The Einstein field equations, twice

Idea. A tensor equation implies its own divergence and its own trace. The divergence of the left side is fixed by geometry; the divergence of the right side is fixed by physics. If they disagree, the equation constrains the matter, and that constraint is the test.

1. Apply \(\nabla^\mu\) to \(R_{\mu\nu} = \kappa T_{\mu\nu}\): \[ \tfrac12\nabla_\nu R = \kappa\nabla^\mu T_{\mu\nu} = 0 . \] So \(\partial_\nu R = 0\) (on a scalar \(\nabla = \partial\)), and on a connected manifold \(R\) is constant. Now trace the proposal with \(g^{\mu\nu}\): \(R = \kappa T\). Hence \(T = R/\kappa\) is constant too.

2. For dust, \(T = g_{\mu\nu}\rho u^\mu u^\nu = -\rho c^2\). Far from an isolated star there is vacuum and \(T = 0\); constancy then forces \(T = 0\) everywhere, so \(\rho = 0\) inside the star as well. The theory admits no isolated massive body: it is not merely restrictive but empty for the case Newton was built for. Electromagnetic radiation has \(T = 0\) identically (the deck's EM slide: \(g_{\mu\nu}T^{\mu\nu}_{\text{EM}} = 0\) since \(g_{\mu\nu}g^{\mu\nu} = 4\)), so the constraint is satisfied for free. That is exactly why Einstein, who in November 1915 briefly hoped all matter was electromagnetic in origin, could entertain the proposal for a week.

3. Using \(\nabla g = 0\), \[ \nabla^\mu(R_{\mu\nu} + \beta Rg_{\mu\nu}) = \tfrac12\nabla_\nu R + \beta\nabla_\nu R = (\tfrac12 + \beta)\nabla_\nu R . \] The field equation then implies \((\tfrac12+\beta)\nabla_\nu R = 0\). Tracing the equation gives \((1 + 4\beta)R = \kappa T\), so for \(\beta \ne -\tfrac14\) this is a condition \((\tfrac12+\beta)\nabla_\nu T = 0\) on the matter — the same disease — unless \(\beta = -\tfrac12\). (For \(\beta = -\tfrac14\) the trace gives \(T = 0\) directly, which is just as fatal.) So \(\beta = -\tfrac12\) uniquely, and the left side is \(G_{\mu\nu}\): the equation of 25 November.

4. With \(T_{\mu\nu} = 0\), the repaired equation traced gives \(R - 2R = 0\), so \(R = 0\), and then \(G_{\mu\nu} = R_{\mu\nu} = 0\). Conversely \(R_{\mu\nu} = 0\) gives \(R = 0\) and \(G_{\mu\nu} = 0\). Both are \(R_{\mu\nu} = 0\).

Check. Dimensions: \([R_{\mu\nu}] = [Rg_{\mu\nu}] = \mathrm{m^{-2}}\), so \(\beta\) is dimensionless, as a pure number fixed by an identity must be ✓. Special case: the vacuum agreement in part 4 is why Einstein's perihelion calculation of 18 November, made with the wrong equation, gave the right 43 arcseconds — the Sun's exterior is vacuum.

Idea. "Divergence-free identically" means for every metric. A computation shows the divergence is a multiple of \(\nabla R\); one explicit metric with non-constant \(R\) then kills the multiple.

1. With \(\nabla g = 0\) and \(\nabla^\mu R_{\mu\nu} = \tfrac12\nabla_\nu R\): \(\nabla^\mu A_{\mu\nu} = \tfrac12\alpha\nabla_\nu R + \beta\nabla_\nu R + 0\).

2. The geodesic Lagrangian \(L = \tfrac12(\dot r^2 + f^2\dot\varphi^2)\) gives Euler-Lagrange equations \(\ddot r - ff'\dot\varphi^2 = 0\) and \(\ddot\varphi + 2(f'/f)\dot r\dot\varphi = 0\). Matching to \(\ddot x^\lambda + \Gamma^\lambda_{\mu\nu}\dot x^\mu\dot x^\nu = 0\): \(\Gamma^r_{\varphi\varphi} = -ff'\), \(\Gamma^\varphi_{r\varphi} = \Gamma^\varphi_{\varphi r} = f'/f\), all others zero. Nothing depends on \(\varphi\). Then \[ R_{rr} = -\partial_r\Big(\frac{f'}{f}\Big) - \Big(\frac{f'}{f}\Big)^2 = -\frac{ff'' - f'^2}{f^2} - \frac{f'^2}{f^2} = -\frac{f''}{f}, \] \[ R_{\varphi\varphi} = R^r{}_{\varphi r\varphi} = \partial_r\Gamma^r_{\varphi\varphi} - \Gamma^r_{\varphi\varphi}\Gamma^\varphi_{r\varphi} = -(f'^2 + ff'') + f'^2 = -ff'' . \] So \(R = g^{rr}R_{rr} + g^{\varphi\varphi}R_{\varphi\varphi} = -f''/f - f''/f = -2f''/f\).

