Part III · General Relativity — Chapter 3.8

Light, Redshift, and What a Horizon Is

Collect the factor of two Chapter 3.1 confessed. Then take the horizon apart, and find that the only thing broken was the chart.

Tools you'll need  — Chapter 3.7 above all: its §3 solution of $R_{\mu\nu}=0$, its §5 Killing charges $E$ and $L$, and the radial equation of its §6, which is the line this chapter's §1 takes a limit of. Chapter 3.5 §9, where a Killing vector $\Rightarrow$ conserved quantity, used three times below. Chapter 2.4 §§5.1 and 6 for the fact that a full contraction is a scalar, which is the instrument §6 leans on. Chapter 3.4 §5 for the symmetries of the Riemann tensor and §5.3 for locally inertial coordinates, and §4 for geodesic deviation. Chapter 3.3 §8.1 for the geodesic equation and the meaning of an affine parameter, and §2, the flat plane in polar coordinates, which is the whole of §6 in miniature. Chapter 3.2 §2 on charts, atlases and transition maps: changing chart is the operation §7 performs. Chapter 3.1 §6 (the cabin, the redshift, and $g_{00}=1+2\Phi/c^{2}$), §7 (the half-sized deflection and the register of debts) and §5 (how big a freely falling laboratory may be). Chapter 3.6 Problem 3, the spatial part of the weak-field metric, which is the missing half in weak-field clothing. Chapter 2.3 Problem 3, the Rindler horizon, compared in §7.5, and §6.2 on conjugate points. Chapter 2.5 §7 for the Doppler formula. Chapter 1.2 §5.1, the conjugate point, which §3.6 collects. Chapter 0.8 §3.1 for the superposition that makes a linear second-order equation tractable and §6 for particular solutions, both used below, and Chapter 0.2 §5 for the improper integrals of §6.

1 · Null geodesics, and what changes

Let's announce the destination before we start. We take Chapter 3.7's radial equation for a massive particle, replace the timelike normalisation by the null one, and find that the two conserved quantities collapse into a single length. That length is the impact parameter, and everything light does outside a spherical mass is a function of it alone.

1.1 · The normalisation, and the parameter that is no longer available

A massive particle's worldline was parametrised by its proper time. Chapter 3.3 §8.1 showed this is legitimate because $g_{\mu\nu}u^{\mu}u^{\nu}$ is constant along any geodesic, so it may be set to $c^{2}$ once and the parameter is thereby fixed. For light that route is closed. The interval along a light ray is zero by construction, which is what "moves at $c$" became in Part II. So the normalisation we have to work with reads

and there is no proper time to divide by. Between the emission of a flash at one end of the galaxy and its absorption at the other, the flash's own clock records nothing at all.

Something weaker does survive, and Chapter 3.3 §8.1 proved it as well: the geodesic equation holds for an affine parameter, and affine parameters are related to one another by

Look at what that freedom means here. For a massive particle, fixing the constant in $g_{\mu\nu}u^{\mu}u^{\nu}=c^{2}$ was what removed the freedom $a$. Here the constant is zero, and setting zero to zero fixes nothing, so $a$ remains free. That leftover freedom is the whole content of this section.

1.2 · Two charges, one number

Chapter 3.7 §5 applied Chapter 3.5 §9, which says that a Killing vector supplies a quantity conserved along every geodesic. It applied that result to the Schwarzschild metric's two obvious symmetries: the metric components do not depend on $t$, and they do not depend on $\varphi$. The derivation there did not use the timelike normalisation anywhere, so it applies verbatim to a null geodesic, and the two charges are

(The reduction to the equatorial plane $\theta=\pi/2$ is legitimate for the same reason Chapter 3.7 §5 gave. Spherical symmetry lets you rotate any single geodesic into that plane, and a geodesic starting there with $\dd\theta/\dd\lambda=0$ stays there, since $\theta=\pi/2$ makes the only source term in the $\theta$ equation vanish.)

Now apply (3.8.2). Rescaling $\lambda\to a\lambda$ divides both $E$ and $L$ by $a$. Neither is therefore a property of the ray. You can make either one anything you like by relabelling the parameter, and no measurement can distinguish the labellings. Precisely one combination is immune, and we name it

which is unchanged by the rescaling because numerator and denominator scale together. Section 3.2 will show that $b$ is the perpendicular distance from the centre at which the ray would have passed had it gone straight, which is the impact parameter. Note the shape of the argument, because it is one the book has made before: two quantities, individually meaningless because of a normalisation nobody can fix, and one ratio that is physical. That is the same move as the impact parameter in a scattering problem, where the incoming flux and the beam normalisation are conventions and the ratio is what the detector sees.

1.3 · The radial equation, as a limit

Chapter 3.7 §6 obtained, for a massive particle, from the normalisation $g_{\mu\nu}u^{\mu}u^{\nu}=c^{2}$ together with (3.8.3),

We want the same line for light, and the good news is that we do not have to redo the work. Every step of that derivation used the normalisation exactly once, in the single place where the constant $c^{2}$ appears on the right. Swapping (3.8.1) in for (3.8.5)'s normalisation therefore amounts to setting that one $c^{2}$ to zero, and to writing $\lambda$ for $\tau$, which is no longer available:

Now spend the leftover freedom, with the aim of getting $E$ and $L$ to appear only in the combination $b$. Divide (3.8.6) by $L^{2}$ and define the rescaled affine parameter $\sigma\equiv L\lambda$, so that $\dd r/\dd\sigma=L^{-1}\dd r/\dd\lambda$. Using $E^{2}/c^{2}L^{2}=1/b^{2}$ from (3.8.4), that gives

One equation, one parameter $b$, and no trace of $E$ or $L$ separately. The same division turns the angular charge into $\dd\varphi/\dd\sigma=1/r^{2}$, which is the form §3 needs.

Let's look at what that last line is actually saying. Compare (3.8.7) with (3.8.5). The Newtonian-looking $-GM/r$ that sat inside $A(r)\cdot c^{2}$ has gone, taking with it every trace of the inverse-square law. Whatever bends light here, it is not the term that makes planets orbit.

2 · The photon sphere

Again, the destination first. Equation (3.8.7) has the form (kinetic term) $=$ (constant) $-$ (potential), so it can be read the way Chapter 3.7 §6 read its timelike counterpart. We locate the maximum of that potential, find a circular orbit for light at $3GM/c^{2}$, show it is unstable, and extract the aim that separates capture from escape.

2.1 · The null effective potential

We want (3.8.7) in the shape of an energy balance, so move the $r$-dependent piece to the left:

The left-hand term is a square and so is never negative. A ray with impact parameter $b$ can therefore only reach radii where $W(r)\le 1/b^{2}$. Think of the graph of $W$ as a barrier, with $1/b^{2}$ the height at which the ray is launched. Everything in this section is read off the shape of one function.

The barrier's top is what decides who gets through, so differentiate and look for it:

The bracket vanishes at exactly one radius, and $W$ rises for smaller $r$ and falls for larger, so that radius is a maximum:

Now let's check that this really is a circular orbit. A circular orbit needs $\dd r/\dd\sigma=0$ and $\dd^{2}r/\dd\sigma^{2}=0$. Differentiating (3.8.8) once with respect to $\sigma$ gives $2(\dd r/\dd\sigma)(\dd^{2}r/\dd\sigma^{2}+\half W'(r))=0$, so the second condition is $W'(r)=0$. There is a circular orbit for light, and it is at $r=3GM/c^{2}$. Because $W$ has a maximum there, with $W''(r_{\text{ph}})=-2c^{8}/81G^{4}M^{4}$ negative, the orbit is unstable. Nudge a ray inward and it falls in. Nudge it outward and it leaves. Nothing sits on the photon sphere for long, which is why it behaves as a boundary rather than as a place.

2.2 · The critical impact parameter

A ray coming in from far away has $\dd r/\dd\sigma\to -1/b$ at large $r$, since $W\to0$. Whether it turns around or is captured depends on whether $1/b^{2}$ clears the top of the barrier. So we need the height of the peak. Evaluate $W$ at $r_{\text{ph}}$:

The critical aim is the one whose launch height sits exactly level with that peak. So set $1/b^{2}=W(r_{\text{ph}})$, take the square root, and read off the impact parameter:

Three cases follow from that one number.

• A ray aimed with $b\gt b_{\text{crit}}$ is turned back.
• A ray with $b\lt b_{\text{crit}}$ goes over the barrier and never returns.
• A ray aimed at exactly $b_{\text{crit}}$ spirals in and asymptotes onto the photon sphere, taking infinitely many turns to arrive. Problem 1 derives the logarithm that says so.

