Part F · Einstein · spine

The classical tests: three numbers, and one of them is 43

Unit 08's orbit equation acquires exactly one term. From it come the 43 arcseconds of Mercury, the 1.75 arcseconds of starlight at the Sun's edge, a radar echo that returns late, and an orbit that no longer has to be stable.

needs 36, 21, 08, 10~3½ h slides + 7 h problemsproblems: ./mr problems 37

Unit 36 found the metric. This unit asks the only question that makes a metric physics: what do things do in it? The answer is the geodesics, and the geodesics of Schwarzschild reduce — exactly, with no approximation anywhere — to unit 08's Binet equation with one extra term. Everything in the unit is that term, read in different regimes.

Three of the four prerequisites are old friends, and each is used for the thing it was built for. Unit 08 supplies the Binet substitution and the effective potential. Unit 10 supplies secular perturbation, which is how the 531 arcseconds of Newtonian planetary perturbation on Mercury are computed. Unit 21 supplies Poincaré–Lindstedt and, more importantly, the argument that the secular term in the orbit is the answer rather than an error. Unit 36 supplies the metric, its Killing fields — four of them, $\partial_t$ and the three rotations, of which this unit uses $\partial_t$ and $\partial_\varphi$ — and the theorem that each one conserves $g(K, \dot x)$ along a geodesic.

How to read this deck

One example runs through everything: Mercury, and alongside it a ray of light grazing the Sun. Section 1 derives the orbit equation both are governed by. Section 2 reads it through the effective potential, which is where the effects with no Newtonian analogue live. Section 3 computes Mercury's 43 arcseconds per century, end to end. Section 4 bends the ray, twice — once as a Newtonian corpuscle, once as general relativity requires — and names the metric component responsible for the difference. Section 5 delays and reddens it. Section 6 places the classical tests among the modern ones and closes the example.

Conventions

Signature $(-,+,+,+)$, $c$ kept explicit, as in unit 32. $r_g = GM/c^2$ is the gravitational radius, half the Schwarzschild radius $r_s = 2r_g$. Unit 36 wrote every formula with $r_s$, as $f = 1 - r_s/r$; here the same function is $1 - 2r_g/r$, because every expansion in this unit is in $r_g/r$ and the factors of 2 land better this way. A dot is $\mathrm d/\mathrm d\tau$ along a timelike geodesic (proper time) or $\mathrm d/\mathrm d\lambda$ along a null one (affine parameter); a prime is $\mathrm d/\mathrm d\varphi$.

$E := -g(\partial_t, \dot x)$ and $L := g(\partial_\varphi, \dot x)$ are unit 36's two Killing constants, unchanged in definition and in sign. For a massive particle, with $\dot x$ normalised by $g(\dot x, \dot x) = -c^2$, they are unit 36's specific energy and angular momentum: per unit rest mass, so that no mass appears anywhere — the equivalence principle of unit 30, in bookkeeping form. Unit 21 called the same $L$ by the name $h$. For light there is no rest mass to divide by and no proper time: $E$ and $L$ are the same two Killing constants, but they now scale with the arbitrary choice of affine parameter, so neither alone means anything. Only their ratio does, and it is the impact parameter $b = cL/E$ — which is why every light formula in this unit is written in $b$ and never in $E$ or $L$ separately.

Constants used everywhere: $GM_\odot = 1.32712440018\times10^{20}\ \mathrm{m^3\,s^{-2}}$, $c = 299\,792\,458$ m/s, $R_\odot = 6.957\times10^{8}$ m, 1 au $= 1.495978707\times10^{11}$ m, one radian $= 206\,264.806''$. Every number in the deck was computed from these.

37 · overview

One orbit, one ray, one extra term

the running example

Mercury: $a = 5.79\times10^{10}$ m, $e = 0.2056$, 415.2 orbits a century. Its perihelion advances by about $5600''$ per century, and Newton accounts for all but $43''$ of it.

A ray grazing the Sun: impact parameter $b = R_\odot$. Newton, treating light as a fast particle, bends it by $0.88''$. The sky says $1.75''$.

Both are geodesics of the Schwarzschild metric, and both obey one equation, in $u = 1/r$ as a function of the angle $\varphi$:

$$u'' + u = \underbrace{\frac{GM}{L^2}}_{\text{unit 08}} + \underbrace{3r_g\,u^2}_{\text{new}}$$

$r_g = GM/c^2$, $L$ the angular momentum per unit mass. The first term comes from the particle's rest mass, so for light it is absent and the ray obeys $u'' + u = 3r_gu^2$ alone.

how the equation is reached · section 1

1. The metric does not involve $t$ or $\varphi$, so each geodesic conserves an energy $E$ and an angular momentum $L$.

2. Symmetry makes the orbit planar, and the tangent's fixed length, $g(\dot x, \dot x) = -c^2$ (zero for light), gives $\dot r^2$ as a function of $r$: an energy equation.

3. Unit 08's trick: $u = 1/r$, $\varphi$ as parameter, $\dot r = -Lu'$.

4. Differentiate in $\varphi$. One cross term, $2r_gL^2u^3$ — the $g_{tt}$ correction times the angular motion — becomes the new $3r_gu^2$.

The plan. 1 the equation · 2 near a black hole · 3 Mercury's $43''$ · 4 $0.88''$ against $1.75''$ · 5 delay and redshift · 6 the tests now.

Dotted underline, like $r_g$: a glossary term. Solid, like Mercury: the running example, on hover. N: notes and long proofs.

The title's three numbers are the perihelion advance ($43''$ per century), the light deflection ($1.75''$) and the Shapiro delay (about $233\ \mu$s for a radar echo from Venus grazing the Sun). The fourth classical test, the gravitational redshift, was closed in unit 30 from the equivalence principle alone, and section 5 only makes it exact. Einstein's own list of 1916 had three tests: perihelion, deflection, redshift. Shapiro added the delay in 1964, and it is now the most precisely measured of them.

The sketch, step by step. Each step is a slide in section 1. Step 1 is Two Killing vectors, two constants: $\partial_t$ and $\partial_\varphi$ are Killing vectors, and each conserves $g(K, \dot x)$. Step 2 is The third constant, and the plane, whose boxed $\dot r^2 = E^2/c^2 - (1 - 2r_g/r)(\sigma + L^2/r^2)$ is the energy equation, with $\sigma = c^2$ for a massive particle and $0$ for light. Steps 3 and 4 are Worked example 1 — the orbit equation, exactly. The new term's origin is visible already in step 2: expanding the product gives $2r_gL^2/r^3 = 2r_gL^2u^3$, a coupling of the $g_{tt}$ correction $2r_g/r$ to the angular kinetic term $L^2/r^2$, and $\frac{\mathrm d}{\mathrm d\varphi}(2r_gL^2u^3) = 6r_gL^2u^2u'$, which after dividing by $2L^2u'$ is $3r_gu^2$. The $GM/L^2$ term comes from $2r_g\sigma u$, the product of the same correction with the rest-mass term $\sigma$, which is why light, with $\sigma = 0$, does without it.

Why these numbers are small. Every effect in this unit is proportional to $r_g/r$ at the place the effect happens. At Mercury's orbit that is $2.5\times10^{-8}$; at the Sun's surface $2.1\times10^{-6}$. That is why first-order perturbation theory is enough in sections 3 to 5, and why section 2, which is about $r$ of a few $r_g$, has nothing to do with the solar system and everything to do with black holes.

section 1

Geodesics of Schwarzschild

Two Killing vectors and a normalisation reduce eight equations to one.

37 · geodesics

Two Killing vectors, two constants

The metric's components do not depend on $t$ or $\varphi$, so $\partial_t$ and $\partial_\varphi$ are Killing vectors — two of the four unit 36 counted, the other two rotations being what makes the orbit planar on the next slide. Along any geodesic with tangent $\dot x$, timelike or null, $g(K, \dot x)$ is constant for a Killing $K$ (unit 36). For a massive particle, per unit rest mass:

the two constants $$E := -g(\partial_t, \dot x) = \Big(1 - \frac{2r_g}{r}\Big)c^2\,\dot t,$$ $$L := g(\partial_\varphi, \dot x) = r^2\sin^2\theta\;\dot\varphi .$$

The specific energy $E$ and the specific angular momentum $L$.

they are what the names say

For a slow particle far away, $E = c^2 + \tfrac12v^2 - GM/r + \dots$: rest energy plus Newton's energy, per unit mass. And $L$ is unit 08's $r^2\dot\varphi$, with proper time for time.

this is Noether, again

$t$ and $\varphi$ are absent from the geodesic Lagrangian $\mathcal L = g_{\mu\nu}\dot x^\mu\dot x^\nu$ (unit 32), so their momenta are conserved: unit 13's theorem for a cyclic coordinate. The Killing vector is the coordinate-free statement of the same symmetry.

Full proof that $g(K,\dot x)$ is constant. Cited from unit 36, where it is proved from Killing's equation $\nabla_\mu K_\nu + \nabla_\nu K_\mu = 0$: $\frac{\mathrm d}{\mathrm d\tau}(K_\mu\dot x^\mu) = \dot x^\nu\nabla_\nu(K_\mu\dot x^\mu) = (\nabla_\nu K_\mu)\dot x^\nu\dot x^\mu + K_\mu\dot x^\nu\nabla_\nu\dot x^\mu$. The second term is zero by the geodesic equation; the first is a symmetric tensor $\dot x^\nu\dot x^\mu$ contracted with the antisymmetric (by Killing's equation) $\nabla_\nu K_\mu$, so zero. For this metric there is a second, entirely elementary proof, and it is the one worth using in practice: the $t$ and $\varphi$ Euler–Lagrange equations of $\mathcal L$ read $\frac{\mathrm d}{\mathrm d\tau}\big(\partial\mathcal L/\partial\dot t\big) = \partial_t\mathcal L = 0$ and likewise for $\varphi$, and $\partial\mathcal L/\partial\dot t = -2(1-2r_g/r)c^2\dot t$, $\partial\mathcal L/\partial\dot\varphi = 2r^2\sin^2\theta\,\dot\varphi$.

The Newtonian reading of $E$. For a slow particle $\dot t = \mathrm dt/\mathrm d\tau \approx 1 + v^2/2c^2 + r_g/r$ (unit 27's time dilation plus unit 30's gravitational one). Then $E = (1 - 2r_g/r)c^2(1 + v^2/2c^2 + r_g/r) = c^2 + \tfrac12v^2 - c^2r_g/r + O(c^{-2})$, and $c^2r_g = GM$. Dimensions: $E$ is an energy per unit mass, $\mathsf{L^2T^{-2}}$ ✓; $L$ is $\mathsf{L^2T^{-1}}$ ✓.

And for light. Unit 36's proof of the conservation theorem never used $g(\dot x, \dot x) = -c^2$, and unit 36 says so explicitly, so the same two constants exist for a null geodesic parametrised by an affine $\lambda$. What does not carry over is the word "specific": a photon has no rest mass to divide by, and rescaling $\lambda$ rescales both $E$ and $L$. Sections 4 and 5 therefore use only the scale-free combination $b = cL/E$, and the deflection and the delay are written in $b$ from the start.

Sign convention. The minus in $E = -g(\partial_t, \dot x)$ is there because $\partial_t$ is timelike and future-pointing outside the horizon, so $g(\partial_t, \dot x) \lt 0$ for any future-pointing $\dot x$. With it, $E \gt 0$. Unit 36's warning applies here too: $g(\partial_t, \partial_t) = -(1 - 2r_g/r)c^2$ is negative only for $r \gt 2r_g$, null at $2r_g$ and positive inside, so "$E$ is an energy" is a statement about the exterior. Everything in this unit lives at $r \gt 2r_g$ — section 2 goes down to the photon sphere at $3r_g$ and no further, and follows a plunging particle only up to the horizon.

37 · geodesics

The third constant, and the plane

the orbit is planar

Every geodesic lies in a plane through the centre; choose coordinates so it is $\theta = \pi/2$.

