Part III · General Relativity — Chapter 3.7

Schwarzschild: The Solution and Its Orbits

Solve Einstein's equations exactly, by hand, in one sitting. Then extract from the answer an orbit Newton has no name for, and a planet's slow turning that nobody could explain for sixty years.

Tools you'll need  — Chapter 3.6 §5.1, which supplies the trace-reversed field equations (3.6.23) and the remark that in vacuum they read $R_{\mu\nu}=0$. Chapter 3.5 §8.1 for Killing's equation and its practical test, which says that every coordinate the metric does not mention supplies a Killing vector, and §9 in full, which is the section this chapter spends. Chapter 3.4 §6.1 for the Ricci tensor and its definition (3.4.47), and §6.3 for Ricci-flat against flat. Chapter 3.3 §7 (the Christoffel formula, (3.3.50)), §8 (geodesics, both derivations, and the constancy of $g_{\mu\nu}u^{\mu}u^{\nu}$ along one) and §3 (proper time). Chapter 2.5 §1 for the normalisation $u\cdot u=c^{2}$, equation (2.5.9), in flat spacetime. Chapter 3.2 §2 for what a chart is and what it is not. Chapter 3.1 §6.5, the derived weak-field component $g_{00}=1+2\Phi/c^{2}$ (3.1.39), which fixes this chapter's mass. Chapter 0.8 §2.2, the integrating factor, which does the one integration in §3, and §4.3 for reading an energy equation as motion in a landscape, a reading Chapter 1.3 §4 turned into phase portraits. Chapter 0.3 §2.4, the binomial series, for the one expansion in §8.4. Chapter 1.4 §3 for energy and angular momentum as Noether charges, and Worked example 2 there, on the Laplace–Runge–Lenz vector, which proved that Kepler ellipses close and predicted that any departure from an exact inverse square would make them precess. This chapter supplies the departure.

1 · What "spherically symmetric and static" buys

Here is where this section is going. We are going to translate two English adjectives into conditions on the metric, using the machinery of Chapter 3.5 rather than a picture. The translation matters because "looks the same from every direction" is a statement about a drawing, and a drawing cannot be substituted into a field equation. "There are three vector fields whose flows leave the metric unchanged" can be.

1.1 · Spherical symmetry, as a statement about the geometry

Chapter 3.5 §8.1 defined a Killing vector field $\xi$ as one whose flow leaves the metric exactly as it found it, $\mathcal{L}_{\xi}g=0$, equivalently $\nabla_{\mu}\xi_{\nu}+\nabla_{\nu}\xi_{\mu}=0$. That chapter also emphasised the distinction this section needs: a Killing vector is not a direction along which the picture looks the same, it is a direction along which every length anybody could measure comes out the same.

A spacetime is spherically symmetric when it admits three Killing vector fields $\xi_{(1)},\xi_{(2)},\xi_{(3)}$, each spacelike, whose commutators reproduce the algebra of rotations:

Two things about this definition are worth pausing on, because it looks more elaborate than the English phrase it replaces.

The bracket is Chapter 3.2's commutator of vector fields, and by Chapter 3.5 §7.2 it is also the Lie derivative of one field along the other. Requiring (3.7.1) rather than merely requiring three Killing fields is what distinguishes rotations from, say, three translations: it is the statement that these three symmetries fail to commute in the particular pattern that turns about three perpendicular axes fail to commute.

You have already met these three fields. Chapter 3.5 §8.3 computed them on a sphere of radius $a$: the obvious one $\partial_{\varphi}$, and two more, $\cos\varphi\,\partial_{\theta}-\cot\theta\sin\varphi\,\partial_{\varphi}$ and $-\sin\varphi\,\partial_{\theta}-\cot\theta\cos\varphi\,\partial_{\varphi}$, which are rotations about the other two axes and are invisible in the usual chart. Those are the $\xi_{(i)}$ of (3.7.1), extended off the sphere.

1.2 · Static, which is stronger than stationary

A spacetime is stationary when it admits a further Killing field $\xi_{(0)}$ that is timelike, meaning $g_{\mu\nu}\xi_{(0)}^{\mu}\xi_{(0)}^{\nu}>0$ in this book's signature. In a chart adapted to it, that says the metric components do not depend on $x^{0}$. Nothing about the geometry changes with time.

A spacetime is static when it is stationary and additionally invariant under reversing the time direction, $x^{0}\to-x^{0}$.

The second condition is not decoration and it is not implied by the first. Here is a case that separates them, and it is the case that matters physically. Take the geometry outside a rotating star. Nothing about it changes with time, so it is stationary. But reversing time reverses the sense of the rotation, which is a different physical situation, so it is not static. The exact solution for a rotating mass is Kerr's, found in 1963, and it is a different and much longer calculation which this book does not perform. What follows is the non-rotating case.

1.3 · These are hypotheses, and they are being chosen

Nothing derived so far says that any actual object is spherically symmetric or static. The Sun rotates. The Earth is oblate. A collapsing star is neither. What §§2–3 solve is a special case chosen because it is exactly solvable, and the honest description of the result is this is the geometry outside a non-rotating spherical body. The accuracy of everything downstream is limited by how well a given body satisfies that.

Section 4 removes one of the two hypotheses at no cost, which is a considerably better outcome than it sounds.

2 · The ansatz, and why it has only two unknown functions

Here is where this section is going. Einstein's equations are ten coupled nonlinear equations for ten unknown functions of four variables. We are going to use §1's symmetries to reduce that to two unknown functions of one variable. Then we will say exactly which step in the reduction was a choice rather than a consequence, because that step is the one Chapter 3.8 has to un-learn.

2.1 · What the symmetries remove

Work in a chart adapted to the symmetries, which exists by Chapter 3.2 §2 and which we build one piece at a time.

Stationarity removes the $x^{0}$ dependence. Chapter 3.5 §8.1's practical test runs in reverse here: if $\partial_{0}$ is a Killing field, there is a chart in which no metric component depends on $x^{0}$. Ten functions of four variables become ten functions of three.

Time reversal removes the cross terms. By §1.2, $g_{0i}=0$. Six functions remain in the spatial block, plus $g_{00}$: seven.

Spherical symmetry removes the angular dependence and most of the spatial block. The three rotational Killing fields act on the spatial slices. Their orbits are the sets of points reachable from a given point by rotating, and those sets are two-dimensional spheres. The metric restricted to any one of them must be invariant under all three rotations. Chapter 3.5 §8.3's calculation on the sphere is exactly the statement that up to overall scale there is one such metric, the round one:

So each sphere contributes $-S\,\dd\Omega^{2}$ for a single positive function $S$, one number per sphere rather than three functions of angle. Rotational invariance also forbids any cross term pairing $\dd\theta$ or $\dd\varphi$ with the remaining radial direction, since such a term would pick out a direction on the sphere and there is none to pick. What is left is

with $A$, $\tilde B$ and $S$ three functions of the radial label $\rho$ alone. Three unknown functions of one variable. One more move takes it to two.

2.2 · The areal radius — the load-bearing choice

The label $\rho$ in (3.7.3) is arbitrary: relabel $\rho\to\tilde\rho(\rho)$ and the same geometry is described by three different functions. That freedom is one function's worth, and we spend it.

