Part F · Einstein · spine

The Einstein field equations, twice

Write down what a law of gravity must satisfy, and discover that in four dimensions exactly one equation satisfies it. Then derive the same equation again from an action, and watch the same constant appear.

needs 33, 30~3 h slides + 6 h problemsproblems: ./mr problems 35

Units 31 to 33 built machinery with a promise attached: that the object no coordinate change removes — curvature — is what gravity is. This unit collects on the promise. It does so twice, because two independent routes reaching the same equation with the same constant is the strongest evidence a physical theory ever gets before an experiment.

Two routes, and which one you can read

Route two (sections 3 and 4) needs only units 33 and 30. It argues that the contracted Bianchi identity hands you a divergence-free symmetric second-order tensor, that Lovelock's theorem says in four dimensions there is no other, and that the Newtonian limit then fixes the single remaining constant. It is complete, and it is the argument Einstein himself did not have.

Route one (section 5) needs unit 29. It varies the Einstein-Hilbert action. It is the route that explains where the right-hand side comes from, because unit 29's definition of the stress-energy tensor is a variational derivative with respect to the metric and this is the action it is a derivative of. If you have not done unit 29, section 5 is readable as narrative but its central definition will be an assertion. Skip it without guilt and come back; nothing later in the course depends on it.

Route two is deliberately given first, even though history and most textbooks go the other way. The reason is that route two answers the harder question — why this equation — and route one, taken first, makes it look as though the answer were "because someone wrote down that action".

How to read this deck

One example runs through every section: the weak static field, $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ with a slow dust source. Section 1 uses it to say what has to come out, section 2 gives it a source tensor, section 3 produces a candidate equation with one unknown constant, section 4 computes the constant, section 5 gets the same constant from an action, and section 6 linearises about it to find the waves. There is a number attached at every stage, and all of them are checked.

Conventions, fixed here and not changed again

Signature $(-,+,+,+)$; $x^0 = ct$; Greek indices run $0\ldots3$, Latin $1\ldots3$; Einstein summation. Unit 33's curvature sign: $R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}$, with $R_{\sigma\nu} = R^\mu{}_{\sigma\mu\nu}$ and $R = g^{\mu\nu}R_{\mu\nu}$. Factors of $c$ are kept everywhere: this course has never set $c = 1$ and will not start in the unit where the numbers matter most.

35 · overview

One equation, two roads, one number

route two · forced (sections 1–4)

1. Matter obeys local conservation, $\nabla_\mu T^{\mu\nu} = 0$, with $T$ the stress-energy tensor. So the left side $A_{\mu\nu}$ of $A_{\mu\nu} = \kappa T_{\mu\nu}$ must be divergence-free for every metric.

2. The Bianchi identity says $G_{\mu\nu} = R_{\mu\nu} - \tfrac12Rg_{\mu\nu}$ is one such tensor, and Lovelock's theorem that in four dimensions the only other is $g_{\mu\nu}$. So $G_{\mu\nu} + \Lambda g_{\mu\nu} = \kappa T_{\mu\nu}$.

3. In the weak static field with dust, the $00$ component reads $\nabla^2\Phi/c^2 = \tfrac12\kappa\rho c^2$.

4. That is Poisson's equation, the Newtonian limit, only if $\kappa = 8\pi G/c^4$.

route one · varied (section 5, needs unit 29)

1. The only scalars built from $g$ with at most two derivatives are $R$ and constants: the Einstein-Hilbert action $\int(R - 2\Lambda)\sqrt{-g}$, plus matter.

2. Varying $\sqrt{-g}$ and $R_{\mu\nu}$ with respect to the metric gives $G_{\mu\nu} + \Lambda g_{\mu\nu}$, plus a divergence.

3. Varying the matter part defines $T_{\mu\nu}$. The same equation falls out, and matching Newton gives the same $\kappa$.

The plan. 1 five requirements · 2 the source, for dust, fluid and light · 3 route two · 4 the constant · 5 route one · 6 what it says: curvature in vacuum; waves, with $10 - 4 - 4 = 2$ polarisations (10 metric functions, minus 4 coordinate choices, minus 4 equations that only constrain the initial data); $\Lambda$; energy; the Cauchy problem.

Words with a dotted underline are glossary terms: click for a definition, G for the index, N for the notes, where the long proofs are. A solid underline like the target shows a piece of the running example on hover.

Where each step is done. Route two: step 1 is Requirement four does all the work; step 2 is One divergence-free tensor, already in hand and Lovelock: there was never a choice; step 3 is The weak, slow, static setting (the source side, $R_{00} = \tfrac12\kappa\rho c^2$, after trace reversal) and The Ricci component, to first order (the geometry side, $R_{00} = \nabla^2\Phi/c^2$); step 4 is Poisson appears, and with it the coupling. Route one: step 1 is Which scalar? The Einstein-Hilbert action; step 2 is the two lemma slides; step 3 is The variation, assembled. The count $10 - 4 - 4 = 2$ is Ten minus four minus four, and is checked independently by Linearise: a wave with two polarisations.

Sections 3 and 4 together are a complete derivation on their own. Section 5 is the second, independent one, and it needs unit 29. A learner who has not done unit 29 should read sections 1-4 and 6 and treat section 5 as a preview; the checkpoint is written so that the part of it depending on unit 29 is clearly marked.

The order within section 1 matters. Every one of the five requirements was available to Einstein in 1913 — none of them needs a result later than Riemann and Bianchi — and the remarkable fact is that together they are almost enough. The word almost is doing one piece of work: Lovelock's theorem, which closes the gap, is from 1971. Einstein had to guess what Lovelock proved.

section 1

What the equation has to be

Five requirements, written down before anything is written down.

35 · requirements

Newton's equation cannot be the law

$\nabla^2\Phi = 4\pi G\rho$ is unit 05's field equation for gravity, and it is right to one part in $10^{6}$ everywhere in the solar system. It is also, as a candidate law, broken in three separate ways.

three failures

It is instantaneous. $\Phi$ at time $t$ depends on $\rho$ at time $t$ everywhere. Unit 26 forbids that: "at the same time" is not an invariant statement.
$\rho$ is not a scalar. Mass density is the $00$ component of something, and unit 27 showed energy and momentum mix under a boost. A scalar source is not available.
$\nabla^2$ is not an invariant operator. It is the spatial part of an operator, and which part is spatial depends on the observer.

what survives

The shape survives: second derivatives of the field on the left, matter on the right, linear in the source, one constant of proportionality. Everything in this unit is the search for the tensor equation with that shape.

Unit 30 closed off the easy repair. One cannot keep flat spacetime and make $\Phi$ a field on it, because tidal forces cannot be transformed away and so there is no global inertial frame to define the field on. Unit 33 then supplied the object that measures exactly the thing that could not be removed. The field is the metric, and its second derivatives are curvature.

The "one part in $10^6$" is the size of $\Phi/c^2$ at the surface of the Sun, $GM_\odot/(R_\odot c^2) = 2.12\times10^{-6}$, which is the natural size of the first relativistic correction to any solar-system quantity. Mercury's perihelion advance of 43 arcseconds per century, computed in unit 37, is a correction of exactly this order, accumulated over many orbits.

On the second failure. It is worth being precise about why a scalar source is not merely inconvenient but unavailable. Unit 27 showed that energy density is not the component of a four-vector either: it is $T^{00}$, one component of a rank-two object, because energy density is a density (one factor of a volume, which Lorentz-contracts) of an energy (which is itself a component of a four-vector). Two boosts' worth of transformation means rank two.

There were scalar theories of gravity. Nordstrom's second theory (1913) is a consistent, fully relativistic scalar theory in which the source is the trace $T$ — the only scalar available — and spacetime is conformally flat. It is not wrong on grounds of consistency; it is wrong on grounds of experiment. Its Weyl tensor vanishes, so it predicts no light deflection at all, and it gives Mercury a perihelion regression of about $-7$ arcseconds per century rather than an advance of 43. Einstein knew this theory well; Einstein and Fokker wrote it in geometric form in 1914, which is one of the routes by which he came to believe the metric had to be the field.

35 · requirements

Five requirements, before any equation

what a field equation for gravity must be

A tensor equation. Both sides tensors of the same type on the spacetime manifold, so the statement holds in every chart or none. Unit 31's warning applies: this is weaker than it sounds, and the notes say how much weaker.
Second order in $g$. At most two derivatives of the metric, so that prescribing $g$ and $\partial_t g$ on a spacelike slice determines the future — the shape of initial-value problem unit 01 taught for $m\ddot x = F$, not some higher-order variant with extra data and, generically, runaway solutions.
Sourced by everything that carries energy and momentum. Because $E = mc^2$: the binding energy of a compressed spring weighs, so the source cannot be rest mass alone.
Automatically compatible with local conservation. Matter obeys $\nabla_\mu T^{\mu\nu} = 0$, and the field equation must not be allowed to contradict it.
Newtonian in the weak, slow, static limit. It must reduce to $\nabla^2\Phi = 4\pi G\rho$.

Requirements 1-3 constrain the shape. Requirement 4 all but determines it. Requirement 5 supplies the only number.

On requirement 1, and how weak it is. Kretschmann objected in 1917 that any physical law can be written as a tensor equation if one is willing to add enough auxiliary structure, and he was right. Newton's gravity itself can be: Cartan (1923) wrote it as a statement about a curved connection on spacetime, with an absolute time function and a degenerate spatial metric carried along as extra fields. So "tensorial" rules out nothing by itself. What gives it teeth is the tacit addition and the metric is the only field describing gravity — no absolute time, no background $\eta$, no preferred frame. That addition is a physical hypothesis, and Lovelock's theorem in section 3 is exactly where it is used: its hypothesis is "built from $g$ and nothing else".

On requirement 2. The objection to higher derivatives is not aesthetic. A field equation of fourth order in time needs $g$, $\dot g$, $\ddot g$ and $\dddot g$ as data, and Ostrogradsky's theorem (1850) says that a non-degenerate Lagrangian with more than first time derivatives has a Hamiltonian unbounded below and linear in one momentum (cited; a short modern proof is R. Woodard, Scholarpedia 10 (2015) 32243) — so the system has no ground state and decays into ever wilder motion. That is a genuine instability, not a stylistic preference. It is also not absolute: theories with higher derivatives that appear only as corrections, treated perturbatively, evade it, and that is how quadratic curvature terms are handled in effective field theory.

On requirement 3. The spring is the standard argument and it is worth doing properly. Compress a spring by storing energy $E$; its inertial mass rises by $E/c^2$ (unit 27). By the weak equivalence principle, tested to one part in $10^{15}$ in unit 30, its gravitational mass rises by exactly the same amount. So elastic energy gravitates. There is nothing special about elastic energy, so all energy does — and momentum too, since a boost turns energy into momentum.

On requirement 4. "Automatically" is the strong word. It would be logically possible to write a field equation and then impose $\nabla_\mu T^{\mu\nu} = 0$ as a separate condition. The system would be overdetermined: ten field equations plus four constraints for ten metric components and the matter variables. Generic overdetermined systems have no solutions. The only safe design is that the left-hand side be divergence-free as an identity, so that the field equation implies the conservation law instead of competing with it. That is the observation that picks out the answer, and it is section 3.

35 · requirements

Requirement four does all the work

Suppose the equation has the shape $A_{\mu\nu} = \kappa\,T_{\mu\nu}$, with $A_{\mu\nu}$ built from the metric and $T_{\mu\nu}$ the source.

the constraint

Matter satisfies $\nabla^\mu T_{\mu\nu} = 0$. For the equation to be consistent for every matter distribution, $A_{\mu\nu}$ must satisfy

$$\nabla^\mu A_{\mu\nu} = 0 \quad\text{identically},$$

that is, for every metric whatever, not only for those solving the equation.

why identically

If $\nabla^\mu A_{\mu\nu}$ vanished only on solutions, the four equations $\nabla^\mu A_{\mu\nu} = 0$ would be four extra conditions on $g$ beyond the ten already imposed. Fourteen conditions on ten functions: no solutions in general. $\square$

So the search narrows sharply. We need a tensor that is

symmetric, because $T_{\mu\nu}$ is (visibly so for the sources of section 2; unit 29 proves it in general);
built from $g$ and its first two derivatives;
identically divergence-free.

we already have one

Unit 33 proved the contracted second Bianchi identity, $\nabla^\mu G_{\mu\nu} = 0$, where $G_{\mu\nu} = R_{\mu\nu} - \tfrac12 R\,g_{\mu\nu}$. It was proved there as a fact about curvature with no physics attached. It is the whole of section 3.

Why $\nabla_\mu T^{\mu\nu} = 0$ is not negotiable. In flat spacetime it is unit 13's Noether theorem applied to the four translations — conservation of energy and of the three components of momentum, one component per translation — and unit 29 proves it for fields. The section 2 slide on the non-relativistic shadow checks it for dust directly. In curved spacetime it is what the equivalence principle demands. Go to normal coordinates at a point (unit 32): there $\Gamma = 0$, the laws of special relativity hold, and $\partial_\mu T^{\mu\nu} = 0$. That statement is not tensorial; the tensorial statement that reduces to it is $\nabla_\mu T^{\mu\nu} = 0$, and that is the one that must hold everywhere. This is minimal coupling, unit 32's comma-goes-to-semicolon rule, applied to the one equation it cannot fail on.

The counting argument, stated carefully. The system is: ten equations $A_{\mu\nu} = \kappa T_{\mu\nu}$ for the ten $g_{\mu\nu}$, plus whatever equations the matter obeys. If $\nabla^\mu A_{\mu\nu}$ were some nonzero tensor $B_\nu$ depending on $g$, the field equation would force $\kappa\nabla^\mu T_{\mu\nu} = B_\nu$, contradicting $\nabla^\mu T_{\mu\nu} = 0$ unless $B_\nu$ happened to vanish on that particular metric. Since we may choose the matter freely, we may arrange the metric to be almost anything, so $B_\nu$ must vanish for all metrics. That is the meaning of "identically".

There is a pleasant converse worth flagging now and proving in unit 36's neighbourhood: because $\nabla_\mu T^{\mu\nu} = 0$ follows from the field equations rather than being imposed, the motion of matter is not an independent postulate. Unit 32 mentioned this: a small enough body whose stress-energy is conserved moves on a geodesic. Geodesic motion is a theorem of the field equations (Einstein-Infeld-Hoffmann, 1938), not an axiom beside them. No other classical field theory does this: in electromagnetism the Lorentz force law is separate from Maxwell's equations.

35 · requirements

The running example, and the numbers in it

the weak static field

The example this unit runs on:

$$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu},\qquad |h_{\mu\nu}|\ll1,$$

with $\partial_0 h_{\mu\nu} = 0$, the source slow dust of rest-mass density $\rho$, and unit 32's geodesic result

$$g_{00} = -\Bigl(1 + \frac{2\Phi}{c^2}\Bigr),\qquad h_{00} = -\frac{2\Phi}{c^2}.$$

Is $|h|\ll1$ ever a real approximation? The numbers:

The error of working to first order in $h$ is of relative size $|\Phi|/c^2$: the solar system sits deep in the green band.

First order is accurate to six figures wherever the classical tests are done — why unit 37's numbers can be computed at all, and why neutron stars are a different subject.

The two computed entries: $GM_\oplus/(R_\oplus c^2)$ with $GM_\oplus = 3.986004418\times10^{14}\ \mathrm{m^3s^{-2}}$ and $R_\oplus = 6.378137\times10^6$ m gives $6.953\times10^{-10}$; $GM_\odot/(R_\odot c^2)$ with $GM_\odot = 1.32712440018\times10^{20}$ and $R_\odot = 6.957\times10^{8}$ m gives $2.1225\times10^{-6}$. The white-dwarf figure uses $M = 0.6\,M_\odot$, $R = 9000$ km, giving $9.8\times10^{-5}$, quoted to one figure; the neutron-star figure uses $M = 1.4\,M_\odot$, $R = 12$ km, giving $0.17$, quoted to one figure because the radius is not known better than that.

Why this example and not a more interesting one. It has no symmetry to exploit, no exact solution, and nothing dramatic happens in it. That is the point: it is the one regime in which we independently know the answer — Newton's — so it is the only place a derivation can be checked rather than merely admired. Every constant in general relativity is fixed here. Unit 36's Schwarzschild solution takes its integration constant from this limit; unit 37's three numbers are corrections to it.

Reading the figure. The horizontal scale is logarithmic, ten decades across. The neglected terms in a first-order expansion are of order $h^2$, so relative to what is kept they are of order $|h| \sim |\Phi|/c^2$: one part in $10^{9}$ at the Earth's surface, two in $10^{6}$ at the Sun's. The two bodies on the left are computed in the notes above; the two on the right are quoted to one figure.