Proof. The rotations are isometries of the metric (unit 36), and an isometry carries geodesics to geodesics (unit 32), so it is enough to prove it for a rotated copy: rotate until $x(0)$ and $\dot x(0)$ have $\theta = \pi/2$, $\dot\theta = 0$. The $\theta$-equation, $\frac{\mathrm d}{\mathrm d\tau}(r^2\dot\theta) = r^2\sin\theta\cos\theta\,\dot\varphi^2$, is solved by $\theta \equiv \pi/2$, and by uniqueness (unit 01) that is the solution. $\square$

the normalisation

$g(\dot x, \dot x) = -\sigma$, with $\sigma = c^2$ for a massive particle and proper time as parameter (unit 27), and $\sigma = 0$ for light. In the plane, with $f = 1 - 2r_g/r$:

$$-f\,c^2\dot t^2 + \frac{\dot r^2}{f} + r^2\dot\varphi^2 = -\sigma .$$

Substitute $\dot t = E/(fc^2)$ and $\dot\varphi = L/r^2$, multiply by $f$:

$$\boxed{\ \dot r^2 = \frac{E^2}{c^2} - \big(1 - 2r_g/r\big)\big(\sigma + L^2/r^2\big)\ }$$

The substitution, line by line. Write $f = 1 - 2r_g/r$. Then $-fc^2\dot t^2 = -fc^2E^2/(f^2c^4) = -E^2/(fc^2)$ and $r^2\dot\varphi^2 = L^2/r^2$. So the normalisation is $-E^2/(fc^2) + \dot r^2/f + L^2/r^2 = -\sigma$. Multiply by $f$: $-E^2/c^2 + \dot r^2 + fL^2/r^2 = -f\sigma$, which rearranges to the box.

What has happened to the eight equations. Four second-order geodesic equations are eight first-order ones. The plane uses two ($\theta$ and $\dot\theta$), the two Killing constants use two more ($\dot t$ and $\dot\varphi$ are now known functions of $r$), and the normalisation turns the $r$-equation into a first-order one. What remains is one first-order ODE for $r(\tau)$ plus quadratures for $t$ and $\varphi$ — exactly the structure of unit 08's reduction from six to two, and for exactly the same reason: symmetries.

Why the normalisation is a constant of motion. Along a geodesic $\frac{\mathrm d}{\mathrm d\tau}g(\dot x, \dot x) = 2g(\nabla_{\dot x}\dot x, \dot x) = 0$ by metric compatibility and the geodesic equation (unit 32). So a geodesic that starts timelike stays timelike, and the parameter can be chosen so the constant is $-c^2$.

Why $\sigma$. Many texts call this constant $\kappa$ or $\epsilon$. Neither is free in this course: $\kappa$ is Einstein's $8\pi G/c^4$ in unit 35 and a horizon's surface gravity in unit 40, and $\varepsilon$ is section 3's small parameter $3r_g/p$. So $\sigma$, with this one meaning throughout the unit: $\sigma = c^2$ for a massive particle, $\sigma = 0$ for light.

Planarity. Unit 08 needed the cross product to see that central-force motion is planar; here it is a symmetry plus Picard–Lindelöf.

Checks. Dimensions: every term is $\mathsf{L^2T^{-2}}$ when $\sigma = c^2$ ✓. The limit $r_g \to 0$ gives $\dot r^2 = E^2/c^2 - c^2 - L^2/r^2$: free motion in Minkowski space in polar coordinates, with $E^2/c^2 - c^2 = \gamma^2v^2$ ✓.

37 · geodesics

The effective potential

For a massive particle ($\sigma = c^2$), halve the box and expand the product:

$$\tfrac12\dot r^2 + V_{\rm eff}(r) = \tfrac12\Big(\frac{E^2}{c^2} - c^2\Big),$$

the relativistic effective potential $$V_{\rm eff}(r) = -\frac{GM}{r} + \frac{L^2}{2r^2} - \frac{GM L^2}{c^2r^3}$$

Unit 08's $V_{\rm eff}$ — gravity plus the centrifugal barrier — and one new term, attractive, $\propto r^{-3}$. Exact.

two things that look Newtonian and are not

The time is proper time $\tau$, not $t$. And the "energy" on the right is $\tfrac12(E^2/c^2 - c^2)$, which is Newton's $E_N$ only when $E \approx c^2 + E_N$.

checks

Dimensions: $GML^2/c^2r^3$ is $\mathsf{L^3T^{-2}\cdot L^4T^{-2}/(L^2T^{-2}\cdot L^3)} = \mathsf{L^2T^{-2}}$, like $GM/r$ ✓.
Limit $c\to\infty$: the new term vanishes and unit 08 returns ✓.
Size at Mercury: new term over centrifugal term is $2r_g/r = 5.1\times10^{-8}$ at mean distance.

The expansion. $(1 - 2r_g/r)(c^2 + L^2/r^2) = c^2 + L^2/r^2 - 2r_gc^2/r - 2r_gL^2/r^3$. With $r_gc^2 = GM$, the box on the previous slide is $\dot r^2 = E^2/c^2 - c^2 - L^2/r^2 + 2GM/r + 2GML^2/(c^2r^3)$. Halving and moving the $r$-dependent terms to the left gives the slide.

Where the new term comes from. It is the product of $g_{tt}$'s correction $-2r_g/r$ with the $L^2/r^2$ that came from $g_{\varphi\varphi}$: a coupling of the gravitational potential to the kinetic energy of angular motion. In Newton, kinetic energy does not gravitate; here it does, as $E = mc^2$ says it must. The "$c^2$ in the denominator" is the cost of converting a kinetic energy per unit mass into a mass.

The ratio quoted. $\frac{GML^2/c^2r^3}{L^2/2r^2} = 2GM/(c^2r) = 2r_g/r$. With $r_g = 1476.6$ m and Mercury's $a = 5.7909\times10^{10}$ m: $5.10\times10^{-8}$.

Why the proper-time caveat does not bite the orbit. The next slide eliminates the parameter entirely in favour of $\varphi$, so the shape of the orbit never knew which clock was used. Only the period does, and no test in this unit uses the period to the precision at which the difference matters.

37 · geodesics · example

Worked example 1 — the orbit equation, exactly

from the radial equation to the orbit equation

Step 1 — change variable. $u = 1/r$ and $\varphi$ as parameter. Since $\dot\varphi = Lu^2$, $$\dot r = \frac{\mathrm dr}{\mathrm d\varphi}\dot\varphi = -\frac{u'}{u^2}\,Lu^2 = -Lu'.$$ Unit 08's Binet trick.

Step 2 — substitute into $\dot r^2 = E^2/c^2 - (1 - 2r_gu)(\sigma + L^2u^2)$: $$L^2u'^2 = \frac{E^2}{c^2} - \sigma - L^2u^2 + 2r_g\sigma u + 2r_gL^2u^3 .$$

Step 3 — differentiate in $\varphi$ and divide by $2L^2u'$ (away from turning points, $u' \ne 0$): $$u'' = -u + \frac{r_g\sigma}{L^2} + 3r_gu^2 .$$

Step 4. Massive ($\sigma = c^2$) and massless ($\sigma = 0$):

$$\boxed{\ u'' + u = \frac{GM}{L^2} + 3r_gu^2\ }$$ $$\boxed{\ u'' + u = 3r_gu^2\ }$$

Step 5 — check. $r_g \to 0$: unit 08's Kepler equation ✓, and a straight line for light ✓.

Step 3 in full. $\frac{\mathrm d}{\mathrm d\varphi}(L^2u'^2) = 2L^2u'u''$, and the right side differentiates to $-2L^2uu' + 2r_g\sigma u' + 6r_gL^2u^2u'$. Divide by $2L^2u'$.

The turning points. Dividing by $u'$ fails where $u' = 0$, at perihelion and aphelion. The second-order equation obtained is nonetheless valid there too: both sides are continuous in $\varphi$, and they agree on a dense set, so they agree everywhere. (The same point arises in unit 00's energy method, which was where "multiply by $\dot x$ and integrate" first lost the equilibria.) The circular orbits, where $u' \equiv 0$, are recovered as the constant solutions of the second-order equation, which is exactly why one prefers it to the first-order one.

Dimensions. $[u''] = [u] = \mathsf{L^{-1}}$; $[GM/L^2] = \mathsf{L^3T^{-2}/L^4T^{-2}} = \mathsf{L^{-1}}$ ✓; $[r_gu^2] = \mathsf{L\cdot L^{-2}} = \mathsf{L^{-1}}$ ✓.

How exact is "exactly". No approximation has been made: this is the equatorial geodesic equation of the Schwarzschild metric, rewritten. All the approximating happens later, when it is solved. That is the reason to prefer this route to the weak-field one: the equation is right at every $r \gt 2r_g$, so section 2 can use it near a black hole and section 3 at Mercury with the same confidence.

Unit 08 predicted this. Its notes say "unit 37 gets $u'' + u = GM\mu^2/L^2 + 3GMu^2/c^2$". There $L$ was the total angular momentum of a body of mass $\mu$; here it is per unit mass, so $\mu$ has gone.

37 · geodesics

What the extra term is

Unit 08's Binet equation for a force $f(r)$ per unit mass is $u'' + u = -f(1/u)/(L^2u^2)$. So the orbit equation is Newton's, for the force

$$f(r) = -\frac{GM}{r^2} - \frac{3GML^2}{c^2r^4}.$$

An inverse-square law plus an inverse-fourth-power correction, attractive, proportional to $L^2$ — to the square of the transverse speed.

a reading, not a mechanism

No such force acts: the particle is in free fall. The reading is exact only for the shape $r(\varphi)$, and only with proper time as the clock.

the size, at Mercury

The ratio of the new term to Newton's is $3r_gu^2/(GM/L^2) = 3r_g\,p/r^2$, with $p = L^2/GM$:

where	$r$ (m)	$3r_gp/r^2$
perihelion	$4.600\times10^{10}$	$1.16\times10^{-7}$
mean	$5.791\times10^{10}$	$7.33\times10^{-8}$
aphelion	$6.982\times10^{10}$	$5.04\times10^{-8}$

A part in ten million. Section 3 shows why it is still visible.

The force reading, derived. Set $-f(1/u)/(L^2u^2) = GM/L^2 + 3r_gu^2$; then $f = -GMu^2 - 3r_gL^2u^4 = -GM/r^2 - 3GML^2/(c^2r^4)$.

Why "proportional to $L^2$" matters. A light ray has large $L$ relative to its energy scale, and the same term does all the bending in section 4. A planet moving slowly has small $L^2/c^2$ relative to $r^2$, and the term is tiny. The $u^2$-term therefore measures how fast things are moving sideways compared with light, as well as how deep they are in the potential: the ratio $3r_gp/r^2$ is of order $3(v/c)^2$ for a near-circular orbit, since $GM/r \approx v^2$.

The numbers. $r_g = 1476.625$ m, $p = 5.54604\times10^{10}$ m. At perihelion $r = a(1 - e) = 4.60012\times10^{10}$ m, ratio $3\times1476.625\times5.54604\times10^{10}/(4.60012\times10^{10})^2 = 1.161\times10^{-7}$. At $r = a$: $7.33\times10^{-8}$. At aphelion $a(1 + e) = 6.98169\times10^{10}$ m: $5.04\times10^{-8}$.

Bertrand's theorem. Unit 08 cited it: the only central forces all of whose bounded orbits close are the inverse square and the linear spring. The force here is neither, so bounded orbits do not close — the qualitative content of section 3 before a single number is computed.

section 2

What the new term does close in

Circular orbits that stop being stable, a plunge, and a sphere of light.

37 · the potential

Circular orbits come in pairs

Write $\ell = L/c$, a length. Then $V_{\rm eff} = c^2\big(-r_g/r + \ell^2/2r^2 - r_g\ell^2/r^3\big)$ and

$$V_{\rm eff}'(r) = \frac{c^2}{r^4}\big(r_g\,r^2 - \ell^2\,r + 3r_g\ell^2\big).$$

A circular orbit is a zero of the quadratic:

$$r_\pm = \frac{\ell^2 \pm \sqrt{\ell^4 - 12\,r_g^2\ell^2}}{2r_g}.$$

Newton's $V_{\rm eff}'$ has the bracket $r_gr - \ell^2$: one root, always a minimum.

circular orbits of Schwarzschild

(i) Circular orbits exist iff $\ell^2 \ge 12r_g^2$.
(ii) $r_+$ is a minimum of $V_{\rm eff}$ (stable), $r_-$ a maximum (unstable).
(iii) As $\ell$ runs from $\sqrt{12}\,r_g$ to $\infty$, $r_+$ increases from $6r_g$ to $\infty$ and $r_-$ decreases from $6r_g$ to $3r_g$.

full proof in the notes (N)

The next slide plots both roots against $\ell^2$.

Full proof

(i) The quadratic $r_gr^2 - \ell^2r + 3r_g\ell^2$ has real roots iff its discriminant $\ell^4 - 12r_g^2\ell^2 = \ell^2(\ell^2 - 12r_g^2)$ is non-negative. Both roots are positive (their sum $\ell^2/r_g$ and product $3\ell^2$ are), and both exceed $3r_g$, as (iii) shows, so they lie outside the horizon $2r_g$.

