Part III · General Relativity — Chapter 3.4

Curvature

A test for curvature that can be run from inside, with no outside to look in from. It measures the tide.

Tools you'll need.  Chapter 3.3 throughout: the covariant derivative of its §5, its extension to tensors (3.3.31) and (3.3.32), the inhomogeneous transformation law (3.3.26) of the connection, parallel transport (3.3.35), the Christoffel formula (3.3.50), the geodesic equation (3.3.57) and the sphere's connection coefficients from §9. Chapter 3.1 §4 in detail, meaning the tidal equation, the point-mass tidal matrix, its traceless character in vacuum, and the ring that deforms into an ellipse. Section 4 below is where all of that is collected. Chapter 3.2 §8, the commutator of vector fields, whose second-derivative cancellation is the template for §2, and §7 for the tensor transformation law. Chapter 2.4 §5.1 (contraction), §6 (a tensor equation true in one chart is true in all) and §7.2 (counting independent components of a symmetric or antisymmetric array). Chapter 0.6 §6.1, Clairaut's theorem, used four separate times below. Chapter 0.5 §6, the spectral theorem, used once in §5.

1 · The loop test, and the sphere

Here is the test in full. Take a closed path, plant a vector at a point on it, carry the vector all the way round by the parallel-transport equation (3.3.35), and compare the arrival with the departure. Both vectors live in the same tangent space, so the comparison is legitimate in the strict sense of Chapter 3.2 §5.1. Transport preserves length by metric compatibility, so the only thing that can have happened along the way is a rotation. The angle of that rotation is the measurement.

Check first that the test gives nothing where it should. In flat space with Cartesian coordinates the connection vanishes, so (3.3.35) reads $\dd V^{\mu}/\dd\lambda=0$ and the vector comes back identical. The angle is zero for every loop. Any non-zero answer is therefore a fact about the space and not about the chart, provided we also check that flat space still gives zero when the chart is curvilinear. Section 8 runs that check.

1.1 · A loop on a sphere, computed by hand

Here is where we are going, announced before we set off. We choose a closed path on a sphere of radius $a$, integrate the transport equation along it in three pieces, and find that the vector returns rotated through an angle exactly equal to the area enclosed divided by $a^{2}$. Everything we need comes from the Christoffel symbols computed in Chapter 3.3 §9, which were

with $\theta$ the angle from the north pole and $\phi$ the azimuth.

A convenient pair of numbers. Before integrating anything we should choose components that are easy to read. Components carry the length of their basis vector, and here the two basis vectors have different lengths: $\abs{\partial_{\theta}}=a$ and $\abs{\partial_{\phi}}=a\sin\theta$. So define the orthonormal components

That last equality is just $g_{\mu\nu}V^{\mu}V^{\nu}=a^{2}(V^{\theta})^{2}+a^{2}\sin^{2}\theta\,(V^{\phi})^{2}$ regrouped. In this pair, $P$ is how much of the vector points south and $U$ how much points east, each measured in ordinary length.

Leg one, along a circle of latitude $\theta=\theta_{0}$, from $\phi=0$ to $\phi=\Delta\phi$. The tangent to this leg has only a $\phi$ component, so in (3.3.35) the sum over $\nu$ keeps only the term $\nu=\phi$:

Those are equations for $V^{\theta}$ and $V^{\phi}$, and what we want are equations for $P$ and $U$, so rewrite them. Since $\theta_{0}$ is constant along this leg, the factors $a$ and $a\sin\theta_{0}$ are constants and pass straight through the derivative:

In each line the constant was regrouped to build the other orthonormal component. In the first, $a\sin\theta_{0}\cdot V^{\phi}=U$. In the second, $a\cos\theta_{0}\cdot V^{\theta}$ splits as $\cos\theta_{0}\cdot(aV^{\theta})=\cos\theta_{0}P$, using $\sin\theta_{0}\cot\theta_{0}=\cos\theta_{0}$.

Let's take two readings of (3.4.4) before integrating it. The first reading is that $P^{2}+U^{2}$ is constant, since $\dd(P^{2}+U^{2})/\dd\phi = 2P\cos\theta_{0}U - 2U\cos\theta_{0}P = 0$. That is metric compatibility showing up as a conserved quantity, exactly as Chapter 3.3 §7.1 said it would.

The second reading is that the pair $(P,U)$ obeys precisely the equations of a rotation. To see it, write a rotation through angle $\alpha$ as $P\to P\cos\alpha-U\sin\alpha$, $U\to P\sin\alpha+U\cos\alpha$ and differentiate, which gives $\dd P/\dd\alpha=-U$ and $\dd U/\dd\alpha=P$. Matching that against (3.4.4) gives $\dd\alpha/\dd\phi=-\cos\theta_{0}$, so over the whole leg the vector turns through

a rotation measured positively from the southward direction toward the eastward one.

Legs two and three, along meridians. Now the tangent has only a $\theta$ component, so (3.3.35) keeps only the term $\nu=\theta$:

The first of those vanishes because both $\Gamma^{\theta}{}_{\theta\theta}$ and $\Gamma^{\theta}{}_{\theta\phi}$ are zero in (3.4.1). So $P=aV^{\theta}$ is constant outright. The second equation separates into $\dd V^{\phi}/V^{\phi}=-\cot\theta\,\dd\theta$, and integrating both sides gives $\ln V^{\phi}=-\ln\sin\theta+\text{const}$, that is $V^{\phi}\propto1/\sin\theta$. Hence $U=a\sin\theta\,V^{\phi}$ is constant too. Along a meridian, nothing rotates, and that is what we should expect. A meridian is a great circle and therefore a geodesic, and Chapter 3.3 §8 built geodesics to be the curves along which the angle to the direction of travel is held fixed.

The one subtlety, which is the frames at the pole. If the loop passes through the north pole we must be careful, because $(\hat e_{\theta},\hat e_{\phi})$ is not a single frame there. Watch what happens to it. As $\theta\to0$ along the meridian of azimuth $\phi$, the southward direction $\hat e_{\theta}$ becomes the horizontal direction pointing away from the pole at azimuth $\phi$. In the ambient description, used here only to name directions, $\hat e_{\theta}\to(\cos\phi,\sin\phi,0)$ and $\hat e_{\phi}\to(-\sin\phi,\cos\phi,0)$. So the frame belonging to azimuth $\phi$ is the frame belonging to azimuth $0$ turned through $+\phi$. The consequence is the piece we need. A single fixed vector at the pole, described first in the frame of azimuth $\Delta\phi$ and then in the frame of azimuth $0$, has its component pair rotated by $+\Delta\phi$ between the two descriptions.