Here is what we spend it on. The sphere labelled $\rho$ has, by (3.7.3), metric $S\,\dd\Omega^{2}$, so its area is $S\int\sin\theta\,\dd\theta\,\dd\varphi=4\pi S$. Define a new radial label $r$ by

This is the areal radius. One condition is being assumed and it should be said out loud: $S$ must be strictly monotonic in the old radial label, so that $r=\sqrt{S}$ is invertible and can serve as a coordinate at all. Where it is not, this relabelling is unavailable, and the solution derived below does not describe that region. The place where it fails has a name, a throat, meaning a place where the spheres stop growing as you move outward. Nothing in this chapter meets such a place, but the assumption is real and it is being spent here rather than later.

With the areal radius, (3.7.3) becomes

Two unknown functions of one variable, $A$ and $B$, both dimensionless. That is the ansatz §3 solves.

2.3 · Ten equations, two unknowns, and why that is not yet a contradiction

Substituting (3.7.5) into $R_{\mu\nu}=0$ gives ten equations for two functions. That is a wildly overdetermined system and it would be reasonable to expect it to have no solution at all. Section 3 finds that six of the ten equations are $0=0$ identically, by the symmetry, that a seventh is a multiple of another, and that the surviving three are consistent. The consistency is not automatic and Chapter 3.6 §7.1 says where it comes from: the contracted Bianchi identity supplies four relations among the ten field equations for every metric, so the count of genuinely independent equations was never ten.

3 · Solving $R_{\mu\nu}=0$

Here is the route, because there are five steps and each one is small.

• Compute the connection for (3.7.5).
• Compute the Ricci tensor from it.
• Find that one particular combination of two of the equations collapses to the statement that $(AB)'=0$.
• Fix the resulting constant by demanding that the geometry become flat far away.
• Substitute $B=1/A$ into the angular equation, which becomes $(rA)'=1$, and integrate that once.

Two constants of integration appear along the way, and each one is fixed by naming a physical requirement. Out comes an exact solution.

Chapter 3.6 §5.1 supplies the equation we are solving. The trace-reversed field equations (3.6.23) with $T_{\mu\nu}=0$ and $\lambda=0$ read

These are the equations for empty space. They apply outside the star, not inside it, and Chapter 3.4 §6.3 already warned that they do not say the geometry is flat.

3.1 · The connection

Chapter 3.3 §7 derived the Christoffel formula (3.3.50),

and the metric (3.7.5) is diagonal, so the inverse metric is diagonal too, with entries $g^{00}=1/A$, $g^{11}=-1/B$, $g^{22}=-1/r^{2}$ and $g^{33}=-1/(r^{2}\sin^{2}\theta)$. That collapses the sum over $\sigma$ in (3.7.7) to a single term every time. Grind box A does the arithmetic. The answer is nine non-zero components, counting each symmetric pair once:

Four of the nine are exactly the connection coefficients of flat space in spherical coordinates: the two carrying $1/r$, the one carrying $\cot\theta$ and the one carrying $\sin\theta\cos\theta$. Chapter 3.3 §2 computed those four as the standing warning that varying components do not mean curvature. They are here because the chart is spherical, not because the spacetime is curved. The other five carry $A$ and $B$, and they are where the physics lives.

Grind box A — the nine Christoffel symbols, from (3.3.50)

Write $x^{\mu}=(x^{0},r,\theta,\varphi)$ and recall that no metric component depends on $x^{0}$ or on $\varphi$, and that only $g_{33}=-r^{2}\sin^{2}\theta$ depends on $\theta$.

$\Gamma^{0}{}_{01}$. Take $\lambda=0$, so $\sigma=0$ and $g^{00}=1/A$. With $\mu=0$, $\nu=1$ the bracket in (3.7.7) is $\partial_{0}g_{01}+\partial_{1}g_{00}-\partial_{0}g_{01}=\partial_{r}A=A'$. Hence $\Gamma^{0}{}_{01}=\half(1/A)A'=A'/2A$.

$\Gamma^{1}{}_{00}$. Take $\lambda=1$, so $g^{11}=-1/B$. With $\mu=\nu=0$ the bracket is $2\partial_{0}g_{10}-\partial_{1}g_{00}=-A'$. Hence $\Gamma^{1}{}_{00}=\half(-1/B)(-A')=A'/2B$.

$\Gamma^{1}{}_{11}$. Same $\lambda$, and the bracket is $\partial_{1}g_{11}=-B'$, giving $\half(-1/B)(-B')=B'/2B$.

$\Gamma^{1}{}_{22}$. Same $\lambda$ again. With $\mu=\nu=2$ the bracket is $2\partial_{2}g_{12}-\partial_{1}g_{22}=+2r$, giving $\half(-1/B)(2r)=-r/B$. The same computation with $33$ in place of $22$ carries the extra $\sin^{2}\theta$ through and gives $\Gamma^{1}{}_{33}=-r\sin^{2}\theta/B$.

$\Gamma^{2}{}_{12}$. Take $\lambda=2$, $g^{22}=-1/r^{2}$. With $\mu=1$, $\nu=2$ the bracket is $\partial_{1}g_{22}=-2r$, giving $\half(-1/r^{2})(-2r)=1/r$. Identically for $\Gamma^{3}{}_{13}$, the extra $\sin^{2}\theta$ cancelling between $g^{33}$ and $\partial_{1}g_{33}$.

$\Gamma^{2}{}_{33}$. $\lambda=2$, $\mu=\nu=3$: the bracket is $-\partial_{2}g_{33}=+2r^{2}\sin\theta\cos\theta$, giving $\half(-1/r^{2})(2r^{2}\sin\theta\cos\theta)=-\sin\theta\cos\theta$.

$\Gamma^{3}{}_{23}$. $\lambda=3$, $\mu=2$, $\nu=3$: the bracket is $\partial_{2}g_{33}=-2r^{2}\sin\theta\cos\theta$, and $\half\big(-1/(r^{2}\sin^{2}\theta)\big)(-2r^{2}\sin\theta\cos\theta)=\cot\theta$.

Everything else vanishes. A component $\Gamma^{\lambda}{}_{\mu\nu}$ is non-zero only if the bracket contains a surviving derivative, and the only non-vanishing derivatives of metric components are $\partial_{r}$ of $g_{00},g_{11},g_{22},g_{33}$ and $\partial_{\theta}$ of $g_{33}$. Running through the possibilities leaves the nine above.

One contraction, for later. Chapter 3.5 §6.4 derived $\Gamma^{\lambda}{}_{\lambda\mu}=\partial_{\mu}\ln\sqrt{-g}$ from Jacobi's formula, and here $-g=AB\,r^{4}\sin^{2}\theta$, so $\sqrt{-g}=\sqrt{AB}\,r^{2}\sin\theta$ and

$$\Gamma^{\lambda}{}_{\lambda 1} \;=\; \frac{A'}{2A} + \frac{B'}{2B} + \frac{2}{r}, \qquad \Gamma^{\lambda}{}_{\lambda 2} \;=\; \cot\theta,$$

and the other two vanish. Adding the list above directly gives the same thing, which is the Jacobi formula checked rather than trusted.

(All nine were confirmed symbolically from (3.7.7) for general $A(r)$ and $B(r)$, and no other component is non-zero.)

3.2 · The Ricci tensor, and the count

Chapter 3.4 §6.1 defined the Ricci tensor and wrote it out in terms of the connection, equation (3.4.47):

Feeding (3.7.8) into it is a page of bookkeeping, folded into grind box B. The results are

with every off-diagonal component zero.