(ii) $V_{\rm eff}' = (c^2/r^4)\,q(r)$ with $q$ an upward parabola. For $r$ between the roots $q \lt 0$, outside them $q \gt 0$. So $V_{\rm eff}$ increases, decreases, increases: $r_-$ is a local maximum and $r_+$ a local minimum. When $\ell^2 = 12r_g^2$ the roots merge into a double root, $q \ge 0$ everywhere, and $V_{\rm eff}$ is monotone with an inflection at $6r_g$.

(iii) Put $s = \ell^2/r_g^2 \ge 12$. Then $r_\pm/r_g = \tfrac12\big(s \pm \sqrt{s^2 - 12s}\big)$. At $s = 12$ both equal 6. $r_+/r_g$ is a sum of increasing functions of $s$, so increases to $\infty$. For $r_-$: $r_-/r_g = \tfrac12\big(s - \sqrt{s^2 - 12s}\big) = \frac{6s}{s + \sqrt{s^2 - 12s}} = \frac{6}{1 + \sqrt{1 - 12/s}}$ (rationalise), which decreases from 6 at $s = 12$ towards 3 as $s \to \infty$. Every $r \gt 6r_g$ is some $r_+$ and every $r \in (3r_g, 6r_g)$ some $r_-$, by continuity and monotonicity. $\square$

37 · the potential

The two branches, plotted

Both roots $r_\pm$, in units of $r_g$, against $\ell^2/r_g^2$. They are born together at the ISCO and part:

$r_+$, stable, climbs towards Newton's line. Large $\ell$: $r_+ \approx \ell^2/r_g = L^2/GM$, unit 08's circular radius ✓.

$r_-$, unstable, falls towards $3r_g$, a radius Newton does not have: the photon sphere, later in this section, reached only as $\ell \to \infty$, that is, at the speed of light.

Left of $\ell^2 = 12r_g^2$ the band is empty: there is no circular orbit at all.

Computed from the formula for $r_\pm$. The two circles at $\ell = 4r_g$, radii 4 and 12, are the peak and the well of the next slide's curve. Shaded: no circular orbit (left), inside the horizon (bottom).

What to look for. The two branches meet at a vertical tangent at $\ell^2 = 12r_g^2$: that is the double root of $r_gr^2 - \ell^2r + 3r_g\ell^2$, and it is the whole content of the ISCO. Reading the plot sideways, a horizontal line at radius $r$ meets the stable branch once if $r \gt 6r_g$, the unstable branch once if $3r_g \lt r \lt 6r_g$, and nothing if $r \le 3r_g$ — which is part (iii) of the theorem, drawn.

The large-$\ell$ limit. $\sqrt{\ell^4 - 12r_g^2\ell^2} = \ell^2(1 - 6r_g^2/\ell^2 + \dots)$, so $r_+ = \ell^2/r_g - 3r_g + \dots$ and $r_- = 3r_g + \dots$ With $\ell^2/r_g = L^2/(c^2r_g) = L^2/GM$ ✓. The offset $-3r_g$ is visible in the figure: the stable branch runs parallel to Newton's line, a little below it.

Why $r_- \to 3r_g$ needs the speed of light. $L$ is per unit rest mass, and an orbit with $\ell \to \infty$ at fixed radius is one whose speed tends to $c$. Light itself orbits at exactly $3r_g$, as the photon-sphere slide shows from its own equation.

37 · the potential

Six gravitational radii, and the plunge

$V_{\rm eff}/c^2$ against $r/r_g$, computed. Dashed: Newton, whose barrier rises for ever. Red: general relativity with the same $\ell$ — the barrier has a top, at $r_-$, and beyond it the potential falls away. Blue: the critical $\ell$, whose well has closed into an inflection.

the innermost stable circular orbit

No stable circular orbit has $r \lt 6r_g = 3r_s$. At the ISCO, $\ell^2 = 12r_g^2$.

the plunge

Newton's barrier $\ell^2/2r^2$ beats $-r_g/r$ as $r \to 0$ for any $\ell \ne 0$: nothing with angular momentum reaches the centre. Here $-r_g\ell^2/r^3$ beats both. A particle with $\ell^2 \lt 12r_g^2$, or with energy above the barrier's top, falls in.

The figure. Units $r_g = c = 1$. Newton: $-1/r + 8/r^2$ ($\ell = 4$), minimum at $r = 16$. General relativity with the same $\ell$: $-1/r + 8/r^2 - 16/r^3$; the bracket $r^2 - 16r + 48 = (r - 4)(r - 12)$ gives the stable orbit at 12 and the unstable one at 4, where $V_{\rm eff} = -\tfrac14 + \tfrac12 - \tfrac14 = 0$ exactly. The critical curve has $\ell^2 = 12$: $-1/r + 6/r^2 - 12/r^3$, with a double root of the bracket at 6 and value $-1/18 = -0.0556$ there.

Why the plunge has no Newtonian analogue. In Newton, the centrifugal barrier is $L^2/2r^2$ and gravity is $-GM/r$: at small $r$ the barrier wins, always. Only exactly radial motion ($L = 0$) reaches the centre. In Schwarzschild the extra term $-GML^2/c^2r^3$ is also proportional to $L^2$ and wins at small $r$, so more angular momentum helps the fall at small radius. Whether a particle survives is decided by whether its energy is below the top of the barrier, at $r_-$.

Where the ISCO matters. A thin accretion disc is a sequence of nearly circular orbits; gas spirals slowly inwards, radiating, until it reaches $6r_g$ and then plunges in a few orbits, radiating little more. The energy radiated is therefore the binding energy at the ISCO: $1 - \sqrt{8/9} = 5.72\%$ of the rest mass. The problem sheet's P3 derives it and compares it with what accreting black holes are observed to radiate.

Sizes. $6r_g = 8.86$ km for one solar mass: far inside the Sun, which is why none of this touches the solar system.

37 · the potential

Nearly circular orbits precess, exactly

Put $u = u_c + \delta$ in the orbit equation, $u_c = 1/r_c$ a circular orbit, to first order in $\delta$:

$$\delta'' + \big(1 - 6r_g u_c\big)\,\delta = 0 .$$

So $r$ oscillates with frequency, in $\varphi$,

the apsidal frequency $$\nu = \sqrt{1 - \frac{6r_g}{r_c}},$$ $$\Delta\varphi_{\rm peri} = 2\pi\Big(\frac1\nu - 1\Big)\ \text{per orbit.}$$

three limits, one formula

$r_c \gg r_g$: $\Delta\varphi \approx 6\pi r_g/r_c$, section 3's answer at $e = 0$.
$r_c \to 6r_g$: $\nu \to 0$: at the ISCO the orbit stops returning to perihelion.
$r_c \lt 6r_g$: $\nu^2 \lt 0$: unstable.

The next slide plots the exact advance against its first-order approximation.

The linearisation. $(u_c + \delta)'' + u_c + \delta = GM/L^2 + 3r_g(u_c^2 + 2u_c\delta + \delta^2)$. The zeroth-order terms cancel by the circular-orbit condition, and $\delta^2$ is dropped: $\delta'' + \delta = 6r_gu_c\delta$. This is unit 02's linearisation about an equilibrium, and unit 08's "which power laws close" problem done for this force: there the frequency was $\sqrt{3 + n}$ for $f \propto r^n$.

From frequency to precession. $\delta = A\cos(\nu\varphi)$, so successive perihelia (maxima of $u$) are $\Delta\varphi = 2\pi/\nu$ apart, not $2\pi$. The excess is $2\pi(1/\nu - 1)$. For small $x = 6r_g/r_c$: $1/\sqrt{1 - x} = 1 + x/2 + \dots$, giving $\pi x = 6\pi r_g/r_c$.

Why this is worth having beside section 3's calculation: section 3 is first order in $r_g$ and all orders in $e$; this is all orders in $r_g$ and first order in $e$. They overlap in the corner $r_g \to 0$, $e \to 0$, and agree there. Two expansions agreeing where both are valid is the check unit 21 taught.

37 · the potential

The advance, at every radius

$\Delta\varphi_{\rm peri} = 2\pi(1/\nu - 1)$ with $\nu = \sqrt{1 - 6r_g/r_c}$, against its first-order form $6\pi r_g/r_c$.

At $r_c = 10r_g$ the perihelion advances $209^\circ$ an orbit, nearly twice the first-order $108^\circ$. At $100r_g$ it is $11.3^\circ$ against $10.8^\circ$. At Mercury, $r/r_g \approx 4\times10^7$, the two agree to a part in $10^7$.

the ISCO, seen from the orbit

As $r_c \to 6r_g$ the advance per orbit grows without bound: the particle circles many times between one perihelion and the next. The first-order curve sees nothing there.

Computed. The two curves converge as $r_c$ grows, with a gap of relative size about $r_g/r_c$; they part completely near the ISCO, where only the exact formula knows that $\nu \to 0$.

The numbers. $r_c = 10r_g$: $\nu = \sqrt{0.4} = 0.63246$, $2\pi(1/\nu - 1) = 3.6514$ rad $= 209.2^\circ$, against $6\pi/10 = 1.8850$ rad $= 108.0^\circ$. $r_c = 100r_g$: $\nu = \sqrt{0.94} = 0.96954$, excess $0.19743$ rad $= 11.31^\circ$, against $6\pi/100 = 0.18850$ rad $= 10.80^\circ$: the exact value is $4.7\%$ above the first-order one, a relative error of order $r_g/r$ as it should be. For Mercury the relative error is $\sim 10^{-7}$.

Reading the divergence. $1/\nu - 1 \to \infty$ as $\nu \to 0$: the radial oscillation's period in $\varphi$, $2\pi/\nu$, becomes infinite, so the orbit winds round many times per radial oscillation. At exactly $6r_g$ the linearised equation $\delta'' = 0$ has the solution $\delta = A\varphi$: a small displacement grows linearly instead of oscillating, the marginal case between the oscillation outside and the exponential growth inside.

37 · the potential

Light has a potential too: the photon sphere

For light ($\sigma = 0$), divide the radial equation by $L^2$:

$$u'^2 = \frac1{b^2} - W(u),$$ $$W(u) = u^2\big(1 - 2r_gu\big),$$

$b = cL/E$ is the impact parameter, the distance of the incoming line. Rescaling the affine parameter rescales $E$ and $L$ together, so $b$ is the only thing about a ray that either of them says. $W$ peaks at $u = 1/3r_g$, where $W = 1/27r_g^2$.

the photon sphere

Light can orbit — unstably — at $r = 3r_g$. A ray from infinity with $b \lt b_c = 3\sqrt3\,r_g$ is captured; with $b \gt b_c$ it escapes.

$W$ against $r = 1/u$, computed, with three rays' levels $1/b^2$. A ray moves where its level is above $W$, and turns where they meet. Above the peak, at $1/27r_g^2$, nothing stops it.

The null radial equation. With $\sigma = 0$, $\dot r^2 = E^2/c^2 - (1 - 2r_g/r)L^2/r^2$. Divide by $L^2$ and use $\dot r = -Lu'$ from worked example 1: $u'^2 = E^2/(c^2L^2) - u^2(1 - 2r_gu)$, and $E^2/(c^2L^2) = 1/b^2$. Far away, $u \to 0$ and $u'^2 \to 1/b^2$; the straight line $u = \sin\varphi/b$ of flat space has exactly this, so $b$ is the perpendicular distance of the incoming line.

Capture, proved. $u'^2 \ge 0$ forces $W(u) \le 1/b^2$ along the ray. $W$ rises from 0 at $u = 0$ to its maximum $1/27r_g^2$ at $u = 1/3r_g$ and falls to zero at the horizon $u = 1/2r_g$. If $1/b^2 \gt 1/27r_g^2$, i.e. $b \lt 3\sqrt3\,r_g$, no turning point exists on the way in and the ray crosses $r = 3r_g$ and reaches the horizon. If $b \gt 3\sqrt3\,r_g$ it turns at the first root of $W(u) = 1/b^2$, which is at $r \gt 3r_g$, and escapes. The circular orbit at the maximum of $W$ is unstable by the same argument as for $r_-$. Note $b_c \ne 3r_g$: the ray that just grazes the photon sphere came in from further out, because it was bent on the way.

What to look for in the figure. The equation $u'^2 = 1/b^2 - W(u)$ is a one-dimensional energy equation, read exactly like unit 08's effective potential: the level $1/b^2$ plays the energy, $W$ the potential. Coming in from $r = \infty$, where $W = 0$, the ray moves left along its level. The blue ray, $b = 7r_g$, meets $W$ at $r = 5.62r_g$ and goes back out. The red one, $b = 4r_g$, is above the peak everywhere and reaches the horizon. The dashed critical level touches the peak: a ray with exactly $b = b_c$ spirals towards the photon sphere for ever.