So the object presents a capture disc of radius $b_{\text{crit}}$. To a distant observer the region from which no light returns is therefore an apparent disc of that radius, larger than $r_{s}$ by a factor of $2.598$, and larger than the photon sphere too. The reason is that a ray arriving from far away is bent inward on the way in, and so gets closer than its aim suggests.

Numbers, for two objects whose masses are known from the orbits of things around them.

Object	$M$	$r_{s}$	$b_{\text{crit}}$	distance	$2b_{\text{crit}}/D$
Galactic centre	$4.3\times10^{6}M_{\odot}$	$1.27\times10^{10}\ \mathrm{m}$	$3.30\times10^{10}\ \mathrm{m}$	$8.2\ \mathrm{kpc}$	$54\ \mu\mathrm{as}$
M87 centre	$6.5\times10^{9}M_{\odot}$	$1.92\times10^{13}\ \mathrm{m}$	$4.99\times10^{13}\ \mathrm{m}$	$16.8\ \mathrm{Mpc}$	$40\ \mu\mathrm{as}$

Forty microarcseconds is $1.9\times10^{-10}$ radians. ⚑ An aperture resolves an angle $\theta\approx\lambda/D_{\text{ap}}$, a result quoted here rather than derived, since this book has not built diffraction anywhere. Resolving that angle at a wavelength of $1.3\ \mathrm{mm}$ needs $D_{\text{ap}}\approx1.3\times10^{-3}/1.9\times10^{-10}=6.8\times10^{6}\ \mathrm{m}$, comparable to the radius of the Earth. No single instrument will do, and the aperture has to be synthesised from stations spread across the planet. So (3.8.12) is not a decorative result. It is the number that says what instrument is required, and it came from maximising one function of one variable.

3 · Deflection: the integral, done

Here is the destination, and the route with it, because there are four steps and each is small. We convert (3.8.7) into an equation for the shape of the orbit. We notice that its unperturbed solution is a straight line. We solve it to first order in the mass, read the deflection off the asymptotes, and put in numbers for the Sun.

3.1 · The orbit equation

We do not want the radius as a function of the parameter. We want the shape of the path, so we should trade the parameter for the angle. That is Chapter 3.7 §8's trick, used there for planets, and it works best in the variable $u\equiv1/r$ rather than $r$. From §1.3, $\dd\varphi/\dd\sigma=1/r^{2}=u^{2}$, so the chain rule gives

the second from the chain rule applied to $1/u$, since $\dd(1/u)/\dd u=-1/u^{2}$. Multiplying these two gives $\dd r/\dd\sigma=-\dd u/\dd\varphi$, so squaring kills the sign and (3.8.7) becomes

That is a first-order equation with a square in it, and squares are awkward to solve. We would rather have a linear second-order equation, so differentiate with respect to $\varphi$. Every term on the right carries $\dd u/\dd\varphi$ out by the chain rule, and the left gives $2(\dd u/\dd\varphi)(\dd^{2}u/\dd\varphi^{2})$:

The common factor is now begging to be cancelled. Divide by $2\,\dd u/\dd\varphi$, which is legitimate wherever the ray is not at a turning point, and the resulting equation then holds everywhere by continuity:

Compare this with the planetary case. Chapter 3.7 §8 obtained $\dd^{2}u/\dd\varphi^{2}+u=GM/L^{2}+3GMu^{2}/c^{2}$. There the constant source term is Newton's inverse-square law in disguise and gives closed ellipses, and the small $u^{2}$ correction is what precesses them. For light the constant term is absent altogether. The entire deflection of starlight is the term that for a planet is a correction of one part in $10^{7}$.

3.2 · The unperturbed solution is a straight line

Before solving the full equation, let's see what it gives when the mass is switched off. Set $M=0$ in (3.8.16). What is left is $u''+u=0$, the equation Chapter 0.8 §4 solved in its sleep, with general solution $u=\alpha\sin\varphi+\beta\cos\varphi$. Choose the origin of $\varphi$ so that $\beta=0$ and the amplitude so that the closest approach is at $r=b$:

Now let's check what that is. In the plane, $x=r\cos\varphi$ and $y=r\sin\varphi$, so (3.8.17) says $y=b$. That is a straight line at perpendicular distance $b$ from the origin, traversed from $\varphi=0$ (far away on one side) through $\varphi=\pi/2$ (closest approach) to $\varphi=\pi$ (far away on the other). It confirms the promise of §1.2, namely that $b$ is the impact parameter. It confirms something else worth naming too. With no mass present, light goes straight. The metric is then flat, and Chapter 3.3 §2 warned that a flat geometry in curvilinear coordinates has non-constant metric components and non-zero connection coefficients anyway. Here that warning is cashed as a formula that looks curved and is not.

3.3 · First order in the mass

Name the approximation before making it. For a ray grazing the Sun, $3GM_{\odot}/c^{2}b\approx6\times10^{-6}$, so the term on the right of (3.8.16) is a millionth of the terms on the left. Write $u=u_{0}+u_{1}$ with $u_{1}$ of first order in $GM/c^{2}$, substitute, and keep only first-order terms. Since $u_{0}$ satisfies the equation exactly with the right-hand side dropped, what is left is

and the discarded pieces are $3GM(2u_{0}u_{1}+u_{1}^{2})/c^{2}$, second order in the mass. This is first-order perturbation theory in exactly the sense Chapter 3.1 §7.2 used it. You evaluate the small term along the uncorrected path, so that the neglected piece is the correction to the correction. Grind box A solves (3.8.18) in four lines, and the answer is

and substituting it back into (3.8.18) gives residual exactly zero, not approximately zero. The solution is exact for the equation it solves.

Grind box A — solving the first-order equation, and the one trigonometric identity used

Line 1. Get rid of the square. The source in (3.8.18) is $\sin^{2}\varphi$, and a squared sine is not a natural right-hand side for a linear equation with resonant frequency $1$. Use the double-angle identity of Chapter 0.3, $\cos2\varphi=1-2\sin^{2}\varphi$, rearranged:

$$\sin^{2}\varphi \;=\; \tfrac12\big(1-\cos2\varphi\big), \qquad\text{so}\qquad \frac{\dd^{2}u_{1}}{\dd\varphi^{2}}+u_{1} \;=\; \frac{3GM}{2c^{2}b^{2}}\Big(1-\cos2\varphi\Big).$$

Line 2. The source has two pieces and the equation is linear, so solve for each and add (Chapter 0.8 §3.1). Write $k\equiv 3GM/2c^{2}b^{2}$ for brevity.

Line 3. The constant piece. For $u''+u=k$, try $u=$ constant $=P$. Then $u''=0$ and the equation reads $P=k$. So $P=k$.

Line 4. The oscillating piece. For $u''+u=-k\cos2\varphi$, try $u=Q\cos2\varphi$. Then $u''=-4Q\cos2\varphi$, so the left-hand side is $(-4Q+Q)\cos2\varphi=-3Q\cos2\varphi$. Matching, $-3Q=-k$, so $Q=k/3$. Note why this works and would not have worked for a $\cos\varphi$ source: the driving frequency is $2$ and the natural frequency is $1$, so there is no resonance and no secular term. Chapter 0.8 §6 is the general statement.

Adding. $u_{1}=k+\tfrac{k}{3}\cos2\varphi=k\big(1+\tfrac13\cos2\varphi\big)$, which is (3.8.19).

The homogeneous solution has been set to zero, and that is a choice of boundary condition rather than an omission. Adding $\alpha\sin\varphi+\beta\cos\varphi$ to $u_{1}$ amounts to shifting the impact parameter and rotating the axis, which is a redefinition of $b$ and of where $\varphi=0$ sits, not a different ray.

(Checked symbolically: substituting (3.8.19) into (3.8.18) returns exactly $0$, and substituting $u_{0}+u_{1}$ into the full equation (3.8.16) leaves a residual whose coefficient of $M^{1}$ vanishes identically, so the error really is second order.)

3.4 · Reading off the deflection

We have the shape of the path. The deflection is the difference between where the ray comes in and where it goes out, so we need the two angles at which the ray is infinitely far away. The ray is at infinity when $r\to\infty$, that is when $u\to0$. Without the mass that happened at $\varphi=0$ and $\varphi=\pi$, and the total swing was exactly $\pi$, which is a straight line. With the mass, solve $u(\varphi)=0$ near each end. Put $\varphi=-\varepsilon$ with $\varepsilon$ small, expand $\sin(-\varepsilon)=-\varepsilon+O(\varepsilon^{3})$ and $\cos(-2\varepsilon)=1+O(\varepsilon^{2})$, and keep first order in both small quantities:

That is one end. By the symmetry of (3.8.19) under $\varphi\to\pi-\varphi$ the other end is displaced by the same $\varepsilon$, so the total swing is $\pi+2\varepsilon$ instead of $\pi$ and the ray has been turned through

3.5 · The number

For a ray grazing the Sun, $b=R_{\odot}=6.957\times10^{8}\ \mathrm{m}$ and $GM_{\odot}=1.32712\times10^{20}\ \mathrm{m^{3}\,s^{-2}}$, with $c^{2}=8.98755\times10^{16}\ \mathrm{m^{2}\,s^{-2}}$. Every conversion on the page:

Radians are not the unit astronomers quote, so convert to arcseconds. One radian is $180/\pi$ degrees and one degree is $3600$ arcseconds, so one radian is $(180/\pi)\times3600=2.062648\times10^{5}$ arcseconds:

3.6 · Lensing, and two chapters' promise collected

Equation (3.8.21) has a consequence that Chapters 1.2 and 2.3 both pointed at and neither could develop. Put a source far behind a mass and consider all the null geodesics running from source to observer. If the source is exactly behind, every ray leaving it in a cone of the right opening angle is bent by the same amount and arrives, so the image is a ring. If it is slightly off-axis the ring breaks into two or more separate images. There is more than one geodesic between the same two events.

