The equivalence principle, and why flat spacetime cannot hold gravity

Idea. One formula, \(\Delta\nu/\nu = \Delta\Phi/c^2\) with \(\Delta\Phi = \Phi_{\text{emit}} - \Phi_{\text{receive}}\), applied six times. The whole exercise is in noticing that \(gh\) is a special case of it, valid when the field does not change appreciably over \(h\), and that the range of the answer is thirteen orders of magnitude.

1–3, uniform field. \(\Delta\nu/\nu = -gh/c^2\) with \(c^2 = 8.987552\times10^{16}\ \mathrm{m^2\,s^{-2}}\):

• \(h = 10^{-3}\) m: \(-1.091\times10^{-19}\).
• \(h = 0.33\) m: \(-3.601\times10^{-17}\).
• \(h = 22.5\) m: \(-2.455\times10^{-15}\).

4, GPS. Emitter on the ground, receiver in orbit: \[\frac{\Delta\nu}{\nu} = \frac{1}{c^2}\Big(-\frac{GM_\oplus}{R_\oplus} + \frac{GM_\oplus}{a}\Big) = -\frac{GM_\oplus}{c^2}\Big(\frac1{R_\oplus} - \frac1a\Big).\] With \(1/R_\oplus = 1.567855\times10^{-7}\) and \(1/a = 3.764919\times10^{-8}\), the difference is \(1.191363\times10^{-7}\ \mathrm{m^{-1}}\); times \(GM_\oplus\) that is \(4.74858\times10^{7}\ \mathrm{m^2\,s^{-2}}\), so \(\Delta\nu/\nu = -5.284\times10^{-10}\). (Read the other way: the satellite clock gains at that rate, which is P9.)

5, the Sun. \(\Delta\nu/\nu = -GM_\odot/(R_\odot c^2) = -1.32712\times10^{20}/(6.957\times10^{8}\times8.987552\times10^{16}) = -2.123\times10^{-6}\), an equivalent \(636\ \mathrm{m\,s^{-1}}\). It is measured, and it is a nuisance: it must be separated from convective Doppler shifts in solar spectra, which is why the solar redshift was not a clean test.

6, Sirius B. \(GM = 1.018\times1.32712\times10^{20} = 1.35101\times10^{20}\ \mathrm{m^3\,s^{-2}}\); \(R = 0.0084\times6.957\times10^{8} = 5.844\times10^{6}\) m. Then \[\Big|\frac{\Delta\nu}{\nu}\Big| = \frac{1.35101\times10^{20}}{5.844\times10^{6}\times8.987552\times10^{16}} = 2.572\times10^{-4},\] an equivalent \(c\times2.572\times10^{-4} = 77.1\ \mathrm{km\,s^{-1}}\), against the measured \(80.65\pm0.77\ \mathrm{km\,s^{-1}}\). The \(4\%\) discrepancy is not a failure of the physics: it is the uncertainty in the star's mass and radius, and the measurement is nowadays used backwards, assuming the redshift and solving for \(M/R\).

Check. Dimensions: \(GM/Rc^2\) is \(\mathrm{(m^3s^{-2})/(m\cdot m^2s^{-2})}\) — dimensionless ✓. Limit: part 4's formula with \(a\to\infty\) gives \(-GM_\oplus/R_\oplus c^2 = -6.95\times10^{-10}\), the Earth's own surface redshift, and the orbital answer is correctly smaller ✓. Consistency: part 3's uniform-field answer against the exact \(GM_\oplus h/(R_\oplus(R_\oplus+h)c^2) = 2.453\times10^{-15}\) — a \(0.09\%\) difference, which is the size of \(g - GM_\oplus/R_\oplus^2\) and confirms the uniform-field approximation ✓. Ordering: \(10^{-19}\) to \(10^{-4}\), monotone in \(\Delta\Phi\) ✓.

Another way

Parts 5 and 6 can be quoted as \(z = \Delta\lambda/\lambda = +GM/Rc^2\) to first order, since a fractional frequency drop is a fractional wavelength rise of the same size. Astronomers publish \(z\), and often as \(cz\) in \(\mathrm{km\,s^{-1}}\), which is where the \(80.65\) comes from: it is not a velocity and nothing is moving.

Idea. The equivalence principle is not an identity but an inequality: it holds to the accuracy with which the tidal term is invisible. Putting numbers on that term turns "locally" from a hedge into a specification.

1. \(R_\oplus^3 = (6.378137\times10^6)^3 = 2.5946\times10^{20}\ \mathrm{m^3}\), so \[\frac{GM_\oplus}{R_\oplus^3} = \frac{3.986004\times10^{14}}{2.5946\times10^{20}} = 1.5362\times10^{-6}\ \mathrm{s^{-2}} .\] Dimensions: \(\mathrm{(m^3s^{-2})/m^3 = s^{-2}}\) — the square of a frequency, as unit 05 noted. The corresponding period \(2\pi/\sqrt{GM_\oplus/R_\oplus^3} = 5070\ \mathrm s = 84.5\) minutes is unit 05's tunnel time, and it is not a coincidence.

2. Vertical: \(\ddot\xi = 2(GM_\oplus/R_\oplus^3)\xi = +3.072\times10^{-6}\ \mathrm{m\,s^{-2}}\) for \(\xi = 1\) m — separating, because the lower ball is nearer the centre and falls harder. Horizontal: \(\ddot\xi = -1.536\times10^{-6}\ \mathrm{m\,s^{-2}}\) — converging, because the two balls fall along radii that meet at the centre. The magnitudes are in the ratio \(2:1\) and the signs are opposite, which is the eigenvalue pattern \((-2,1,1)\) and the trace-free condition.

3. Over \(1\) s the acceleration is constant to a part in \(10^{6}\), so the extra separation is \[\tfrac12(3.072\times10^{-6})(1)^2 = 1.536\times10^{-6}\ \mathrm m = 1.54\ \mu\mathrm m ,\] while both balls have fallen \(\tfrac12g(1)^2 = 4.903\) m. The tidal residue is \(1.536\times10^{-6}/4.903 = 3.13\times10^{-7}\) of the fall.

4. \(\tfrac12(3.072\times10^{-6})t^2 = 10^{-15}\) gives \[t = \sqrt{\frac{2\times10^{-15}}{3.072\times10^{-6}}} = 2.55\times10^{-5}\ \mathrm s = 26\ \mu\mathrm s .\] Twenty-six microseconds. The elevator's cover is not hard to blow.

