Part III · General Relativity — Chapter 3.6

The Einstein Field Equations

The law of gravity, derived three times: cornered by a conservation requirement, extremised from an action, and its one free constant fixed by demanding that apples fall.

Tools you'll need  — Chapter 3.4 §7 above all: the second Bianchi identity, its double contraction (3.4.57), and the Einstein tensor (3.4.58) with $\nabla^{\mu}G_{\mu\nu}=0$. Also 3.4 §6 (Ricci and the scalar), §5.3 (locally inertial coordinates) and §4.5's $R_{00}=\nabla^{2}\Phi/c^{2}$, equation (3.4.33). Chapter 2.6 §10 for the energy–momentum tensor, its components, and $\partial_{\mu}T^{\mu\nu}=0$. Chapter 3.5 §6 for the volume element, Jacobi's formula (3.5.47) and the divergence identity (3.5.51), and §5 for Stokes' theorem. Chapter 3.3 §7 (the Christoffel formula and metric compatibility) and §8.3 (the Newtonian limit, and its $\Gamma^{i}{}_{00}=\partial_{i}\Phi/c^{2}$, equation (3.3.71)). Chapter 3.1 §6.5, the derived weak-field component $g_{00}=1+2\Phi/c^{2}$, equation (3.1.39). Chapter 1.2 §3 (varying an action) and §8.1 (the table this chapter's fourth line comes from). Chapter 0.7 §7.5 for Poisson's equation.

1 · What sits on the right-hand side, and why it is not mass

Here is where this section is going. We identify the source of gravity as the energy–momentum tensor of Chapter 2.6. We promote its conservation law to a curved manifold, and we write down its form for the two kinds of matter this book needs. The important consequence is that pressure is a source of gravity, and no Newtonian intuition supplies that.

1.1 · The candidate, and why nothing simpler will do

In Newtonian gravity the source is the mass density $\rho$, a single function. That cannot survive into a relativistic theory, and Chapter 2.5 §9 already established the reason. Mass is not additive, and it is not separately conserved.

Energy is both of those things, so energy is the natural replacement. But Chapter 2.5 §3 showed that energy is one component of a four-vector. A theory that sources gravity with energy alone would therefore source it with a frame-dependent quantity, and two observers in relative motion would compute different geometries for the same physical situation. That is not a difficulty to be worked around. It is a contradiction.

Chapter 2.6 §10 built the object that fixes this. The energy–momentum tensor $T^{\mu\nu}$ arose there as the Noether current of spacetime translations. Its components were computed there too, and they are worth having in front of us again:

Component	What it is
$T^{00}$	energy density
$T^{0i}$	$c\times$ momentum density, equivalently energy flux $/c$
$T^{ij}$	the stress: flux of $i$-momentum across a surface of constant $x^{j}$

The tensor is symmetric, $T^{\mu\nu}=T^{\nu\mu}$. Chapter 2.6 §10.1 checked that explicitly for the electromagnetic field, after fixing the Noether current with an improvement term. The same chapter, in §10.3, went on to prove

in flat spacetime for a closed system. That is four equations, one for each value of $\nu$. Together they express local conservation of energy and of each component of momentum.

1.2 · The same statement on a curved manifold

Equation (3.6.1) is not yet a statement about a manifold. The trouble is that $\partial_{\mu}$ of a tensor is not itself a tensor, as Chapter 3.3 §4 showed. We need a replacement that is, and it is this:

and the reason for the replacement is not an appeal to elegance. It is Chapter 3.4 §5.3's locally inertial coordinates.

At any chosen point there is a chart in which $g_{\mu\nu}=\eta_{\mu\nu}$ and $\Gamma^{\lambda}{}_{\mu\nu}=0$. In that chart $\nabla$ and $\partial$ coincide, so (3.6.2) reduces to (3.6.1), and the equivalence principle says the flat law must hold there. Now use the second half of the argument. Equation (3.6.2) is an equation between tensors, so by Chapter 2.4 §6, holding in one chart at a point means holding in every chart at that point. And the point was arbitrary.

1.3 · Dust, and then a fluid with pressure

Dust. Take non-interacting particles all moving together, with rest-mass density $\rho$ measured in their own rest frame and four-velocity field $u^{\mu}$. In the rest frame the only thing present is energy density $\rho c^{2}$, so $T^{00}=\rho c^{2}$ and every other component vanishes. There is exactly one symmetric tensor built from $\rho$ and $u$ with that property. At rest $u^{\mu}=(c,0,0,0)$, so the tensor we want is this one:

It gives $T^{00}=\rho c^{2}$ and nothing else, which is what was required, and it is built from tensors, so it is one. Its trace is worth recording while we are here. Using $u\cdot u=c^{2}$ from Chapter 2.5, $T\equiv g_{\mu\nu}T^{\mu\nu}=\rho c^{2}$.

Perfect fluid. Now allow isotropic pressure $p$. In the rest frame the stress block is $T^{ij}=p\,\delta^{ij}$. That is what the word "pressure" means: a flux of $i$-momentum in the $i$ direction, the same in every direction. The energy density is still $\rho c^{2}$, so in the rest frame $T^{\mu\nu}=\mathrm{diag}(\rho c^{2},p,p,p)$. We want a tensor that reproduces that, and only two objects are available to build it out of, $u^{\mu}$ and $g^{\mu\nu}$:

Let's check it component by component, in the rest frame, where $g^{\mu\nu}=\eta^{\mu\nu}$ and $u^{\mu}u^{\nu}$ has only a $00$ entry, equal to $c^{2}$. The $00$ component is $(\rho+p/c^{2})c^{2}-p\cdot1=\rho c^{2}+p-p=\rho c^{2}$ ✓. The $11$ component is $0-p\cdot(-1)=+p$ ✓, the minus sign coming from $\eta^{11}=-1$, and likewise for $22$ and $33$. Off-diagonal entries vanish on both sides. ✓ Setting $p=0$ returns (3.6.3).

