Part III · General Relativity — Chapter 3.9

Cosmology, and a Loose Thread

Put the largest possible source on the right-hand side. Then collect the oldest promise in the book, and leave one number deliberately unpaid.

Tools you'll need  — Chapter 3.6 above all. Section 1.3 for the perfect fluid $T^{\mu\nu}=(\rho+p/c^{2})u^{\mu}u^{\nu}-pg^{\mu\nu}$, and §1.4 for the radiation equation of state $p=\rho c^{2}/3$. Section 5.5 for $\nabla^{2}\Phi=4\pi G(\rho+3p/c^{2})$, whose bracket is the one this chapter's §3.2 reproduces exactly. Section 6 for the cosmological term as a fluid with $p=-\rho c^{2}$. And §3, the cornering, which is the structural fact §3.3 meets again. Chapter 3.5 §8 for Killing vectors as the statement of a symmetry, and §9 for the conserved charge, which §6 needs precisely because it is unavailable. Chapter 3.4 §6 for the Ricci tensor and scalar, §7 for $\nabla^{\mu}G_{\mu\nu}=0$, and §8 for the flat-and-yet-not-a-plane distinction that §2.4 turns on the universe. Chapter 3.3 §7 for the Christoffel formula and §8 for geodesics, timelike and null. Chapter 3.7 §2 for the areal-radius convention, reused in §2.2, and §1 for symmetry stated as a Killing-vector condition. Chapter 3.8 §5.1 for the rule that the frequency an observer measures is the photon momentum contracted with that observer's four-velocity, used again in §5.1 here. Chapter 3.2 §1.1, which promised that this chapter would take the global shape of spacetime seriously, and §5 on vectors at different points. Chapter 1.4 §4.3, the honest note. Also its §1 for Noether's theorem itself, and §7.2 for the converse, which §6 leans on and without which this chapter's central argument would run the theorem backwards. Chapter 1.1 §2, where the promise was first made. Chapter 0.8 §2.1 for separable first-order equations, which is what every integration in §4 is. Chapter 0.5 §6 for the spectral theorem, used once in §2.2 to say what a rotationally invariant matrix must be, and §8 for commuting operators, which §7.4 hands forward.

1 · The cosmological principle as an assumption, and its evidence

Here is where this section is going. We write down the two symmetry assumptions that make cosmology possible. They are stated in the same language Chapter 3.7 used for the star, as conditions on Killing vectors. Then we say exactly what observation supports them, and where they are false.

1.1 · Two symmetries, and the difference from a star

Chapter 3.7 §1 refused to state spherical symmetry as a picture and stated it as a fact about Killing vectors: there exist three spacelike Killing fields whose commutators reproduce the algebra of rotations. That is isotropy about one point, and the point is the centre. Everything in Chapters 3.7 and 3.8 followed from that, plus staticity.

Cosmology assumes two things, and the first is what makes it a different subject.

Homogeneity. There is a family of spacelike slices such that, for any two points $p$ and $q$ on one slice, some isometry of the geometry carries $p$ to $q$. Infinitesimally, that means three independent spacelike Killing fields at every point, spanning the directions along the slice. No place is special.

Isotropy about every point. At each point of a slice there are three further Killing fields vanishing at that point and rotating the directions around it, again with the algebra of rotations. No direction is special, and this holds wherever you stand, not merely at a centre.

Six Killing fields, then, where the Schwarzschild solution had four (three rotations and one time translation). Note what has been traded.

• The star's geometry has a timelike Killing field, and isotropy at a single point.
• The universe's has isotropy at every point, and no timelike Killing field at all. Section 6 proves that second half.

The two assumptions taken together are called the cosmological principle, and the whole of this chapter is the consequence of assuming it.

One clarification, because anyone who has done Part II will already be uneasy. Isotropy about every point does not follow from isotropy about one, and homogeneity does not follow from isotropy either. Two examples make the point. An infinite cylinder is homogeneous along its axis without being isotropic, since the axis is a direction the geometry can tell apart from the others. A spherically symmetric star is isotropic about its centre and homogeneous nowhere.

Isotropy about every point does imply homogeneity, and that is the version usually assumed. Stating the two separately keeps visible which one each later step uses.

1.2 · What is actually measured

Say plainly where the principle fails. It is false at the scale of your body, of the Earth, of the Solar System, and of the Galaxy, all of which are conspicuous departures from a uniform fluid. The object being described in this chapter is an average, taken over regions large enough that the lumps stop mattering, and the equations below govern that average and nothing smaller. Section 5.5 returns to this, because the commonest error in reading cosmology is applying a result derived for the average to something bound and small.

1.3 · The preferred slicing, and why it is not a betrayal of Part II

Homogeneity picks out a family of slices, namely those on which conditions are the same everywhere. That in turn picks out a preferred notion of simultaneity and a preferred time coordinate. After six chapters insisting that no observer's slicing is privileged, that deserves an explicit answer.

The answer is that the privilege belongs to a solution rather than to the theory. Nothing in the field equations distinguishes a slicing. What distinguishes one here is the matter, which happens to be distributed in a way that is uniform on one family of slices and not on others. A room full of air at rest picks out a frame too, and no one takes that as evidence against relativity.

The invariant content of the statement is that the matter has a four-velocity field $u^{\mu}$, and "cosmic time" is proper time measured along it. Any observer may use any coordinates. The ones below are chosen because they make the symmetry visible, exactly as Chapter 3.7's were.

2 · The FLRW metric

Here is the plan for this section. We convert the two assumptions into a line element with one unknown function in it. The time part is fixed immediately. The spatial part takes more work. We have to show that a homogeneous, isotropic three-geometry has constant curvature, and then solve one separable equation to find every such geometry. There are exactly three, up to scale.

2.1 · The time part, fixed by inspection

Use the matter's own four-velocity to build coordinates, as §1.3 described. Label each homogeneous slice by the proper time $t$ read on a clock carried by the matter, and give the matter fixed spatial coordinates $x^{i}$. That second step is legitimate because the matter has no preferred direction to move in. If it moved relative to these labels, the motion would pick one out. Such coordinates are called comoving.

Two components are then fixed with no computation.

$g_{00}=1$. A comoving worldline has $\dd x^{i}=0$, and along it $\dd s^{2}=g_{00}(\dd x^{0})^{2}=g_{00}c^{2}\dd t^{2}$. But $t$ was defined as proper time along exactly that worldline, and Chapter 3.3 §3 defines proper time by $c^{2}\dd\tau^{2}=\dd s^{2}$. So $g_{00}=1$.

$g_{0i}=0$. The three quantities $g_{0i}$ form a spatial vector at each point, and isotropy says no spatial direction is distinguishable from any other. A non-zero vector distinguishes the direction it points in. So it vanishes.

Hence the line element is $\dd s^{2}=c^{2}\dd t^{2}+g_{ij}\dd x^{i}\dd x^{j}$, with the spatial part negative-definite and, by homogeneity, the same geometry on every slice up to an overall time-dependent factor. Write that factor out:

with $\tilde g_{ij}$ a fixed positive-definite three-geometry carrying no time dependence and $a(t)\gt0$ the scale factor.

Pause on why the time dependence can be pulled out as one overall factor. That is homogeneity doing its work. If two slices differed by anything other than an overall scale, comparing them would define a position-dependent quantity, and there are no position-dependent quantities here.

2.2 · Which three-geometries qualify

Now the only real work in the section. We need every three-geometry that is homogeneous and isotropic. Do it in two steps, and note that the first step is an argument about a matrix, not about geometry.

Step 1: isotropy forces the Ricci tensor to be a multiple of the metric. In three dimensions the Riemann tensor has $3^{2}(3^{2}-1)/12=6$ independent components, by Chapter 3.4 §5.2's count. The Ricci tensor $\tilde R_{ij}$, being symmetric, has exactly six as well. In three dimensions, then, the Ricci tensor carries the whole of the curvature and discards nothing. So it is enough to constrain $\tilde R_{ij}$.

At a chosen point, $\tilde R_{ij}$ is a real symmetric $3\times3$ matrix, and by isotropy it must be unchanged by every rotation about that point. Chapter 0.5 §6's spectral theorem says a real symmetric matrix has an orthonormal eigenbasis with real eigenvalues. If two eigenvalues differed, the corresponding eigenspaces would be different, and rotating the eigenvectors of one into those of another would change the matrix. Put the other way round, the eigenvector directions would be distinguishable, which is exactly what isotropy forbids. So all three eigenvalues are equal and

the factor $2$ being a convention chosen to make what follows tidy. Homogeneity then says $K$ is the same number at every point: it is built from the geometry, and no place is special.

Step 2: solve (3.9.2). Isotropy about the origin lets us use exactly Chapter 3.7 §2's ansatz, minus the time part. That ansatz is spherical symmetry about a point together with the areal-radius convention, which labels spheres so that the sphere carrying label $r$ has area $4\pi r^{2}$. It gives

with a single unknown function $\beta(r)$. The same caution Chapter 3.7 §2 gave applies verbatim: $r$ is a label chosen for the area of a sphere, not the distance from the origin, and the two differ by exactly the factor $\ee^{\beta}$.