5. \(2(1.5362\times10^{-6})L \le 10^{-9}\times9.80665\) gives \(L \le 3.19\times10^{-3}\) m: about three millimetres. To keep the equivalence principle good to a part in \(10^{9}\) on the Earth's surface you need a laboratory the size of a pea.

Check. Dimensions throughout: \(\mathrm{s^{-2}}\) times a length is an acceleration ✓. Limit: \(r\to\infty\) sends \(GM/r^3\to0\) and every answer to the deep-space one ✓. Sanity: the vertical relative acceleration \(3.07\times10^{-6}\ \mathrm{m\,s^{-2}}\) should equal \(g(r) - g(r+1\,\mathrm m) = 2GM/R^3\), and \(2g/R_\oplus = 2\times9.80665/6.378137\times10^6 = 3.075\times10^{-6}\) — agreeing to the three figures that \(g \ne GM/R^2\) allows ✓.

Idea. The chart is not guessed: it is unit 27's Born-rigid family of hyperbolae, relabelled. Once the metric is in hand, the horizon is a statement about one null coordinate being monotone, which is three lines and no calculation.

1. On \(X = \) const, \(x^2 - c^2t^2 = X^2(\cosh^2 - \sinh^2) = X^2\), so the worldline is unit 27's hyperbola with \(c^2/\alpha = X\), i.e. proper acceleration \(\alpha = c^2/X\). Its four-velocity is \(u = \mathrm d(ct,x)/\mathrm d\tau\); since unit 27 showed \(\theta = \alpha\tau/c\) for the rapidity, and here \(\theta = gT/c\), we get \(\tau = c\theta/\alpha = (X/c)\,\theta = (X/c^2)\,gT\). For \(X = c^2/g\) that is \(\tau = T\): the reference observer's proper time is the coordinate \(T\).

2. With \(\theta = gT/c\), \[\mathrm d(ct) = \sinh\theta\,\mathrm dX + \frac{gX}{c}\cosh\theta\,\mathrm dT,\qquad \mathrm dx = \cosh\theta\,\mathrm dX + \frac{gX}{c}\sinh\theta\,\mathrm dT .\] Then \[-\mathrm d(ct)^2 + \mathrm dx^2 = (\cosh^2\theta - \sinh^2\theta)\,\mathrm dX^2 + \Big(\frac{gX}{c}\Big)^2(\sinh^2\theta - \cosh^2\theta)\,\mathrm dT^2,\] the cross terms cancelling because they are \(2(gX/c)(-\sinh\theta\cosh\theta + \cosh\theta\sinh\theta)\,\mathrm dX\,\mathrm dT = 0\). Hence \[\mathrm ds^2 = -\Big(\frac{gX}{c}\Big)^2\mathrm dT^2 + \mathrm dX^2 .\]

3. Into the wedge. \(x = X\cosh\theta \gt 0\) since \(X \gt 0\), and \(x^2 - c^2t^2 = X^2 \gt 0\); together these give \(x \gt |ct|\). Onto. Given \(x \gt |ct|\), put \(X := \sqrt{x^2 - c^2t^2} \gt 0\) and \(\theta := \operatorname{artanh}(ct/x)\), legitimate because \(|ct/x| \lt 1\); then \(X\cosh\theta = x\) and \(X\sinh\theta = ct\), using \(\cosh(\operatorname{artanh}s) = 1/\sqrt{1-s^2}\). Injective. \(X\) and \(\theta\) are determined by \((ct,x)\) through those two formulas, and \(T = c\theta/g\). So the map is a bijection with the stated inverse. \(\square\)

4. A null curve has \(\mathrm ds^2 = 0\), so \((gX/c)^2\mathrm dT^2 = \mathrm dX^2\), i.e. \(\mathrm dT/\mathrm dX = \pm c/(gX)\). The outgoing sign integrates to \(T = (c/g)\ln X + \) const, so \(T(X) = T_1 + (c/g)\ln(X/X_1)\) and the coordinate flight time from \(X_1\) to \(X_2\) is \((c/g)\ln(X_2/X_1)\). Note that it depends only on the ratio, and diverges as \(X_1\to0\).

5. Use the null coordinate \(u := ct - x\). Along any future-directed causal curve \(\mathrm d(ct) \ge |\mathrm dx|\), so \(\mathrm du = \mathrm d(ct) - \mathrm dx \ge 0\): \(u\) is non-decreasing. At \(E\), \(u_E = ct_E - x_E \ge 0\). But on the observer's worldline \[u = X(\sinh\theta - \cosh\theta) = -X e^{-\theta} \lt 0\] for every \(\theta\in\mathbb R\), since \(X \gt 0\). A signal from \(E\) has \(u \ge u_E \ge 0\) for ever, so it never meets a point with \(u \lt 0\). Hence it never reaches the observer, at any \(X\), ever. \(\square\)

Check. Dimensions: \(gX/c\) is \(\mathrm{m\,s^{-1}}\) and multiplies \(\mathrm dT\) in seconds, giving metres, matching \(\mathrm dX\) ✓. Limit: at \(X = c^2/g\) and \(|gT/c| \ll 1\), \(x \approx c^2/g\) and \(ct \approx cT\), and the metric is \(-c^2\mathrm dT^2 + \mathrm dX^2\), Minkowski ✓. A number: \(u = -Xe^{-\theta}\) tends to \(0^-\) as \(\theta\to+\infty\), so the observer approaches the horizon asymptotically and never touches it — consistent with part 4's divergent flight time ✓.

Another way

Part 5 also follows from part 4. A signal emitted at \(X_1\) arrives at \(X_2\) after coordinate time \((c/g)\ln(X_2/X_1)\), which diverges as \(X_1\to0^+\). Pushing the emitter down to the horizon pushes the arrival to \(T = +\infty\), and events beyond it are worse than that. The null-coordinate argument is cleaner because it never has to take a limit.

Idea. A static metric hands you clock rates by inspection. The instructive part is that the "approximate" elevator answer turns out to be exact, once you are careful about whose acceleration \(g\) is.

1. Static means \(\mathrm dX = 0\), so \(-c^2\mathrm d\tau^2 = \mathrm ds^2 = -(gX/c)^2\mathrm dT^2\), giving \(\mathrm d\tau = (gX/c^2)\,\mathrm dT = (X/X_0)\,\mathrm dT\) with \(X_0 = c^2/g\). At fixed \(T\), \(\mathrm ds^2 = \mathrm dX^2\), so the proper distance is simply \(X_2 - X_1\): the coordinate \(X\) is proper distance along a line of simultaneity.