Assemble the loop. Start at $A=(\theta_{0},\,\phi=0)$. Go east along the latitude circle to $(\theta_{0},\,\Delta\phi)$, then north up that meridian to the pole, then south down the meridian of azimuth $0$ back to $A$. Now add the rotations in the order they were picked up:

The area enclosed. To make anything of that angle we need the area of the region the loop bounds, so compute it next. The region is the wedge of the polar cap between azimuths $0$ and $\Delta\phi$, running from the pole down to colatitude $\theta_{0}$. The area element on the sphere is $\sqrt{g}\,\dd\theta\,\dd\phi = a^{2}\sin\theta\,\dd\theta\,\dd\phi$, since the metric is diagonal with $g_{\theta\theta}g_{\phi\phi}=a^{4}\sin^{2}\theta$. Integrating over the wedge,

Both numbers are now in hand, and each of them is the same product $\Delta\phi\,(1-\cos\theta_{0})$, once with a factor $a^{2}$ in front and once without. So the ratio of (3.4.7) to (3.4.8) carries no trace of the loop at all, and rearranging that ratio gives the result in its usual form:

colatitude θ₀

60.00°

azimuth Δφ

90.0°

halve the loop

reset

measured rotation of the returned vector: 44.999998°

enclosed area / a² (theory, §1.1): 45.000000° [7.8540e-1 sr]

difference: -1.73e-6° ratio angle/area = 1.000000 = curvature of the unit sphere

checks: |V| drift 3.3e-14 tangency at the base point 1.5e-15

2 · The commutator, and the tensor it produces

This is the longest computation in the book so far. The plan comes first, so that you can follow the whole argument without opening a single grind box.

2.1 · The computation

Step one. Write $W^{\rho}{}_{\nu}\equiv\nabla_{\nu}V^{\rho}=\partial_{\nu}V^{\rho}+\Gamma^{\rho}{}_{\nu\sigma}V^{\sigma}$. This is a type $(1,1)$ tensor by the construction of Chapter 3.3 §5.

Step two. Now differentiate that object covariantly. Equation (3.3.32) gives one $+\Gamma$ for the upper index and one $-\Gamma$ for the lower one:

Step three. Put $W$ back in. In the first term the product rule of Chapter 0.1 §4 acts on $\Gamma^{\rho}{}_{\nu\sigma}V^{\sigma}$ and produces two pieces. In the third term the bracket $W^{\rho}{}_{\lambda}$ is exactly $\nabla_{\lambda}V^{\rho}$, and we leave it that way, which keeps the expression short and makes the cancellation visible. Labelling the six terms for reference:

The grind box below writes that out with nothing folded away, for anyone who wants to see every index in place. The argument does not need it.

Grind box — the same six terms, with every substitution written out

Start from (3.4.10) and substitute $W^{\rho}{}_{\nu}=\partial_{\nu}V^{\rho}+\Gamma^{\rho}{}_{\nu\sigma}V^{\sigma}$ in all three places, being careful to rename the summed index in each so that no letter does two jobs.

The first term.

$$\partial_{\mu}W^{\rho}{}_{\nu} \;=\; \partial_{\mu}\Big(\partial_{\nu}V^{\rho}+\Gamma^{\rho}{}_{\nu\sigma}V^{\sigma}\Big) \;=\; \partial_{\mu}\partial_{\nu}V^{\rho} \;+\; \big(\partial_{\mu}\Gamma^{\rho}{}_{\nu\sigma}\big)V^{\sigma} \;+\; \Gamma^{\rho}{}_{\nu\sigma}\,\partial_{\mu}V^{\sigma},$$

where the product rule was applied to $\Gamma^{\rho}{}_{\nu\sigma}V^{\sigma}$, both factors being functions of position. These are $T_{1}$, $T_{2}$ and $T_{3}$.

The second term. Here $W^{\lambda}{}_{\nu}=\partial_{\nu}V^{\lambda}+\Gamma^{\lambda}{}_{\nu\sigma}V^{\sigma}$, so

$$\Gamma^{\rho}{}_{\mu\lambda}W^{\lambda}{}_{\nu} \;=\; \Gamma^{\rho}{}_{\mu\lambda}\,\partial_{\nu}V^{\lambda} \;+\; \Gamma^{\rho}{}_{\mu\lambda}\Gamma^{\lambda}{}_{\nu\sigma}V^{\sigma},$$

which are $T_{4}$ and $T_{5}$. Note that $\lambda$ is summed inside the double product. It is the index shared by the two connection factors, and it is what makes $T_{5}$ a matrix product rather than an entry-by-entry one.

The third term. $W^{\rho}{}_{\lambda}$ is by definition $\nabla_{\lambda}V^{\rho}$, and we deliberately do not expand it, because the whole term is about to vanish for a reason that has nothing to do with its contents. Written out it would be $-\Gamma^{\lambda}{}_{\mu\nu}\partial_{\lambda}V^{\rho}-\Gamma^{\lambda}{}_{\mu\nu}\Gamma^{\rho}{}_{\lambda\sigma}V^{\sigma}$, which is seven terms in total rather than six. Folding that pair back into $\nabla_{\lambda}V^{\rho}$ is the only compression made anywhere in this computation.

Collecting gives (3.4.11) exactly.

2.2 · Antisymmetrising, one term at a time

Now write the same expression with $\mu$ and $\nu$ exchanged and subtract. We go through the six terms in order. Each one is a separate small argument, and each is named.

$T_{1}$ dies by Clairaut. The term is $\partial_{\mu}\partial_{\nu}V^{\rho}$, an ordinary second partial derivative of a set of smooth functions. Clairaut's theorem (Chapter 0.6 §6.1) says mixed partials commute, so exchanging $\mu$ and $\nu$ leaves it unchanged and the subtraction kills it. This is the same cancellation that Chapter 3.2 §8.3 used to build the Lie bracket.