37 · the potential

The shadow, and the ring the EHT saw

Rays traced from the null orbit equation, $u'' + u = 3r_gu^2$, at eight impact parameters. Red: $b \lt b_c$ (shaded band), captured. Blue: $b \gt b_c$, bent and released — the closer to $b_c$, the further round the hole. Dashed circle: the photon sphere. To scale.

the ring in the Event Horizon Telescope image

Captured rays leave a dark disc of angular diameter $2b_c/D$, the shadow. M87*: $M = 6.5\times10^9M_\odot$, $D = 16.8$ Mpc, so $r_g = 9.598\times10^{12}$ m and $b_c = 5.196\,r_g = 4.987\times10^{13}$ m:

$$\frac{2b_c}{D} = \frac{9.975\times10^{13}}{5.184\times10^{23}} = 1.924\times10^{-10}\ \text{rad},$$

which is $39.7\ \mu$as.

EHT, 2019: a ring of $42 \pm 3\ \mu$as.

What the figure shows. A distant observer looking at the hole receives, from each direction within angular radius $b_c/D$ of its centre, only rays that would have come from the horizon — none, so that disc is dark. That is the shadow: its size is set by $b_c = 3\sqrt3\,r_g = 5.196\,r_g$, not by the horizon's $2r_g$, because the photon sphere bends rays inwards that a straight-line picture would have let pass. The rays just outside $b_c$ are the ones that wind partly round the hole before escaping, and they pile up at the shadow's edge: that is the bright ring.

How the rays were drawn. Each starts at $r = 40r_g$ on the line at height $b$, with $u'$ from the first integral $u'^2 = 1/b^2 - u^2 + 2r_gu^3$, and the second-order equation is integrated in $\varphi$ until the ray reaches the horizon or recedes. The shapes are exact Schwarzschild null geodesics, in the coordinates $(r, \varphi)$ drawn as if they were polar coordinates in a flat plane.

Honest caveats about the image. The EHT's bright ring is emission from hot gas lensed around the hole, and simulations put it slightly outside the shadow's edge; the mass $6.5\times10^9M_\odot$ was in fact inferred by the EHT from the ring with that calibration. The independent stellar-dynamics mass (Gebhardt et al. 2011, $6.6\times10^9$ at a comparable distance) agrees — that agreement, not the ring alone, is the test. Sgr A*, imaged in 2022, gave a ring of $51.8 \pm 2.3\ \mu$as for a mass known independently from the S2 orbit.

Arithmetic. $D = 16.8\times10^6\times3.0857\times10^{16}$ m $= 5.184\times10^{23}$ m. $1.924\times10^{-10}\times206\,264.806 = 3.969\times10^{-5}{}'' = 39.7\ \mu$as.

section 3

Mercury: the 43

A ledger with one line nobody could fill, and the secular term that fills it.

37 · Mercury

The ledger

The longitude of Mercury's perihelion, measured from the equinox, advances by about $5600''$ per century. Where it goes, in Clemence's 1947 accounting:

cause	$''$ per century
observed	$5599.7 \pm 0.4$
general precession (the equinox moves)	$5025.6$
the other planets	$531.5$
left over	$42.6 \pm 0.9$

The modern value of the anomalous perihelion advance, from radar ranging and spacecraft: $42.98 \pm 0.04''$.

5025 of it is not Mercury at all

The reference direction, the equinox, is where the Earth's equator meets its orbit. The Sun and Moon pull on the Earth's equatorial bulge — unit 10's $J_2$ — and the axis precesses, once in about 25 800 years. Every longitude drifts with it.

1859

Le Verrier, who had found Neptune from the anomalies of Uranus in 1846, found $38''$ unexplained and proposed an inner planet, Vulcan. It was searched for at eclipses for decades and never found. Newcomb (1882) revised the residual to $43''$.

The table, and what it cites. G. M. Clemence, Rev. Mod. Phys. 19, 361 (1947), whose itemisation is: observed $5599.74 \pm 0.41$; general precession $5025.645 \pm 0.50$ (Newcomb's constant); Venus $277.856$, Earth $90.038$, Mars $2.536$, Jupiter $153.584$, Saturn $7.302$, Uranus $0.141$, Neptune $0.042$, and solar oblateness $0.010$; residual $42.56 \pm 0.94$. The planets sum to $531.50$. The next slide computes most of that by unit 10's method. The modern value $42.98 \pm 0.04''$ comes from fitting planetary ephemerides to radar and spacecraft ranging, as reviewed by C. M. Will, Living Reviews in Relativity 17, 4 (2014).

The precession of the equinoxes. $1\,296\,000''/50.29''$ per year $\approx 25\,770$ years for one circuit. It is a coordinate effect in the sense of unit 30: it would vanish if longitudes were measured from a fixed direction in space. It is subtracted, not explained away.

Why the planets' $531''$ can be trusted to the needed accuracy. The masses of Venus and the Earth dominate it, and both were well known by 1900 from their own mutual perturbations. An error of $43''$ in $531''$ would need an 8% error in those masses. None of the Newtonian explanations — Vulcan, a ring of dust, a slightly oblate Sun, an exponent very slightly above 2 in the force law (Hall, 1894) — survived: each either failed a search or disturbed some other orbit that was observed not to be disturbed.

37 · Mercury

Where the 531 comes from: unit 10's averaging

Unit 10's recipe: average the perturbation over the fast angles, and keep what does not cancel. Averaged over its orbit, an outer planet of mass $m'$ at radius $a'$ is a ring. Its potential in the plane, from unit 10's expansion with $\langle P_2(\cos\psi)\rangle = \tfrac14$:

$$\Phi_{\rm ring}(r) = -\frac{Gm'}{a'}\Big(1 + \frac{r^2}{4a'^2} + \dots\Big).$$

A small outward force $Gm'r/2a'^3$. Put it into Binet and linearise as on "Nearly circular orbits precess, exactly":

$$\Delta\varphi_{\rm peri} \approx \frac{3\pi}{2}\,\frac{m'}{M}\Big(\frac{a}{a'}\Big)^3\ \text{per orbit.}$$

planet	ring, all orders in $a/a'$	Clemence
Venus	286.3	277.9
Earth	96.6	90.0
Mars	2.4	2.5
Jupiter	160.3	153.6
Saturn	7.7	7.3
Uranus, Neptune	0.2	0.2
total	553.5	531.5

The ring model ignores Mercury's eccentricity and the other planets' — it is 4% high, and 4% of 531 is half of 43. Clemence's full secular theory is what makes the residual trustworthy.

The ring potential. Unit 10's expansion $1/|\vec r - \vec r\,{}'| = \sum_n r^n P_n(\cos\psi)/a'^{n+1}$ for $r \lt a'$, averaged over a uniform ring in the plane of the orbit (so $\psi$ is uniform on $[0, 2\pi)$): $\langle P_0\rangle = 1$, $\langle P_1\rangle = \langle\cos\psi\rangle = 0$, $\langle P_2\rangle = \tfrac32\langle\cos^2\psi\rangle - \tfrac12 = \tfrac34 - \tfrac12 = \tfrac14$.

The precession, in full. The ring adds $f = +\kappa r$, $\kappa = Gm'/2a'^3$, to the Sun's $-GM/r^2$. Binet: $u'' + u = GM/L^2 - \kappa/(L^2u^3)$. About a circular orbit $u_c$: $\delta'' + \delta = (3\kappa/L^2u_c^4)\delta$, and with $L^2 \approx GM/u_c$ the coefficient is $3\kappa r^3/GM = \tfrac32(m'/M)(r/a')^3 =: 3\eta$. So $\nu = \sqrt{1 - 3\eta}$ and the advance is $2\pi(1/\nu - 1) \approx 3\pi\eta$, as on the slide. For Jupiter: $m'/M = 1/1047.35$, $a/a' = 0.387098/5.20336 = 0.074395$, advance $0.3821''$ per orbit, $158.7''$ per century — 3% above Clemence's $153.6$. That is problem P8.

The table's first column keeps every multipole of the ring, not just the quadrupole. Summing them gives the classical Laplace–Lagrange rate for an inner orbit perturbed by an outer one, $\dot\varpi = \tfrac14 n\,(m'/M)\,\alpha^2 b^{(1)}_{3/2}(\alpha)$ with $\alpha = a/a'$ and the Laplace coefficient $b^{(1)}_{3/2}(\alpha) = \frac1\pi\int_0^{2\pi}\cos\psi\,(1 - 2\alpha\cos\psi + \alpha^2)^{-3/2}\,\mathrm d\psi$ (cited: Murray and Dermott, Solar System Dynamics, ch. 7; for small $\alpha$, $b^{(1)}_{3/2} \approx 3\alpha$ and it reduces to the formula on the slide). The column was computed from that integral numerically; for Venus, at $\alpha = 0.535$, the quadrupole alone would give only $151''$, which is why all orders are needed there.

What the full theory adds. Mercury's $e = 0.206$ is not small, and the other planets' eccentricities couple into Mercury's perihelion through off-diagonal terms. Clemence kept those; the ring model does not. The slide's honest conclusion is that unit 10's averaging gets the size and the ranking of the contributions right and leaves the last few per cent to the full secular theory.

37 · Mercury

Solving the orbit equation: Poincaré–Lindstedt

Unit 21's scaling, $p = L^2/GM$ and $w = pu$:

$$w'' + w = 1 + \varepsilon w^2,\qquad \varepsilon = \frac{3r_g}{p}.$$

Unit 21's method: give the radial oscillation its own frequency, $\psi = \nu\varphi$ with $\nu = 1 + \varepsilon\nu_1 + \dots$, and demand $w$ be $2\pi$-periodic in $\psi$.

Order 1: $w_0 = 1 + e\cos\psi$, unit 08's ellipse.

Order $\varepsilon$: $\ddot w_1 + w_1 = w_0^2 - 2\nu_1\ddot w_0$, whose $\cos\psi$ coefficient is $2e + 2\nu_1e$.

A periodic $w_1$ exists only if that resonant coefficient vanishes (unit 21's solvability condition). So $\nu_1 = -1$.

the perihelion advance $$\nu = 1 - \varepsilon,\qquad \Delta\varpi = \frac{2\pi}{\nu} - 2\pi \approx 2\pi\varepsilon$$ $$\Delta\varpi = \frac{6\pi GM}{c^2a(1 - e^2)}\ \text{per orbit}$$

The order-$\varepsilon$ equation in full. Substitute $w = w_0 + \varepsilon w_1$ and $\nu^2 = 1 + 2\varepsilon\nu_1 + O(\varepsilon^2)$; a dot here is $\mathrm d/\mathrm d\psi$. Order $\varepsilon$: $\ddot w_1 + 2\nu_1\ddot w_0 + w_1 = w_0^2$. Now $w_0^2 = 1 + \tfrac12e^2 + 2e\cos\psi + \tfrac12e^2\cos2\psi$ and $-2\nu_1\ddot w_0 = 2\nu_1e\cos\psi$. So the $\cos\psi$ coefficient on the right is $2e + 2\nu_1e$. The solvability condition (unit 21, "$u'' + u = f$ has a $2\pi$-periodic solution iff $f$ is orthogonal to $\cos$ and $\sin$") forces $\nu_1 = -1$ whenever $e \ne 0$.

Then $w_1 = 1 + \tfrac12e^2 - \tfrac16e^2\cos2\psi$ plus a homogeneous solution, which only relabels $e$. The full first-order orbit is $w = 1 + e\cos\big((1 - \varepsilon)\varphi\big) + \varepsilon\big(1 + \tfrac12e^2 - \tfrac16e^2\cos2\psi\big)$: an ellipse, very slightly fattened, whose major axis turns.

From $\nu$ to $\Delta\varpi$. Perihelion is a maximum of $w$, at $\psi = 2\pi k$, i.e. $\varphi = 2\pi k/(1 - \varepsilon)$. Consecutive perihelia are $2\pi/(1 - \varepsilon) = 2\pi(1 + \varepsilon + O(\varepsilon^2))$ apart. And $2\pi\varepsilon = 6\pi r_g/p$ with $p = a(1 - e^2)$ — which is unit 08's semi-latus rectum, $L^2/GM$, for the osculating ellipse.

Dimensions and limits. $r_g/p$ is a ratio of lengths ✓. $c \to \infty$: no advance ✓. $e \to 0$: $6\pi r_g/a$, matching the near-circular formula of section 2 ✓. Sign: $\nu \lt 1$, so the perihelion moves forward, in the direction of the orbit ✓ — an extra attraction at small $r$ makes the planet swing round a little further before climbing out.