2. Since the metric components do not depend on \(T\), two signals emitted \(\Delta T\) apart arrive \(\Delta T\) apart (P3 part 4: the null curves differ by a \(T\)-translation). Converting to proper time at each end, \[\frac{\mathrm d\tau_2}{\mathrm d\tau_1} = \frac{gX_2/c^2}{gX_1/c^2} = \frac{X_2}{X_1},\] exactly, and therefore \(\nu_2/\nu_1 = X_1/X_2\). The ratio is unbounded: as \(X_1\to0^+\) at fixed \(X_2\) it grows without limit, so a Rindler observer near the horizon ages arbitrarily slowly compared with one far from it — in flat spacetime, with no matter anywhere.

3. \(X_1 = 8.987552\times10^{16}/9.80665 = 9.16475\times10^{15}\) m, which is \(9.16475\times10^{15}/9.4607\times10^{15} = 0.96872\) ly. Then \[\frac{\nu_2}{\nu_1} = \frac{X_1}{X_1+h} = \frac{1}{1 + h/X_1},\qquad \frac{h}{X_1} = \frac{gh}{c^2} = \frac{9.80665\times22.5}{8.987552\times10^{16}} = 2.45506\times10^{-15}.\] So \(\Delta\nu/\nu = -2.45506\times10^{-15}/(1 + 2.45506\times10^{-15}) = -2.45506\times10^{-15}\) to fifteen significant figures: the correction is \((gh/c^2)^2 = 6.03\times10^{-30}\), which first shows up in the thirtieth decimal place.

4. \(\nu_2/\nu_1 = X_1/X_2 = (X_2 - h)/X_2 = 1 - h/X_2 = 1 - g_2h/c^2\). So \(\Delta\nu/\nu = -g_2h/c^2\) exactly, with the ceiling's proper acceleration, while \(-g_1h/c^2\) is exact only to first order. The two differ by \((gh/c^2)^2 \approx 6\times10^{-30}\), which is why nobody has ever had to care. The statement that is exact both ways is the ratio \(\nu_1/\nu_2 = X_2/X_1 = 1 + g_1h/c^2\).

5. Rindler: \(a(X) = c^2/X\), so \(|\Delta a|/a = h/X_1 = 2.455\times10^{-15}\) over the tower. Earth: \(a(r) = GM/r^2\), so \(|\Delta a|/a = 2h/R_\oplus = 45/6.378137\times10^6 = 7.06\times10^{-6}\). The Earth's field is nine orders of magnitude less uniform than the Rindler field over the same height. That difference is the tidal tensor, and it is what P2 measured and P5 proves is irremovable.

Check. Dimensions: \(gX/c^2\) is \(\mathrm{(m\,s^{-2})(m)/(m^2s^{-2})}\), dimensionless ✓. Limits: \(h\to0\) gives ratio \(1\) ✓; \(X_1\to\infty\) at fixed \(h\) gives \(g_1\to0\) and no shift ✓. A number: \(0.96872\) ly is unit 27's \(c^2/g\), unchanged, as it must be — this is the same hyperbola ✓.

Idea. Gauge freedom is a vector space and so is the thing being gauged away; comparing dimensions settles what must remain, before anyone knows how to compute it. The Newtonian part is the same statement in a case where the whole calculation fits in three lines, and it names the survivor: the tidal tensor.

1. The targets. \(g_{\mu\nu}(p)\): symmetric in \(\mu\nu\), so \(\binom{5}{2} = 10\). \(\partial_\alpha g_{\mu\nu}(p)\): one free index times \(10\), so \(4\times10 = 40\). \(\partial_\alpha\partial_\beta g_{\mu\nu}(p)\): symmetric in \(\alpha\beta\) by Clairaut, so \(\binom{5}{2} = 10\) derivative pairs, times \(10\) metric pairs, \(= 100\).

2. The freedom. \(A^\mu{}_\nu\): \(4\times4 = 16\). \(B^\mu{}_{\nu\rho}\): symmetric lower pair, \(4\times\binom52 = 4\times10 = 40\). \(C^\mu{}_{\nu\rho\sigma}\): totally symmetric lower triple, \(4\times\binom{6}{3} = 4\times20 = 80\).

3. The deduction. Setting \(g_{\mu\nu}(p) = \eta_{\mu\nu}\) imposes \(10\) conditions on \(A\)'s \(16\) parameters, leaving \(6\): precisely \(\dim O(1,3) = \binom42 = 6\), the three boosts and three rotations of unit 26 — the residual freedom to choose which inertial frame the falling observer uses. At first-derivative order, \(40\) parameters face \(40\) targets: the counts match, and unit 32's construction of normal coordinates shows they really can all be removed. At second-derivative order, the transformed \(\partial\partial g\) depends on \(C\) affinely: it is (something independent of \(C\)) plus a linear function \(L(C)\). Removing a given configuration means solving \(L(C) = v\) for \(v\) in a \(100\)-dimensional space, with \(C\) ranging over an \(80\)-dimensional one. The image \(L(\mathbb R^{80})\) has dimension at most \(80\), so the quotient \(\mathbb R^{100}/\operatorname{im}L\) has dimension at least \(20\): at least twenty combinations of second derivatives are unchanged by every coordinate change. That is rigorous, and it needs nothing about what \(L\) is. \(\square\)

(The converse — that exactly \(20\) survive, so that \(L\) has full rank \(80\) — requires exhibiting normal coordinates, which unit 32 does. Unit 33 then identifies the \(20\) as the components of the Riemann tensor.)

4. General \(n\). Second derivatives: \(\binom{n+1}{2}^2 = \frac{n^2(n+1)^2}{4}\). Free \(C\): \(n\binom{n+2}{3} = \frac{n^2(n+1)(n+2)}{6}\). The difference is \[\frac{n^2(n+1)^2}{4} - \frac{n^2(n+1)(n+2)}{6} = \frac{n^2(n+1)}{12}\big[3(n+1) - 2(n+2)\big] = \frac{n^2(n+1)(n-1)}{12} = \frac{n^2(n^2-1)}{12}.\] Values: \(n=1\) gives \(0\) — a curve has no intrinsic geometry, which is why a line and a circle are locally indistinguishable to an ant confined to them. \(n=2\) gives \(1\) — one invariant, the Gaussian curvature, and Gauss's Theorema Egregium is the statement that it is intrinsic. \(n=3\) gives \(6\). \(n=4\) gives \(20\), of which unit 35's field equations constrain ten, leaving ten free to propagate as unit 38's gravitational waves.