35 · requirements

The running example, drawn

Model the Sun as a uniform ball of mass $M$ and radius $R$. Unit 05's Poisson equation gives

$$\Phi = \begin{cases} -\dfrac{GM}{2R^3}\,(3R^2 - r^2), & r \le R,\\[1ex] -\dfrac{GM}{r}, & r \ge R,\end{cases}$$

and the running example's metric component is $h_{00} = -2\Phi/c^2$: the same curve, flipped and doubled.

Inside the curve bends: $\nabla^2\Phi = 4\pi G\rho \ne 0$. Outside it is harmonic, $\nabla^2\Phi = 0$, yet not constant. Section 4 shows $\nabla^2\Phi/c^2$ is a component of the Ricci tensor, so that difference is the whole story of section 6's vacuum slide.

$\Phi/c^2$ for a uniform ball with the Sun's mass and radius. The well is $3\times10^{-6}$ deep — this is the "dip" in the metric that the whole unit perturbs about. The real Sun is centrally condensed, so its true well is deeper at the centre; the surface value is exact.

The profile, checked. Outside, $\Phi = -GM/r$ is unit 05's point-mass potential. Inside, $\Phi' = GMr/R^3$ (the mass inside $r$ is $Mr^3/R^3$), and integrating from the surface value $-GM/R$ gives $-GM(3R^2 - r^2)/2R^3$. Then $\nabla^2\Phi = r^{-2}(r^2\Phi')' = 3GM/R^3 = 4\pi G\rho$ with $\rho = 3M/4\pi R^3$ ✓, and both $\Phi$ and $\Phi'$ are continuous at $r = R$ ✓. The centre value is $\tfrac32$ of the surface value: $1.5\times2.1225\times10^{-6} = 3.18\times10^{-6}$.

A warning that unit 30 gave and that must survive into this unit: $h_{\mu\nu} \ne 0$ is not by itself gravity. Rindler coordinates on flat spacetime have a non-constant metric and even a horizon. What distinguishes the present example is that its curvature is nonzero, and we shall see that in section 4 when $\nabla^2\Phi$ turns out to be a component of the Ricci tensor.

section 2

The source

What goes on the right-hand side, defined as a tensor and then made concrete.

35 · source

What a stress-energy tensor is

definition

A stress-energy tensor on a spacetime $(M,g)$ is a symmetric $(0,2)$-tensor field $T$: at each point a symmetric bilinear map $$T_p : T_pM \times T_pM \longrightarrow \mathbb R$$ with $\nabla_\mu T^{\mu\nu} = 0$.

That is the mathematics: unit 16's definition of a $(0,2)$-tensor, applied at each tangent space by unit 31, symmetric, divergence-free. The physics is in how it is read. Feed it the four-velocity of an observer and you get what that observer measures.

the reading

For an observer with four-velocity $u$ (so $u\cdot u = -c^2$, unit 27), the energy density measured is $$\varepsilon = \frac{1}{c^2}\,T(u,u) = \frac{1}{c^2}T_{\mu\nu}u^\mu u^\nu.$$

In an orthonormal frame of that observer ($\mu, \nu = 0,1,2,3$; $i, j = 1,2,3$), the components read:

Four blocks, one object. The top row and left column carry the same numbers, because $T$ is symmetric; the $3\times3$ block is the subject of the next slide.

Check the reading. In the observer's own rest frame $u^\mu = (c,0,0,0)$, so $T(u,u)/c^2 = T_{00}c^2/c^2 = T_{00}$, the energy density. Dimensions: $[T_{\mu\nu}] = \mathrm{J\,m^{-3}} = \mathrm{kg\,m^{-1}s^{-2}}$ throughout, in every slot, because the index positions are raised and lowered with a dimensionless metric in these coordinates.

Why the off-diagonal entries carry those factors of $c$. With $x^0 = ct$, the flux of the quantity $T^{\mu 0}$ through a surface is measured per unit $x^0$, not per unit $t$. Energy flux per unit time is $c\,T^{0i}$; momentum density is $T^{0i}/c$. The symmetry $T^{0i} = T^{i0}$ therefore says energy flux $= c^2\times$ momentum density, which in flat spacetime is exactly $E = mc^2$ applied to a flowing fluid, and is the relativistic content of a statement non-relativistic continuum mechanics could only assert.

35 · source

The space-space block is stress

The bottom-right block is the momentum flux: $T^{ij}$ is the $i$th component of momentum crossing unit area normal to $j$, per unit time.

It is minus the stress tensor $\sigma$ of continuum mechanics, which, fed unit vectors $\vec m$, $\vec n$, returns the $\vec m$-component of the force per unit area exerted across a surface by the material on the side $\vec n$ points to.

A fluid at rest has $\sigma = -pI$ and $T^{ij} = p\,\delta^{ij}$. Unit 22, a branch off this route, builds $\sigma$ from scratch.

Left: the force per area across one surface has a part along $\vec n$ (tension or pressure) and a part along the surface (shear); $\sigma(\vec m,\vec n)$ is its component along $\vec m$. Right: a fluid at rest has no shear, and every face is pushed inward by $p$, so $\sigma(\cdot,\vec n) = -p\,\vec n$.

Why "minus". Across a surface with normal $\vec n$, the material on the $\vec n$ side exerts force $\sigma\vec n$ per unit area on the material behind it; by the third law the material behind exerts $-\sigma\vec n$ on the material in front, which is momentum delivered in the direction of $\vec n$ per unit area per unit time. Momentum delivered across a surface per area per time is exactly the momentum flux, so $T^{ij} = -\sigma_{ij}$. For the fluid: $\sigma = -pI$ gives $T^{ij} = p\,\delta^{ij}$, a positive flux of $x$-momentum in the $x$ direction — the fluid behind pushes the fluid in front.

Reading the figure. The arrow on the left surface is drawn with both a normal and a tangential part to show that $\sigma$ is a full $3\times3$ object, not a number: shear is the off-diagonal part. A fluid is defined by having none at rest, which is why its block is $p\,\delta^{ij}$ and why the perfect fluid of the next slide needs only one extra number.

Why symmetric. Three answers, converging. (i) For a continuum, the balance of angular momentum on a small cube forces $\sigma_{ij} = \sigma_{ji}$ absent intrinsic spin: an antisymmetric part would exert a torque of order $\ell^3$ on a cube of side $\ell$ whose moment of inertia is of order $\ell^5$, so its angular acceleration would diverge as $\ell\to0$ (unit 22 does this carefully). (ii) Unit 29: the canonical Noether tensor is not symmetric, but the Belinfante-Rosenfeld improvement makes it so without changing any conserved charge. (iii) The definition by metric variation, $T_{\mu\nu} \propto \delta S_{\text{matter}}/\delta g^{\mu\nu}$, is symmetric automatically because $g^{\mu\nu}$ is. Route one takes (iii); route two takes symmetry as given and notes that the left-hand side is symmetric too, so nothing else would be consistent.

35 · source

Dust, and a fluid

worked example 1 · the two sources that matter

Dust: non-interacting particles, rest-mass density $\rho$ in their own rest frame, four-velocity field $u^\mu$. The only symmetric tensor available from $\rho$ and $u$ alone is $$T^{\mu\nu} = \rho\,u^\mu u^\nu.$$

Check, rest frame. $u^\mu = (c,0,0,0)$, so $T^{00} = \rho c^2$ — the energy density, as it must be — and every other component vanishes. Dust has no pressure, which is what "non-interacting" means.

Perfect fluid: add isotropic pressure $p$ and nothing else. The extra tensor available is $g^{\mu\nu}$: $$T^{\mu\nu} = \Bigl(\rho + \frac{p}{c^2}\Bigr)u^\mu u^\nu + p\,g^{\mu\nu}.$$

Check, rest frame. $T^{00} = (\rho + p/c^2)c^2 + p\,\eta^{00} = \rho c^2 + p - p = \rho c^2$ ✓, and $T^{ij} = p\,\delta^{ij}$ — isotropic pressure ✓. Setting $p = 0$ returns dust ✓.

Dimensions. $[\rho u^\mu u^\nu] = \mathrm{kg\,m^{-3}}\cdot\mathrm{m^2s^{-2}} = \mathrm{kg\,m^{-1}s^{-2}} = \mathrm{J\,m^{-3}}$ ✓, and $[p] = \mathrm{Pa} = \mathrm{J\,m^{-3}}$ ✓ — the same, which is why pressure can appear in an energy-momentum tensor at all.

Why these two forms and no others. The argument is a counting argument in the spirit of unit 16. At a point, the only objects available are $u^\mu$, the scalars $\rho$ and $p$, and $g^{\mu\nu}$. A symmetric $(2,0)$-tensor built from them is $a\,u^\mu u^\nu + b\,g^{\mu\nu}$ with $a,b$ scalars. Requiring the rest-frame energy density to be $\rho c^2$ and the rest-frame stress to be $p\,\delta^{ij}$ fixes $b = p$ and $a = \rho + p/c^2$. So the perfect fluid is not a model; it is the most general isotropic source, and anisotropy — shear stress, heat flux, viscosity — is exactly what a third tensor structure would describe.

The trace. $T = g_{\mu\nu}T^{\mu\nu}$. For dust: $\rho\,u\cdot u = -\rho c^2$. For a perfect fluid: $(\rho + p/c^2)(-c^2) + 4p = -\rho c^2 + 3p$. Both are needed in section 4, where the equation is used in trace-reversed form. Note that radiation, with $p = \tfrac13\rho c^2$, has $T = 0$ — the electromagnetic stress-energy tensor is trace-free, and the electromagnetic-field slide says why that is a structural fact rather than a coincidence.

The sign of $p$. Nothing above forbids $p \lt 0$. A negative isotropic pressure is exactly what the cosmological term will turn out to look like, with $p = -\rho c^2$, and that is the equation of state of dark energy. That is section 6, and the reason it is worth noticing here is that no new mathematics is needed to write it — only a sign nobody expected.

35 · source

Its non-relativistic shadow

Take dust in flat spacetime, $\nabla \to \partial$, and expand $\partial_\mu T^{\mu\nu} = 0$ to lowest order in $v/c$, writing $u^\mu \approx (c, v^i)$:

$\nu = 0$ $$\begin{aligned}&\partial_0(\rho c^2) + \partial_i(\rho c\,v^i) = 0\\ \Rightarrow\;&\partial_t\rho + \nabla\!\cdot\!(\rho\vec v) = 0.\end{aligned}$$

The continuity equation: mass is conserved.

$\nu = j$ $$\begin{aligned}&\partial_0(\rho c\,v^j) + \partial_i(\rho v^iv^j) = 0\\ \Rightarrow\;&\partial_t(\rho v^j) + \partial_i(\rho v^iv^j) = 0.\end{aligned}$$

With pressure the flux is $\rho v^iv^j + p\,\delta^{ij}$: momentum balance for a fluid (unit 22's, if you took that branch).

what this buys

Two laws that continuum mechanics derives separately are one equation here: the four components of $\partial_\mu T^{\mu\nu} = 0$. Unit 27 made energy and momentum one four-vector; here their conservation laws become one divergence.

what is lost

"Mass is conserved" is an approximation. The exact statement, $\nabla_\mu T^{\mu 0} = 0$, lets rest mass convert to kinetic and internal energy at higher order.

The expansion, with the orders visible. For dust $T^{00} = \rho\gamma^2c^2$, $T^{0i} = \rho\gamma^2cv^i$, $T^{ij} = \rho\gamma^2v^iv^j$, where $\rho$ is the rest-frame density. To lowest order $\gamma = 1$. The $\nu = 0$ equation is $\partial_0 T^{00} + \partial_i T^{i0} = 0$, and $\partial_0 = c^{-1}\partial_t$, so $c^{-1}\partial_t(\rho c^2) + \partial_i(\rho cv^i) = 0$; divide by $c$ to get the continuity equation. The $\nu = j$ equation is $c^{-1}\partial_t(\rho cv^j) + \partial_i(\rho v^iv^j) = 0$, which is Euler's equation for pressureless dust once the product rule and continuity are used: $\rho(\partial_t v^j + v^i\partial_i v^j) = 0$, where $\partial_t + v^i\partial_i$ is the rate of change following a particle of the dust. It says each particle moves in a straight line at constant speed, which is exactly what free, non-interacting dust in flat spacetime should do.

Where the factors of $c$ went. They cancelled, every time, which is the check that the index placement is right. If a factor of $c$ survives in a non-relativistic limit, an index has been raised with the wrong sign or a $\partial_0$ has been mistaken for $\partial_t$.

This slide is also an answer to a question a careful reader will already have: why the source of gravity should be a tensor whose components are exactly the quantities of continuum mechanics. The answer is that continuum mechanics was always the non-relativistic shadow of the same object. The stress tensor was not chosen for convenience; it is $T^{ij}$, and the other six components were simply invisible at $v \ll c$. A reader who has taken unit 22 will recognise its two balance laws; one who has not has now seen both, derived.

Dimensions of the momentum equation. $[\partial_t(\rho v)] = \mathrm{kg\,m^{-2}s^{-2}}$ and $[\partial_i(\rho v^iv^j)] = \mathrm{m^{-1}}\cdot\mathrm{kg\,m^{-3}}\cdot\mathrm{m^2s^{-2}}$, the same ✓; $[\partial_i p] = \mathrm{Pa\,m^{-1}} = \mathrm{kg\,m^{-2}s^{-2}}$ ✓.

35 · source

The electromagnetic field, if you have unit 29

cited: unit 29

For the electromagnetic field with field tensor $F_{\mu\nu}$ (unit 28),

$$T^{\mu\nu}_{\text{EM}} = \frac{1}{\mu_0}\Bigl(F^{\mu\alpha}F^\nu{}_\alpha - \tfrac14 g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\Bigr).$$

Derived in unit 29 by varying Maxwell's action with respect to the metric, and shown there to equal the canonical Noether tensor plus a Belinfante improvement term.

it is trace-free

$g_{\mu\nu}T^{\mu\nu} = \frac{1}{\mu_0}(F^{\mu\alpha}F_{\mu\alpha} - \tfrac14\cdot4\cdot F_{\alpha\beta}F^{\alpha\beta}) = 0$, using $g_{\mu\nu}g^{\mu\nu} = 4$ in four dimensions. One line, and it will matter twice.

$T^{00}_{\text{EM}} = \tfrac12(\varepsilon_0E^2 + B^2/\mu_0)$, the field energy density of unit 23; $cT^{0i}$ is the Poynting vector of unit 24; $T^{ij}$ is the Maxwell stress tensor. Three objects that looked unrelated, one tensor.

for route-two readers

Without unit 29 this slide is a citation, and nothing later depends on being able to derive it. What is needed downstream is only that some symmetric divergence-free $T_{\mu\nu}$ describes each kind of matter, and that for dust and fluids we have written it down explicitly.

Why trace-free matters twice. First: it is the reason Einstein's 1913 guess $R_{\mu\nu} = \kappa T_{\mu\nu}$ cannot be repaired, as the next section shows — that guess forces $T$ to be constant, and a universe containing both electromagnetic radiation ($T = 0$) and dust ($T = -\rho c^2$) cannot have constant $T$. Second: Nordström's scalar theory, sourced by the trace $T$ alone, is blind to anything trace-free. In it light neither gravitates nor — its metric being conformally flat, so that null curves are those of flat spacetime — is deflected. General relativity is sourced by the whole tensor, and light both gravitates and bends.

Where the minimal coupling happened. The formula above is unit 29's flat-space formula with $\eta^{\mu\nu}$ replaced by $g^{\mu\nu}$ and indices raised with $g$. That is unit 32's comma-goes-to-semicolon rule, and for this object it is unambiguous because no derivatives appear: $T_{\text{EM}}$ is algebraic in $F$. The ordering ambiguity unit 32 warned about only bites when two covariant derivatives appear and one must choose their order.

Dimensions. $[F_{\mu\nu}]$ in these conventions is that of $E/c$, i.e. $\mathrm{V\,s\,m^{-2}} = \mathrm{T}$; $[1/\mu_0]\cdot\mathrm{T}^2 = \mathrm{J\,m^{-3}}$ ✓. A quick check on the $00$ component: $F^{0i} = E^i/c$ up to sign, so $F^{0\alpha}F^0{}_\alpha = \sum_i(F^{0i})^2 = E^2/c^2$ (lowering a spatial index with $\eta_{ii} = +1$ changes no sign), while $F_{\alpha\beta}F^{\alpha\beta} = 2(B^2 - E^2/c^2)$ and $\eta^{00} = -1$, so the two terms combine to $\tfrac12(\varepsilon_0E^2 + B^2/\mu_0)$ using $\varepsilon_0\mu_0 = 1/c^2$.

section 3

Route two: nothing else was available

Needs only units 33 and 30. The equation is forced, not chosen.

35 · forced

One divergence-free tensor, already in hand

unit 33, restated

The second Bianchi identity $\nabla_{[\lambda}R_{\mu\nu]\rho\sigma} = 0$, contracted twice, gives the contracted Bianchi identity

$$\nabla^\mu G_{\mu\nu} = 0,\qquad G_{\mu\nu} \equiv R_{\mu\nu} - \tfrac12 R\,g_{\mu\nu},$$

for every metric on every manifold. It is an identity of Riemannian geometry, proved before any physics was mentioned.

