Part F · unit 37 · problems

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

the slides · the curriculum

Checkpoint. 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.

Work each problem on paper first. Under it, Hints opens a ladder one rung at a time — each rung appears only once the one before it is open, and the last gives the first line of the working — and Worked solution has the whole thing: the idea, the working, then a check. Read the solution after your attempt, including when you got it right. F folds everything back; D switches dark and light.

Eleven problems, about seven hours. P2 is the checkpoint's number and P5 its factor of two; P6 is its clause about the secular term. Signature \((-,+,+,+)\), \(c\) explicit, a dot is \(\mathrm d/\mathrm d\tau\) (or an affine parameter for light), 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: per unit rest mass for a massive particle, and for light merely proportional to the affine parameter's scale, so that only \(b = cL/E\) is physical. \(u = 1/r\), and \(r_g = GM/c^2\) is the gravitational radius — half the \(r_s = 2GM/c^2\) unit 36 wrote its formulas with. The Schwarzschild metric is

\[ \mathrm ds^2 = -\Big(1 - \frac{2r_g}{r}\Big)c^2\mathrm dt^2 + \Big(1 - \frac{2r_g}{r}\Big)^{-1}\mathrm dr^2 + r^2\big(\mathrm d\theta^2 + \sin^2\theta\,\mathrm d\varphi^2\big). \]

Constants: \(GM_\odot = 1.32712440018\times10^{20}\ \mathrm{m^3\,s^{-2}}\), \(c = 299\,792\,458\ \mathrm{m\,s^{-1}}\), so \(r_{g\odot} = 1476.625\) m and \(GM_\odot/c^3 = 4.92549\ \mu\)s; \(R_\odot = 6.957\times10^{8}\) m; 1 au \(= 1.495978707\times10^{11}\) m; 1 pc \(= 3.0857\times10^{16}\) m; 1 rad \(= 206\,264.806''\); a Julian century is \(36\,525\) days.

Geodesics and the orbit equation

P1 The orbit equation from the geodesic Lagrangian routine

exercises: the geodesic-Lagrangian trick, cyclic coordinates as Killing vectors, the Binet substitution, a Newtonian limit

Take the geodesic Lagrangian \(\mathcal L = g_{\mu\nu}\dot x^\mu\dot x^\nu\) of the Schwarzschild metric, restricted to the equatorial plane \(\theta = \pi/2\).

  1. Write \(\mathcal L\) out. Name its cyclic coordinates, and from the corresponding Euler–Lagrange equations obtain the conserved quantities \(E = (1 - 2r_g/r)c^2\dot t\) and \(L = r^2\dot\varphi\). Say which unit-13 theorem this is and which Killing vectors (unit 36) are behind it.
  2. Using the normalisation \(g(\dot x, \dot x) = -c^2\), show that \(\dot r^2 = E^2/c^2 - (1 - 2r_g/r)(c^2 + L^2/r^2)\).
  3. Substitute \(u = 1/r\) with \(\varphi\) as parameter and derive \(u'' + u = GM/L^2 + 3r_gu^2\). Where did you divide by something that can vanish, and why is the result still valid there?
  4. For a slow particle far away, show \(E = c^2 + \tfrac12 v^2 - GM/r + \dots\), using \(\mathrm dt/\mathrm d\tau \approx 1 + v^2/2c^2 + r_g/r\).
Hints · 3 rungs
Rung 1

Two constants, one first-order ODE for \(r(\tau)\), one second-order ODE for \(u(\varphi)\), and an expansion. Each step is a substitution; nothing needs solving.

Rung 2

A coordinate that does not appear in \(\mathcal L\) has a conserved conjugate momentum \(\partial\mathcal L/\partial\dot q\). For part 3, the chain rule gives \(\dot r\) in terms of \(u'\) and \(\dot\varphi\), and \(\dot\varphi = Lu^2\).

Rung 3

First line of part 3: \(\dot r = \dfrac{\mathrm dr}{\mathrm d\varphi}\dot\varphi = -\dfrac{u'}{u^2}\,Lu^2 = -Lu'\), so the radial equation becomes \(L^2u'^2 = E^2/c^2 - (1 - 2r_gu)(c^2 + L^2u^2)\). Now differentiate in \(\varphi\).

Worked solution

Idea. Symmetries give two first integrals, the normalisation gives a third, and the Binet substitution trades time for angle. This is unit 08's reduction with a metric in place of a force.

1. With \(f = 1 - 2r_g/r\), \[ \mathcal L = -f c^2\dot t^2 + f^{-1}\dot r^2 + r^2\dot\varphi^2 . \] \(t\) and \(\varphi\) do not appear, so \(\frac{\mathrm d}{\mathrm d\tau}(\partial\mathcal L/\partial\dot t) = 0\) and likewise for \(\varphi\): \(-2fc^2\dot t\) and \(2r^2\dot\varphi\) are constant. Halving and fixing signs so that \(E \gt 0\) gives \(E = fc^2\dot t\), \(L = r^2\dot\varphi\). This is Noether's theorem for a cyclic coordinate (unit 13); the coordinate-free statement is that \(\partial_t\) and \(\partial_\varphi\) are Killing vectors, and \(E = -g(\partial_t, \dot x)\), \(L = g(\partial_\varphi, \dot x)\).

2. The normalisation is \(-fc^2\dot t^2 + f^{-1}\dot r^2 + r^2\dot\varphi^2 = -c^2\). Substitute \(\dot t = E/(fc^2)\), \(\dot\varphi = L/r^2\): \(-E^2/(fc^2) + \dot r^2/f + L^2/r^2 = -c^2\). Multiply by \(f\): \(\dot r^2 = E^2/c^2 - f\,(c^2 + L^2/r^2)\).

3. With \(\dot r = -Lu'\), expand the product: \[ L^2u'^2 = \frac{E^2}{c^2} - c^2 - L^2u^2 + 2r_gc^2u + 2r_gL^2u^3 . \] Differentiate in \(\varphi\): \(2L^2u'u'' = -2L^2uu' + 2r_gc^2u' + 6r_gL^2u^2u'\). Divide by \(2L^2u'\) and use \(r_gc^2 = GM\): \[ u'' + u = \frac{GM}{L^2} + 3r_gu^2 . \] The division fails where \(u' = 0\), at the turning points. But both sides of the final equation are continuous in \(\varphi\), and they agree wherever \(u' \ne 0\); on a non-circular orbit the turning points are isolated, so the two sides agree on a dense set and hence everywhere. Circular orbits (\(u' \equiv 0\)) are exactly the constant solutions of the second-order equation, which is why it, not the first-order one, is the equation to keep.

4. \(E = (1 - 2r_g/r)c^2(1 + v^2/2c^2 + r_g/r) = c^2 + \tfrac12v^2 + c^2r_g/r - 2c^2r_g/r + O(c^{-2}) = c^2 + \tfrac12v^2 - GM/r + O(c^{-2})\).

Check. Dimensions: \([GM/L^2] = \mathsf{L^3T^{-2}}/\mathsf{L^4T^{-2}} = \mathsf{L^{-1}}\) and \([r_gu^2] = \mathsf{L^{-1}}\), like \(u\) ✓. Limit \(r_g \to 0\) (that is, \(c \to \infty\) at fixed \(GM\)): unit 08's \(u'' + u = GM/L^2\) ✓, whose solution is the conic \(u = (1 + e\cos\varphi)GM/L^2\).