Now the count, which stays out here in the main text. Ten equations went in. Six of them, the off-diagonal ones, are $0=0$ and say nothing. Of the remaining four, the last two are proportional, so $R_{33}=0$ carries no information beyond $R_{22}=0$. That proportionality is spherical symmetry showing up in the answer, since $\theta$ and $\varphi$ are on the same footing. What is left is three equations for two unknown functions.

Three equations for two unknowns is one equation too many, and a system in that position has no right to possess a solution. It does, and the reason was supplied a chapter ago: by Chapter 3.6 §7.1 the contracted Bianchi identity imposes four differential relations among the ten field equations for every metric whatever, so the three surviving equations are not independent of one another. That is worth a sentence rather than a shrug, because it is the same identity that fixed the factor of $\half$ in the Einstein tensor, doing a second job here.

Grind box B — the Ricci components, term by term

$R_{00}$, all four terms of (3.7.9).

Term 1. $\partial_{\lambda}\Gamma^{\lambda}{}_{00}$. The only non-zero $\Gamma^{\lambda}{}_{00}$ is $\Gamma^{1}{}_{00}=A'/2B$, so this is

$$\partial_{r}\!\left(\frac{A'}{2B}\right) \;=\; \frac{A''}{2B} \;-\; \frac{A'B'}{2B^{2}}.$$

Term 2. $-\partial_{0}\Gamma^{\lambda}{}_{\lambda 0}=0$: nothing depends on $x^{0}$.

Term 3. $\Gamma^{\lambda}{}_{\lambda\rho}\Gamma^{\rho}{}_{00}$. Only $\rho=1$ contributes, and grind box A's contraction gives

$$\left(\frac{A'}{2A}+\frac{B'}{2B}+\frac{2}{r}\right)\frac{A'}{2B} \;=\; \frac{(A')^{2}}{4AB}+\frac{A'B'}{4B^{2}}+\frac{A'}{rB}.$$

Term 4. $-\Gamma^{\lambda}{}_{0\rho}\Gamma^{\rho}{}_{\lambda 0}$. The only connection coefficients with a $0$ among their indices are $\Gamma^{0}{}_{01}$ and $\Gamma^{1}{}_{00}$, so the double sum has exactly two surviving terms, $(\lambda,\rho)=(0,1)$ and $(1,0)$, and they are equal:

$$-2\cdot\frac{A'}{2A}\cdot\frac{A'}{2B} \;=\; -\frac{(A')^{2}}{2AB}.$$

Add. The two $A'B'$ terms combine as $-\tfrac12+\tfrac14=-\tfrac14$, and the two $(A')^{2}$ terms as $+\tfrac14-\tfrac12=-\tfrac14$, giving the first line of (3.7.10).

$R_{11}$. Same four terms. Term 1 is $\partial_{r}(B'/2B)=B''/2B-(B')^{2}/2B^{2}$. Term 2 is $-\partial_{r}\big[A'/2A+B'/2B+2/r\big]$, which cancels the $B''$ of term 1 exactly and leaves $-A''/2A+(A')^{2}/2A^{2}+(B')^{2}/2B^{2}+2/r^{2}$. Term 3 is $\big[A'/2A+B'/2B+2/r\big](B'/2B)$. Term 4 sums the squares $-(A')^{2}/4A^{2}-(B')^{2}/4B^{2}-2/r^{2}$, the last from the two coefficients equal to $1/r$. Collecting gives the second line of (3.7.10). Note that the $2/r^{2}$ cancels between terms 2 and 4, which is why no such term appears in the answer.

$R_{22}$. Term 1 is $\partial_{r}(-r/B)=-1/B+rB'/B^{2}$. Term 2 is $-\partial_{\theta}\cot\theta=+1+\cot^{2}\theta$. Term 3 is $\big[A'/2A+B'/2B+2/r\big](-r/B)$. Term 4 is $-\big[2(-r/B)(1/r)+\cot^{2}\theta\big]=2/B-\cot^{2}\theta$. The $\cot^{2}\theta$ cancels between terms 2 and 4, which it must, since $R_{22}$ cannot depend on $\theta$. What survives is the third line of (3.7.10).

$R_{33}$ and the off-diagonal components. $R_{33}$ repeats the $R_{22}$ computation with the extra $\sin^{2}\theta$ carried along, giving $\sin^{2}\theta\,R_{22}$. Every off-diagonal component vanishes: any $R_{\mu\nu}$ with $\mu\neq\nu$ requires a connection coefficient carrying both a $\mu$ and a $\nu$, and inspection of (3.7.8) shows the surviving sums are empty or cancel in pairs.

(All of (3.7.10) was reproduced symbolically from (3.7.9) for general $A(r)$, $B(r)$, together with the vanishing of all six off-diagonal components and the identity $R_{33}=\sin^{2}\theta\,R_{22}$.)

3.3 · The line the whole solution turns on

Look at $R_{00}$ and $R_{11}$ in (3.7.10). Each contains $A''$, and each contains the same three other structures. Weight the first by $B$ and the second by $A$ and add:

Every second derivative has gone, and so has every quadratic term. The two survivors are the two halves of a product rule:

Setting the vacuum equations (3.7.6) to zero makes the left-hand side vanish, and $r$ and $B$ are finite and non-zero in the region where we are solving, so

That single line is why the Schwarzschild solution can be found by hand. A pair of coupled second-order equations has been replaced, exactly and with no approximation, by an algebraic relation between the two unknowns. It is worth noticing why the combination works: $R_{00}$ and $R_{11}$ differ from one another only in which of $A$ and $B$ sits in each denominator, so weighting by the opposite function makes the three complicated structures identical up to sign. Nothing was clever about the choice. It was forced by the shape of (3.7.10).

3.4 · The first constant, fixed by a physical demand

Equation (3.7.13) gives a constant, and algebra cannot say which one. Name the physical input that does: asymptotic flatness. Far from an isolated body the geometry should approach that of empty flat spacetime, so that a distant observer's clocks and rulers are the ordinary ones of Part II. In the chart (3.7.5) that means

since with $A=B=1$ the line element becomes $c^{2}\dd t^{2}-\dd r^{2}-r^{2}\dd\Omega^{2}$, which is Minkowski spacetime in spherical coordinates: non-constant components and no curvature whatever, which is the lesson Chapter 3.3 §2 delivered early using the polar plane. The product $AB$ therefore tends to $1$, and being constant it equals $1$ everywhere:

One unknown function remains. Note carefully what was spent: not a calculation, but a decision about the physical situation being described. A different boundary condition would describe a different situation, and Problem 1 asks what a different constant would have meant.

3.5 · One integration

Put $B=1/A$ into the third line of (3.7.10). Each of the four terms simplifies, and it is worth doing them one at a time. With $B=1/A$ we have $B'=-A'/A^{2}$ and $B^{2}=1/A^{2}$, so

The two derivative terms are equal, so they add rather than cancel, and

Name the technique. The left-hand side is already a product rule: $rA'+A=(rA)'$. So

This is a first-order linear equation of the kind Chapter 0.8 §2.2 handled with an integrating factor, and here the integrating factor has already done its work: $r$ is exactly the factor that turns $A'+A/r=1/r$ into an exact derivative. One integration, and one constant of integration $C_{1}$:

Notice that (3.7.14) is satisfied automatically for any $C_{1}$: the solution is asymptotically flat whatever the constant is, so asymptotic flatness has nothing left to say and something else must fix $C_{1}$.