$T_{2}$ survives. Exchanging $\mu$ and $\nu$ turns $(\partial_{\mu}\Gamma^{\rho}{}_{\nu\sigma})V^{\sigma}$ into $(\partial_{\nu}\Gamma^{\rho}{}_{\mu\sigma})V^{\sigma}$, which is a genuinely different object. The difference is $\big(\partial_{\mu}\Gamma^{\rho}{}_{\nu\sigma}-\partial_{\nu}\Gamma^{\rho}{}_{\mu\sigma}\big)V^{\sigma}$.

$T_{3}$ and $T_{4}$ die together. This is the one that needs a moment. Individually neither is symmetric, but their sum is. Exchange $\mu$ and $\nu$ in $T_{3}=\Gamma^{\rho}{}_{\nu\sigma}\partial_{\mu}V^{\sigma}$ to get $\Gamma^{\rho}{}_{\mu\sigma}\partial_{\nu}V^{\sigma}$. Now relabel the summed index $\sigma\to\lambda$, which changes nothing since it is a dummy, and the result is $\Gamma^{\rho}{}_{\mu\lambda}\partial_{\nu}V^{\lambda}$, which is $T_{4}$ exactly. By the same two moves, exchanging $\mu\leftrightarrow\nu$ in $T_{4}$ gives $T_{3}$. So the pair $T_{3}+T_{4}$ maps to $T_{4}+T_{3}$ and is symmetric, and the subtraction removes it.

$T_{5}$ survives. Exchanging gives $\Gamma^{\rho}{}_{\nu\lambda}\Gamma^{\lambda}{}_{\mu\sigma}V^{\sigma}$, and the two connection factors do not commute as index objects, so the difference is $\big(\Gamma^{\rho}{}_{\mu\lambda}\Gamma^{\lambda}{}_{\nu\sigma}-\Gamma^{\rho}{}_{\nu\lambda}\Gamma^{\lambda}{}_{\mu\sigma}\big)V^{\sigma}$.

$T_{6}$ dies by vanishing torsion. The term is $-\Gamma^{\lambda}{}_{\mu\nu}\nabla_{\lambda}V^{\rho}$. Exchanging $\mu$ and $\nu$ affects only the two lower indices of the connection, and Chapter 3.3's Demand 2 made those symmetric: $\Gamma^{\lambda}{}_{\mu\nu}=\Gamma^{\lambda}{}_{\nu\mu}$. So the term is unchanged and the subtraction kills it. Had we allowed torsion, this term would have survived and contributed $-T^{\lambda}{}_{\mu\nu}\nabla_{\lambda}V^{\rho}$ to the answer, which is exactly how the general formula reads in books that keep torsion.

Two terms have survived, $T_{2}$ and $T_{5}$. Collect them, and give the array they multiply a name:

That equation defines the symbol $R^{\rho}{}_{\sigma\mu\nu}$ by what it does. The two terms that survived the subtraction say what its components are:

This is the Riemann curvature tensor. Before doing anything else with (3.4.13), notice its shape. It is antisymmetric in $\mu\nu$ by construction, since swapping those two letters exchanges the first two terms with each other and the last two with each other, reversing the sign of the whole. And it is built from the connection and its first derivatives, hence from the metric and its first and second derivatives.

3 · Why the result is a tensor

Now the payoff for §2's length. Look at what (3.4.13) is made of. Every ingredient on the right-hand side is a badly behaved object.

• The connection is not a tensor, by Chapter 3.3 §5.3.
• A derivative of a non-tensor is not a tensor either.
• A product of two non-tensors is not a tensor.

Yet the combination of them is perfectly well behaved. Here is the argument, which takes three lines because §2 did the work.

The left side is a tensor. Chapter 3.3 §5 established that $\nabla$ maps tensors to tensors, one rank higher. So $\nabla_{\nu}V^{\rho}$ is a $(1,1)$ tensor, and $\nabla_{\mu}\nabla_{\nu}V^{\rho}$ is a $(1,2)$ tensor. A difference of two tensors of the same type is a tensor of that type (Chapter 2.4 §5.1). Hence $[\nabla_{\mu},\nabla_{\nu}]V^{\rho}$ is a $(1,2)$ tensor, whatever $V$ is.

The right side has no derivatives of $V$. That is what §2's four cancellations bought. If even one of $T_{1}$, $T_{3}$, $T_{4}$, $T_{6}$ had survived, the right side would depend on how $V$ varies near the point and the next step would be unavailable.

Therefore the array is a tensor. We have a quantity $R^{\rho}{}_{\sigma\mu\nu}$ such that $R^{\rho}{}_{\sigma\mu\nu}V^{\sigma}$ is a $(1,2)$ tensor for every vector field $V$. That is the hypothesis of the quotient theorem of Chapter 2.4 §5.1, and its conclusion is that $R^{\rho}{}_{\sigma\mu\nu}$ is itself a tensor, of type $(1,3)$. The same thing can be said in the language of Chapter 3.2's Problem 3. The map $V\mapsto[\nabla_{\mu},\nabla_{\nu}]V$ is function-linear, meaning that if you multiply $V$ by any function $h$ the answer is multiplied by $h$, with no leftover term in $\partial h$. Function-linearity is exactly what distinguishes a tensor from a differential operator.

3.1 · A concrete demonstration: the flat plane in polar coordinates

Argument by quotient theorem is airtight and slightly bloodless. Here is the cancellation happening in a case where every number is known. Chapter 3.3 §7.8 found that the flat plane in polar coordinates has

and Chapter 3.3 §9's second worked example showed that every one of those numbers is the inhomogeneous term of the transformation law, since the connection vanishes in the Cartesian chart. So this connection is entirely made of the part that is not tensorial. If §3's claim is right, the curvature must be exactly zero. Compute it.