5. The Newtonian shadow. Substituting \(\vec x = \vec x_0(t) + \vec x'\) into \(\ddot{\vec x} = -\nabla\Phi(\vec x)\): \[\ddot{\vec x}_0 + \ddot{\vec x}' = -\nabla\Phi(\vec x_0 + \vec x') \quad\Longrightarrow\quad \ddot{\vec x}' = -\nabla\Phi(\vec x_0 + \vec x') - \ddot{\vec x}_0 .\] Since \(\nabla'(\ddot{\vec x}_0\cdot\vec x') = \ddot{\vec x}_0\) (the vector \(\ddot{\vec x}_0\) does not depend on \(\vec x'\)), the right-hand side is \(-\nabla'\Phi'\) with \(\Phi'(\vec x',t) := \Phi(\vec x_0+\vec x') + \ddot{\vec x}_0\cdot\vec x'\). Killing the field. \(\nabla'\Phi'(\vec 0,t) = \nabla\Phi(\vec x_0(t)) + \ddot{\vec x}_0(t)\), which vanishes for all \(t\) exactly when \(\ddot{\vec x}_0 = -\nabla\Phi(\vec x_0)\) — that is, when \(\vec x_0\) is itself a free-fall worldline. Such a curve exists by Picard–Lindelöf (unit 01). What survives. \[\partial'_i\partial'_j\Phi' = \partial_i\partial_j\Phi(\vec x_0+\vec x') + \partial'_i\partial'_j\big(\ddot x_{0k}x'_k\big) = \partial_i\partial_j\Phi + 0,\] because the added term is linear in \(\vec x'\) and its second derivatives vanish identically. So the tidal tensor \(T_{ij} = \partial_i\partial_j\Phi\) is unchanged by every such frame change, whatever \(\vec x_0\) is. \(\square\) It is the surviving object, and unit 05's deviation equation \(\ddot\xi_i = -T_{ij}\xi_j\) is what the falling observer still measures.

Check. Dimensions: \(T_{ij}\) is \(\mathrm s^{-2}\), the same before and after, as an invariant must be ✓. Special case: a genuinely uniform field \(\Phi = gz\) has \(T = 0\), and indeed the choice \(\ddot{\vec x}_0 = -g\hat z\) removes it everywhere at once — the elevator, working exactly ✓. Counting check: at \(n=2\), \(\binom32^2 = 9\) second derivatives against \(2\binom43 = 8\) free coefficients leaves \(1 = 2^2(2^2-1)/12\) ✓.

Idea. "Gravity is a force on a flat background" is the claim that a single frame can be inertial for a whole region. Written out, that claim says \(\nabla\Phi\) is independent of position, which says the tidal tensor vanishes. So the existence of tides and the existence of a global inertial frame are literally the same question with opposite answers.

1. Proof. (\(\Leftarrow\)) Suppose \(\partial_i\partial_j\Phi \equiv 0\) on \(U\), which is connected, so \(\nabla\Phi\) is a function \(\vec a(t)\) of \(t\) alone there. Pick any solution \(\vec x_0\) of \(\ddot{\vec x}_0 = -\vec a(t)\); it exists by integrating twice. Then for any free-fall worldline in \(U\), by P5, \(\ddot{\vec x}' = -\nabla\Phi(\vec x_0 + \vec x') - \ddot{\vec x}_0 = -\vec a + \vec a = \vec 0\). Every such worldline is straight at constant speed: a global inertial frame.
(\(\Rightarrow\)) Suppose a global inertial frame \(\vec x_0\) exists. Fix \(t\) and fix any \(\vec x\in U\). Picard–Lindelöf (unit 01) supplies a free-fall worldline through \(\vec x\) at time \(t\) with any initial velocity, and it stays in \(U\) for a short interval; by hypothesis it satisfies \(\ddot{\vec x}' = \vec 0\) there. Therefore \(\nabla\Phi(\vec x) + \ddot{\vec x}_0(t) = \vec 0\), i.e. \(\nabla\Phi(\vec x) = -\ddot{\vec x}_0(t)\), and this holds for every \(\vec x\in U\) with the same right-hand side. Differentiating with respect to \(x_j\) and using that the right-hand side has no \(\vec x\)-dependence, \(\partial_j\partial_i\Phi(\vec x) = 0\) for all \(i,j\) and all \(\vec x\in U\). \(\square\)

2. Outside a spherical mass \(M\), \(\Phi = -GM/r\) and \(T_{ij} = (GM/r^3)(\delta_{ij} - 3n_in_j)\) with eigenvalues \((-2,1,1)GM/r^3\), none of which is zero for \(r \lt \infty\). So on any open set \(U\) outside the mass, \(\partial_i\partial_j\Phi\) is nowhere the zero matrix, and part 1 says no global inertial frame exists on \(U\) — however small \(U\) is. Shrinking the region shrinks the size of the tidal effect (P2), never its presence.

3. With \(\vec x'' = R(t)\vec x'\) the transformed equation acquires unit 03's centrifugal, Coriolis and Euler terms: \(\ddot{\vec x}'' = R(-\nabla\Phi - \ddot{\vec x}_0) - 2\vec\omega\times\dot{\vec x}'' - \vec\omega\times(\vec\omega\times\vec x'') - \dot{\vec\omega}\times\vec x''\). Demanding \(\ddot{\vec x}'' = \vec 0\) for every initial velocity at a fixed point forces the Coriolis coefficient to vanish, since it is the only velocity-dependent term and the others are not: hence \(\vec\omega = \vec 0\), and the rest reduces to part 1. The rotation adds a term linear in \(\vec x''\) — centrifugal — which is exactly the kind of term P5 showed cannot touch \(\partial_i\partial_j\Phi\): its own second derivatives vanish. Rotating the frame cannot manufacture a second derivative, so it cannot cancel one.

4. The spacetime claim: there is a chart on a region in which \(g_{\mu\nu} = \eta_{\mu\nu}\) throughout if and only if the \(20\) surviving second-derivative combinations of P5 vanish throughout. The Newtonian proof does not carry over as written at two points. (i) "Free-fall worldlines are straight lines" must become "free-fall worldlines are the straight lines of the chart", which needs the notion of a geodesic — unit 32. (ii) The step "a function of \(\vec x\) equals a function of \(t\) alone, so differentiate" becomes an integrability condition on a system of PDEs for the coordinate change, and the object whose vanishing makes the system integrable is the Riemann tensor — unit 33. What does carry over verbatim is the shape of the argument, and the conclusion.

