read as in unit 00: a horizontal line at the energy $-kc^2/2$ (here in units of $H_0^2$), and motion only where it lies above $V$. Arrows show which way $a$ can go; dots on the lower line are turning points.

Reading the figure. $\tfrac12\dot a^2 = E - V(a) \ge 0$, so a universe of energy $E$ can only be where the horizontal line at $E$ lies above the curve, and $\dot a = 0$ exactly where they meet. The curve is the running example's $V/H_0^2 = -\Omega_{\mathrm m}/2a - \Omega_\Lambda a^2/2$, computed. The line at $0$ is the flat case, the running example itself. The lower line, at $-0.6$, is a closed universe with the same matter and $\Lambda$; it has two disjoint allowed regions, and which one a universe is in depends on where it started, not on its energy.

Bouncing and loitering. Energies slightly above the top give a long "loitering" phase near the top (Lemaître's 1927–31 hesitating universe). Bouncing solutions with no Big Bang require $\Omega_\Lambda$ so large relative to $\Omega_{\mathrm m}$ that the observed redshifts of the most distant objects would be impossible; they are excluded.

39 · energy

The Einstein static universe

Einstein static universe

$\dot a = \ddot a = 0$ with dust. The acceleration equation gives $\Lambda = 4\pi G\rho/c^2$; the first then gives $k/a^2 = 4\pi G\rho/c^2 = \Lambda$: closed, with curvature radius $1/\sqrt\Lambda$.

Einstein (1917) introduced $\Lambda$ to make this possible. It sits at a maximum of $V$, so it is unstable: a perturbation grows as $e^{c\sqrt\Lambda\,t}$ (P7). With today's $\Lambda$ the e-folding time is $10.1$ Gyr.

The sign of $k$ is decided by whether the energy exceeds $0$, not whether it exceeds the top: with $\Lambda > 0$, a closed universe can expand for ever.

That is the previous figure again: any line between the top and $0$ is closed ($k > 0$) and still clears the top.

The Einstein static numbers. With $\ddot a = 0$ and $p = 0$: $\frac{4\pi G}{3}\rho = \frac{\Lambda c^2}{3}$, so $\rho = \Lambda c^2/4\pi G$. With $\dot a = 0$: $0 = \frac{8\pi G}{3}\rho - kc^2/a^2 + \frac{\Lambda c^2}{3} = \Lambda c^2 - kc^2/a^2$, so $k/a^2 = \Lambda$. With $\Lambda = 1.091\times10^{-52}\ \mathrm m^{-2}$: $\rho = 1.17\times10^{-26}\ \mathrm{kg\,m^{-3}}$, radius $1/\sqrt\Lambda = 9.57\times10^{25}$ m $= 10.1$ Gly, and $1/(c\sqrt\Lambda) = 10.1$ Gyr.

The instability, sketched (P7 does it in full). Perturb $a = a_E + \delta$ keeping the dust's $\rho a^3$ fixed. Linearising $\ddot a = -V'(a)$ gives $\ddot\delta = -V''(a_E)\delta$ and $-V''(a_E) = \Lambda c^2/3 + 8\pi G\rho_E/3 = \Lambda c^2$, positive: exponential growth at rate $c\sqrt\Lambda$. Eddington pointed this out in 1930, which is part of why Einstein abandoned the model once Hubble's redshifts appeared.

"Biggest blunder". The remark is reported second-hand by Gamow and its authenticity is doubted by historians. What is documented is that Einstein wrote to Weyl in 1923 that "if there is no quasi-static world, then away with the cosmological term", and dropped it after 1931.

39 · energy

Critical density and the density parameters

critical density

The density that makes $k = 0$ with $\Lambda$ counted as a component: $\rho_{\mathrm c} = 3H^2/8\pi G$.

density parameters

$\Omega_i = \rho_{i,0}/\rho_{\mathrm c,0}$ for each component, $\Omega_\Lambda = \Lambda c^2/3H_0^2$, $\Omega_k = -kc^2/H_0^2$. The first Friedmann equation becomes $$\frac{H(a)^2}{H_0^2} = \Omega_{\mathrm r}a^{-4} + \Omega_{\mathrm m}a^{-3} + \Omega_k a^{-2} + \Omega_\Lambda.$$ At $a = 1$: $\ \Omega_{\mathrm r} + \Omega_{\mathrm m} + \Omega_k + \Omega_\Lambda = 1$.

for the running example

$\rho_{\mathrm c,0} = \dfrac{3(2.184\times10^{-18})^2}{8\pi\times6.674\times10^{-11}} = 8.53\times10^{-27}\ \mathrm{kg\,m^{-3}}$ — five protons per cubic metre.

$\Omega_{\mathrm m} + \Omega_\Lambda = 1$, so $\Omega_k = 0$. Measured: $|\Omega_k| \lesssim 0.002$.

The sum rule. Divide $H^2 = \frac{8\pi G}{3}\sum\rho_i - kc^2/a^2 + \Lambda c^2/3$ by $H_0^2$ and write $\rho_i = \rho_{i,0}a^{-3(1+w_i)}$. Each density term becomes $\Omega_i a^{-3(1+w_i)}$ because $8\pi G\rho_{i,0}/3H_0^2 = \rho_{i,0}/\rho_{\mathrm c,0}$. At $a = 1$ the left side is $1$. So measuring the densities measures the curvature: the sign of $k$ is the sign of $\sum\Omega_i - 1$ (with $\Omega_\Lambda$ included).

Dimensions. $[3H^2/8\pi G] = \mathrm{s^{-2}}/(\mathrm{m^3kg^{-1}s^{-2}}) = \mathrm{kg\,m^{-3}}$ ✓. Each $\Omega$ is dimensionless ✓.

The numbers. $\rho_{\mathrm c,0} = 8.533\times10^{-27}\ \mathrm{kg\,m^{-3}}$ with $G = 6.6743\times10^{-11}$; divided by $m_p = 1.6726\times10^{-27}$ kg, $5.10$ protons per $\mathrm m^3$. Of the $\Omega_{\mathrm m} = 0.315$, only $0.049$ is ordinary (baryonic) matter; the rest is dark matter, inferred gravitationally (galaxy rotation curves, cluster dynamics, lensing, and the CMB peaks). In this unit it is all dust and the distinction does not enter.

The curvature bound. Planck 2018 combined with baryon acoustic oscillations gives $\Omega_k = 0.0007 \pm 0.0019$; "$|\Omega_k| \lesssim 0.002$" is that, rounded. Setting $k = 0$ in the running example is therefore a choice consistent with data, not a derivation.

39 · energy

Matter against Λ: the plane of models

deceleration parameter

$q = -\ddot a a/\dot a^2$. From the acceleration equation, $q_0 = \tfrac12\Omega_{\mathrm m} + \Omega_{\mathrm r} - \Omega_\Lambda$; for the running example $q_0 = -0.53$: accelerating.

With radiation neglected, a dust-plus-$\Lambda$ universe today is one point $(\Omega_{\mathrm m}, \Omega_\Lambda)$. Two lines cut the plane: the sum rule's $\Omega_{\mathrm m} + \Omega_\Lambda = 1$ (curvature), and $q_0 = 0$, i.e. $\Omega_\Lambda = \tfrac12\Omega_{\mathrm m}$ (acceleration).

They are different questions: a closed universe can accelerate, and a flat one can decelerate.

Every dust-plus-$\Lambda$ model today. Above the solid line space is closed, below it open; above the dashed line the expansion accelerates. The Planck universe sits on the first line and well above the second.