P2 The 43, from scratch routine

exercises: the perihelion formula, careful arithmetic, a comparison with data

Mercury has \(a = 5.7909\times10^{10}\) m, \(e = 0.20563\) and sidereal period \(87.9691\) days.

  1. Starting from \(\Delta\varpi = 6\pi GM/(c^2a(1 - e^2))\) per orbit, compute Mercury's perihelion advance in arcseconds per orbit and per Julian century. Compare with the observed anomalous advance \(42.98 \pm 0.04''\) per century.
  2. A common slip is to use \(a\) where \(p = a(1 - e^2)\) belongs. What does that give, and would the checkpoint's "within \(1''\)" catch it?
  3. Repeat for Venus (\(a = 0.723332\) au, \(e = 0.006773\), \(224.701\) d) and the Earth (\(a = 1.000001\) au, \(e = 0.016709\), \(365.256\) d). Show that for near-circular orbits the advance per century scales as \(a^{-5/2}\), and check your three numbers roughly obey it.
Hints · 3 rungs
Rung 1

Three numbers per planet: a dimensionless angle per orbit, a count of orbits per century, their product. Keep five significant figures until the end.

Rung 2

Compute \(r_g = GM_\odot/c^2\) first; then the per-orbit angle is \(2\pi\varepsilon\) with \(\varepsilon = 3r_g/p\). For the scaling, the number of orbits per century goes as \(1/T \propto a^{-3/2}\) by Kepler's third law.

Rung 3

First line: \(r_g = 1.32712\times10^{20}/(2.99792\times10^8)^2 = 1476.6\) m, and \(1 - e^2 = 1 - 0.042284 = 0.95772\).

Worked solution

Idea. Everything is fixed: one constant in the field equations, one in the solution, and the orbit's measured elements. There is nothing to adjust.

1. \(p = 5.7909\times10^{10}\times0.95772 = 5.5460\times10^{10}\) m. \(\varepsilon = 3r_g/p = 4429.9/5.5460\times10^{10} = 7.9875\times10^{-8}\). Per orbit \(2\pi\varepsilon = 5.0187\times10^{-7}\) rad \(= 0.10352''\). Orbits per century \(36\,525/87.9691 = 415.20\). So \[ \Delta\varpi = 415.20\times0.10352'' = 42.98''\ \text{per century}, \] in agreement with \(42.98 \pm 0.04''\) — well within the error bar.

2. With \(a\) in place of \(p\), everything is multiplied by \(1 - e^2 = 0.95772\): \(41.16''\). That is \(1.8''\) low, so the checkpoint's \(1''\) tolerance catches it — which is why the tolerance is \(1''\) and not \(2''\).

3.

  • Venus: \(p = 1.08204\times10^{11}\) m, \(\varepsilon = 4.094\times10^{-8}\), \(0.05306''\) per orbit, \(162.55\) orbits per century, \(8.62''\) per century.
  • Earth: \(p = 1.49556\times10^{11}\) m, \(\varepsilon = 2.962\times10^{-8}\), \(0.03839''\) per orbit, \(100.00\) orbits, \(3.84''\) per century.

Scaling: per orbit \(\propto 1/a\), orbits per century \(\propto a^{-3/2}\), product \(\propto a^{-5/2}\). From Mercury: \(42.98\times(0.387098/0.723332)^{5/2} = 9.0''\) for Venus and \(42.98\times0.387098^{5/2} = 4.0''\) for the Earth. The actual \(8.62\) and \(3.84\) are 4% lower, because Mercury's \(1/(1 - e^2) = 1.044\) boosts it and the others' barely.

Check. \(\varepsilon\) is dimensionless ✓; \(c \to \infty\) gives zero ✓; the rounded inputs \(a = 5.79\times10^{10}\), \(e = 0.2056\) give \(42.99''\), so the answer is not fragile ✓.

P3 The ISCO, and what an accretion disc can radiate medium

exercises: circular orbits from the effective potential, stability, a binding energy, comparison with observed efficiencies

Use \(\tfrac12\dot r^2 + V_{\rm eff}(r) = \tfrac12(E^2/c^2 - c^2)\) with \(V_{\rm eff} = -GM/r + L^2/2r^2 - GML^2/c^2r^3\), and write \(x = r_g/r\).

  1. Show that a circular orbit at radius \(r\) has \[ \frac{L^2}{c^2} = \frac{r_g\,r}{1 - 3x},\qquad \frac{E}{c^2} = \frac{1 - 2x}{\sqrt{1 - 3x}} . \] Where do these stop making sense, and what is there?
  2. Show that circular orbits are stable exactly when \(r \gt 6r_g\), so the innermost stable circular orbit is at \(r = 6r_g\).
  3. Gas in a thin accretion disc spirals slowly inwards through near-circular orbits, radiating as it goes, and plunges without further radiation once it passes the ISCO. Show that the fraction of rest energy radiated is \(1 - \sqrt{8/9}\), and evaluate it.
  4. Compare with hydrogen fusion (four protons of \(1.007825\) u each to one helium-4 atom of \(4.002603\) u), and with the mean radiative efficiency of about \(0.1\) inferred for accreting black holes by comparing the quasar light in the universe with the mass now in black holes (Soltan's argument; Yu and Tremaine 2002). What does the comparison suggest?
Hints · 3 rungs
Rung 1

Circular means \(\dot r = 0\) and \(V_{\rm eff}' = 0\) at once; stable means \(V_{\rm eff}'' \gt 0\) there. The efficiency is \((c^2 - E)/c^2\), because the gas started at rest far away with \(E = c^2\).

Rung 2

\(V_{\rm eff}' = 0\) is linear in \(L^2\), so solve it for \(L^2\). Then \(\dot r = 0\) in the exact radial equation \(\dot r^2 = E^2/c^2 - (1 - 2x)(c^2 + L^2/r^2)\) gives \(E^2\). For stability, it is easiest to ask where \(L^2(r)\) has its minimum: that is where the two roots of \(V_{\rm eff}' = 0\) merge.

Rung 3

First line: \(r^4V_{\rm eff}'/c^2 = r_gr^2 - \ell^2r + 3r_g\ell^2\) with \(\ell = L/c\), and setting it to zero gives \(\ell^2(r - 3r_g) = r_gr^2\).

Worked solution

Idea. A circular orbit is an equilibrium of the one-dimensional problem; stability is the sign of the second derivative; and the energy a disc can release is the depth of the last stable well.

1. \(V_{\rm eff}'(r) = (c^2/r^4)(r_gr^2 - \ell^2r + 3r_g\ell^2)\). Setting it to zero, \(\ell^2 = r_gr^2/(r - 3r_g) = r_gr/(1 - 3x)\). Then \(\dot r = 0\) gives \[ \frac{E^2}{c^4} = (1 - 2x)\Big(1 + \frac{\ell^2}{r^2}\Big) = (1 - 2x)\Big(1 + \frac{x}{1 - 3x}\Big) = \frac{(1 - 2x)^2}{1 - 3x}. \] Both blow up as \(r \to 3r_g\): circular orbits need infinite \(L\) and \(E\) there. That limit is the photon sphere, where only light can orbit: a photon's own \(E\) and \(L\) are finite, but they are per unit affine parameter rather than per unit rest mass, and the finite quantity that survives the limit is the ratio \(b = cL/E \to 3\sqrt3\,r_g\). Between \(3r_g\) and \(2r_g\) there are no circular orbits at all.