So the Einstein tensor $G_{\mu\nu}$ meets every structural requirement at once: it is a symmetric $(0,2)$-tensor, built from $g$ and its first two derivatives, linear in the second derivatives, and identically divergence-free.

the candidate $$G_{\mu\nu} = \kappa\,T_{\mu\nu}$$

with $\kappa$ an unknown constant. Nothing yet says it is the only candidate, and nothing yet fixes $\kappa$. Those are the next two slides and the next section.

read the logic in the right direction

$\nabla^\mu G_{\mu\nu} = 0$ is not a consequence of the field equation. It is a consequence of geometry, and it is what makes the field equation possible. Had Riemannian geometry supplied no such identity, no equation of this shape could have been consistent.

Full proof of the contracted identity, since it is the load-bearing step and unit 33 proved it in a different notation. The second Bianchi identity is $\nabla_\lambda R_{\mu\nu\rho\sigma} + \nabla_\rho R_{\mu\nu\sigma\lambda} + \nabla_\sigma R_{\mu\nu\lambda\rho} = 0$. Contract with $g^{\mu\rho}$: using the symmetries $R_{\mu\nu\sigma\lambda} = -R_{\nu\mu\sigma\lambda}$ and metric compatibility $\nabla g = 0$ so that the contraction passes through the derivative, this gives $\nabla_\lambda R_{\nu\sigma} - \nabla_\sigma R_{\nu\lambda} + \nabla^\mu R_{\mu\nu\sigma\lambda} = 0$. Contract again with $g^{\nu\sigma}$: $\nabla_\lambda R - \nabla^\nu R_{\nu\lambda} - \nabla^\mu R_{\mu\lambda} = 0$, that is $\nabla^\mu R_{\mu\lambda} = \tfrac12\nabla_\lambda R$. Hence $\nabla^\mu(R_{\mu\lambda} - \tfrac12 Rg_{\mu\lambda}) = \tfrac12\nabla_\lambda R - \tfrac12\nabla_\lambda R = 0$. $\square$ The only inputs are the Riemann symmetries and $\nabla g = 0$; the second Bianchi identity itself is a one-line computation in normal coordinates, where $\Gamma = 0$ at the point and $R \sim \partial\Gamma$.

History. Einstein did not know this identity in 1915. Bianchi published it in 1902 (Aurel Voss had found it in 1880 and Ricci around 1889, and both were forgotten), and it did not reach Einstein until after November 1915, through the work of Hilbert, Klein and Weyl in 1916-18. Einstein arrived at $G_{\mu\nu}$ by a different and much more painful route — trial, error, and the demand that the equation reproduce Mercury's 43 arcseconds — over the three years 1912-1915. That the identity existed all along, waiting, is the single strongest piece of evidence that the equation is not an invention.

35 · forced

Einstein's first guess, and how it died

worked example 2 · reconstructing November 1915

The obvious guess — considered by Einstein in 1912, dropped, and briefly adopted in November 1915 — is the simplest tensor equation of the right shape: $$R_{\mu\nu} = \kappa\,T_{\mu\nu}.$$ Step 1. Take the divergence. The contracted Bianchi identity, in the intermediate form proved in the notes to the Bianchi slide, says $\nabla^\mu R_{\mu\nu} = \tfrac12\nabla_\nu R$. The guess and $\nabla^\mu T_{\mu\nu} = 0$ therefore give $$\tfrac12\nabla_\nu R = \kappa\,\nabla^\mu T_{\mu\nu} = 0 \;\Longrightarrow\; R = \text{const}.$$ Step 2. Take the trace of the guess. $g^{\mu\nu}R_{\mu\nu} = R$ and $g^{\mu\nu}T_{\mu\nu} = T$, so $R = \kappa T$, and hence $$T = \text{const}.$$ Step 3. Contradiction. $T$ is the trace of the matter's stress-energy. For dust it is $-\rho c^2$. A constant $T$ means constant density everywhere in the universe — and if $T\to0$ at spatial infinity, as it must for an isolated star, then $T\equiv0$ everywhere, so the star has no mass. The guess is not merely restrictive; it is empty.

The repair. The failure is precisely that $\nabla^\mu R_{\mu\nu} \ne 0$. Subtract off exactly what spoils it: since $\nabla^\mu(\tfrac12 Rg_{\mu\nu}) = \tfrac12\nabla_\nu R$ too, the difference is divergence-free. That difference is $G_{\mu\nu}$.

The history, accurately. Einstein's Zurich notebook of 1912-13 already considers the Ricci tensor as a left-hand side and sets it aside, because he could not see how to recover Newton from it. The "Entwurf" theory of 1913, with Grossmann, then had a left-hand side that was not generally covariant, and he defended its non-covariance for two years with the "hole argument". In November 1915 he published four papers in four weeks. The paper of 11 November proposed $R_{\mu\nu} = \kappa T_{\mu\nu}$ — the guess above — with the additional assumption $\sqrt{-g} = 1$ and the claim that $T = 0$ for all matter, which he briefly believed. The paper of 18 November computed Mercury's perihelion and got 43 arcseconds per century, which is unit 37. The paper of 25 November has $R_{\mu\nu} - \tfrac12 Rg_{\mu\nu} = \kappa T_{\mu\nu}$. Hilbert presented the action derivation on 20 November, five days earlier, which has caused a century of argument; the consensus now is that Hilbert's paper as printed cites Einstein's field equations and that neither derived the other's result.

An equivalent form worth having. The guess is not wrong for vacuum. Taking the trace of $G_{\mu\nu} = \kappa T_{\mu\nu}$ gives $R - 2R = \kappa T$, so $R = -\kappa T$, and substituting back gives the trace-reversed field equation $$R_{\mu\nu} = \kappa\Bigl(T_{\mu\nu} - \tfrac12 T g_{\mu\nu}\Bigr).$$ This is the form used in section 4 and in every weak-field computation in the rest of the course, because its left-hand side is a single Ricci component rather than a difference. When $T_{\mu\nu} = 0$ it reduces to $R_{\mu\nu} = 0$: Einstein's guess and the correct equation agree exactly where there is no matter, which is why unit 36's Schwarzschild derivation could be done with either.

Why "trace reversed". The map $T_{\mu\nu}\mapsto T_{\mu\nu} - \tfrac12 Tg_{\mu\nu}$ sends the trace $T$ to $T - \tfrac12\cdot4\cdot T = -T$, and it is an involution on symmetric tensors in four dimensions: apply it twice and you are back. That is worth checking on paper once; it is problem P4.

35 · forced

Lovelock: there was never a choice

cited · Lovelock 1971

Let $A_{\mu\nu}$ be a symmetric $(0,2)$-tensor field on a four-dimensional manifold, constructed pointwise from the metric and its first and second partial derivatives, and satisfying $\nabla^\mu A_{\mu\nu} = 0$ identically. Then there are constants $a,b$ with $$A_{\mu\nu} = a\,G_{\mu\nu} + b\,g_{\mu\nu}.$$

No linearity in the second derivatives is assumed: in four dimensions it follows. Proved in D. Lovelock, J. Math. Phys. 12 (1971) 498 and 13 (1972) 874. Vermeil (1917), Cartan (1922) and Weyl (1922) had it assuming linearity.

every hypothesis is load-bearing

Drop four dimensions: in $D\ge5$ the Gauss-Bonnet combination $R^2 - 4R_{\mu\nu}R^{\mu\nu} + R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}$ contributes a new divergence-free tensor. In $D = 4$ it is a topological density and contributes nothing.
Drop second derivatives only: $\nabla^2 G_{\mu\nu}$ and friends reappear, and with them Ostrogradsky's instability.
Drop metric only: add a scalar field and you have Brans-Dicke, which is not excluded and is still being tested.

the conclusion

Given the five requirements, the field equation must be $$G_{\mu\nu} + \Lambda\,g_{\mu\nu} = \kappa\,T_{\mu\nu}$$ with $\Lambda = b/a$ and $\kappa = 1/a$. Two numbers remain undetermined, and geometry has nothing more to say about either.

What the theorem is not. It is not a proof that general relativity is true. It is a proof that general relativity is the only theory satisfying a list of requirements, so that an experiment contradicting general relativity must contradict one of the listed requirements — and the list is short enough that the experiment would tell you which. That is a far more useful thing to know than a derivation, and it is why this route is given first.

The part of the proof that fits here. Grant that $A_{\mu\nu}$ is linear in the second derivatives of $g$. Then, since unit 32's normal coordinates make $\partial g = 0$ at a point and the second derivatives there are carried by Riemann, $A$ is a linear function of Riemann plus a multiple of $g$; and unit 33's trace count says the only symmetric $(0,2)$-tensors linear in Riemann are $R_{\mu\nu}$ and $Rg_{\mu\nu}$. So $A = \alpha R_{\mu\nu} + \beta Rg_{\mu\nu} + \gamma g_{\mu\nu}$, and the divergence is $(\tfrac12\alpha + \beta)\nabla_\nu R$ (using $\nabla^\mu R_{\mu\nu} = \tfrac12\nabla_\nu R$), which vanishes for every metric only if $\beta = -\tfrac12\alpha$. That is problem P2. Lovelock's real contribution is removing the linearity hypothesis, and that part is cited.

The Gauss-Bonnet remark, in more detail (cited: the Chern-Gauss-Bonnet theorem, Chern 1944). In four dimensions the integral $\int\sqrt{-g}\,(R^2 - 4R_{\mu\nu}R^{\mu\nu} + R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma})\, \mathrm{d}^4x$ is, for a closed manifold, $32\pi^2$ times the Euler characteristic — a topological invariant. Its metric variation therefore vanishes identically, and the corresponding tensor is zero rather than merely conserved. In five or more dimensions the same expression is not topological, its variation is a genuine new divergence-free symmetric tensor, and Lovelock's theorem produces a tower of them: this is Lovelock gravity, and it is what string-theoretic corrections to Einstein gravity look like. The accident that four dimensions is the highest dimension in which the tower has only two rungs is not explained by anything in this course.

Brans-Dicke, since it is the live alternative. Replace $G$ by a dynamical scalar field $\phi$ with its own kinetic term governed by a dimensionless parameter $\omega$. General relativity is the limit $\omega\to\infty$. The Cassini spacecraft's measurement of the Shapiro delay (unit 37) bounds $\omega \gt 4\times10^4$. The theory is not excluded, merely squeezed, and every such bound is an experimental test of one of Lovelock's hypotheses.

35 · forced

What is left free, and what fixes it

Route two in three boxes. The first two produce the form and leave two constants; only the last box produces a number, and it fixes $\kappa$ alone — $\Lambda$ is left to measurement.

$\kappa$ is fixed by requirement 5 and nothing else. Section 4 computes it: $\kappa = 8\pi G/c^4$. It is the only place Newton's constant enters the theory.

$\Lambda$ is not fixed by anything in the theory. It must be measured. Kept through section 4's computation (the notes do it), it adds a term to the Newtonian limit, $\nabla^2\Phi = 4\pi G\rho - \Lambda c^2$, which is undetectable in the solar system — at 1 AU the extra acceleration $\Lambda c^2 r/3$ is $4.9\times10^{-25}\ \mathrm{m\,s^{-2}}$ against the Sun's $5.9\times10^{-3}$, a ratio of $8\times10^{-23}$.

the shape of the argument

Structure from geometry, numbers from experiment, and a clean line between them. That line is what makes the theory falsifiable rather than adjustable.

Where $-\Lambda c^2$ comes from. Redo section 4 with $\Lambda$ kept. Trace $G_{\mu\nu} + \Lambda g_{\mu\nu} = \kappa T_{\mu\nu}$: $R - 2R + 4\Lambda = \kappa T$, so $R = 4\Lambda - \kappa T$. Substituting back, $R_{\mu\nu} = \kappa(T_{\mu\nu} - \tfrac12Tg_{\mu\nu}) + \Lambda g_{\mu\nu}$. For static dust the $00$ component is $R_{00} = \tfrac12\kappa\rho c^2 + \Lambda g_{00} = \tfrac12\kappa\rho c^2 - \Lambda$ to lowest order. Section 4's geometry side, $R_{00} = \nabla^2\Phi/c^2$, is unchanged, so $\nabla^2\Phi = \tfrac12\kappa c^4\rho - \Lambda c^2 = 4\pi G\rho - \Lambda c^2$. Dimensions: $[\Lambda c^2] = \mathrm{m^{-2}\,m^2s^{-2}} = \mathrm{s^{-2}}$, and $[4\pi G\rho] = \mathrm{m^3kg^{-1}s^{-2}\,kg\,m^{-3}} = \mathrm{s^{-2}}$ ✓.

The $\Lambda$ numbers, computed. Taking $\Lambda = 1.1\times10^{-52}\ \mathrm{m^{-2}}$ (section 6 says where that comes from), the vacuum Poisson equation $\nabla^2\Phi = -\Lambda c^2$ has the spherically symmetric solution $\Phi = -\Lambda c^2r^2/6$, so the outward acceleration is $\mathrm{d}\Phi/\mathrm{d}r$ in magnitude $\Lambda c^2 r/3$. At $r = 1\,\mathrm{AU} = 1.496\times10^{11}$ m this is $1.1\times10^{-52}\times8.988\times10^{16}\times1.496\times10^{11}/3 = 4.93\times10^{-25}\ \mathrm{m\,s^{-2}}$. The Sun's Newtonian acceleration there is $GM_\odot/r^2 = 1.327\times10^{20}/(2.238\times10^{22}) = 5.93\times10^{-3}\ \mathrm{m\,s^{-2}}$. The ratio is $8.31\times10^{-23}$. No solar-system measurement will ever see it. The ratio does not depend on $r$ at all once written as $\Lambda c^2/(4\pi G\bar\rho)$, with $\bar\rho = 3M/4\pi r^3$ the mean density inside $r$: at 1 AU $\bar\rho = 1.42\times10^{-4}\ \mathrm{kg\,m^{-3}}$, and $\Lambda c^2/4\pi G = 1.18\times10^{-26}\ \mathrm{kg\,m^{-3}}$. Cosmology sees $\Lambda$ easily because there the mean density of matter is itself of order $10^{-26}\ \mathrm{kg\,m^{-3}}$, and the two terms compete on equal terms.

Why $\Lambda$ cannot be argued away. Every derivation in this unit produces it. The Bianchi route produces it because $g_{\mu\nu}$ is itself divergence-free ($\nabla_\lambda g_{\mu\nu} = 0$ is metric compatibility), so $b\,g_{\mu\nu}$ is always an allowed addition. The action route produces it because a constant is a perfectly good scalar to add to $R$ in the Lagrangian. Setting $\Lambda = 0$ is a choice, not a derivation, and it is a choice the observations of 1998 rejected.

section 4

The Newtonian limit fixes the constant

The running example, worked from the beginning with nothing skipped.

35 · limit

The weak, slow, static setting

three assumptions, stated separately

Weak. $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ with $|h_{\mu\nu}|\ll1$; all results to first order in $h$. This is the weak-field expansion.
Static. $\partial_0 h_{\mu\nu} = 0$.
Slow. The source is dust at rest: $T_{00} = \rho c^2$, all other components zero to leading order, and $p \ll \rho c^2$.

Each is a different physical statement and each is used in a different line below. Keeping them apart is how one finds out which assumption a result really needs.

Work with the trace-reversed field equation. From $G_{\mu\nu} = \kappa T_{\mu\nu}$, take the trace: $R - 2R = \kappa T$, so $R = -\kappa T$, and

equivalent form $$R_{\mu\nu} = \kappa\Bigl(T_{\mu\nu} - \tfrac12 T\,g_{\mu\nu}\Bigr).$$

For our dust: $T = g^{\mu\nu}T_{\mu\nu} \approx \eta^{00}T_{00} = -\rho c^2$, so $$T_{00} - \tfrac12 Tg_{00} = \rho c^2 - \tfrac12(-\rho c^2)(-1) = \tfrac12\rho c^2,$$ giving the one equation we need:

the right-hand side $$R_{00} = \tfrac12\,\kappa\,\rho c^2.$$

Why trace-reverse first. $G_{00} = R_{00} - \tfrac12 Rg_{00}$ mixes $R_{00}$ with the full scalar curvature, and $R$ at first order involves spatial components of $h$ that we have not specified. The trace-reversed form isolates a single Ricci component, and $R_{00}$ at first order turns out to depend on $h_{00}$ alone. That is the whole reason the computation is short.

The sign of $g_{00}$, checked. $\eta_{00} = -1$ in signature $(-,+,+,+)$. Then $-\tfrac12 Tg_{00} = -\tfrac12(-\rho c^2)(-1) = -\tfrac12\rho c^2$, so $T_{00} - \tfrac12 Tg_{00} = \rho c^2 - \tfrac12\rho c^2 = \tfrac12\rho c^2$. A reader using $(+,-,-,-)$ will find $\eta_{00} = +1$, $T = +\rho c^2$ and the same $\tfrac12\rho c^2$: the combination is signature-independent, as it must be, since it is going to equal something measurable.