That is precisely the phenomenon Chapter 1.2 §5.1 named a conjugate point and illustrated with the two arcs of a great circle joining two cities, both of them stationary points of arc length and only one of them the shortest. Chapter 2.3 §§6.1 and 6.4 met it again from the other side. In flat spacetime the straight worldline is not merely locally but globally the longest, and that chapter noted that in a curved spacetime the global statement can fail. Here is the failure, and it is observed. The several geodesics are the several images. The surface on which their number changes is the caustic, and the caustic is where the second variation of Chapter 1.2 stops being definite.

4 · Where the missing half was

This is the section Chapter 3.1 §7.3 named in writing. Entry (a) of its register of debts reads: "The factor of two in $\alpha_{\text{EP}}=2GM/bc^{2}$ — Chapter 3.8 §4, which derives $\alpha=4GM/bc^{2}$ from the Schwarzschild solution and names the missing half as spatial curvature." Half of that is now done, because §3 derived the $4GM/c^{2}b$. What remains is to identify the missing half rather than assert it, and then to explain why light collects both halves and a planet does not.

Here is the destination. We mutilate the metric in two different ways. The first keeps only the distortion of time, the second only the distortion of space. We run §3's calculation on each, and get exactly $2GM/c^{2}b$ from each. Then we run the same pair of calculations for a particle moving at arbitrary speed $v$ and find that the space half is independent of $v$ while the time half grows as $1/v^{2}$. Their ratio is $v^{2}/c^{2}$, which is the sentence Chapter 3.1 promised.

4.1 · Why the two halves may be computed separately

Before splitting anything, we should check that splitting is allowed. The deflection was obtained to first order in $GM/c^{2}b$. At that order the first-order equation (3.8.18) is linear in the perturbation. Splitting the metric perturbation into two pieces gives a source that is the sum of two sources, and a linear equation with a summed source has the sum of the two solutions. So the contributions add, and computing them separately is not a heuristic. It is a consequence of linearity, and §4.3 checks it explicitly by showing that the two first-order solutions sum to (3.8.19) term by term.

Name the two mutilations precisely. Write the Schwarzschild line element as

in the equatorial plane, with $A(r)=1-2GM/c^{2}r$ as before. The angular term $r^{2}\dd\varphi^{2}$ carries no mass and belongs to neither. Mutilation one keeps $A$ in the time slot and replaces $1/A$ by $1$. That is Newtonian clocks-only gravity, which is precisely what Chapter 3.1's cabin argument had access to. Mutilation two does the reverse, with $1$ in the time slot and $1/A$ in the radial slot.

4.2 · Mutilation one: keep only the distortion of time

Repeat §1.3 with $g_{rr}=-1$. The Killing charges are unchanged in form, $E=c^{2}A\,\dd t/\dd\lambda$ and $L=r^{2}\dd\varphi/\dd\lambda$, since neither used $g_{rr}$. The null condition $A c^{2}(\dd t/\dd\lambda)^{2}-(\dd r/\dd\lambda)^{2}-r^{2}(\dd\varphi/\dd\lambda)^{2}=0$ now gives, after substituting $\dd t/\dd\lambda = E/c^{2}A$ and rescaling to $\sigma=L\lambda$ as before,

The factor $A$ has moved. It now sits under the $1/b^{2}$ instead of multiplying $1/r^{2}$. We want the orbit equation again, so convert to $u=1/r$ as in §3.1 and expand $1/A=1/(1-r_{s}u)=1+r_{s}u+O(u^{2})$, keeping first order in the mass:

the second expression obtained by differentiating and dividing by $2\,\dd u/\dd\varphi$ exactly as in (3.8.15). Now look at what the right-hand side has become. The source is a constant, not a $u^{2}$. Its particular solution is that same constant, $u_{1}=GM/c^{2}b^{2}$, so $u=\sin\varphi/b+GM/c^{2}b^{2}$. Putting $\varphi=-\varepsilon$ and $u=0$ as in (3.8.20) gives the deflection:

That is Chapter 3.1 §7.2's $\alpha_{\text{EP}}=2GM/bc^{2}$ exactly, obtained here from a metric rather than from a cabin, and it is $0.8756''$ at the solar limb. It is also the prediction Einstein published in 1911, with the solar constants of his day, and superseded in 1915. The equivalence principle sees the time part of the geometry and nothing else, and this calculation is what that sentence means.

4.3 · Mutilation two: keep only the distortion of space

Now $g_{00}=1$ and $g_{rr}=-1/A$. The time charge becomes $E=c^{2}\dd t/\dd\lambda$ with no $A$ in it. The null condition $c^{2}(\dd t/\dd\lambda)^{2}-A^{-1}(\dd r/\dd\lambda)^{2}-r^{2}(\dd\varphi/\dd\lambda)^{2}=0$ rearranges to $(\dd r/\dd\lambda)^{2}=A\big(E^{2}/c^{2}-L^{2}/r^{2}\big)$, and rescaling as before,

We want this in the same second-order shape as §3.1's orbit equation, so differentiate with respect to $\varphi$ and divide by $2\,\dd u/\dd\varphi$ once more:

Two source terms this time. Substituting $u_{0}=\sin\varphi/b$ on the right and using $\sin^{2}\varphi=\half(1-\cos2\varphi)$ as in grind box A, the source becomes $\dfrac{GM}{2c^{2}b^{2}}\big(1-3\cos2\varphi\big)$, whose particular solution by the same two-line method is

At $\varphi=-\varepsilon$ this equals $GM/c^{2}b^{2}$, the same value as the time part's constant, so the asymptote calculation of (3.8.20) runs identically and delivers

Now the check §4.1 promised. If linearity really lets us compute the halves separately, adding the two first-order solutions must reproduce §3's answer. Add (3.8.26)'s constant to (3.8.30):

which is (3.8.19) exactly, not merely at the asymptotes but at every angle. The debt is now identified rather than asserted: the missing half of Chapter 3.1's deflection is the $\dd r^{2}/A$ in (3.8.24), the curvature of space itself. Chapter 3.6 Problem 3 met the same object in weak-field clothing, deriving $\dd s^{2}=(1+2\Phi/c^{2})c^{2}\dd t^{2}-(1-2\Phi/c^{2})(\dd x^{2}+\dd y^{2}+\dd z^{2})$ and predicting that its spatial factor would be the missing half. It is.

4.4 · Why light collects both halves and a planet collects one

The two halves are equal for light. They are not equal for anything else, and computing the inequality is what turns the slogan into an explanation. So run §§4.2 and 4.3 again for a massive particle arriving from far away with speed $v$ and impact parameter $b$. At infinity the metric is flat, so Chapter 2.5's kinematics applies there and the two charges are

the first because $A\to1$ and $\dd t/\dd\tau\to\gamma$ far away. The second holds because angular momentum per unit mass is $\gamma\,\abs{\vv r\times\vv v}$, evaluated on the incoming asymptote, where the speed is $v$ and the perpendicular distance from the centre to that asymptote is what $b$ means. (Not at closest approach: there the speed has been increased by the fall inward, and the distance is not $b$.) These give $(E^{2}/c^{2}-c^{2})/L^{2}=c^{2}(\gamma^{2}-1)/\gamma^{2}v^{2}b^{2}=1/b^{2}$ using $\gamma^{2}-1=\gamma^{2}v^{2}/c^{2}$, so the unperturbed orbit is again $u_{0}=\sin\varphi/b$ and $b$ is again the impact parameter. Everything is set up as before.