2. \(\ell^2(r) = r_gr^2/(r - 3r_g)\) has derivative \(r_g r(r - 6r_g)/(r - 3r_g)^2\): it decreases on \((3r_g, 6r_g)\) and increases for \(r \gt 6r_g\), with minimum \(\ell^2 = 12r_g^2\) at \(6r_g\). For a given \(\ell^2 \gt 12r_g^2\) the bracket \(q(r) = r_gr^2 - \ell^2r + 3r_g\ell^2\) has two roots, one on each side of \(6r_g\). \(q\) is an upward parabola, so \(V_{\rm eff}' = c^2q/r^4\) is positive outside the roots and negative between: the outer root is a minimum of \(V_{\rm eff}\) (stable) and the inner one a maximum (unstable). So circular orbits at \(r \gt 6r_g\) are stable and those in \((3r_g, 6r_g)\) are not; at \(6r_g\) the roots merge into an inflection, marginally stable.

3. At \(x = 1/6\): \(E/c^2 = (2/3)/\sqrt{1/2} = \sqrt{8/9} = 0.942809\). Gas that started at rest far away had \(E = c^2\). The disc radiated the difference: \[ \eta = 1 - \sqrt{8/9} = 0.0572 = 5.72\%. \]

4. Fusion: \(4\times1.007825 - 4.002603 = 0.028697\) u out of \(4.031300\), that is \(0.71\%\). Accretion onto a non-rotating hole is eight times better than burning hydrogen completely — which is why quasars, not galaxies of stars, are the most luminous steady objects. But Soltan's average of about \(0.1\) is nearly twice \(5.7\%\). The natural reading is that typical accreting holes rotate: for a rotating (Kerr) hole with prograde orbits the ISCO moves inwards and the efficiency rises, to \(42\%\) in the extreme case. That is unit 40's business, cited here, not derived.

Check. \(L^2/c^2\) has dimensions \(\mathsf{L\cdot L} = \mathsf{L^2}\) ✓. Large \(r\) (\(x \to 0\)): \(\ell^2 \to r_gr\), i.e. \(L^2 = GMr\), Newton's circular orbit ✓; and \(E/c^2 \approx 1 - 2x + \tfrac32x = 1 - \tfrac12x\), i.e. \(E - c^2 \approx -GM/2r\), Newton's binding energy per unit mass ✓.

Light and clocks

P4 Shapiro from Cassini medium

exercises: integrating a null line element, the Shapiro formula, a measurement with an error bar

Use the weak-field metric \(\mathrm ds^2 = -(1 - 2r_g/r)c^2\mathrm dt^2 + (1 + 2\gamma r_g/r)|\mathrm d\vec x|^2\), with \(\gamma = 1\) in general relativity.

  1. Show that a radar signal from a point at distance \(r_1\) from the Sun to one at \(r_2\) and back, passing the Sun at closest distance \(b \ll r_1, r_2\), is delayed relative to flat space by \[ \Delta t = (1 + \gamma)\frac{2GM}{c^3}\ln\frac{4r_1r_2}{b^2}. \] Say why integrating along the undeflected straight line is legitimate.
  2. In June 2002 the Earth was \(1.0162\) au from the Sun, the Cassini spacecraft \(7.4216\) au on the far side, and the beam passed at \(b = 1.6R_\odot\). Evaluate \(\Delta t\) for \(\gamma = 1\), and the part of it due to \(g_{ij}\).
  3. Bertotti, Iess and Tortora (2003) found \(\gamma - 1 = (2.1 \pm 2.3)\times10^{-5}\). Convert this into a statement about the delay at closest approach, in nanoseconds. Is general relativity consistent with it?
  4. What was actually measured was the fractional frequency shift \(y = \mathrm d\Delta t/\mathrm dt\). Model the line of sight as sweeping past the Sun with \(b(t) = \sqrt{b_0^2 + v^2t^2}\), where \(v\) is the Earth's orbital speed \(29.78\) km/s scaled by \(r_2/(r_1 + r_2)\). Find the largest \(|y|\), and evaluate it.
Hints · 3 rungs
Rung 1

Part 1 is a one-dimensional integral along a line; part 2 a number of order hundreds of microseconds; part 3 a scaling; part 4 a maximum of a function of \(t\).

Rung 2

Set \(\mathrm ds = 0\) and solve for \(c\,\mathrm dt\) to first order in \(r_g\). Put the line at \(y = b\), parametrised by \(x \in [-x_1, x_2]\), so \(r = \sqrt{x^2 + b^2}\). For part 3, note which factor in the formula carries \(\gamma\).

Rung 3

First line: \(c\,\mathrm dt = \big(1 + (1 + \gamma)r_g/r\big)\,\mathrm dx\), so the one-way excess is \((1 + \gamma)\dfrac{r_g}{c}\displaystyle\int_{-x_1}^{x_2}\frac{\mathrm dx}{\sqrt{x^2 + b^2}}\), and the integrand's antiderivative is \(\operatorname{arsinh}(x/b) = \ln\big((x + \sqrt{x^2 + b^2})/b\big)\).

Worked solution

Idea. Light moves at \(c\) locally, but in these coordinates both the clocks (\(g_{00}\)) and the rulers (\(g_{ij}\)) are off by order \(r_g/r\), so the coordinate speed is less than \(c\) near the Sun. The delay is the integral of the excess.

1. \(\mathrm ds = 0\) gives \(c^2\mathrm dt^2 = \dfrac{1 + 2\gamma r_g/r}{1 - 2r_g/r}|\mathrm d\vec x|^2\), and the square root is \(1 + (1 + \gamma)r_g/r + O(r_g^2)\). Along the line: \[ c\,\Delta t_{\rm one} = (1 + \gamma)r_g\Big[\operatorname{arsinh}\frac xb\Big]_{-x_1}^{x_2} = (1 + \gamma)r_g\ln\frac{(x_2 + r_2)(x_1 + r_1)}{b^2}, \] using \(\operatorname{arsinh}(-s) = -\operatorname{arsinh}s\) and \(\ln\big((x_1 + r_1)/b\big) = -\ln\big((r_1 - x_1)/b\big)\) — or simply \((\sqrt{x_1^2 + b^2} - x_1)(\sqrt{x_1^2 + b^2} + x_1) = b^2\). For \(b \ll r_i\), \(x_i \approx r_i\), so the argument is \(4r_1r_2/b^2\). Doubling for the round trip and writing \(r_g/c = GM/c^3\) gives the formula. The straight line is legitimate by Fermat's principle: the true ray makes the coordinate travel time stationary among nearby paths — unit 11's principle, proved for this metric in the notes of the deck's slide "The Shapiro delay" — so a path that differs from it at order \(r_g\) changes the time only at order \(r_g^2\).

2. \(r_1 = 1.5202\times10^{11}\) m, \(r_2 = 1.1103\times10^{12}\) m, \(b = 1.1131\times10^{9}\) m. \[ \frac{4r_1r_2}{b^2} = 5.449\times10^{5},\qquad \ln = 13.208, \] \[ \Delta t = 4\times4.92549\ \mu\text{s}\times13.208 = 260.2\ \mu\text{s}. \] Half of it, \(130.1\ \mu\)s, is the \(\gamma\) part: the spatial metric.

3. The delay is proportional to \((1 + \gamma)/2\), so a fractional change \((\gamma - 1)/2 = (1.05 \pm 1.15)\times10^{-5}\). At closest approach: \(260.2\ \mu\text{s}\times(1.05 \pm 1.15)\times10^{-5} = 2.7 \pm 3.0\) ns. The experiment says the delay is general relativity's \(260.2\ \mu\)s to within about 3 ns, and \(\gamma = 1\) sits \(0.9\sigma\) from the central value: consistent. The \(g_{ij}\) half, \(130.1\ \mu\)s, is present in full to about 2 parts in \(10^5\).