Take $\rho=r$, $\sigma=\theta$, $\mu=r$, $\nu=\theta$ in (3.4.13) and write the four terms out:

Evaluate the four in turn, using (3.4.14). The first is $\partial_{r}(-r)=-1$. The second is zero, because $\Gamma^{r}{}_{r\theta}=0$. The third is zero, because both $\Gamma^{r}{}_{rr}$ and $\Gamma^{r}{}_{r\theta}$ vanish. The fourth has two values of $\lambda$: $\lambda=r$ gives $\Gamma^{r}{}_{\theta r}\Gamma^{r}{}_{r\theta}=0$, and $\lambda=\theta$ gives $\Gamma^{r}{}_{\theta\theta}\Gamma^{\theta}{}_{r\theta}=(-r)(1/r)=-1$, entering with the minus sign in front. So

The derivative term and the quadratic term are equal and opposite. Do it once more with the other index assignment, $\rho=\theta$, $\sigma=r$, $\mu=\theta$, $\nu=r$:

Again exactly. Every remaining component vanishes term by term for the same kind of reason, and the whole tensor is identically zero. (Confirmed by computer algebra: all sixteen components of $R^{\rho}{}_{\sigma\mu\nu}$ for this metric are identically zero as functions of $r$ and $\theta$.) A connection made entirely of chart-dependent junk produces exactly no curvature, which is what a tensor built to discard chart-dependent junk had better do.

4 · Geodesic deviation, and where Part III's thesis lands

Everything so far has been mathematics. This section is where it becomes physics, and it is the section the whole of Part III has been walking toward, so it goes slowly.

4.1 · The setup, and one identity

Consider a one-parameter family of geodesics $x^{\mu}(\tau,s)$. For each fixed $s$, the curve traced out as $\tau$ varies is a geodesic parametrised by its own proper time, and $s$ is the label saying which geodesic we are on. Two vector fields live naturally on the two-dimensional surface those curves sweep out, so define them both now:

Because $\tau$ and $s$ are coordinates on that surface, $u$ and $\xi$ are coordinate vector fields, and Chapter 3.2 §8.4 proved that coordinate vector fields commute:

One identity is needed before starting, and it is short. Compute the components of $\nabla_{u}\xi-\nabla_{\xi}u$ directly:

In the last term, relabel the two summed indices, exchanging the names $\nu$ and $\rho$. That is free, since both are dummies. It becomes $\Gamma^{\mu}{}_{\rho\nu}\xi^{\rho}u^{\nu}$, and now use the symmetry of the connection in its lower indices to write $\Gamma^{\mu}{}_{\rho\nu}=\Gamma^{\mu}{}_{\nu\rho}$. The last two terms of (3.4.20) are then identical and cancel. What remains is the Lie bracket of Chapter 3.2 (3.2.32), so

4.2 · The derivation, in four lines

We want the second derivative of the separation along the geodesic, which is what "relative acceleration" means. Write $D/\dd\tau\equiv\nabla_{u}$ for the derivative along the curve.

Line 1. Apply $\nabla_{u}$ twice and use (3.4.21) to swap the inner pair:

Line 2. What we want on the right is §2's commutator, so rewrite that expression as the opposite ordering plus the difference between the two orderings:

Line 3. The first term on the right is zero. That is the only place the geodesic property is used, and it is used exactly once: $\nabla_{u}u^{\mu}=0$ is the geodesic equation (3.3.57), so differentiating it in any direction still gives zero.

Line 4. Convert the remaining commutator of directional derivatives into the commutator of coordinate derivatives. Expanding $\nabla_{u}\nabla_{\xi}V=u^{\mu}\nabla_{\mu}(\xi^{\nu}\nabla_{\nu}V)$ with the Leibniz rule and subtracting the other ordering gives

and the first bracket vanishes by (3.4.21). Setting $V=u$ and using the definition (3.4.12) of the Riemann tensor:

Tidy the indices. Rearrange the factors, which are just numbers, and then use the antisymmetry of $R$ in its last two indices, $R^{\mu}{}_{\gamma\alpha\beta}=-R^{\mu}{}_{\gamma\beta\alpha}$, noted under (3.4.13). Renaming $\gamma\to\nu$, $\beta\to\rho$, $\alpha\to\sigma$ gives the standard form:

This is the equation of geodesic deviation. Read it before comparing it with anything. Take two particles, each moving as freely as it is possible to move, with no forces acting and no potentials anywhere, each following the straightest path available to it. They nevertheless accelerate relative to one another, and the rate is fixed by the curvature of the space they are in, linearly in their separation. In flat space $R=0$, and neighbouring straight lines drift apart at a constant rate and no faster, which is what straight lines do.

4.3 · Setting it beside Newton

Now take (3.4.26) to the Newtonian limit, under exactly the three assumptions of Chapter 3.3 §8.3: slow motion, a static field, and a weak one.

Slow motion makes $u^{\mu}\approx(c,0,0,0)$, since $u^{0}=c\,\dd t/\dd\tau\approx c$ and the spatial components are smaller by $v/c$. It also makes $\tau\approx t$. The separation between two particles at the same coordinate time is purely spatial, $\xi^{\mu}=(0,\xi^{j})$.

Substituting into (3.4.26), the sums over $\nu$ and $\sigma$ collapse to the single value $0$ and the sum over $\rho$ to spatial values:

And here is Chapter 3.1's equation (3.1.14), derived there from Newton's law of gravitation and one linearisation, with no relativity anywhere in it:

Put (3.4.27) and (3.4.28) side by side. They are the same equation. The separation vector plays the same role, the second time derivative plays the same role, and in the place where Newton has the matrix of second derivatives of the potential, general relativity has

4.4 · The identification, checked rather than assumed

Two equations having the same shape is suggestive, not conclusive. So compute $R^{i}{}_{0j0}$ directly from the weak-field metric and see whether (3.4.29) comes out. Everything needed is already derived: Chapter 3.1 §6.5 obtained $g_{00}=1+2\Phi/c^{2}$ from the accelerating cabin and the Doppler effect, using no general relativity, and Chapter 3.3 §8.3 turned that into

With that connection coefficient in hand we can compute the curvature component directly. Take (3.4.13), the definition from §2, and set $\rho=i$, $\sigma=0$, $\mu=j$ and $\nu=0$:

The second term vanishes because nothing depends on time. The two quadratic terms are products of two connection coefficients, each of which is first order in $\Phi/c^{2}$, so they are second order and are dropped at this accuracy. What survives is the first term, and by (3.4.30),

which is (3.4.29) exactly. (Confirmed symbolically for a general potential $\Phi(x,y,z)$, all nine components at once.) The identification was not fitted to make the two equations match. It was computed, from a metric component that Chapter 3.1 obtained without using any of this machinery.