Now run the two mutilations through the same three steps, which were normalisation, change of variable and differentiation. They give

and the second of these is identical to (3.8.29). The space half does not know how fast the particle is going. Reading off both deflections as before,

Three checks on that. Setting $v=c$ gives $\alpha=4GM/c^{2}b$, recovering (3.8.21). Letting $c\to\infty$ gives $\alpha=2GM/v^{2}b$, which is the Newtonian deflection of a fast projectile by an inverse-square force and can be checked against Chapter 1.4's Kepler machinery. And the ratio of the two contributions is

That is the sentence Chapter 3.1 promised, and it is now a formula. Let's say it in words. Both parts of the geometry are distorted by the same fractional amount, of order $GM/c^{2}r$. What differs is how much of each a moving body samples. A body's motion through spacetime is divided between the time direction and the space directions, and for a slow body the division is grotesquely lopsided: in one second of its own time it covers $3\times10^{8}$ metres of the time direction and $v$ metres of space. The spatial distortion therefore enters its trajectory suppressed by $v^{2}/c^{2}$.

Light divides its motion equally. It covers exactly $c\,\dd t$ of space in every $\dd t$ of time, which is what moving at $c$ means, and so it samples both distortions with the same weight. Two equal contributions, and hence a factor of two.

Body	$v$	$v^{2}/c^{2}$	space share of the deflection
Mercury at perihelion	$5.90\times10^{4}\ \mathrm{m\,s^{-1}}$	$3.9\times10^{-8}$	$3.9\times10^{-8}$
Earth in orbit	$2.98\times10^{4}\ \mathrm{m\,s^{-1}}$	$9.9\times10^{-9}$	$9.9\times10^{-9}$
A $10\ \mathrm{MeV}$ electron	$0.9988\,c$	$0.9976$	$0.4994$
Light	$c$	$1$	$0.5$

The last column is $\alpha_{\text{space}}/\alpha$, that is $(v^{2}/c^{2})/(1+v^{2}/c^{2})$. Read the table downward. The entire content of the factor of two is that this column runs from zero to a half. Newtonian gravity is the top row, and it is right there because that is where the planets are.

Four rows is what a page holds. The figure below runs the two mutilated calculations at every speed between them, by integrating the orbit equation rather than evaluating (3.8.35). So the boxed formula is something the picture checks rather than something it draws.

speed v/c

0.501187 c

impact parameter b

1.00 R⊙

Mercury

a 10 MeV electron

light

reset

α_time = 3.4858071″ against 2GM/v²b = 3.4858071″ α_space = 0.87559736″ against 2GM/c²b = 0.87559516″

α_space/α_time = 2.511892729e-1 v²/c² = 2.511886432e-1 they differ by 2.51e-6

α with both parts kept = 4.36143935″ α_time + α_space = 4.36140450″ the halves add to the whole to within 7.99e-6, which is §4.1's linearity claim, measured

b = 1.000 R⊙ = 2.356e+5 r_s GM/c²b = 2.1225e-6 lower panel: transverse scale magnified ×2.28e+4

first order in GM/c²b is good here: the integrated α exceeds the closed-form 2GM/v²b + 2GM/c²b = 4.36140229″ by 8.50e-6, a residual proportional to GM/c²b = 2.123e-6 — drag the second slider and watch the two scale together.

5 · Gravitational redshift, three ways

The destination first. A signal climbing out of a gravitational well arrives with its frequency reduced by the factor $\sqrt{A(r)}$. We derive this three times: from the conserved Killing charge of §1.2, from the relation between a static clock and coordinate time, and from Chapter 3.1's cabin, recovered as the weak-field limit. Then we do the satellite-navigation arithmetic on one metric instead of two effects added by hand.

5.1 · Route (a): the conserved charge, against a local clock

This is where §1.2's warning is cashed. A photon has a four-momentum $p^{\mu}$ proportional to its tangent $\dd x^{\mu}/\dd\lambda$. Chapter 3.5 §9 says $\xi_{\mu}p^{\mu}$ is constant along the ray for every Killing vector $\xi$, and for the time-translation Killing vector $\xi^{\mu}=(1,0,0,0)$ in $x^{0}=ct$ coordinates that constant is

That is the bookkeeping quantity. What we want is what somebody actually measures, so put an observer in. An observer sitting still at radius $r$ has four-velocity with only a time component, and normalisation $g_{\mu\nu}u^{\mu}u^{\nu}=c^{2}$ gives $A(u^{0})^{2}=c^{2}$, so $u^{0}=c/\sqrt{A}$. The energy they measure is the projection of the photon's momentum on their own four-velocity, which is Chapter 2.5 §4's rule, unchanged:

The bracket is the constant of (3.8.37). So the locally measured frequency is a fixed number divided by $\sqrt{A(r)}$, and comparing an emitter at $r$ with a receiver at infinity where $A=1$,

which is less than one, so the signal is reddened. Notice what did the work here. The conserved quantity did not change, and the local clock did. That is why §1.2 insisted that $E$ is bookkeeping and the observer's measurement is the physics.

5.2 · Route (b): counting crests, with the metric only

Forget photons. Chapter 3.3 §3 wrote proper time along a worldline at rest as $\dd\tau=\sqrt{g_{00}}\,\dd t$, which here is

Now repeat Chapter 3.1 §6.3's crest-counting argument verbatim, because every word of it still applies. The geometry is static, meaning the metric components do not depend on $t$. A crest leaving the emitter at coordinate time $t$ arrives at the receiver at coordinate time $t+\Delta t$, and because nothing in the geometry depends on $t$, the same $\Delta t$ applies to the next crest. So the coordinate-time interval between crests is the same at both ends, and no crests are created or destroyed in between. Converting each end to its own proper time with (3.8.40),

which reduces to (3.8.39) when the receiver is at infinity. The two routes share not one step. Route (a) used a Killing vector and a four-momentum. Route (b) used only the $00$ component of the metric and the fact that nothing depends on $t$.

5.3 · Route (c): the cabin, recovered

The third route is to check that the exact answer reduces to the one Chapter 3.1 already had. So expand (3.8.39) for $2GM/c^{2}r\ll1$, using $\sqrt{1-x}=1-x/2+O(x^{2})$:

with $\Phi(r)=-GM/r$ the Newtonian potential. Chapter 3.1 §6.4 obtained $\Delta\nu/\nu=-\Delta\Phi/c^{2}$ from an accelerating cabin, the Doppler formula and the equivalence principle, with no general relativity in it whatever. With the receiver at infinity, $\Delta\Phi=0-\Phi(r)=+GM/r$ and that formula gives $\Delta\nu/\nu=-GM/c^{2}r$, which is (3.8.42). Three derivations, one answer. The third is the one that matters historically, because it shows that the whole effect was available in 1907 from special relativity plus one principle, and that the exact solution merely resums it.

Numbers, from (3.8.39) directly. The fourth column converts the fractional shift into the speed a source would need in order to produce the same shift by ordinary Doppler, which is how spectroscopists usually report it.

Surface	$r_{s}$	$r_{s}/R$	$\sqrt{1-r_{s}/R}$	equivalent speed
Earth	$8.87\ \mathrm{mm}$	$1.39\times10^{-9}$	$1-6.96\times10^{-10}$	$0.21\ \mathrm{m\,s^{-1}}$
Sun	$2953\ \mathrm{m}$	$4.25\times10^{-6}$	$1-2.12\times10^{-6}$	$636\ \mathrm{m\,s^{-1}}$
White dwarf, $1M_{\odot}$, $7000\ \mathrm{km}$	$2953\ \mathrm{m}$	$4.22\times10^{-4}$	$1-2.11\times10^{-4}$	$63.3\ \mathrm{km\,s^{-1}}$
Neutron star, $1.4M_{\odot}$, $12\ \mathrm{km}$	$4135\ \mathrm{m}$	$0.345$	$0.810$	—

The last row has no equivalent-speed entry because the shift is $19\%$ and the expansion that produced (3.8.42) has stopped being useful. That is the row where the cabin argument dies and (3.8.39) keeps working.

5.4 · Satellite navigation, on one metric

Chapter 3.1's Worked example 2 computed the clock-rate difference between a navigation satellite and the ground, and was explicit that it was gluing together two effects from two different chapters: "adding is legitimate here only because both effects are of order $10^{-10}$ and their product is of order $10^{-20}$ … the place where Chapter 3.8 §5 does it properly with a single metric rather than two separate small corrections." Here is the single metric.