Check. Dimensions: \(\partial_i\partial_j\Phi\) in \(\mathrm s^{-2}\), and the claim is that it vanishes — dimensionally trivial, which it should be, since an "if and only if" between a geometric statement and a vanishing tensor cannot depend on units ✓. A limit: as the source mass \(M\to0\), \(T\to0\) and the global inertial frame reappears, which is unit 26's Minkowski space ✓. A number: at the Earth's surface the largest eigenvalue is \(2GM_\oplus/R_\oplus^3 = 3.07\times10^{-6}\ \mathrm{s^{-2}}\), so the obstruction is small but exactly as nonzero as P2 measured ✓.

Idea. Gravity's source is energy; gravity's field has energy; therefore gravity sources itself. That one sentence makes a linear theory impossible, and everything after it is bookkeeping — up to the point where the bookkeeping, done relativistically, eats the flat background it was written on.

1. Substitute \(\rho = \nabla^2\Phi/4\pi G\): \[U = \tfrac12\int\rho\Phi\,\mathrm dV = \frac{1}{8\pi G}\int\Phi\,\nabla^2\Phi\,\mathrm dV .\] Now use the identity \(\nabla\cdot(\Phi\nabla\Phi) = |\nabla\Phi|^2 + \Phi\nabla^2\Phi\) and the divergence theorem (unit 05) over a ball of radius \(r\): \[\int_{B_r}\Phi\nabla^2\Phi\,\mathrm dV = \oint_{S_r}\Phi\,\nabla\Phi\cdot\mathrm d\vec S - \int_{B_r}|\nabla\Phi|^2\,\mathrm dV .\] Outside the support of \(\rho\), \(\Phi \sim -GM/r\) and \(|\nabla\Phi| \sim GM/r^2\), while the sphere's area is \(4\pi r^2\); so the surface term is \(O(1/r)\) and vanishes as \(r\to\infty\). (This is where compact support and \(\Phi\to0\) are used, and both are needed: a mass distribution extending to infinity can make the surface term finite or divergent.) Hence \[U = -\frac{1}{8\pi G}\int|\nabla\Phi|^2\,\mathrm dV ,\] which is manifestly \(\le 0\). Reading the integrand as a local energy density gives \(u = -|\nabla\Phi|^2/8\pi G\), negative everywhere: unlike the electrostatic case, where \(u = \epsilon_0|\vec E|^2/2 \gt 0\), gravitational field energy is negative, because gravity attracts and assembling a body releases energy.

2. Assembling a uniform sphere shell by shell gives \(U = -\int_0^R \frac{G m(r)}{r}\,4\pi r^2\rho\,\mathrm dr\) with \(m(r) = M r^3/R^3\), which integrates to \(U = -\frac{3GM^2}{5R}\). So \[\frac{|U|}{Mc^2} = \frac{3GM}{5Rc^2} .\] Earth: \(3\times3.986004\times10^{14}/(5\times6.378137\times10^6\times8.987552\times10^{16}) = 4.17\times10^{-10}\). Sun: \(3\times1.327124\times10^{20}/(5\times6.957\times10^{8}\times8.987552\times10^{16}) = 1.27\times10^{-6}\). (The Sun is centrally condensed, so the true figure is a few times larger; the uniform-sphere value is the right order.)

3. By \(E = mc^2\) the field energy \(u\) is equivalent to a mass density \(u/c^2\). The weak equivalence principle says gravitational mass equals inertial mass for every form of energy, so \(u/c^2\) must source the field like any other mass density: \[\nabla^2\Phi = 4\pi G\Big(\rho - \frac{|\nabla\Phi|^2}{8\pi Gc^2}\Big) .\] This is nonlinear. Solving it perturbatively, \(\Phi = \Phi_0 + \Phi_1 + \dots\) with \(\nabla^2\Phi_0 = 4\pi G\rho\) and \(\nabla^2\Phi_1 = -|\nabla\Phi_0|^2/2c^2\), the second solution changes \(|\nabla\Phi|^2\), which changes the source again, for ever: an infinite iteration, each term smaller by roughly \(GM/Rc^2\). For the Sun the first correction is of relative size \(\sim1.3\times10^{-6}\) — small, and of exactly the order of the perihelion and light-bending effects, which is not a coincidence.

4. Do the same relativistically: let \(h_{\mu\nu}\) be a massless spin-2 field on Minkowski space whose linear field equation is sourced by the matter stress-energy tensor. Consistency of the linear equation requires the source to be conserved, \(\partial_\mu T^{\mu\nu}_{\text{matter}} = 0\); but the field does work on matter, so matter's stress-energy alone is not conserved once the coupling is switched on. The repair is to source the field with the total stress-energy, matter plus field. Deser (1970) showed that this repair, carried out consistently, has for its endpoint the Einstein–Hilbert action of \(g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}\). The premise is then destroyed: in the resulting theory every rod, clock and light ray responds to \(g_{\mu\nu}\), and no experiment measures \(\eta_{\mu\nu}\). A background that no measurement can reach is not a background; the split \(g = \eta + h\) is a change of variables inside general relativity, not a theory of gravity on flat spacetime.

5. The weak equivalence principle enters at exactly one place: the claim that the mass-equivalent of the field energy gravitates with the same constant of proportionality as ordinary matter. That is the statement \(m_{\mathrm g} = m_{\mathrm i}\) applied to a form of energy that is neither a particle nor a material body. Strictly, the principle being used is the strong one: gravitational field energy is precisely the self-energy that WEP and even EEP exclude from their statements, both of which speak only of test bodies and of non-gravitational experiments. If only WEP held, the argument would give no constraint at all on how field energy gravitates, and theories in which it gravitates differently — Brans–Dicke, for instance — are exactly the ones the Nordtvedt effect tests.

Check. Dimensions: \(|\nabla\Phi|^2/G\) is \(\mathrm{(m\,s^{-2})^2/(m^3kg^{-1}s^{-2})} = \mathrm{kg\,m^{-1}s^{-2}} = \mathrm{J\,m^{-3}}\) ✓, an energy density. Special case: a point mass gives a divergent \(\int|\nabla\Phi|^2\) near the origin, matching \(U = -3GM^2/5R \to -\infty\) as \(R\to0\) — the classical self-energy divergence, and a reminder that "point mass" is an idealisation ✓. A number: for the Sun \(3GM/5Rc^2 = 1.27\times10^{-6}\), and the observed solar gravitational redshift is \(GM/Rc^2 = 2.12\times10^{-6}\) — the same order, as both measure \(GM/Rc^2\) ✓.