Grind box A computes the Ricci tensor of (3.9.3). Its $rr$ component is $2\beta'/r$, so (3.9.2) reads

What we want is $\beta$ itself, and that is a separable first-order equation of the kind Chapter 0.8 §2.1 handles. So integrate both sides:

One constant, one physical condition, exactly as in Chapter 3.7. Here the condition is regularity at the origin: at $r=0$ the geometry must look like ordinary flat space to a local observer, since a deficit there would be a conical point and would make the origin a special place. That requires $\ee^{2\beta}\to1$ as $r\to0$, so $-2C=1$, and

Grind box A also checks the $\theta\theta$ component of (3.9.2) and finds it satisfied by the same $\beta$, which is not automatic: two equations, one unknown function, and the system is consistent. That consistency is isotropy paying for itself.

Grind box A — the Ricci tensor of the isotropic three-geometry, and the consistency check

Metric $\tilde g_{ij}=\mathrm{diag}\big(\ee^{2\beta},\,r^{2},\,r^{2}\sin^{2}\theta\big)$ in coordinates $(r,\theta,\varphi)$, all signs positive because this is the three-geometry itself and not its embedding in (3.9.1). From Chapter 3.3's formula, the non-zero connection coefficients are

$$\tilde\Gamma^{r}{}_{rr}=\beta', \quad \tilde\Gamma^{r}{}_{\theta\theta}=-r\ee^{-2\beta}, \quad \tilde\Gamma^{r}{}_{\varphi\varphi}=-r\ee^{-2\beta}\sin^{2}\theta,$$ $$\tilde\Gamma^{\theta}{}_{r\theta}=\tilde\Gamma^{\varphi}{}_{r\varphi}=\frac1r, \quad \tilde\Gamma^{\theta}{}_{\varphi\varphi}=-\sin\theta\cos\theta, \quad \tilde\Gamma^{\varphi}{}_{\theta\varphi}=\cot\theta.$$

Feeding these through Chapter 3.4's definition of the Riemann tensor and contracting gives a diagonal Ricci tensor with

$$\tilde R_{rr} \;=\; \frac{2\beta'}{r}, \qquad \tilde R_{\theta\theta} \;=\; \ee^{-2\beta}\big(r\beta' + \ee^{2\beta} - 1\big), \qquad \tilde R_{\varphi\varphi} \;=\; \tilde R_{\theta\theta}\sin^{2}\theta.$$

The check. With $\ee^{2\beta}=(1-Kr^{2})^{-1}$ we have $\beta'=Kr/(1-Kr^{2})$, so $r\beta'=Kr^{2}/(1-Kr^{2})$ and $\ee^{2\beta}-1=Kr^{2}/(1-Kr^{2})$. Adding and multiplying by $\ee^{-2\beta}=1-Kr^{2}$,

$$\tilde R_{\theta\theta} \;=\; \big(1-Kr^{2}\big)\cdot\frac{2Kr^{2}}{1-Kr^{2}} \;=\; 2Kr^{2} \;=\; 2K\,\tilde g_{\theta\theta}. \;\checkmark$$

(Verified symbolically for the general $\beta$, and the resulting metric checked to satisfy the full four-index condition $\tilde R_{ijkl}=K(\tilde g_{ik}\tilde g_{jl}-\tilde g_{il}\tilde g_{jk})$, with three-dimensional Ricci scalar $\tilde R=6K$.)

2.3 · The metric, and what the scale factor is not

Rescale the radial label to absorb the size of $K$. If $K\neq0$, put $\bar r=\sqrt{\abs K}\,r$ and move the factor $\abs{K}^{-1/2}$ into $a$. The geometry is unchanged, and the constant is reduced to its sign. Dropping the bar and calling the sign $k$,

This is the Friedmann–Lemaître–Robertson–Walker metric. One function, $a(t)$, and one discrete choice, $k$. Everything else was forced. The curvature of a spatial slice at time $t$ is $K=k/a^{2}$, so a positively curved universe has three-dimensional Ricci scalar $6/a^{2}$ and grows flatter as it expands.

Three warnings about $a$, because all three are commonly got wrong.

It is not the radius of the universe. For $k=+1$ the slices are three-spheres and $a$ is their radius. For $k=0$ and $k=-1$ they are infinite in extent, and there is no radius for $a$ to be.

Its absolute value carries no information when $k=0$. Multiplying $a$ by any constant while dividing every $r$ by the same constant leaves (3.9.7) unchanged. Only the ratio $a(t_{1})/a(t_{2})$ is measurable, and that is exactly the quantity §5 shows a redshift delivers. For $k=\pm1$ the rescaling freedom was already spent normalising $k$ to $\pm1$, so there $a$ does have an absolute meaning.

It is not a substance and it does not stretch anything. $a$ is a component of a metric. Section 5.3 makes this precise and §5.5 says what does not expand.

2.4 · Flat is not the same as infinite

Chapter 3.2 §1.1 said that whether the universe closes up on itself is "a question to be answered rather than assumed" and named this chapter as where it would be asked seriously. Chapter 3.4 §8's flagged converse and Chapter 3.5 §4.3 both pointed here too. Here is the answer, and it is a distinction rather than a number.

Everything derived above is local. Equation (3.9.7) was obtained by solving a differential equation for the metric components, and a differential equation knows only about a neighbourhood. The sign $k$ is a statement about curvature, measurable in principle by summing the angles of a large enough triangle. It is not a statement about whether space is finite.

The counterexample is one Chapter 3.4 §8 pointed here for. Take flat $k=0$ space and identify $x\sim x+L$ in each of the three directions. Every local measurement is unchanged. The metric is still $\delta_{ij}$, the curvature is still zero, and triangles still have angles summing to $\pi$. But the resulting space is a three-torus, of finite volume $L^{3}$. Travel far enough in one direction and you return to where you started.

Chapter 3.3's Problem 1 built the two-dimensional version of the same mismatch, a cone, and observed that flatness away from the apex did not make it a plane. The same mismatch between local and global appears here at cosmic scale, and it is the fifth time this book has met it.

So the measurement reported in §4 as $\Omega_{k}\approx0$ says the curvature of space is small. It says nothing whatever about the topology, which no local measurement can reach. Settling that would take a different kind of observation altogether: one would look for the pattern an identification forces, in which a single feature appears twice on the sky because two lines of sight wrap round and meet. A null result there bounds the size of any identification from below. It cannot exclude one, and it is not what a curvature measurement reports.

3 · The Friedmann equations

Here is where this section is going, and the route matters as much as the results. We compute the Einstein tensor of (3.9.7), set it equal to the perfect fluid of Chapter 3.6 §1.3, and read off two equations. The $00$ component gives one directly. The $11$ component gives a second, but only after the first has been substituted into it. That is not an inconvenience. It is the announcement that the two are not independent. We then show that differentiating the first and subtracting a multiple of the second leaves exactly the fluid's own equation of motion, which is Chapter 3.4 §7's contracted Bianchi identity wearing cosmological clothes. Finally we perturb the one static solution and watch it fall over.

3.1 · The $00$ component, and Friedmann I

Grind box B does the computation. Its result, in the primed notation of the conventions, is

Grind box B — the connection, the Ricci tensor and the Einstein tensor for FLRW

Coordinates $(x^{0},r,\theta,\varphi)$ with $x^{0}=ct$, metric $g_{\mu\nu}=\mathrm{diag}\big(1,\,-a^{2}/(1-kr^{2}),\,-a^{2}r^{2},\,-a^{2}r^{2}\sin^{2}\theta\big)$. Primes are $\dd/\dd x^{0}$.

Step 1, the connection, from Chapter 3.3's formula. Thirteen non-zero components, counting each symmetric pair once:

$$\Gamma^{0}{}_{rr}=\frac{aa'}{1-kr^{2}}, \qquad \Gamma^{0}{}_{\theta\theta}=r^{2}aa', \qquad \Gamma^{0}{}_{\varphi\varphi}=r^{2}aa'\sin^{2}\theta,$$ $$\Gamma^{r}{}_{0r}=\Gamma^{\theta}{}_{0\theta}=\Gamma^{\varphi}{}_{0\varphi}=\frac{a'}{a}, \qquad \Gamma^{r}{}_{rr}=\frac{kr}{1-kr^{2}},$$ $$\Gamma^{r}{}_{\theta\theta}=-r\big(1-kr^{2}\big), \qquad \Gamma^{r}{}_{\varphi\varphi}=-r\big(1-kr^{2}\big)\sin^{2}\theta,$$ $$\Gamma^{\theta}{}_{r\theta}=\Gamma^{\varphi}{}_{r\varphi}=\frac1r, \qquad \Gamma^{\theta}{}_{\varphi\varphi}=-\sin\theta\cos\theta, \qquad \Gamma^{\varphi}{}_{\theta\varphi}=\cot\theta.$$

Two of these are worth pausing on. Every $\Gamma^{i}{}_{00}$ vanishes. Section 5.3 uses that, and it is the statement that comoving matter is in free fall. And $\Gamma^{0}{}_{ij}=-(a'/a)\,g_{ij}$ for every spatial pair, which is the single fact §5.1's redshift derivation needs. Check it against the three entries above, remembering that $g_{ij}$ carries a minus sign.