What "slow" buys and what it costs. $T_{0i} = 0$ discards gravitomagnetic effects — frame dragging, the Lense-Thirring precession — which are first order in $v/c$ relative to what we keep. They are real, and Gravity Probe B measured the Earth's at 37.2 milliarcseconds per year in 2011. They are simply not what fixes $\kappa$.

35 · limit

The Ricci component, to first order

worked example 3 · computing $R_{00}$, every step

Step 1. The Christoffel symbols to first order. $\Gamma^\mu_{\nu\rho} = \tfrac12 g^{\mu\sigma}(\partial_\nu g_{\sigma\rho} + \partial_\rho g_{\sigma\nu} - \partial_\sigma g_{\nu\rho})$. Since $\partial g = \partial h$ is already first order, the prefactor may be replaced by $\tfrac12\eta^{\mu\sigma}$. With $\nu = \rho = 0$ and $\partial_0 h = 0$: $$\Gamma^\mu_{00} = \tfrac12\eta^{\mu\sigma}\bigl(2\,\partial_0h_{\sigma0} - \partial_\sigma h_{00}\bigr) = -\tfrac12\eta^{\mu\sigma}\partial_\sigma h_{00}.$$ So $\Gamma^0_{00} = -\tfrac12\eta^{00}\partial_0h_{00} = 0$ and $\Gamma^i_{00} = -\tfrac12\,\partial_i h_{00}$.

Step 2. The Ricci tensor to first order. $R_{00} = \partial_\mu\Gamma^\mu_{00} - \partial_0\Gamma^\mu_{\mu0} + \Gamma^\mu_{\mu\lambda}\Gamma^\lambda_{00} - \Gamma^\mu_{0\lambda}\Gamma^\lambda_{\mu0}$. The last two terms are quadratic in $h$: drop them. The second vanishes because the field is static. So $$R_{00} = \partial_\mu\Gamma^\mu_{00} = \partial_i\Gamma^i_{00} = -\tfrac12\,\partial_i\partial_i h_{00} = -\tfrac12\nabla^2h_{00}.$$

Step 3. Insert unit 32's identification. $h_{00} = -2\Phi/c^2$, so $$\boxed{\;R_{00} = \frac{1}{c^2}\nabla^2\Phi\;}$$

Check. Dimensions: $[R_{00}] = \mathrm{m^{-2}}$ and $[\nabla^2\Phi/c^2] = (\mathrm{m^2s^{-2}})\,\mathrm{m^{-2}}/(\mathrm{m^2s^{-2}}) = \mathrm{m^{-2}}$ ✓. Special case: $\Phi$ constant gives $R_{00} = 0$, and a constant potential is no field ✓. Special case: for a point mass $\Phi = -GM/r$ is harmonic away from the origin, so $R_{00} = 0$ in vacuum ✓ — which is exactly what unit 36 will solve.

The step that is usually skipped. Replacing $g^{\mu\sigma}$ by $\eta^{\mu\sigma}$ in step 1 is legitimate only because the bracket it multiplies is already $O(h)$; the correction is $O(h^2)$. Note also $g^{\mu\nu} = \eta^{\mu\nu} - h^{\mu\nu} + O(h^2)$ with indices on $h$ raised by $\eta$ — a minus sign, from $g^{\mu\lambda}g_{\lambda\nu} = \delta^\mu_\nu$ — which does not matter here but does in section 6.

The second term of $R_{00}$, in full. $\partial_0\Gamma^\mu_{\mu0}$ vanishes by staticity, but it is worth noting what it is: $\Gamma^\mu_{\mu0} = \tfrac12 g^{\mu\sigma}\partial_0g_{\sigma\mu} = \tfrac12\partial_0\ln|g|$, the logarithmic time derivative of the metric determinant. In cosmology (unit 39) this is the term that does not vanish and that produces the Hubble expansion, so the assumption being used here is exactly the one cosmology drops.

Why $R_{00}$ and not $G_{00}$. One can do it the other way. $G_{00}$ at first order for a static metric is $\nabla^2(h_{ii} - \text{trace terms})$, involving the spatial metric perturbation, and one then needs the $ij$ equations to determine $h_{ij}$ before the $00$ equation says anything. The answer is the same, and the slide "The weak field, a third time" finds $h_{ij} = -2\Phi\delta_{ij}/c^2$ in Lorenz gauge, but the route through $R_{00}$ needs none of that. The lesson is general: trace-reverse before linearising.

35 · limit

Poisson appears, and with it the coupling

Two expressions for the same component. From the source side, $R_{00} = \tfrac12\kappa\rho c^2$; from the geometry side, $R_{00} = \nabla^2\Phi/c^2$. Equate:

$$\frac{\nabla^2\Phi}{c^2} = \tfrac12\,\kappa\,\rho c^2 \quad\Longrightarrow\quad \nabla^2\Phi = \tfrac12\,\kappa\,c^4\rho.$$

This is Poisson's equation if and only if $\tfrac12\kappa c^4 = 4\pi G$:

the coupling constant $$\kappa = \frac{8\pi G}{c^4} = 2.0766\times10^{-43}\ \mathrm{m^{-1}kg^{-1}s^{2}}$$

and the field equations in full: $$R_{\mu\nu} - \tfrac12 R\,g_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu\nu}.$$

checks

Dimensions. $[G_{\mu\nu}] = \mathrm{m^{-2}}$. $[\kappa T_{\mu\nu}] = \mathrm{m^{-1}kg^{-1}s^2}\cdot\mathrm{kg\,m^{-1}s^{-2}} = \mathrm{m^{-2}}$ ✓.

Limit. $G\to0$ gives $\kappa\to0$ and $G_{\mu\nu} = 0$: matter stops curving spacetime and, by unit 33, Minkowski space is a solution ✓.

Limit. $c\to\infty$ at fixed $G$ gives $\kappa\to0$ likewise — gravity is a relativistic effect whose Newtonian shadow survives only because $\rho c^2$ grows at the same rate ✓.

Every number in the rest of this course descends from this one equation. Unit 36's Schwarzschild radius $2GM/c^2$, unit 37's 43 arcseconds, unit 38's wave amplitudes, unit 39's Friedmann equations: all of them carry $8\pi G/c^4$.

Where the $4\pi$ went. It did not go anywhere: $4\pi G$ on the Newtonian side becomes $8\pi G$ on the relativistic side, and the factor of two is the $\tfrac12$ from the trace reversal. It is the same factor of two that unit 32 had to get right in $g_{00} = -(1 + 2\Phi/c^2)$, and it is the factor of two that separates Einstein's light deflection from a Newtonian corpuscle's, which is unit 37. Three appearances of the same two; they are not independent, and tracking them is a good way to catch an error.

The coupling as a stiffness. Write the equation as $T_{\mu\nu} = (c^4/8\pi G)\,G_{\mu\nu}$. The prefactor $c^4/8\pi G = 4.815\times10^{42}$ N is a force, and it is enormous — it is why spacetime is so nearly flat. Inverting: an energy density of $1\ \mathrm{J\,m^{-3}}$ produces curvature $2.077\times10^{-43}\ \mathrm{m^{-2}}$, a radius of curvature of $2.19\times10^{21}$ m, about $2.3\times10^5$ light years. The next slide makes that concrete.

Geometrised units, since the literature uses them. Setting $G = c = 1$ makes $\kappa = 8\pi$ and the equation $G_{\mu\nu} + \Lambda g_{\mu\nu} = 8\pi T_{\mu\nu}$. Masses then carry units of length via $GM/c^2$: the Sun is 1.48 km, the Earth 4.44 mm. This course keeps $G$ and $c$ visible because every dimension check in it depends on them, but one should be able to read either convention.

35 · limit

How stiff is spacetime?

From $R_{00} = \tfrac12\kappa\rho c^2$, a static body of density $\rho$ has a curvature scale $\ell = (\kappa\rho c^2)^{-1/2}$ — the radius at which curvature would matter. The coupling makes it absurd.

material	$\rho$ / kg m$^{-3}$	$\ell$
water	$10^{3}$	1.5 AU
the Earth, mean	$5.5\times10^{3}$	0.66 AU
nuclear matter	$2.3\times10^{17}$	15 km

A curvature radius of one metre needs $\rho = 5.4\times10^{25}\ \mathrm{kg\,m^{-3}}$ — $2\times10^{8}$ times nuclear density.

the last row is the point

At nuclear density the curvature scale is 15 km — the size of a neutron star. That is not a coincidence and it is not an accident of the arithmetic: it is the statement that neutron stars are the objects for which gravity and nuclear physics are the same size, and it is why they are the laboratory for strong-field gravity.

what the table does not say

$\ell$ is a curvature scale, not a horizon. A ball of water of radius 1.5 AU would collapse long before it reached that size, and the Schwarzschild radius — unit 36 — is a different length built from $M$ rather than $\rho$.

The arithmetic. $\ell = (\kappa\rho c^2)^{-1/2}$ with $\kappa = 2.0766\times10^{-43}$. Water, $\rho = 1000$: $\kappa\rho c^2 = 1.866\times10^{-23}\ \mathrm{m^{-2}}$, $\ell = 2.315\times10^{11}$ m $= 1.547$ AU. Earth, $\rho = 5513$: $\kappa\rho c^2 = 1.029\times10^{-22}$, $\ell = 9.858\times10^{10}$ m $= 0.659$ AU. Nuclear matter, $\rho = 2.3\times10^{17}$: $\kappa\rho c^2 = 4.29\times10^{-9}\ \mathrm{m^{-2}}$, $\ell = 1.526\times10^4$ m $= 15.3$ km. For $\ell = 1$ m one needs $\rho = 1/(\kappa c^2) = 5.358\times10^{25}\ \mathrm{kg\,m^{-3}}$, which is $5\times10^{22}$ times water and $2\times10^{8}$ times nuclear density. Nothing in the universe holds that density in bulk except matter already inside a horizon.

The relation between $\ell$ and the Schwarzschild radius. For a uniform ball of radius $R$ and density $\rho$, $r_s = 2GM/c^2 = 8\pi G\rho R^3/(3c^2)$. The ratio $r_s/R = 8\pi G\rho R^2/(3c^2) = \tfrac23(R/\ell)^2$. So $R \sim \ell$ is precisely the condition $r_s \sim R$: the curvature scale and the horizon scale coincide when the object is as compact as it can be. A neutron star has $r_s/R \approx 0.34$, twice the $|\Phi|/c^2 \approx 0.17$ of the figure on the running-example slide, since $r_s/R = 2GM/(Rc^2) = 2|\Phi|/c^2$.

35 · limit

Pressure gravitates

Redo the limit with a perfect fluid instead of dust. Now $T = -\rho c^2 + 3p$, and $$T_{00} - \tfrac12 Tg_{00} = \rho c^2 - \tfrac12(-\rho c^2 + 3p)(-1) = \tfrac12(\rho c^2 + 3p).$$ So $R_{00} = \tfrac12\kappa(\rho c^2 + 3p)$, and with $R_{00} = \nabla^2\Phi/c^2$ and $\kappa = 8\pi G/c^4$:

Poisson, corrected $$\nabla^2\Phi = 4\pi G\Bigl(\rho + \frac{3p}{c^2}\Bigr)$$

Checks. Dimensions: $p/c^2$ is $\mathrm{Pa\,m^{-2}s^{2}} = \mathrm{kg\,m^{-3}}$ ✓. Limit $p\to0$: dust ✓. Special case $p = \tfrac13\rho c^2$ (radiation, trace-free): $\nabla^2\Phi = 8\pi G\rho$, twice as strong ✓.

a genuinely non-Newtonian statement

Pressure is a source of gravity in its own right, not only through the energy it stores. In a box of gas the walls' tension cancels it in the total; inside a star nothing cancels it.

The effect is minute for ordinary matter: at the centre of the Sun, $p \approx 2.34\times10^{16}$ Pa and $\rho\approx1.5\times10^{5}\ \mathrm{kg\,m^{-3}}$, so $3p/c^2 = 0.78\ \mathrm{kg\,m^{-3}}$, five parts in a million.

It is not minute for a star at the end of its life. The pressure that holds the star up also pulls it down, and above a critical mass no equation of state can win. That is why massive stars must collapse, and it is a statement with no Newtonian counterpart whatever.

The solar-centre number. $p_c = 2.34\times10^{16}$ Pa, $\rho_c = 1.50\times10^{5}\ \mathrm{kg\,m^{-3}}$ from the standard solar model. $3p_c/c^2 = 3\times2.34\times10^{16}/8.988\times10^{16} = 0.781\ \mathrm{kg\,m^{-3}}$, which is $0.78/1.5\times10^5 = 5.2\times10^{-6}$ of the density. Quoted as "five parts in a million".

The box of gas, and why it does not weigh more. A sealed box of hot gas has $p \gt 0$ inside, but its walls are under tension — negative stress — and for any static, self-contained body the volume integral of the total stress vanishes (von Laue's theorem: integrate $\partial_i(x^j T^{ik}) = \delta^j_i T^{ik}$ over a region containing the body, using $\partial_iT^{ik} = 0$ for a static flat-space $T$). So the box's active gravitational mass is its total energy over $c^2$ and no more. In a star the pressure is held not by walls but by gravity, the flat-space theorem does not apply, and the $3p/c^2$ survives.

Where this leads. Put the same correction into the equation of hydrostatic equilibrium and one gets the Tolman-Oppenheimer-Volkoff equation, whose right-hand side has three separate places where pressure appears and makes matters worse. Integrating it with any equation of state satisfying $\mathrm{d}p/\mathrm{d}\rho \le c^2$ gives a maximum mass. For neutron stars the bound is near $2.2\,M_\odot$ observationally, and the heaviest confirmed neutron star is about $2.1\,M_\odot$. Beyond that, unit 36.

The trace-free check, again. Radiation has $T = 0$, so the trace-reversal does nothing and $R_{\mu\nu} = \kappa T_{\mu\nu}$ — Einstein's failed guess — is correct for a pure radiation source. That is why the guess survived as long as it did: in November 1915 Einstein hoped, briefly and wrongly, that all matter was ultimately electromagnetic, and for electromagnetic matter the guess is right.

section 5 · needs unit 29

Route one: vary an action

The same equation, from a principle rather than a list. And the source becomes a definition.

35 · action

Which scalar? The Einstein-Hilbert action

Unit 11 taught the calculus of variations for curves; unit 29 extended it to fields on spacetime, with actions $\int\mathcal L\sqrt{-g}\,\mathrm d^4x$, with $\sqrt{-g}\,\mathrm d^4x$ the invariant volume element. Take $g_{\mu\nu}$ itself as the field. What scalar can the Lagrangian be?

the candidates, by order

Zero derivatives of $g$: only constants. That is $\Lambda$.
One derivative: none. $\partial g$ is not a tensor, and unit 32's normal coordinates kill it at any chosen point.
Two derivatives: exactly one, up to scale — the scalar curvature $R$.

the Einstein-Hilbert action $$S = \frac1c\int_{\mathcal M}\Bigl[\frac{c^4}{16\pi G}(R - 2\Lambda) + \mathcal L_{\text{m}}\Bigr]\sqrt{-g}\,\mathrm d^4x$$

with $\mathrm d^4x = \mathrm d x^0\mathrm dx^1\mathrm dx^2\mathrm dx^3$ and $x^0 = ct$.

Dimensions. $[c^4/G] = \mathrm{kg\,m\,s^{-2}} = \mathrm N$; $[R] = \mathrm{m^{-2}}$; $[\mathrm d^4x] = \mathrm{m^4}$; $[1/c] = \mathrm{s\,m^{-1}}$. Product: $\mathrm{N\cdot m^{-2}\cdot m^{4}\cdot s\,m^{-1}} = \mathrm{J\,s}$ ✓ — an action.

The prefactor is written as $c^4/16\pi G$ in advance, which is cheating in the pedagogical sense only: call it $1/2\alpha$, derive the equations, and match to section 4 to find $\alpha$. Nothing is assumed.

"Exactly one, up to scale" needs a word. The scalars available from $g$ and two derivatives are built by contracting the Riemann tensor, and by its symmetries the only full contraction is $R$ (up to sign convention). Scalars quadratic in curvature — $R^2$, $R_{\mu\nu}R^{\mu\nu}$, $R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}$ — are available but contain four derivatives of $g$, so their Euler-Lagrange equations are fourth order and excluded by requirement 2. Thus Lovelock's theorem and the action argument exclude the same things for the same reason, arriving from different directions. That is the consistency worth noticing between the two routes.