Idea. Every step is a substitution, and the surprise is at the end: a family of observers in flat spacetime measures a space in which circles have too much circumference. Nothing has gone wrong; "space" was never defined until the radar construction defined it.

1. \(\rho^2\mathrm d\phi^2 = \rho^2\mathrm d\phi'^2 + 2\omega\rho^2\mathrm d\phi'\mathrm dt + \omega^2\rho^2\mathrm dt^2\), so \[\mathrm ds^2 = -\Big(1-\frac{\omega^2\rho^2}{c^2}\Big)c^2\mathrm dt^2 + 2\omega\rho^2\,\mathrm d\phi'\,\mathrm dt + \mathrm d\rho^2 + \rho^2\mathrm d\phi'^2 + \mathrm dz^2 .\] With \(x^0 = ct\): \(g_{00} = -(1-\omega^2\rho^2/c^2)\), \(g_{0\phi'} = \omega\rho^2/c\), \(g_{\phi'\phi'} = \rho^2\), \(g_{\rho\rho} = g_{zz} = 1\).

2. \(g_{00} = 0\) at \(\rho = c/\omega\), the light cylinder. There the curve "fixed \(\rho,\phi',z\)" is null, and beyond it spacelike: no observer can co-rotate, because doing so would require moving faster than light. For \(716\) Hz, \(\omega = 2\pi\times716 = 4499\ \mathrm{s^{-1}}\) and \(c/\omega = 2.998\times10^{8}/4499 = 6.66\times10^{4}\ \mathrm m = 66.6\) km — a few radii outside a \(12\) km neutron star, which is why the light cylinder bounds a pulsar's co-rotating magnetosphere. (The real star's spacetime is not flat; this is an order-of-magnitude statement about the flat-space idealisation.)

3. Radially \(g_{0\rho} = 0\), so \(\gamma_{\rho\rho} = g_{\rho\rho} = 1\) and the measured radius is \(\int_0^\rho\mathrm d\rho = \rho\). Around the rim, \[\gamma_{\phi'\phi'} = \rho^2 - \frac{(\omega\rho^2/c)^2}{-(1-\omega^2\rho^2/c^2)} = \rho^2\Big[1 + \frac{\omega^2\rho^2/c^2}{1-\omega^2\rho^2/c^2}\Big] = \frac{\rho^2}{1-\omega^2\rho^2/c^2} = \gamma^2\rho^2 .\] So the measured circumference is \(C = \int_0^{2\pi}\gamma\rho\,\mathrm d\phi' = 2\pi\gamma\rho\) and the measured radius is \(R = \rho\).

4. Therefore \(C/2R = \pi\gamma \gt \pi\). Rim speed \(0.1c\): \(\gamma = (1-0.01)^{-1/2} = 1.005038\), a \(0.50\%\) excess. Ultracentrifuge rim at \(10^3\ \mathrm{m\,s^{-1}}\): \(\beta^2 = 1.113\times10^{-11}\) and \(\gamma - 1 = 5.56\times10^{-12}\) — real, and hopelessly below any mechanical measurement, though the associated time dilation was measured directly with Mössbauer rotors in the 1960s.

5. No: the spacetime is Minkowski space, exactly, and every event in the chart is an ordinary event of it. The non-Euclidean geometry belongs to the family of co-rotating observers — it is the geometry of the distances they measure with radar, not the geometry of a slice of spacetime. (Strictly, their simultaneity does not even close up around the rim; that failure is the Sagnac effect, and it is why "the disc's space" is a quotient rather than a hypersurface.)

Check. Dimensions: \(\omega\rho/c\) dimensionless ✓; \(g_{0\phi'} = \omega\rho^2/c\) has the units of \(\rho^2\) divided by nothing, matching \(g_{\phi'\phi'}\), since \(\phi'\) and \(x^0\)'s metric slots must pair consistently ✓. Limit: \(\omega\to0\) gives \(\gamma\to1\), \(C = 2\pi R\), Euclid back ✓. Second limit: \(\rho\to c/\omega\) sends \(C\to\infty\) at finite \(R\), consistent with the light cylinder being unreachable by a co-rotating ruler ✓.

Another way

Without the \(\gamma_{ij}\) formula: in the lab frame the rim moves at \(v = \omega\rho\), so a co-rotating ruler of proper length \(\ell\) is seen as \(\ell/\gamma\). Laying rulers end to end around a lab circle of circumference \(2\pi\rho\) therefore takes \(2\pi\gamma\rho\) of them, and the riders count \(2\pi\gamma\rho\) of proper length. Radial rulers move perpendicular to their own length and are uncontracted, so \(R = \rho\). Same answer, less machinery — but it does not generalise, and it hides the fact that "the circumference the riders measure" required a definition of simultaneity.

Idea. Two first-order corrections of opposite sign, differing by a factor of six, in a system that stops working if either is dropped. The arithmetic is short; the lesson is that the two halves of relativity are one subject.

1. \(-v^2/2c^2\) is unit 26's time dilation, \(\mathrm d\tau/\mathrm dt = \sqrt{1-v^2/c^2}\), expanded to first order. \(+\Phi/c^2\) is this unit's gravitational redshift, read as a statement about clock rates — the box on the rocket slide, with the reference clock at \(\Phi = 0\).

2. Potential. \(1/R_\oplus = 1.567855\times10^{-7}\), \(1/a = 3.764919\times10^{-8}\), difference \(1.191363\times10^{-7}\ \mathrm{m^{-1}}\); times \(GM_\oplus\) gives \(4.74858\times10^{7}\ \mathrm{m^2\,s^{-2}}\); divided by \(c^2\), \[\frac{\Phi_s - \Phi_g}{c^2} = +5.2837\times10^{-10} \;\Rightarrow\; +5.2837\times10^{-10}\times86400 = +45.65\ \mu\mathrm{s/day}.\] Speed. \(v_s^2 = GM_\oplus/a = 1.50070\times10^{7}\ \mathrm{m^2\,s^{-2}}\), so \(v_s = 3873.9\ \mathrm{m\,s^{-1}}\) and \(-v_s^2/2c^2 = -8.3488\times10^{-11} \Rightarrow -7.21\ \mu\mathrm{s/day}\).