4.5 · Three things that follow immediately

(i) The trace, and Laplace's equation. Contract (3.4.29) on $i$ and $j$. On the left, the sum $R^{i}{}_{0i0}$ is nearly the Ricci tensor: by the definition $R_{\mu\nu}=R^{\lambda}{}_{\mu\lambda\nu}$ stated in this chapter's conventions,

where $R^{0}{}_{000}$ vanishes because $R$ is antisymmetric in its last two indices and they are equal. Poisson's equation $\nabla^{2}\Phi=4\pi G\rho$ (Chapter 0.7 §7) then gives

Chapter 3.1 §4.4 promised that its tracelessness result "is the covariant descendant of $\nabla^{2}\Phi=0$". Equation (3.4.34) is that descendant, and Chapter 3.6 will promote it to the full field equations.

(ii) The ellipse. Chapter 3.1 §4.5 said: "that shape is what a tidal field is, and Chapter 3.4 will recover this very ellipse from the curvature tensor." Worked example 2 below does exactly that, computing $R^{i}{}_{0j0}$ for a point mass and finding the eigenvalues $(-2GM/r^{3},\,+GM/r^{3},\,+GM/r^{3})/c^{2}$ that produce the stretch-and-squeeze.

(iii) The units are right. $\Phi_{,ij}$ has dimensions $\mathrm{s^{-2}}$, so $R^{i}{}_{0j0}$ has dimensions $\mathrm{m^{-2}}$. Curvature is one over a length squared, which is what §1's $1/a^{2}$ also was. For the Earth's surface field, $GM/r^{3}=1.54\times10^{-6}\,\mathrm{s^{-2}}$ and dividing by $c^{2}$ gives a curvature of about $1.7\times10^{-23}\,\mathrm{m^{-2}}$: a radius of curvature of roughly $2\times10^{11}$ metres, which is why nobody noticed.

5 · The symmetries, and how many numbers there really are

The array $R^{\rho}{}_{\sigma\mu\nu}$ has $4^{4}=256$ components in four dimensions. Almost all of them are redundant. This section derives the symmetries, then counts.

5.1 · The four symmetries, derived

Work with the fully lowered version $R_{\rho\sigma\mu\nu}\equiv g_{\rho\lambda}R^{\lambda}{}_{\sigma\mu\nu}$, which is legitimate and unambiguous by Chapter 3.3's Problem 2.

(S1) Antisymmetry in the last pair: $R_{\rho\sigma\mu\nu}=-R_{\rho\sigma\nu\mu}$. Immediate from (3.4.13), as noted there: exchanging $\mu$ and $\nu$ exchanges the first two terms and the last two, flipping the overall sign. Lowering an index does not touch $\mu\nu$.

(S2) Antisymmetry in the first pair: $R_{\rho\sigma\mu\nu}=-R_{\sigma\rho\mu\nu}$. This one needs an argument, and the argument is pretty. We get at it by first extending the commutator to covectors. For any covector $\omega$ and vector $V$, the contraction $\omega_{\rho}V^{\rho}$ is a scalar, and $[\nabla_{\mu},\nabla_{\nu}]$ annihilates scalars, which is Chapter 3.3 (3.3.40) with vanishing torsion. Apply Leibniz to that vanishing expression:

We want to strip $V$ off both terms, so relabel the summed indices in the second one, exchanging $\rho$ and $\sigma$, and both terms then carry $V^{\rho}$. Since $V$ is arbitrary, the coefficients must match:

Iterating Leibniz once more gives the rule for a $(0,2)$ tensor: one such term per lower index,

Now put $T=g$. The left side is zero, because $\nabla_{\rho}g_{\mu\nu}=0$ identically and so is any further derivative of it. The right side becomes, on lowering,

which is (S2). So metric compatibility, the very demand that fixed the connection in Chapter 3.3, is what makes the curvature antisymmetric in its first pair.

(S3) The first Bianchi identity: $R_{\rho\sigma\mu\nu}+R_{\rho\mu\nu\sigma}+R_{\rho\nu\sigma\mu}=0$. Cyclic in the last three indices. Derive it directly from (3.4.13). Write the three terms out with the upper index kept:

Add the three lines and go through the six derivative terms first. Take $\partial_{\mu}\Gamma^{\rho}{}_{\nu\sigma}$ from the first line and $-\partial_{\mu}\Gamma^{\rho}{}_{\sigma\nu}$ from the third: the two connections have their lower indices in opposite orders, and Chapter 3.3's Demand 2 says the order does not matter, so the pair cancels. The same argument pairs $-\partial_{\nu}\Gamma^{\rho}{}_{\mu\sigma}$ from the first line with $+\partial_{\nu}\Gamma^{\rho}{}_{\sigma\mu}$ from the second, and $-\partial_{\sigma}\Gamma^{\rho}{}_{\nu\mu}$ from the second with $+\partial_{\sigma}\Gamma^{\rho}{}_{\mu\nu}$ from the third. All six derivative terms are gone, in three pairs.

Now the six quadratic terms, by exactly the same mechanism. Pair $+\Gamma^{\rho}{}_{\mu\lambda}\Gamma^{\lambda}{}_{\nu\sigma}$ from the first line with $-\Gamma^{\rho}{}_{\mu\lambda}\Gamma^{\lambda}{}_{\sigma\nu}$ from the third: identical after using the symmetry of $\Gamma^{\lambda}{}_{\nu\sigma}$, so they cancel. Pair $-\Gamma^{\rho}{}_{\nu\lambda}\Gamma^{\lambda}{}_{\mu\sigma}$ with $+\Gamma^{\rho}{}_{\nu\lambda}\Gamma^{\lambda}{}_{\sigma\mu}$, and $-\Gamma^{\rho}{}_{\sigma\lambda}\Gamma^{\lambda}{}_{\nu\mu}$ with $+\Gamma^{\rho}{}_{\sigma\lambda}\Gamma^{\lambda}{}_{\mu\nu}$. All six vanish. The cyclic sum is zero, and every single cancellation used the same one fact, that the connection is symmetric in its lower indices.