A satellite in a circular equatorial orbit of radius $r$ has $\dd r=0$ and $\dd\varphi/\dd t=\Omega$. We want its proper time against coordinate time, so divide the line element by $c^{2}$:

with $v\equiv r\Omega$. One square root, containing both terms. Nothing has been added by hand. The height dependence and the speed dependence are two entries of the same metric evaluated on the same worldline. The ground station, at rest at $R_{\oplus}$, has $\dd\tau/\dd t=\sqrt{1-2GM/c^{2}R_{\oplus}}$ with the rotation of the Earth neglected. What we want is the ratio of the two rates, so expand both square roots to first order and divide:

The two small terms are the two lines of Chapter 3.1's calculation, and now they are visibly two terms of one expansion rather than two competing physical effects. With $GM_{\oplus}=3.9860\times10^{14}\ \mathrm{m^{3}s^{-2}}$, $R_{\oplus}=6.371\times10^{6}\ \mathrm{m}$, $r=2.656\times10^{7}\ \mathrm{m}$ and $v=\sqrt{GM_{\oplus}/r}=3874\ \mathrm{m\,s^{-1}}$:

Both of those are fractional rates, so to get an accumulated time we multiply each by the $86400\ \mathrm{s}$ in a day. That gives $+45.7\ \mu\mathrm{s}$ and $-7.2\ \mu\mathrm{s}$, netting

And now the justification for dropping the cross terms is a calculation rather than a promise. The product of the two small quantities is $(5.29\times10^{-10})(8.35\times10^{-11})=4.4\times10^{-20}$, which over a day amounts to $3.8\times10^{-15}$ seconds. That is four femtoseconds, about $10^{-10}$ of the effect being kept, and far below the stability of any clock that has been flown. Note also that the gravitational term is six times the kinematic one, so a working navigation system is predominantly a test of this chapter and only secondarily of Part II.

6 · The horizon is not where the metric blows up

Look at the Schwarzschild metric again and read off what happens at $r=r_{s}=2GM/c^{2}$:

One component vanishes and another diverges. Stop here and decide what you think that means, before reading on. You have the tools to settle it and the question is worth answering for yourself first: is (3.8.47) a statement about spacetime, or a statement about the labels somebody chose for it?

Chapter 3.3 §2 is the reason to hesitate. There, the flat plane in polar coordinates was worked early and deliberately. Its metric is $\dd s^{2}=\dd\rho^{2}+\rho^{2}\dd\phi^{2}$, the components are not constant, one of them vanishes at $\rho=0$, the inverse metric component $g^{\phi\phi}=1/\rho^{2}$ diverges there, and the plane is the plane. Nothing is wrong with the paper. The moral was stated in that chapter and has not been used since: badly behaved components are not evidence of badly behaved geometry.

Here is a first hint that the same thing is happening. Compute the determinant of the Schwarzschild metric, in the very coordinates where (3.8.47) misbehaves:

Look at what happened to the two offending factors. They cancel exactly. The determinant is the same function of $r$ and $\theta$ as it is in flat spherical coordinates, it is finite and non-zero at $r=r_{s}$, and its only zeros are at $r=0$ and on the polar axis. That last one is the familiar artefact of spherical coordinates that Chapter 3.2 §2 discussed. This is suggestive, not conclusive, because a determinant is not an invariant. It changes under a change of chart by the square of a Jacobian. Three real calculations follow, and the third is decisive.

6.1 · Calculation one: the distance down to the surface is finite

"$g_{rr}\to-\infty$" is often read as "the horizon is infinitely far away". Let's test it. The proper radial distance between two radii at one instant of $t$ is obtained by setting $\dd t=\dd\theta=\dd\varphi=0$ in the line element and integrating $\sqrt{-\dd s^{2}}$:

Grind box — the proper radial distance, done

Substitute $r=r_{s}\cosh^{2}u$, so that $\dd r=2r_{s}\cosh u\sinh u\,\dd u$ and $\sqrt{r/(r-r_{s})}=\cosh u/\sinh u$. The two hyperbolic sines cancel and what is left is elementary:

$$\int 2r_{s}\cosh^{2}u\,\dd u \;=\; r_{s}\int\big(1+\cosh 2u\big)\,\dd u \;=\; r_{s}\big(u+\sinh u\cosh u\big),$$

using $\cosh^{2}u=\tfrac12(1+\cosh2u)$ from Chapter 0.3 and $\int\cosh 2u\,\dd u=\tfrac12\sinh2u=\sinh u\cosh u$. Undo the substitution with $\cosh u=\sqrt{r/r_{s}}$ and $\sinh u=\sqrt{r/r_{s}-1}$, so that $u=\mathrm{arcosh}\sqrt{r/r_{s}}$. At the lower limit $r=r_{s}$ we have $u=0$ and both terms vanish, which is why only the upper limit survives.

The integrand diverges at the lower limit like $(r-r_{s})^{-1/2}$, and Chapter 0.2 §5 classified exactly this case. An inverse power $p\lt1$ is integrable, so the improper integral converges. For $r_{0}=2r_{s}$ the answer is $\ell=2.296\,r_{s}$, a perfectly ordinary length, slightly larger than the coordinate difference $r_{0}-r_{s}=r_{s}$. Being larger is the correct behaviour, and it is the reason $r$ is called the areal radius rather than the distance to the centre. A diverging metric component has produced a convergent distance.

6.2 · Calculation two: two clocks disagree about the crossing

Drop something in radially from rest a long way out and follow it. Radial means $L=0$. Released from rest at infinity means $\dd t/\dd\tau\to1$ there, so $E=c^{2}$ by (3.8.3). Substituting both into (3.8.5),

That is remarkable on its own account. It is exactly the Newtonian free-fall speed $\sqrt{2GM/r}$, written in relativistic clothing, and it is perfectly finite and smooth at $r=r_{s}$, where it equals $c$. Nothing in (3.8.50) notices that anything special is happening. What we want next is the time this takes on the faller's own watch, so integrate, as grind box B does:

which is finite for every $r_{1}$, including $r_{1}=r_{s}$ and including $r_{1}=0$. The faller's own watch records an ordinary number of seconds and keeps running.

Now the same journey in coordinate time. From (3.8.3), $\dd t/\dd\tau=E/c^{2}A=1/A$, so $\dd t/\dd r=(\dd t/\dd\tau)(\dd\tau/\dd r)$ and

The extra factor $1/A$ is the whole difference between the two clocks. Near $r=r_{s}$ the integrand behaves as $r_{s}/(r-r_{s})$, an inverse first power. Chapter 0.2 §5 classified that as the borderline case that diverges, logarithmically. Grind box B does the integral exactly and extracts

Grind box B — both integrals, done exactly

Proper time. Invert (3.8.50) to get $\dd\tau=-\dd r\sqrt{r/r_{s}}/c$ and integrate from $r_{0}$ down to $r_{1}$, flipping the limits to absorb the minus sign:

$$\tau \;=\; \frac{1}{c\sqrt{r_{s}}}\int_{r_{1}}^{r_{0}} r^{1/2}\,\dd r \;=\; \frac{1}{c\sqrt{r_{s}}}\cdot\frac23\Big[r^{3/2}\Big]_{r_{1}}^{r_{0}} \;=\; \frac{2\big(r_{0}^{3/2}-r_{1}^{3/2}\big)}{3c\sqrt{r_{s}}}.$$

Nothing in the integrand is singular anywhere in $[0,r_{0}]$, so there is nothing to discuss. Setting $r_{1}=0$ gives $\tau=2r_{0}^{3/2}/3c\sqrt{r_{s}}$ for the whole journey to the centre, and setting $r_{0}=r_{s}$, $r_{1}=0$ gives the celebrated $\tau=\tfrac23\,r_{s}/c$ from horizon to centre.

Coordinate time. Integrate (3.8.52). Substitute $r=r_{s}x$, so $\dd r=r_{s}\dd x$ and $r-r_{s}=r_{s}(x-1)$:

$$t \;=\; \frac{1}{c\sqrt{r_{s}}}\int_{r_{1}}^{r_{0}}\frac{r^{3/2}\,\dd r}{r-r_{s}} \;=\; \frac{r_{s}}{c}\int_{x_{1}}^{x_{0}}\frac{x^{3/2}}{x-1}\,\dd x.$$

Split the integrand. Add and subtract $x^{1/2}$ in the numerator:

$$\frac{x^{3/2}}{x-1} \;=\; \frac{x^{1/2}(x-1)+x^{1/2}}{x-1} \;=\; \sqrt{x} \;+\; \frac{\sqrt{x}}{x-1}.$$

The second piece. Put $w=\sqrt{x}$, so $x=w^{2}$ and $\dd x=2w\,\dd w$:

$$\int\frac{\sqrt x}{x-1}\,\dd x \;=\; \int\frac{w}{w^{2}-1}\,2w\,\dd w \;=\; 2\int\left(1+\frac{1}{w^{2}-1}\right)\dd w \;=\; 2w + \ln\left|\frac{w-1}{w+1}\right|,$$

the middle step being polynomial division, $w^{2}/(w^{2}-1)=1+1/(w^{2}-1)$, and the last using the partial fractions of Chapter 0.8 §2.1. Collecting, and writing $F(x)\equiv\tfrac23x^{3/2}+2\sqrt x+\ln\big|(\sqrt x-1)/(\sqrt x+1)\big|$,

$$t \;=\; \frac{r_{s}}{c}\Big[F(x_{0})-F(x_{1})\Big], \qquad x=\frac{r}{r_{s}}.$$

(Differentiating $F$ returns $x^{3/2}/(x-1)$ exactly; checked symbolically.)