Step 2, the Ricci tensor, by contracting Chapter 3.4's Riemann tensor. It is diagonal, with

$$R_{00}=-\frac{3a''}{a}, \qquad R_{rr}=\frac{aa''+2a'^{2}+2k}{1-kr^{2}},$$ $$R_{\theta\theta}=r^{2}\big(aa''+2a'^{2}+2k\big), \qquad R_{\varphi\varphi}=R_{\theta\theta}\sin^{2}\theta,$$

and the Ricci scalar, contracting once more with $g^{\mu\nu}$, is

$$R \;=\; -\,\frac{6\big(aa''+a'^{2}+k\big)}{a^{2}}.$$

Step 3, the Einstein tensor $G_{\mu\nu}=R_{\mu\nu}-\half Rg_{\mu\nu}$. The $00$ entry is $-3a''/a-\half\cdot\big({-6(aa''+a'^{2}+k)/a^{2}}\big)$, and the $a''$ terms cancel:

$$G_{00}=\frac{3\big(k+a'^{2}\big)}{a^{2}}, \qquad G_{rr}=-\,\frac{k+2aa''+a'^{2}}{1-kr^{2}},$$

with $G_{\theta\theta}$ and $G_{\varphi\varphi}$ carrying the same bracket times $-r^{2}$ and $-r^{2}\sin^{2}\theta$. That the three spatial components are the same equation is isotropy again: they could not have differed.

(Verified symbolically: the connection, the full Riemann tensor, the Ricci tensor and the Einstein tensor were computed from the metric with nothing assumed, returning exactly the components above.)

The source. Chapter 3.6 §1.3 built the perfect fluid $T^{\mu\nu}=(\rho+p/c^{2})u^{\mu}u^{\nu}-pg^{\mu\nu}$. Comoving matter has $u^{\mu}=(c,0,0,0)$, which satisfies the normalisation $g_{\mu\nu}u^{\mu}u^{\nu}=c^{2}$ because $g_{00}=1$, and lowering gives $u_{\mu}=(c,0,0,0)$ as well. So

Now set $G_{\mu\nu}=\kappa T_{\mu\nu}$ with $\kappa=8\pi G/c^{4}$, from Chapter 3.6 §5. The $00$ component reads $3(k+a'^{2})/a^{2}=8\pi G\rho/c^{2}$. Multiply through by $c^{2}$ and use $a'=\dot a/c$:

That is Friedmann's first equation, and it is one component of Einstein's equation with the symmetry already imposed. Define the Hubble parameter $H(t)\equiv\dot a/a$, the fractional rate of growth, whose reciprocal is a time and whose present value is written $H_{0}$. Read (3.9.10) term by term: the square of the growth rate is set by the density, reduced by a curvature term that is negligible when $a$ is small and can matter when $a$ is large. Nothing in it is second order in time. This is a constraint on $\dot a$ given $\rho$ rather than an equation of motion, which is the shape Chapter 3.6 §7.2 predicted for the $00$ component of any solution.

3.2 · The $11$ component, and why it needs Friedmann I

Take the $rr$ component with the same substitutions. From grind box B and (3.9.9), cancelling the common factor $(1-kr^{2})^{-1}$ and its sign,

This is not yet a statement about $\ddot a$ alone: it contains $a'^{2}$ and $k$ as well. Divide by $a^{2}$ and isolate them,

and now notice that the second term on the left is precisely the combination (3.9.10) evaluates: $(a'^{2}+k)/a^{2}=8\pi G\rho/3c^{2}$. Substituting Friedmann I into the $11$ equation and rearranging,

having multiplied through by $c^{2}$ to convert primes to dots. This is the acceleration equation, and three things about it deserve to be said out loud.

The substitution was not optional. Equation (3.9.11) on its own does not give (3.9.13). It gives it only in combination with (3.9.10). That is the first evidence that these are not two independent statements about the universe, and §3.3 turns the evidence into a proof.

The bracket is Chapter 3.6's, exactly. That chapter's §5.5 derived the Newtonian limit of the field equations as $\nabla^{2}\Phi=4\pi G(\rho+3p/c^{2})$ and observed that pressure gravitates, three times over. Here is the same bracket, with no weak-field approximation anywhere, governing the acceleration of the whole universe. Everything that chapter said about the bracket applies verbatim, and it comes down to two cases.

• Radiation, with $p=\rho c^{2}/3$, gives $2\rho$, and so pulls twice as hard as dust of the same energy density.
• A fluid with $p\lt-\rho c^{2}/3$ gives a negative bracket, and so pushes.

With ordinary matter the universe cannot be static, and cannot always have been expanding. For any fluid with $\rho\gt0$ and $p\ge0$, (3.9.13) gives $\ddot a\lt0$, so the expansion always decelerates. Now run that backwards. If $\dot a\gt0$ now, then $\dot a$ was larger in the past, so the graph of $a(t)$ lies below its own tangent at the present. Extrapolating that tangent backwards therefore reaches $a=0$ later than the true curve does. The tangent hits zero after a time $a_{0}/\dot a_{0}=1/H_{0}$, so

with $a\to0$ at a finite time in the past. That single line, requiring nothing but the sign of the bracket, is the argument that a universe of ordinary matter has a beginning, and §4.4 puts a number on it.

3.3 · Two equations, not three — and it is the Bianchi identity again

There is a third equation available, and it has not been used. The fluid must obey its own equation of motion, $\nabla_{\mu}T^{\mu\nu}=0$, which Chapter 3.6 §1.2 promoted to curved spacetime. Work it out for (3.9.7). The three spatial components vanish identically, because there is nowhere for momentum to flow when nothing is special. The time component gives

So there are three equations, (3.9.10), (3.9.13) and (3.9.15), for two unknown functions $a(t)$ and $\rho(t)$ once an equation of state relates $p$ to $\rho$. One equation too many, and a system in that position has no business being consistent.

Here is why it is. Take (3.9.10) in the form $\mathrm{I}\equiv a'^{2}-\tfrac{8\pi G}{3c^{2}}\rho a^{2}+k$, which vanishes, and (3.9.13) in the form $\mathrm{II}\equiv a''+\tfrac{4\pi G}{3c^{2}}\big(\rho+3p/c^{2}\big)a$, which also vanishes. Differentiate the first with respect to $x^{0}$ and subtract $2a'$ times the second. The whole computation is four lines and every term cancels except one:

Grind box C — the cancellation, line by line

Line 1, differentiate I. Using the product rule on both $\rho a^{2}$ terms,

$$\dv{\mathrm{I}}{x^{0}} \;=\; 2a'a'' \;-\; \frac{8\pi G}{3c^{2}}\Big(\rho'a^{2} + 2\rho a a'\Big).$$

Line 2, multiply II by $2a'$. Our aim is a combination in which the second-derivative terms drop out, and this is the multiple that arranges it.

$$2a'\,\mathrm{II} \;=\; 2a'a'' \;+\; \frac{8\pi G}{3c^{2}}\Big(\rho + \frac{3p}{c^{2}}\Big)a a'.$$

Line 3, subtract. The $2a'a''$ terms cancel, which is the point of the factor $2a'$, and the rest collects:

$$\dv{\mathrm{I}}{x^{0}} - 2a'\,\mathrm{II} \;=\; -\frac{8\pi G}{3c^{2}}\left[\rho'a^{2} + 2\rho aa' + \Big(\rho+\frac{3p}{c^{2}}\Big)aa'\right].$$

Line 4, collect the $aa'$ terms. $2\rho aa'+\rho aa'=3\rho aa'$, so the bracket is $\rho'a^{2}+3\big(\rho+p/c^{2}\big)aa'$, which is $a^{2}$ times $\rho'+3(a'/a)(\rho+p/c^{2})$. Hence

$$\dv{\mathrm{I}}{x^{0}} - 2a'\,\mathrm{II} \;=\; -\,\frac{8\pi G a^{2}}{3c^{2}}\left[\rho' + 3\frac{a'}{a}\Big(\rho+\frac{p}{c^{2}}\Big)\right],$$

and the bracket is (3.9.15) with primes for dots, which is the same equation since converting $\dd/\dd x^{0}$ to $\dd/\dd t$ multiplies both terms by $c$.

(Verified symbolically, including the coefficient: the combination is exactly $-8\pi Ga^{2}/3c^{2}$ times the fluid equation, with no remainder. Also verified that $\nabla_{\mu}T^{\mu 0}=c^{2}\times$ the fluid equation and $\nabla_{\mu}T^{\mu i}=0$ identically.)