Why $\sqrt{-g}$. Under $x\to\tilde x$ the coordinate volume $\mathrm d^4x$ picks up the Jacobian determinant $J$, and $\det g$ picks up $J^{-2}$, so $\sqrt{-g}\,\mathrm d^4x$ is invariant. $-g$ rather than $g$ because the signature is Lorentzian and $\det g \lt 0$: with $\eta = \mathrm{diag}(-1,1,1,1)$, $\det\eta = -1$. Unit 32 met the same object as the metric volume form, and unit 29 used it in field actions.

History. Hilbert presented this action to the Göttingen Academy on 20 November 1915. He was not doing general relativity for its own sake; he was pursuing a unified theory of gravity and Mie's electromagnetism, and the gravitational part is a few pages of a longer and mostly abandoned programme. The action is universally called Einstein-Hilbert, which is roughly fair: Hilbert wrote it, Einstein had the field equations. The printer's proofs of Hilbert's paper, found in 1997, do not show the field equations in Einstein's explicit form; part of one page is missing, and how much that matters is still argued.

35 · action

Lemma: varying the determinant

lemma 1 $$\delta\sqrt{-g} = -\tfrac12\sqrt{-g}\;g_{\mu\nu}\,\delta g^{\mu\nu}$$

proof

Write $g = \det(g_{\mu\nu})$. Jacobi's formula for the derivative of a determinant gives $\delta\ln|g| = \operatorname{tr}(g^{-1}\delta g) = g^{\mu\nu}\delta g_{\mu\nu}$, so $$\delta g = g\,g^{\mu\nu}\,\delta g_{\mu\nu}.$$ Differentiating $g^{\mu\nu}g_{\nu\rho} = \delta^\mu_\rho$ gives $g^{\mu\nu}\delta g_{\nu\rho} = -g_{\nu\rho}\delta g^{\mu\nu}$, hence $g^{\mu\nu}\delta g_{\mu\nu} = -g_{\mu\nu}\delta g^{\mu\nu}$. So $\delta g = -g\,g_{\mu\nu}\delta g^{\mu\nu}$, and $$\delta\sqrt{-g} = \frac{-\delta g}{2\sqrt{-g}} = \frac{g\,g_{\mu\nu}\delta g^{\mu\nu}}{2\sqrt{-g}} = -\tfrac12\sqrt{-g}\,g_{\mu\nu}\delta g^{\mu\nu},$$ the last step using $g = -(-g) = -(\sqrt{-g})^2$. $\square$

the sign trap

$\delta g^{\mu\nu}$ is not obtained by raising the indices of $\delta g_{\mu\nu}$. It is minus that: $$\delta g^{\mu\nu} = -g^{\mu\alpha}g^{\nu\beta}\,\delta g_{\alpha\beta}.$$ Every sign error in this section comes from forgetting it. Varying with respect to $g^{\mu\nu}$ rather than $g_{\mu\nu}$, as we do, means the trap is sprung only once — here.

Check in two dimensions. For $g = \mathrm{diag}(a,b)$ with $a,b\gt0$, $\sqrt{g} = \sqrt{ab}$ and $\delta\sqrt{g} = \tfrac12\sqrt{ab}\,(\delta a/a + \delta b/b)$. The lemma gives $-\tfrac12\sqrt{ab}\,(a\,\delta a^{-1} + b\,\delta b^{-1}) = -\tfrac12\sqrt{ab}\,(-\delta a/a - \delta b/b)$ ✓.

Full proof of Jacobi's formula, since it is the only piece of linear algebra here that unit 16 did not do. For an invertible matrix $A$ and small $X$, $\det(A + X) = \det A\,\det(I + A^{-1}X)$. Expanding the second determinant in the entries of $A^{-1}X$ and keeping first order, $\det(I + Y) = 1 + \operatorname{tr}Y + O(Y^2)$ — because the determinant is the sum over permutations and only the identity permutation contributes a term linear in $Y$. Hence $\delta\det A = \det A\operatorname{tr}(A^{-1}\delta A)$, which is Jacobi's formula, equivalently $\delta\ln|\det A| = \operatorname{tr}(A^{-1}\delta A)$. $\square$

Why $\delta g^{\mu\nu}$ and not $\delta g_{\mu\nu}$. Either works and they give the same field equations, since one is an invertible linear function of the other at each point. Varying with respect to the inverse metric produces $R_{\mu\nu} - \tfrac12Rg_{\mu\nu}$ with lower indices directly and matches unit 29's definition of $T_{\mu\nu}$ as a derivative with respect to $g^{\mu\nu}$, so the two units lock together with no index gymnastics. The other choice gives $-G^{\mu\nu}$ and one then raises indices; the physics is identical and the sign errors are more numerous.

On the two-dimensional check. $\delta a^{-1} = -\delta a/a^2$, so $a\,\delta a^{-1} = -\delta a/a$; and the metric being diagonal, $g_{\mu\nu}\delta g^{\mu\nu} = a\,\delta a^{-1} + b\,\delta b^{-1}$. The signature is Euclidean here, so $\sqrt g$ rather than $\sqrt{-g}$; the algebra is unaffected.

35 · action

Lemma: the Palatini identity

lemma 2 · the Palatini identity $$\delta R_{\mu\nu} = \nabla_\lambda\bigl(\delta\Gamma^\lambda_{\mu\nu}\bigr) - \nabla_\nu\bigl(\delta\Gamma^\lambda_{\mu\lambda}\bigr)$$

valid for any variation of the metric, with $\delta\Gamma^\lambda_{\mu\nu}$ a genuine $(1,2)$-tensor even though $\Gamma$ is not.

proof sketch · full version in the notes

$\delta\Gamma$ is the difference of two connections, hence tensorial. Evaluate both sides in normal coordinates at a point $p$ (unit 32): there $\Gamma(p) = 0$, so $R_{\mu\nu} = \partial_\lambda\Gamma^\lambda_{\mu\nu} - \partial_\nu\Gamma^\lambda_{\mu\lambda}$ and $\nabla = \partial$ on tensors at $p$. The identity is then immediate. Both sides being tensors, it holds in every chart. $\square$

the corollary that matters

Contract with $g^{\mu\nu}$ and use $\nabla g = 0$ to move it inside: $$g^{\mu\nu}\delta R_{\mu\nu} = \nabla_\lambda v^\lambda,$$ $$v^\lambda \equiv g^{\mu\nu}\delta\Gamma^\lambda_{\mu\nu} - g^{\lambda\nu}\delta\Gamma^\mu_{\nu\mu}.$$ A total divergence. It contributes nothing to the field equations and everything to the boundary term.

Named for Attilio Palatini, who derived it in 1919. The "Palatini formulation", in which $g$ and $\Gamma$ are varied independently and the Levi-Civita connection comes out as an equation of motion, is in fact Einstein's (1925).

Full proof. First, tensoriality of $\delta\Gamma$: if $\Gamma$ and $\Gamma + \delta\Gamma$ are both connections, their difference transforms homogeneously, because the inhomogeneous term in a connection's transformation law depends only on the coordinate change and cancels. So $\delta\Gamma^\lambda_{\mu\nu}$ is a $(1,2)$-tensor.

Second, the identity. Fix $p$ and choose normal coordinates there, so $\Gamma^\lambda_{\mu\nu}(p) = 0$ (unit 32's local flatness theorem). The Ricci tensor is $R_{\mu\nu} = \partial_\lambda\Gamma^\lambda_{\mu\nu} - \partial_\nu\Gamma^\lambda_{\mu\lambda} + \Gamma^\lambda_{\lambda\sigma}\Gamma^\sigma_{\mu\nu} - \Gamma^\lambda_{\nu\sigma}\Gamma^\sigma_{\mu\lambda}$. Varying, the two quadratic terms each produce something proportional to $\Gamma(p) = 0$, so at $p$, $\delta R_{\mu\nu} = \partial_\lambda\delta\Gamma^\lambda_{\mu\nu} - \partial_\nu\delta\Gamma^\lambda_{\mu\lambda}$. At $p$, for any tensor, $\nabla = \partial$ since the connection coefficients vanish there. So at $p$ the claimed identity holds. Both sides are tensors and $p$ was arbitrary, so it holds everywhere. $\square$

The explicit form of $\delta\Gamma$, for the reader who wants to see $v^\lambda$ concretely: $\delta\Gamma^\lambda_{\mu\nu} = \tfrac12 g^{\lambda\sigma} (\nabla_\mu\delta g_{\sigma\nu} + \nabla_\nu\delta g_{\sigma\mu} - \nabla_\sigma\delta g_{\mu\nu})$ — the Christoffel formula with partial derivatives replaced by covariant ones. It follows by the same normal-coordinate argument. From it, $v^\lambda = \nabla_\mu(\delta g)^{\lambda\mu} - \nabla^\lambda(g^{\mu\nu}\delta g_{\mu\nu})$, where $(\delta g)^{\lambda\mu} = g^{\lambda\alpha}g^{\mu\beta}\delta g_{\alpha\beta}$ is $\delta g_{\alpha\beta}$ with its indices raised — minus the variation of the inverse metric, by the sign trap on the determinant slide. Check with a conformal variation $\delta g_{\mu\nu} = 2\epsilon\,g_{\mu\nu}$: $v^\lambda = 2\nabla^\lambda\epsilon - 8\nabla^\lambda\epsilon = -6\nabla^\lambda\epsilon$, so $g^{\mu\nu}\delta R_{\mu\nu} = -6\,g^{\alpha\beta}\nabla_\alpha\nabla_\beta\epsilon$, which is the standard first-order change of the scalar curvature under $g\to e^{2\epsilon}g$ in four dimensions, $\delta R = -2\epsilon R - 6\,g^{\alpha\beta}\nabla_\alpha\nabla_\beta\epsilon$, once the $-2\epsilon R$ from $R_{\mu\nu}\delta g^{\mu\nu}$ is set aside. This $v^\lambda$ visibly contains derivatives of $\delta g$ normal to the boundary — the fact that makes The boundary term nobody can ignore necessary.

35 · action

The variation, assembled

worked example 4 · varying the Einstein-Hilbert action

Step 1. Write $R = g^{\mu\nu}R_{\mu\nu}$ and vary the gravitational part term by term: $$\delta\bigl(\sqrt{-g}\,R\bigr) = \sqrt{-g}\,R_{\mu\nu}\,\delta g^{\mu\nu} \;+\; \sqrt{-g}\,g^{\mu\nu}\,\delta R_{\mu\nu} \;+\; R\,\delta\sqrt{-g}.$$

Step 2. Lemma 1 turns the third term into $-\tfrac12\sqrt{-g}\,R\,g_{\mu\nu}\,\delta g^{\mu\nu}$; lemma 2 turns the second into $\sqrt{-g}\,\nabla_\lambda v^\lambda$. So $$\delta\bigl(\sqrt{-g}R\bigr) = \sqrt{-g}\Bigl(R_{\mu\nu} - \tfrac12 Rg_{\mu\nu}\Bigr)\delta g^{\mu\nu} + \sqrt{-g}\,\nabla_\lambda v^\lambda.$$

Step 3. The $\Lambda$ term: $\delta(\sqrt{-g}\,(-2\Lambda)) = -2\Lambda\cdot(-\tfrac12\sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu}) = +\sqrt{-g}\,\Lambda g_{\mu\nu}\,\delta g^{\mu\nu}$.

Step 4. The matter side. Unit 29's definition, which is the definition because of this calculation: $$\delta\bigl(\sqrt{-g}\,\mathcal L_{\text m}\bigr) = -\tfrac12\sqrt{-g}\;T_{\mu\nu}\,\delta g^{\mu\nu}.$$

Step 5. Collect, discard the divergence for now, and demand $\delta S = 0$ for arbitrary compactly supported $\delta g^{\mu\nu}$. Unit 11's fundamental lemma then kills the integral sign: $$\frac{c^4}{16\pi G}\Bigl(G_{\mu\nu} + \Lambda g_{\mu\nu}\Bigr) - \tfrac12 T_{\mu\nu} = 0 \;\Longrightarrow\; G_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu\nu}.$$

Step 4 unpacked. Unit 29 defined $T_{\mu\nu} = -2\,\delta\mathcal L_{\text m}/ \delta g^{\mu\nu} + g_{\mu\nu}\mathcal L_{\text m}$. Check that this is the same statement: $\delta(\sqrt{-g}\mathcal L_{\text m}) = \sqrt{-g}\,\delta\mathcal L_{\text m} + \mathcal L_{\text m}\delta\sqrt{-g} = \sqrt{-g}\bigl[\delta\mathcal L_{\text m}/\delta g^{\mu\nu} - \tfrac12 g_{\mu\nu}\mathcal L_{\text m}\bigr]\delta g^{\mu\nu}$, which equals $-\tfrac12\sqrt{-g}T_{\mu\nu}\delta g^{\mu\nu}$ exactly when $T_{\mu\nu}$ is as unit 29 defined it. Equivalently and more memorably, $T_{\mu\nu} = -\dfrac{2}{\sqrt{-g}}\dfrac{\delta(\sqrt{-g}\,\mathcal L_{\text m})} {\delta g^{\mu\nu}}$.

Why this is the sharp point of route one. In route two, $T_{\mu\nu}$ is given from outside: we assert that matter has a symmetric divergence-free stress-energy tensor and write it down case by case. In route one it is defined — as the response of the matter action to a change of the metric — and its symmetry is automatic, its gauge invariance is automatic, and its conservation is automatic too, because diffeomorphism invariance of $S_{\text m}$ implies $\nabla^\mu T_{\mu\nu} = 0$ whenever the matter's own field equations hold — the argument by which unit 29 showed gauge invariance forces current conservation, applied with diffeomorphisms as the gauge group (Noether's second theorem). Route one thus derives what route two assumes. That is the price of needing unit 29.

What "arbitrary compactly supported" is doing. Exactly what it did in unit 11's fundamental lemma of the calculus of variations: if $\int f_{\mu\nu}\,\delta g^{\mu\nu} \sqrt{-g}\,\mathrm d^4x = 0$ for every smooth $\delta g^{\mu\nu}$ vanishing near $\partial \mathcal M$, and $f$ is continuous and symmetric, then $f = 0$. Compact support is also what lets step 5 discard the divergence; the next slide is about what happens when it may not be discarded.

35 · action

The boundary term nobody can ignore

Since $\sqrt{-g}\,\nabla_\lambda v^\lambda = \partial_\lambda(\sqrt{-g}\,v^\lambda)$ (problem P10), the discarded term is an ordinary divergence, and unit 05's divergence theorem, in the four-dimensional form unit 29 used, turns it into a boundary integral: $$\int_{\mathcal M}\!\sqrt{-g}\,\nabla_\lambda v^\lambda\,\mathrm d^4x = \oint_{\partial\mathcal M}\! v^\lambda n_\lambda\,\sqrt{|\gamma|}\;\mathrm d^3x.$$

why it does not just vanish

In ordinary variational problems one imposes $\delta g = 0$ on $\partial\mathcal M$ and boundary terms die. Here they do not: as the notes to lemma 2 show, $v^\lambda$ contains derivatives of $\delta g$ normal to the boundary. Fixing $g$ on $\partial\mathcal M$ does not fix its normal derivative, and fixing both overdetermines the problem.

A region $\mathcal M$ of spacetime and its boundary, time upward. The outward normal $n$ is timelike on the spacelike top and bottom ($\varepsilon = n\cdot n = -1$) and spacelike on the timelike sides ($\varepsilon = +1$). The boundary integral is a sum over the pieces, each with its own induced metric $\gamma$.

An analogy that makes it concrete. Consider $S[y] = \int_0^1 y\,y''\,\mathrm dx$. Varying, $\delta S = \int(\delta y\,y'' + y\,\delta y'')$, and integrating the second term by parts twice gives $\int 2y''\delta y + [y\,\delta y' - y'\,\delta y]_0^1$. Fixing $y$ at the endpoints kills the second bracket but not the first, which involves $\delta y'$. The repair is to add $-[yy']_0^1$ to the action — that is, to use $S' = -\int (y')^2$, which differs from $S$ by exactly that boundary term and has a clean variational problem. The Gibbons-Hawking-York term is the same manoeuvre in four dimensions. This analogy is problem P9.

The pieces of the boundary formula. $\sqrt{|\gamma|}\,\mathrm d^3x$ is the volume element the metric induces on the boundary, and $n_\lambda$ is the outward conormal normalised with the inverse metric. The coordinate proof: choose coordinates in which a boundary piece is $x^3 = $ const with $\mathcal M$ on the side $x^3$ smaller; then $\int\partial_\lambda(\sqrt{-g}v^\lambda)\,\mathrm d^4x$ reduces by the fundamental theorem of calculus in $x^3$ to $\int\sqrt{-g}\,v^3\,\mathrm d^3x$ over that piece. Put $n_\lambda = \delta^3_\lambda/|g^{33}|^{1/2}$, so $v^\lambda n_\lambda = v^3/|g^{33}|^{1/2}$; and the cofactor formula for the inverse matrix gives $g^{33} = \gamma/g$, with $\gamma$ the determinant of the $3\times3$ block $g_{ab}$, $a,b\ne3$ — the induced metric. Hence $\sqrt{-g} = \sqrt{|\gamma|}/|g^{33}|^{1/2}$ and $\sqrt{-g}\,v^3 = \sqrt{|\gamma|}\,v^\lambda n_\lambda$. $\square$ In the GHY term, $\varepsilon = n^\lambda n_\lambda = \pm1$ records whether that piece of boundary is timelike ($+1$) or spacelike ($-1$). Unit 34 states the same theorem as Stokes' theorem for forms.