The divergence, isolated. Only the logarithm misbehaves as $x_{1}\to1^{+}$. There $\sqrt x-1\approx\tfrac12(x-1)$ and $\sqrt x+1\approx2$, so $\ln\big|(\sqrt x-1)/(\sqrt x+1)\big|\to\ln\big[(x-1)/4\big]\to-\infty$, and since it enters with a minus sign, $t\to+\infty$. Solving for $x-1$ gives $x-1\propto\ee^{-ct/r_{s}}$, which is (3.8.53). The approach is exponential, with the length $r_{s}$ setting the scale.

The contrast, stated as plainly as possible. The faller crosses $r=r_{s}$ after a finite, unremarkable interval on their own watch, computed by (3.8.51), and carries on. The coordinate label $t$ assigned to the crossing is $+\infty$. Both statements are correct, and they are statements about different things. One is about the traveller. The other is about a labelling scheme anchored to observers who are at rest far away, and §5's warning already established that no such observer exists at $r_{s}$ at all. Only one of these two numbers is about anybody's experience.

6.3 · What a distant observer actually sees, and how briefly

"The infalling object appears frozen at the horizon" is the usual gloss on (3.8.53), and it is misleading in a way worth correcting with two lines of arithmetic, because the object does not hang there visibly: it winks out.

Take the falling body to emit a steady signal. Its four-velocity is $u^{\mu}=(c/A,\,-c\sqrt{r_{s}/r},\,0,\,0)$, the first entry from $\dd t/\dd\tau=1/A$ and the second from (3.8.50). An outgoing radial photon has $p_{\mu}=(p_{0},\,-p_{0}/A,\,0,\,0)$, the second entry forced by the null condition $g^{\mu\nu}p_{\mu}p_{\nu}=A^{-1}p_{0}^{2}-A\,p_{1}^{2}=0$ with the sign chosen so that $p^{1}\gt0$. The frequency measured by the emitter is $p_{\mu}u^{\mu}$, and the frequency at infinity is $c\,p_{0}$, so writing $y\equiv\sqrt{r_{s}/r}$,

using $1-y^{2}=(1-y)(1+y)$. That is a tidy exact result, and it vanishes at $r=r_{s}$. Near the horizon it is $\approx(r-r_{s})/2r_{s}$, which is proportional to the very quantity (3.8.53) says decays exponentially.

One correction is needed before the exponential can be quoted, and skipping it costs a factor of two. The signal emitted at coordinate time $t$ takes time to climb out, and that climb time also diverges. Section 7.1 constructs the function $r_{*}$ whose difference gives the travel time, and $r_{*}\to-\infty$ logarithmically at the horizon in exactly the same way. Adding the two logarithms doubles the coefficient, so in terms of the time $t_{\text{obs}}$ at which the light arrives,

The characteristic time is $4GM/c^{3}=2r_{s}/c$, which is $0.20\ \mathrm{ms}$ for a $10M_{\odot}$ hole and $85\ \mathrm{s}$ for the one at the galactic centre. After a few multiples of that the signal is redshifted out of any detector, and the last photon has been emitted. The image does not linger. It fades on that timescale and is gone.

6.4 · Calculation three: the invariant, which cannot be argued with

Distances and clock readings are physical but chart-dependent in their bookkeeping, and the argument above therefore relies on interpretation. Chapter 2.4 §§5.1 and 6 supplied the instrument that does not. The Riemann tensor is a tensor, so a full contraction of it with itself is a scalar: every observer, in every chart, computes the same number at the same event. The simplest such quantity quadratic in the curvature is the Kretschmann scalar

Why not something simpler? The Ricci scalar is useless here. Chapter 3.7 solved $R_{\mu\nu}=0$, so $R$ and $R_{\mu\nu}$ vanish identically outside the mass and tell us nothing. What survives is the Weyl part that Chapter 3.4 §6 named and set aside, and $K$ is how you get at it. Grind box C computes it, and the answer is

Grind box C — the Kretschmann scalar for Schwarzschild, in three steps

The computation is mechanical and long, so here is the route with every intermediate result. The metric is $g_{\mu\nu}=\mathrm{diag}\big(A,\,-1/A,\,-r^{2},\,-r^{2}\sin^{2}\theta\big)$ in coordinates $(x^{0},r,\theta,\varphi)$ with $x^{0}=ct$, $A=1-r_{s}/r$ and $A'\equiv\dd A/\dd r=r_{s}/r^{2}$.

Step 1. The connection, from Chapter 3.3's formula (3.3.50). Nine non-zero components, counting each symmetric pair once, and these are the same nine Chapter 3.7 §3 listed while solving the field equations:

$$\Gamma^{0}{}_{0r}=\frac{A'}{2A}, \qquad \Gamma^{r}{}_{00}=\frac{AA'}{2}, \qquad \Gamma^{r}{}_{rr}=-\frac{A'}{2A}, \qquad \Gamma^{r}{}_{\theta\theta}=-rA,$$ $$\Gamma^{r}{}_{\varphi\varphi}=-rA\sin^{2}\theta, \qquad \Gamma^{\theta}{}_{r\theta}=\Gamma^{\varphi}{}_{r\varphi}=\frac1r, \qquad \Gamma^{\theta}{}_{\varphi\varphi}=-\sin\theta\cos\theta, \qquad \Gamma^{\varphi}{}_{\theta\varphi}=\cot\theta.$$

Step 2. The Riemann tensor, from Chapter 3.4 (3.4.13), then lowered with the metric. After the symmetries of Chapter 3.4 §5 are used, six independent components survive and every other one is either zero or one of these up to a sign:

$$R_{0r0r}=\frac{r_{s}}{r^{3}}, \qquad R_{0\theta0\theta}=-\frac{r_{s}\big(r-r_{s}\big)}{2r^{2}}, \qquad R_{0\varphi0\varphi}=R_{0\theta0\theta}\,\sin^{2}\theta,$$ $$R_{r\theta r\theta}=\frac{r_{s}}{2\big(r-r_{s}\big)}, \qquad R_{r\varphi r\varphi}=R_{r\theta r\theta}\,\sin^{2}\theta, \qquad R_{\theta\varphi\theta\varphi}=-\,r\,r_{s}\sin^{2}\theta.$$

Two things follow without further work. Every component is linear in $r_{s}$, so $K$, being quadratic, carries $r_{s}^{2}$. And $K$ has dimensions of (length)$^{-4}$, so with only $r_{s}$ and $r$ available it must be a pure number times $r_{s}^{2}/r^{6}$. Step 3 only fixes the pure number.

Step 3. The contraction, and the counting. All the surviving components have the same pair of index pairs on both sides, $R_{abab}$ with $a\lt b$, and every component with $(ab)\neq(cd)$ vanishes here. Each such component appears in four index assignments, namely $abab$, $abba$, $baab$ and $baba$, and the two sign flips cancel in the square, so

$$K \;=\; 4\sum_{a\lt b} R_{abab}\,R^{abab}, \qquad R^{abab}=g^{aa}g^{aa}g^{bb}g^{bb}R_{abab},$$

the last because the metric is diagonal, so raising an index is multiplication by one entry of $g^{-1}$. Take the $0\theta$ term as the model. There $g^{00}g^{\theta\theta}=(1/A)(-1/r^{2})$ and $A=(r-r_{s})/r$, so

$$R_{0\theta0\theta}R^{0\theta0\theta} \;=\; \big(g^{00}g^{\theta\theta}\big)^{2}\big(R_{0\theta0\theta}\big)^{2} \;=\; \frac{r^{2}}{\big(r-r_{s}\big)^{2}r^{4}}\cdot\frac{r_{s}^{2}\big(r-r_{s}\big)^{2}}{4r^{4}} \;=\; \frac{r_{s}^{2}}{4r^{6}}.$$

Every factor of $A$ has cancelled between the lowered component and the raising factors, which is the structural reason the answer contains no $A$ and therefore nothing that could misbehave where $A=0$. Doing the same for the other five and multiplying each by four, in units of $r_{s}^{2}/r^{6}$:

$$\underbrace{4}_{0r0r} + \underbrace{1}_{0\theta0\theta} + \underbrace{1}_{0\varphi0\varphi} + \underbrace{1}_{r\theta r\theta} + \underbrace{1}_{r\varphi r\varphi} + \underbrace{4}_{\theta\varphi\theta\varphi} \;=\; 12,$$ $$K \;=\; \frac{12\,r_{s}^{2}}{r^{6}} \;=\; \frac{12}{r^{6}}\cdot\frac{4G^{2}M^{2}}{c^{4}} \;=\; \frac{48\,G^{2}M^{2}}{c^{4}r^{6}}.$$

(Verified symbolically: the connection, the full four-index Riemann tensor and the complete contraction were computed from the metric above with nothing assumed, returning $48G^{2}M^{2}/c^{4}r^{6}$ exactly, and $R_{\mu\nu}=0$ in all sixteen components, which is the check that the metric being differentiated really is the vacuum solution Chapter 3.7 derived.)