So any two of the three imply the third. Take Friedmann I and the fluid equation, and the acceleration equation follows. Take the acceleration equation and the fluid equation, and Friedmann I follows up to one constant of integration, which is $k$. Only two are independent, and which two you choose is a matter of convenience.

3.4 · The critical density, and a convenient bookkeeping

Equation (3.9.10) has a distinguished value of the density: the one for which the curvature term vanishes. Setting $k=0$ and solving for $\rho$ at time $t$ defines the critical density

and dividing (3.9.10) through by $H^{2}$ turns it into a statement about ratios. Writing $\Omega_{i}\equiv\rho_{i}/\rho_{\text{c}}$ for each contribution to the density, and $\Omega_{k}\equiv-kc^{2}/a^{2}H^{2}$ for the curvature term treated the same way,

This is not a new equation. It is (3.9.10) divided by its own left-hand side. Its usefulness is that the sign of $k$ is now readable off a measurement of densities. A total density above critical makes $\Omega_{k}\lt0$, and therefore $k=+1$. A total density below critical gives $k=-1$. The subscripts anticipate §4, where the three fluids are named.

3.5 · The Einstein static universe, and the pencil on its point

Chapter 3.6 §6.3 told the history. Einstein added the cosmological term in 1917 to permit a static universe, and that chapter's Problem 4(d) argued the balance is unstable by inspecting the signs. It then said in writing that "Chapter 3.9 §3 shows the same thing by integrating the equations rather than arguing from them". Here is the integration.

Find the solution. Take dust of density $\rho_{\text{m}}$, with $p_{\text{m}}=0$, plus the cosmological term, which Chapter 3.6 §6.1 showed is a fluid with $\rho_{\Lambda}=\Lambda c^{2}/8\pi G$ and $p_{\Lambda}=-\rho_{\Lambda}c^{2}$. Static means $\dot a=0$ and $\ddot a=0$, and both Friedmann equations must hold. The acceleration equation (3.9.13) requires the bracket to vanish:

Then Friedmann I with $\dot a=0$ requires $0=\tfrac{8\pi G}{3}(\rho_{\text{m}}+\rho_{\Lambda})-kc^{2}/a^{2}=8\pi G\rho_{\Lambda}-kc^{2}/a^{2}$, which forces $k=+1$. That is a closed universe, with no choice left in the matter. It also fixes the size:

Now perturb it. Write $a(t)=a_{\text{E}}\big(1+\varepsilon(t)\big)$ with $\abs{\varepsilon}\ll1$. Two densities respond differently, and that difference is the whole argument. The dust dilutes with volume, $\rho_{\text{m}}=2\rho_{\Lambda}(a_{\text{E}}/a)^{3}=2\rho_{\Lambda} (1-3\varepsilon)$ to first order, using $(1+\varepsilon)^{-3}\approx1-3\varepsilon$. The cosmological term does not dilute at all, because Chapter 3.6 §6.1(iii) showed its density is built from constants. Substituting both into (3.9.13), whose left side is $\ddot\varepsilon/(1+\varepsilon)\approx\ddot\varepsilon$,

the zeroth-order terms cancelling by (3.9.19), which is the statement that the unperturbed solution really is a solution. The result is $\ddot\varepsilon=+\Lambda c^{2}\varepsilon$, with a plus sign, whose solutions are

Any disturbance grows exponentially. Compare the sign with the one that would have given $\ddot\varepsilon=-\omega^{2}\varepsilon$ and oscillation. Nothing here restores. Physically the runaway is easy to state, and it runs the same way whichever way you push.

• Squeeze the universe slightly. The dust density rises while the cosmological term does not, so gravity wins and the squeeze accelerates.
• Expand it slightly. The dust thins while the cosmological term holds steady, so the repulsion wins and the expansion runs away.

That is Chapter 3.6's Problem 4(d) in words, and (3.9.21) in symbols.

Put a number on it with the measured $\Lambda=1.1\times10^{-52}\ \mathrm{m^{-2}}$ from Chapter 3.6 §6.3. Then $\sqrt{\Lambda}=1.05\times10^{-26}\ \mathrm{m^{-1}}$ and

A disturbance of one part in $10^{9}$ therefore grows to order one in about $20\tau$, some $200\ \mathrm{Gyr}$. That is a long time, and not remotely long enough. The universe is a fifteenth of that age, and its density fluctuations were never that small.

4 · The three fluids, and what each does to $a(t)$

Here is what this section sets out to do. Everything a universe does is decided by one number per constituent: the ratio of its pressure to its energy density. Put that ratio into the fluid equation and it tells us how each constituent dilutes. Put the result into Friedmann I and it gives $a(t)$ for each constituent acting alone. Then we find that because the three dilute at different rates, the order in which they take charge is fixed no matter how much of each there is.

4.1 · One number per fluid, and how each dilutes

Define the equation-of-state parameter

and note that all three values this book needs were derived elsewhere rather than chosen here.

• Dust, meaning matter whose particles move slowly enough that their pressure is negligible beside their rest energy, has $w=0$. Chapter 3.6 §1.3 built it as the $p\to0$ limit of the perfect fluid.
• Radiation has $w=1/3$, which Chapter 3.6 §1.4 derived from the tracelessness of the electromagnetic energy–momentum tensor of Chapter 2.6, and from nothing else.
• The cosmological term has $w=-1$, which Chapter 3.6 §6.1 derived by demanding that $T^{\text{vac}}_{\mu\nu}\propto g_{\mu\nu}$ take the perfect-fluid form.

Put (3.9.24) into the fluid equation (3.9.15) with $w$ constant:

That is separable, Chapter 0.8 §2.1, and what we want out of it is $\rho$ as a function of $a$. Both sides integrate to logarithms, giving $\ln\rho=-3(1+w)\ln a+\text{const}$. Exponentiating,

Fluid	$w$	$\rho\propto$	where $w$ came from
dust (cold matter)	$0$	$a^{-3}$	negligible pressure, 3.6 §1.3
radiation	$+\tfrac13$	$a^{-4}$	tracelessness of $T^{\mu\nu}_{\text{EM}}$, 3.6 §1.4
cosmological term	$-1$	$a^{0}$, constant	$T_{\mu\nu}\propto g_{\mu\nu}$, 3.6 §6.1

Read the three exponents. Dust falls as $a^{-3}$, which is arithmetic and not physics: the same particles in a volume that has grown by $a^{3}$. Radiation falls one power faster, and that extra factor of $1/a$ is a genuinely relativistic effect. Each photon's energy falls as $1/a$ as well, which §5.1 derives from a null geodesic and which is the cosmological redshift. The number of photons per unit volume falls as $a^{-3}$ like anything else. Their individual energies supply the fourth power. And the cosmological term does not dilute at all, because Chapter 3.6 §6.1 showed its density is $\Lambda c^{2}/8\pi G$, built from constants.

4.2 · $a(t)$ for one fluid at a time

Take $k=0$ for the moment, since §4.3 explains why that is the right first case. What we are after is $a(t)$, so put (3.9.26) into Friedmann I:

We want $a$ on one side and $t$ on the other, so take the square root and separate. Writing $\alpha^{2}$ for the constant,

choosing the integration constant so that $a=0$ at $t=0$, which is a choice of the origin of time and not an extra assumption. Raising both sides to the power $2/(3+3w)$,

Two of the three cases follow at once: $w=0$ gives $a\propto t^{2/3}$ and $w=1/3$ gives $a\propto t^{1/2}$. The third case has to be done separately, because $w=-1$ makes the exponent infinite. There $\rho$ is a constant $\rho_{\Lambda}$, so Friedmann I reads $\dot a/a=\sqrt{8\pi G\rho_{\Lambda}/3}\equiv H_{\Lambda}$, a constant, which is the equation of Chapter 0.8 §2.1's very first example:

This is de Sitter expansion: growth by a fixed factor per unit time, never slowing, never stopping, with no beginning at $a=0$ at any finite time in the past.

Fluid	$w$	$\rho\propto$	$a(t)\propto$	$\ddot a$	at $t\to0$
radiation	$+\tfrac13$	$a^{-4}$	$t^{1/2}$	negative	$a\to0$
dust	$0$	$a^{-3}$	$t^{2/3}$	negative	$a\to0$
stiff fluid, $p=\rho c^{2}$	$+1$	$a^{-6}$	$t^{1/3}$	negative	$a\to0$
cosmological term	$-1$	constant	$\ee^{H_{\Lambda}t}$	positive	no zero

4.3 · Which fluid is in charge, and when the expansion accelerates

A real universe contains all three, and the total density is the sum. Because they dilute at different rates, the sum is dominated by different terms at different times, and the order is fixed:

As $a\to0$ the steepest term wins, so radiation dominates the early universe whatever the proportions are. As $a\to\infty$ the shallowest wins, so the cosmological term dominates the late universe whatever the proportions are, provided it is not zero. Matter has its era in between. That running order is not an observation. It is arithmetic, and no measurement could have come out otherwise.