4. Only the logarithm depends on \(t\): \(y = \dfrac{\mathrm d}{\mathrm dt}\Big(-\dfrac{8GM}{c^3}\ln b\Big) = -\dfrac{8GM}{c^3}\dfrac{\dot b}{b}\), and \(\dot b/b = v^2t/(b_0^2 + v^2t^2)\), which is largest at \(vt = b_0\) with value \(v/2b_0\). So \[ |y|_{\max} = \frac{4GM}{c^3}\,\frac{v}{b_0}. \] \(v = 29.78\times7.4216/8.4378 = 26.19\) km/s (about \(3.3R_\odot\) a day), and \(|y|_{\max} = 19.702\times10^{-6}\times26\,190/1.1131\times10^9 = 4.6\times10^{-10}\).

Check. \([GM/c^3] = \mathsf T\) ✓, and \(y\) is dimensionless ✓. \(M \to 0\): no delay ✓. The exact argument \((x_2 + r_2)(x_1 + r_1)/b^2 = 5.4488\times10^5\) differs from \(4r_1r_2/b^2\) in the fifth figure, so the approximation \(b \ll r_i\) costs nothing ✓.

P5 Proof: the factor of two is \(g_{rr}\) medium

exercises: null geodesics of a general static metric, first-order perturbation, the discriminating prediction, a number

Let \(\mathrm ds^2 = -A(r)c^2\mathrm dt^2 + B(r)\mathrm dr^2 + r^2\mathrm d\Omega^2\) with, to first order, \(A = 1 - 2r_g/r\) and \(B = 1 + 2\gamma r_g/r\).

  1. Prove that null geodesics in the equatorial plane obey, to first order in \(r_g\), \[ u'' + u = \frac{(1 - \gamma)\,r_g}{b^2} + 3\gamma\,r_gu^2, \qquad b = cL/E. \]
  2. Prove that the total deflection of a ray with impact parameter \(b\) is \(\Delta\theta = 2(1 + \gamma)r_g/b\).
  3. Show that \(\gamma = 0\) reproduces, identically, the equation of a Newtonian particle moving at \(c\) far away, so that general relativity's deflection is exactly twice the corpuscle's. Which metric component supplies the difference?
  4. Evaluate both deflections for a ray grazing the Sun, and set them against Eddington's 1919 results \(1.98 \pm 0.12''\) (Sobral) and \(1.61 \pm 0.30''\) (Príncipe), and VLBI's \(1.7510 \pm 0.0004''\).
Hints · 3 rungs
Rung 1

Same three constants as always — \(E\), \(L\), and a normalisation, now null. The answer to part 2 is a sum of two pieces, one from each metric function.

Rung 2

The null normalisation with \(\dot r = -Lu'\) gives \(Bu'^2 = 1/(Ab^2) - u^2\). Expand \(1/A\) and \(1/B\) to first order and differentiate. For part 2, solve with \(u_0 = \sin\varphi/b\) and find where the first-order solution returns to \(u = 0\).

Rung 3

First line: \(u'^2 = (1 - 2\gamma r_gu)\big(\tfrac1{b^2}(1 + 2r_gu) - u^2\big) = \tfrac1{b^2} - u^2 + \tfrac{2(1 - \gamma)r_gu}{b^2} + 2\gamma r_gu^3 + O(r_g^2)\).

Worked solution

Idea. Newton, dressed as a metric, is \(g_{tt}\) with flat space. Put a parameter on the spatial part and watch the deflection split into a \(g_{tt}\) piece and a \(g_{rr}\) piece of equal size.

1. \(E = Ac^2\dot t\) and \(L = r^2\dot\varphi\) are conserved as before (the metric is still independent of \(t\) and \(\varphi\)). The null condition \(-Ac^2\dot t^2 + B\dot r^2 + r^2\dot\varphi^2 = 0\) with \(\dot r = -Lu'\) gives \(BL^2u'^2 = E^2/(Ac^2) - L^2u^2\), i.e. \(Bu'^2 = 1/(Ab^2) - u^2\). With \(1/A = 1 + 2r_gu\), \(1/B = 1 - 2\gamma r_gu\), the expansion in Rung 3 follows. Differentiate in \(\varphi\) and divide by \(2u'\) (valid off turning points, and then everywhere by continuity as in P1): \(u'' = -u + (1 - \gamma)r_g/b^2 + 3\gamma r_gu^2\). \(\square\)

2. Zeroth order: \(u_0 = \sin\varphi/b\). First order: \(u_1'' + u_1 = (1 - \gamma)r_g/b^2 + \tfrac{3\gamma r_g}{2b^2}(1 - \cos2\varphi)\). The constant part is solved by a constant; \(-\tfrac{3\gamma r_g}{2b^2}\cos2\varphi\) by \(C\cos2\varphi\) with \(-3C = -\tfrac{3\gamma r_g}{2b^2}\). So \[ u_1 = \frac{(1 - \gamma)r_g}{b^2} + \frac{\gamma r_g}{b^2}\big(1 + \cos^2\varphi\big). \] No homogeneous term is added: \(A\cos\varphi\) would break the symmetry of the ray about closest approach, and \(B\sin\varphi\) would change the slope \(u' = 1/b\) that part 1's first integral fixes where \(u = 0\). At the incoming end, put \(\varphi = -\delta\): \(0 = -\delta/b + \big[(1 - \gamma) + 2\gamma\big]r_g/b^2\), so \(\delta = (1 + \gamma)r_g/b\). The outgoing end gives the same by symmetry about \(\varphi = \pi/2\). Total \(\Delta\theta = 2\delta = 2(1 + \gamma)r_g/b\). \(\square\)

3. \(\gamma = 0\): \(u'' + u = r_g/b^2 = GM/(b^2c^2)\). Unit 08's Binet equation for \(-GM/r\) is \(u'' + u = GM/L_N^2\) with \(L_N = bc\) for a particle of speed \(c\) far away — the same equation, not merely the same answer. So \(\Delta\theta_{\rm N} = 2r_g/b\) and \(\Delta\theta_{\rm GR} = 4r_g/b = 2\Delta\theta_{\rm N}\), exactly at this order. The difference, the term \(2\gamma r_g/b\), exists only because \(\gamma \ne 0\): it is \(g_{rr}\), the curvature of space.

4. \(r_g/R_\odot = 1476.625/6.957\times10^8 = 2.1225\times10^{-6}\). \[ \Delta\theta_{\rm N} = 4.2450\times10^{-6}\ \text{rad} = 0.8756'',\qquad \Delta\theta_{\rm GR} = 8.4900\times10^{-6}\ \text{rad} = 1.7512''. \] Sobral is \(1.98 - 0.88 = 1.10''\) from Newton, nine of its probable errors, and \(0.23''\), about two, from general relativity. Príncipe is \(2.4\) probable errors from Newton and \(0.5\) from general relativity. VLBI is \((1.7510 - 0.8756)/0.0004 \approx 2200\) standard errors from Newton, and within half of one of \(1.7512''\).

Check. Dimensionless ✓. \(b \to \infty\): no deflection ✓. \(\gamma = 1\) gives \(u'' + u = 3r_gu^2\), the Schwarzschild null orbit equation ✓. The deflection is independent of frequency, as it must be for geodesic motion ✓.