Now evaluate (3.8.57) at the two places in question. At the horizon, substituting $r=r_{s}=2GM/c^{2}$,

a perfectly finite number. And $K\to\infty$ at exactly one radius, namely $r=0$.

That settles it, and it settles it in a way no change of chart can touch. $K$ is a scalar. If it is finite at an event in one chart it is finite at that event in every chart, by Chapter 2.4 §6. So there is no chart in which the curvature at $r=r_{s}$ is infinite, because there is nothing infinite there to display. Whatever (3.8.47) is telling us, it is not that the geometry has failed. Since 2.3 the word invariant has meant exactly this: a number all observers agree on, which therefore cannot be an artefact of any of them.

6.5 · How gentle the edge is, and why bigger is gentler

A number being finite is not the same as its being small, so put a scale on it. Chapter 3.4 §4 derived geodesic deviation and identified it with Chapter 3.1's tidal acceleration: two freely falling bodies separated radially by $\ell$ accelerate apart at $2GM\ell/r^{3}$. Evaluate that at the horizon, using $r_{s}=2GM/c^{2}$:

The same scaling is visible in (3.8.58), where $\sqrt{K(r_{s})}\propto M^{-2}$. Indeed $2GM/r^{3}=\big(c^{2}/2\sqrt3\big)\sqrt{K}$ identically, so the tidal coefficient is the square root of the invariant up to a fixed factor. Chapter 3.1's Worked example 1 remarked on this scaling and pointed here for the reason. Here is the reason, and here are the numbers for a person of height $\ell=1.8\ \mathrm{m}$:

Mass	$r_{s}$	$K(r_{s})^{1/2}$	head-to-toe stretch at $r_{s}$	as a multiple of $g$
$10\,M_{\odot}$	$29.5\ \mathrm{km}$	$4.0\times10^{-9}\ \mathrm{m^{-2}}$	$1.9\times10^{8}\ \mathrm{m\,s^{-2}}$	$1.9\times10^{7}$
$4.3\times10^{6}\,M_{\odot}$	$1.27\times10^{7}\ \mathrm{km}$	$2.1\times10^{-20}\ \mathrm{m^{-2}}$	$1.0\times10^{-3}\ \mathrm{m\,s^{-2}}$	$1.0\times10^{-4}$
$6.5\times10^{9}\,M_{\odot}$	$1.92\times10^{10}\ \mathrm{km}$	$9.4\times10^{-27}\ \mathrm{m^{-2}}$	$4.4\times10^{-10}\ \mathrm{m\,s^{-2}}$	$4.5\times10^{-11}$

A bigger hole is gentler at its edge. That is counter-intuitive and it follows from elementary scaling: the tidal field goes as $M/r^{3}$ and the horizon radius as $M$, so the tidal field at the horizon goes as $M/M^{3}$. It also joins up with Chapter 3.1 §5, which bounded the size of a freely falling laboratory by requiring tidal effects to stay below the measurement precision. Small $K$ means a large local inertial frame, and for the last row of the table the local inertial frame at the horizon is enormous. An observer crossing it in a sealed cabin has no local experiment that will tell them anything has happened. That is precisely the equivalence principle doing its job, and precisely why §7 is needed to say what has happened.

7 · Eddington–Finkelstein, and crossing

Section 6 showed that the geometry is fine at $r=r_{s}$ and that the chart is not. Chapter 3.2 §2 said what to do about a chart that fails on part of a manifold: find another chart, check that the two agree on the overlap, and read the geometry in whichever one is well behaved where you are working. That is the entire operation performed here. It is worth noticing that no new physics enters. All that changes is the labels, and by Chapter 2.4 §6 a change of labels cannot change anything a tensor says.

Here is the destination. We build a new radial coordinate by integrating the condition for a radial light ray, use the arrival label of an ingoing ray as a new time coordinate, rewrite the metric, check its determinant, and then compute the two radial null directions at every radius. Inside $r_{s}$ both of them have $\dd r\lt0$. That computation is what a horizon is.

7.1 · Following the light: the tortoise coordinate

The construction is dictated by the problem rather than chosen. What misbehaves in the Schwarzschild chart is the description of light. As $r\to r_{s}$, radial rays take infinite coordinate time to get anywhere, which is why $t$ is a bad label. So build a coordinate for which they do not.

A radial null curve has $\dd\theta=\dd\varphi=0$ and $\dd s^{2}=0$, so from the line element

The right-hand side is a function of $r$ alone, so it can be integrated once and for all. Define the tortoise coordinate $r_{*}$ by $\dd r_{*}/\dd r = 1/A$:

where the third step is the polynomial division $r/(r-r_{s})=1+r_{s}/(r-r_{s})$ and the constant of integration has been chosen to make the logarithm's argument dimensionless. Differentiate the answer back to check: $\dd r_{*}/\dd r = 1+r_{s}/(r-r_{s}) = r/(r-r_{s}) = 1/A$ ✓. Notice what $r_{*}$ does: it is $\approx r$ far away, and it runs to $-\infty$ as $r\to r_{s}^{+}$, stretching the last kilometre above the horizon into an infinite coordinate range. That is exactly the divergence §6.2 found, isolated into one function of $r$ where it can be dealt with.

Now we can use it to label the rays themselves. With (3.8.61) in hand, radial null curves are $ct=\mp r_{*}+\text{const}$, so consider the two constants

which are respectively constant along ingoing and outgoing radial rays. Use $v$, the label of the ingoing ray that passes through an event, as the new time coordinate, keeping $r$, $\theta$ and $\varphi$ as they were. This is the ingoing Eddington–Finkelstein chart, and both $v$ and $r$ have dimensions of length.

7.2 · The metric in the new chart

From (3.8.62), $\dd v = c\,\dd t + \dd r_{*} = c\,\dd t + \dd r/A$, so $c\,\dd t = \dd v - \dd r/A$. Substitute into the line element and expand the square:

and now watch the two $\dd r^{2}/A$ terms, which are the only place $1/A$ appeared. They cancel outright:

Every component is finite at $r=r_{s}$, and the metric is perfectly ordinary there. The $vv$ component vanishes at $r_{s}$, but that on its own is not a failure. The $\theta\theta$ component of a flat plane's metric vanishes at the origin too. What would be a failure is degeneracy, meaning the metric becoming non-invertible, so that raising an index is impossible. Let's test it. In the ordering $(v,r,\theta,\varphi)$ the matrix is block-diagonal with a $2\times2$ block $\begin{pmatrix}A&-1\\-1&0\end{pmatrix}$, whose determinant is $A\cdot0-(-1)(-1)=-1$, so

which is non-zero for every $r\gt0$ off the polar axis, and in particular is $-r_{s}^{4}\sin^{2}\theta$ at the horizon. The determinant does not merely fail to vanish. It never notices that $r=r_{s}$ exists. And the $2\times2$ block's determinant is $-1$ independently of $A$, which is the algebraic reason: the off-diagonal $-1$ carries the invertibility, and it is there precisely because the new time coordinate was built out of the light rays rather than out of the static observers.

7.3 · The two radial null directions, computed

Now the calculation that makes a horizon a horizon. Set $\dd s^{2}=0$ with $\dd\theta=\dd\varphi=0$ in (3.8.64):

A product is zero when a factor is, so there are exactly two families, as there must be, because a light cone has two radial generators:

Branch (i) is the ingoing family, by construction: $v$ was defined to be constant along ingoing rays. Branch (ii) is everything else. To see what it does, plot against a time-like label rather than against $v$. Put $\tilde t\equiv v-r$, which is the natural choice because it makes the ingoing rays run at $45^{\circ}$ as they do in flat spacetime. Then $\dd\tilde t = \dd v - \dd r$ and

One thing has to be fixed before those signs mean anything: which way is the future. It is settled once and for all outside the horizon, where $v=ct+r_{*}$ plainly increases with $t$. A time orientation is a continuous choice on a connected spacetime, so it carries inward unchanged rather than being re-chosen at $r_{s}$. Check it survives: an infalling particle has $\dd v/\dd\lambda\gt0$ and $\dd r/\dd\lambda\lt0$, so $\dd\tilde t/\dd\lambda=\dd v/\dd\lambda-\dd r/\dd\lambda\gt0$. Increasing $\tilde t$ is the future everywhere, inside $r_{s}$ as much as outside.