37 · Mercury

The perihelion advance, drawn

The exact orbit equation, integrated from perihelion $r = p/(1 + e)$ with $p = 40r_g$, $e = 0.5$, over ten radial periods. Blue dots: the eleven perihelia, each $33.8^\circ$ past the last. Dashed: Newton's ellipse from the same start.

The radial oscillation has frequency $\nu \lt 1$ in $\varphi$, so $r$ returns to its minimum only after the orbit has turned through $2\pi/\nu \gt 2\pi$. Each perihelion is a fixed angle ahead of the last, and the orbit traces a rosette, not an ellipse.

the drawing, against the formula

Here $\varepsilon = 3r_g/p = 0.075$, a million times Mercury's. First order predicts $2\pi\varepsilon = 27.0^\circ$ per orbit; the integration gives $33.8^\circ$. The excess, 25%, is the $O(\varepsilon^2)$ that Poincaré–Lindstedt dropped — for Mercury, a few parts in $10^7$.

The orbit also swings in less far than Newton's: the extra attraction pulls the aphelion in from $80r_g$ to $58r_g$.

How it was drawn. In units of $r_g$, $u'' + u = 1/p + 3u^2$ with $p = 40$, from $u = 1.5/40$, $u' = 0$ at $\varphi = 0$, integrated by fourth-order Runge–Kutta in $\varphi$, and plotted as $(r\cos\varphi, r\sin\varphi)$. The perihelia are the maxima of $u$, located where $u'$ changes sign; the four measured gaps all equal $360^\circ + 33.79^\circ$, as they must for an autonomous equation — the advance is the same every orbit.

Why not draw Mercury. Its advance is $0.1035''$ per orbit; after a century of orbits the perihelion has turned $43''$, about a hundredth of a degree. Any honest drawing of Mercury is an ellipse. The exaggeration is in $\varepsilon$ only: the mechanism, a radial frequency slightly below the angular one, is the same.

Is this orbit still in the solar-system regime? No: $p = 40r_g$ is just a few times the ISCO, and the second-order terms are visible. That is the point of the comparison box — the first-order formula is an expansion in $\varepsilon$, and here one can see its first correction.

37 · Mercury

Why the secular term must not be removed

Expand naively instead, with $w_0 = 1 + e\cos\varphi$. The resonant forcing $2e\cos\varphi$ puts $e\,\varphi\sin\varphi$ into $w_1$: unit 21's secular term. And

$$\begin{aligned}&1 + e\cos\varphi + \varepsilon e\varphi\sin\varphi\\ &\quad= 1 + e\cos\big((1 - \varepsilon)\varphi\big) + O(\varepsilon^2\varphi^2).\end{aligned}$$

The secular term is the frequency shift, written as the first term of a Taylor series.

unit 21's warning, in its intended use

Deleting $\varepsilon e\varphi\sin\varphi$ "because secular terms are unphysical" predicts zero precession, and is inconsistent: at order $\varepsilon$ there is no $2\pi$-periodic solution when $e \ne 0$. The equation forbids the orbit to close after one turn. Poincaré–Lindstedt does not remove the term; it rewrites it as $\nu - 1$.

The identity. $\cos((1 - \varepsilon)\varphi) = \cos\varphi\cos\varepsilon\varphi + \sin\varphi\sin\varepsilon\varphi = \cos\varphi + \varepsilon\varphi\sin\varphi + O(\varepsilon^2\varphi^2)$. The naive series is the Taylor expansion in $\varepsilon$ at fixed $\varphi$ of a function whose $\varepsilon$-dependence is through $\varepsilon\varphi$; it is accurate while $\varepsilon\varphi$ is small and useless after.

Why this matters here more than for Duffing. In unit 21's Duffing oscillator the secular term encoded a frequency shift whose accumulated phase nobody measures. Here the accumulated phase is the observable: astronomers have measured the longitude of Mercury's perihelion for three centuries. The distinction is between removing a non-uniformity, which is good mathematics, and removing a physical effect, which is an error. This is the clause in the checkpoint.

The numbers. $\varepsilon = 7.9875\times10^{-8}$, so $1/\varepsilon = 1.252\times10^7$ rad, that is $1.99\times10^6$ orbits of $87.969$ days, $4.80\times10^5$ years. One century is $415.2$ orbits, $\varphi = 2609$ rad, $\varepsilon\varphi = 2.08\times10^{-4}$. The neglected term, $\tfrac12 e(\varepsilon\varphi)^2\cos\varphi$, is at most $4.5\times10^{-9}$ in $w$: a part in $10^4$ of the secular term, whose ratio to it is $\tfrac12\varepsilon\varphi$.

37 · Mercury

The secular term, plotted

The same exaggerated $\varepsilon = 0.075$, over eight orbits.

Red, Poincaré–Lindstedt: $\cos\big((1 - \varepsilon)\varphi\big)$. Its amplitude never changes; it slips behind $\cos\varphi$ by $2\pi\varepsilon$ each orbit. That slip is the precession.

Blue, naive: $\cos\varphi + \varepsilon\varphi\sin\varphi$. It agrees while $\varepsilon\varphi \ll 1$, then its amplitude grows like $\sqrt{1 + (\varepsilon\varphi)^2}$ — an orbit that swells without limit, which no solution does.

and here the naive form is enough

It fails when $\varepsilon\varphi \sim 1$: for Mercury, after $2.0$ million orbits. Over a century $\varepsilon\varphi = 2.1\times10^{-4}$.

$(w - 1)/e$ against $\varphi/2\pi$, computed. Dotted: the naive curve's envelope, $\pm\sqrt{1 + (\varepsilon\varphi)^2}$. Up to the line $\varepsilon\varphi = 1$ the two curves are hard to tell apart; beyond it they share nothing but the zeros of $\sin$.

What to look for. In the first orbit the red and blue curves coincide to the eye: the secular term $\varepsilon\varphi\sin\varphi$ is exactly the first-order effect of the phase slip, and there it is small. By $\varepsilon\varphi = 1$ the naive curve has the right phase drift but an amplitude 41% too large, and after that it is wrong in kind. Deleting the secular term would leave plain $\cos\varphi$, which is right in amplitude and wrong in the one thing measured — the phase.

The envelope. $\cos\varphi + x\sin\varphi$ with $x = \varepsilon\varphi$ treated as slowly varying has amplitude $\sqrt{1 + x^2}$, while $\cos((1 - \varepsilon)\varphi)$ has amplitude 1. They agree to order $x$ and disagree at order $x^2$ — the $O(\varepsilon^2\varphi^2)$ of the identity on the previous slide.

37 · Mercury · example

Worked example 2 — Mercury, end to end

from the metric to arcseconds, for Mercury

Step 1. Worked example 1: $u'' + u = GM/L^2 + 3r_gu^2$.

Step 2 — the scale. $r_g = GM_\odot/c^2 = 1.32712\times10^{20}/8.98755\times10^{16} = 1476.6$ m.

Step 3 — the orbit. $a = 5.7909\times10^{10}$ m, $e = 0.20563$, $1 - e^2 = 0.95772$, so $p = 5.5460\times10^{10}$ m.

Step 4 — the small parameter. $\varepsilon = 3r_g/p = 4429.9/5.5460\times10^{10} = 7.9875\times10^{-8}$.

Step 5 — per orbit. $2\pi\varepsilon = 5.0187\times10^{-7}$ rad $\times\ 206\,264.8 = 0.10352''$.

Step 6 — per century. $36\,525/87.9691 = 415.20$ orbits a century:

$$\Delta\varpi = 415.20\times0.10352'' = \mathbf{42.98''}.$$

Step 7 — compare. Observed $42.98 \pm 0.04''$, with no free parameter used.

Step 8 — check. Dimensionless ✓; $c \to \infty$ gives 0 ✓; rounded inputs $a = 5.79\times10^{10}$, $e = 0.2056$ give $42.99''$ ✓.

Step 3. $e^2 = 0.0422837$; $p = 5.790905\times10^{10}\times0.9577163 = 5.546044\times10^{10}$ m.

Step 4. $3r_g = 4429.875$ m. Divided by $p$: $7.98745\times10^{-8}$.

Step 5. $2\pi\times7.98745\times10^{-8} = 5.01866\times10^{-7}$ rad; times $206\,264.806$ gives $0.1035173''$.

Step 6. The sidereal period is $87.9691$ days; a Julian century is $36\,525$ days; $36\,525/87.9691 = 415.2026$. Then $415.2026\times0.1035173 = 42.9807''$.

What "no free parameter" means, precisely. Einstein's field equations contain one constant, fixed in unit 35 by the Newtonian limit. The Schwarzschild solution contains one constant, fixed in unit 36 by matching to $-GM/r$ far away. Neither was chosen with Mercury in mind, and nothing in this calculation could be adjusted to make $43$ come out. This is why the result, announced to the Prussian Academy on 18 November 1915, carried such weight: a prediction that could have come out as $40$ or $86$ and did not. Einstein wrote to Ehrenfest that for some days he was beside himself with joy.

An uncertainty to be honest about. The comparison is really of the full model, including the Sun's quadrupole moment $J_{2\odot} \approx 2\times10^{-7}$, which contributes about $0.03''$ per century. That is comparable to the observational uncertainty and is part of why the error bar is $0.04''$ and not smaller.

37 · Mercury

The same formula, far from the Sun

$\Delta\varpi = 6\pi r_g/a(1 - e^2)$ per orbit applies to any orbit with $\varepsilon = 3r_g/p \ll 1$.

orbit	$\varepsilon$	advance
Mercury	$8.0\times10^{-8}$	$42.98''$/century
Venus	$4.1\times10^{-8}$	$8.62''$/century
Earth	$3.0\times10^{-8}$	$3.84''$/century
S2 around Sgr A*	$5.6\times10^{-4}$	$12.2'$ per orbit

S2, the star closest to our Galaxy's black hole

$M = 4.26\times10^6M_\odot$, $a = 1031$ au, $e = 0.8847$, period $16.05$ yr. Pericentre $119$ au $\approx 1400$ Schwarzschild radii. First order in $\varepsilon$ is still excellent.

GRAVITY (2020) measured the precession as $f = 1.10 \pm 0.19$ times the prediction.

Double pulsars go further: PSR B1913+16's periastron advances $4.2266^\circ$ a year, used not to test the formula but, trusting it, to weigh the two neutron stars.

The table. Venus: $a = 0.723332$ au, $e = 0.006773$, period $224.701$ d, giving $0.05306''$ per orbit and $8.62''$ per century. Earth: $a = 1.000001$ au, $e = 0.016709$, $365.256$ d, giving $0.03839''$ and $3.84''$. Mercury wins because $\varepsilon$ is largest and it goes round most often: the advance per century scales as $a^{-5/2}$ for near-circular orbits.

S2. $r_g = 4.26\times10^6\times1476.6$ m $= 6.29\times10^9$ m; $p = 1031.3\times1.49598\times10^{11}\times(1 - 0.8847^2) = 3.36\times10^{13}$ m; $\varepsilon = 3r_g/p = 5.63\times10^{-4}$; $2\pi\varepsilon = 3.54\times10^{-3}$ rad $= 12.2'$ per orbit. GRAVITY Collaboration, Astron. Astrophys. 636, L5 (2020), where the prediction is quoted as $12.1'$ with their best-fit orbit. The same star's gravitational redshift at pericentre was detected in 2018; problem P11 computes both effects.

PSR B1913+16. Periastron advance $4.226598 \pm 0.000005$ degrees per year (Weisberg, Nice and Taylor 2010). Unit 21's problem sheet computed it to first order. When a formula is trusted enough to be used as a scale, it has stopped being a test and become an instrument.

section 4

Light: the factor of two

Bend the ray as a Newtonian particle, bend it as a null geodesic, and find which metric component supplies the difference.

37 · light

Bending by general relativity

The null orbit equation (worked example 1, $\sigma = 0$):

$$u'' + u = 3r_g\,u^2 .$$

No $GM/L^2$ term: light has no rest mass for Newton's $-GM/r$ to act on. All the bending is in the new term.

Order 0: $u_0 = \sin\varphi/b$, a line at distance $b$, coming in at $\varphi = 0$, leaving at $\varphi = \pi$.

Order $r_g$: $u_1'' + u_1 = 3r_g\sin^2\varphi/b^2 = \tfrac{3r_g}{2b^2}(1 - \cos2\varphi)$, solved by

$$u_1 = \frac{r_g}{b^2}\big(1 + \cos^2\varphi\big).$$

The ray $u_0 + u_1$ drawn with $r_g/b = 0.12$, fifty thousand times the Sun's, so the bend can be seen. Far away it runs along two straight lines, dashed: the asymptotes, each at $\delta$ to the undeflected line $u_0$. The next slide reads $\delta$ off $u_1$.