Mercury's ledger

P6 Proof: the secular term is the answer hard

exercises: naive perturbation, the solvability condition, Poincaré–Lindstedt, when an expansion fails

In the scaled form \(w'' + w = 1 + \varepsilon w^2\), with \(w = pu\), \(p = L^2/GM\) and \(\varepsilon = 3r_g/p\):

  1. Expand naively, \(w = w_0 + \varepsilon w_1 + O(\varepsilon^2)\) with \(w_0 = 1 + e\cos\varphi\). Find \(w_1\) explicitly, up to homogeneous terms.
  2. Prove that when \(e \ne 0\) the order-\(\varepsilon\) equation has no \(2\pi\)-periodic solution. Conclude that no orbit near the ellipse closes after one revolution.
  3. Show that \(w_0 + \varepsilon w_1\) agrees, to the order it is valid, with an ellipse whose perihelion turns by \(2\pi\varepsilon\) per orbit.
  4. For Mercury (\(\varepsilon = 7.99\times10^{-8}\)) find after how many orbits the naive expansion stops being accurate, and say why that does not matter for the century of data but would matter over geological time.
  5. Someone deletes the secular term "because secular terms are unphysical". What do they predict, and what is the precise error?
Hints · 3 rungs
Rung 1

Part 1 is a forced linear oscillator with a resonant forcing term. Part 2 is an integral identity. Part 3 is a trigonometric Taylor expansion. Part 4 is where \(\varepsilon\varphi\) stops being small.

Rung 2

Expand \(w_0^2\) into constant, \(\cos\varphi\) and \(\cos2\varphi\) parts, and solve each separately; the \(\cos\varphi\) part resonates with the free oscillation. For part 2 multiply the equation by \(\cos\varphi\) and integrate over a period.

Rung 3

First line: \(w_1'' + w_1 = 1 + \tfrac12e^2 + 2e\cos\varphi + \tfrac12e^2\cos2\varphi\), and \((\varphi\sin\varphi)'' + \varphi\sin\varphi = 2\cos\varphi\).

Worked solution

Idea. The forcing at the free frequency cannot be absorbed by any periodic function; it produces a term growing like \(\varphi\), and that term is the first Taylor coefficient of a shift in frequency. Removing it removes the physics.

1. Order \(\varepsilon\): \(w_1'' + w_1 = w_0^2 = 1 + \tfrac12e^2 + 2e\cos\varphi + \tfrac12e^2\cos2\varphi\). Term by term: the constant gives \(1 + \tfrac12e^2\); \(\tfrac12e^2\cos2\varphi\) gives \(C\cos2\varphi\) with \(-3C = \tfrac12e^2\), so \(-\tfrac16e^2\cos2\varphi\); and \(2e\cos\varphi\), resonant, gives \(e\varphi\sin\varphi\) by the identity in Rung 3 (check: \((\varphi\sin\varphi)' = \sin\varphi + \varphi\cos\varphi\), \((\varphi\sin\varphi)'' = 2\cos\varphi - \varphi\sin\varphi\)). So \[ w_1 = 1 + \tfrac12e^2 - \tfrac16e^2\cos2\varphi + e\,\varphi\sin\varphi . \]

2. Suppose \(w_1\) is \(2\pi\)-periodic and \(C^2\). Integrating by parts twice, \(\int_0^{2\pi}(w_1'' + w_1)\cos\varphi\,\mathrm d\varphi = \int_0^{2\pi}w_1(-\cos\varphi + \cos\varphi)\,\mathrm d\varphi = 0\), the boundary terms cancelling by periodicity. But the right side gives \(\int_0^{2\pi}w_0^2\cos\varphi\,\mathrm d\varphi = 2e\pi \ne 0\). Contradiction. So no periodic \(w_1\) exists: this is unit 21's solvability condition. An orbit that closed after one turn would be a \(2\pi\)-periodic \(w\), and its order-\(\varepsilon\) coefficient would be a periodic \(w_1\); so near the ellipse, for \(e \ne 0\), none closes after one turn. (An orbit may still close after many turns, if its radial frequency \(\nu\) happens to be rational.) \(\square\)

3. \(\cos\big((1 - \varepsilon)\varphi\big) = \cos\varphi\cos\varepsilon\varphi + \sin\varphi\sin\varepsilon\varphi = \cos\varphi + \varepsilon\varphi\sin\varphi + O(\varepsilon^2\varphi^2)\). Hence \(1 + e\cos\varphi + \varepsilon e\varphi\sin\varphi = 1 + e\cos\big((1 - \varepsilon)\varphi\big) + O(\varepsilon^2\varphi^2)\): an ellipse whose argument runs at rate \(1 - \varepsilon\). Perihelia sit at \((1 - \varepsilon)\varphi = 2\pi k\), spaced \(2\pi/(1 - \varepsilon) = 2\pi + 2\pi\varepsilon + O(\varepsilon^2)\) apart. Poincaré–Lindstedt gets this directly, by allowing \(\nu \ne 1\) and choosing \(\nu_1 = -1\) to satisfy the solvability condition.

4. The naive form fails when \(\varepsilon\varphi \sim 1\): \(\varphi \sim 1.25\times10^7\) rad, \(1.99\times10^6\) orbits, \(4.8\times10^5\) years. A century is \(415\) orbits, \(\varepsilon\varphi = 2.1\times10^{-4}\), and the neglected term is \(10^{-4}\) of the secular one: the naive series is perfectly good over the observational record. Over \(10^6\) years it would predict a perihelion "oscillation" of amplitude growing without bound in \(e\varphi\sin\varphi\) — nonsense — while the uniform form just keeps turning.

5. Deleting \(\varepsilon e\varphi\sin\varphi\) leaves a closed, fattened ellipse and predicts zero precession. The error is not an approximation but a contradiction: by part 2, the truncated series with a periodic \(w_1\) does not satisfy the equation at order \(\varepsilon\). Unit 21's method removes a non-uniformity by re-expressing it as a frequency shift; it never removes the effect. Here the accumulated phase is the observable — the \(43''\).

Check. \(e \to 0\): the secular term vanishes, and the circular orbit simply has a slightly different radius, \(w = 1 + \varepsilon\) ✓. Units: \(w\), \(\varphi\), \(\varepsilon\) are all dimensionless ✓. Number: \(2\pi\varepsilon = 5.02\times10^{-7}\) rad per orbit, P2's \(0.1035''\) ✓.

P7 The shadow of Sagittarius A* routine

exercises: the null effective potential, the critical impact parameter, a number against an image

  1. From \(u'^2 = 1/b^2 - W(u)\), \(W(u) = u^2(1 - 2r_gu)\), show that light can orbit only at \(r = 3r_g\) and that a ray from infinity is captured if and only if \(b \lt 3\sqrt3\,r_g\).
  2. Why is the critical impact parameter \(3\sqrt3\,r_g \approx 5.20r_g\) larger than the radius \(3r_g\) of the photon sphere?
  3. Sagittarius A* has \(M = 4.297\times10^6M_\odot\) at \(D = 8.277\) kpc (both from stellar orbits, independent of any image). Compute the angular diameter \(2b_c/D\) of its shadow in microarcseconds, and compare with the Event Horizon Telescope's 2022 values: ring diameter \(51.8 \pm 2.3\ \mu\)as, inferred shadow diameter \(48.7 \pm 7.0\ \mu\)as.
Hints · 3 rungs
Rung 1

A one-dimensional motion in a potential \(W\): the ray turns where \(W(u) = 1/b^2\), and fails to turn if the barrier is too low.