3.6 · The second constant, and where the mass comes in

Nothing in (3.7.6) mentions a mass. The equations are the equations of empty space, and they are the same equations whether the object at the centre is the Sun or a grain of dust. The mass can therefore enter in only one place: as the constant of integration, fixed by matching to something known.

The known thing is the Newtonian limit. Chapter 3.1 §6.5 derived, from the gravitational redshift in an accelerating cabin and with no general relativity used at all, that a weak static field has

equation (3.1.39), with the plus sign belonging to this book's signature, and Chapter 3.6 §5 used the same component to fix $\kappa$. Far from a mass $M$ the Newtonian potential is $\Phi=-GM/r$, so

Both sides carry a term falling off as $1/r$ and matching their coefficients fixes the constant. And $M$ is defined by this equation: it is the mass a distant observer would infer from the orbits of test bodies, and nothing in the derivation refers to what is inside. Substituting into (3.7.5) with (3.7.15),

That is the Schwarzschild metric, and it is exact. No expansion was made, no small parameter was introduced, and the only approximation anywhere in the derivation was in the matching of (3.7.21), which was performed at large $r$ where it is valid and which fixed a constant that is then exact everywhere.

3.7 · The Schwarzschild radius, as a length

The metric (3.7.22) contains one length, the value of $r$ at which the coefficient of $c^{2}\dd t^{2}$ vanishes:

Put numbers into it, because as a symbol it is easy to mistake for something enormous. With $c^{2}=8.988\times10^{16}\ \mathrm{m^{2}\,s^{-2}}$ throughout:

The Sun. $GM_{\odot}=1.3271\times10^{20}\ \mathrm{m^{3}\,s^{-2}}$, so

The Earth. Now run the same arithmetic with the Earth's measured value in place of the Sun's, $GM_{\oplus}=3.986\times10^{14}\ \mathrm{m^{3}\,s^{-2}}$, so

Nine millimetres. That is the number worth carrying, because it converts a symbol into a length you could hold: compress the entire Earth inside a sphere the size of a small grape and the coefficient of $c^{2}\dd t^{2}$ at its surface reaches zero. And because $r_{s}$ is linear in $M$, the number scales straight up with the mass. A hole of $10^{9}$ solar masses, the class imaged at the centre of M87, has $r_{s}=2.95\times10^{9}\ \mathrm{km}$, which is $19.7$ astronomical units, comparable to the orbit of Uranus.

The ratio that governs how big the relativistic corrections are is $r_{s}/r$. At the Sun's surface, $r_{s,\odot}/R_{\odot}=2953/(6.957\times10^{8})=4.2\times10^{-6}$: four parts in a million. That number is why Newtonian gravity survived three centuries of astronomy, and §8 shows what it takes to detect the difference anyway.

4 · Birkhoff's theorem, and what it says about a collapsing star

Here is where this section is going. Section 1 assumed staticity and §3 used it in two places: to drop the $x^{0}$ dependence and to drop the cross terms. We now throw the assumption away, keep only spherical symmetry, and redo the calculation. The result is that staticity comes back on its own, because the field equations force it. That is Birkhoff's theorem, and it is derived here rather than quoted.

4.1 · Redo it with time allowed in

Keep spherical symmetry, which by §2.1 still forces every sphere to be round, still forbids angular cross terms, and still permits the areal radius (3.7.4). Drop staticity entirely and allow both remaining functions to depend on time:

Two remarks on what this ansatz does and does not assume, since the whole force of the theorem rests on them. It does not assume the matter is static: the source can be pulsing, collapsing or exploding, provided it stays spherical. It does still exclude a cross term $g_{01}$, and that is not a further assumption about the geometry. With $g_{01}\neq0$ the time part of the line element reads $A(\dd x^{0})^{2}+2g_{01}\dd x^{0}\dd r$. Complete the square and supply an integrating factor, which exists because only one other variable is present. A function $T(x^{0},r)$ then appears for which the whole of it is $\tilde A\,c^{2}\dd T^{2}$. That is a change of chart, precisely the operation Chapter 3.2 §2 defined, and it costs nothing.

Recompute. Every component of (3.7.8) that survives acquires possible time derivatives, three new connection coefficients appear carrying $\partial_{0}A$ and $\partial_{0}B$, and one Ricci component that was identically zero before is no longer so. Grind box C computes it. The result is remarkably clean: writing $\dot B\equiv\partial_{0}B$,

4.2 · The theorem

Setting (3.7.27) to zero, and noting again that $r$ and $B$ are finite and non-zero where we are solving, gives

One component of the vacuum equations has forbidden the radial function to depend on time, with no assumption made. The rest follows in three moves.

Move 1. With $B$ a function of $r$ alone, the combination of §3.3 goes through unchanged. Grind box C confirms that $B R_{00}+A R_{11}$ is still $\partial_{r}(AB)/(rB)$ even when $A$ depends on time. So $\partial_{r}(AB)=0$ and

for some function of time alone. Note what has changed: the "constant" of §3.3 was constant in $r$, and nothing said it could not vary with $x^{0}$.

Move 2. The time dependence is not in the geometry, it is in the clock. From (3.7.29), $A=f(x^{0})/B(r)$, so the time part of the line element is $f(x^{0})(\dd x^{0})^{2}/B(r)$. Define a new time coordinate $T$ by

which is legitimate because $f$ depends on $x^{0}$ alone, so the right-hand side is an exact differential and $T$ is a genuine coordinate. Then $f(x^{0})(\dd x^{0})^{2}=c^{2}\dd T^{2}$ and

which has no time dependence at all and no cross terms: static. The whole of the apparent time dependence was a statement about how a distant observer had chosen to run their clock.

Move 3. Equation (3.7.31) is the ansatz (3.7.5) with $A=1/B$ already imposed, so §§3.5–3.6 apply verbatim: $(rA)'=1$, one integration, and the constant fixed by the Newtonian limit. The solution is (3.7.22) again.

Grind box C — the time-dependent components, and $R_{01}$

The new connection coefficients. With $A$ and $B$ both functions of $(x^{0},r)$, three components join the list (3.7.8) and two acquire new pieces. From (3.7.7), with $\dot{\phantom{x}}\equiv\partial_{0}$:

$$\Gamma^{0}{}_{00}=\frac{\dot A}{2A}, \qquad \Gamma^{0}{}_{11}=\frac{\dot B}{2A}, \qquad \Gamma^{1}{}_{01}=\frac{\dot B}{2B},$$

each obtained the same way as in grind box A, the only difference being which derivative survives in the bracket. For instance $\Gamma^{1}{}_{01}$ takes $\lambda=1$, $g^{11}=-1/B$, and the bracket $\partial_{0}g_{11}+\partial_{1}g_{10}-\partial_{1}g_{01}=-\dot B$, giving $\half(-1/B)(-\dot B)=\dot B/2B$. The nine of (3.7.8) keep their forms with $A$ and $B$ now functions of two variables.