The deceleration parameter. $-\ddot a/(aH^2) = \frac{4\pi G}{3H^2}\sum(\rho_i + 3p_i/c^2) - \frac{\Lambda c^2}{3H^2}$; at $t_0$ this is $\tfrac12\sum\Omega_i(1 + 3w_i)$ over matter and radiation, minus $\Omega_\Lambda$. With $\Omega_{\mathrm r}$ neglected: $0.1575 - 0.685 = -0.5275$. The discovery that $q_0 \lt 0$ from Type Ia supernovae (Riess et al. 1998, Perlmutter et al. 1999) earned the 2011 Nobel Prize.

The four marked points. Einstein-de Sitter $(1, 0)$, flat and decelerating with $q_0 = \tfrac12$; de Sitter $(0, 1)$, flat with $q_0 = -1$; Milne $(0, 0)$, empty, open, coasting with $q_0 = 0$ (the empty universe of the Einstein-tensor checks); and the running example $(0.315, 0.685)$. Section 4 solves the first two and the running example exactly.

What the plane leaves out. Whether a given point had a Big Bang or bounces, and whether it recollapses, are further curves in this plane, found from the potential of the previous section: they need $V$'s top compared with the energy, and are not drawn. Supernova data alone constrain a band, $0.8\,\Omega_{\mathrm m} - 0.6\,\Omega_\Lambda \approx -0.2 \pm 0.1$ (cited: Perlmutter et al. 1999), which the Planck point satisfies ($-0.16$); the CMB constrains $\Omega_{\mathrm m} + \Omega_\Lambda$; they cross near the Planck point.

section 4

Solving it

Three pure cases in closed form, then the running example's age.

39 · solutions

Three pure universes

worked example 2 · one component, flat

General $w > -1$. $H^2 = H_0^2a^{-3(1+w)}$, so $\dot a = H_0a^{1 - \frac32(1+w)}$. Separate (unit 00): $a^{\frac32(1+w)-1}\,\mathrm da = H_0\,\mathrm dt$. Integrate from $a = 0$ at $t = 0$: $$\frac{a^{\frac32(1+w)}}{\tfrac32(1+w)} = H_0t \ \Longrightarrow\ a = \Bigl(\tfrac32(1+w)H_0t\Bigr)^{\frac{2}{3(1+w)}}.$$ Today $a = 1$: $\ t_0 = \dfrac{2}{3(1+w)H_0}$.

universe	$a(t)$	$t_0$	Gyr
matter, $w = 0$ Einstein-de Sitter	$\bigl(\tfrac32H_0t\bigr)^{2/3}$	$\frac{2}{3H_0}$	9.67
radiation, $w = \tfrac13$	$(2H_0t)^{1/2}$	$\frac{1}{2H_0}$	7.25
$\Lambda$, $w = -1$ de Sitter universe	$e^{H_0(t - t_0)}$	none	$\infty$

Numbers with $H_0 = 67.4$, so $1/H_0 = 14.51$ Gyr.

Checks. $H_0t$ is dimensionless ✓. $H = \dot a/a = \frac{2}{3(1+w)t}$, which equals $H_0$ at $t_0$ ✓. As $w \to -1$, $t_0 \to \infty$ and the power law steepens without bound, consistent with the exponential ✓. Every $w > -\tfrac13$ decelerates: $\ddot a \lt 0$ since the exponent $\frac{2}{3(1+w)} \lt 1$ ✓.

The $w = -1$ case separately. $H = H_0$ constant, so $\dot a = H_0a$, unit 00's linear ODE, $a = e^{H_0(t - t_0)}$. It has no $a = 0$ at any finite time: de Sitter space in these coordinates has no beginning. (In fact these coordinates cover only half of the full de Sitter hyperboloid, which is geodesically complete; cited, Hawking and Ellis §5.2.)

Why the age is less than the Hubble time for decelerating universes. If $\ddot a \lt 0$ always, the graph of $a(t)$ lies below its tangent line at $t_0$, which reaches $a = 0$ at $t_0 - 1/H_0$. So the Big Bang is later than that: $t_0 \lt 1/H_0$. Matter gives $\tfrac23$ of the Hubble time, radiation $\tfrac12$. Acceleration at late times is what lets the running example reach $0.95$ of it.

The historical age problem. In the 1990s, with $H_0 \approx 70$ and a matter-only universe, $t_0 = 2/(3H_0) = 9.3$ Gyr, younger than the oldest globular-cluster stars (then dated at 12–16 Gyr). The universe could not be younger than its stars; $\Lambda$ resolved it.

Mixed and curved cases. With more than one component the integral $t = \int\mathrm da/ (aH(a))$ rarely has a closed form. Two that do: flat matter plus $\Lambda$ (the age slide), and matter with $k > 0$, whose solution is a cycloid, $a \propto 1 - \cos\theta$, $t \propto \theta - \sin\theta$, with $\theta$ proportional to conformal time (P8). Matter plus radiation, flat, also has a closed form in conformal time (P5).

39 · solutions

Four histories through today

Each curve passes through $a = 1$ with slope $H_0$ today. Where it meets $a = 0$ is its Big Bang; the axis labels there are the three finite ages, in Hubble times.

Every curve has the same $a$ and $\dot a$ today; they differ in $\ddot a$, and so in their pasts.

The Planck universe (thick) followed the radiation curve's shape at first, the matter curve's in the middle, and is turning towards de Sitter now.

The eras, from $\Omega_i a^{-3(1+w_i)}$:

matter–radiation equality: $a = \Omega_{\mathrm r}/\Omega_{\mathrm m}$, $z = 3420$, $t = 50\,000$ yr;
matter–Λ equality: $z = 0.30$ (dated on We live near matter–Λ equality).

How the figure was made. The horizontal axis is $H_0(t - t_0)$ and the vertical $a$ from $0$ to $2.2$. Matter: $a = (\tfrac32 s)^{2/3}$ with $s = H_0t$, shifted by $-\tfrac23$. Radiation: $a = \sqrt{2s}$, shifted by $-\tfrac12$. De Sitter: $a = e^{s}$. Planck: $t(a)$ from the closed form on the next slide, shifted by $-0.951$. All four were computed, not sketched.

Radiation's density parameter. Photons at $T_0 = 2.7255$ K have $\Omega_\gamma h^2 = 2.47\times10^{-5}$ (Stefan-Boltzmann, cited: this course has no statistical mechanics). Three neutrino species add a factor $1 + 0.2271\times3.046 = 1.692$ (cited). With $h = 0.674$, $\Omega_{\mathrm r} = 9.21\times10^{-5}$.

Matter–radiation equality. $\Omega_{\mathrm r}a^{-4} = \Omega_{\mathrm m}a^{-3}$ gives $a_{\mathrm{eq}} = \Omega_{\mathrm r}/\Omega_{\mathrm m} = 2.92\times10^{-4}$, $z_{\mathrm{eq}} = 3420$, and integrating $\mathrm dt = \mathrm da/(aH)$ with all three components gives $t_{\mathrm {eq}} = 5.05\times10^4$ yr. Recombination, at $z = 1090$, is after equality — so the CMB was released in the matter era, but only just ($a_*/a_{\mathrm{eq}} = 3.1$), and radiation still matters there. The horizon problem will need that.