3. The \((t,z)\) components of \(g\) are constant and do not depend on \(r,\varphi\), and the \((r,\varphi)\) components do not depend on \(t,z\); every Christoffel symbol with a mixed index therefore vanishes, the \((t,z)\) block contributes no curvature, and the scalar curvature is that of the surface. With \(f = r + r^3\), \(f'' = 6r\): \[ R = -\frac{12r}{r + r^3} = -\frac{12}{1 + r^2}, \qquad \partial_rR = \frac{24r}{(1+r^2)^2} \ne 0 \text{ for } r \gt 0 . \]

4. If \(\beta = -\tfrac12\alpha\), part 1 gives zero for every metric. If \(\beta \ne -\tfrac12\alpha\), the metric of part 3 has \(\nabla^\mu A_{\mu r} = (\tfrac12\alpha + \beta)\partial_rR \ne 0\). So divergence-freeness for every metric holds exactly when \(\beta = -\tfrac12\alpha\), and then \(A = \alpha(R_{\mu\nu} - \tfrac12Rg_{\mu\nu}) + \gamma g_{\mu\nu} = \alpha G_{\mu\nu} + \gamma g_{\mu\nu}\). \(\square\)

Check. Special case \(f = a\sin(r/a)\), the round sphere of radius \(a\): \(f''/f = -1/a^2\) and \(R = 2/a^2\), twice unit 33's Gaussian curvature \(1/a^2\) ✓. Limit \(f = r\), the flat plane in polar coordinates: \(R = 0\) ✓. Dimensions: \(f''/f\) is \(\mathrm{m^{-2}}\) ✓.

Another way

Part 3 does not need the product structure spelt out: \(R\) is a scalar, and in any chart adapted to a product of Riemannian and Lorentzian factors it is the sum of the factors' scalar curvatures. The flat factor contributes nothing.

Idea. The tensor is one object; its components depend on the frame, its values on given vectors and its trace do not.

1. \(u^\mu = \gamma(c, v, 0, 0)\), \(\gamma = (1 - v^2/c^2)^{-1/2}\). Nonzero components: \(T^{00} = \gamma^2\rho c^2\), \(T^{01} = T^{10} = \gamma^2\rho cv\), \(T^{11} = \gamma^2\rho v^2\).

2. \(O\)'s orthonormal frame is the coordinate frame, so its energy density is \(T^{00} = \gamma^2\rho c^2\). One \(\gamma\): each particle's energy is \(\gamma mc^2\) (unit 27). The other: a box that holds \(N\) particles in the dust frame is Lorentz-contracted by \(\gamma\) in \(O\)'s frame, so the number density is \(\gamma n\).

3. \(u_O = (c,0,0,0)\), \(u\cdot u_O = \eta_{00}\gamma c\cdot c = -\gamma c^2\), so \(T(u_O,u_O)/c^2 = \rho(\gamma c^2)^2/c^2 = \gamma^2\rho c^2\) ✓. Evaluated in the dust frame instead, \(u = (c,\vec0)\) and \(u_O = \gamma(c,-\vec v)\): the same product \(-\gamma c^2\), as a scalar must give.

4. \(T = \rho\,u\cdot u = -\rho c^2\) in every frame; componentwise in \(O\)'s: \(-T^{00} + T^{11} = -\gamma^2\rho(c^2 - v^2) = -\rho c^2\) ✓. At \(v = 0.6c\), \(\gamma^2 = 1/(1 - 0.36) = 1.5625\): \(O\) measures \(1.5625\) times the rest-frame energy density.

5. \(T(u_O,u_O) = (\rho + p/c^2)(u\cdot u_O)^2 + p\,u_O\cdot u_O = (\rho + p/c^2)\gamma^2c^4 - pc^2\); divide by \(c^2\): \(\gamma^2(\rho c^2 + p) - p\). A moving fluid carries momentum, and pressure does work on the fluid's moving boundaries: the momentum flux \(T^{11}\) contains \(p\), and a boost mixes \(T^{11}\) into \(T^{00}\) with weight \(\gamma^2v^2/c^2\). Indeed \(\gamma^2(\rho c^2 + p) - p = \gamma^2\rho c^2 + \gamma^2(v^2/c^2)p\).

Check. \(v\to0\): \(\gamma\to1\), energy density \(\rho c^2\) ✓; \(p\to0\) in part 5 returns part 2 ✓. Dimensions: \(\gamma^2\rho c^2\) and \(p\) are both \(\mathrm{J\,m^{-3}}\) ✓.

Idea. On traces, \(\tau\) acts as multiplication by \(1 - n/2\); on trace-free tensors it is the identity. An involution must square to one on both pieces.

1. \(\operatorname{tr}\tau(S) = \operatorname{tr}S - \tfrac n2\operatorname{tr}S = (1-\tfrac n2)\operatorname{tr}S\). Then \[ \tau(\tau(S)) = \tau(S) - \tfrac12(1-\tfrac n2)(\operatorname{tr}S)g = S - \tfrac12\big(2 - \tfrac n2\big)(\operatorname{tr}S)\,g . \] This is \(S\) for every \(S\) iff \(2 - n/2 = 0\) iff \(n = 4\). (Take \(S = g\) to see the coefficient must vanish.) In four dimensions \(\operatorname{tr}\tau(S) = -\operatorname{tr}S\): the name "trace reversal" is literal.