Reading the figure. A boundary in spacetime is not a sphere around a ball: it is typically two spacelike slices — the "initial" and "final" times — joined by a timelike tube far away. The normalisation $n\cdot n = \pm1$ fails on a null piece, where $n\cdot n = 0$; null boundaries need separate treatment, which is one of the things this course does not do.

35 · action

The Gibbons-Hawking-York term

the Gibbons-Hawking-York term $$S_{\text{GHY}} = \frac1c\,\frac{c^4}{8\pi G}\oint_{\partial\mathcal M}\! \varepsilon K\sqrt{|\gamma|}\;\mathrm d^3x$$

with $\gamma$ the metric induced on $\partial\mathcal M$ (the boundary's own lengths; $\gamma$, not $h$, since $h$ is the weak-field perturbation), $n$ its outward unit normal, $\varepsilon = n\cdot n = \pm1$, and $K$ the trace of its extrinsic curvature — how the boundary bends inside spacetime, which a cylinder's surface does though it is intrinsically flat. Its variation cancels the previous slide's leftover boundary integral exactly, so that $S_{\text{EH}} + S_{\text{GHY}}$ has a well-posed variational problem with only $g|_{\partial\mathcal M}$ fixed. York 1972, Gibbons-Hawking 1977; the cancellation is cited, since the extrinsic curvature is not in this course.

For the field equations themselves it changes nothing — that is the point of a boundary term. It matters for black-hole thermodynamics, where the action's value, not merely its stationary points, is the physics.

The leftover from the previous slide contains $n^\lambda\partial_\lambda\,\delta g$, a normal derivative. $K$ contains first derivatives of the metric (through $\nabla n$), so $\delta K$ contains normal derivatives of $\delta g$ too; with the coefficient $c^4/8\pi G$, twice the bulk's $c^4/16\pi G$, they cancel (cited).

Dimensions of $S_{\text{GHY}}$. $[K] = \mathrm{m^{-1}}$, $[\mathrm d^3x] = \mathrm{m^3}$, $[c^4/G] = \mathrm N$, $[1/c] = \mathrm{s\,m^{-1}}$: product $\mathrm{N\,m\,s} = \mathrm{J\,s}$ ✓, the same as the bulk term, as it must be to cancel against it.

Why anyone cares. In the Euclidean path-integral approach to black-hole thermodynamics, the gravitational entropy comes out as a boundary integral, and for a Schwarzschild black hole the Einstein-Hilbert bulk term vanishes ($R = 0$ in vacuum) so the entire on-shell action is the Gibbons-Hawking-York term. Getting $S = A/4$ in Planck units depends on the coefficient above being exactly right. That is well beyond this course, but it is the answer to "why does a term that cannot affect the equations of motion have a name".

Extrinsic curvature, in one sentence, since it is cited. For a hypersurface with unit normal $n$, $K_{\mu\nu} = \nabla_\mu n_\nu$ projected onto the surface, and $K = \gamma^{\mu\nu}K_{\mu\nu}$ measures how the surface bends inside the larger space — as opposed to the intrinsic curvature of unit 33, which a surface has on its own. A cylinder has zero intrinsic and nonzero extrinsic curvature.

35 · action

The same constant, by other mathematics

the two routes, side by side

	route two	route one
input	Bianchi, Lovelock	an action principle
needs	33, 30	29 as well
$T_{\mu\nu}$ is	assumed	defined
$\Lambda$ from	$b\,g_{\mu\nu}$	a constant in $\mathcal L$
$\kappa$ from	the Newtonian limit	the Newtonian limit

The last row is not a coincidence and it is not circular: neither route can produce a number, because neither contains $G$. Both hand over an equation with one free constant, and the same Newtonian limit fixes it to $8\pi G/c^4$ both times.

what would falsify it

Once Newton has fixed $\kappa$, nothing is left to adjust. The same equation then fixes the spatial metric too — the slide "The weak field, a third time" finds $h_{ij} = -2\Phi\delta_{ij}/c^2$ — and that predicts 1.75 arcseconds of light deflection at the Sun's limb. The same $h_{00}$ with flat space, which is all a Newtonian corpuscle feels, predicts 0.88. Eddington's 1919 measurement excluded the second. One experiment fixes the constant; a different one tests the equation.

Two derivations from unrelated premises, one equation, one constant. That is what it takes before a theory is worth measuring.

The 1.75 arcseconds, since it is quoted. Unit 37 derives it; the value at the solar limb is $4GM_\odot/(c^2R_\odot) = 8.49\times10^{-6}$ rad $= 1.75''$. The Newtonian-corpuscle prediction is half of it, $0.88''$, and the factor of two is the presence of the spatial metric perturbation $h_{ij}$ alongside $h_{00}$ — which is to say, the factor of two is the whole content of general relativity as distinct from a relativistic patch on Newton. Eddington reported $1.98''\pm0.12''$ at Sobral and $1.61''\pm0.30''$ at Príncipe (probable errors, as published in 1920), which excluded $0.88''$ and was consistent with $1.75''$.

Is the agreement of the two routes surprising? Less than it looks. Lovelock's theorem says any divergence-free symmetric second-order tensor is $aG + bg$, and the variation of a diffeomorphism-invariant action is automatically divergence-free by Noether's second theorem, so the action route was always going to land inside the Lovelock family. What is not automatic is that the coefficient works out to the same $a$ once both are matched to Newton — but that too is forced, since both are matched to the same Newton. The honest statement is that the two routes are independent in their premises and not in their conclusions, and that the value of doing both is that a mistake in either is visible.

section 6

What the equations say

Vacuum, counting, waves, the cosmological term, and one thing that genuinely fails.

35 · consequences

Vacuum is not empty of geometry

Set $T_{\mu\nu} = 0$ and $\Lambda = 0$. Taking the trace of $G_{\mu\nu} = 0$ gives $R - 2R = 0$, so $R = 0$, and the vacuum field equations are simply $$R_{\mu\nu} = 0.$$

ten equations, not twenty

$R_{\mu\nu} = 0$ does not say $R_{\mu\nu\rho\sigma} = 0$. Unit 33 counted: Riemann has $n^2(n^2-1)/12 = 20$ independent components in four dimensions, Ricci has 10. Ten components of curvature are entirely unconstrained by the vacuum equations.

Those ten are the Weyl tensor $C_{\mu\nu\rho\sigma}$, unit 33's totally trace-free part of Riemann. In vacuum, Riemann is Weyl.

In three dimensions the count is $3^2\cdot8/12 = 6$, the same as Ricci's, so Weyl vanishes identically: $R_{\mu\nu} = 0$ forces flatness. No vacuum gravitational waves in 3D. The theory is sensitive to the number of dimensions in a way Newton's is not.

what Weyl carries

Tidal distortion at constant volume: the part of geodesic deviation that stretches in one direction and squeezes in another. Ricci controls the volume change, and in vacuum there is none — a falling sphere of dust in vacuum deforms into an ellipsoid of the same volume. The next slide draws it.

The volume statement, made precise. Unit 33's geodesic deviation for a congruence of freely falling dust gives, for the fractional rate of change of the volume of a small ball, $\ddot V/V|_{t=0} = -R_{\mu\nu}u^\mu u^\nu$. In vacuum this is zero. So the sphere deforms — Weyl does that — without changing volume. Inside matter it does change, at the rate $-\tfrac12\kappa c^2(\rho c^2 + 3p)$, which is the same $\rho + 3p/c^2$ as the corrected Poisson equation. This one statement, "$\ddot V/V = -4\pi G(\rho + 3p/c^2)$ for a small ball of test particles momentarily at rest relative to one another", is equivalent to the full field equations given that it holds for every observer — J. Baez and E. Bunn's formulation (Am. J. Phys. 73 (2005) 644), proved as problem P7, and the most compact honest summary of the theory in words.

Why 1 in two dimensions, 6 in three, 20 in four. Unit 33's formula $n^2(n^2-1)/12$ gives 1, 6, 20 for $n = 2,3,4$. Ricci, being a symmetric $n\times n$ matrix, has $n(n+1)/2 = 3, 6, 10$. Weyl is the difference only in dimension four and above: $20 - 10 = 10$ in four dimensions, $6 - 6 = 0$ in three, and in two dimensions the formula for Weyl does not even make sense. Four is the smallest dimension in which general relativity has anything to propagate.

Where this bites next. Unit 36 solves $R_{\mu\nu} = 0$ with a spherically symmetric ansatz and finds Schwarzschild. The solution is not flat despite satisfying the vacuum equations, and the reason is exactly this slide: its Riemann tensor is pure Weyl and nonzero. A reader who has not internalised this slide will find the Schwarzschild solution paradoxical.

35 · consequences

Weyl squeezes, Ricci shrinks

Newton already has both pieces. The relative acceleration of two nearby free particles a displacement $\vec\xi$ apart is $-(\partial_i\partial_j\Phi)\,\xi^j$, and the matrix $\partial_i\partial_j\Phi$ splits into its trace, $\nabla^2\Phi = 4\pi G\rho$, and a trace-free rest.

The trace is the Ricci part — it is $c^2R_{00}$, section 4 — and it changes volume. The trace-free part is the Weyl part, and it only changes shape.

Two consequences. Gravity acts across empty space: the Sun's field is pure Weyl at the Earth. And gravitational waves exist, since they are vacuum curvature.

A small ball of dust, dashed, and the same ball a moment later, solid (exaggerated); arrows are the tidal accelerations. Beside a mass the stretch rates along the three axes are $2, -1, -1$ times $GM/r^3$ — the third axis, out of the page, is squeezed like the vertical one — and they sum to zero. Inside uniform dust all three are $-\tfrac{4\pi}{3}G\rho$.

The two matrices, computed. Outside a point mass, $\Phi = -GM/r$, and with the $x$ axis pointing away from the mass, $-\partial_i\partial_j\Phi = (GM/r^3)\, \mathrm{diag}(2,-1,-1)$ at a point on that axis: stretch along the line to the mass, squeeze in both transverse directions. The trace is $0$, as $\nabla^2\Phi = 0$ in vacuum requires. Inside a uniform ball, the profile of The running example, drawn gives $\Phi = \tfrac{2\pi}{3}G\rho\,r^2 + \text{const}$, so $-\partial_i\partial_j\Phi = -\tfrac{4\pi}{3}G\rho\,\delta_{ij}$: equal squeeze in every direction, trace $-4\pi G\rho$, and no trace-free part at all. Real situations are mixtures; these two are the pure cases.

What to look for in the figure. On the left the ellipse looks larger in area than the circle, because the page shows only two of the three axes; with the third axis squeezed too, the volume to first order is $(1 + 2\epsilon)(1 - \epsilon)^2 = 1 + O(\epsilon^2)$. On the right nothing changes shape. Volume change is Ricci; shape change is Weyl.

35 · consequences

Ten minus four minus four

ten

$g_{\mu\nu}$ is symmetric $4\times4$: 10 functions of four coordinates, and the field equations are 10 equations.

minus four: gauge freedom

A diffeomorphism $x\to\tilde x(x)$ is four arbitrary functions and changes no physics. Two metrics related by one are the same solution. 4 of the 10 are pure coordinate choice.

minus four: constraint equations

The four equations $G^{0\mu} = \kappa T^{0\mu}$ contain no second time derivative, so they are conditions on the initial data, not evolution equations. 4 more.

the count

$10 - 4 - 4 = \mathbf2$ propagating degrees of freedom per point. Compare electromagnetism: $A_\mu$ has 4, minus 1 gauge, minus 1 constraint (Gauss's law), leaves 2 — the two polarisations of light. Gravity's two are the plus and cross polarisations of worked example 5.

Why the constraints contain no second time derivative — proved. The contracted Bianchi identity $\nabla_\mu G^{\mu\nu} = 0$ expands to $\partial_0 G^{0\nu} = -\partial_i G^{i\nu} - \Gamma\text{-terms}$. The right-hand side contains at most two time derivatives of the metric (since $G$ itself does), so the left-hand side, $\partial_0G^{0\nu}$, contains at most two — which means $G^{0\nu}$ itself contains at most one. A quantity with at most one time derivative of $g$ is a condition on $(g,\dot g)$, that is, on the initial data. So four of the ten field equations are constraints, and the same identity that made the equations consistent is what makes four of them non-dynamical. $\square$

Propagation of the constraints. A natural worry: if the constraints are imposed at $t = 0$, do the six evolution equations preserve them? Yes, and the proof is again the Bianchi identity: it gives $\partial_0(G^{0\nu} - \kappa T^{0\nu})$ in terms of quantities that vanish when the evolution equations and $\nabla_\mu T^{\mu\nu} = 0$ hold. So the constrained surface is preserved by the flow. This is exactly the structure of Maxwell's equations, where $\nabla\cdot\vec B = 0$ and Gauss's law are constraints preserved by the curl equations.

Counting is not the same as solving. The count says how many functions' worth of freedom a solution has, not that finding them is easy. The constraint equations are a coupled elliptic system, and constructing initial data satisfying them is a research subject in its own right — Lichnerowicz's conformal method, and the York decomposition, are its standard tools.

35 · consequences

The wave operator, built here

definition

On Minkowski space, the wave operator is the second-order operator obtained by contracting two derivatives with the inverse metric: $$\Box f = \eta^{\mu\nu}\partial_\mu\partial_\nu f = -\frac{1}{c^2}\frac{\partial^2 f}{\partial t^2} + \nabla^2 f.$$

lemma · it is Lorentz invariant

If $\tilde x = Lx$ with $L$ in the Lorentz group, then $\tilde\Box = \Box$.

proof

By unit 03's chain rule $\partial_\mu = L^\alpha{}_\mu\tilde\partial_\alpha$, so $\Box = \eta^{\mu\nu}L^\alpha{}_\mu L^\beta{}_\nu\,\tilde\partial_\alpha\tilde\partial_\beta = \eta^{\alpha\beta}\tilde\partial_\alpha\tilde\partial_\beta$, the last step being unit 26's defining property of $L$, $L^{\mathsf T}\eta L = \eta$, inverted to $L\eta^{-1}L^{\mathsf T} = \eta^{-1}$. $\square$

lemma · plane waves travel at $c$

For $F\in C^2(\mathbb R)$ and a constant covector $k$, $f = F(k_\mu x^\mu)$ has $\Box f = \eta^{\mu\nu}k_\mu k_\nu\,F''$. So $f$ solves $\Box f = 0$ for every profile $F$ iff $k$ is null.

Read it. $k = (-\omega/c,\,0,0,\,\omega/c)$ is null, and $F(k_\mu x^\mu) = F(\tfrac{\omega}{c}(z - ct))$: any shape, moving along $z$ at speed $c$, unchanged.

Check. $[\Box] = \mathrm{m^{-2}}$, both terms ✓. Static limit: $\partial_t = 0$ gives $\Box = \nabla^2$, unit 05's Laplacian ✓.

Why this slide exists. Nothing on this unit's route has met the wave operator — units 18 and 24, which study it properly, are branches this unit does not need. So the three facts about it that section 6 uses are proved here, and the one it uses without proof is labelled.

Proof of the plane-wave lemma. $\partial_\nu F(k\cdot x) = k_\nu F'$ by the chain rule, and again $\partial_\mu\partial_\nu F(k\cdot x) = k_\mu k_\nu F''$. Contract with $\eta^{\mu\nu}$. If $\eta^{\mu\nu}k_\mu k_\nu \ne 0$, take $F(s) = s^2$ to get $\Box f \ne 0$. $\square$

Cited, not proved. That the inhomogeneous equation $\Box f = s$ has a solution for every smooth source $s$ — needed on the linearisation slide to reach Lorenz gauge. Unit 24 constructs it, the retarded solution $f(t,\vec x) = -\tfrac{1}{4\pi}\int s(t - |\vec x - \vec y|/c,\,\vec y)\, |\vec x - \vec y|^{-1}\,\mathrm d^3y$, using unit 20's contour integrals; units 18 and 24 are the place to see it proved. In the static case, with $s = 4\pi G\rho$, it reduces to unit 05's superposition formula $\Phi(\vec x) = -G\int\rho(\vec y)\,|\vec x - \vec y|^{-1}\,\mathrm d^3y$, which is on this unit's route.

The curved version. On a spacetime $(M,g)$ the same contraction with covariant derivatives, $g^{\alpha\beta}\nabla_\alpha\nabla_\beta$, is the curved wave operator. It is what "harmonic coordinates" refers to on the well-posedness slide.