39 · solutions

The age of the Planck universe

worked example 3 · the age integral, in closed form

$\mathrm dt = \mathrm da/\dot a = \mathrm da/(aH)$, and for the running example $H = H_0\sqrt{\Omega_{\mathrm m}a^{-3} + \Omega_\Lambda}$: $$t_0 = \frac1{H_0}\int_0^1\frac{\mathrm da}{a\sqrt{\Omega_{\mathrm m}a^{-3} + \Omega_\Lambda}} = \frac1{H_0}\int_0^1\frac{\sqrt a\,\mathrm da}{\sqrt{\Omega_{\mathrm m} + \Omega_\Lambda a^3}}.$$ Substitute $u = a^{3/2}$, $\mathrm du = \tfrac32\sqrt a\,\mathrm da$: $$t_0 = \frac{2}{3H_0}\int_0^1\frac{\mathrm du}{\sqrt{\Omega_{\mathrm m} + \Omega_\Lambda u^2}} = \frac{2}{3H_0\sqrt{\Omega_\Lambda}}\operatorname{arsinh}\sqrt{\frac{\Omega_\Lambda}{\Omega_{\mathrm m}}}.$$

Numbers. $\sqrt{0.685/0.315} = 1.4747$; $\operatorname{arsinh}1.4747 = \ln(1.4747 + \sqrt{1 + 1.4747^2}) = 1.1806$. $\ \tfrac{2}{3\sqrt{0.685}} = 0.8055$. Product: $0.9510$ Hubble times.

$$t_0 = 0.9510 \times 14.51\ \text{Gyr} = \mathbf{13.80\ Gyr}.$$

Adding radiation changes it to $13.79$ Gyr; Planck's own fit quotes $13.787 \pm 0.020$.

Checks. $\Omega_\Lambda \to 0$: $\operatorname{arsinh}x \approx x$, giving $\frac{2}{3H_0\sqrt{\Omega_{\mathrm m}}} \to \frac{2}{3H_0}$ at $\Omega_{\mathrm m} = 1$, Einstein-de Sitter ✓. $[t_0] = [1/H_0] = $ s ✓.

The integral. $\int\mathrm du/\sqrt{\alpha^2 + \beta^2u^2} = \beta^{-1}\operatorname{arsinh} (\beta u/\alpha)$, by differentiating $\operatorname{arsinh}x = \ln(x + \sqrt{1 + x^2})$, whose derivative is $1/\sqrt{1+x^2}$. With $\alpha = \sqrt{\Omega_{\mathrm m}}$, $\beta = \sqrt{\Omega_\Lambda}$, evaluated from $0$ to $1$ ✓.

The whole history, for free. Integrating from $0$ to $a$ instead of $0$ to $1$ and inverting, $$a(t) = \Bigl(\frac{\Omega_{\mathrm m}}{\Omega_\Lambda}\Bigr)^{1/3}\sinh^{2/3}\Bigl(\tfrac32 \sqrt{\Omega_\Lambda}\,H_0t\Bigr).$$ Early, $\sinh x \approx x$ and $a \approx (\tfrac32 \sqrt{\Omega_{\mathrm m}}H_0t)^{2/3}$, matter-like ✓; late, $\sinh x \approx \tfrac12e^x$ and $a \propto e^{\sqrt{\Omega_\Lambda}H_0t}$, de Sitter with $H_\infty = \sqrt{\Omega_\Lambda}H_0$ ✓.

Verification. The closed form was checked against direct numerical quadrature of the first integral ($13.7962$ Gyr both ways), and the radiation correction was computed by quadrature with $\Omega_{\mathrm r} = 9.21\times10^{-5}$ and $\Omega_\Lambda = 1 - \Omega_{\mathrm m} - \Omega_{\mathrm r}$: $13.7907$ Gyr. Radiation dominates only the first $50\,000$ years, which is why it moves the age by only $5.5$ Myr.

Sensitivity. $t_0 \propto 1/H_0$ at fixed $\Omega$s, so a one per cent error in $H_0$ is a one per cent error in the age. At fixed $H_0$, raising $\Omega_{\mathrm m}$ from $0.308$ to $0.322$ (Planck's $\pm1\sigma$) lowers $t_0$ by $0.17$ Gyr. The checkpoint's tolerance, $0.3$ Gyr, is chosen to be larger than any rounding a correct computation makes, and smaller than the error from forgetting $\Lambda$ ($4.1$ Gyr).

39 · solutions

We live near matter–Λ equality

worked example 3, continued · two redshifts

Equality: $\Omega_{\mathrm m}a^{-3} = \Omega_\Lambda$, so $$a_{\mathrm{m}\Lambda} = \Bigl(\frac{\Omega_{\mathrm m}}{\Omega_\Lambda}\Bigr)^{1/3} = 0.772,\quad z = 0.296.$$

Acceleration begins: $\ddot a = 0$ when $\tfrac12\Omega_{\mathrm m}a^{-3} = \Omega_\Lambda$, so $$a = \Bigl(\frac{\Omega_{\mathrm m}}{2\Omega_\Lambda}\Bigr)^{1/3} = 0.613,\quad z = 0.632.$$

When, from $t(a)$ in closed form: equality at $10.30$ Gyr, $3.50$ Gyr ago; acceleration from $7.69$ Gyr, $6.10$ Gyr ago — the top of the potential.

the coincidence

$\rho_{\mathrm m}/\rho_\Lambda \propto a^{-3}$ changes by a factor of $10^{9}$ between $z = 1090$ and $z = 0$ alone, and by $10^{29}$ since nucleosynthesis. We happen to observe about a quarter of the universe's age after the moment it equalled $1$. Whether this needs explaining is disputed; it is not explained.

Dimensions: $a$ and $z$ are ratios ✓. Limit: $\Omega_\Lambda \to 0$ sends both redshifts to $-1$ — equality and acceleration recede into the infinite future, $a \to \infty$ ✓.

Why acceleration starts at half the density. The acceleration equation with $p = 0$ for matter and $\rho_\Lambda$, $p_\Lambda = -\rho_\Lambda c^2$: $\ddot a/a = -\frac{4\pi G}{3}(\rho_{\mathrm m} + \rho_\Lambda - 3\rho_\Lambda) = -\frac{4\pi G}{3}(\rho_{\mathrm m} - 2\rho_\Lambda)$. Zero when $\rho_{\mathrm m} = 2\rho_\Lambda$, i.e. $\Omega_{\mathrm m}a^{-3} = 2\Omega_\Lambda$. This is the maximum of $V$ found on the effective-potential slide ✓, since $\ddot a = -V'$.

The times. $t(a) = \frac{2}{3H_0\sqrt{\Omega_\Lambda}}\operatorname{arsinh}\bigl( \sqrt{\Omega_\Lambda/\Omega_{\mathrm m}}\,a^{3/2}\bigr)$. At $a = 0.772$: $a^{3/2} = 0.678$, argument $1.000$ (it must be exactly $1$: at equality $\Omega_\Lambda a^3 = \Omega_{\mathrm m}$), $\operatorname{arsinh}1 = 0.8814$, and $t = 0.8055\times0.8814\times14.51 = 10.30$ Gyr. At $a = 0.613$ the argument is $1/\sqrt2$, $\operatorname{arsinh}(0.7071) = 0.6585$, $t = 7.69$ Gyr.

The ratios. $(1091)^3 = 1.30\times10^9$ from recombination. From nucleosynthesis at $a \approx 2.35\times10^{-10}$ ($kT = 1$ MeV), $a^{-3} = 7.7\times10^{28}$, so "$10^{29}$" is the order of magnitude; a quarter: $3.50/13.80 = 0.254$.