The $01$ component of the Ricci tensor. Put $\mu=0$, $\nu=1$ into (3.7.9):

$$R_{01} \;=\; \partial_{\lambda}\Gamma^{\lambda}{}_{10} \;-\; \partial_{1}\Gamma^{\lambda}{}_{\lambda 0} \;+\; \Gamma^{\lambda}{}_{\lambda\rho}\Gamma^{\rho}{}_{10} \;-\; \Gamma^{\lambda}{}_{1\rho}\Gamma^{\rho}{}_{\lambda 0}.$$

Take them in turn. The only non-zero $\Gamma^{\lambda}{}_{10}$ are $\Gamma^{0}{}_{01}=A'/2A$ and $\Gamma^{1}{}_{01}=\dot B/2B$, so term 1 is $\partial_{0}(A'/2A)+\partial_{1}(\dot B/2B)$. Using the contraction of grind box A, now reading $\Gamma^{\lambda}{}_{\lambda 0}=\dot A/2A+\dot B/2B$, term 2 is $-\partial_{r}\big(\dot A/2A+\dot B/2B\big)$. The two $\dot B/2B$ pieces cancel outright between terms 1 and 2, and the two remaining pieces, $\partial_{0}(A'/2A)$ and $-\partial_{r}(\dot A/2A)$, cancel as well. The quotient rule turns the first into $\partial_{0}\partial_{r}A/2A-A'\dot A/2A^{2}$ and the second into $\partial_{r}\partial_{0}A/2A-\dot A A'/2A^{2}$, and these are the same expression, because mixed partial derivatives commute. Terms 1 and 2 together give zero.

Term 3 is $\Gamma^{\lambda}{}_{\lambda\rho}\Gamma^{\rho}{}_{10}$ summed over $\rho=0,1$:

$$\left(\frac{\dot A}{2A}+\frac{\dot B}{2B}\right)\frac{A'}{2A} \;+\; \left(\frac{A'}{2A}+\frac{B'}{2B}+\frac{2}{r}\right)\frac{\dot B}{2B}.$$

Term 4 is $-\Gamma^{\lambda}{}_{1\rho}\Gamma^{\rho}{}_{\lambda 0}$, whose surviving contributions come from $(\lambda,\rho)=(0,0),(0,1),(1,0),(1,1)$ and the two angular pairs, the latter giving nothing because $\Gamma^{2}{}_{20}=\Gamma^{3}{}_{30}=0$:

$$-\left[\underbrace{\frac{A'}{2A}\frac{\dot A}{2A}}_{(0,0)} + \underbrace{\frac{\dot B}{2A}\frac{A'}{2B}}_{(0,1)} + \underbrace{\frac{\dot B}{2B}\frac{A'}{2A}}_{(1,0)} + \underbrace{\frac{B'}{2B}\frac{\dot B}{2B}}_{(1,1)}\right] ,$$

the two angular pairs contributing nothing because $\Gamma^{2}{}_{20}=\Gamma^{3}{}_{30}=0$. Expand term 3 and set it beside this: the four products match one for one and cancel, and the only survivor is the piece term 3 carries with the $2/r$ in it. Hence

$$R_{01} \;=\; \frac{2}{r}\cdot\frac{\dot B}{2B} \;=\; \frac{\dot B}{rB},$$

which is (3.7.27).

(Confirmed symbolically for general $A(x^{0},r)$, $B(x^{0},r)$: $R_{01}$ came out exactly $\dot B/(rB)$, and with $B$ restricted to depend on $r$ alone the combination $BR_{00}+AR_{11}$ came out exactly $\partial_{r}(AB)/(rB)$, as Move 1 claims.)

4.3 · What the theorem says about a star

The consequences are large and they are not obvious from the algebra.

A pulsating star has a static exterior. Take a spherical star that expands and contracts. A Cepheid variable does exactly this, and so does a star ringing after a disturbance. Outside it, the geometry is (3.7.22) with a fixed $M$, unchanging, for all time. A planet orbiting it feels nothing at all.

There is no spherically symmetric gravitational radiation. A radiating system must have a time-dependent field outside it, and Birkhoff says a spherical one cannot. Electromagnetism forbids the corresponding thing, since a spherically pulsing charge does not radiate. But it forbids it by a weaker route, conservation of charge, which says nothing about the field elsewhere. Here the statement is stronger: the entire exterior geometry is unchanged, because it is the only solution the equations possess. That is why the gravitational waves detected in this century come from pairs of compact objects spiralling about each other, which are as far from spherical as a system can be, rather than from single objects ringing.

The field of a collapsing star does not change as it collapses. This one is worth stating slowly. A spherical star collapses through its own Schwarzschild radius. During the collapse the matter distribution changes violently. Outside the star's surface, the geometry is (3.7.22) with the same $M$ throughout, before, during and after. Nothing propagates outward, nothing announces the collapse, and a distant orbit is undisturbed. Whatever a horizon is, its formation is not an event that anything outside can feel. That makes the question of what it is considerably more interesting, and Chapter 3.8 asks it.

And a debt of Chapter 3.4 is collected. That chapter's §6.3 insisted that Ricci-flat does not mean flat, and named the Weyl part of the curvature as what survives contraction. Here is the example. Equation (3.7.22) has $R_{\mu\nu}=0$ everywhere it applies, since it was constructed to. And it is not flat: it bends light, holds planets in orbit, and §7 shows it does something no flat geometry could. Every bit of that is Weyl curvature. The tidal deviation of Chapter 3.1, which Chapter 3.4 §4 identified with curvature, is entirely carried by the part the Ricci tensor throws away.

5 · Two Killing vectors, two conserved quantities

The solution exists. Now spend it, which means asking how a body moves in it. Here is where this section is going. Free motion in a metric is four coupled second-order equations for $t(\tau),r(\tau),\theta(\tau),\varphi(\tau)$, with position-dependent coefficients, and that is not a problem anybody solves by hand. We reduce it to one first-order equation in one variable, using symmetry three times: once to fix the plane, and twice to integrate an equation once for free. Chapter 3.5 §9 promised exactly this, twice and by name, and this section is where the promise falls due.

5.1 · The promise, and the two charges

Chapter 3.5 §8.1 gave the practical test: every coordinate the metric does not mention supplies a Killing vector. Look at (3.7.22). It does not mention $t$, and it does not mention $\varphi$. So

are Killing vector fields. And Chapter 3.5 §9.1 proved, in four lines, that for any Killing field $\xi$ and any geodesic with tangent $u^{\mu}=\dd x^{\mu}/\dd\tau$, the number $\xi_{\mu}u^{\mu}$ is the same at every point of the geodesic. That is equation (3.5.73). Two Killing fields, two constants.

Compute them. Lowering an index means contracting with the metric, and the metric is diagonal, so $\xi_{\mu}=g_{\mu\nu}\xi^{\nu}$ picks out one component each time.

The time charge. In the coordinates $x^{\mu}=(x^{0},r,\theta,\varphi)$ with $x^{0}=ct$, the field $\partial_{t}$ has components $\xi^{\mu}=(c,0,0,0)$, since $\partial_{t}=c\,\partial_{0}$. Then $\xi_{0}=g_{00}\,c=cA$ and

The angular charge. Now do the same for the rotational Killing field. The field $\partial_{\varphi}$ has components $\xi^{\mu}=(0,0,0,1)$, so lowering the index gives $\xi_{3}=g_{33}=-r^{2}\sin^{2}\theta$, and

The minus sign is a convention chosen so that $L$ is positive for motion in the direction of increasing $\varphi$. The conserved quantity is $\xi_{\mu}u^{\mu}$ either way.