Both slopes are now explicit functions of $r$ and there is nothing left to interpret. Read branch (ii) off:

Radius	$\dd r/\dd\tilde t$ for branch (ii)	What that means
$r\to\infty$	$\to+1$	outgoing at the speed of light, as in flat spacetime
$r=2r_{s}$	$+1/3$	still outgoing, but slowed in these coordinates
$r=r_{s}$	$0$	the ray stays at $r=r_{s}$ forever
$r=r_{s}/2$	$-1/3$	the "outgoing" ray moves inward
$r\to0$	$\to-1$	both branches fall at the same rate, and the cone has closed

The middle row is the horizon, arrived at by evaluating a formula and not by drawing anything. The row below it is the content of the whole chapter's second half:

Inside $r_{s}$ every future-directed light ray moves to smaller $r$. A material particle's worldline lies strictly inside the light cone, by Chapter 2.3 §4, and that statement is unchanged by the move to a manifold because it is a statement about the local light cone. So every material particle inside $r_{s}$ also moves to smaller $r$. Not because anything pushes it. Because the only directions available in which its own future lies all point inward. That is what the surface $r=r_{s}$ is: a one-way membrane, and the word horizon means precisely this and nothing more.

One more fact falls out of (3.8.68) for free. Both branches can be integrated in closed form. Branch (i) gives $\tilde t = -r + \text{const}$. For branch (ii), integrating $\dd\tilde t/\dd r = (r+r_{s})/(r-r_{s}) = 1 + 2r_{s}/(r-r_{s})$ by the same polynomial division as (3.8.61),

Both families are therefore drawable exactly, with no numerical integration, and that is what the figure below draws.

radius r / r_s

2.20

sit on the horizon

go inside

go far away

ingoing edge dr/dt̃ = −1.000000 other edge dr/dt̃ = (r−r_s)/(r+r_s) = +0.375000

A(r) = 1 − r_s/r = 0.545455 dr/dv on the second branch = A/2 = 0.272727

outside: one edge points to larger r, so escape is possible; the cone straddles the dashed line.

7.4 · What has and has not been shown about escape

It is worth separating two statements that are easy to run together. (3.8.69) is a local fact: at any event with $r\lt r_{s}$, the future light cone lies entirely in the direction of decreasing $r$. What follows from it is that a particle inside cannot stop, cannot turn round and cannot hover, since all three would need a future direction with $\dd r\ge0$ and there is none. How long it has left is then §6.2's integral. For a body falling from rest far away, (3.8.51) with $r_{0}=r_{s}$ and $r_{1}=0$ gives a proper time of $\tfrac23\,r_{s}/c$ from the horizon to the centre. That is $66\ \mu\mathrm{s}$ for a $10M_{\odot}$ hole and $28\ \mathrm{s}$ for the one at the galactic centre.

What has not been shown is that the surface $r=r_{s}$ is a horizon in the general sense that would apply to a spacetime with no symmetry at all. That statement needs a definition, and the definition needs machinery beyond this book.

7.5 · The Rindler horizon, compared honestly

Chapter 2.3 Problem 3 met a horizon already, in flat spacetime with no matter in it anywhere. A rocket with constant proper acceleration $a$ follows the hyperbola $x^{2}-c^{2}t^{2}=(c^{2}/a)^{2}$, whose asymptote is the null line $x=ct$. That problem showed by explicit calculation that a flash emitted from beyond that line never catches the rocket, however long it chases. The dividing surface is the Rindler horizon, and Chapter 2.3 promised that this chapter would meet the same object again.

What is the same. Both are one-way surfaces with no local marker on them: nothing measurable happens there, and an observer crossing either can perform no local experiment that detects the crossing. Both are an artefact of a chart in one specific sense. A coordinate system adapted to a family of observers, the static ones here and the uniformly accelerated ones there, assigns infinite coordinate time to the crossing, while the crossing takes finite proper time. Both produce an exponentially growing redshift with a characteristic time set by the surface gravity. For Rindler that is $c/a$, which is about a year for $a=g$. Chapter 3.1 §6.1 noted the companion length $c^{2}/g$, about a light-year, which is where the horizon sits. For Schwarzschild (3.8.55) gives $2r_{s}/c$. And in both cases the repair is a change of chart, which was the point of §7.2.

And now the honest difference, which matters more than the similarity. The Rindler horizon belongs to an observer. Different accelerations give different horizons. An inertial observer sails across it noticing nothing and has no horizon at all. If the rocket cuts its engines its horizon dissolves and every delayed signal arrives. The Schwarzschild horizon belongs to nobody. It is at $r=2GM/c^{2}$ for every observer, however they move and whatever chart they use, because $r$ is the areal radius and the area of a sphere is a geometric fact. Stop accelerating and it does not go away. The reason for the difference is exactly the one §6.4 established. Flat spacetime has $K=0$ everywhere and there is nothing for a horizon to attach to, while here $K(r_{s})=3c^{8}/4G^{4}M^{4}$ is a definite non-zero number that marks the surface unambiguously even though it is finite.

The correct summary is that the two horizons have the same local character and different global status, and the vocabulary for saying so precisely is the flagged definition of §7.4. Chapter 2.3's claim that they are "the same kind of object" is right about the local part and overstated about the rest. This is the sharper statement it pointed forward to.

8 · What is actually singular

Everything so far has been about a place where the geometry is fine and the chart is not. This section is about the other place, and the whole point of the previous two sections is that the distinction is now a computation rather than a matter of taste.

8.1 · The centre, and why no chart can repair it

Return to the Kretschmann scalar (3.8.57), $K=48G^{2}M^{2}/c^{4}r^{6}$, and read it in the other direction:

Set the two statements side by side, because they are the chapter in miniature. At $r=r_{s}$ an invariant is finite, so no observer computes anything infinite there, so §7 could and did repair the chart. At $r=0$ an invariant is infinite, and there is nothing to repair. A change of chart is a relabelling. A scalar's value at an event does not depend on labels, by Chapter 2.4 §6. So no relabelling can make $K$ finite at $r=0$. Every clever coordinate system anybody could invent has already been ruled out, in one line, without inventing any of them.

This is what the word curvature singularity means and it is all it means: an invariant built from the curvature grows without bound as an event is approached. It is not a claim about infinite density. That would be a statement about $T_{\mu\nu}$, and there is no matter here, because Chapter 3.7 solved the vacuum equations. It is a statement about the geometry alone.

8.2 · What general relativity says, and where it stops

Take the falling observer of §6.2 and follow them all the way. Equation (3.8.51) with $r_{1}=0$ gives a finite proper time. Section 7.3 showed that once inside $r_{s}$ they have no choice about going. And $K$ diverges on arrival, so the tidal forces that were gentle at the horizon of a large hole grow without bound as $r^{-3}$. Then the worldline ends. Not "reaches a point of infinite density", because there is no point there to reach. $r=0$ is not part of the manifold: a manifold is a set on which the metric is defined and smooth, and the metric is defined nowhere on $r=0$.

The technical name for what has happened is geodesic incompleteness: there is a geodesic which, parametrised by its own proper time, cannot be extended past a finite parameter value. And the honest statement of the situation is short.

General relativity does not predict what happens at $r=0$. It predicts that it does not know. The theory's own equations produce a place where its own equations stop making sense, after a finite and computable interval of somebody's watch, and they do so from perfectly innocuous initial data, namely a star, sitting there. That is not a paradox and not a scandal. It is a theory reporting the boundary of its own domain of validity, which is the most useful thing an incomplete theory can do, and one should trust the report precisely because the equations making it are the ones being invalidated.

What is expected to matter there is quantum mechanics. The curvature radius $K^{-1/4}$ falls through the Planck length $\sqrt{\hbar G/c^{3}}=1.6\times10^{-35}\ \mathrm{m}$ at a radius that is still finite, and at that scale a classical smooth metric is not a description anybody has reason to believe. Chapter 7.1 is where this book asks what replaces it, and Chapter 7.9 is where the honest accounting of the answer is done.

8.3 · Is it the symmetry's fault?

One reasonable objection remains, and it deserves an answer rather than a reassurance. Everything above was computed for a perfectly spherical vacuum solution. Perfect spheres do not occur. Perhaps the singularity is an artefact of the idealisation, in the way that the infinite field at the centre of a point charge is an artefact of pretending a charge is a point. Give the star a little rotation or a little lumpiness and perhaps the collapse misses the centre and comes back out.

So the situation is this. A star more massive than any pressure can support collapses. A trapped surface forms. The theorems then guarantee that some observer's worldline ends after finite proper time. And general relativity, which predicted all of that from its own equations, has nothing to say about the ending. That is where this chapter stops, and it is where the theory stops. The next chapter takes the same equations and the same machinery and puts the largest available source on the right-hand side, which turns out to raise the same question about the beginning.

9 · Worked examples

10 · Your turn