The anthropic argument, stated without endorsement. Observers need galaxies, galaxies need matter domination long enough to collapse, and a much larger $\Lambda$ would have prevented it (Weinberg 1987). Whether this explains the value or merely constrains it is not a question the mathematics in this unit can settle.

section 5

What light tells us

Null geodesics in FLRW: redshift, horizons, and the measurement of $H_0$.

39 · light

Cosmological redshift

cosmological redshift

Light emitted by a comoving source at $t_e$ and received by a comoving observer at $t_0$ has $$1 + z \equiv \frac{\lambda_0}{\lambda_e} = \frac{a(t_0)}{a(t_e)} = \frac{1}{a(t_e)}.$$

proof

Put the observer at $\chi = 0$. A radial null geodesic has $\mathrm ds^2 = 0$: $c\,\mathrm dt = a\,\mathrm d\chi$. A crest leaving $\chi_e$ at $t_e$ arrives at $t_0$, the next leaves at $t_e + \delta t_e$ and arrives at $t_0 + \delta t_0$: $$\int_{t_e}^{t_0}\frac{c\,\mathrm dt}{a} = \chi_e = \int_{t_e+\delta t_e}^{t_0+\delta t_0}\frac{c\,\mathrm dt}{a}.$$ Subtract: $\delta t_0/a(t_0) = \delta t_e/a(t_e)$ to first order. Periods scale as $a$, and so do wavelengths. $\square$

checks

$a$ constant: $z = 0$ ✓. Dimensionless ✓. Small distances: $a(t_e) \approx 1 - H_0(t_0 - t_e)$, so $z \approx H_0 d/c$ — Hubble's 1929 law, $v = cz = H_0 d$ ✓.

The CMB: $z = 1090$, so it left at $T = 2.7255 \times 1091 = 2974$ K, when hydrogen recombined, $3.7\times10^5$ yr after the Big Bang.

Why a radial null curve is a geodesic. Unit 32's geodesic equation for $\theta$ is $\ddot\theta + \Gamma^\theta_{\mu\nu}\dot x^\mu\dot x^\nu = 0$. On a curve with $\theta, \phi$ constant only $\dot x^0, \dot x^r$ are nonzero, and from worked example 1 every $\Gamma^\theta_{\mu \nu}$ and $\Gamma^\phi_{\mu\nu}$ with $\mu, \nu \in \{0, r\}$ vanishes. So $\theta = $ const, $\phi = $ const is consistent, and the remaining two equations are solved by $\mathrm ds^2 = 0$ with a suitable affine parameter (unit 32's null case). By homogeneity any comoving observer can be put at $\chi = 0$.

First order. $\int_{t_0}^{t_0+\delta t_0}c\,\mathrm dt/a - \int_{t_e}^{t_e+\delta t_e} c\,\mathrm dt/a = 0$ after cancelling the common range, and each integral is its integrand times its length plus $O(\delta t^2)$. For light, $\delta t \sim 10^{-15}$ s and $a$ changes on a scale of $10^{17}$ s, so the approximation is exact to $30$ figures.

Temperatures. A black-body spectrum stays black-body under uniform stretching of all wavelengths, with $T \propto 1/a$ (cited: the Planck distribution is a function of $\lambda T$ only). This is why the CMB is still a perfect black body, and why $T(z) = T_0(1+z)$. The release time $3.7\times10^5$ yr was computed by integrating $\mathrm dt = \mathrm da/(aH)$ with radiation included, to $a = 1/1091$; the commonly quoted "380 000 years" differs only in rounding and in the precise definition of the moment.

Massive particles too. The same $\Gamma^0_{ij}$ that stretches wavelengths makes the momentum of a free massive particle fall as $1/a$ (P2's last part): "peculiar" velocities decay, which is why galaxies are, to good approximation, comoving.

39 · light

Redshift, drawn

Two crests from a galaxy at $z = 1$, computed for the running example, the gap between them exaggerated. Left: in conformal time both travel on $45°$ lines, so the gap $\delta\eta$ is the same at both ends. Right: $\delta t = a\,\delta\eta$, and $a$ doubled while the light was in flight, so the period — and the wavelength — doubled.

Conformal time says the proof at once: both crests travel on $45°$ lines in $(\eta, \chi)$, so $\delta\eta$ is the same at both ends, and $\delta t = a\,\delta\eta$ gives $\delta t_0/\delta t_e = a(t_0)/a(t_e) = 1 + z$.

How the figure was made. The source is the comoving galaxy whose light, received today, left at $a = \tfrac12$: $z = 1$, comoving distance $11.1$ Gly, emitted $5.8$ Gyr after the Big Bang (radiation included). The two crests leave $\delta\eta = 1.2$ Gyr apart, absurdly long for light but large enough to see. In the right panel each crest is $\chi = \chi_e - c(\eta - \eta_e)$ converted to cosmic time through the computed $t(\eta)$. The gap goes from $0.62$ Gyr at emission to $1.25$ Gyr at reception, a ratio of $2.01$; the $0.01$ is the price of the exaggeration, since $a$ changes slightly within one gap, and it vanishes as $\delta\eta \to 0$.

Why the left panel proves it. Two parallel lines at $45°$ are the same vertical distance apart everywhere, which is all the redshift proof's integral identity says. The right panel is the same picture in the coordinate clocks read: the bending of the rays is not a new effect, only the stretching $\mathrm dt = a\,\mathrm d\eta$.

39 · light

Why it is not a Doppler shift

Unit 26's Doppler shift compares two four-velocities at one event. Source and observer here are at different events in curved spacetime, and unit 32 says vectors at different points cannot be compared without choosing a path. There is no "relative velocity" for a Doppler formula to take.

The redshift depends only on $a(t_0)/a(t_e)$: how much space stretched while the light was in flight — not on how fast anything was moving at emission or reception.

in numbers, for the running example

A galaxy at $z = 1$. Unit 26's formula would say $\beta = \frac{(1+z)^2 - 1}{(1+z)^2 + 1} = 0.600$. Its proper distance grows now at $H_0d = 0.765c$, and grew at $0.684c$ when the light left. Three numbers, none equal.

The CMB. Doppler would need $\beta = 1 - 1.7\times10^{-6}$. The last-scattering surface's proper distance grows today at $3.1c$.

but locally, yes

Over a short step between neighbouring comoving observers the shift is a Doppler shift, to first order; the total is a product of these along the ray. The global shift is not a single Doppler shift.

The numbers. For $z = 1$ the comoving distance is $\chi = (c/H_0)\int_{0.5}^1\mathrm da /(a^2E(a))$ with $E = \sqrt{\Omega_{\mathrm m}a^{-3} + \Omega_\Lambda}$; numerically $0.7647\,c/H_0 = 3401$ Mpc. Today $\dot d = H_0\chi = 0.765c$. At emission the proper distance was $\chi/2$ and $H = 1.790H_0$, so $\dot d = 1.790\times0.5\times0.7647\,c = 0.684c$. For the CMB, $\chi_{\mathrm{ls}} = 3.117\,c/H_0 = 13\,864$ Mpc (radiation included), and $H_0\chi_{\mathrm{ls}} = 3.12c$. The special- relativistic $\beta$ for $1 + z = 1091$ is $(1091^2 - 1)/(1091^2 + 1) = 1 - 1.68\times10^{-6}$.