What they are. Chapter 3.5 §9.2 ran Chapter 1.4's Noether theorem on the same system and found that the charge belonging to $\xi$ is $m\,\xi_{\mu}u^{\mu}$, which is equation (3.5.79). So $mE$ is the Noether charge of time-translation invariance and $mL$ is the Noether charge of rotational invariance: $E$ is the energy per unit mass and $L$ the angular momentum per unit mass, and they are the same two quantities Chapter 1.4 §3 obtained for planets long before any of this geometry existed. Their names were not assigned by analogy. They were derived and then recognised.

5.2 · The equatorial plane, and why taking it costs nothing

Both charges still contain $\theta$. Dispose of it, and say why the disposal is legitimate rather than convenient.

The argument is a symmetry argument and it has two steps. First, take any geodesic and any point $p$ on it. The tangent $u$ at $p$ and the radial direction at $p$ span a plane through the centre. By spherical symmetry there is a rotation carrying that plane to the equatorial plane $\theta=\pi/2$, and a rotation is an isometry, so it carries geodesics to geodesics. Therefore every geodesic can be rotated so that at $p$ it lies in the equatorial plane with no motion out of it, $\dd\theta/\dd\tau=0$.

Second, check that it stays there. The $\theta$ component of the geodesic equation, using $\Gamma^{2}{}_{12}=1/r$ and $\Gamma^{2}{}_{33}=-\sin\theta\cos\theta$ from (3.7.8), is

At $\theta=\pi/2$ the last term vanishes because $\cos(\pi/2)=0$, and if $\dd\theta/\dd\tau=0$ there as well then the middle term vanishes too, leaving $\dd^{2}\theta/\dd\tau^{2}=0$. Zero value, zero first derivative and zero second derivative: by the uniqueness theorem for the initial-value problem, Chapter 0.8 §1.1, the solution is $\theta\equiv\pi/2$ for all $\tau$.

This is exactly the statement that a planet's orbit lies in a plane, and the reason is exactly the Newtonian one: Chapter 1.4 §3.3's conservation of the angular momentum vector, whose fixed direction is the normal to the orbital plane. From here on, $\theta=\pi/2$ and (3.7.35) reads $L=r^{2}\dd\varphi/\dd\tau$.

5.3 · The count, before and after

	Unknowns	Equations
Geodesic equation as written	$t,r,\theta,\varphi$	four, coupled, second order
after spherical symmetry (§5.2)	$t,r,\varphi$	three, second order
after the two Killing charges	$r$	one, first order

That is the whole of the reduction, and it was bought entirely with symmetry. Chapter 3.5 §9 said it in advance: "Without this section, Chapter 3.7 would be a wall of algebra with no visible structure. With it, that chapter is mostly Chapter 1.3's phase-plane reasoning applied to one variable." The next section supplies the one remaining equation.

6 · The effective potential

Here is where this section is going. One line of algebra turns the normalisation of the four-velocity into an energy equation for $r(\tau)$. We then read that equation as a particle moving in a landscape, compare the landscape with Newton's term by term, and find that exactly one term is new. Everything in §§7 and 8 is that term's doing.

6.1 · One line of algebra

The four-velocity of a massive body satisfies

which is Chapter 2.5 §1's $u\cdot u=c^{2}$, equation (2.5.9), carried onto a manifold in Chapter 3.3 §8. There it was shown to be constant along any geodesic, so that imposing it once at one point imposes it everywhere. Write it out for (3.7.22) in the equatorial plane, with $f(r)\equiv1-2GM/c^{2}r$ as shorthand:

Now substitute the two charges. From (3.7.34), $u^{0}=c\,\dd t/\dd\tau=E/(cf)$, so the first term is $f\cdot E^{2}/(c^{2}f^{2})=E^{2}/(c^{2}f)$. From (3.7.35), $\dd\varphi/\dd\tau=L/r^{2}$, so the third term is $L^{2}/r^{2}$. Hence

We want the radial derivative on its own, since $r(\tau)$ is the one unknown we are chasing. So multiply through by $f$ and rearrange for the radial term:

That is the whole of orbital dynamics in general relativity. One equation, first order, one unknown function, two constants. Everything from here to the end of the chapter is extracted from it.

6.2 · Reading it as a landscape

Expand the product in (3.7.40), one term at a time. The four products are $c^{2}$, $L^{2}/r^{2}$, $-2GM/r$ and $-2GML^{2}/(c^{2}r^{3})$, so

Divide by two and move everything except the derivative to the right:

Everything depending on position alone has been gathered into the single function $V_{\text{eff}}$. Here is what that function is:

Equation (3.7.42) is the statement that a quantity quadratic in a velocity, plus a function of position alone, is constant. That is the form Chapter 1.3 §4 used to draw phase portraits, and the form Chapter 0.8 §4.3 used for the oscillator. The technique is read off the turning points: motion is possible only where $\mathcal{E}\ge V_{\text{eff}}(r)$, the body turns around where the two are equal, and it orbits in a circle where $V_{\text{eff}}$ is stationary.

Two cautions about what $\mathcal{E}$ is, since it looks like an energy and is not quite one. It is built from $E$, which already includes the rest energy per unit mass $c^{2}$. The combination $(E^{2}-c^{4})/2c^{2}$ removes it, and for a slow body $E\simeq c^{2}+\varepsilon$ gives $\mathcal{E}\simeq\varepsilon$, the Newtonian energy per unit mass. And the "time" in $\dd r/\dd\tau$ is the traveller's proper time, not the distant observer's $t$. Both facts matter and neither affects the shape of $V_{\text{eff}}$.

6.3 · Term by term against Newton

Build the Newtonian comparison from scratch, in two lines, so that nothing is being taken on trust. Newtonian energy per unit mass for a body in a plane is $\varepsilon=\half\big(\dot r^{2}+r^{2}\dot\varphi^{2}\big)-GM/r$, and Chapter 1.4 §3.3's angular momentum per unit mass is $L=r^{2}\dot\varphi$, so $r^{2}\dot\varphi^{2}=L^{2}/r^{2}$. Eliminating $\dot\varphi$,

The second term is the centrifugal barrier, and it is a piece of kinetic energy wearing a potential's clothes. Chapter 1.2's Problem 3 on the bead on a rotating hoop already made that point, and it is the reason no fictitious force has to be invoked to explain the barrier.

Set the two side by side.

Term	Newton	Schwarzschild	What it does
attraction	$-GM/r$	$-GM/r$	identical, and it pulls inward
centrifugal barrier	$+L^{2}/2r^{2}$	$+L^{2}/2r^{2}$	identical, and it keeps a rotating body out
the new term	—	$-GML^{2}/c^{2}r^{3}$	attractive, and it beats the barrier at small $r$

The first two terms are not approximately the same as Newton's. They are the same, with no correction of any order. The entire difference between the two theories, for a massive body in orbit, is the third term.

Three things about it deserve naming before §7 exploits them.

It is attractive. The minus sign means it deepens the well, and it deepens it in the region where the barrier was supposed to be doing its job.

It carries $L^{2}$. A body falling straight in, $L=0$, does not feel it at all. Problem 2 shows that radial infall in this metric obeys the Newtonian equation exactly. The new physics is entirely about bodies with sideways motion.

It falls off as $1/r^{3}$, so it wins at small $r$. Compare it with the barrier: the ratio is

independent of $L$ altogether. So the new term is a fraction $r_{s}/r$ of the barrier: negligible far out, comparable at $r\sim r_{s}$, dominant inside. Take Mercury's perihelion, a distance $a(1-e)=4.60\times10^{10}\ \mathrm{m}$ from the orbital elements §8.5 lists. There $r_{s,\odot}/r=2953/(4.60\times10^{10})=6.4\times10^{-8}$, which is the size of the effect §8 has to dig out of the data.