3. Net \(+45.65 - 7.21 = +38.44\ \mu\mathrm{s/day}\). Ignored, the accumulated range error in one day is \(c\times3.844\times10^{-5}\ \mathrm s = 1.152\times10^{4}\ \mathrm m \approx 11.5\) km.

4. \(v_g = \omega_\oplus R_\oplus = 7.292115\times10^{-5}\times 6.378137\times10^{6} = 465.1\ \mathrm{m\,s^{-1}}\), so \(+v_g^2/2c^2 = +1.203\times10^{-12} = +0.104\ \mu\mathrm{s/day}\): the ground clock is itself slowed, so the satellite gains a little more. The net becomes \(+38.54\ \mu\mathrm{s/day}\). That \(0.1\ \mu\mathrm{s/day}\) is exactly the scale of the spread in published figures, and the rest of the spread comes from whether the ground clock is placed on the equator or on the geoid (whose defining potential already folds in the centrifugal term) and whether the Earth's quadrupole \(J_2\) is included. "About \(38.5\ \mu\mathrm{s/day}\)" is the honest statement.

5. Setting the satellite's total equal to the ground clock's: \[-\frac{GM}{r} - \frac{GM}{2r} = -\frac{GM}{R_\oplus} \;\Longrightarrow\; \frac{3}{2r} = \frac1{R_\oplus} \;\Longrightarrow\; r = \tfrac32R_\oplus = 9567.2\ \mathrm{km},\] an altitude of \(\tfrac12R_\oplus = 3189.1\) km. A clock there keeps the same rate as one on the ground — not because relativity has switched off, but because two effects of opposite sign happen to be equal. Below that altitude the kinematic term wins and orbiting clocks run slow; the Space Station, at about \(410\) km, is one of them.

Check. Dimensions: \(GM/rc^2\) dimensionless ✓; \(c\,\delta t\) in metres ✓. Limit: \(a\to\infty\) gives \(+GM/R_\oplus c^2 = 6.95\times10^{-10}\) with no kinematic term, the Earth's own surface redshift ✓. A number: the GPS satellites are built with a fractional rate offset of \(-4.4647\times10^{-10}\), which is \(38.57\ \mu\mathrm{s/day}\) — the design figure, agreeing with part 4 ✓.

Another way

Part 5 without arithmetic: for a circular orbit the virial theorem (unit 08) gives \(\langle T\rangle = -\tfrac12\langle V\rangle\) per unit mass, i.e. \(v^2/2 = GM/2r\), so the total rate correction is \(-GM/r - GM/2r = -\tfrac32 GM/r\): the orbital clock behaves as if it sat at rest at \(\tfrac23 r\). Setting \(\tfrac23r = R_\oplus\) gives \(r = \tfrac32R_\oplus\) immediately, and explains the factor cleanly.

Idea. If a photon's energy were unchanged by climbing, the conversion between mass and radiation would let you ratchet mass upward for free. Requiring the ratchet to fail determines the redshift exactly — no fields, no metrics, no geometry, only bookkeeping. This is Einstein's argument of 1911.

1. The cycle. (i) A body of rest mass \(m\) sits at the top of the tower; its energy is \(mc^2\). (ii) Lower it to the bottom, extracting the work \(mgh\) and storing it. (Here the WEP is used: the work is \(m_{\mathrm g}gh\) and the body's inertia is \(m_{\mathrm i}\), and the argument needs them equal — see part 5.) At the bottom its energy is \(mc^2\), with \(mgh\) now in the store. (iii) At the bottom, convert the body and the stored work entirely into radiation of energy \(mc^2 + mgh\), and send it up the tower. (iv) At the top, by hypothesis the radiation still has energy \(mc^2 + mgh\); convert it back into a body at rest, of rest mass \(m' = m + mgh/c^2\). The apparatus is back where it began, and the mass at the top has grown.

2. The gain per cycle is \(\Delta E = (m'-m)c^2 = mgh\): exactly the work extracted in step (ii), now available again and for ever. Running the cycle \(N\) times yields \(Nmgh\) from nothing, which contradicts conservation of energy.

3. Break-even requires \(m' = m\), i.e. the radiation must arrive at the top with energy \(mc^2\) having left with \(mc^2 + mgh\): \[\frac{E_{\text{top}}}{E_{\text{bottom}}} = \frac{mc^2}{mc^2 + mgh} = \frac{1}{1 + gh/c^2} = 1 - \frac{gh}{c^2} + O\!\Big(\frac{g^2h^2}{c^4}\Big).\] Since \(E = h_{\mathrm P}\nu\) with \(h_{\mathrm P}\) Planck's constant, the same ratio holds for frequencies: \(\Delta\nu/\nu = -gh/c^2\) to first order, which is the deck's boxed result. \(\square\)

4. This argument assumes the mass–energy interconversion of unit 27 and global conservation of energy in a static field — neither of which the rocket derivation needs. It does not need any statement about Doppler shifts, inertial frames or propagation: no kinematics at all appears. The two derivations are genuinely independent, which is why Einstein gave both, and their agreement is the first evidence that the redshift is forced rather than modelled.

5. The WEP enters in step (ii): the work recovered by lowering the body is \(m_{\mathrm g}gh\), while the mass that is later converted to radiation is the rest mass \(m_{\mathrm i}\) (inertia is what \(E = mc^2\) refers to). If \(m_{\mathrm g} = (1+\eta)m_{\mathrm i}\), the cycle would break even only if \(E_{\text{top}}/E_{\text{bottom}} = 1/(1+(1+\eta)gh/c^2)\), so the redshift would be composition-dependent — different for a tower loaded with beryllium and one loaded with titanium. A composition-dependent redshift is exactly what a clock-comparison experiment does not see, and it is why a bound on \(\eta\) and a redshift measurement are testing related but distinct things.

Check. Dimensions: \(gh/c^2\) dimensionless ✓; \(mgh\) an energy ✓. Limit: \(h\to0\) gives ratio \(1\), no shift, and the cycle degenerates ✓. A number: for the Harvard tower the ratio is \(1/(1+2.455\times10^{-15})\), and the energy that could be ratcheted per kilogram per cycle would be \(mgh = 220.6\) J — small, but free, and repeatable, which is what makes the contradiction fatal rather than negligible ✓.

Idea. The elevator argument gives light a transverse acceleration; adding it up along the ray is a one-line integral. The factor of two at the end is the sharpest statement in the course of what the equivalence principle is and is not for.