The local statement, with its proof sketched. Two comoving observers a proper distance $\delta d$ apart are, to first order, in one local inertial frame (unit 32's normal coordinates) with relative velocity $H\,\delta d$. Light crossing between them takes $\delta t = \delta d/c$ and is Doppler-shifted by $\delta\lambda/\lambda = H\,\delta d/c = H\,\delta t = \delta a/a$. Integrating $\mathrm d\lambda/\lambda = \mathrm da/a$ recovers $\lambda \propto a$ ✓. So "infinitesimal Doppler shifts, parallel-transported" and "stretching" are the same computation (Bunn and Hogg, Am. J. Phys. 77, 688 (2009), argue for the first reading). What fails is a single Doppler shift with a single velocity, and that is what the slide denies.

Superluminal recession, again. We receive light today from objects whose proper distance has always grown faster than $c$ (everything beyond $z \approx 1.48$ is now receding faster than $c$, where $H_0\chi = c$). No contradiction: that light was, at emission, moving towards us through space that was carrying it away, and the Hubble sphere has since grown past it.

39 · light

Horizons

particle horizon

The comoving distance light has travelled since $t = 0$: $$\chi_{\mathrm p}(t) = \int_0^t\frac{c\,\mathrm dt'}{a(t')} = c\,\eta(t).$$ Nothing beyond it can have influenced us yet. Finite if the integral converges at $0$.

event horizon

The comoving distance light emitted now will ever travel: $$\chi_{\mathrm e}(t) = \int_t^\infty\frac{c\,\mathrm dt'}{a(t')}.$$ Events beyond it now can never influence us. Finite only if expansion accelerates enough.

two cases in closed form

Einstein-de Sitter: $\chi_{\mathrm p} = \int_0^a\frac{c\,\mathrm da}{a^2H} = \frac{c}{H_0}\int_0^a a^{-1/2}\mathrm da = \frac{2c}{H_0}\sqrt a$. Today $2c/H_0 = 3ct_0$. No event horizon: $\int^\infty a^{-1/2}\mathrm da$ diverges.

de Sitter: $\chi_{\mathrm e} = \int_t^\infty c\,e^{-H(t'-t_0)}\mathrm dt'$; proper $a\chi_{\mathrm e} = c/H$, constant.

Both horizons of the running example are drawn on the next slide.

Why the integrals. A radial light ray has $\mathrm d\chi = c\,\mathrm dt/a$, so the comoving distance covered between $t_1$ and $t_2$ is $\int_{t_1}^{t_2}c\,\mathrm dt/a$. Changing variable to $a$ with $\mathrm dt = \mathrm da/(aH)$ gives the $\int\mathrm da/(a^2H)$ forms used for computation. Dimensions: $[c\,\mathrm dt/a] = $ m ✓.

Convergence at $t = 0$. Near the Big Bang with $a \propto t^n$, $\int_0 t^{-n}\mathrm dt$ converges iff $n \lt 1$, i.e. iff the expansion decelerates. So every decelerating early universe has a finite particle horizon — which is the root of the horizon problem in section 6, and what inflation (accelerating, $n > 1$ effectively) is designed to change.

The $3ct_0$. With $t_0 = 2/(3H_0)$, $2c/H_0 = 3ct_0$: in Einstein-de Sitter the most distant matter we could in principle see is now three light-travel-times away, because the space the light crossed has stretched since.

39 · light

The Planck universe's horizons

Computed for the running example: comoving distance across, conformal time up, light at $45°$. Our past light cone meets the Big Bang at the particle horizon; our future light cone meets $a \to \infty$, at finite $\eta$, at the event horizon.

Today, by quadrature: particle horizon at proper distance $46.1$ Gly; event horizon $16.7$ Gly.

The top line is the whole point. With $\Lambda$, $\eta$ stays finite as $a \to \infty$, so the future light cone ends at a finite distance. Galaxies now beyond $z = 1.87$ are already beyond the event horizon: we still see their past, and never will see their present.

Without $\Lambda$ — Einstein-de Sitter — $\eta \to \infty$, the top line moves off to infinity, and there is no event horizon.

How the figure was made. $\eta(a) = \int_0^a\mathrm da'/(a'^2H)$, integrated numerically with $\Omega_{\mathrm r} = 9.21\times10^{-5}$: $\eta_0 = 46.1$ Gyr today and $\eta_\infty = 62.8$ Gyr at $a \to \infty$. Light cones are $45°$ lines in $(\chi, c\eta)$ by the conformal-time slide. The last-scattering line is at $\eta_* = 0.91$ Gyr, barely above the Big Bang; the CMB we see was emitted where our past cone crosses it, the dot at its foot.

The Planck numbers. Particle horizon $\chi_{\mathrm p}(t_0) = (c/H_0)\int_0^1\mathrm da/ (a^2E)$ with radiation included $= 14\,144$ Mpc $= 46.1$ Gly; event horizon $(c/H_0)\int_1^\infty \mathrm da/(a^2E) = 5114$ Mpc $= 16.7$ Gly, less than the de Sitter limit $c/(H_0\sqrt{\Omega_\Lambda}) = 17.5$ Gly it tends to as matter dilutes. The redshift $1.87$ is where $\chi(z) = \chi_{\mathrm e}$.

Rindler, again. The de Sitter event horizon is structurally unit 27's and unit 30's Rindler horizon: a boundary set by acceleration, not by any singularity, beyond which signals can never catch up.

The event horizon is not a place. Nothing happens to a galaxy as it crosses it; the horizon is defined by us, relative to our worldline, and every observer has their own.

39 · light

The distance ladder

$H_0$ needs distances, and no single method reaches far enough. The distance ladder calibrates each rung on the one below:

Parallax (Gaia): geometry, to a few kpc.
Cepheid variables: pulsation period fixes luminosity (Leavitt, 1912), calibrated by parallax; visible to about 40 Mpc.
Type Ia supernovae: standard candles, after a light-curve correction, calibrated on galaxies that also host Cepheids; seen to $z > 1$.
The Hubble flow: at $z \sim 0.02$–$0.15$, $cz \approx H_0 d$.

Schematic ranges on a log axis, ten decades of distance. Each rung is calibrated where it overlaps the one below (arrows), so an error in any rung is inherited by every rung above it.

Reading the figure. The ranges are rough and the edges soft: parallax from the Sun's neighbours out to a few kiloparsecs; Cepheids from the nearest, a few hundred parsecs away and reachable by parallax, out to about $40$ Mpc; Type Ia supernovae from the nearest hosts with Cepheids, near $7$ Mpc, out past $z = 1$; the Hubble flow from $z = 0.02$ to $0.15$, $cz/H_0 = 89$ to $670$ Mpc. The dashed line is $c/H_0 = 4.4$ Gpc. Only the top rung measures $H_0$; the lower three exist to give it an absolute luminosity.

History. Hubble's 1929 value was about $500\ \mathrm{km\,s^{-1}Mpc^{-1}}$, too large by a factor of seven, mostly because two kinds of Cepheids were then conflated (Baade, 1952). The ladder's history is a history of systematic errors found one rung at a time — which is why the Hubble tension, two slides on, is taken seriously rather than averaged away.

39 · light

Why the ladder must reach far

why one must go far

Galaxies have peculiar velocities of order $300\ \mathrm{km\,s^{-1}}$. At $100$ Mpc, $H_0d = 6740\ \mathrm{km\,s^{-1}}$, so the contamination is $4.5\%$; at $10$ Mpc it would be $45\%$. The ladder exists because the nearby universe is too lumpy to measure the expansion.

flux and distance

A standard candle of luminosity $L$ at small $z$ has flux $F = L/4\pi d^2$ (unit 05's flux law). At larger $z$ the FLRW version needs two factors of $1+z$: $F = L/[4\pi\chi^2(1+z)^2]$ in flat space (notes).