35 · consequences

Waves at c, drawn

In one space dimension the wave equation can be solved completely. With $u = z - ct$ and $v = z + ct$ the chain rule turns it into

$$\partial_z^2 f - \frac{1}{c^2}\partial_t^2 f = 4\,\partial_u\partial_v f = 0,$$

so $\partial_v f$ does not depend on $u$, and

$$f = A(z - ct) + B(z + ct).$$

Two arbitrary profiles, one moving each way at $c$, and nothing else. The plane-wave lemma is the three-dimensional version of $A$.

Each profile is constant along a $45^\circ$ line: $A(z - ct)$ moves right, $B(z + ct)$ left, and they pass through each other. In one space dimension that is every solution (notes).

The general solution in one space dimension, d'Alembert's. Put $u = z - ct$, $v = z + ct$. The chain rule gives $\partial_z = \partial_u + \partial_v$ and $c^{-1}\partial_t = -\partial_u + \partial_v$, so $\partial_z^2 - c^{-2}\partial_t^2 = (\partial_u + \partial_v)^2 - (\partial_v - \partial_u)^2 = 4\,\partial_u\partial_v$ (Clairaut lets the mixed partials commute). Then $\partial_u\partial_v f = 0$ says $\partial_v f$ is independent of $u$, so $f = A(u) + B(v)$: one wave moving right at $c$, one moving left. Nothing else. The figure draws exactly this: each pulse is a function of one of $u$, $v$ alone, so it keeps its shape along the $45^\circ$ lines $u = $ const or $v = $ const, and because the equation is linear the two simply add where they cross.

What to look for. On a spacetime diagram with equal scales for $z$ and $ct$, a signal at speed $c$ is a $45^\circ$ line. Each pulse keeps its shape along such a line — that is the content of "$A$ depends on $z - ct$ only". The next slide's linearised equation says that every component of $\bar h_{\mu\nu}$ in vacuum behaves like this.

35 · consequences

Linearise: a wave with two polarisations

worked example 5 · the vacuum equation at first order

Step 1. Take the weak field but drop staticity: $g = \eta + h$, first order, $T = 0$. Define the trace-reversed perturbation $\bar h_{\mu\nu} = h_{\mu\nu} - \tfrac12\eta_{\mu\nu}h$ with $h = \eta^{\mu\nu}h_{\mu\nu}$.

Step 2. Use the gauge freedom. Under $x^\mu\to x^\mu + \xi^\mu$, $h_{\mu\nu}\to h_{\mu\nu} - \partial_\mu\xi_\nu - \partial_\nu\xi_\mu$. Then $\partial^\mu\bar h_{\mu\nu}\to\partial^\mu\bar h_{\mu\nu} - \Box\xi_\nu$, so solving $\Box\xi_\nu = \partial^\mu\bar h_{\mu\nu}$ (cited solvability, the wave-operator slide) imposes the four Lorenz gauge conditions $\partial^\mu\bar h_{\mu\nu} = 0$.

Step 3. In that gauge the linearised Einstein tensor collapses to $G^{(1)}_{\mu\nu} = -\tfrac12\Box\bar h_{\mu\nu}$, so the field equations become $$\Box\,\bar h_{\mu\nu} = -2\kappa\,T_{\mu\nu} = -\frac{16\pi G}{c^4}T_{\mu\nu},\qquad \Box\,\bar h_{\mu\nu} = 0 \text{ in vacuum}.$$ The wave equation of the wave-operator slide, one copy per component: vacuum perturbations travel at $c$.

Step 4. Count the polarisations. Lorenz gauge does not use up all the freedom: any $\xi$ with $\Box\xi^\mu = 0$ preserves it, four more functions. Fixing them gives transverse-traceless gauge, in which $\bar h = h$ and, for a wave along $z$, $$h_{\mu\nu} = \begin{pmatrix}0&0&0&0\\0&h_+&h_\times&0\\0&h_\times&-h_+&0\\0&0&0&0\end{pmatrix}.$$ Two free functions: plus and cross polarisations. $10 - 4 - 4 = 2$, confirmed by a second and independent route.

Step 3, in full. The linearised Ricci tensor is $R^{(1)}_{\mu\nu} = \tfrac12 (\partial_\alpha\partial_\mu h^\alpha{}_\nu + \partial_\alpha\partial_\nu h^\alpha{}_\mu - \Box h_{\mu\nu} - \partial_\mu\partial_\nu h)$, obtained exactly as $R_{00}$ was in section 4 but without dropping time derivatives. Forming $G^{(1)}_{\mu\nu} = R^{(1)}_{\mu\nu} - \tfrac12\eta_{\mu\nu}R^{(1)}$ and substituting $h_{\mu\nu} = \bar h_{\mu\nu} - \tfrac12\eta_{\mu\nu}\bar h$ (trace reversal is an involution, so this inverts the definition) gives $G^{(1)}_{\mu\nu} = -\tfrac12\bigl(\Box\bar h_{\mu\nu} + \eta_{\mu\nu}\partial^\alpha\partial^\beta\bar h_{\alpha\beta} - \partial_\mu\partial^\alpha\bar h_{\nu\alpha} - \partial_\nu\partial^\alpha\bar h_{\mu\alpha}\bigr)$. Every term but the first carries a factor $\partial^\alpha\bar h_{\alpha\beta}$, which the Lorenz condition kills. That is why the gauge is chosen: the operator becomes the wave operator on the nose.

Where the gauge transformation comes from. A diffeomorphism generated by $\xi$ changes the components of the metric, to first order, by minus its Lie derivative (unit 31). Unit 31's coordinate formula for $\mathcal L_\xi g$, applied to the constant $\eta$, gives $\eta_{\lambda\nu}\partial_\mu\xi^\lambda + \eta_{\mu\lambda}\partial_\nu\xi^\lambda = \partial_\mu\xi_\nu + \partial_\nu\xi_\mu$; the part acting on $h$ is second order. Hence $h_{\mu\nu}\to h_{\mu\nu} - \partial_\mu\xi_\nu - \partial_\nu\xi_\mu$. Its trace changes by $-2\partial^\lambda\xi_\lambda$, so $\bar h_{\mu\nu}$ changes by $-\partial_\mu\xi_\nu - \partial_\nu\xi_\mu + \eta_{\mu\nu}\partial^\lambda\xi_\lambda$, and taking $\partial^\mu$ of that gives $-\Box\xi_\nu$, as step 2 says.

The trace reversal, twice over. $\bar h$ is the trace reversal of $h$, and $\bar{\bar h} = h$ because the operation is an involution in four dimensions. In transverse- traceless gauge $h = 0$, so $\bar h = h$ and the bar may be dropped — which is why the literature drops it there without comment.

Why the residual gauge is exactly four functions. A gauge transformation preserves $\partial^\mu\bar h_{\mu\nu} = 0$ if and only if $\Box\xi_\nu = 0$. That is a wave equation for each of four functions, so its solution space is parametrised by four functions of three variables — the initial data $(\xi,\dot\xi)$ on a slice, modulo the equation. Using them, one may impose $h_{0\mu} = 0$ and $h^i{}_i = 0$: that is $4 + 1 = 5$ conditions, but one is redundant given the Lorenz condition, so exactly four. The result is the matrix above.

A number to hold on to for unit 38. The strain $h$ from the first detected black-hole merger, GW150914, peaked at about $1.0\times10^{-21}$ — a length change of $4\times10^{-18}$ m over LIGO's 4 km arms, about a two-hundredth of a proton radius. The smallness is $\kappa$ again: the coupling is $2.08\times10^{-43}$, and spacetime is stiff.

35 · consequences

Plus and cross, drawn

Free particles at rest keep their transverse-traceless coordinates (notes), so the wave shows up in the proper distances: two particles a coordinate distance $L$ apart along $x$ are $\int_0^L\sqrt{1 + h_+}\,\mathrm dx \approx L(1 + \tfrac12h_+)$ apart, and along $y$, $L(1 - \tfrac12 h_+)$.

A ring of free particles in the $x$–$y$ plane as a wave $h_+ = h\sin\omega t$ (top) or $h_\times = h\sin\omega t$ (bottom) passes along $z$, out of the page ($h$ exaggerated to $0.5$). Cross is plus turned by $45^\circ$; nothing moves along $z$.

Why the particles keep their coordinates. A particle at rest has $\mathrm dx^\mu/\mathrm d\tau = (c,0,0,0)$, so the geodesic equation gives $\mathrm d^2x^i/\mathrm d\tau^2 = -c^2\Gamma^i_{00}$. To first order $\Gamma^i_{00} = \tfrac12(2\partial_0h_{i0} - \partial_ih_{00})$, and in transverse-traceless gauge $h_{0\mu} = 0$, so $\Gamma^i_{00} = 0$: the particle stays at fixed $x^i$. The coordinates are glued to the free particles, and all the physics is in the metric between them. Unit 38 does this with geodesic deviation, which is gauge-free, and gets the same pattern.

How the rings were drawn. A particle at coordinates $(x, y)$ is drawn at its proper position: for plus, $(x(1 + \tfrac12h_+),\ y(1 - \tfrac12h_+))$; for cross, the metric has $2h_\times\,\mathrm dx\,\mathrm dy$, giving $(x + \tfrac12h_\times y,\ y + \tfrac12h_\times x)$, which stretches along the diagonal $x = y$ and squeezes along $x = -y$. With $h = 0.5$ the stretch is 25 per cent; a real wave at the Earth has $h \sim 10^{-21}$.

Why a rotation by 45° and not 90°. Rotating the plus pattern by $90^\circ$ swaps stretch and squeeze, which is the same as $h_+\to-h_+$: the same polarisation, half a period later. So the independent one sits halfway, at $45^\circ$. For light, rotating a polarisation by $180^\circ$ gives its negative and the independent one is at $90^\circ$.

35 · consequences

The weak field, a third time

Put the running example back: static, dust at rest, so $T_{00} = \rho c^2$ and every other component zero. The linearised equation $\Box\bar h_{\mu\nu} = -2\kappa T_{\mu\nu}$ becomes, with $\Box = \nabla^2$ for a static field,

$$\nabla^2\bar h_{00} = -2\kappa\rho c^2,\qquad \nabla^2\bar h_{0i} = 0,\qquad \nabla^2\bar h_{ij} = 0.$$

Fields that vanish at infinity and are harmonic everywhere are zero (unit 05's uniqueness), so only $\bar h_{00}$ survives. Undo the trace reversal: $\bar h = \eta^{\mu\nu}\bar h_{\mu\nu} = -\bar h_{00}$, and $h_{\mu\nu} = \bar h_{\mu\nu} - \tfrac12\eta_{\mu\nu}\bar h$ gives

$$h_{00} = \tfrac12\bar h_{00},\qquad h_{ij} = \tfrac12\bar h_{00}\,\delta_{ij}.$$

two results

The constant, again. With $h_{00} = -2\Phi/c^2$, $\bar h_{00} = -4\Phi/c^2$, and the first equation reads $-4\nabla^2\Phi/c^2 = -2\kappa\rho c^2$, i.e. $\nabla^2\Phi = \tfrac12\kappa c^4\rho$. Poisson again iff $\kappa = 8\pi G/c^4$ — a third derivation, through the linearised equation.

Something new. $h_{ij} = -\dfrac{2\Phi}{c^2}\delta_{ij}$. Space is curved too, by exactly as much as time. That is the factor of two in light deflection, and unit 37 cashes it.

Check. $\Phi = -GM/r$ at the Sun's surface gives $|h_{ij}| = 4.2\times10^{-6}$, dimensionless and small ✓.

Why only $\bar h_{00}$ survives, carefully. Each of $\bar h_{0i}$, $\bar h_{ij}$ satisfies Laplace's equation on all of $\mathbb R^3$ and tends to zero at infinity (an isolated source, far away the metric is flat). By the maximum principle for harmonic functions, proved with the uniqueness theorem in unit 05, such a function is identically zero. That is where "static and isolated" is used; a gravitomagnetic $\bar h_{0i}$ needs a source $T_{0i}\ne0$, moving matter, which the slow limit discards.

Is this gauge the one unit 32 used? Unit 32 identified $h_{00} = -2\Phi/c^2$ from slow geodesics, and that identification is gauge-independent at this order: a static gauge transformation $\xi$ changes $h_{00}$ by $-2\partial_0\xi_0 = 0$. So the $h_{00}$ here and unit 32's are the same function. The spatial part $h_{ij}$ is gauge-dependent, and $-2\Phi\delta_{ij}/c^2$ is its form in Lorenz gauge — the "isotropic" form, which is the far field of the Schwarzschild metric when that is written in isotropic coordinates.

The number. $2GM_\odot/(R_\odot c^2) = 2\times2.1225\times10^{-6} = 4.245\times10^{-6}$.

Three derivations of one constant. Section 4 used the trace-reversed equation and the $00$ Ricci component; section 5 an action; this slide the linearised Einstein tensor in Lorenz gauge. The last two share section 4's Newtonian input, so they are checks of algebra rather than of physics — but checks of algebra are what a learner working alone needs.

35 · consequences

The cosmological term

The cosmological constant term $\Lambda g_{\mu\nu}$ may be moved to the right-hand side and read as a source: $$T^{(\Lambda)}_{\mu\nu} = -\frac{\Lambda c^4}{8\pi G}\,g_{\mu\nu},$$ which is a perfect fluid with $\rho_\Lambda = \Lambda c^2/8\pi G$ and $p_\Lambda = -\rho_\Lambda c^2$. Negative pressure, and by "pressure gravitates" a total $\rho + 3p/c^2 = -2\rho_\Lambda$: it pushes.

Section 4's "pressure gravitates" says the source of attraction is $\rho + 3p/c^2$, not $\rho$. The figure puts the three sources of this unit on that one scale.

The gravitating combination $\rho + 3p/c^2$, in units of $\rho$. Radiation's pressure doubles its pull; the cosmological term's negative pressure more than cancels its positive density, and it is the only source that pushes.

The three bars. Dust: $p = 0$, so $1$. Radiation: $p = \tfrac13\rho c^2$, so $1 + 1 = 2$, the "twice as strong" of the pressure slide. The cosmological term: $p = -\rho c^2$, so $1 - 3 = -2$. A ball of test particles inside each would, by the Weyl-and-Ricci slide's volume rule $\ddot V/V = -4\pi G(\rho + 3p/c^2)$, shrink, shrink twice as fast, and grow. That growth, at cosmological scale, is the accelerating expansion.

The fluid, checked. A perfect fluid has $T_{\mu\nu} = (\rho + p/c^2)u_\mu u_\nu + p\,g_{\mu\nu}$. With $p = -\rho c^2$ the first bracket vanishes and $T_{\mu\nu} = -\rho c^2g_{\mu\nu}$, which is $-(\Lambda c^4/8\pi G)g_{\mu\nu}$ when $\rho = \Lambda c^2/8\pi G$ ✓. Because $u$ dropped out, every observer sees the same density: it is the only fluid with no rest frame, which is what one expects of a property of empty space.

Is $\Lambda$ geometry or matter? The equation cannot tell. On the left it is a property of spacetime; on the right it is a fluid with a strange equation of state. The two are the same equation, and no measurement distinguishes them unless $\Lambda$ turns out to vary — which is what "dark energy" as opposed to "cosmological constant" means, and what the surveys are looking for. Unit 39 returns to it.

35 · consequences

The cosmological term, measured

the measured value · cited

Cosmology (unit 39) measures two numbers: the present expansion rate of the universe, the Hubble constant $H_0 = 67.4\ \mathrm{km\,s^{-1}Mpc^{-1}}$, and the share of the cosmological term in the universe's total density, $\Omega_\Lambda = 0.685$. Unit 39's Friedmann equation turns them into $\Lambda = 3\Omega_\Lambda H_0^2/c^2 = 1.09\times10^{-52}\ \mathrm{m^{-2}}$, so $\rho_\Lambda = 5.85\times10^{-27}\ \mathrm{kg\,m^{-3}}$ — about 3.5 hydrogen atoms per cubic metre, or $5.3\times10^{-10}\ \mathrm{J\,m^{-3}}$.

History, without editorial. Einstein introduced $\Lambda$ in 1917 to permit a static universe. Friedmann (1922) and Lemaître (1927) found expanding solutions; Hubble (1929) found the expansion. Einstein then dropped the term. In 1998 two supernova surveys found the expansion accelerating, which $\Lambda\gt0$ explains, and it returned under the name dark energy.

the open problem, stated plainly

Quantum field theory expects a vacuum energy density; a naive cutoff at the Planck scale gives $c^5/\hbar G^2 = 5.15\times10^{96}\ \mathrm{kg\,m^{-3}}$, which exceeds the measured $\rho_\Lambda$ by a factor $8.8\times10^{122}$. No accepted resolution exists. This is a statement about the size of a mismatch, not about what causes it.