7 · The innermost stable circular orbit

Here is where this section is going, and it is the chapter's thesis. Circular orbits sit where $V_{\text{eff}}$ is stationary. We solve $\dd V_{\text{eff}}/\dd r=0$ and find a quadratic, which therefore has two roots, one root or none. In Newtonian gravity the same question has exactly one root for every radius. The discriminant tells us when the roots exist, and the answer is that below a certain angular momentum there is no circular orbit at any radius whatever. The two roots merge at $r=6GM/c^{2}$, and that radius has a name and an observational consequence.

7.1 · Where the landscape is level

Differentiate (3.7.43) term by term:

Set it to zero and multiply by $r^{4}$, which is legitimate for $r>0$ and clears every denominator at once:

A quadratic in $r$. Stop and compare with Newton before solving it. Differentiating (3.7.44) gives $GM/r^{2}-L^{2}/r^{3}=0$, which after multiplying by $r^{3}$ is linear:

One root, always, for every $L>0$. Turn it round: name any radius you like and $L=\sqrt{GMr}$ puts a circular orbit there. Newtonian gravity has a circular orbit at every radius, and each is a minimum of $V_{\text{N}}$, hence stable. Equation (3.7.47) is a different kind of object, and the difference is the term carrying $1/c^{2}$.

7.2 · The discriminant, and a floor on the angular momentum

Solve (3.7.47) by the quadratic formula, with $a=GM$, $b=-L^{2}$ and $\;c_{0}=3GML^{2}/c^{2}$:

the second form obtained by taking one factor of $L$ out of the square root, which is legal since $L>0$, and then a factor of $1/c$ out of what remains.

Real roots require the quantity under the square root to be non-negative:

Read that as a physical statement, because it is a startling one.

7.3 · The two roots, and where they merge

When both roots exist they are different orbits with different characters. Factorise (3.7.46) using them: since the numerator of $\dd V_{\text{eff}}/\dd r$ is $GM(r-r_{-})(r-r_{+})$ divided by $r^{4}$, with $r_{-}\lt r_{+}$, the derivative is positive for $r\lt r_{-}$, negative between the roots and positive again for $r>r_{+}$. So

• $r_{-}$ is a maximum of $V_{\text{eff}}$, so it is an unstable circular orbit, destroyed by any disturbance.
• $r_{+}$ is a minimum, so it is a stable circular orbit, which a nudged body oscillates about.

As $L$ decreases toward $L_{\min}$ the square root in (3.7.49) shrinks and the two roots approach each other. At $L=L_{\min}$ it vanishes and they coincide:

This is the innermost stable circular orbit, the ISCO. At exactly $r_{\text{ISCO}}$ both $\dd V_{\text{eff}}/\dd r$ and $\dd^{2}V_{\text{eff}}/\dd r^{2}$ vanish, because the maximum and the minimum have annihilated into an inflection. So the orbit there is marginally stable: neither restored nor expelled at leading order. Inside it there are no stable circular orbits at all, and Problem 3 shows that the unstable branch $r_{-}$ runs from $r_{\text{ISCO}}$ inward and never reaches below $3GM/c^{2}$.

7.4 · The number, and what it is the inner edge of

Put a mass in. Take a black hole of ten solar masses, the class produced by the collapse of a massive star. Its ISCO sits at

with $r_{s}=29.5\ \mathrm{km}$ for comparison. And $L_{\min}=2\sqrt3\,GM/c=3.464\,GM/c$, which for the same hole is $3.464\times(1.3271\times10^{21})/(2.998\times10^{8})=1.53\times10^{13}\ \mathrm{m^{2}\,s^{-1}}$.

The observational hook. Gas falling toward a compact object does not fall straight in. It has angular momentum, so it settles into a disc and spirals slowly inward as friction removes that angular momentum. At each stage it occupies the stable circular orbit appropriate to its current $L$. When it reaches $r_{\text{ISCO}}$ there is nowhere further to go: the next orbit inward does not exist, and the gas leaves the disc and plunges. An accretion disc therefore has an inner edge, at $6GM/c^{2}$, and since the temperature and the emitted spectrum depend on how deep the gas gets, the inner edge is a measurable feature and it is one of the ways the mass of a black hole is estimated. Worked example 2 computes how much energy is released on the way in, and the answer is startling.

angular momentum L  [units of GM/c]

4.100

L = 2√3 exactly (the ISCO)

below it: no orbit anywhere

L = 4.40, two orbits

L = 4.1000 GM/c → discriminant L² − 12 = 4.81000 ≥ 0 : circular orbits exist

unstable circular orbit at r = 3.90900 · stable circular orbit at r = 12.90100 · barrier height V(r₋) − V(r₊) = 0.047647

closed form of §7.2 : r₋ = 3.90900 , r₊ = 12.90100 |numerical − formula| = 5.7e-10 , 1.0e-9

8 · The orbit equation, and perihelion precession

Here is where this section is going. Section 6's equation describes $r$ against proper time, which is not what an astronomer measures. We change variables so that it describes the shape of the orbit, meaning $r$ against $\varphi$. We derive the Newtonian version first, where the answer is a closed ellipse. The relativistic version differs by one term. Linearising about a circular orbit turns that term into a change of frequency rather than of shape, so the ellipse still repeats but slightly slower than once per turn, and the whole figure rotates. Then we compute the rate for Mercury, with every conversion on the page.

8.1 · From motion to shape

Two changes of variable, and both are standard for central-force problems because both remove a nuisance.

Change one: use $\varphi$ as the independent variable instead of $\tau$. The chain rule and (3.7.35) give the conversion

which is available precisely because $L$ is conserved. That is a Killing vector paying for itself a second time.

Change two: use $u\equiv1/r$ instead of $r$. The motivation is that the attraction goes as $1/r^{2}$ and the barrier as $1/r^{2}$, so in terms of $u$ the equation will have polynomial coefficients. Then $r=1/u$, and

the middle step using $\dd r/\dd\varphi=\dd(1/u)/\dd\varphi=-u^{-2}\dd u/\dd\varphi$. The factors of $u^{2}$ cancel exactly, which is why the substitution is worth making.

Put (3.7.54) into (3.7.41) and divide by $L^{2}$:

We want a second-order equation for the shape rather than a first-order one carrying a square, because the second-order kind is the one we can recognise on sight. So differentiate with respect to $\varphi$ and divide through by $2\,\dd u/\dd\varphi$:

The division by $\dd u/\dd\varphi$ is legitimate wherever that derivative is non-zero, which is everywhere except at the turning points and on exactly circular orbits. And a circular orbit, having $u$ constant, satisfies (3.7.56) directly, so nothing has been lost.

8.2 · Newton first, and why an ellipse closes

Delete the term carrying $1/c^{2}$. That is exactly the Newtonian limit, since the deleted term is the only place $c$ appears. Then (3.7.56) becomes

This is the simplest differential equation in physics. It is the harmonic oscillator of Chapter 0.8 §4 with a constant driving term, and its general solution is a particular constant plus the homogeneous solution:

with the two integration constants written as an amplitude $e$ and a choice of where $\varphi=0$ sits, taken at the point of closest approach. That is the polar equation of a conic, an ellipse for $e\lt1$. It is the same expression Chapter 1.4's Worked example 2 obtained by an entirely different route, from the Laplace–Runge–Lenz vector, with $p=L^{2}/mk$ in that chapter's variables.