(S4) Pair exchange: $R_{\rho\sigma\mu\nu}=R_{\mu\nu\rho\sigma}$. This is a consequence of the first three rather than an independent fact, and the proof is a four-line combinatorial manoeuvre with no new ingredient in it. It sits in the grind box, and the conclusion is what we use.

Grind box — pair exchange from (S1), (S2) and (S3)

Write the first Bianchi identity four times, each time cycling the last three indices of a different starting arrangement. Abbreviate the four indices as $a,b,c,d$.

$$\begin{aligned} \text{(i)}\quad & R_{abcd}+R_{acdb}+R_{adbc}=0,\\ \text{(ii)}\quad & R_{bcda}+R_{bdac}+R_{bacd}=0,\\ \text{(iii)}\quad & R_{cdab}+R_{cabd}+R_{cbda}=0,\\ \text{(iv)}\quad & R_{dabc}+R_{dbca}+R_{dcab}=0. \end{aligned}$$

Form the combination (i) − (ii) − (iii) + (iv). Each of the four is separately zero, so the combination is zero. Now sort its twelve terms into two groups, using only (S1) and (S2) to move indices about. The first Bianchi identity is not used again from here.

Group one: four terms that cancel in pairs.

• $+R_{acdb}$ from (i) against $-R_{cabd}$ from (iii). Apply (S2) to the first, $R_{acdb}=-R_{cadb}$, and then (S1) to the last pair, $-R_{cadb}=+R_{cabd}$. The two terms are equal and oppositely signed, so they cancel.
• $+R_{adbc}$ from (i) against $+R_{dabc}$ from (iv). Here (S2) alone gives $R_{adbc}=-R_{dabc}$, so the sum is zero.
• $-R_{bcda}$ from (ii) against $-R_{cbda}$ from (iii). Again (S2): $R_{bcda}=-R_{cbda}$, so the two add to zero.
• $-R_{bdac}$ from (ii) against $+R_{dbca}$ from (iv). Apply (S2), $R_{bdac}=-R_{dbac}$, then (S1) on the last pair, $-R_{dbac}=+R_{dbca}$. Equal and oppositely signed, so they cancel.

Group two: four terms that double up.

• $+R_{abcd}$ from (i) and $-R_{bacd}$ from (ii). By (S2), $-R_{bacd}=+R_{abcd}$, so together they give $2R_{abcd}$.
• $-R_{cdab}$ from (iii) and $+R_{dcab}$ from (iv). By (S2), $R_{dcab}=-R_{cdab}$, so together they give $-2R_{cdab}$.

Everything has now been accounted for, and what the vanishing combination says is

$$2R_{abcd} \;-\; 2R_{cdab} \;=\; 0, \qquad\text{that is}\qquad R_{abcd} \;=\; R_{cdab}. \qquad\blacksquare$$

Notice that the three inputs were used in different places: the first Bianchi identity supplied the four vanishing sums, and the two antisymmetries did all the rearranging. That is why (S4) is a consequence rather than an independent assumption.

Numerical confirmation of the algebra. Taking a random four-index array with (S1) and (S2) imposed and nothing else, the combination (i) − (ii) − (iii) + (iv) was evaluated for all $256$ index assignments and compared with $2\big(R_{abcd}-R_{cdab}\big)$. The two agreed to $9\times10^{-16}$. So the rearrangement above is an identity in (S1) and (S2) alone, exactly as claimed.

Numerical confirmation. A deliberately generic four-dimensional metric was used, one with no symmetry at all, with off-diagonal entries and all four coordinates appearing. The Riemann tensor was computed from (3.4.13) and all four identities checked at a generic point. The largest violation of any of them was $1.2\times10^{-13}$ against components of size $0.19$, which is round-off.

5.2 · The count, derived

Now count, in $n$ dimensions, and then set $n=4$. The argument has three stages and each is a sentence.

Stage 1. Treat the index pairs as single labels. By (S1) and (S2), $R_{abcd}$ is antisymmetric within $ab$ and within $cd$. An antisymmetric pair of indices in $n$ dimensions takes

independent values, which is Chapter 2.4 §7.2's count of an antisymmetric array. In four dimensions $N=6$. So $R$ may be regarded as an $N\times N$ array $M_{[ab],[cd]}$, which is $36$ numbers in four dimensions rather than $256$.

Stage 2. Pair exchange makes that array symmetric. By (S4), $M_{AB}=M_{BA}$ where $A$ and $B$ are the pair labels. A symmetric $N\times N$ array has $N(N+1)/2$ independent entries, which for $N=6$ is $21$.

Stage 3. The first Bianchi identity removes exactly one more. Here is the step that has to be done carefully, because most of the cyclic identity is already implied by the other symmetries. The claim is that, given (S1), (S2) and (S4), the cyclic identity (S3) is equivalent to the single statement

the vanishing of the completely antisymmetric part. The reason is that the cyclic sum $R_{abcd}+R_{acdb}+R_{adbc}$, once the pair symmetries are imposed, is itself totally antisymmetric in all four indices, and any totally antisymmetric four-index array in $n$ dimensions has

independent components, again by Chapter 2.4 §7.2. In four dimensions that count is $\binom44=1$. There is exactly one totally antisymmetric four-index array up to scale, namely the Levi-Civita symbol of Chapter 2.4 §8.1, so (S3) imposes precisely one further condition.

The answer.

The boxed expression is what that difference collapses to once the binomials are written out and simplified. Now put in the number of dimensions we actually live in:

independent components, or directly, $21-1=20$.

5.3 · The same twenty, counted a completely different way

A number derived once is a number. Derived twice by independent routes, it is a fact. Here is the second route, which also settles two words that will be used together for the rest of Part III. A local inertial frame is a region small enough that gravity has no detectable effect across it. The locally inertial coordinates are the particular coordinates, constructed below, that are adapted to such a region. The frame is the physical situation, and the coordinates are the bookkeeping. Both will be needed again in §7.

Both halves are constructions rather than assertions.