Solving for $u_1$. The constant part $3r_g/2b^2$ is solved by the constant $3r_g/2b^2$. The part $-\tfrac{3r_g}{2b^2}\cos2\varphi$ is solved by $C\cos2\varphi$ with $C(1 - 4) = -\tfrac{3r_g}{2b^2}$, so $C = r_g/2b^2$. Together: $\tfrac{r_g}{2b^2}(3 + \cos2\varphi) = \tfrac{r_g}{b^2}(1 + \cos^2\varphi)$. Check: $\big(\cos^2\varphi\big)'' = -2\cos2\varphi$, so $u_1'' + u_1 = \tfrac{r_g}{b^2}(1 + \cos^2\varphi - 2\cos2\varphi) = \tfrac{r_g}{b^2}(1 + \tfrac12 + \tfrac12\cos2\varphi - 2\cos2\varphi) = \tfrac{3r_g}{2b^2}(1 - \cos2\varphi)$ ✓.

Why no homogeneous piece is added. A homogeneous solution $A\cos\varphi + B\sin\varphi$ would either break the symmetry about closest approach at $\varphi = \pi/2$ ($A$) or change the slope at infinity ($B$). The ray is symmetric about its point of closest approach — the metric is time-reversal invariant and spherically symmetric — so $A = 0$. And the first integral $u'^2 = 1/b^2 - u^2 + 2r_gu^3$ fixes $u' = 1/b$ where $u = 0$, which $u_0$ already has and $B\sin\varphi$ would spoil at order $r_g$, so $B = 0$. The distance of closest approach, at $\varphi = \pi/2$, is $1/u = b - r_g + O(r_g^2/b)$: it differs from $b$ by $r_g$, which moves the deflection only at second order.

What to look for. The order-0 ray $u_0 = \sin\varphi/b$ is the horizontal line at height $b$; $u_1$ is everywhere positive, so it raises $u$, lowers $r$ and pulls the ray towards the mass all along. Where $u_0$ alone would put the ends of the ray at $\varphi = 0$ and $\pi$, the corrected $u$ vanishes a little beyond them, at $\varphi = -\delta$ and $\pi + \delta$ — which is what turns the ends.

37 · light

Reading the angle off the asymptotes

The asymptotes. $u = 0$ far away. Near $\varphi = 0$ put $\varphi = -\delta$:

$$0 = -\frac{\delta}{b} + \frac{2r_g}{b^2}\ \Rightarrow\ \delta = \frac{2r_g}{b},$$

and the same at the outgoing end, by symmetry about $\varphi = \pi/2$.

light deflection $$\Delta\theta_{\rm GR} = 2\delta = \frac{4GM}{c^2b}$$

checks

Dimensionless ✓. $b \to \infty$ or $M \to 0$: no bending ✓. Independent of the light's frequency, as the equivalence principle demands ✓.

On the figure of the previous slide, $\delta$ is the angle each dashed asymptote makes with the horizontal, and $\Delta\theta = 2\delta$ the angle between them, measured where they cross above the mass.

For the ray at the limb, $\delta = 2r_g/R_\odot = 0.876''$ at each end.

37 · light

Bending by Newton

Treat light as a particle that moves at $c$ far away, attracted by $-GM/r$ like anything else — Soldner's calculation of 1801, and Cavendish's before him. Unit 08's Binet equation with $L = bc$:

$$u'' + u = \frac{GM}{b^2c^2} = \frac{r_g}{b^2} .$$

Exact solution through the same order-0 line:

$$u = \frac{\sin\varphi}{b} + \frac{r_g}{b^2}.$$

That is a hyperbola, unit 08's unbound orbit. Asymptote: $-\delta/b + r_g/b^2 = 0$, $\delta = r_g/b$.

the Newtonian corpuscle $$\Delta\theta_{\rm N} = \frac{2GM}{c^2b} = \tfrac12\,\Delta\theta_{\rm GR}$$

To first order in $r_g/b$; the exact hyperbola gives $2\arcsin\big(1/\sqrt{1 + b^2/r_g^2}\big)$.

the same as unit 30

Unit 30 got $2GM/c^2b$ from the equivalence principle alone, with no corpuscle. Both see only $g_{tt}$. The next slide proves that is exactly what they have in common.

The exact Newtonian deflection. Unit 08's hyperbola has eccentricity $e = \sqrt{1 + 2E_NL^2/(GM)^2}$ with $E_N = \tfrac12c^2$ and $L = bc$, so $e = \sqrt{1 + b^2c^4/(GM)^2} = \sqrt{1 + b^2/r_g^2}$; the deflection is $2\arcsin(1/e)$ (unit 08's Rutherford problem). For $b \gg r_g$ that is $2r_g/b$. At the Sun's limb $b/r_g = 4.71\times10^5$, and the first-order formula is accurate to about a part in $10^{12}$.

Why "$L = bc$". For a Newtonian particle $L = |\vec r\times\vec v|$, which far away is $b$ times the speed there, $c$.

History. Johann Georg von Soldner published $0.84''$ in 1801 (with the constants of his day); Cavendish had done the calculation in an unpublished note around 1784. Einstein's 1911 paper, from the equivalence principle, got the same half-value, $0.83''$, and proposed an eclipse measurement. The First World War prevented the 1914 expedition. Had it gone ahead and succeeded it would have refuted the 1911 theory, and by 1915 Einstein had the factor of two.

37 · light

Which metric component supplies the other half

Take any static spherical metric $-A(r)c^2\mathrm dt^2 + B(r)\mathrm dr^2 + r^2\mathrm d\Omega^2$, where $\mathrm d\Omega^2 = \mathrm d\theta^2 + \sin^2\theta\,\mathrm d\varphi^2$, and keep first order:

$$A = 1 - \frac{2r_g}{r},\qquad B = 1 + \gamma\,\frac{2r_g}{r}.$$

$A$ is fixed by the Newtonian limit (unit 32). $\gamma$ labels the spatial part: Schwarzschild has $\gamma = 1$, flat space $\gamma = 0$.

The null radial equation is now $B\,u'^2 = 1/(Ab^2) - u^2$. Expand, differentiate as in worked example 1:

$$u'' + u = \frac{(1 - \gamma)\,r_g}{b^2} + 3\gamma\,r_gu^2 .$$

deflection, component by component $$\Delta\theta = \underbrace{\frac{2r_g}{b}}_{g_{tt}} + \underbrace{\gamma\,\frac{2r_g}{b}}_{g_{rr}} = \frac{1 + \gamma}{2}\cdot\frac{4GM}{c^2b}$$

$\gamma = 0$ is exactly Newton's equation of the previous slide. $\gamma = 1$ is exactly general relativity's. The missing half is $g_{rr}$: radial distances near the Sun are longer than $\mathrm dr$, and light pays for it.

Full proof

The constants $E = Ac^2\dot t$ and $L = r^2\dot\varphi$ are unchanged (same Killing vectors). The null normalisation $-Ac^2\dot t^2 + B\dot r^2 + r^2\dot\varphi^2 = 0$ with $\dot r = -Lu'$ gives $BL^2u'^2 = E^2/(Ac^2) - L^2u^2$, i.e. $Bu'^2 = 1/(Ab^2) - u^2$. To first order in $r_g$: $1/A = 1 + 2r_gu$ and $1/B = 1 - 2\gamma r_gu$, so $u'^2 = \big(1/b^2 - u^2\big) + 2r_gu/b^2 - 2\gamma r_gu\big(1/b^2 - u^2\big) = 1/b^2 - u^2 + 2(1 - \gamma)r_gu/b^2 + 2\gamma r_gu^3$. Differentiate and divide by $2u'$: $u'' = -u + (1 - \gamma)r_g/b^2 + 3\gamma r_gu^2$.

Solve with $u_0 = \sin\varphi/b$: the constant forcing $(1 - \gamma)r_g/b^2$ gives the constant $(1 - \gamma)r_g/b^2$, and the $u^2$ forcing gives $\gamma\cdot\tfrac{r_g}{b^2}(1 + \cos^2\varphi)$ by the slide "Bending by general relativity". At $\varphi = -\delta$: $-\delta/b + \big[(1 - \gamma) + 2\gamma\big]r_g/b^2 = 0$, so $\delta = (1 + \gamma)r_g/b$ and $\Delta\theta = 2(1 + \gamma)r_g/b$. $\square$

Checks. $\gamma = 0$: $u'' + u = r_g/b^2$, which is the Newtonian corpuscle's equation identically, not just its answer ✓. $\gamma = 1$: $u'' + u = 3r_gu^2$ ✓.

37 · light

One incoming line, two rays

The solution of the previous slide's equation for $\gamma = 0$ and for $\gamma = 1$, each rotated so that it arrives along the same line at distance $b$.

$\gamma = 0$, flat space: Newton's corpuscle, bent by $2r_g/b$, all of it from $g_{tt}$.

$\gamma = 1$, Schwarzschild: the same pull from $g_{tt}$, plus as much again from $g_{rr}$, $4r_g/b$ in all.

$\gamma$ is the PPN parameter γ of the parametrised post-Newtonian framework — "how much space curvature per unit mass".

First-order solutions drawn with $r_g/b = 0.12$, as on the asymptotes figure. The deflections measured from the drawing, $13.8^\circ$ and $27.0^\circ$, are $2r_g/b$ and $4r_g/b$ to within the second-order terms the drawing also contains.

What to look for. The two rays come in together and part near the mass: the $\gamma = 1$ ray dips closer and leaves at twice the angle. Nothing about the incoming line tells them apart; only what space does near the mass does. A ray is the one probe for which $g_{rr}$ matters as much as $g_{tt}$, because it moves at $c$: its spatial path length is as large as its time interval, in the units the metric uses.

How it was drawn. For each $\gamma$, $u = \sin\varphi/b + (1 - \gamma)r_g/b^2 + \gamma(r_g/b^2)(1 + \cos^2\varphi)$, the first-order solution of the full-proof notes on the previous slide, with $r_g/b = 0.12$. Each curve is rotated by its own $\delta$ so its incoming asymptote is horizontal. The deflections measured between the asymptotes are $0.2406$ and $0.4713$ rad, against the first-order $0.24$ and $0.48$.

Is "the $g_{rr}$ half" coordinate-dependent? The split of the total into two halves is stated in these (Schwarzschild, areal-radius) coordinates. In unit 35's isotropic weak-field form the spatial correction sits as $(1 + 2\gamma r_g/r)\delta_{ij}$ on all of space, and the same total $(1 + \gamma)$ emerges with the same split between time and space parts. The total is invariant; the statement "one half from the time part, one from the space part" is true in every coordinate system in which a static observer's clock and rulers are separated this way. This is what unit 35's "same $h_{00}$ with flat space predicts $0.88''$" meant.

The PPN framework, cited: C. M. Will, Theory and Experiment in Gravitational Physics (2nd ed., 2018), ch. 4 and 7. In it $A$ also carries a second parameter $\beta$ at order $r_g^2$, and the perihelion advance becomes $\tfrac13(2 + 2\gamma - \beta)$ times section 3's value. General relativity has $\beta = \gamma = 1$.

37 · light · example

Worked example 3 — the ray at the limb, both ways

a ray grazing the Sun, $b = R_\odot$

Step 1 — the ratio. $r_g/b = 1476.6/6.957\times10^8 = 2.1225\times10^{-6}$.

Step 2 — Newton ($\gamma = 0$). $2r_g/b = 4.2450\times10^{-6}$ rad; $\times\,206\,264.8$:

$$\Delta\theta_{\rm N} = 0.8756''.$$

Step 3 — general relativity ($\gamma = 1$). $4r_g/b = 8.4900\times10^{-6}$ rad:

$$\Delta\theta_{\rm GR} = 1.7512''.$$

Step 4 — the responsible component. The difference, $0.8756''$, is the $\gamma\,2r_g/b$ term: $g_{rr}$.

Step 5 — the data.

measurement	limb deflection
Eddington, Sobral 1919	$1.98 \pm 0.12''$
Eddington, Príncipe 1919	$1.61 \pm 0.30''$
VLBI, quasars (2004)	$1.7510 \pm 0.0004''$

Step 6 — verdict. $0.8756''$ is excluded by the 1919 data at several probable errors, and by VLBI at 2000 standard errors. $1.7512''$ agrees with both.

Step 7 — check. Deflection $\propto 1/b$: a star seen at two solar radii is bent $0.876''$ in general relativity ✓, which is what VLBI fits across the sky.