The curvature term is worth putting on the same footing. Written as a contribution to (3.9.10) it behaves as $a^{-2}$, which puts it between matter and $\Lambda$. That is why §4.2's $k=0$ assumption is the right first case: whatever $k$ is, at small enough $a$ the curvature term is negligible beside radiation and matter.

When does the expansion accelerate? Read (3.9.13): $\ddot a$ has the sign of $-(\rho+3p/c^{2})$, and with $p=w\rho c^{2}$ that bracket is $\rho(1+3w)$. So a single fluid accelerates the expansion exactly when

which is precisely Chapter 3.6 §5.5's condition for a substance to repel rather than attract, arriving here with no weak-field approximation. That chapter's figure had a shaded region below $w=-1/3$ and said the observed universe sits at $w\approx-1$. Here is what sitting there does. Note also what (3.9.32) says about the other end of the table: radiation, with $w=+1/3$, has a bracket of $2\rho$ and decelerates the expansion twice as hard as dust of the same energy density. That is Chapter 3.6 §5.5's "a gas of light gravitates twice as strongly", which that section flagged as something this chapter would need. It is needed here.

4.4 · The numbers, the exact solution, and the age

The critical density, from (3.9.17), with every conversion on the page. One megaparsec is $3.0857\times10^{22}\ \mathrm{m}$, so

using $1\ \mathrm{Gyr}=3.156\times10^{16}\ \mathrm{s}$. Now put that value of $H_{0}$ into the definition (3.9.17) to get the critical density itself in kilograms per cubic metre:

which is $5.1$ hydrogen atoms per cubic metre. The whole universe, averaged, is a vacuum far better than any laboratory has made.

The transitions. Set two terms of (3.9.31) equal to each other, and use $1+z=a_{0}/a$ from §5.1. Three moments come out of it.

• Radiation gave way to matter at $a/a_{0}=\Omega_{\text{r}}/\Omega_{\text{m}}=2.9\times10^{-4}$, that is $z\approx3400$.
• Matter gave way to $\Lambda$ at $a/a_{0}=(\Omega_{\text{m}}/\Omega_{\Lambda})^{1/3}=0.77$, that is $z=0.30$.
• By (3.9.32) applied to the mixture, the expansion stopped decelerating when $\rho_{\text{m}}=2\rho_{\Lambda}$, at $a/a_{0}=(\Omega_{\text{m}}/2\Omega_{\Lambda})^{1/3}=0.61$, that is $z=0.63$.

The acceleration is recent, and the coincidence that it began at all near the present epoch has no explanation in this book or any other.

The exact solution, since two of the three fluids can be handled together. Neglect radiation, which matters only for the first few thousand years, and take $k=0$. Friedmann I becomes

which looks unpleasant until one substitution tidies it. Put $u\equiv a^{3/2}$, so that $\dot u=\tfrac32a^{1/2}\dot a$ and $\dot u^{2}=\tfrac94a\dot a^{2}$. Multiplying (3.9.35) by $\tfrac94 a$ turns the awkward $C/a$ into a constant:

a separable equation. Separating gives $\dd u/\sqrt{H_{\Lambda}^{2}u^{2}+C}=\tfrac32\,\dd t$, and the substitution $u=\big(\sqrt C/H_{\Lambda}\big)\sinh\psi$ turns the square root into $\sqrt{C}\cosh\psi$ by $\cosh^{2}-\sinh^{2}=1$, while $\dd u$ supplies a matching $\big(\sqrt C/H_{\Lambda}\big)\cosh\psi\,\dd\psi$. The two cancel, and the left side is $\dd\psi/H_{\Lambda}$. So $\psi=\tfrac32H_{\Lambda}t$ with the origin of $t$ at $u=0$. Undoing both substitutions,

Check both ends. For small argument $\sinh x\approx x$, so $a\propto t^{2/3}$, which is the matter era of (3.9.29). For large argument $\sinh x\approx\ee^{x}/2$, so $a\propto\ee^{H_{\Lambda}t}$, which is the de Sitter era of (3.9.30). One formula contains both, and the crossover is where the argument is of order one.

The age. Set $a=a_{0}$ in (3.9.37). Then $\sinh^{2}(3H_{\Lambda}t_{0}/2)=\rho_{\Lambda}/\rho_{\text{m},0}=\Omega_{\Lambda}/\Omega_{\text{m}}$, and with $H_{\Lambda}=H_{0}\sqrt{\Omega_{\Lambda}}$ from (3.9.30) divided by (3.9.17),

Now the arithmetic, which is what turns that formula into a number of years. We have $\sqrt{0.685/0.315}=1.475$, and $\mathrm{arsinh}(1.475)=\ln\big(1.475+\sqrt{1+1.475^{2}}\big)=\ln(3.257)=1.181$. Substituting those, together with $\sqrt{\Omega_{\Lambda}}=0.828$,

which sits below $1/H_{0}=14.5\ \mathrm{Gyr}$. That is a coincidence here rather than a guarantee, because (3.9.14) assumed $p\ge0$ and this universe is $68.5\%$ cosmological term, for which $p\lt0$. Hold the universe flat and push $\Omega_{\Lambda}$ above about $0.74$, and $t_{0}$ exceeds $1/H_{0}$. The bound is real, and this universe is not one of the universes it covers.

Now repeat the whole calculation with $H_{0}=73.0$, and the answer becomes $12.7\ \mathrm{Gyr}$. Set that against the oldest globular clusters, which are dated at around $13\ \mathrm{Gyr}$ by stellar-evolution modelling, ⚑ quoted. A universe younger than the stars inside it is one reason the higher value is uncomfortable, and one reason the disagreement is taken seriously rather than averaged away.

4.5 · The whole history, on one plot

The three eras are hard to see on a linear plot, because the scale factor changes by a factor of $10^{9}$ over the interval where anything interesting happens. On logarithmic axes each single-fluid solution of (3.9.29) is a straight line of slope $2/3(1+w)$, so the eras appear as straight segments and the transitions as bends. The figure integrates (3.9.10) numerically with all four terms of (3.9.31) present, and reports the local slope $\dd\ln a/\dd\ln t$ at a marker you can drag.

Ω_m

0.315

Ω_Λ

0.685

marker a / a₀

10^-0.30

concordance

matter only, critical

no Λ

closed, recollapsing

Omega_k = 1 - Or - Om - OL = -0.0001 expands forever.

marker: a/a0 = 5.01e-1 t = 5.860e+9 yr d ln a / d ln t = 0.721

in charge there: matter (radiation 0.0%, matter 78.5%, curvature 0.0%, Lambda 21.5%) age at a = a0: 13.79 Gyr

5 · Redshift, distance, and what "expanding" does not mean

Here is where this section is heading. We follow a photon along an actual null geodesic of (3.9.7) and find how the frequency a comoving observer measures changes. Then we do it a second way, by counting wave crests, and the two agree. Then we say what the result does not mean. That last part occupies more space than the derivation itself, because it repairs more damage.

5.1 · The redshift, from the geodesic equation

A photon has a four-momentum $p^{\mu}=\dd x^{\mu}/\dd\lambda$ with $\lambda$ an affine parameter, which Chapter 3.8 §1.1 explained is what stands in for proper time on a null curve. It obeys the geodesic equation of Chapter 3.3 §8,

and we need only its $0$ component. From grind box B, $\Gamma^{0}{}_{00}=0$ and $\Gamma^{0}{}_{0i}=0$, so the sum runs over spatial indices alone, and there $\Gamma^{0}{}_{ij}=-(a'/a)g_{ij}$. Hence

Now use the one property of a photon that distinguishes it. The null condition $g_{\mu\nu}p^{\mu}p^{\nu}=0$ reads $(p^{0})^{2}+g_{ij}p^{i}p^{j}=0$, since $g_{00}=1$, so the spatial sum is $-\,(p^{0})^{2}$. Substituting into (3.9.41) and then into (3.9.40),

Trade the affine parameter for the coordinate, which is legitimate because $p^{0}=\dd x^{0}/\dd\lambda$ and $p^{0}\neq0$ for a photon that is going anywhere: by the chain rule $\dd p^{0}/\dd\lambda=(\dd p^{0}/\dd x^{0})\,p^{0}$, and one factor of $p^{0}$ cancels from each side:

Now convert that into something measurable, using Chapter 3.8 §5.1's rule verbatim: the energy an observer measures is the photon's momentum contracted with that observer's four-velocity. A comoving observer has $u^{\mu}=(c,0,0,0)$, so

Defining the redshift $z$ by $1+z\equiv\lambda_{\text{obs}}/\lambda_{\text{em}}$, and using $\lambda\nu=c$ for both,

A measured redshift is a direct reading of how much the universe has grown since the light left. Light reaching us at $z=1$ set out when everything was half its present separation. The microwave background, at $z\approx1100$ ⚑, set out when everything was a thousandth of its present separation. That figure is a measurement of when the plasma recombined. It needs atomic physics Part IV has not reached, so it is not derived here.