Its trace is worth computing now, because §5 will need it. Use $g_{\mu\nu}u^{\mu}u^{\nu}=c^{2}$ together with $g_{\mu\nu}g^{\mu\nu}=\delta^{\mu}{}_{\mu}=4$ in four dimensions, and the trace comes out as

1.4 · Three consequences worth naming before we start

(i) Pressure gravitates. $T^{\mu\nu}$ contains $p$, so pressure appears in the source. This has no Newtonian counterpart at all. In $\nabla^{2}\Phi=4\pi G\rho$ the pressure of a gas is invisible. Section 5 will find exactly how it enters, and the answer is that the combination doing the gravitating is $\rho+3p/c^{2}$.

(ii) A hot body weighs more than a cold one. Not because its atoms have gained mass, but because $T^{00}$ counts every form of energy, including the kinetic energy of thermal motion and the energy stored in the fields binding it. Chapter 2.5 §9 showed the same arithmetic from the other side, in the mass deficit of a bound system.

(iii) A gas of radiation has a fixed pressure. Chapter 2.6 §10 built the electromagnetic energy–momentum tensor, and its closing summary recorded that the tensor is traceless, $\eta_{\mu\nu}T^{\mu\nu}=0$. So set (3.6.5) to zero and solve for the pressure, which gives

and that result is derived rather than quoted. The equation of state of light follows from the tracelessness of its stress tensor and from nothing else. Section 5 and Chapter 3.9 both use it, and Chapter 7.3 builds a symmetry principle on the tracelessness itself.

2 · What the equation is allowed to look like

Before writing anything down, let's list what the answer has to satisfy. Each item below is a constraint with a stated origin rather than a matter of taste. Section 3 will find that together they leave almost nothing to choose.

(C1) It must be an equation between tensors. The authority is Chapter 3.2 §7 together with Chapter 2.4 §6. A tensor equation true in one chart is true in all of them, and an equation that is not tensorial holds only in some charts. On a manifold there are no preferred charts for it to hold in. So a non-tensorial gravitational law would be a law about coordinates rather than about the world.

(C2) The source is a symmetric rank-2 tensor, so the geometry side must be one too. That is the content of §1. Since $T^{\mu\nu}$ is symmetric with ten independent components, whatever equals it has the same type. This immediately rules out the Riemann tensor itself, which has four indices, and the Ricci scalar alone, which has none.

(C3) It must contain no derivatives of the metric beyond the second. There are two independent reasons for this, and they are worth keeping separate.

The first is the Newtonian limit, and it is the honest one. Newtonian gravity is $\nabla^{2}\Phi=4\pi G\rho$, which is second order in the potential. Chapter 3.1 §6.5 derived $g_{00}=1+2\Phi/c^{2}$, so the potential lives inside the metric. An equation that reproduces Poisson must therefore be second order in $g$. Third or fourth derivatives would produce extra terms that Newtonian gravity does not have and that experiment does not see.

The second is the structure of every field equation so far in this book. Chapter 1.2 §8 showed that an action built from a field and its first derivatives yields an equation of motion with second derivatives, and every entry of that chapter's table of actions has that form. Section 4 below finds that the gravitational action is an exception in one specific and interesting way. The exception costs exactly one boundary term.

(C4) Its divergence must vanish identically. This is the constraint that does the work. Equation (3.6.2) says $\nabla_{\mu}T^{\mu\nu}=0$ for every physical matter configuration. Suppose the geometry side of the field equation had a divergence that vanished only sometimes. Then setting the two sides equal would impose a further condition on the matter, and that is a condition no experiment supports and no principle supplies.

So the geometrical side must have vanishing divergence as an identity. That means automatically, for every metric anybody could write down, whether or not it solves anything.

(C5) It must reduce to Newton where Newton works. Chapter 3.3 §8.3 established the correspondence limit for the motion of a body. The same limit for the field equation is §5's business, and it is what fixes the one remaining constant.

3 · The cornering — why $G_{\mu\nu}$ and nothing else

Here is where this section is going. We list every symmetric rank-2 tensor that can be built from the metric and its first two derivatives. We impose (C4), and we find that a one-parameter family survives, up to overall scale. That family is the Einstein tensor plus a multiple of the metric, and there is nothing else.

3.1 · The list of candidates

Start with what is available. By (C3) the ingredients are $g_{\mu\nu}$, its inverse, and its derivatives up to second order. Chapter 3.4 §2 showed that the only tensor that can be built from the metric and its first two derivatives is the Riemann tensor. The first derivatives alone give the connection, which is not a tensor, and the second derivatives enter tensorially only in the combination (3.4.13). So the ingredients are $g_{\mu\nu}$ and $R^{\rho}{}_{\sigma\mu\nu}$.

Now impose (C2), which asks for something symmetric with two indices. From $R^{\rho}{}_{\sigma\mu\nu}$ the only contraction producing a rank-2 tensor is the Ricci tensor. Chapter 3.4 §6.1 showed that it is essentially unique, and §6 of that chapter showed that it is symmetric. Contracting once more gives the Ricci scalar $R$, which can be multiplied by $g_{\mu\nu}$ to make a symmetric rank-2 object. And $g_{\mu\nu}$ itself qualifies, since it needs no derivatives at all. The complete list is therefore three terms, and the most general candidate we can write down is

with $A$, $B$, $C$ and $\kappa$ constants. One caveat about that list should be said out loud now. If terms quadratic in the curvature were allowed, such as $R^{2}g_{\mu\nu}$ or $R_{\mu\alpha}R^{\alpha}{}_{\nu}$, the list would be longer. Those terms contain squares of second derivatives, and (C3) as it was stated asked for the geometry side to be linear in the second derivatives, so they are out. Section 3.3 says what happens if that restriction is relaxed.