Arithmetic. $r_g = 1476.625$ m; $R_\odot = 6.957\times10^8$ m (IAU nominal); $r_g/R_\odot = 2.12250\times10^{-6}$. $4.24500\times10^{-6}\times206\,264.806 = 0.87560''$ and twice that is $1.75119''$. These agree with unit 30's and unit 35's figures, as they must: the same two formulas.

The 1919 numbers are as published in 1920, with probable errors (unit 35 quotes them the same way). The Sobral astrographic plates, which gave about $0.93''$ and were discarded for focus problems, have been argued over ever since; a 1979 re-reduction of those astrographic plates with modern methods gave $1.55 \pm 0.34''$, so the discard did not manufacture the agreement. Honest summary: 1919 favoured general relativity and was not decisive by modern standards.

VLBI. S. S. Shapiro, J. L. Davis, D. E. Lebach and J. S. Gregory, Phys. Rev. Lett. 92, 121101 (2004): 1.7 million delay measurements of 541 radio sources over twenty years gave $\gamma = 0.99983 \pm 0.00045$. The deflection scales as $(1 + \gamma)/2 = 0.99992 \pm 0.00023$, so the limb value is $1.75104 \pm 0.00039''$, quoted on the slide as $1.7510 \pm 0.0004''$. The radio measurements do not need an eclipse: the Sun is dark at radio wavelengths near the limb only in a limited sense, and the sources are observed at many solar elongations, the $1/b$ law being fitted across all of them.

"2000 standard errors". $(1.7510 - 0.8756)/0.0004 \approx 2200$.

section 5

Clocks: the delay and the redshift

Light that arrives late, and light that arrives redder.

37 · clocks

The Shapiro delay

Unit 35's weak field, with unit 32's $g_{00}$ and the spatial part labelled by $\gamma$:

$$\mathrm ds^2 = -\Big(1 - \frac{2r_g}{r}\Big)c^2\mathrm dt^2 + \Big(1 + \gamma\frac{2r_g}{r}\Big)|\mathrm d\vec x|^2 .$$

Light has $\mathrm ds = 0$, so to first order

$$c\,\mathrm dt = \Big(1 + (1 + \gamma)\frac{r_g}{r}\Big)|\mathrm d\vec x| .$$

Integrate along the undeflected line $x \in [-x_1, x_2]$ at distance $b$, with $r = \sqrt{x^2 + b^2}$ and $\int\mathrm dx/\sqrt{x^2 + b^2} = \operatorname{arsinh}(x/b)$.

Schematic, not to scale. The integral runs along the straight line from Earth to Venus; $r$ is the distance from the Sun to the point $x$. Above: the integrand $\propto 1/r$, peaked at closest approach, where most of the delay is gathered.

Full proof

From $\mathrm ds^2 = 0$: $c^2\mathrm dt^2 = \frac{1 + 2\gamma r_g/r}{1 - 2r_g/r}|\mathrm d\vec x|^2$, and the square root of the ratio is $1 + (1 + \gamma)r_g/r + O(r_g^2)$. The path differs from the straight line at order $r_g$, but by Fermat's principle — the coordinate travel time is stationary with respect to the path, proved for this metric below — a first-order change of path changes the time only at second order. So integrate along the line. One way: $\Delta t_{\rm one} = (1 + \gamma)\frac{r_g}{c}\int_{-x_1}^{x_2}\frac{\mathrm dx}{\sqrt{x^2 + b^2}} = (1 + \gamma)\frac{r_g}{c}\ln\frac{(x_2 + r_2)(x_1 + r_1)}{b^2}$, using $\operatorname{arsinh}(x/b) = \ln\big((x + \sqrt{x^2 + b^2})/b\big)$. For $b \ll r_i$, $x_i \approx r_i$ and the argument is $4r_1r_2/b^2$. Round trip doubles it, and $r_g/c = GM/c^3$. $\square$

Fermat's principle for this metric, proved. Write the metric as $-Ac^2\mathrm dt^2 + B|\mathrm d\vec x|^2$ and put $n = \sqrt{B/A}$. A null curve has $c\,\mathrm dt = n|\mathrm d\vec x|$, so its coordinate travel time is $T = \frac1c\int n\,|\dot{\vec x}|\,\mathrm d\lambda$, a functional of the spatial path alone: unit 11's optical functional with refractive index $n$. Its Euler–Lagrange equations are $\frac{\mathrm d}{\mathrm d\lambda}\big(n\dot x_i/|\dot{\vec x}|\big) = |\dot{\vec x}|\,\partial_in$. A null geodesic satisfies them. With $E = Ac^2\dot t$ and the null condition $c\dot t = n|\dot{\vec x}|$, $n\dot x_i/|\dot{\vec x}| = B\dot x_i/(Ac\dot t) = (c/E)\,B\dot x_i$. The geodesic Lagrangian's $x^i$-equation (unit 32) is $\frac{\mathrm d}{\mathrm d\lambda}(B\dot x_i) = \tfrac12\big(-c^2\dot t^2\,\partial_iA + |\dot{\vec x}|^2\partial_iB\big)$; multiply by the constant $c/E$ and use $B|\dot{\vec x}|^2 = Ac^2\dot t^2$ to get $\tfrac12c\dot t\,(\partial_iB/B - \partial_iA/A)$, which is $|\dot{\vec x}|\,\partial_in$ because $\partial_in = \tfrac n2(\partial_iB/B - \partial_iA/A)$ and $n|\dot{\vec x}| = c\dot t$. So the ray is a stationary point of $T$. $\square$

What to look for in the figure. The integrand $1/\sqrt{x^2 + b^2}$ is $1/b$ at closest approach and falls to half that by $|x| = \sqrt3\,b$; but its tails fall only like $1/|x|$, so the integral grows like $\ln$ of the path length. Most of the delay is gathered within a few $b$ of the Sun, and the rest — the logarithm's slow growth — over the whole path. The dashed bent ray is there to show what is not integrated along: by Fermat's principle, below, it would change the answer only at second order.

37 · clocks

The delay, and how it falls off with $b$

the Shapiro delay, round trip $$\Delta t = (1 + \gamma)\frac{2GM}{c^3}\ln\frac{4r_1r_2}{b^2}$$

for $b \ll r_1, r_2$, the distances of the two ends from the Sun.

Same $(1 + \gamma)$ as the deflection: half from the clocks ($g_{00}$), half from the rulers ($g_{ij}$). $2GM/c^3 = 9.85\ \mu$s for the Sun.

checks

$[GM/c^3] = \mathsf{T}$ ✓. $M \to 0$: no delay ✓. The delay grows only logarithmically as $b$ shrinks: the extra time is gathered near closest approach.

The round-trip delay for the Earth–Venus geometry of worked example 4, computed from the box. A logarithm is a straight line on a log axis: doubling $b$ always removes the same $27.3\ \mu$s. Dashed: flat space for light, $\gamma = 0$, half as much.

The figure. The formula with $\gamma = 1$ and with $\gamma = 0$, for the Earth–Venus geometry of worked example 4, $r_1 = 1$ au and $r_2 = 0.72333$ au. On a logarithmic $b$ axis the delay is a straight line, of slope $-2\cdot4GM/c^3 = -39.4\ \mu$s per factor $e$ in $b$; doubling $b$ removes $27.3\ \mu$s. That slope, not the absolute delay, is what the experiments fit.

What is measured. The coordinate time $t$ is the proper time of a static clock far from the Sun; the Earth's clock differs by a constant factor $1 - O(10^{-8})$, which rescales $\Delta t$ negligibly. The constant "4" inside the logarithm depends on coordinates; what is not coordinate-dependent, and what experiments fit, is how the delay changes as $b$ sweeps through the conjunction — by $\frac{4GM}{c^3}\cdot2\ln(b_1/b_2)$ between two impact parameters.

The weak-field form's origin. Unit 35 found $h_{00} = -2\Phi/c^2$ and $h_{ij} = -2\Phi\delta_{ij}/c^2$ for a static source, which is this metric with $\Phi = -GM/r$ and $\gamma = 1$. It is Schwarzschild in isotropic coordinates, to first order; the radial coordinates differ at order $r_g$, which moves the delay only by a constant.

Arithmetic. $GM_\odot/c^3 = 4.9255\ \mu$s, so $2GM/c^3 = 9.851\ \mu$s and $4GM/c^3 = 19.702\ \mu$s.

37 · clocks · example

Worked example 4 — Shapiro's proposal: radar to Venus

a radar echo from Venus at superior conjunction, grazing the Sun's limb

Step 1 — geometry. Earth at $r_1 = 1$ au $= 1.4960\times10^{11}$ m; Venus beyond the Sun at $r_2 = 0.72333$ au $= 1.0821\times10^{11}$ m; superior conjunction with the beam grazing, $b = R_\odot = 6.957\times10^8$ m.

Step 2 — the logarithm. $$\frac{4r_1r_2}{b^2} = \frac{4\times1.4960\times10^{11}\times1.0821\times10^{11}}{(6.957\times10^8)^2} = 1.3378\times10^5,$$ whose natural logarithm is $11.804$.

Step 3 — the delay ($\gamma = 1$). $$\Delta t = \frac{4GM}{c^3}\times11.804 = 19.702\ \mu\text{s}\times11.804 = \mathbf{232.6\ \mu s}.$$

Step 4 — the Newtonian-light answer ($\gamma = 0$) is half: $116.3\ \mu$s. So the delay, like the deflection, measures $g_{rr}$.

Step 5 — as a length. $c\Delta t = 69.7$ km, in a round trip of $2(r_1 + r_2) = 5.2\times10^{11}$ m: a part in $7\times10^6$. What is timed is how the echo's delay changes as $b$ sweeps.

Step 6 — check. At $b = 2R_\odot$ the delay drops by $19.702\times\ln4 = 27.3\ \mu$s: the logarithmic sweep an experiment fits ✓.

Step 2. Numerator $4\times1.495979\times10^{11}\times1.082089\times10^{11} = 6.4750\times10^{22}$; denominator $4.8400\times10^{17}$; ratio $1.33784\times10^5$; $\ln = 11.8040$.

Step 3. $19.70196\times11.80398 = 232.56\ \mu$s.

Step 5. $299\,792\,458\times232.56\times10^{-6} = 6.972\times10^4$ m. The round trip is $2(r_1 + r_2) = 3.4467$ au $= 5.156\times10^{11}$ m, and $6.972\times10^4/5.156\times10^{11} = 1.35\times10^{-7}$, one part in $7.4\times10^6$. Nobody knows the Earth–Venus distance to that accuracy in advance; what the experiment measures is the change in the echo time as $b$ sweeps through the conjunction, fitted together with the orbits.

History. Irwin Shapiro proposed the test in 1964 (Phys. Rev. Lett. 13, 789), noting that it was the first new test of general relativity suggested since 1915. The first measurements, with radar echoes from Mercury and Venus using the Haystack and Arecibo antennas, came in 1967–1971 and agreed with general relativity to within about 5%. Radar cannot actually penetrate to $b = R_\odot$ at those frequencies because of the solar corona's plasma, which adds its own frequency-dependent delay; the experiments used larger $b$ and fitted the logarithm. The Viking landers on Mars (1976–77) reached $0.1\%$.

37 · clocks

Cassini, 2002: one part in a hundred thousand

In June 2002 the Cassini spacecraft, on its way to Saturn, passed behind the Sun as seen from Earth:

Earth at $1.016$ au from the Sun, Cassini at $7.42$ au;
the radio beam's closest approach $b \approx 1.6\,R_\odot$, on 21 June;
two-way links at X band (8.4 GHz) and Ka band (32 GHz), so the plasma's $\nu^{-2}$ delay could be measured and removed.

Problem P4 computes the delay for this geometry: about $260\ \mu$s at closest approach.

the result

Bertotti, Iess and Tortora (2003):

$$\gamma - 1 = (2.1 \pm 2.3)\times10^{-5}.$$

So the $g_{rr}$ half of the delay is present in full to one part in $10^5$. It is the tightest bound on $\gamma$ from any experiment, and it is why unit 35 could say that Brans–Dicke theory has been squeezed.

What was measured was the rate of change of the delay — a fractional frequency shift of order $10^{-10}$ — not the delay itself.

Where the geometry comes from. The distances were computed from JPL ephemerides for 21 June 2002: Earth $1.0162$ au, Cassini $7.4216$ au from the Sun. The minimum impact parameter $1.6R_\odot$ is the figure given in the paper, B. Bertotti, L. Iess and P. Tortora, Nature 425, 374 (2003).