Why it closes. The right-hand side of (3.7.58) has period exactly $2\pi$ in $\varphi$, because the coefficient of $u$ in (3.7.57) is exactly $1$. So after one full turn the body is at the same radius moving the same way, and the orbit is a closed curve traced over and over. Chapter 1.4 said the same thing in the language of symmetry: the Laplace–Runge–Lenz vector points from the focus to the perihelion, it is conserved for an exact inverse-square law and for nothing else, and its conservation is the statement that the axis of the ellipse never moves. That chapter then wrote, in advance: "Adding any deviation from exact $1/r$ destroys $\vv A$ and the orbit precesses. That is exactly what the $\lambda$ dial did to the closed orbit in §4.4's figure, and exactly what the Sun's other planets, and then general relativity, do to Mercury (Chapter 3.7)." Here is the deviation.

8.3 · The new term, linearised

Restore it. The full equation (3.7.56) is nonlinear, and we are not going to solve it exactly. Instead we do what Chapter 3.6 §7.3 warned would become the normal state of affairs once the field equations turned out to be nonlinear: expand about a simple solution and keep the first correction, naming the small parameter as we go.

Step 1. The circular solution. A constant $u=u_{c}$ has $\dd^{2}u/\dd\varphi^{2}=0$, so (3.7.56) requires

which is §7's circular-orbit condition wearing different symbols. Multiply by $r_{c}^{2}=1/u_{c}^{2}$ and it becomes (3.7.47) exactly.

Step 2. Perturb. Write $u=u_{c}+w$ with $\abs{w}\ll u_{c}$. Substituting into (3.7.56), the square on the right expands as $(u_{c}+w)^{2}=u_{c}^{2}+2u_{c}w+w^{2}$:

The terms with no $w$ in them cancel by (3.7.59). The term in $w^{2}$ is dropped, and this is the only approximation in the section. Be careful about why it is safe, because the obvious reason is the wrong one.

The obvious reason would be that $w/u_{c}$, the fractional variation of $1/r$ around the orbit, is small. For Mercury that quantity is the eccentricity $e=0.206$, which is not small at all. A $10\%$ error in $\Delta\varphi$ would be four arcseconds per century, a hundred times the precision §8.5 claims. So that reason will not do.

What actually makes the neglect harmless is that the dropped term is quadratic. Let's take that slowly, because the chapter's headline number rests on it. Square a sinusoid and you get a constant plus a term at twice the frequency, and neither of those is at the frequency of the oscillation itself. A term that carries nothing at the fundamental frequency cannot shift that frequency at first order in the amplitude: it only displaces the orbit slightly and adds a small wobble at double the rate. The shift in $\omega^{2}$ therefore starts at second order, at $O\big((3GMu_{c}/c^{2})^{2}\big)$, and that is second order in exactly the small parameter §8.4 names.

Now put a number on it. For Mercury that is a relative change in $\Delta\varphi$ of order $10^{-8}$, a millionth of an arcsecond per century. The comparison in §8.5 is against a measurement good to one part in a thousand, so the neglect is legitimate there for any eccentricity a planet actually has. What is left is

Step 3. Read it. Equation (3.7.61) is the harmonic oscillator again, but the coefficient of $w$ is no longer $1$. Its solution $w=w_{0}\cos(\omega\varphi)$ has period $2\pi/\omega$ in $\varphi$ rather than $2\pi$, and $\omega\lt1$ because the correction is subtracted. So the radial distance returns to its minimum later than one full turn.

Finally, replace $u_{c}$ by its leading value. From (3.7.59), $u_{c}=GM/L^{2}$ up to a relative correction of order $3GMu_{c}/c^{2}$, which we have already agreed to neglect at the same order:

8.4 · The advance per orbit

The perihelion recurs when $\omega\varphi$ has advanced by $2\pi$, that is after $\Delta\varphi_{\text{total}}=2\pi/\omega$. Subtract the full turn to get the advance:

using the binomial expansion $(1-x)^{-1/2}\simeq1+x/2$ of Chapter 0.3 §2.4, valid because the small parameter is

which is $1.6\times10^{-7}$ for Mercury. Now convert $L$ into something an astronomer reports.

With $L^{2}=GMp$, equation (3.7.63) becomes

The second relation is pure conic geometry and takes two lines. From (3.7.58), the closest and furthest distances are $r_{\min}=p/(1+e)$ at $\varphi=0$ and $r_{\max}=p/(1-e)$ at $\varphi=\pi$. Their sum is the major axis $2a$, so

8.5 · Mercury, with every conversion shown

Mercury's orbit, as measured: semi-major axis $a=5.7909\times10^{10}\ \mathrm{m}$, eccentricity $e=0.20563$, orbital period $P=87.969$ days. And $GM_{\odot}=1.3271\times10^{20}\ \mathrm{m^{3}\,s^{-2}}$, $c^{2}=8.9876\times10^{16}\ \mathrm{m^{2}\,s^{-2}}$.

Step 1. The semi-latus rectum. $e^{2}=0.042284$, so

Step 2. Radians per orbit. Substitute into (3.7.65). The numerator is $6\pi GM_{\odot}=18.8496\times1.3271\times10^{20}=2.5016\times10^{21}\ \mathrm{m^{3}\,s^{-2}}$. The denominator is $c^{2}p=\big(8.9876\times10^{16}\big)\big(5.5460\times10^{10}\big)=4.9845\times10^{27}\ \mathrm{m^{3}\,s^{-2}}$. So

Half a microradian. Per orbit, this is undetectable. The effect is extracted by waiting.

Step 3. Orbits per century. A Julian century is $100\times365.25=36525$ days, so

Step 4. Multiply, and convert. The advances add, since each orbit turns the ellipse by the same angle:

One radian is $180/\pi$ degrees and each degree is $3600$ arcseconds, so $1\ \mathrm{rad}=(180/\pi)(3600)=2.06265\times10^{5}$ arcseconds. Hence

8.6 · Why Mercury, and not the Earth

Run the same arithmetic for the Earth: $a=1.4960\times10^{11}\ \mathrm{m}$, $e=0.01671$, $P=365.26$ days. Then $p=a(1-e^{2})=1.4956\times10^{11}\ \mathrm{m}$, and

and with $N=36525/365.26=100.0$ orbits per century the total is $1.861\times10^{-5}\ \mathrm{rad}=3.84$ arcseconds per century, smaller than Mercury's by a factor of $11.2$.

The reason is visible in the formula and it explains the whole history of the problem. Two factors work in the same direction:

where Kepler's third law $P\propto a^{3/2}$, from Chapter 1.4 §3, supplied the second step. The inner planet wins twice. Its orbit is smaller, so each turn shifts the ellipse further, and its year is shorter, so it takes more turns per century. Eccentricity helps a third time, weakly, and it helps in another way the formula does not show: a nearly circular orbit has no well-defined perihelion to track. Venus, whose eccentricity is $0.0068$, is a poor target for that reason despite being closer to the Sun than the Earth. Mercury is the innermost planet and much the most eccentric. It is the only place in the Solar System where the effect was ever going to be seen first.

9 · Worked examples

10 · Your turn