The arithmetic. $H_0 = 67.4\ \mathrm{km\,s^{-1}Mpc^{-1}} = 67.4\times10^3/ 3.0857\times10^{22} = 2.184\times10^{-18}\ \mathrm{s^{-1}}$. Then $\Lambda = 3\Omega_\Lambda H_0^2/c^2 = 3\times0.685\times(2.184\times10^{-18})^2/(8.988\times10^{16}) = 1.091\times10^{-52}\ \mathrm{m^{-2}}$. And $\rho_\Lambda = \Lambda c^2/8\pi G = 1.091\times10^{-52}\times8.988\times10^{16}/1.6774\times10^{-9} = 5.85\times10^{-27}\ \mathrm{kg\,m^{-3}}$; dividing by the hydrogen mass $1.674\times10^{-27}$ kg gives 3.49 atoms per cubic metre, and multiplying by $c^2$ gives $5.25\times10^{-10}\ \mathrm{J\,m^{-3}}$, about 3.3 GeV per cubic metre. The Planck density is $c^5/\hbar G^2 = 5.15\times10^{96}\ \mathrm{kg\,m^{-3}}$ and the ratio is $8.8\times10^{122}$. Quoted values of "the worst prediction in physics" range from $10^{60}$ to $10^{123}$ depending on which cutoff and which density is compared; the figure above is the plainest version, Planck density over measured density, and it is computed rather than remembered.

"Einstein's greatest blunder". The phrase is attributed to Einstein by Gamow, in Gamow's 1970 autobiography, and there is no contemporaneous record of Einstein saying it. It is quoted constantly and its provenance is thin. What is documented is that Einstein dropped the term after 1931 and wrote that it was "theoretically unsatisfactory anyway". Separately, Eddington showed in 1930 that the static solution is unstable: $\Lambda$ balanced against matter is a maximum of the effective potential, so the tiniest perturbation grows. That is an objection to the 1917 model independent of any observation.

35 · consequences

Energy is not conserved, and that is not a gap

$\nabla_\mu T^{\mu\nu} = 0$ is local. Expanding the covariant divergence: $$\nabla_\mu T^{\mu\nu} = \frac{1}{\sqrt{-g}}\partial_\mu\bigl(\sqrt{-g}\,T^{\mu\nu}\bigr) + \Gamma^\nu_{\mu\lambda}T^{\mu\lambda}.$$

the obstruction

Only the first term is an ordinary divergence, integrable by the divergence theorem into a statement about a total. The $\Gamma$ term is not, and it does not vanish. So there is no integral conservation law for energy-momentum in a general spacetime.

Contrast a current: $\nabla_\mu J^\mu = (1/\sqrt{-g})\,\partial_\mu(\sqrt{-g}J^\mu)$, with no extra term, because $J$ has one index. Electric charge is globally conserved. The problem is the index, not the curvature.

Between two slices $\Sigma_1$ and $\Sigma_2$, schematic. For a current every flow line that enters through $\Sigma_1$ leaves through $\Sigma_2$, so the totals agree. For $T^{\mu\nu}$ the $\Gamma$ term acts like sources and sinks spread through the region, and the totals need not agree.

Reading the figure. Integrate $\partial_\mu(\sqrt{-g}J^\mu) = 0$ over the region between the slices, with $J$ vanishing far away. The divergence theorem turns it into the flux through $\Sigma_2$ minus the flux through $\Sigma_1$, so the charge $Q(\Sigma) = \int_\Sigma\sqrt{-g}J^0\,\mathrm d^3x$ is the same on both. For $T^{\mu\nu}$, fix $\nu$ and treat $T^{\mu\nu}$ as if it were a current: now $\partial_\mu(\sqrt{-g}T^{\mu\nu}) = -\sqrt{-g}\,\Gamma^\nu_{\mu\lambda}T^{\mu\lambda}$, a source term inside the region, and the difference of the two fluxes is its integral. Worse, "fix $\nu$" is not a coordinate-free instruction, so even the four totals are not well defined. The right-hand panel draws the source term as lines that begin or end in the bulk.

Do not say "energy is conserved if you include the energy of the gravitational field". It is a popular repair and it is not a theorem. One can define pseudotensors — Einstein's, Landau-Lifshitz's, and others — whose sum with $T^{\mu\nu}$ has vanishing ordinary divergence, but they are not tensors: their values at a point can be set to zero by a coordinate change, precisely because the equivalence principle says the gravitational field can be transformed away locally. So there is no local gravitational energy density. What does exist are global notions of energy for spacetimes with suitable asymptotic symmetry — the ADM mass for asymptotically flat spacetimes, and the Bondi mass at null infinity, whose decrease is how a merging binary's radiated energy is defined. Those are theorems; the pseudotensor is a bookkeeping device.

35 · consequences

When energy is conserved: a symmetry

when it does work

If $\xi$ is a Killing field (unit 31, $\mathcal L_\xi g = 0$, equivalently $\nabla_{(\mu}\xi_{\nu)} = 0$), then $J^\mu = T^{\mu\nu}\xi_\nu$ satisfies

$$\begin{aligned}\nabla_\mu J^\mu &= (\nabla_\mu T^{\mu\nu})\,\xi_\nu\\ &\quad + T^{\mu\nu}\,\nabla_\mu\xi_\nu = 0,\end{aligned}$$

the first term by local conservation, the second because $T$ is symmetric and $\nabla_\mu\xi_\nu$ antisymmetric. So a symmetry gives a conserved charge — Noether, exactly as in units 13 and 29.

The repair is to manufacture a current. A vector field $\xi$ turns the two-index $T^{\mu\nu}$ into the one-index $J^\mu = T^{\mu\nu}\xi_\nu$, and a one-index current has no $\Gamma$ term — the left panel of the previous figure. What it costs is that $\xi$ must be a symmetry of the metric.

Energy is conserved when spacetime has a timelike Killing field. Schwarzschild does (unit 36). An expanding universe does not, and cosmological redshift is the visible consequence.

Why $\mathcal L_\xi g = 0$ is Killing's equation. Unit 31's coordinate formula is $(\mathcal L_\xi g)_{\mu\nu} = \xi^\lambda\partial_\lambda g_{\mu\nu} + g_{\lambda\nu}\partial_\mu\xi^\lambda + g_{\mu\lambda}\partial_\nu\xi^\lambda$. Every partial may be replaced by a covariant derivative: the $\Gamma$-terms this introduces cancel in pairs because $\Gamma$ is symmetric (no torsion). Then $\nabla g = 0$ kills the first term and the other two are $\nabla_\mu\xi_\nu + \nabla_\nu\xi_\mu$. $\square$

Proof of the antisymmetry step. Killing's equation is $\nabla_\mu\xi_\nu + \nabla_\nu\xi_\mu = 0$, i.e. $\nabla_\mu\xi_\nu$ is antisymmetric. Contracting an antisymmetric tensor with the symmetric $T^{\mu\nu}$ gives zero: $T^{\mu\nu}A_{\mu\nu} = T^{\nu\mu}A_{\nu\mu} = -T^{\mu\nu}A_{\mu\nu}$, so it equals its own negative. $\square$ The identity $\nabla_\mu J^\mu = (1/\sqrt{-g})\partial_\mu(\sqrt{-g}J^\mu)$ for a vector follows from $\Gamma^\mu_{\mu\lambda} = \partial_\lambda\ln\sqrt{-g}$, itself Jacobi's formula from the determinant lemma; it is problem P10.

The redshift, concretely. In an expanding universe a photon emitted with wavelength $\lambda_e$ arrives with $\lambda_o = \lambda_e(1+z)$, so its energy has fallen by a factor $1+z$. The cosmic microwave background was emitted at $z\approx1090$ and has lost that factor: its energy density today is about $4.2\times10^{-14}\ \mathrm{J\,m^{-3}}$, roughly $10^{-4}$ of $\rho_\Lambda c^2$. Asking where the energy went is asking for a conserved quantity that does not exist, because FLRW has no timelike Killing field.

35 · consequences

Is it even well posed?

Ten coupled nonlinear second-order PDEs for ten functions, four of which are pure gauge. Whether such a system has solutions at all, determined by data on a spacelike slice, is not obvious — and was not settled for forty years.

The obstacle is the gauge freedom. If one solution exists, every diffeomorphism of it is another, so data on a slice can never fix the solution's components. The cure is to spend the four free functions on a choice of coordinates that turns the equations into wave equations.

Below, $\Box$ is the curved wave operator $g^{\alpha\beta}\nabla_\alpha\nabla_\beta$: the wave-operator slide's $\Box$ with $\eta \to g$ and $\partial \to \nabla$.

harmonic coordinates

Impose $\Box x^\mu = 0$, equivalently $g^{\alpha\beta}\Gamma^\mu_{\alpha\beta} = 0$: four conditions, using up exactly the four functions of gauge freedom. In them, $$R_{\mu\nu} = -\tfrac12 g^{\alpha\beta}\partial_\alpha\partial_\beta g_{\mu\nu} + (\text{first derivatives}),$$ manifestly a quasilinear wave operator, one per metric component.

The harmonic-gauge Ricci tensor, justified. For any function, $g^{\alpha\beta}\nabla_\alpha\nabla_\beta f = g^{\alpha\beta}(\partial_\alpha\partial_\beta f - \Gamma^\lambda_{\alpha\beta}\partial_\lambda f)$; with $f = x^\mu$ the first term vanishes, giving $-\Gamma^\mu$ where $\Gamma^\mu \equiv g^{\alpha\beta}\Gamma^\mu_{\alpha\beta}$. Write out $R_{\mu\nu}$ from unit 33's formula, keeping only terms with two derivatives of $g$: $R_{\mu\nu} = -\tfrac12 g^{\alpha\beta}\partial_\alpha\partial_\beta g_{\mu\nu} + \tfrac12(g_{\lambda\mu}\partial_\nu\Gamma^\lambda + g_{\lambda\nu}\partial_\mu\Gamma^\lambda) + (\text{terms quadratic in }\partial g)$. The middle bracket is the whole obstruction to hyperbolicity, and harmonic coordinates set $\Gamma^\lambda = 0$. The algebra is a page and a good exercise; it is done in full in Wald, General Relativity, §10.2.

That the vacuum system $-\tfrac12g^{\alpha\beta}\partial_\alpha\partial_\beta g_{\mu\nu} + \ldots = 0$ then has local solutions is a theorem about quasilinear hyperbolic systems — cited, with the proof in Choquet-Bruhat's paper and in Ringström, The Cauchy Problem in General Relativity (2009), which uses Fourier-analytic energy estimates beyond this course.

Why it took forty years. Three genuine difficulties. First, the gauge freedom means the equations are not hyperbolic as they stand — they cannot be, since solutions are never unique, only unique up to diffeomorphism. Harmonic coordinates repair this, but one must then prove the gauge is preserved by the evolution, which is where the contracted Bianchi identity returns for the third time in this unit. Second, the constraints must be shown to propagate. Third, the local result must be patched into a global one, which needed Geroch's Zorn's-lemma argument on the set of all developments — a 1969 paper that had to wait for the language of global hyperbolicity to exist.

35 · consequences

Data on a slice, and what it determines

cited · Choquet-Bruhat 1952

Given initial data $(\gamma_{ij}, K_{ij})$ on a spacelike slice satisfying the four constraint equations, there is a solution of the vacuum Einstein equations inducing that data, unique up to diffeomorphism in a neighbourhood of the slice.

Here $\gamma_{ij}$ is the metric the slice inherits — its own lengths — and $K_{ij}$, its extrinsic curvature, measures how the slice bends inside spacetime; it plays the part of $\partial_t\gamma_{ij}$. Y. Fourès-Bruhat (later Choquet-Bruhat), Acta Math. 88 (1952) 141.

Choquet-Bruhat and Geroch (1969) extended it to one largest solution, unique up to diffeomorphism, in which every causal curve that cannot be extended further crosses the slice exactly once — a globally hyperbolic spacetime, one where the slice sees everything.

Data on a piece $S$ of the slice fix the solution in the region $D(S)$ above it, bounded by light rays from the ends of $S$. At $p$ every past-directed signal meets $S$; at $q$ some meet the slice outside $S$, so $S$ alone cannot decide what happens there.

So the initial-value formulation has the shape of unit 01's: data on a slice determine the evolution — here up to diffeomorphism, with signals moving no faster than $c$. General relativity is a field theory in the ordinary sense, and this is the theorem that says so.

The data, made concrete. In coordinates where the slice is $t = 0$ and $g_{00} = -1$, $g_{0i} = 0$ near it, $\gamma_{ij}$ is simply $g_{ij}$ on the slice and $K_{ij} = \pm\tfrac{1}{2c}\partial_t g_{ij}$, the sign being a convention that differs between books. So $(\gamma, K)$ is "$g$ and $\dot g$" — unit 01's position and velocity — for the spatial metric. The remaining components $g_{00}$, $g_{0i}$ carry no data: they are the gauge. The covariant definition, $K_{\mu\nu}$ as $\nabla_\mu n_\nu$ projected onto the slice, is in the notes to The Gibbons-Hawking-York term, and is cited.

Reading the figure. $D(S)$ is the domain of dependence of $S$: the set of points every past-directed causal curve of which meets $S$. The 45° boundaries are the statement that nothing travels faster than light, which the theorem proves rather than assumes — the solution's light cones are part of the solution. Globally hyperbolic means $D(\Sigma)$ is the whole spacetime.

Why anyone should care beyond tidiness. Numerical relativity — the simulations whose waveform templates LIGO matched against GW150914 — is the numerical solution of this initial-value problem. It did not work for decades, because naive formulations are only weakly hyperbolic and constraint violations grow exponentially. The breakthroughs of 2005 (Pretorius; Campanelli et al.; Baker et al.) were, in essence, reformulations with better hyperbolicity. The abstract theorem and the working simulations are about the same difficulty.

Choquet-Bruhat. Yvonne Choquet-Bruhat (1923-2025) proved the local theorem in 1952, aged 28 and publishing as Y. Fourès-Bruhat, and it remains the foundation of the subject. She was the first woman elected to the French Academy of Sciences, in 1979.

35 · consequences

What the running example bought

one example, six uses

§1 — it said what has to come out: Poisson's equation, and it gave the numbers showing $|h|\ll1$ is not a convenience.
§2 — it fixed the source: dust, $T_{00} = \rho c^2$.
§3 — it was what the Bianchi candidate had to reproduce.
§4 — it produced $\kappa = 8\pi G/c^4$, the only number geometry could not supply.
§5 — it checked the action route against the same constant.
§6 — dropping only staticity turned it into the wave equation; restoring it gave $\kappa$ a third time and $h_{ij} = -2\Phi\delta_{ij}/c^2$.

What is next. Unit 36 solves $R_{\mu\nu} = 0$ exactly for a spherically symmetric field and finds the integration constant by matching to this unit's weak field. Unit 37 expands that solution and gets 43 arcseconds, 1.75 arcseconds and the Shapiro delay. Unit 38 solves $\Box\bar h_{\mu\nu} = -2\kappa T_{\mu\nu}$ with unit 24's retarded Green's function. Unit 39 drops staticity and weakness and finds the Friedmann equations. Unit 40 says where the methods stop.

one honest limitation

Nothing in this unit proves general relativity is true. It proves that given five requirements the equation is unique, and that two routes agree. Truth is unit 37's business, and it is measured in arcseconds.

35 · checkpoint

Checkpoint

you can now do this, or you cannot

You argue from the contracted Bianchi identity and Lovelock's theorem why the Einstein tensor is forced, fix the coupling at $8\pi G/c^4$ from the Newtonian limit, and linearise the equations to obtain the wave equation with the correct count of propagating degrees of freedom — and, if you have unit 29, you obtain the same field equations independently by varying the Einstein-Hilbert action.

Failing it is unambiguous. If you cannot say in one sentence why $R_{\mu\nu} = \kappa T_{\mu\nu}$ is wrong, you have not got the Bianchi argument. If your $\kappa$ has a stray factor of two, redo worked example 3 with the signs written out. If you cannot get to 2 from 10 by two different routes, redo worked example 5.

Thirteen problems, about six hours: ./mr problems 35. P1 reconstructs November 1915; P8 varies the action with every boundary term kept; P12 obtains the two polarisations and counts them twice. When the checkpoint is true: ./mr done 35.

The four clauses are separable and should be tested separately. Clause one is P1 and P2; clause two is P5 and P6; clause three is P12, with P11 and P13 behind it; clause four is P8, with P9 as its one-dimensional rehearsal, and it requires unit 29. A learner without unit 29 is complete on the first three clauses and should mark the unit done on that basis, returning to P8 after unit 29.

The most common way to think the checkpoint is passed when it is not: reproducing $\kappa = 8\pi G/c^4$ from memory rather than from the computation. The criterion is whether you can get it with the factors of $c$ in place, from $R_{00} = -\tfrac12\nabla^2h_{00}$ and $h_{00} = -2\Phi/c^2$, without looking. If the $\tfrac12$ from the trace reversal is not in your working, the agreement with $4\pi G$ was a coincidence of memory.