The same result a second way, sharing no step. Take a radial ray, $\dd\theta=\dd\varphi=0$, and set $\dd s^{2}=0$ in (3.9.7):

the right-hand side being a fixed number, since the source sits at fixed comoving $r$. Now send the next wave crest a time $\delta t_{\text{em}}$ later. It arrives a time $\delta t_{0}$ later, and the right-hand side of (3.9.46) is the same for both, because it does not contain $t$ at all. Subtracting the two integrals leaves only the end contributions,

which is (3.9.45) again. The first route used the geodesic equation, the connection and a four-velocity. The second used only the line element and the observation that $\chi$ does not depend on when you leave. Two derivations, one answer.

5.2 · It is not a Doppler shift, and here is the evidence

Look at what is absent from (3.9.45). No velocity appears. No relative speed of emitter and observer appears. Nothing about the expansion history between the two events appears either: two universes with wildly different $a(t)$ in between, agreeing only on the ratio of the endpoints, produce identical redshifts. A Doppler shift is none of those things. Chapter 2.5 §7's formula is a function of one relative velocity and nothing else.

The confusion has a respectable origin, and it is worth deriving so that its limits are visible. For a nearby source, expand $a(t_{\text{em}})$ about the present: $a_{\text{em}}\approx a_{0}\big[1-H_{0}(t_{0}-t_{\text{em}})\big]$. For small light-travel times the distance is $d\approx c(t_{0}-t_{\text{em}})$, so from (3.9.45),

That is Hubble's law, and the left-hand side has the units of a velocity, which is why it is universally reported as one. But (3.9.48) is the first term of a Taylor expansion of (3.9.45), and the resemblance to a Doppler formula is a fact about first-order expansions rather than about physics. At $z=7$ there is no velocity of $7c$. There is no velocity at all.

5.3 · Galaxies are not moving through space

Two facts, both computed rather than asserted.

Comoving worldlines are geodesics. Grind box B recorded that every $\Gamma^{i}{}_{00}$ vanishes. The geodesic equation for a curve with $u^{\mu}=(c,0,0,0)$ is $\dd u^{i}/\dd\tau+\Gamma^{i}{}_{\mu\nu}u^{\mu}u^{\nu}=c^{2}\Gamma^{i}{}_{00}=0$, satisfied identically. A galaxy placed at rest in comoving coordinates stays at those coordinates for ever, in free fall, with nothing acting on it. It is not being dragged, pushed or carried.

What grows is the conversion from coordinates to distance. Define the proper distance to a comoving source at one instant of cosmic time by integrating along the slice, which means setting $\dd t=0$ in (3.9.7):

with $\chi$ the fixed comoving distance of (3.9.46). Differentiating, since $\chi$ is a constant,

This is Hubble's law again, and now it is exact rather than a first-order expansion, because it is a statement about proper distance at fixed $t$ rather than about a redshift. And it has a consequence that stops people: for $d\gt c/H$, (3.9.50) gives $\dot d\gt c$. With $H_{0}=67.4$, that threshold is

and galaxies are catalogued well beyond it.

Why that violates nothing. Chapter 2.3 established a speed limit, and Chapter 3.2 §5 established what kind of statement it is. A velocity is a vector in the tangent space at one event. Comparing two velocities means comparing two vectors, and two vectors at different points of a manifold live in different vector spaces with no canonical map between them. That was the single most important sentence of Chapter 3.2, and this is where it is spent. The quantity $\dot d$ in (3.9.50) is not anybody's velocity: it is the rate of change of an integral taken along a curve through many tangent spaces. No local measurement anywhere finds anything moving past anything else faster than $c$, and that is the statement the speed limit makes.

5.4 · The balloon, and the exact point where it fails

The standard picture is dots on an inflating balloon. It gets one thing exactly right, and an analogy this book keeps has to be shown to be tight where it is tight.

What is right, and it is a theorem rather than a resemblance. Two dots on a sphere of radius $R(t)$ at fixed angular separation $\vartheta$ are a geodesic distance $s=R\vartheta$ apart, so $\dot s=\dot R\vartheta=(\dot R/R)s$. Every dot sees every other receding at a rate proportional to its distance, with the same constant of proportionality, and no dot is the centre. That is (3.9.50) exactly, one dimension down, with $R$ playing the part of $a$. The surface has no edge and no special point, and both of those are properties the $k=+1$ case genuinely has.

Where it breaks, and the break is not a detail. The rubber sits in a room. That room is the three-dimensional space the balloon is embedded in, and it is what makes the phrase "expanding into" meaningful. It is also what makes the question "what is outside?" answerable. There is no room. Nothing in §2 referred to an embedding at any point: Chapter 3.2 §1 forbade any reference to an ambient space and built the manifold so that curvature could be measured from inside. The whole of (3.9.7) was obtained by solving equations for the metric components alone. A universe is not a surface in something. The moment you start asking what lies beyond the balloon you have imported an object the mathematics does not contain, and every subsequent question you ask will be unanswerable for that reason and no other.

Two smaller failures follow from the same source.

• The rubber is a material, with tension, and it stretches things stuck to it. The scale factor is a component of a metric and stretches nothing, as §5.3 computed.
• The balloon is a closed surface, so it can only ever depict the $k=+1$ case. For $k=0$ and $k=-1$ the slices are infinite, and there is no balloon to inflate.

5.5 · What does not expand

The commonest error in reading this chapter is applying its results to something small. Everything above was derived from the assumption of §1, and that assumption is violated by many orders of magnitude inside anything bound. The mean density of the Galaxy is around $10^{5}$ times critical. For the Solar System it is vastly more, and for your body it is about $10^{29}$ times.

What governs a bound system is not (3.9.7) but the solution Chapter 3.7 derived, and Chapter 3.7 §4's Birkhoff theorem is what licenses saying so: the geometry outside a spherical mass is Schwarzschild whatever the rest of the universe is doing. Atoms, rulers, planetary orbits and galaxies do not grow. ⚑ The cosmological term does exert a small outward tendency on a bound orbit, and that much is quoted, since it needs a solution this book has not built. Chapter 3.6's Problem 4(b) computed where it starts to matter, which is at around $100\ \mathrm{pc}$ for a solar mass, far outside any bound system. Below that scale the effect exists and is unmeasurable.

6 · Energy is not conserved, and why that is Noether rather than a paradox

This is the section three separate chapters wrote down in advance, and the whole of it is one theorem applied honestly. Here is where it is going. Chapter 1.4 proved that a conserved quantity is the output of an argument whose input is a continuous symmetry. Chapter 3.5 §9 showed that in a curved spacetime the symmetry in question is a Killing vector, and that the two theorems are one theorem. We now prove that an expanding universe containing matter has no timelike Killing vector. The machine is therefore handed nothing and returns nothing, and the photon gas loses energy with nothing to catch it.

6.1 · What the theorem requires, restated

Chapter 1.4 §2 proved Noether's theorem and its §3 specialised it: if the Lagrangian does not depend explicitly on time, then one quantity, the energy, is constant along every solution.

That chapter's §4 was devoted to what happens when a hypothesis fails. Its §4.3, headed "An expanding universe: no conserved total energy", applied the failure to this case in advance. It said in writing: "Chapter 3.5 makes this precise: the symmetries of a spacetime are its Killing vector fields, a conserved energy requires a timelike one, and the expanding solutions of Chapter 3.9 do not have one."

Chapter 3.5 §9 supplied exactly that machinery. Its §9.1 proved in four lines that $Q=\xi_{\mu}u^{\mu}$ is constant along every geodesic when $\xi$ is Killing. Its §9.2 then ran Chapter 1.4's theorem on the same system and got the same charge back, with the change in the Lagrangian under the shift turning out to be the Lie derivative of the metric. The geometric condition and the mechanical one are the same condition. Chapter 3.7 §5 spent that result to get $E$ and $L$ for a planet, and Chapter 3.8 §5.1 spent it to get the gravitational redshift. Both spent it on a geometry with a timelike Killing vector.

So the question is not philosophical. It is: does the FLRW metric have a timelike Killing vector?

6.2 · The obvious candidate, and then the proof that no candidate works

The obvious candidate fails immediately. Try $\xi^{\mu}=(1,0,0,0)$, the vector pointing along cosmic time. Chapter 3.5 §8.1 gives the Lie derivative of the metric as $(\mathcal{L}_{\xi}g)_{\mu\nu}=\xi^{\lambda}\partial_{\lambda}g_{\mu\nu}+g_{\lambda\nu}\partial_{\mu}\xi^{\lambda}+g_{\mu\lambda}\partial_{\nu}\xi^{\lambda}$, and with constant components the last two terms vanish, leaving

which vanishes if and only if $a'=0$. Time translation is a symmetry exactly when the universe is static, and not otherwise. That is Chapter 3.5 §8.1's practical test read the other way round. The test says that a coordinate the metric does not mention gives a Killing vector. Here the metric does mention $t$, through $a(t)$, so it gives none.