Why a frequency. The Cassini experiment tracked the Doppler shift of a stable radio signal. The delay enters as $\mathrm d(\Delta t)/\mathrm dt$: as the Earth moves, $b$ changes by about three solar radii a day, and the delay's logarithm changes with it, giving a fractional frequency shift $y = \frac{\mathrm d\Delta t}{\mathrm dt} = -\frac{8GM}{c^3}\frac{\dot b}{b}$ (for $\gamma = 1$). P4's last part estimates its peak at about $5\times10^{-10}$. Frequency stability of $10^{-14}$ over the relevant times is what made the $10^{-5}$ level reachable.

Why the plasma matters. The solar corona delays radio waves by an amount proportional to $\nu^{-2}$; near the Sun at X band it is far larger than the relativistic effect. Two frequencies allow the $\nu^{-2}$ piece to be solved for and removed. Earlier missions without Ka band were limited by exactly this, at about $10^{-3}$.

Unit 35's statement, made good: Brans–Dicke theory predicts $\gamma = (1 + \omega)/(2 + \omega)$, so $1 - \gamma = 1/(2 + \omega) \gt 0$: Brans–Dicke always has $\gamma \lt 1$. Cassini's two-standard-error lower end is $\gamma - 1 \gt 2.1\times10^{-5} - 4.6\times10^{-5} = -2.5\times10^{-5}$, and $1/(2 + \omega) \lt 2.5\times10^{-5}$ requires $\omega \gt 4.0\times10^4$.

37 · clocks

The redshift, closed exactly

A static observer at $r$ has four-velocity $U = \partial_t/\sqrt{A(r)}$, with $A = 1 - 2r_g/r$, so that $g(U, U) = -c^2$ (unit 27). A photon with wavevector $k$ has conserved $E_k = -g(\partial_t, k)$ (Killing, again). The frequency $\nu$ the observer measures — a light frequency now, not the apsidal $\nu$ of sections 2 and 3 — is proportional to $-g(k, U)$:

$$\nu(r) \propto -g(k, U) = \frac{E_k}{\sqrt{A(r)}} .$$

the exact gravitational redshift $$\frac{\nu_{\rm received}}{\nu_{\rm emitted}} = \sqrt{\frac{1 - 2r_g/r_{\rm e}}{1 - 2r_g/r_{\rm r}}}$$

Light from a static emitter at $r_{\rm e}$ to a receiver far away, computed. Red: the exact ratio, which reaches zero at the horizon. Blue: unit 30's first-order formula, which still gives one half there. Far out they agree, and at the Sun's surface, $r/r_g = 4.7\times10^5$, they agree to $10^{-12}$.

The frequency an observer measures. For a photon of wavevector $k$ and an observer of four-velocity $U$ with $g(U, U) = -c^2$, the measured frequency is a fixed constant times $-g(k, U)$: unit 27's statement that energy is the time component of the four-momentum in the observer's rest frame, written invariantly. Here $g(\partial_t, \partial_t) = -Ac^2$, so $U = \partial_t/\sqrt A$ has $g(U, U) = -c^2$, and $-g(k, U) = -g(k, \partial_t)/\sqrt A = E_k/\sqrt A$. In the ratio of two such measurements, at emitter and receiver, the constant and $E_k$ cancel. $\square$

Limits. $r_{\rm e} = r_{\rm r}$: no shift ✓. $r_{\rm e} \to 2r_g$: infinite redshift — light from the horizon never arrives, as unit 36 found ✓. First order: expand each square root.

What to look for. The two curves part only where $r_g/r_{\rm e}$ is not small: at $r_{\rm e} = 3r_g$ the exact ratio is $0.577$ and the first-order one $0.667$. The exact one goes to zero at the horizon like $\sqrt{1 - 2r_g/r_{\rm e}}$, with a vertical tangent: a clock held just outside the horizon is seen from far away to tick arbitrarily slowly.

37 · clocks

The redshift, measured

unit 30, recovered

To first order, $\Delta\nu/\nu = -r_g(1/r_{\rm e} - 1/r_{\rm r}) = (\Phi_{\rm e} - \Phi_{\rm r})/c^2$. Near the ground this is $-g_\oplus h/c^2$, with $g_\oplus = 9.81$ m/s² the acceleration of gravity: Pound–Rebka's $2.46\times10^{-15}$ ✓.

It involves only $g_{tt}$, so it has no $\gamma$ in it: the equivalence principle alone predicts it, which is why unit 30 could.

Sun's surface to infinity: $r_g/R_\odot = 2.12\times10^{-6}$, equivalent to a recession of $636$ m/s, measured in the solar spectrum.

Gravity Probe A, 1976

A hydrogen maser rocketed to $10\,000$ km and compared with one on the ground confirmed the formula to $7\times10^{-5}$.

section 6

The tests, then and now

What the classical tests measure, what they cannot, and what came after.

37 · the tests

Modern tests, named and dated

test	when	what	result
Gravity Probe B	2004–05 flight, 2011 result	gyroscopes in Earth orbit: geodetic precession and frame dragging	$6601.8 \pm 18.3$ against $6606.1$ mas/yr; $37.2 \pm 7.2$ against $39.2$ mas/yr
GW150914	14 Sep 2015	two black holes, $36$ and $29\,M_\odot$, merging; about $3\,M_\odot c^2$ radiated	waveform matches general relativity; unit 38
S2 at Sgr A*	2018, 2020	a star's orbit about $4.3\times10^6M_\odot$: redshift, then precession	precession $1.10 \pm 0.19$ of the prediction
Event Horizon Telescope	2019, 2022	the photon ring of M87* and of Sgr A*	$42 \pm 3$ and $51.8 \pm 2.3\ \mu$as, as $3\sqrt3\,r_g$ predicts
Cassini	2002	Shapiro delay	$\gamma - 1 = (2.1 \pm 2.3)\times10^{-5}$

Gravity Probe B. C. W. F. Everitt et al., Phys. Rev. Lett. 106, 221101 (2011). The geodetic (de Sitter) precession of a gyroscope carried round a mass is, for a circular orbit, $\Omega = \tfrac32(GM)^{3/2}/(c^2r^{5/2})$ — cited, not derived here: it is parallel transport (unit 32) around the orbit in the Schwarzschild metric, and in PPN form the $\tfrac32$ is $\tfrac12 + \gamma$: a half that comes with $g_{tt}$ and the motion, and $\gamma = 1$ from curved space — the same split as section 4's. At GP-B's radius $7027$ km it gives $6605$ mas/yr; the quoted $6606.1$ includes the Earth's oblateness. Frame dragging needs the Kerr metric's weak-field limit, unit 40.

GW150914. LIGO Scientific and Virgo Collaborations, Phys. Rev. Lett. 116, 061102 (2016): initial masses $36^{+5}_{-4}$ and $29 \pm 4\,M_\odot$, final $62 \pm 4\,M_\odot$, radiated $3.0 \pm 0.5\,M_\odot c^2$. The late inspiral passes through the regime where section 2's effective potential governs — the ISCO is where the adiabatic inspiral ends and the plunge begins.

S2. GRAVITY Collaboration, Astron. Astrophys. 615, L15 (2018) for the redshift, 636, L5 (2020) for the precession.

EHT. Astrophys. J. Lett. 875, L1 (2019) for M87*, and 930, L12 (2022) for Sgr A*.

37 · the tests

Where each test happens

Each test placed at the $r_g/r$ of the place its effect is gathered, computed from the numbers on the earlier slides. Ten decades separate Gravity Probe B from the photon sphere; the solar-system tests all sit in the left half.

what the classical tests cannot see

Mercury, the limb and Cassini all probe $r_g/r \lt 10^{-5}$, first order in $r_g$ with two parameters, $\gamma$ and $\beta$. The ISCO, the plunge, the photon sphere and the merger live at $r_g/r \sim 0.1$–$0.5$. Only GW150914 and the Event Horizon Telescope go there; even S2 at pericentre is at $r_g/r \approx 4\times10^{-4}$.

The numbers placed. Gravity Probe B: the Earth's $r_g = GM_\oplus/c^2 = 4.4$ mm at $7027$ km, $6.3\times10^{-10}$. The Earth's orbit, $9.9\times10^{-9}$; Mercury at $a$, $2.5\times10^{-8}$; Cassini's ray at $1.6R_\odot$, $1.3\times10^{-6}$; the limb, $2.1\times10^{-6}$. S2 at pericentre, $119$ au from a black hole of $4.26\times10^6M_\odot$: $r_g = 6.29\times10^9$ m, ratio $3.5\times10^{-4}$. The ISCO, $1/6$; the photon sphere, $1/3$, which is where the rays forming the EHT's ring pass; the merger, roughly $0.1$ to $0.5$.

Why the gap matters. Everything left of $10^{-5}$ is described, to the accuracy measured, by the metric expanded to first order in $r_g/r$ (second order in $g_{tt}$ for the perihelion), with two numbers, $\gamma$ and $\beta$. A theory that agreed with general relativity there and differed at $r_g/r \sim 0.3$ would pass every classical test. Only the right-hand cluster can rule it out — which is why the 2015 detection and the EHT images were tests of a new kind, not just more decimals.

37 · summary

Three numbers, and what each one measures

test	formula	value	metric part
perihelion	$6\pi r_g/a(1 - e^2)$	$42.98''$/cy	$g_{tt}$ to $O(r_g^2)$, $g_{rr}$
deflection	$4r_g/b$	$1.7512''$	$g_{tt}$ + $g_{rr}$, half each
Shapiro	$\frac{4GM}{c^3}\ln\frac{4r_1r_2}{b^2}$	$233\ \mu$s	$g_{tt}$ + $g_{rr}$, half each
redshift	$\sqrt{A_{\rm e}/A_{\rm r}}$	$2.1\times10^{-6}$	$g_{tt}$ only

Newton, dressed as a metric, is $g_{tt}$ with flat space. Every test with a factor of two in it is a test of $g_{rr}$ — of curved space.

Mercury, closed

The metric gave the constants, the constants gave the orbit equation, the orbit equation had one new term, unit 21's method turned the term's secular effect into $\nu = 1 - 3r_g/p$, and $415.2$ orbits of $0.1035''$ make $42.98''$. Le Verrier's residual, 56 years after he found it, with no parameter to spare.

the ray, closed

$0.8756''$ from $g_{tt}$, $0.8756''$ more from $g_{rr}$, $1.7512''$ in all.

The perihelion row. In the PPN form of the previous section the advance is $\tfrac13(2 + 2\gamma - \beta)$ times $6\pi r_g/p$, so it depends on $g_{rr}$ through $\gamma$ and on the second-order term of $g_{tt}$ through $\beta$ (cited, Will). That is why the perihelion is the one classical test that sees beyond first order in the metric: Newton's $1/r$ already fixes the first-order $g_{tt}$, and the $43''$ comes from what comes after it. With $\gamma = \beta = 1$ the factor is 1.

The Shapiro row uses the Venus geometry of worked example 4. The value depends on the geometry; the formula does not.

What comes next. Unit 38 perturbs the metric and finds waves; unit 39 applies the same field equations to the universe as a whole; unit 40 rotates the black hole and says where this course's methods stop. None of them adds a new parameter; each is the same equation in a new regime.

37 · checkpoint

Checkpoint

you can now do this, or you cannot

From scratch, with no formula looked up, you compute Mercury's perihelion advance to within $1''$ per century of $43$ and the solar light deflection as $1.75''$, show the latter is exactly twice the Newtonian value and name the metric component responsible — and you can say why unit 21 forbade you from removing the secular term.

Failing it is unambiguous. If your orbit equation has a $2$ or a $\tfrac32$ where the $3$ should be, redo worked example 1 from the normalisation. If you get $0.88''$, you have bent a corpuscle, not a null geodesic. If you cannot say what deleting $\varepsilon e\varphi\sin\varphi$ would predict, reread "Why the secular term must not be removed".

Eleven problems, about seven hours: ./mr problems 37. P2 is the checkpoint's number; P5 proves the factor of two; P3 is the ISCO and accretion discs; P4 is Cassini. When the checkpoint is true: ./mr done 37.

The checkpoint has four clauses and they fail in different ways. Clause one fails arithmetically, usually by using $a$ where $p = a(1 - e^2)$ belongs (that gives $41.2''$ — still within a couple of arcseconds, which is why the clause asks for $1''$). Clause two fails on the factor of two. Clause three fails when "the metric component" is answered with "curvature" rather than with $g_{rr}$, or when the answer cannot say why $g_{tt}$ alone gives Newton. Clause four fails when the answer is "because the secular term is small": it is not small, it grows, and that growth is the precession.

The most useful single exercise for this unit is to do worked examples 1, 2 and 3 on paper in one sitting with the slides closed. If that takes under an hour with the right numbers, the unit is done.