3.2 · Imposing (C4), one term at a time

We want to know what (C4) costs, so take the divergence of the left-hand side of (3.6.7), one term at a time. The index is raised with the metric, and the metric passes through $\nabla$ untouched by metric compatibility (Chapter 3.3 §7.1).

Third term. $\nabla^{\mu}\big(C g_{\mu\nu}\big)=C\,\nabla^{\mu}g_{\mu\nu}=0$, again by metric compatibility. So this term is divergence-free whatever $C$ is, and (C4) says nothing about it at all. Remember that. It is §6.

First term. Chapter 3.4 §7.2 contracted the second Bianchi identity twice and obtained (3.4.57):

So the first term contributes $A\cdot\half\nabla_{\nu}R$ to the divergence.

Second term. $\nabla^{\mu}\big(B R g_{\mu\nu}\big)=B\,g_{\mu\nu}\nabla^{\mu}R=B\,\nabla_{\nu}R$, the metric having again passed through the derivative and then lowered the free index on $\nabla^{\mu}R$.

Add them. Putting the three contributions together, the total divergence of the left-hand side is

Now the crucial step. It is a logical one rather than an algebraic one. Constraint (C4) demands that (3.6.9) vanish identically, meaning for every metric and not merely for solutions.

But $\nabla_{\nu}R$ is not identically zero. The sphere of Chapter 3.4 has constant $R$, and a generic metric does not, and one has only to write down a metric with a position-dependent Ricci scalar to see it. So the only way to make the right-hand side vanish for every metric is for the coefficient in front of it to vanish:

That single condition ties $B$ to $A$, so the next move is to put it back where it came from. Substituting the relation into (3.6.7) and factoring out the overall $A$,

The bracket is exactly the Einstein tensor $G_{\mu\nu}$ of Chapter 3.4 (3.4.58). All that is left is to name the constants, so divide through by $A$, absorb it into $\kappa$, and write $\lambda\equiv C/A$:

Two constants remain. Section 5 fixes $\kappa$ by (C5). Section 6 examines $\lambda$, which (C5) constrains only weakly and which the argument of this section cannot exclude at all.

3.3 · What happens if the hypotheses are weakened

4 · The same answer from an action

Chapter 1.2 §8.1 printed a table of eight actions and promised that each would be constructed in its own chapter. Line four read Gravity (Einstein–Hilbert), Chapter 3.6. This section constructs that action and varies it. The equation that comes out is (3.6.12), obtained a second time by a route that shares no step with §3.

Let's lay out the structure before starting, because this is the longest computation in Part III after Chapter 3.4 §2. The integrand is $R\sqrt{-g}$, and the thing being varied is the metric. Write $R=g^{\mu\nu}R_{\mu\nu}$, and the product rule splits the variation into exactly three pieces:

Piece 2 needs no work at all. It is already in the form (something) times $\delta g^{\mu\nu}$, which is exactly what a variational calculation wants. Piece 1 is a determinant identity, and grind box A does it.

Piece 3 looks the worst and is the most interesting. It is a total derivative, so it contributes nothing to the equations of motion and everything to a discussion of boundaries, and it takes two grind boxes, B and C. The reasoning stays here. Only the algebra is folded away.

4.1 · Why $R$, and the action

An action is a number, so what we need this time is a scalar rather than a tensor. Constraint (C3) still says at most two derivatives of the metric, so the question is what scalars can be built under that restriction.

A scalar has no free indices, so every index must be contracted. With one factor of the Riemann tensor there is essentially one way to do that. Contract the first and third indices to get the Ricci tensor, and Chapter 3.4 §6.1 showed that the other choices give zero or the same thing with a sign. Then contract the remaining two indices with the inverse metric to get $R$.

With no factor of Riemann there is only a constant. With two factors of Riemann, meaning $R^{2}$ or $R_{\mu\nu}R^{\mu\nu}$ or $R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}$, one has products of second derivatives, and by the same counting as §3.3 those are excluded by (C3) as stated. So the complete list of admissible scalars is a constant and $R$, and the action is

with $\alpha$ and $\lambda$ constants, and with $S_{\text{m}}$ whatever action describes the matter. The factor $-2$ in front of $\lambda$ is pure convention. It is chosen so that the symbol matches the $\lambda$ of (3.6.12), and we shall see it do so.

The volume element $\sqrt{-g}\,\dd^{4}x$ is Chapter 3.5 §6.2's. It is there because without it the integral would depend on the chart, which by (C1) is not allowed. Chapter 0.6 §8.3 said this six chapters ago and named this action as the reason.

Set $\lambda=0$ until §6, to keep the derivation clean. The constant is restored in three lines at the end of §4.4.

4.2 · Piece 1: the variation of the volume element

Piece 1 asks how the volume element responds to a change in the metric. The claim is

Everything needed for it is already in hand. Chapter 3.5 §6.3 derived Jacobi's formula (3.5.47), which gives the change in a determinant produced by a change in its entries, and it derived the specialisation to the metric, (3.5.48). The only additional ingredient is the relation between varying $g_{\mu\nu}$ and varying $g^{\mu\nu}$, which comes from differentiating the statement that the two are inverse. Grind box A does it in five lines.