But that rules out one candidate, not all of them. A Killing vector need not have constant components, and a timelike one need not point along the cosmic-time coordinate. Here is the general argument, and it takes three steps.

Step 1: a Killing flow cannot change any scalar built from the geometry. Chapter 3.5 §7.1 defined the flow of a vector field, and $\mathcal{L}_{\xi}g=0$ says the metric is unchanged when dragged along $\xi$'s flow. Now take any scalar $S$ computed from the metric by a fixed recipe applied at each point. The Ricci scalar $R$ of Chapter 3.4 §6 will do, or the Kretschmann scalar of Chapter 3.8 §6.4. Now drag the whole configuration along the flow by a parameter $s$. The recipe is the same recipe everywhere, and it is applied to a metric that the dragging left alone, so it returns the value it returned before. The scalar's value at the dragged point therefore equals its value at the original point, for every $s$. Differentiating that at $s=0$,

Step 2: produce a scalar that changes with time. Take the trace of the field equation. In four dimensions $g^{\mu\nu}G_{\mu\nu}=R-\half R\cdot4=-R$, and Chapter 3.6's $T=g_{\mu\nu}T^{\mu\nu}=\rho c^{2}-3p$ gives the other side, so

For dust this is $R=-8\pi G\rho/c^{2}$ with $\rho\propto a^{-3}$. For radiation the bracket vanishes identically, so $R=0$ and this particular scalar says nothing. A second scalar covers that case. The Kretschmann scalar of Chapter 3.8 §6.4, computed from grind box B's connection in exactly the way that chapter computed it for Schwarzschild, is $K=\tfrac32\,(ct)^{-4}$ in the flat radiation era, which is not constant either. Now take the dust case, which is the universe from $z\approx3400$ until recently. There $R$ is a strictly monotonic function of $t$ alone whenever $\dot a\neq0$. It is large early and small late, and it takes a different value on every slice.

Step 3: combine. Apply (3.9.53) with $S=R$. Since $R$ depends on $x^{0}$ alone, $\xi^{\mu}\partial_{\mu}R=\xi^{0}\,\dd R/\dd x^{0}$, and the derivative is non-zero, so

A vector with no time component has norm $g_{\mu\nu}\xi^{\mu}\xi^{\nu}=g_{ij}\xi^{i}\xi^{j}$, and the spatial block of (3.9.7) is negative-definite, so that norm is negative unless $\xi$ vanishes. By Chapter 2.3 §4's classification, every Killing vector of an expanding dust universe is spacelike. There are six of them, as §1.1 assumed, and not one of them is timelike. $\blacksquare$

6.3 · Therefore there is no conserved energy, and this is not a defect

Put the pieces together in the order they were proved. A conserved energy is the output of Chapter 1.4's theorem. That theorem's input is a continuous symmetry of the action. Chapter 3.5 §9 identified the relevant symmetry as a timelike Killing vector. Section 6.2 proved there is none. So no conserved energy can be produced this way.

On its own that would be a shrug rather than a result. What turns it into a result is that the theorem runs backwards as well. Chapter 1.4 §7.2 proved the converse: a conserved $Q$ with no explicit time dependence satisfies $\{Q,H\}=0$ and therefore is a symmetry, viewed as a generator.

Now run the chain in that direction. A conserved energy would generate a time translation. That translation would be a timelike Killing vector. And §6.2 proved there is none. So there is no conserved total energy of an expanding universe, and it is the converse, not the failure of the forward theorem, that says so.

One matter of scope, because it is easy to claim too much here. Even the converse is a statement about the Hamiltonian systems Chapter 1.4 worked with. What rules out a global energy for a whole universe is §6.5's second and third tiers. And it remains true, alongside all of that, that a theorem handed no hypothesis returns no conclusion.

Chapter 1.1 §2 saw this coming, three parts ago, before Noether's theorem had even been stated: "in an expanding universe the laws are time-dependent, and the energy of the cosmological photon gas is correspondingly not conserved… A quantity whose conservation you have been taught is absolute turns out to be contingent on a symmetry that the universe does not exactly possess." That sentence is now a calculation.

6.4 · The photon gas, in numbers

Take a region that expands along with the matter, and call it a comoving box. It has fixed coordinate volume $V_{\text{c}}$, and therefore proper volume $V=a^{3}V_{\text{c}}$. The energy of the radiation inside it is its energy density times its volume:

Do the same sum for the other two fluids and the contrast is the point of the section:

Fluid	$\rho\propto$	energy in a comoving box	as $a$ doubles
dust	$a^{-3}$	constant	unchanged
radiation	$a^{-4}$	$\propto1/a$	halves
cosmological term	constant	$\propto a^{3}$	increases eightfold

Numbers, using §4.4's $\Omega_{\gamma}\approx5.4\times10^{-5}$ and (3.9.34)'s critical density. The energy density of the microwave background today is

Take a comoving box that is one cubic metre now, and run it back to last scattering at $z\approx1100$, that is $a/a_{0}=1/1101$. Its proper volume then was $(1101)^{-3}\ \mathrm{m^{3}}$ and the energy density in it was $(1101)^{4}u_{0}$, so the energy it contained was $1101\,u_{0}$, that is $4.5\times10^{-11}\ \mathrm{J}$ against $4.1\times10^{-14}\ \mathrm{J}$ now. Ninety-nine point nine one per cent of the energy of the cosmic microwave background has gone, and nothing has it.

Three places you will instinctively look for it, and why none of them holds it.

Not the matter. Each non-interacting fluid obeys (3.9.15) separately, and that equation has no source term coupling one to another. Transfer would appear as such a term, and there is none.

Not the gravitational field. There is no local energy density for gravity to store it in, and the reason is the equivalence principle. Chapter 3.4 §5.3 constructed locally inertial coordinates in which the metric is $\eta_{\mu\nu}$ and its first derivatives vanish at a chosen event. Now take any candidate local energy density built from the metric and its first derivatives, and required to vanish where there is no gravitational field at all. In those coordinates, at that event, it is zero. And a scalar that vanishes in one chart at an event vanishes in every chart there. So no such scalar exists to be non-zero anywhere.

Not "the expansion". The expansion is not an object with an energy account. Section 5.3 showed that nothing is moving, nothing is being stretched, and $a$ is a metric component.

6.5 · What does survive, in three tiers

Chapter 1.4 §4.3 set out three claims in decreasing order of confidence and marked all three as quoted forward to here. Each can now be checked.

Tier one, exactly and always true: the local statement. $\nabla_{\mu}T^{\mu\nu}=0$ holds, and in this spacetime it is not an extra assumption but a consequence of the field equations, by §3.3. More than that: you have already written it down. The $\nu=0$ component of $\nabla_{\mu}T^{\mu\nu}=0$ for the FLRW metric is exactly the fluid equation (3.9.15), up to a factor of $c^{2}$, and the spatial components vanish identically. Nothing is created or destroyed at any point.

Tier two: why that cannot be added up. Chapter 3.5 §6.4 proved a divergence identity, $\nabla_{\mu}V^{\mu}=\partial_{\mu}(\sqrt{-g}V^{\mu})/\sqrt{-g}$. It turns the covariant divergence of a vector into an ordinary derivative, so that integrating it gives a boundary term and hence a conservation law. For a tensor with a free index the same manipulation leaves a remainder:

the extra term coming from the second index, which the connection also has to correct. Evaluate that remainder here, for $\nu=0$. Using $\Gamma^{0}{}_{ij}=-(a'/a)g_{ij}$ and $T^{ij}=-p\,g^{ij}$, so that $g_{ij}T^{ij}=-3p$,

The obstruction is the pressure. Set $p=0$ and it vanishes, the divergence becomes a total derivative, and dust's energy in a comoving box is conserved, which is the first row of the table above. With pressure it does not vanish, and there is no boundary term to collect. The second obstruction is Chapter 3.2's: $\nu$ is a free index, so the thing one would like to add up over a region is a vector at each point, and vectors at different points of a manifold cannot be added.

Tier three: when a total energy does exist. ⚑ In an asymptotically flat spacetime, such as an isolated star, or Chapter 3.8's black hole in an otherwise empty universe, the geometry approaches Minkowski far away, the time-translation symmetry is restored there, and a conserved total energy can be defined from the behaviour of the field at infinity. It is called the ADM mass. We quote its existence and do not construct it. The expanding universe is not asymptotically flat, in the strong sense that it is not asymptotically anything: (3.9.7) is the same everywhere, and there is no far away.