Rung 2

Find the maximum of \(W\) on \((0, 1/2r_g)\). A circular orbit needs \(u' = 0\) and \(u'' = 0\) at once. For part 3 the only work is units.

Rung 3

First line: \(W'(u) = 2u - 6r_gu^2 = 2u(1 - 3r_gu)\), zero at \(u = 1/3r_g\), where \(W = \tfrac19r_g^{-2}\cdot\tfrac13 = 1/(27r_g^2)\).

Worked solution

Idea. Light has an effective potential with a single peak. Rays with \(1/b^2\) above the peak go over it and into the hole; those below turn back. The boundary between them is the edge of a dark disc.

1. \(W \ge 0\) on \([0, 1/2r_g]\), zero at both ends, with a single critical point at \(u = 1/3r_g\) where \(W = 1/27r_g^2\): a maximum. A circular orbit needs \(u' = 0\) and, from \(u'' = -\tfrac12W'(u)\), \(W'(u) = 0\): only \(u = 1/3r_g\), \(r = 3r_g\), and it sits on a maximum, so it is unstable. A ray from infinity starts at \(u = 0\) with \(u'^2 = 1/b^2\). If \(1/b^2 \gt 1/27r_g^2\), \(u'^2 \gt 0\) for all \(u\) up to the horizon: no turning point, capture. If \(1/b^2 \lt 1/27r_g^2\), the ray reaches the first root of \(W(u) = 1/b^2\), which lies below \(1/3r_g\), turns and escapes. So capture iff \(b \lt b_c = 3\sqrt3\,r_g\).

2. \(b\) is the perpendicular distance of the ray's straight asymptote far away. The ray that ends up skimming the photon sphere was being pulled inwards all the way in, so its asymptote lies further out than its final radius: \(b_c/3r_g = \sqrt3\). The shadow is therefore a lensed image of the photon sphere, magnified by \(\sqrt3\).

3. \(r_g = 4.297\times10^6\times1476.625 = 6.345\times10^9\) m; \(b_c = 5.196\,r_g = 3.297\times10^{10}\) m; \(D = 8277\times3.0857\times10^{16} = 2.554\times10^{20}\) m. \[ \frac{2b_c}{D} = 2.582\times10^{-10}\ \text{rad} = 53.3\ \mu\text{as}. \] The EHT's ring, \(51.8 \pm 2.3\), is within one standard error of it; the inferred shadow, \(48.7 \pm 7.0\), within one too. No parameter was fitted to the image: mass and distance came from the S2 orbit and its neighbours.

Check. \(b_c\) is a length ✓, \(2b_c/D\) an angle ✓. Scaling: the shadow is \(\propto M/D\), and M87*'s \(M/D\) (\(6.5\times10^9M_\odot\) at \(16.8\) Mpc) is \(0.745\) of Sgr A*'s, giving \(39.7\ \mu\)as — the deck's number ✓.

P8 Jupiter's share of the 531 medium

exercises: unit 10's Legendre expansion and averaging, linearisation of the Binet equation, a Newtonian number

  1. Averaging an outer planet (mass \(m'\), circular orbit of radius \(a'\), coplanar) over its orbit replaces it by a uniform ring. Using the expansion \(1/|\vec r - \vec r\,{}'| = \sum_n r^nP_n(\cos\psi)/a'^{n+1}\) for \(r \lt a'\), show the ring's potential in the plane is \(-\frac{Gm'}{a'}\big(1 + \frac{r^2}{4a'^2} + \dots\big)\).
  2. Keeping only the \(r^2\) term, show that a near-circular orbit of radius \(a\) precesses forward by \(\Delta\varpi = \tfrac{3\pi}{2}\,\frac{m'}{M}\big(\frac a{a'}\big)^3\) per orbit.
  3. Evaluate for Jupiter on Mercury: \(m'/M = 1/1047.35\), \(a' = 5.20336\) au, \(a = 0.387098\) au, \(415.20\) orbits per century. Compare with Clemence's \(153.6''\) per century.
  4. Do the same for Venus (\(m'/M = 1/408\,524\), \(a' = 0.723332\) au) and compare with Clemence's \(277.9''\). Explain the discrepancy.
Hints · 3 rungs
Rung 1

Part 1 is three averages of Legendre polynomials; part 2 is the near-circular precession computation of the deck, for a different extra force; parts 3 and 4 are numbers.

Rung 2

In the plane, \(\psi\) runs uniformly over the ring. The \(r^2\) term gives a force \(+Gm'r/2a'^3\), outward. Put \(f = -GM/r^2 + \kappa r\) into Binet, linearise about the circular orbit, and read off the frequency \(\nu\) of the radial oscillation in \(\varphi\).

Rung 3

First line: \(u'' + u = GM/L^2 - \kappa/(L^2u^3)\), and with \(u = u_c + \delta\), \(\delta'' + \delta = +\dfrac{3\kappa}{L^2u_c^4}\,\delta\).

Worked solution

Idea. Unit 10's recipe: average over the fast angle, keep what survives. What survives from an outer ring is a small outward force growing with \(r\); any force that is not exactly inverse-square makes the ellipse turn.

1. \(\langle P_0\rangle = 1\); \(\langle P_1\rangle = \langle\cos\psi\rangle = 0\); \(\langle P_2\rangle = \tfrac32\langle\cos^2\psi\rangle - \tfrac12 = \tfrac34 - \tfrac12 = \tfrac14\). So \(\Phi = -\frac{Gm'}{a'}\big(1 + \frac14\frac{r^2}{a'^2} + \dots\big)\).

2. Force per unit mass \(-\Phi' = +Gm'r/2a'^3 =: \kappa r\). Binet (\(u'' + u = -f(1/u)/(L^2u^2)\)) gives \(u'' + u = GM/L^2 - \kappa/(L^2u^3)\). About \(u_c\): \(-\kappa/(L^2(u_c + \delta)^3) \approx -\kappa/(L^2u_c^3) + 3\kappa\delta/(L^2u_c^4)\), so \(\delta'' + (1 - 3\kappa/(L^2u_c^4))\delta = 0\). With \(L^2 \approx GM/u_c = GMa\): \(3\kappa/(L^2u_c^4) = 3\kappa a^3/GM = \tfrac32(m'/M)(a/a')^3 =: 3\eta\). Then \(\nu = \sqrt{1 - 3\eta}\) and perihelia are \(2\pi/\nu \approx 2\pi(1 + \tfrac32\eta)\) apart: \(\Delta\varpi = 3\pi\eta = \tfrac{3\pi}2(m'/M)(a/a')^3\), forward since \(\nu \lt 1\). \(\square\)

3. \(a/a' = 0.074394\), cubed \(4.1174\times10^{-4}\). \(\Delta\varpi = 4.7124\times4.1174\times10^{-4}/1047.35 = 1.8525\times10^{-6}\) rad \(= 0.3821''\) per orbit, \(\times415.20 = 158.7''\) per century, against Clemence's \(153.6''\): \(3\%\) high.

4. Venus: \(a/a' = 0.5352\), and the same formula gives \(151.4''\) per century against \(277.9''\) — barely half. The quadrupole term is the first of a series in \(a/a'\), and at \(0.535\) the series converges slowly: the \(r^4\), \(r^6\), \(\dots\) terms matter. Summing all multipoles of the ring gives \(286.3''\) (the deck's table), \(3\%\) high like Jupiter's. The remaining few per cent come from Mercury's own eccentricity and the planets', which the ring model drops.