The FLRW flux law (proved here, flat case). Light emitted at $t_e$ from a comoving source at $\chi$ is spread today over a sphere of proper area $4\pi\chi^2$ (flat slices). Each photon's energy is down by $1+z$, and the arrival rate is down by another $1+z$ (the period stretching of the redshift slide). So $F = L/[4\pi\chi^2(1+z)^2]$, and the luminosity distance defined by $F = L/4\pi d_L^2$ is $d_L = (1+z)\chi$. For $z \ll 1$, $d_L \approx \chi \approx cz/H_0$ ✓. For $k \ne 0$, replace $\chi$ by $S_k(\chi)$.

Supernovae and acceleration. Measuring $d_L(z)$ out to $z \sim 1$ and comparing with the prediction for each $(\Omega_{\mathrm m}, \Omega_\Lambda)$ is exactly how $q_0 \lt 0$ was found in 1998: the distant supernovae were fainter (further) than a decelerating universe allows.

The $4.5\%$. $300/6740 = 0.045$ at $100$ Mpc; $300/674 = 0.45$ at $10$ Mpc. Peculiar velocities of a few hundred $\mathrm{km\,s^{-1}}$ are what the Local Group's own motion (the CMB dipole, $370\ \mathrm{km\,s^{-1}}$ for the Sun) exemplifies.

39 · light

The Hubble tension

method	$H_0$
Planck CMB, assuming flat $\Lambda$ + cold dark matter	$67.4 \pm 0.5$
Cepheids + Type Ia supernovae (SH0ES, 2022)	$73.0 \pm 1.0$

The difference is $5.6$ units against a combined uncertainty of $1.15$ (from the unrounded $\pm1.04$; notes): about $4.9\sigma$. Other ladder calibrations (the tip of the red-giant branch) have given values near $70$.

what the two numbers are

The CMB value is not a distance measurement. It is the $H_0$ that makes the FLRW model fit the CMB's angular scales; it is inferred through the model, from $z = 1090$. The ladder's is local, direct and model-light.

the Hubble tension, stated as an open problem

Either one method has an unrecognised systematic error, or the model connecting $z = 1090$ to $z = 0$ — the running example's — is missing something. As of the mid-2020s neither has been established, and neither is excluded.

For the age: at fixed $\Omega$s, $t_0 \propto 1/H_0$, and $H_0 = 73$ gives $12.74$ Gyr instead of $13.80$. Problem P9 asks how much of that survives when the CMB's constraint is kept.

Sources. Planck 2018 VI: $H_0 = 67.36 \pm 0.54$, usually rounded to $67.4 \pm 0.5$. Riess et al., ApJL 934, L7 (2022): $73.04 \pm 1.04$. The significance: $(73.04 - 67.4)/ \sqrt{1.04^2 + 0.5^2} = 5.64/1.154 = 4.9$. Freedman and collaborators, using the tip of the red- giant branch and later JWST data, have reported values around $70$ with uncertainties that overlap both. The situation has moved every year since 2019; treat the table as a snapshot.

Why the CMB value is model-dependent. What the CMB measures precisely is an angle — the angular size of the sound horizon at last scattering, $\theta_* \approx 0.0104$ rad — together with the physical densities $\Omega_{\mathrm m}h^2$ and $\Omega_{\mathrm b}h^2$. Turning an angle into $H_0$ requires the distance to $z = 1090$, which is the integral $\int\mathrm dz/H(z)$ of this unit, computed in a specified model. Change the model (early dark energy, extra relativistic species, evolving $w$) and the inferred $H_0$ changes.

Candidate explanations, named without endorsement: systematic errors in Cepheid photometry in crowded fields (largely excluded by JWST re-observations, per Riess et al. 2024, contested by Freedman et al.); a local underdensity (constrained to be too small); new physics before recombination that shrinks the sound horizon. None is accepted.

The 12.74. $13.796\times67.4/73.0 = 12.74$ Gyr.

section 6

The early universe: a success and two puzzles

Where the model works quantitatively at three minutes, and what it cannot explain.

39 · early

Big Bang nucleosynthesis

Run $a \to 0$ in the radiation era: $T \propto 1/a$, $a \propto t^{1/2}$. At $T \sim 1$ MeV ($t \sim 1$ s) weak reactions stop converting neutrons and protons; after about three minutes deuterium survives, and almost every free neutron ends in helium-4. This is Big Bang nucleosynthesis.

helium from the neutron fraction

If the neutron-to-proton ratio is $x = n/p$ and every neutron ends in $^4$He, the helium mass fraction is $$Y = \frac{2x}{1 + x}.$$

Why: each $^4$He takes two neutrons and two protons, so its mass is $2\times$ that of the neutrons; per $1 + x$ nucleons, $2x$ are in helium.

the number

Freeze-out at $kT \approx 0.75$ MeV gives $x = e^{-1.293/0.75} = 1/5.6$ (Boltzmann factor, cited); free decay ($\tau_n = 878$ s) until $t \approx 200$ s lowers it to $x \approx 1/7$. Then $$Y = \frac{2/7}{8/7} = 0.25.$$ Measured primordial helium: $Y_p = 0.245 \pm 0.003$.

why it is a success

It predicted a quarter of all ordinary matter is helium, everywhere, before this was measured; and the deuterium abundance fixes the baryon density to agree with the CMB's, measured at $3.7\times10^5$ yr instead of $3$ minutes.

What is cited and what is computed. The course has no statistical mechanics, so the Boltzmann factor $e^{-\Delta mc^2/kT}$ for the equilibrium ratio, the freeze-out temperature and the time-temperature relation are cited (Weinberg, Cosmology (2008) §3.2; Kolb and Turner, The Early Universe, ch. 4). What is computed here: $\Delta mc^2 = 1.293$ MeV, $e^{-1.293/0.75} = 0.178 = 1/5.6$; the neutron fraction $x/(1+x) = 0.151$ decays by $e^{-200/878} = 0.796$ to $0.120$, i.e. $x = 0.137 \approx 1/7.3$, and $Y = 2\times0.120 = 0.24$. Rounding $x$ to $1/7$ gives the textbook $0.25$. The sensitivity is to the freeze-out temperature, and the full network calculation with Planck's baryon density gives $Y_p = 0.247$.

The $Y$ formula. Mass fraction $=$ (mass in helium)/(total baryon mass), neglecting the neutron-proton mass difference and the binding energy (both below one per cent). With $N_n$ neutrons and $N_p$ protons, $N_n/2$ helium nuclei are formed, of total mass $4\times N_n/2 = 2N_n$ nucleon masses, out of $N_n + N_p$: $Y = 2N_n/(N_n + N_p) = 2x/(1+x)$ ✓. Limits: $x = 0$ gives $Y = 0$; $x = 1$ gives $Y = 1$, all helium ✓.

Measurements. $Y_p = 0.2453 \pm 0.0034$ from metal-poor extragalactic H II regions (Aver et al. 2021). Primordial D/H $= (2.527 \pm 0.030)\times10^{-5}$ from absorption in quasar spectra (Cooke et al. 2018), which fixes $\Omega_{\mathrm b}h^2 \approx 0.022$ to within a few per cent, the spread coming from the nuclear rate for deuterium burning; the CMB gives $0.02237 \pm 0.00015$.