Grind box A — $\delta\sqrt{-g}$, from Jacobi's formula

Line 1. The two variations are related. The metric and its inverse satisfy $g^{\mu\alpha}g_{\alpha\nu}=\delta^{\mu}{}_{\nu}$. The right-hand side is a constant array, so varying both sides gives

$$\big(\delta g^{\mu\alpha}\big)g_{\alpha\nu} \;+\; g^{\mu\alpha}\,\delta g_{\alpha\nu} \;=\; 0.$$

Multiply by $g^{\nu\beta}$ and sum over $\nu$, which turns $g_{\alpha\nu}g^{\nu\beta}$ into $\delta^{\beta}{}_{\alpha}$ in the first term:

$$\delta g^{\mu\beta} \;=\; -\,g^{\mu\alpha}g^{\nu\beta}\,\delta g_{\alpha\nu}.$$

Line 2. Contract it. Multiply the previous display by $g_{\mu\beta}$ and sum. On the right, $g_{\mu\beta}g^{\mu\alpha}=\delta^{\alpha}{}_{\beta}$ and then $\delta^{\alpha}{}_{\beta}g^{\nu\beta}=g^{\nu\alpha}$, so

$$g_{\mu\nu}\,\delta g^{\mu\nu} \;=\; -\,g^{\mu\nu}\,\delta g_{\mu\nu}.$$

The two contracted variations are negatives of each other. This is the step that is easiest to get backwards, and getting it backwards costs a sign in the field equations.

Line 3. Jacobi's formula. Chapter 3.5's (3.5.48) applied to the metric reads $\delta g = g\,g^{\mu\nu}\delta g_{\mu\nu}$, where $g=\det g_{\mu\nu}$. Using line 2 to swap which variation appears,

$$\delta g \;=\; -\,g\;g_{\mu\nu}\,\delta g^{\mu\nu}.$$

Line 4. The square root. $\delta\sqrt{-g}=\dfrac{\delta(-g)}{2\sqrt{-g}}=\dfrac{-\delta g}{2\sqrt{-g}}$. Substituting line 3,

$$\delta\sqrt{-g} \;=\; \frac{g\;g_{\mu\nu}\,\delta g^{\mu\nu}}{2\sqrt{-g}}.$$

Line 5. Tidy the determinant. Since $g$ is negative, $-g=(\sqrt{-g})^{2}$, so $g=-(\sqrt{-g})^{2}$ and therefore $g/\sqrt{-g}=-\sqrt{-g}$. Hence

$$\delta\sqrt{-g} \;=\; -\,\half\,\sqrt{-g}\;g_{\mu\nu}\,\delta g^{\mu\nu},$$

which is (3.6.15). (This was checked symbolically against a direct differentiation of $\sqrt{-\det g}$ for a four-dimensional metric family with off-diagonal entries. The two agreed exactly, as did the equivalent form $\delta\sqrt{-g}=+\half\sqrt{-g}\,g^{\mu\nu}\delta g_{\mu\nu}$.)

Now combine that result with piece 2, which was already in the shape a variational calculation wants. Together the two of them produce this:

The Einstein tensor has appeared without being sought. In §3 the factor of $\half$ came from the contracted Bianchi identity. Here it comes from the derivative of a determinant. The two routes have nothing in common, and they produce the same coefficient, which is the strongest kind of check a derivation can have.

4.3 · Piece 3: the Palatini identity, derived

What remains is $g^{\mu\nu}\delta R_{\mu\nu}$, which is piece 3. The strategy is to show that it is a total derivative, and it takes two steps.

Step one is the Palatini identity,

and the reason this identity can be derived rather than quoted is a fact worth stating on its own. Although the connection is not a tensor, the difference of two connections is.

Chapter 3.3 §5.2 derived the transformation law (3.3.26) and pointed out that its offending inhomogeneous term depends only on the change of chart, not on the metric. So take the difference of two connections belonging to two nearby metrics. That term cancels, and what is left transforms as a $(1,2)$ tensor. A variation $\delta\Gamma$ is exactly such a difference. Grind box B uses that fact together with Chapter 3.4 §5.3's locally inertial coordinates.

Grind box B — the Palatini identity, in four lines

Line 1. The definition. Contracting Chapter 3.4's (3.4.13) on the first and third indices gives the Ricci tensor written out in terms of the connection:

$$R_{\mu\nu} \;=\; \partial_{\lambda}\Gamma^{\lambda}{}_{\nu\mu} \;-\; \partial_{\nu}\Gamma^{\lambda}{}_{\lambda\mu} \;+\; \Gamma^{\lambda}{}_{\lambda\rho}\Gamma^{\rho}{}_{\nu\mu} \;-\; \Gamma^{\lambda}{}_{\nu\rho}\Gamma^{\rho}{}_{\lambda\mu}.$$

Line 2. Vary it. The first two terms give $\partial_{\lambda}\delta\Gamma^{\lambda}{}_{\nu\mu}-\partial_{\nu}\delta\Gamma^{\lambda}{}_{\lambda\mu}$. The quadratic terms give four contributions by the product rule, and every one of them carries an undifferentiated $\Gamma$ as a factor.

Line 3. Evaluate at a point in locally inertial coordinates. Chapter 3.4 §5.3 constructed, around any chosen point $p$, a chart in which $\Gamma^{\lambda}{}_{\mu\nu}(p)=0$ while $\partial\Gamma(p)\neq0$. In that chart, at that point, all four quadratic contributions vanish, and

$$\delta R_{\mu\nu}\big|_{p} \;=\; \partial_{\lambda}\,\delta\Gamma^{\lambda}{}_{\nu\mu} \;-\; \partial_{\nu}\,\delta\Gamma^{\lambda}{}_{\lambda\mu}.$$

Note what is not being claimed: $\partial\Gamma$ is not zero at $p$, only $\Gamma$ is, which is exactly the point that made Chapter 3.4 §7.1's proof of the Bianchi identity work.

Line 4. Promote to covariant derivatives, and then to every chart. Since $\delta\Gamma$ is a tensor, its covariant derivative is defined, and at $p$ in this chart every correction term in that covariant derivative carries a factor of $\Gamma(p)=0$. So the partial derivatives above may be replaced by covariant ones without changing anything, giving (3.6.17) at $p$ in this chart. But (3.6.17) is an equation between tensors, so by Chapter 2.4 §6 it holds in every chart. And the point $p$ was arbitrary. $\blacksquare$

(This was checked symbolically. For a four-dimensional metric family with off-diagonal entries, all sixteen components of the two sides of (3.6.17) agreed exactly at randomly chosen points.)