1. Crossing time \(L/c = 10/2.998\times10^{8} = 3.336\times10^{-8}\) s. Drop \[\tfrac12 g\Big(\frac Lc\Big)^2 = \tfrac12(9.80665)(3.336\times10^{-8})^2 = 5.46\times10^{-15}\ \mathrm m ,\] about five femtometres, the scale of an atomic nucleus. Deflection angle \(gL/c^2 = 1.091\times10^{-15}\) rad \(= 2.25\times10^{-10}\) arcseconds. This is why terrestrial light-bending experiments do not exist.

2. Along the unperturbed ray, \(\mathrm dt = \mathrm dx/c\), so \[\Delta\theta = \frac{1}{c^2}\int_{-\infty}^{\infty}\frac{GMb\,\mathrm dx}{(b^2+x^2)^{3/2}} .\] Put \(x = b\tan\psi\), \(\mathrm dx = b\sec^2\psi\,\mathrm d\psi\), \((b^2+x^2)^{3/2} = b^3\sec^3\psi\). Then \[\Delta\theta = \frac{GM}{c^2}\int_{-\pi/2}^{\pi/2}\frac{b\cdot b\sec^2\psi}{b^3\sec^3\psi}\,\mathrm d\psi = \frac{GM}{bc^2}\int_{-\pi/2}^{\pi/2}\cos\psi\,\mathrm d\psi = \frac{2GM}{bc^2}.\]

3. At the Sun's limb, \(b = R_\odot = 6.957\times10^{8}\) m: \[\Delta\theta_{\mathrm{EP}} = \frac{2\times1.327124\times10^{20}} {6.957\times10^{8}\times8.987552\times10^{16}} = 4.2450\times10^{-6}\ \mathrm{rad},\] which is \(4.2450\times10^{-6}\times206264.8 = 0.8756''\).

4. The missing ingredient is the spatial part of the metric, \(g_{ij}\). The equivalence principle fixes only \(g_{00} = -(1+2\Phi/c^2)\), because it is a statement about what a local freely falling frame sees, and the only thing a static observer can compare with a neighbour without leaving the neighbourhood is a clock rate. The deck's clock slide got \(g_{00}\) from the redshift and explicitly left \(g_{ij}\) undetermined; nothing about clocks can determine it. The full calculation (unit 37) finds that the two halves contribute equally, giving \(4GM/c^2b = 1.7512''\).

5. No: the redshift depends only on \(g_{00}\), which is exactly what the equivalence principle does supply, so the equivalence-principle answer is right — as P4 showed, right even beyond first order for the Rindler case.

Check. Dimensions: \(GM/bc^2\) is \(\mathrm{(m^3s^{-2})/(m\cdot m^2s^{-2})}\), a pure number, correctly read as radians ✓. Limit: \(b\to\infty\) gives \(\Delta\theta\to0\) ✓; \(M\to0\) likewise ✓. A number: the ratio of part 3 to part 1 is \(4.245\times10^{-6}/1.091\times10^{-15} = 3.9\times10^{9}\) — the Sun bends light four billion times harder than a ten-metre laboratory, which is why the measurement had to wait for an eclipse ✓.

Idea. A bound on a dimensionless number becomes a bound on an angle, and the angle turns out to be measurable but tiny — which explains why the whole art of the experiment is in modulation, not in sensitivity.

1. \(I = 2md^2 = 2(0.01)(0.02)^2 = 8.00\times10^{-6}\ \mathrm{kg\,m^2}\). From \(T = 2\pi\sqrt{I/\kappa}\), \[\kappa = I\Big(\frac{2\pi}{T}\Big)^2 = 8.00\times10^{-6}\times \Big(\frac{2\pi}{800}\Big)^2 = 4.935\times10^{-10}\ \mathrm{N\,m\,rad^{-1}} .\]

2. \(\eta\) is the difference over the mean, so the two accelerations differ by \(\eta a_\perp\) and each mass feels \(\tfrac12\eta a_\perp\) relative to the mean, in opposite senses. Each contributes a force \(\tfrac12 m\eta a_\perp\) at a lever arm \(d\) and the torques add: \(\tau = 2\cdot\tfrac12 m\,\eta a_\perp\,d = m\,\eta a_\perp\,d\).

3. For \(\eta = 10^{-13}\): \(\eta a_\perp = 10^{-13}\times1.69\times10^{-2} = 1.69\times10^{-15}\ \mathrm{m\,s^{-2}}\), so \[\tau = (0.01)(1.69\times10^{-15})(0.02) = 3.38\times10^{-19}\ \mathrm{N\,m},\qquad \theta = \frac{\tau}{\kappa} = 6.85\times10^{-10}\ \mathrm{rad}.\] About \(0.69\) nanoradians — a \(0.7\ \mu\mathrm m\) displacement of a spot on a screen a kilometre away, or, with an optical lever folded a few times, a readily measurable fringe shift.

4. Orbit buys, first, a much larger and steadier signal source: in free fall around the Earth the full \(8\ \mathrm{m\,s^{-2}}\) of the Earth's field is the driving acceleration, against \(1.69\times10^{-2}\ \mathrm{m\,s^{-2}}\) on the ground — nearly three orders of magnitude — because on the ground the vertical component is taken up by the suspension. Second, it removes the ground's noise: seismic tilt, thermal gradients, and above all the time-varying gravity gradients of nearby mass (people, groundwater, the experimenter's car). The ground experiment is not obsolete because it can be repeated, rebuilt and reconfigured — different material pairs, different source masses (including laboratory attractors, which test short-range deviations from the inverse-square law) — at a cost that is not a satellite's.

5. Because the signal is otherwise at zero frequency, where every drift lives: fibre creep, thermal expansion, slow tilt. Rotating the apparatus at a known slow rate moves the signal to a known nonzero frequency — twice the turntable rate, since the effect reverses when the beam reverses — where the noise is far quieter and a matched filter can pull the amplitude out of hours of data. It is the mechanical version of a lock-in amplifier, and it is what separates the modern generation of balances from Eötvös's.

Check. Dimensions: \(\kappa\) as \(I\omega^2\) is \(\mathrm{kg\,m^2\,s^{-2}} = \mathrm{N\,m}\) per radian ✓; \(\theta = \tau/\kappa\) is a pure number ✓. Limit: \(\eta\to0\) gives \(\theta\to0\), no signal ✓. Scaling: \(\theta\) is linear in \(\eta\), so MICROSCOPE's \(1.5\times10^{-15}\) on this instrument would be \(1\times10^{-11}\) rad — below what a ground balance can do, which is the quantitative reason the satellite was flown ✓.