So neither slogan is right. "Energy conservation fails in general relativity" is too strong, because Chapter 3.8's black hole has a perfectly good conserved energy. "Energy is always conserved, you have to look harder" is false, because §6.2 is a proof and not an admission of ignorance. The correct statement is the one Noether's theorem hands you directly: you get a conserved energy exactly when you have a time-translation symmetry, and not otherwise.

7 · The loose thread

This is the last section of Part III, and it does two things that no other section of this book is in a position to do.

The first is to name the one number the book leaves deliberately unpaid, and to say exactly what is owed and who collects it. The second is to tell you what the next three parts are going to do with everything Part III has built. That second thing is said here, in advance, rather than offered later as an apology.

7.1 · One number, owed

Put a number on it before saying what is strange, because the number is part of what is strange. For a hole of one solar mass, Chapter 3.7 gave $r_{s}=2GM_{\odot}/c^{2}=2953\ \mathrm{m}$ using the measured $GM_{\odot}=1.32712440018\times10^{20}\ \mathrm{m^{3}s^{-2}}$, so $A=4\pi r_{s}^{2}=1.096\times10^{8}\ \mathrm{m^{2}}$. With $\ell_{\text{P}}^{2}=\hbar G/c^{3}=2.612\times10^{-70}\ \mathrm{m^{2}}$,

Set that beside the number of particles involved. Divide the measured $GM_{\odot}$ by the measured $G=6.674\times10^{-11}$, noting in passing that it is $G$ that limits the precision, as Chapter 3.7 warned. That gives $M_{\odot}=1.99\times10^{30}\ \mathrm{kg}$, which is $1.19\times10^{57}$ protons' worth of mass. So the entropy is about $10^{20}$ times Boltzmann's constant per nucleon. Chapter 0.6's Problem 3, which derived $S=k_{\text{B}}\ln W$ from a maximisation, then says the hole has around $\ee^{10^{77}}$ distinguishable microscopic configurations.

Now what is strange, in three parts.

It is an area. Entropy counts configurations, configurations are built from degrees of freedom, and the degrees of freedom of everything else in this book scale with volume: double the linear size of a box of gas and you get eight times the entropy, not four. Here the answer scales with the boundary of the region rather than its interior. Whatever the microscopic description of a gravitating region is, it has as many degrees of freedom as a sheet, not as a solid.

It contains $\hbar$ and $G$ together. Set $\hbar\to0$ and $S$ diverges. Set $G\to0$ and it diverges too. So it belongs to neither quantum mechanics nor general relativity separately, and it is one of very few numbers in physics of which that is true. This chapter's equations and Chapter 3.8's have no $\hbar$ in them anywhere.

And general relativity's own answer is zero. This is the sharpest way to say what is owed. Chapter 3.7 §4 proved Birkhoff's theorem: the geometry outside a spherical mass is determined completely by one number. Chapter 3.8 §6 found nothing else to measure. So classically a black hole of mass $M$ has exactly one configuration, $W=1$, and $S=k_{\text{B}}\ln1=0$. The debt is therefore not a missing constant but a missing list: something has $10^{77}$ states, and general relativity says it has one.

7.2 · Who pays, and with what

Two pointers, both already written into this book.

Chapter 3.6 §4.5, discarding the boundary term in the variation of the Einstein–Hilbert action, flagged the Gibbons–Hawking–York term and said in writing: "It stops being ignorable exactly when the value of the action itself is the thing one wants… the outstanding case is the thermodynamics of black holes, where the numerical value of the gravitational action supplies the entropy. Chapter 3.9 §7 quotes that entropy, $S=A/4$ in appropriate units, and says the derivation is beyond this book." Here is the quotation, and here is the confirmation. The classical part of the calculation does run through that boundary term. This book did not build the extrinsic curvature that term needs, because Chapter 3.2 §1 forbade any reference to an ambient space.

The other half is worse. Getting $A/4\ell_{\text{P}}^{2}$ out of $\ln W$ requires a list of microscopic states to count, and no theory in Parts 0 to VI supplies one. Chapter 7.9 is where this is collected, and the count is done in string theory. It is made for a particular family of charged black holes, by counting the configurations of a bound system of branes at weak coupling and following the count to strong coupling, where the same object is a black hole. It produces $A/4\ell_{\text{P}}^{2}$ with the factor of four included and nothing adjusted. That is the single most persuasive quantitative result the subject has, and it is why Part VII exists in this book at all. GAPS.md lists this as the book's one deliberate loose thread, and it is deliberate: the number is quoted here rather than hidden, so that you can see the size of what is missing.

7.3 · What the next three parts do with gravity, said in advance

Here is the thing that ought to be said before it happens rather than excused afterwards. Parts IV, V and VI are done entirely on flat spacetime. Gravity does not appear again until Chapter 7.1. Thirty chapters, after nine spent establishing that spacetime is dynamical and that the metric is a field with an equation of motion of its own.

That is not the book quietly abandoning its own result. It is an accurate report of the state of physics, and it can be justified with a number.

Chapter 3.4 §4.5 computed the curvature at the Earth's surface as $GM_{\oplus}/c^{2}r^{3}=1.7\times10^{-23}\ \mathrm{m^{-2}}$. That is a radius of curvature of $2\times10^{11}$ metres, which is why nobody noticed it for three centuries. A curvature is one over a length squared, so its effect on a system of size $\ell$ is of order (curvature)$\times\ell^{2}$. For a hydrogen atom, $\ell\approx5.3\times10^{-11}\ \mathrm{m}$, and

Set that beside the precision of the best-tested prediction in Part V, the magnetic moment of the electron, which ⚑ agrees with experiment to about one part in $10^{13}$. The correction from spacetime curvature is thirty orders of magnitude below the error bars. The same comparison made with forces rather than lengths: the gravitational attraction between two protons, divided by their electrostatic repulsion, is

Flat spacetime is therefore not an approximation of convenience. It is an approximation with a computed error, and the error is smaller than any measurement made on a particle has ever been able to resolve. Part II's Minkowski spacetime is the arena for Parts IV to VI because that is what the measurements say. It is also easier mathematics, but that is not the reason.

And now say where it stops being legitimate, since a bound is worth nothing without the place it fails. Four places, three of which this part has already reached.

(i) The centre of a black hole. Chapter 3.8 §8.2 computed that the curvature radius $K^{-1/4}$ falls through the Planck length $\ell_{\text{P}}=1.6\times10^{-35}\ \mathrm{m}$ at a finite radius, and said that a classical smooth metric is not a description anybody has reason to believe below that scale.

(ii) The first instants of the expansion. Run (3.9.29) backwards. As $a\to0$ the density diverges and the curvature with it, and the same Planck scale is reached at a cosmic time of order $\sqrt{\hbar G/c^{5}}=5.4\times10^{-44}\ \mathrm{s}$. Everything this chapter derived assumed a classical metric, so everything this chapter derived stops applying there. The equations produce their own boundary of validity, exactly as Chapter 3.8's did, and by the same mechanism.

(iii) Black-hole entropy. Section 7.1 is itself an instance: a classical object turns out to have a thermodynamics whose formula contains $\hbar$.

(iv) The cosmological constant. Chapter 3.6 §6.3 called its measured value the single most embarrassing number in physics: $\Lambda\approx1.1\times10^{-52}\ \mathrm{m^{-2}}$, fantastically small and not zero. Part V will compute the vacuum energy of a quantum field, and the answer will not be that number. Chapter 7.9 accounts for the failure without fixing it.

7.4 · Three things handed forward

Finally, three connections to state now so that Part IV recognises them rather than meeting them cold.

The action you have been extremising is the one that gets exponentiated. Chapter 3.3 §3 kept $S=-mc^{2}\int\dd\tau$ from Part II and changed only how $\dd\tau$ is computed. Every geodesic in Chapters 3.7, 3.8 and this one is a stationary point of it. Chapter 1.2 §5 already told you what happens to it next. A quantum particle going from one event to another contributes an amplitude $\ee^{\ii S/\hbar}$, and the total amplitude is the sum over all paths. The classical path survives because it is where $S$ is stationary, so neighbouring histories stop cancelling there. Chapter 5.6 builds that sum properly for fields. Nothing about the action changes. Only what is done with it does.

Your Killing vectors become quantum numbers. Chapter 3.5 §9 turned a symmetry into a label that does not change along a trajectory. Chapter 0.5 §8 proved that commuting Hermitian operators share an eigenbasis, so that a symmetry of a quantum system supplies a label that does not change either. These are the same move, and Part IV's angular-momentum quantum numbers are Part III's rotational Killing vectors in a different costume. The $\mathfrak{su}(2)$ of Chapter 4.11 is the algebra §1.1 of this chapter used to state isotropy.

The source term becomes a computation. The perfect fluid of Chapter 3.6 §1.3 was a stand-in: a two-parameter caricature of matter, adequate for a universe and useless for anything smaller. Part V computes $T^{\mu\nu}$ for real fields from the Noether construction of Chapter 2.6. When gravity returns in Chapter 7.1, its right-hand side is that object and not this one.

8 · Worked examples

9 · Your turn