Step two contracts the Palatini identity with the inverse metric and recognises a divergence. The move that makes it work is metric compatibility. Since $g^{\mu\nu}$ passes through $\nabla_{\lambda}$ untouched, it can be taken inside the derivative, and then the whole expression is the divergence of something. Grind box C does the index bookkeeping.

Grind box C — the contraction, and the vector whose divergence it is

Contract (3.6.17) with $g^{\mu\nu}$:

$$g^{\mu\nu}\,\delta R_{\mu\nu} \;=\; g^{\mu\nu}\,\nabla_{\lambda}\delta\Gamma^{\lambda}{}_{\nu\mu} \;-\; g^{\mu\nu}\,\nabla_{\nu}\delta\Gamma^{\lambda}{}_{\lambda\mu}.$$

Take the metric inside. By metric compatibility (Chapter 3.3 §7.1), $\nabla_{\lambda}g^{\mu\nu}=0$, so $g^{\mu\nu}\nabla_{\lambda}X=\nabla_{\lambda}(g^{\mu\nu}X)$ for anything $X$. Applying that to both terms:

$$g^{\mu\nu}\,\delta R_{\mu\nu} \;=\; \nabla_{\lambda}\Big(g^{\mu\nu}\,\delta\Gamma^{\lambda}{}_{\nu\mu}\Big) \;-\; \nabla_{\nu}\Big(g^{\mu\nu}\,\delta\Gamma^{\lambda}{}_{\lambda\mu}\Big).$$

Give the two terms the same derivative index. The first differentiates with respect to $\lambda$ and the second with respect to $\nu$. Both are summed, so both names are private. Rename $\nu\to\lambda$ in the second term, which forces its other $\nu$ to become $\lambda$ as well, and then rename its internal dummy $\lambda\to\mu$ and its $\mu\to\nu$ to avoid a clash. The result is that both terms are $\nabla_{\lambda}$ of something, and

$$g^{\mu\nu}\,\delta R_{\mu\nu} \;=\; \nabla_{\lambda}v^{\lambda}, \qquad v^{\lambda} \;\equiv\; g^{\mu\nu}\,\delta\Gamma^{\lambda}{}_{\mu\nu} \;-\; g^{\lambda\nu}\,\delta\Gamma^{\mu}{}_{\mu\nu},$$

where we have used the symmetry of $\delta\Gamma$ in its two lower indices to write $\delta\Gamma^{\lambda}{}_{\nu\mu}$ as $\delta\Gamma^{\lambda}{}_{\mu\nu}$. That symmetry is inherited from the symmetry of $\Gamma$ itself, Chapter 3.3 §7.2.

Then the volume factor. Chapter 3.5 §6.4's identity (3.5.51) says $\sqrt{-g}\,\nabla_{\lambda}v^{\lambda}=\partial_{\lambda}\big(\sqrt{-g}\,v^{\lambda}\big)$, an ordinary partial derivative of an ordinary product. So piece 3 of (3.6.13) is exactly

$$\sqrt{-g}\;g^{\mu\nu}\,\delta R_{\mu\nu} \;=\; \partial_{\lambda}\Big(\sqrt{-g}\;v^{\lambda}\Big).$$

(The whole identity was checked symbolically. For four-dimensional and three-dimensional metric families, $\delta(\sqrt{-g}R)$ computed by direct differentiation agreed exactly, at random points, with $\sqrt{-g}\,G_{\mu\nu}\delta g^{\mu\nu}+\partial_{\lambda}(\sqrt{-g}\,v^{\lambda})$ using the $v^{\lambda}$ displayed above.)

4.4 · Assembling, and the field equations

All three pieces are now in hand, so put them together. Using (3.6.16) for pieces 1 and 2, and grind box C for piece 3,

The second term is an ordinary divergence, so by Chapter 3.5 §5's theorem it equals an integral over the boundary of the region. Adopt the same boundary condition Chapter 1.2 §3.3 adopted for the pendulum: the variation is required to vanish, along with its first derivatives, outside some bounded region. Then $v^{\lambda}$ vanishes on the boundary and the term contributes nothing. Section 4.5 asks what that condition costs, because it is not free.

The matter side. Define the energy–momentum tensor by how the matter action responds to a change in the geometry:

This is a definition, so by itself it proves nothing. What makes it legitimate is that it reproduces the tensor Chapter 2.6 built by an entirely different argument. Worked example 1 checks exactly that, by feeding Chapter 2.6's electromagnetic Lagrangian (2.6.72) into (3.6.19) and recovering (2.6.80), sign included. The definition also delivers a symmetric tensor automatically, because $\delta g^{\mu\nu}$ is symmetric. Chapter 2.6 had to arrange that by hand with an improvement term.

Stationarity. Demanding $\delta S=\delta S_{\text{EH}}+\delta S_{\text{m}}=0$ for every allowed $\delta g^{\mu\nu}$, and applying the fundamental lemma of the calculus of variations (Chapter 1.2 §3.4) to strip off the arbitrary variation,

That is (3.6.12) with $\lambda=0$, obtained a second time. Note that $\alpha$ is still undetermined. The action principle fixes the form of the law and cannot fix its coupling, exactly as Chapter 2.6 §9 could not fix $\mu_{0}$ from the shape of the electromagnetic Lagrangian alone.

Restoring $\lambda$, in three lines. Put the constant back in (3.6.14). Its variation involves only $\delta\sqrt{-g}$, so by (3.6.15),

which adds $\alpha\lambda\,g_{\mu\nu}$ to the left of (3.6.20), and hence $\lambda g_{\mu\nu}$ to the left of the field equation. That reproduces (3.6.12) in full, and it justifies the $-2$ in the action. The cosmological term is the constant that can be added to any Lagrangian. Section 6 takes that sentence seriously.