Check. \(\eta\) is dimensionless ✓. \(m' \to 0\) or \(a' \to \infty\): no precession ✓. The sign — forward — agrees with Clemence's tabulated contributions, all positive ✓. And the relativistic \(43''\) is \(27\%\) of Jupiter's share: a small residual, but larger than the whole of Saturn's, Mars's, Uranus's and Neptune's put together.

P9 What the perihelion measures: \(\beta\) and \(\gamma\) hard

exercises: post-Newtonian bookkeeping, the orbit equation of a general static metric, bounding a parameter from data

The parametrised post-Newtonian metric of a static spherical mass, written in areal coordinates (cited: Will, Theory and Experiment in Gravitational Physics, ch. 4), is \[ A(r) = 1 - \frac{2r_g}{r} + 2(\beta - \gamma)\frac{r_g^2}{r^2},\qquad B(r) = 1 + 2\gamma\frac{r_g}{r}, \] with general relativity at \(\beta = \gamma = 1\), where \(A\) reduces exactly to Schwarzschild's.

  1. Explain why, for a planet, the \(r_g^2\) term in \(A\) must be kept while \(r_g^2\) terms in \(B\) may be dropped. (Count powers of \(v^2/c^2\), using \(r_g/r \sim v^2/c^2\).)
  2. Derive the timelike orbit equation to this order and show that it has the form \(u'' + (1 - \delta)u = K + 3\gamma r_gu^2\) with \(K \approx GM/L^2\) and \(\delta\) a small constant; find \(\delta\) in terms of \(\beta\), \(\gamma\), \(r_g\) and \(p = L^2/GM\).
  3. Show that the perihelion advance per orbit is \(\tfrac13(2 + 2\gamma - \beta)\cdot6\pi r_g/p\).
  4. Given Cassini's \(\gamma = 1\) to \(2\times10^{-5}\) and Mercury's \(42.98 \pm 0.04''\) against a prediction of \(42.98''\), bound \(|\beta - 1|\).
Hints · 3 rungs
Rung 1

The answer to part 2 is the relativistic orbit equation with one extra piece: a small change in the coefficient of the linear term \(u\). Both it and the \(u^2\) term shift the radial frequency, and the shifts add at first order.

Rung 2

Redo P1 with general \(A\), \(B\): \(BL^2u'^2 = E^2/(Ac^2) - c^2 - L^2u^2\). Write \(x = r_gu\), \(E^2/c^4 = 1 + \eta\) with \(\eta \sim v^2/c^2\), and keep every term of \(c^2\times\) (second order in small quantities). You need \(1/A = 1 + 2x + (4 - 2\beta + 2\gamma)x^2\).

Rung 3

First line: \(L^2u'^2 = (1 - 2\gamma x)\big[c^2\eta + 2c^2x(1 + \eta) + (4 - 2\beta + 2\gamma)c^2x^2 - L^2u^2\big]\) up to terms of third order, and \(c^2x = GMu\), \(c^2x^2 = (GM)^2u^2/c^2\).

Worked solution

Idea. The perihelion is the only classical test that sees beyond first order in \(g_{tt}\). In the energy equation, \(g_{tt}\) is multiplied by \(c^2\), so its \(r_g^2\) term competes with \(g_{rr}\)'s \(r_g\) term multiplied by the kinetic energy. Both shift the radial frequency.

1. In \(-Ac^2\dot t^2\) the metric function multiplies \(c^2\), so a term \(x^2\) in \(A\) contributes \(c^2x^2 \sim c^2(v/c)^4 = v^4/c^2\): the same size as the first relativistic correction to Newton's \(v^2\). In \(B\dot r^2\) the metric multiplies \(v^2\), so the \(x\) term is already \(v^4/c^2\), and an \(x^2\) term would be \(v^6/c^4\): next order.

2. \(1/A\): with \(y = 2x - 2(\beta - \gamma)x^2\), \(1/(1 - y) = 1 + y + y^2 + \dots = 1 + 2x + (4 - 2\beta + 2\gamma)x^2\). Multiply out Rung 3, dropping third-order terms: \[ L^2u'^2 = c^2\eta + 2GMu(1 + \eta - \gamma\eta) + (4 - 2\beta + 2\gamma - 4\gamma)\frac{(GM)^2}{c^2}u^2 - L^2u^2 + 2\gamma r_gL^2u^3 . \] Differentiate and divide by \(2L^2u'\): \[ u'' + \Big(1 - (4 - 2\beta - 2\gamma)\frac{r_gGM}{L^2}\Big)u = \frac{GM}{L^2}\big(1 + (1 - \gamma)\eta\big) + 3\gamma r_gu^2 . \] So \(\delta = (4 - 2\beta - 2\gamma)\,r_g/p\), and \(K = (GM/L^2)(1 + O(\eta))\). At \(\beta = \gamma = 1\), \(\delta = 0\) and P1's equation returns ✓.

3. About the near-circular orbit \(u \approx 1/p\), the \(u^2\) term contributes \(-6\gamma r_g/p\) to the coefficient of \(\delta u\) (as in the deck's "Nearly circular orbits precess, exactly", with \(3r_g \to 3\gamma r_g\)); Poincaré–Lindstedt shows the same shift holds for any \(e\), as in P6. So \[ \nu^2 = 1 - \delta - \frac{6\gamma r_g}{p} = 1 - (4 - 2\beta + 4\gamma)\frac{r_g}{p},\qquad \nu = 1 - (2 - \beta + 2\gamma)\frac{r_g}{p}. \] Advance per orbit \(2\pi(1/\nu - 1) = 2\pi(2 + 2\gamma - \beta)r_g/p = \tfrac13(2 + 2\gamma - \beta)\cdot6\pi r_g/p\). \(\square\)

4. Observed over predicted is \(1.0000 \pm 0.00093\), so \(\tfrac13(2 + 2\gamma - \beta) = 1 \pm 0.00093\), i.e. \(\beta - 1 = 2(\gamma - 1) \mp 0.0028\). With \(|\gamma - 1| \lt 2\times10^{-5}\), the first term is negligible: \(|\beta - 1| \lesssim 3\times10^{-3}\). (Modern ephemeris fits, which also solve for the Sun's quadrupole moment, do better; this crude bound already says the second-order part of \(g_{tt}\) is general relativity's to three parts in a thousand.)

Check. \(\beta = \gamma = 1\): factor 1, the \(43''\) ✓. Pure "Newton in a metric", \(\gamma = 0\) and no \(x^2\) term in isotropic coordinates (\(\beta = 0\)): factor \(\tfrac23\), so \(28.7''\) — curved space is needed for the perihelion too ✓. \(\delta\) and \(r_g/p\) are dimensionless ✓.

Near the hole, and far from the Sun

P10 Kepler's third law survives, and the last orbit before the plunge medium

exercises: coordinate time against proper time on a geodesic, circular orbits, a frequency for a gravitational-wave source

  1. Using P3's circular-orbit \(E\) and \(L\), show that for every circular geodesic of Schwarzschild, exactly, \[ \Big(\frac{\mathrm d\varphi}{\mathrm dt}\Big)^2 = \frac{GM}{r^3}, \] and that \(\mathrm d\tau/\mathrm dt = \sqrt{1 - 3r_g/r}\). Evaluate the second at the ISCO.
  2. Explain why the first result does not contradict the statement that the orbit equation differs from Newton's.
  3. Compute the orbital frequency \(f = (1/2\pi)\,\mathrm d\varphi/\mathrm dt\) at the ISCO, seen from far away, as a function of \(M/M_\odot\), and the gravitational-wave frequency \(2f\) for a total mass of \(65\,M_\odot\). GW150914's signal swept from \(35\) Hz to about \(150\) Hz at peak amplitude. Where does your number fall, and why is it only indicative?
Hints · 3 rungs
Rung 1

\(\mathrm d\varphi/\mathrm dt = \dot\varphi/\dot t\), and both are known in terms of \(E\), \(L\) and \(r\).