Getting $g=\eta$ at $p$. $g_{\mu\nu}(p)$ is a real symmetric non-degenerate matrix, so by the spectral theorem (Chapter 0.5 §6) there is an orthogonal change of basis diagonalising it, and rescaling each basis vector by the square root of the modulus of its eigenvalue turns the diagonal entries into $\pm1$. The signature fixes the pattern to $(+,-,-,-)$. That is a linear change of coordinates, so it is a legitimate chart change.

Getting $\Gamma=0$ at $p$. Put $p$ at the origin and define new coordinates

Two facts about that substitution matter. At the origin $\partial y^{\mu}/\partial x^{\alpha}=\delta^{\mu}{}_{\alpha}$, so the linear part is untouched and $g(p)$ is still $\eta$. And $\partial^{2}y^{\mu}/\partial x^{\alpha}\partial x^{\beta}=\Gamma^{\mu}{}_{\alpha\beta}(p)$ everywhere. Feed those two into Chapter 3.3's transformation law (3.3.26). At $p$ every Jacobian is a Kronecker delta, so the law collapses to

This is precisely why $\Gamma$ had to be a non-tensor: a tensor vanishing in one chart vanishes in all, and no such construction would be possible. And $\partial_{\rho}g_{\mu\nu}(p)=0$ follows at once, since metric compatibility reads $\partial_{\rho}g_{\mu\nu}=\Gamma^{\lambda}{}_{\rho\mu}g_{\lambda\nu}+\Gamma^{\lambda}{}_{\rho\nu}g_{\mu\lambda}$ and the right side is zero at $p$.

Why the second derivatives cannot all be removed, and where the count comes from. Push the construction to the next order, with a cubic term $\tfrac16 C^{\mu}{}_{\alpha\beta\gamma}x^{\alpha}x^{\beta}x^{\gamma}$ whose coefficients are symmetric in the three lower indices. Then count what is available against what is demanded, in four dimensions:

• Available. A symmetric triple of indices from four values takes $\binom{4+2}{3}=20$ values, and $\mu$ takes $4$, so $C$ has $4\times20=80$ free numbers.
• Demanded. Setting $\partial_{\alpha}\partial_{\beta}g_{\mu\nu}(p)=0$ is one condition for each symmetric pair $\alpha\beta$ (10 values) and each symmetric pair $\mu\nu$ (10 values), so $10\times10=100$ conditions.

Eighty knobs, one hundred demands. Twenty combinations of second derivatives of the metric survive every possible change of coordinates. That is the same twenty as before, arrived at with no mention of Bianchi identities or antisymmetric pairs. It also says what the twenty are. They are the irreducible second-derivative content of the metric, the part no observer can transform away.

The same counting at the two lower orders is worth doing as well, because both answers are ones you have seen. The linear part $A^{\mu}{}_{\alpha}$ has $16$ free numbers and $g(p)=\eta$ is $10$ conditions, leaving $16-10=6$. That is the dimension of the Lorentz group, three boosts and three rotations, exactly as Part II found. The quadratic part has $4\times10=40$ numbers against $\partial_{\rho}g_{\mu\nu}(p)=0$'s $4\times10=40$ conditions, matching exactly, which is why $\Gamma(p)$ can always be removed and never more than that.

6 · Ricci, the scalar, and what contraction throws away

Twenty numbers per point is a lot to carry into a field equation. Contracting reduces them, and the question is which contraction and at what cost.

6.1 · There is essentially only one contraction

A $(1,3)$ tensor can be contracted by summing its upper index against any one of its three lower ones. Take them in turn.

Against the first lower index. $R^{\lambda}{}_{\lambda\mu\nu}=g^{\lambda\rho}R_{\rho\lambda\mu\nu}$. The array $R_{\rho\lambda\mu\nu}$ is antisymmetric in $\rho\lambda$ by (S2) and $g^{\lambda\rho}$ is symmetric, and contracting a symmetric array with an antisymmetric one over both indices gives zero. That is Chapter 2.4 §7's standard argument, since relabelling the two summed indices reproduces the expression with the opposite sign. So this contraction vanishes identically.

Against the third lower index. $R^{\lambda}{}_{\mu\nu\lambda}=-R^{\lambda}{}_{\mu\lambda\nu}$ by (S1), which is minus the remaining case.

Against the second lower index. This is the only one that is both non-zero and not a duplicate:

That object is the Ricci tensor. Its sign is a convention, and this book's is stated in the opening callout.

It is symmetric. Lower the contracted index and use pair exchange:

where the third step relabelled the two summed indices $\lambda\leftrightarrow\sigma$ and used the symmetry of $g^{\lambda\sigma}$. So Ricci has $10$ independent components in four dimensions, not $16$.

Contracting once more, with the inverse metric, gives a single number at each point:

the Ricci scalar or scalar curvature. Being a scalar, it is the same number for every observer, which makes it the simplest honest statement about curvature available anywhere.

6.2 · What was discarded

Riemann has $20$ components and Ricci has $10$. So contraction throws away exactly $10$ components of information in four dimensions. That leftover has a name.

6.3 · Why this matters immediately: vacuum is not flat

Here is the reason the discarded part cannot be dismissed. Section 4.5 derived $R_{00}=4\pi G\rho/c^{2}$, so outside a mass, where $\rho=0$, the Ricci tensor has a vanishing component, and Chapter 3.6 will show the whole of it vanishes there. Yet the tidal field outside a mass is emphatically not zero. Chapter 3.1 computed it, and §4 identified it with $R^{i}{}_{0j0}$, which by (3.4.29) is $\Phi_{,ij}/c^{2}$ and is non-zero at every point outside the body.

Both statements hold because Chapter 3.1 §4.4's tidal matrix is traceless in vacuum and non-zero. The trace is the Ricci part, and the traceless remainder is the Weyl part. So:

Ricci flat is not flat. If it were, there would be no gravity outside any body, no orbits, no tides and no gravitational waves. The whole of Chapter 3.7 lives in a region where $R_{\mu\nu}=0$ and the curvature is what makes Mercury precess.

7 · The second Bianchi identity, and a tensor with no divergence

This section is short and its result is the most consequential single line in the chapter, so its importance is flagged in advance.

7.1 · The identity

The identity we are after says something about how the curvature varies from point to point. Here is the claim:

cyclic in the three lower indices $\lambda\mu\nu$, which are the derivative index and the antisymmetric pair together. The proof uses §5.3's locally inertial coordinates and is four lines.