4.5 · The boundary term is not free

Two things were quietly discarded above and both deserve to be named.

Why the equations are second order at all. Constraint (C3) said the field equations must not contain derivatives of $g$ beyond the second, and Chapter 1.2 §8 established that an action built from a field and its first derivatives delivers exactly that. But $R$ contains second derivatives of the metric, so (3.6.14) is not of that form, and one would naively expect fourth-order equations.

It does not happen, and grind box C says why. Every second derivative of the metric in $\delta(\sqrt{-g}R)$ sits inside $\partial_{\lambda}(\sqrt{-g}v^{\lambda})$, which is a total derivative and drops out of the equations of motion entirely. The Einstein–Hilbert action is second order in disguise, and that is a special property of $R$ rather than a general feature of curvature scalars. It is precisely why adding $R^{2}$ to the action does give fourth-order equations.

What the discarded term costs. Because the second derivatives live in the boundary term, making $\delta S$ vanish required more than fixing the metric on the boundary. It required fixing the metric's normal derivative there too, and that is one condition too many. The standard fix is to add to (3.6.14) a boundary integral whose variation cancels the unwanted piece. What is left is a variational principle in which only $g_{\mu\nu}$ itself is held fixed on the boundary.

5 · The constant, fixed by demanding that apples fall

This is the section where the whole construction is put on trial. Everything so far has been structural. The equation has the form it has because of conservation, because of symmetry, and because of the number of derivatives allowed. None of that touches the world. Here it does.

The route has five steps and each one is small, so here they are before we start. We take a weak, static gravitational field and slow-moving matter. We rearrange the field equation into a form where the Ricci tensor stands alone. We evaluate its $00$ component on the left from the metric, and then the same component on the right from the matter. Finally we compare the result with Poisson's equation. Out comes $\kappa=8\pi G/c^{4}$.

5.1 · Step 1 — trace-reverse the field equation

The field equation (3.6.20) has the Ricci scalar buried inside $G_{\mu\nu}$, and that is inconvenient. We want the scalar out where we can see it, so contract the equation with $g^{\mu\nu}$. Using $g^{\mu\nu}R_{\mu\nu}=R$ and $g^{\mu\nu}g_{\mu\nu}=4$,

with $T\equiv g^{\mu\nu}T_{\mu\nu}$. That is an expression for $R$ in terms of the matter, so we can now eliminate the scalar entirely. Substitute $R=-\kappa T$ back into $R_{\mu\nu}=G_{\mu\nu}+\half R\,g_{\mu\nu}$, and the result is

This is the trace-reversed form, and it is exactly equivalent to (3.6.20). Contract it and the previous step runs backwards. It is the form to use whenever the Ricci tensor is easier to compute than the Einstein tensor, which is most of the time, and Chapter 3.7 uses it from its first line.

One consequence comes free and matters a great deal. In vacuum, where $T_{\mu\nu}=0$, the field equations reduce to $R_{\mu\nu}=0$. Chapter 3.4 §6.3 already showed that this does not mean flat, because the Weyl part of the curvature survives. Chapter 3.7 solves $R_{\mu\nu}=0$.

5.2 · Step 2 — the assumptions, each named as it is made

Assumption 1, weak field. We write $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ with every $\abs{h_{\mu\nu}}\ll1$, and we keep only terms linear in $h$. In particular $h_{00}=2\Phi/c^{2}$.

Assumption 2, static. Nothing depends on time, so $\partial_{0}g_{\mu\nu}=0$.

Assumption 3, slow matter. The matter's four-velocity is $u^{\mu}\approx(c,0,0,0)$, so in (3.6.4) we keep the leading term only. Pressure is taken to be small compared with rest-energy density, $p\ll\rho c^{2}$. Section 5.5 removes this one.

5.3 · Step 3 — the left-hand side, $R_{00}$

We want the $00$ component of the left-hand side, so write out the Ricci tensor in terms of the connection, exactly as grind box B did, and then set $\mu=\nu=0$:

Now take those four terms one at a time. Three of them turn out to be nothing.

The two quadratic terms are second order. Each $\Gamma$ is built from one derivative of the metric, so by Assumption 1 each one is first order in $h$. A product of two of them is second order, and it is dropped.

The second term vanishes. It carries $\partial_{0}$, and by Assumption 2 nothing depends on $x^{0}$.

The first term is the whole answer. Split the sum over $\lambda$ into its time and space parts. The time part is $\partial_{0}\Gamma^{0}{}_{00}$, which vanishes by Assumption 2 again. What is left is the spatial sum, and Chapter 3.3 §8.3 has already computed the connection coefficient that sum needs. It is equation (3.3.71), obtained there from exactly these three assumptions:

Now put that coefficient into the surviving term and carry out the remaining differentiation, which finishes this side of the equation:

where the last step recognises the Laplacian of Chapter 0.7 §7.5. At this order indices are raised and lowered with $\delta_{ij}$, so the distinction between $\partial_{i}$ and $\partial^{i}$ costs only a sign that appears twice and cancels.

5.4 · Step 4 — the right-hand side, and the comparison

Now for the right-hand side. Use the trace-reversed form (3.6.23) with $\mu=\nu=0$. By Assumption 3 the matter is dust at rest, so from (3.6.3) and $u_{\mu}=g_{\mu\nu}u^{\nu}\approx(c,0,0,0)$,

Those are the two ingredients the right-hand side asks for, so assemble the bracket that appears in (3.6.23):

where the last step drops $\rho\Phi$, because it is smaller than $\half\rho c^{2}$ by the factor $2\Phi/c^{2}$, and Assumption 1 declared that negligible.