History. Alpher, Bethe and Gamow (1948) and Alpher and Herman's prediction of a relic radiation at about 5 K; Hoyle and Tayler (1964) for the helium fraction; Wagoner, Fowler and Hoyle (1967) for the network. Lithium-7 is over-predicted by a factor of about three, an unresolved discrepancy usually attributed to stellar astrophysics.

39 · early

The horizon problem

worked example 4 · the angle a causal patch subtends, Einstein-de Sitter version

In EdS, $\chi_{\mathrm p}(a) = \frac{2c}{H_0}\sqrt a$. At recombination, $a_* = 1/1091$: $$\chi_{\mathrm p}(a_*) = \frac{2c}{H_0}\sqrt{a_*}.$$ The comoving distance from us to the last-scattering surface is what light covered since: $$\chi_{\mathrm{ls}} = \chi_{\mathrm p}(1) - \chi_{\mathrm p}(a_*) = \frac{2c}{H_0}(1 - \sqrt{a_*}).$$

In flat space a comoving length $\ell$ at comoving distance $\chi$ subtends $\theta = \ell/\chi$: $$\theta_{\mathrm H} = \frac{\sqrt{a_*}}{1 - \sqrt{a_*}} = \frac{1}{\sqrt{1091} - 1} = 0.0312 = 1.79°.$$

Checks. $H_0$ cancels, as it must for an angle ✓. $a_* \to 1$ (recombination today) gives $\theta \to \infty$: everything in causal contact ✓. $a_* \to 0$: $\theta \to 0$ ✓.

So a causal patch on the CMB sky is under two degrees across, in a sky of $4\pi$ steradians. The next slide draws why that is a problem.

39 · early

The horizon problem, drawn

Conformal diagram of the running example, light at $45°$, except that last scattering — and with it the cones of A and B — is drawn $8$ times larger than to scale; to scale it sits at $2\%$ of the height. A and B, on opposite sides of our sky, have past light cones that never meet.

Reading. Two points on the CMB sky more than about $2\theta_{\mathrm H}$ apart had past light cones that never overlapped: no event lies in both pasts. The sky holds some $4/\theta_{\mathrm H}^2 \approx 4000$ such patches, and they have the same temperature to $10^{-5}$.

The Planck universe, with radiation and $\Lambda$ (P10): $\theta_{\mathrm H} = 1.16°$, about $10^4$ patches.

How the figure was made. For the running example with radiation, $\eta_0 = 3.180/H_0$, $\eta_* = 0.063/H_0$ at $a = 1/1091$, and the last-scattering surface is at comoving distance $3.117\,c/H_0$. A and B are where our past light cone meets $\eta = \eta_*$; their own past light cones are triangles of half-width $c\eta_*$. To scale, $\eta_*$ is $2\%$ of the height and the two triangles would be specks, so $\eta_*$ is drawn $8$ times larger; the gap between the triangles is real and enormous either way. The angle a triangle subtends, seen from us, is $\theta_{\mathrm H} = \eta_*/(\eta_0 - \eta_*) = 0.0202$ rad.

Why $2\theta_{\mathrm H}$. Two points share a past event if their past light cones intersect, which in comoving coordinates means their separation is less than $2\chi_{\mathrm p}$. One patch of radius $\chi_{\mathrm p}$ around a point is the region that point has heard from.

The count. A cap of angular radius $\theta$ has solid angle $2\pi(1 - \cos\theta) \approx \pi\theta^2$; the sky has $4\pi$; ratio $4/\theta^2 = 4\times(32.03)^2 = 4104$ for EdS. For the Planck model's $\theta_{\mathrm H} = 0.02020$ rad, $9800$.

Why the realistic angle is smaller. Radiation made the early expansion faster in $a$ per unit time, so less comoving distance was covered: $\chi_{\mathrm p}(a_*)$ falls from $0.108$ to $0.063$ in units of $c/H_0$ (the EdS formula with $\Omega_{\mathrm m} = 0.315$ gives the first). P10 does it with pencil and paper, using the closed form for matter plus radiation.

39 · early

The flatness problem

flatness is unstable

With $\Omega(t) = \rho/\rho_{\mathrm c}(t)$ (all components), the first Friedmann equation is $$\Omega(t) - 1 = \frac{kc^2}{a^2H^2}.$$ In an era dominated by one component with $w$, $a^2H^2 \propto a^{-(1+3w)}$, so $$|\Omega - 1| \propto a^{1+3w}:$$ growing as $a$ for matter, as $a^2$ for radiation.

An unstable equilibrium: $k = 0$ keeps $\Omega = 1$ exactly, but any $k \ne 0$ is amplified in every era with $1 + 3w > 0$ — every decelerating era. The next slide runs today's bound backwards.

This is the flatness problem; the one before was the horizon problem. Both are fine-tuning of initial data, not contradictions.

Proof. Divide $H^2 = \frac{8\pi G}{3}\rho - kc^2/a^2$ (with $\Lambda$ absorbed into $\rho$ as $\rho_\Lambda$) by $H^2$: $1 = \Omega - kc^2/(a^2H^2)$ ✓. With one component, $H^2 \propto \rho \propto a^{-3(1+w)}$ (the curvature term being subdominant, which is the regime of interest), so $a^2H^2 \propto a^{-1-3w}$ ✓. Dimensions: $[kc^2/(a^2H^2)] = \mathrm{m^{-2}m^2s^{-2}/s^{-2}}$, dimensionless ✓. Special case: $k = 0$ gives $\Omega = 1$ for all time ✓ — flatness is an equilibrium, but an unstable one whenever $1 + 3w > 0$.

The dynamical-systems reading. Unit 02's language: $\Omega = 1$ is an equilibrium of the flow of $\Omega$ in $\ln a$, and for $w > -\tfrac13$ it is a repeller. Observing a system very near a repeller after a long time requires it to have started extraordinarily near. For $w \lt -\tfrac13$ — accelerating expansion — it becomes an attractor. That is the observation inflation is built on.

An honest dissent. Some cosmologists (Penrose, and later Carroll and others) argue that "fine-tuning" requires a measure on initial conditions, which nobody has, so the flatness problem may not be a well-posed problem at all. The horizon problem is harder to dismiss, because it is about correlations between causally disconnected regions rather than about a single number.

39 · early

Flatness, run backwards

run it backwards

Today $|\Omega - 1| = |\Omega_k| \lesssim 0.002$. Evolving back with all three components:

recombination: $\lesssim 4.4\times10^{-6}$;
nucleosynthesis ($kT = 1$ MeV): $\lesssim 10^{-18}$.

Nothing in the model requires $\Omega = 1$. It must have started within $10^{-18}$ of $1$ for no stated reason.

Computed: $|\Omega - 1|$ with the running example's densities and today's bound $\Omega_k = 0.002$, radiation, matter and $\Lambda$ all kept. Slope $2$, then slope $1$: fifteen decades of growth between nucleosynthesis and today.

39 · early

Inflation, and its status

inflation

A hypothesised early era with $\ddot a > 0$, nearly de Sitter, driven by a field with $w \approx -1$, ending well before nucleosynthesis.

the mechanism in one line

$\frac{\mathrm d}{\mathrm dt}(aH) = \ddot a$. So $\ddot a > 0$ exactly when the comoving Hubble length $c/(aH)$ shrinks.

Horizon: regions now far apart were inside one Hubble length before inflation; $\chi_{\mathrm p}$ becomes large.
Flatness: $|\Omega - 1| = |k|c^2/(aH)^2$ is driven towards $0$ as $aH$ grows.