Line 1. Fix a point $p$ and adopt coordinates with $\Gamma^{\lambda}{}_{\mu\nu}(p)=0$, which §5.3 constructed. At $p$, and only at $p$, the definition (3.4.13) loses its quadratic terms:

Line 2. The covariant derivative of anything reduces to the ordinary derivative at $p$, since every correction term carries a factor of $\Gamma(p)=0$. Note carefully what is not being claimed: $\partial\Gamma$ is not zero at $p$, only $\Gamma$ itself is, which is why (3.4.52) has content. Differentiating,

the derivatives of the quadratic terms also vanishing at $p$ because each is $(\partial\Gamma)\Gamma+\Gamma(\partial\Gamma)$ and every term carries an undifferentiated $\Gamma$.

Line 3. Write the cyclic sum. Each of the three terms contributes two pieces, so there are six, and they cancel in pairs by Clairaut's theorem:

The first term of each line cancels the second term of the line below it. For instance, $\partial_{\lambda}\partial_{\mu}\Gamma^{\rho}{}_{\nu\sigma}$ goes against $-\partial_{\mu}\partial_{\lambda}\Gamma^{\rho}{}_{\nu\sigma}$, and those two are equal because mixed partials commute. The third line's second term goes against the first line's first term after the same move.

Line 4. Equation (3.4.51) is an equation between tensors, since every term in it is a covariant derivative of a tensor. It has just been shown true at $p$ in one particular chart. By Chapter 2.4 §6, a tensor equation true in one chart is true in every chart. And $p$ was arbitrary. Hence (3.4.51) holds everywhere, in every chart. $\blacksquare$

(Confirmed numerically: for the generic four-dimensional metric of §5.1's grind box, the largest value of the cyclic sum at a generic point was $1.3\times10^{-11}$, against individual terms of size $0.07$.)

7.2 · Contract it twice

Lower the first index of (3.4.51), which is legitimate because $\nabla g=0$ lets the metric pass through the derivative:

First contraction. Multiply by $g^{\rho\mu}$ and sum. Take the three terms one at a time, saying what each contraction produces.

Term one: $g^{\rho\mu}R_{\rho\sigma\mu\nu}=R^{\mu}{}_{\sigma\mu\nu}=R_{\sigma\nu}$, by the definition (3.4.47). So it becomes $\nabla_{\lambda}R_{\sigma\nu}$.

Term two: the $g^{\rho\mu}$ meets the derivative index, giving $g^{\rho\mu}\nabla_{\mu}R_{\rho\sigma\nu\lambda}=\nabla^{\rho}R_{\rho\sigma\nu\lambda}$.

Term three: $g^{\rho\mu}R_{\rho\sigma\lambda\mu}=R^{\mu}{}_{\sigma\lambda\mu}$, which by (S1) is $-R^{\mu}{}_{\sigma\mu\lambda}=-R_{\sigma\lambda}$. So it becomes $-\nabla_{\nu}R_{\sigma\lambda}$.

Second contraction. Multiply by $g^{\sigma\nu}$ and sum, again one term at a time.

Term one: $g^{\sigma\nu}\nabla_{\lambda}R_{\sigma\nu}=\nabla_{\lambda}R$, the Ricci scalar.

Term two: $g^{\sigma\nu}R_{\rho\sigma\nu\lambda}$. Use (S2) to write $R_{\rho\sigma\nu\lambda}=-R_{\sigma\rho\nu\lambda}$, and then the contraction of the first and third indices is the Ricci tensor by (3.4.47), giving $-R_{\rho\lambda}$. So the term is $-\nabla^{\rho}R_{\rho\lambda}$.

Term three: $-g^{\sigma\nu}\nabla_{\nu}R_{\sigma\lambda}=-\nabla^{\sigma}R_{\sigma\lambda}$.

The last step added the two identical terms. They are identical because $\rho$ and $\sigma$ are both summed, and the name of a dummy index is private.

7.3 · The Einstein tensor

Our goal now is to write (3.4.57) as a single divergence, so that the whole content sits inside one $\nabla$. The metric passes through the derivative once more, so $\nabla_{\lambda}R=\nabla^{\rho}\big(g_{\rho\lambda}R\big)$, and the equation reads $\nabla^{\rho}\big(R_{\rho\lambda}-\half g_{\rho\lambda}R\big)=0$. Define

This is the Einstein tensor, and three things about it are worth stating separately.

• It is symmetric, because both $R_{\mu\nu}$ and $g_{\mu\nu}$ are.
• It is built from the metric and its first two derivatives and nothing else.
• Its divergence vanishes identically, for every metric, as a consequence of (3.4.51) and not of any equation of motion.

(Confirmed numerically for the generic metric: all four components of $\nabla^{\mu}G_{\mu\nu}$ came out below $4\times10^{-11}$ against derivatives of Ricci of size $0.11$.)

8 · Flat if and only if the curvature vanishes

Chapter 3.3 §2 promised a genuine test and warned that neither varying metric components nor non-zero connection coefficients was one. Here is the test, in the form of a theorem with one easy half and one hard half.

The easy half, proved. Suppose there exist coordinates in which $g_{\mu\nu}$ is constant, say equal to $\eta_{\mu\nu}$. Then $\partial_{\sigma}g_{\mu\nu}=0$ everywhere in that chart. By the Christoffel formula (3.3.50), every $\Gamma^{\lambda}{}_{\mu\nu}$ therefore vanishes everywhere in that chart. By (3.4.13), every component of $R^{\rho}{}_{\sigma\mu\nu}$ then vanishes in that chart too, each of the four terms being either a derivative of zero or a product containing zero. And $R$ is a tensor by §3, so a tensor vanishing in one chart vanishes in every chart (Chapter 2.4 §6). Hence $R^{\rho}{}_{\sigma\mu\nu}=0$ in all charts. $\blacksquare$

Note what that argument settles. The flat plane in polar coordinates has non-constant metric components and non-zero connection coefficients, yet it must have zero curvature, because Cartesian coordinates exist for it. Section 3.1's explicit computation was the verification. This is the reason.

9 · Worked examples

10 · Your turn