Let's look at what that line is actually saying. Note the arithmetic: $1-\half=\half$. The full energy density enters, the trace-reversal removes half of it, and that surviving factor of one half is exactly what turns the $4\pi$ of Poisson's equation into the $8\pi$ of the final answer.

Both sides of the $00$ equation are now in hand, so set them equal. That means (3.6.26) on the left and $\kappa$ times (3.6.28) on the right:

That is a Poisson equation with an undetermined coefficient in it, so compare it against the real Poisson equation, which Chapter 0.7 §7.5 derived from Newton's inverse-square law and the divergence theorem:

The two agree for every density only if the coefficients of $\rho$ agree, which requires $\half\kappa c^{4}=4\pi G$. Solving that for $\kappa$,

Put that value of $\kappa$ back into the boxed result of §3, and there, at last, are the field equations of general relativity:

And by (3.6.20), $\alpha=-1/(2\kappa)=-c^{4}/(16\pi G)$, so the action is fixed too:

5.5 · What the number means, and one generalisation

The size of $\kappa$. Numerically $\kappa=8\pi G/c^{4}=2.08\times10^{-43}\ \mathrm{m^{-1}\,kg^{-1}\,s^{2}}$. The reciprocal is more telling: $1/\kappa=c^{4}/8\pi G=4.8\times10^{42}\ \mathrm{N}$.

Read (3.6.32) as (curvature) $=\kappa\times$(stress), and $1/\kappa$ becomes the stiffness of spacetime, meaning the stress needed to produce unit curvature. It is about $10^{42}$ newtons. That is why an object as massive as the Earth bends spacetime by only the $1.7\times10^{-23}\ \mathrm{m^{-2}}$ that Chapter 3.4 §4.5 computed, and why gravity looked for three centuries like the weakest thing in physics rather than like geometry.

Restoring the pressure. Assumption 3 dropped $p$, and we can now afford to put it back. Use the perfect fluid (3.6.4) instead of dust, and take its trace from (3.6.5), which is $T=\rho c^{2}-3p$. Then, still at rest and to leading order, the bracket becomes

Nothing else in the calculation changes, so run steps 3 and 4 again with this bracket in place of $\half\rho c^{2}$. What comes out is

This is the promise of §1.4 kept. Pressure gravitates, and it does so three times over. Squeeze a gas without adding anything to it and the source of its gravitational field grows.

The effect is invisible in ordinary matter. There $p/c^{2}$ is smaller than $\rho$ by the square of the ratio of thermal speeds to $c$, which for the Sun's centre is about $10^{-6}$. But it is not invisible everywhere. In a neutron star the pressure term is a sizeable fraction of the source, and it works against the star. Adding pressure to resist collapse also adds to the gravity doing the collapsing. That is part of the reason a sufficiently massive star has nothing left that can hold it up.

And for radiation, where (3.6.6) gives $p=\rho c^{2}/3$, the bracket becomes $2\rho$. A gas of light gravitates twice as strongly as the same energy density in cold dust. Chapter 3.9 needs that.

The same expression also contains the seed of §6, and it takes one line to see. Suppose a substance could have $p<-\rho c^{2}/3$. Then the bracket would be negative, $\nabla^{2}\Phi$ would have the wrong sign, and the gravity of that substance would push rather than pull. Nothing in (3.6.35) forbids it.

equation of state w = p/ρc²

0.00

cold dust

radiation

vacuum

w = 0.00 → source = ρ(1 + 3w) = 1.000 ρ (cold dust — pressureless matter)

attractive, 1.00× the Newtonian value for the same energy density

6 · The one term the argument cannot exclude

Now return to $\lambda$. Section 3 found it and could say nothing about it, because constraint (C4) was silent: $\nabla^{\mu}g_{\mu\nu}=0$ automatically. Section 4 found it again from the other direction, as the constant that may be added to any Lagrangian. This section asks what it is.

6.1 · It is a substance with negative pressure

The way to find out what a term means is usually to move it to the other side and read it as matter. So move it from the geometry side of (3.6.32), which costs a sign:

Now read that bracket as a total energy–momentum tensor. Whatever $\lambda$ turns out to be, its contribution to the source is

We want to know what substance would produce that, so compare it with the perfect fluid (3.6.4), with both indices lowered: $T_{\mu\nu}=(\rho+p/c^{2})u_{\mu}u_{\nu}-p\,g_{\mu\nu}$. For (3.6.37) to have this form, two conditions must hold, and we take them one at a time. The $u_{\mu}u_{\nu}$ term is absent from (3.6.37), so its coefficient must vanish:

That is the first condition. Matching the remaining term, $-p\,g_{\mu\nu}=-(\lambda/\kappa)g_{\mu\nu}$, gives $p=\lambda/\kappa$, and then (3.6.38) converts that into a density:

Three things follow immediately, and not one of them was put in by hand.

(i) The equation of state is fixed, not chosen. A cosmological term is a fluid with $p=-\rho c^{2}$, which is $w=-1$ in the figure's notation. It is the far-left point of that plot. By (3.6.35) its source term is $\rho+3p/c^{2}=-2\rho$. So with a positive energy density it repels, and it repels twice as hard as the same energy density in dust attracts.

(ii) It is the only fluid that looks the same to everybody. Equation (3.6.37) is proportional to $g_{\mu\nu}$, and the metric is the one tensor whose components are the same in every local inertial frame. Every other fluid singles out a rest frame, through $u^{\mu}$. This one does not. That is exactly what one would demand of the energy of empty space, since there is no such thing as moving relative to the vacuum.

(iii) Its density is constant. That is not an assumption. It follows because $\lambda$ is a constant and $\kappa$ is a constant. Expand the universe and ordinary matter thins out while this does not, which is why a term negligible in the early universe can come to dominate the late one. Chapter 3.9 works that out.

6.2 · Why nothing forbids it

The quickest way to see that nothing rules the term out is to go back through the chapter and check every constraint in §2 against $\Lambda g_{\mu\nu}$ in turn.