Rung 2

\(\dot\varphi = L/r^2\), \(\dot t = E/((1 - 2x)c^2)\), with \(x = r_g/r\). Substitute P3's \(L^2 = c^2r_gr/(1 - 3x)\) and \(E/c^2 = (1 - 2x)/\sqrt{1 - 3x}\).

Rung 3

First line: \(\dfrac{\mathrm d\varphi}{\mathrm dt} = \dfrac{L}{r^2}\cdot\dfrac{(1 - 2x)c^2}{E} = \dfrac{L}{r^2}\sqrt{1 - 3x}\).

Worked solution

Idea. Two clocks, one orbit. The far-away observer's clock makes Kepler's third law exact for circles; the orbiting clock runs slow by both special- and general-relativistic factors.

1. \((\mathrm d\varphi/\mathrm dt)^2 = \dfrac{L^2}{r^4}(1 - 3x) = \dfrac{c^2r_gr}{r^4} = \dfrac{GM}{r^3}\). And \(\mathrm d\tau/\mathrm dt = 1/\dot t = (1 - 2x)c^2/E = \sqrt{1 - 3x}\). At the ISCO, \(x = 1/6\): \(\sqrt{1/2} = 0.7071\). An orbiting clock there ticks \(71\%\) as fast as a distant one.

2. The orbit equation is about the shape \(r(\varphi)\) of non-circular orbits, which precess; for a circle there is no shape to get wrong. And "\(r\)" here is the areal coordinate, "\(t\)" the far-away clock: the exact Keplerian form is a coincidence of these coordinates, not a statement that local physics is Newtonian. In proper time the orbiter measures \((\mathrm d\varphi/\mathrm d\tau)^2 = GM/(r^3(1 - 3x))\).

3. At \(r = 6r_g\): \(f = \dfrac{1}{2\pi}\sqrt{\dfrac{GM}{216\,r_g^3}} = \dfrac{c^3}{2\pi\cdot6^{3/2}GM} = 2198.6\ \text{Hz}\times\dfrac{M_\odot}{M}\). For \(65M_\odot\): \(f = 33.8\) Hz, gravitational-wave frequency \(2f = 67.6\) Hz — inside the observed \(35\)–\(150\) Hz sweep, which is where the orbit should stop being a slow inspiral. Only indicative: the ISCO is a property of a test particle orbiting a fixed hole, and GW150914 was two nearly equal masses; the true transition is smeared and shifted, which is why the merger peak is at higher frequency. The honest computation is numerical relativity, cited in unit 38.

Check. \([GM/r^3] = \mathsf{T^{-2}}\) ✓. \(x \to 0\): \(\mathrm d\tau/\mathrm dt \to 1 - \tfrac32x = 1 - \tfrac12v^2/c^2 - GM/(rc^2)\), time dilation (unit 27) plus gravitational redshift (unit 30) with \(v^2 = GM/r\) ✓. \(x \to 1/3\): \(\mathrm d\tau/\mathrm dt \to 0\): at the photon sphere only light can orbit ✓.

P11 S2: precession and redshift at a black hole medium

exercises: the perihelion formula far from the Sun, the exact redshift to first order, adding two clock effects, comparing with data

The star S2 orbits Sagittarius A* (\(M = 4.26\times10^6M_\odot\)) with \(a = 1031\) au and \(e = 0.8847\).

  1. Check the period against \(16.05\) years using Kepler's third law.
  2. Compute \(\varepsilon = 3r_g/p\) and the pericentre advance per orbit, in arcminutes. Is first order in \(\varepsilon\) enough? The GRAVITY collaboration (2020) measured the advance as \(1.10 \pm 0.19\) times this prediction.
  3. At pericentre, compute the gravitational redshift \(r_g/r_p\) and the transverse-Doppler redshift \(v_p^2/2c^2\), each as an equivalent velocity \(cz\) in km/s. GRAVITY (2018) measured the sum as \(f = 0.90 \pm 0.09\,(\text{stat}) \pm 0.15\,(\text{sys})\) times the prediction. What does \(f = 0\) mean, physically?
Hints · 3 rungs
Rung 1

Four numbers: a period, an angle, two velocities. Each is a formula from the deck with new inputs.

Rung 2

\(r_g\) scales with \(M\): \(4.26\times10^6\times1476.6\) m. \(r_p = a(1 - e)\) and \(v_p^2 = GM(2/r_p - 1/a)\) (vis-viva, unit 08). The two redshifts add at first order because the clock rate factor is \(\sqrt{1 - 2x}\,\sqrt{1 - v^2/c^2} \approx 1 - x - v^2/2c^2\).

Rung 3

First line: \(r_g = 6.290\times10^9\) m, \(p = 1031\times1.49598\times10^{11}\times(1 - 0.8847^2)\) m.

Worked solution

Idea. The same formulas as for Mercury and the Sun, with a mass a million times larger and a pericentre only \(1400\) Schwarzschild radii out. Every effect is \(10^4\) times bigger, which makes one orbit enough.

1. \(T = 2\pi\sqrt{a^3/GM}\) with \(a = 1.5424\times10^{14}\) m and \(GM = 5.654\times10^{26}\ \mathrm{m^3\,s^{-2}}\): \(T = 16.04\) yr ✓.

2. \(1 - e^2 = 0.21731\), \(p = 3.352\times10^{13}\) m, \(\varepsilon = 3\times6.290\times10^9/3.352\times10^{13} = 5.63\times10^{-4}\). \(2\pi\varepsilon = 3.538\times10^{-3}\) rad \(= 12.2'\) per orbit. The neglected terms are of relative order \(\varepsilon\), a few parts in \(10^4\), far below the \(17\%\) measurement error: first order is plenty. The measured \(1.10 \pm 0.19\) is half a standard error from 1.

3. \(r_p = 1031\times0.1153\) au \(= 118.9\) au \(= 1.778\times10^{13}\) m, i.e. \(1414\) Schwarzschild radii. Gravitational: \(r_g/r_p = 3.537\times10^{-4}\), \(cz = 106.0\) km/s. Speed: \(v_p^2 = 5.654\times10^{26}(2/1.778\times10^{13} - 1/1.5424\times10^{14})\), \(v_p = 7741\) km/s, and \(v_p^2/2c^2 = 3.333\times10^{-4}\), \(cz = 99.9\) km/s. Total about \(206\) km/s, on top of the ordinary Doppler shift of the orbit (thousands of km/s along the line of sight, which the fit models). \(f = 0\) would be pure Newton with Galilean clocks; \(f = 1\) is relativity. The measurement excludes \(0\) at about five combined standard errors and is consistent with \(1\). Like the Sun's redshift, this tests the equivalence principle and special relativity, not the field equations.

Check. Both redshifts dimensionless ✓. Virial-like: for a circular orbit \(v^2/2c^2 = x/2\), half the gravitational term; at pericentre of an eccentric orbit \(v^2\) is larger, \(v_p^2/c^2 = x(1 + e)\), so the two terms are nearly equal: \((1 + e)/2 = 0.94\), and \(99.9/106.0 = 0.94\) ✓.