The comoving Hubble length, computed under the notes' assumptions ($60$ e-folds, factors of $e$ in $a$, at constant $H$, ending at $10^{15}$ GeV). A comoving scale is a horizontal line: inside the curve before inflation, pushed outside during it, back in long after.

Proof of the one line. $aH = \dot a$, so $\frac{\mathrm d}{\mathrm dt}(aH) = \ddot a$ ✓. The comoving Hubble length $c/(aH)$ is the comoving distance over which expansion is "fast", and in a decelerating universe it grows, so regions keep entering causal contact for the first time. During inflation it shrinks, so a region that was causally connected is stretched beyond it. After inflation it grows again and regions re-enter, already correlated.

How many e-folds. The horizon problem is solved if today's comoving Hubble length was inside the Hubble length when inflation began: $(a H)_{\mathrm{begin}} \le a_0H_0$. During near-de Sitter inflation $H$ is almost constant, so $aH$ grows by $e^N$ over $N$ e-folds; we need $e^N \ge (a_{\mathrm{end}}H_{\mathrm{end}})/(a_0H_0)$. In the radiation era $H = H_0\sqrt{\Omega_{\mathrm r}}a^{-2}$, so $aH = H_0\sqrt{\Omega_{\mathrm r}}/a$. If inflation ends at $kT = 10^{15}$ GeV, then with $T \propto 1/a$ (ignoring order-one factors from particle counting), $a_{\mathrm{end}} = kT_0/10^{24}\ \mathrm{eV} = 2.35\times10^{-28}$, and $(a_{\mathrm{end}}H_{\mathrm{end}})/(a_0H_0) = \sqrt{\Omega_{\mathrm r}}/a_{\mathrm{end}} = 4.1\times10^{25}$: $N \ge \ln(4.1\times10^{25}) = 59$. The standard "about 60" is this computation; a lower reheating temperature lowers it.

How the figure was made. After inflation, $c/(aH) = (c/H_0)/(aE(a))$ with all three components. During it, $H$ is held at its value at the end, $H_{\mathrm{end}} = H_0\sqrt{\Omega_{\mathrm r}}/a_{\mathrm{end}}^2$, so $c/(aH) \propto 1/a$, for exactly $60$ e-folds before $a_{\mathrm{end}} = 2.35\times10^{-28}$. Both are idealisations: real inflation has $H$ slowly falling, and the joins are smooth. The dashed line is a fixed comoving length, today's Hubble length $c/H_0$: the largest scale we can now see as one causally connected region. It was inside the Hubble length when inflation began, so causal processes had time to make it uniform before inflation pushed it out; it is only now coming back in. Any smaller scale — the size of a causal patch at recombination, $0.063\,c/H_0$, sits about a decade lower — left later and returned earlier. The Hubble length now shrinks again: under $\Lambda$, scales are leaving once more.

39 · early

Inflation: the evidence and what is open

evidence for

Its generic predictions are borne out: $|\Omega_k| \lesssim 0.002$; nearly scale-invariant fluctuations with spectral index $n_s = 0.965 \pm 0.004$, slightly below $1$ as predicted; correlations on scales beyond $\theta_{\mathrm H}$.

unresolved

No identified field; the required $\sim 60$ e-folds depend on an unknown energy scale; gravitational waves from inflation not detected ($r \lt 0.036$); initial conditions and a measure are open. It is the leading hypothesis, not an established part of the model.

Numbers cited. $n_s = 0.9649 \pm 0.0042$ (Planck 2018 X). $r \lt 0.036$ at 95% (BICEP/Keck 2021). Guth (1981) proposed inflation to solve the monopole problem and the horizon and flatness problems; Linde (1982) and Albrecht and Steinhardt (1982) repaired its exit. The prediction of the fluctuation spectrum (Mukhanov and Chibisov 1981; several groups 1982) is the part with the most evidential weight, and it needs quantum field theory, which this course does not reach.

Why "unresolved" is the honest word. Inflation makes generic predictions that are confirmed, but alternatives (bouncing cosmologies, varying speed of light proposals) have been constructed to match the same data, and the inflationary framework can accommodate a wide range of outcomes. Whether it is falsifiable in its general form is itself debated (Ijjas, Steinhardt and Loeb 2017, and the reply signed by Guth, Linde and some thirty others).

39 · summary

The Planck universe, closed

derived in this unit

§1 Metric: FLRW with $k = 0$, forced by the cosmological principle.
§2 Equations: Friedmann, from $G_{00}$ and $G_{ij}$; dust $\rho \propto a^{-3}$, $\rho_\Lambda$ constant.
§3 A ball that passed the top of $V$ at $a = 0.613$ and now accelerates; $q_0 = -0.53$.
§4 Age $13.80$ Gyr; matter–$\Lambda$ equality at $z = 0.296$, $3.50$ Gyr ago.
§5 $z = 1$ is $7.95$ Gyr of lookback; horizons $46.1$ and $16.7$ Gly.
§6 A causal patch at recombination spans $1.16°$.

assumed, not derived

The cosmological principle.
The values $0.315$, $0.685$, $67.4$ — measured, and the last is contested.
Dark matter and $\Lambda$ as named components with no identified physics.
Initial conditions that are flat and uniform beyond the horizon.

Next. Unit 40 returns to single objects: rotating and charged holes, and where this course's methods stop.

The lookback to $z = 1$. $t(a = 0.5) = 0.8055\times\operatorname{arsinh}(1.4747\times 0.3536)\times14.51 = 0.8055\times0.5002\times14.51 = 5.85$ Gyr, so $13.80 - 5.85 = 7.95$ Gyr ago.

If one thing is to survive a year, it should be the structure of section 2 and section 3 together: symmetry reduces Einstein's equation to one first-order ODE, $\tfrac12\dot a^2 + V(a) = -\tfrac12kc^2$, which is unit 08's energy equation; the matter content chooses $V$; and every quantitative statement in cosmology — ages, distances, horizons — is a one-dimensional integral of $1/H(a)$ along it.

Further reading. Ryden, Introduction to Cosmology (2nd ed., 2016), chs. 2–6 and 9–11, at exactly this level and without tensors; Weinberg, Cosmology (2008), ch. 1, for the geometry done carefully. Skip Ryden's thermal-history chapters' statistical mechanics on first reading, since the course has not taught it; the results are quoted there the way this deck quotes them.

39 · checkpoint

Checkpoint

you can now do this, or you cannot

You derive the Friedmann equations from the FLRW metric and Einstein's equation without consulting a reference, solve for the scale factor in the matter-, radiation- and $\Lambda$-dominated eras, and compute the age of the universe for the Planck parameters ($\Omega_{\mathrm m} = 0.315$, $\Omega_\Lambda = 0.685$, $H_0 = 67.4\ \mathrm{km\,s^{-1}Mpc^{-1}}$) to within $0.3$ Gyr of $13.8$.

Failing it is unambiguous. If your $G_{00}$ contains $\ddot a$, a sign in $R_{00}$ or $R$ is wrong — redo worked example 1. If your age is $9.7$ Gyr you forgot $\Lambda$; if it is near $14.5$ you computed the Hubble time. If $a(t)$ for radiation is not $\propto t^{1/2}$, redo worked example 2.

Thirteen problems, about six hours: ./mr problems 39. P9 is the age with both values of $H_0$; P10 quantifies the horizon problem; P5 redoes the Friedmann derivation in conformal time. When the checkpoint is true: ./mr done 39.