Constraint	Does $\Lambda g_{\mu\nu}$ satisfy it?
(C1) tensorial	yes, since $g_{\mu\nu}$ is a tensor
(C2) symmetric, rank 2	yes
(C3) at most two derivatives of $g$	yes, it has none
(C4) identically divergence-free	yes, by metric compatibility, $\nabla^{\mu}g_{\mu\nu}=0$
(C5) Newtonian limit	yes, provided $\Lambda$ is small enough, as below

Every row passes. And §4 makes the point from the other side. In the action, the cosmological term is a constant added to the Lagrangian, and nothing forbids adding a constant to a Lagrangian.

In ordinary mechanics, adding a constant to the Lagrangian changes nothing, because the extra contribution to the action is itself a fixed number, and a fixed number has no variation. Here it is not fixed. The constant is multiplied by $\sqrt{-g}$, which depends on the very thing being varied. So gravity, alone among the theories in this book, notices a constant in its Lagrangian. What it notices is this term.

What (C5) says is a bound rather than a prohibition. Redo §5 keeping $\Lambda$. The extra term contributes $-\Lambda g_{00}\approx-\Lambda$ to the left of the field equation, so (3.6.29) becomes $\nabla^{2}\Phi=4\pi G\rho-\Lambda c^{2}$. For that to be indistinguishable from Poisson's equation in the Solar System, $\Lambda c^{2}$ must be tiny compared with $4\pi G\rho$ there, which it is, by an enormous margin. The constant is not excluded. It is bounded.

6.3 · What it is measured to be, and the problem that leaves

One historical note, because it is usually told wrongly. Einstein introduced this term in 1917 to permit a static universe, since with $\Lambda=0$ the equations have no static solution containing matter. The balance he found is unstable, as Chapter 3.9 §3 shows by integrating the equations. Once the expansion of the universe was established, the motivation evaporated.

What did not evaporate is the term. It was never inserted. It was always allowed. Removing it requires an extra assumption, and that assumption turned out to be false.

7 · Ten equations, four identities, and why that is exactly right

The field equations are written. Before spending them, let's count them, because the count reveals something the equations do not say out loud.

7.1 · The count

Unknowns. The metric $g_{\mu\nu}$ is symmetric with two indices in four dimensions, so it has $4\cdot5/2=10$ independent components. That is ten functions of four variables.

Equations. Both sides of (3.6.32) are symmetric rank-2 tensors, so it is also ten equations. Ten equations for ten unknowns looks like a well-posed problem, and looks determined.

But four of the ten are not independent. Chapter 3.4 §7.3 proved $\nabla^{\mu}G_{\mu\nu}=0$ identically, and $\nabla^{\mu}(\Lambda g_{\mu\nu})=0$ as well, so the divergence of the left-hand side of (3.6.32) is zero whatever the metric. That is four differential relations among the ten equations, one for each value of $\nu$. Only six of the ten carry independent information about how the metric evolves.

And that shortfall is exactly right. A solution $g_{\mu\nu}(x)$ can be relabelled by any smooth change of chart $x\to x'(x)$, which is four arbitrary functions. The relabelled metric describes the same geometry, so it must also be a solution.

So the equations cannot determine all ten components. If they did, they would forbid the relabelling, and by (C1) the relabelling is not physical. Four functions' worth of freedom has to remain undetermined, and four is exactly what the Bianchi identities leave undetermined.

7.2 · Which four are the constraints

Let's look at where the four identities bite. Write out $\nabla_{\mu}G^{\mu 0}=0$. It contains $\partial_{0}G^{00}$ and $\partial_{i}G^{i0}$, so it expresses the time derivative of $G^{00}$ in terms of quantities involving one fewer time derivative. Since $G_{\mu\nu}$ contains second derivatives of the metric, the consequence is that the four equations $G^{0\nu}=\kappa T^{0\nu}$ contain no second time derivatives at all.

So they are not evolution equations. They are conditions that the initial data must satisfy, and the identities then guarantee that if the conditions hold at one time they hold at every later time.

Once again electromagnetism did this first. In Chapter 2.6, Gauss's law $\nabla\cdot\vv E=\rho/\epsilon_{0}$ contains no time derivative of $\vv E$. It is a constraint on initial data, preserved by the other equations because charge is conserved. Gravity has six evolution equations and four constraints. Electromagnetism has three and one.

7.3 · The equations are nonlinear, and that is physics

One last structural fact, and it is the one that makes Part III hard and Chapter 7.1 harder. $G_{\mu\nu}$ is built from $\Gamma$ and its derivatives, and $\Gamma$ contains $g^{-1}\partial g$. So the Einstein tensor contains the inverse metric multiplied by derivatives of the metric, over and over. It is nonlinear in $g_{\mu\nu}$, badly so.

What the nonlinearity means physically. Superposition fails, so the field of two masses is not the sum of their separate fields. The reason is not technical. Gravitational fields carry energy, and by §1 everything carrying energy sources gravity, so gravity gravitates. There is no way to write a linear theory with that property, because linearity means the source is independent of the field.

Contrast electromagnetism, which is linear. Chapter 2.6's $\partial_{\mu}F^{\mu\nu}=\mu_{0}j^{\nu}$ is linear in $A$, and two solutions superpose. The reason is that the electromagnetic field is not itself charged. Chapter 6.4 builds a theory where the field does carry the charge it responds to, and finds equations that look startlingly like these.

Two consequences are worth filing. Exact solutions are rare and precious, which is why Chapter 3.7's Schwarzschild solution is a landmark rather than an exercise. And the standard technique of physics, which is to expand about a simple solution and keep the first correction, becomes the only technique available. That is why Chapter 3.1's warning that nearly everything is an approximation applies here with particular force.

8 · Worked examples

9 · Your turn