2. Tracing, \(R(1 - n/2) = \kappa T\), so \(R = -2\kappa T/(n-2)\). Then \[ R_{\mu\nu} = \kappa T_{\mu\nu} + \tfrac12Rg_{\mu\nu} = \kappa\Big(T_{\mu\nu} - \frac{1}{n-2}\,T\,g_{\mu\nu}\Big). \] At \(n = 4\) this is the deck's \(R_{\mu\nu} = \kappa(T_{\mu\nu} - \tfrac12Tg_{\mu\nu})\) ✓.

3. In two dimensions \(G_{\mu\nu} = R_{\mu\nu} - \tfrac12Rg_{\mu\nu} = 0\) for every metric. The field equation becomes \(0 = \kappa T_{\mu\nu}\): it says nothing about the geometry and forbids all matter. Two-dimensional general relativity does not exist, and the formula of part 2 already warned of it with its \(1/(n-2)\).

Check. Dimensions: \(T/(n-2)\) has the dimensions of \(T\) ✓. Limit: in vacuum part 2 gives \(R_{\mu\nu} = 0\) in every dimension \(n \ge 3\) ✓.

Idea. \(\Lambda\) enters the Newtonian limit as a uniform negative mass density \(-\Lambda c^2/4\pi G\), and a uniform density's effect grows linearly with distance while a point mass's falls as \(1/r^2\). So they cross, and the crossing is far away.

1. Trace with \(g^{\mu\nu}\), using \(g^{\mu\nu}g_{\mu\nu} = 4\): \(R - 2R + 4\Lambda = \kappa T\), so \(R = 4\Lambda - \kappa T\). Then \[ R_{\mu\nu} = \kappa T_{\mu\nu} + \tfrac12Rg_{\mu\nu} - \Lambda g_{\mu\nu} = \kappa\big(T_{\mu\nu} - \tfrac12Tg_{\mu\nu}\big) + \Lambda g_{\mu\nu}. \] For static slow dust, as in the deck, \(T_{00} - \tfrac12Tg_{00} = \tfrac12\rho c^2\), and \(g_{00} \approx -1\): \(R_{00} = \tfrac12\kappa\rho c^2 - \Lambda\). Worked example 3 gave \(R_{00} = \nabla^2\Phi/c^2\), so with \(\kappa = 8\pi G/c^4\), \[ \nabla^2\Phi = 4\pi G\rho - \Lambda c^2 . \] (\(\Lambda g_{00}\) is evaluated with \(g_{00} = -1\); the correction \(\Lambda h_{00}\) is a product of two small quantities and is dropped.)

2. Outside the mass, \(\nabla^2\Phi = -\Lambda c^2\). A particular solution is \(-\Lambda c^2r^2/6\) since \(\nabla^2 r^2 = 6\); the homogeneous radial solutions are \(a + b/r\), and matching to the point mass fixes \(b = -GM\): \[ \Phi = -\frac{GM}{r} - \frac{\Lambda c^2r^2}{6} . \] The radial acceleration is \(-\Phi'(r) = -GM/r^2 + \Lambda c^2r/3\). It vanishes at \[ r_\Lambda = \Big(\frac{3GM}{\Lambda c^2}\Big)^{1/3}. \]

3. Sun: \(3\times1.32712\times10^{20}/(1.09\times10^{-52}\times8.988\times10^{16}) = 4.06\times10^{55}\ \mathrm{m^3}\), cube root \(3.44\times10^{18}\) m \(= 111\) pc — about 85 times the distance to the nearest star, and far outside any orbit the Sun controls. For \(10^{12}M_\odot\), \(r_\Lambda\) scales as \(M^{1/3}\), up by \(10^4\): \(3.44\times10^{22}\) m \(= 1.11\) Mpc. That is the scale of the Local Group, where Andromeda sits at about 0.77 Mpc: \(\Lambda\) is invisible in the solar system and a real effect between galaxies.

Check. Dimensions: \([\Lambda c^2] = \mathrm{m^{-2}\,m^2s^{-2}} = \mathrm{s^{-2}}\), and \([4\pi G\rho] = \mathrm{m^3kg^{-1}s^{-2}\,kg\,m^{-3}} = \mathrm{s^{-2}}\) ✓; \([GM/\Lambda c^2] = \mathrm{m^3s^{-2}}/\mathrm{s^{-2}} = \mathrm{m^3}\) ✓. Limit \(\Lambda\to0\): \(r_\Lambda\to\infty\) and Poisson returns ✓.

Idea. The geometric side of the Newtonian limit is dimension-blind; the source side carries a factor \((n-3)/(n-2)\) from the trace. That factor fixes the coupling for \(n \ge 4\) and destroys the Newtonian limit at \(n = 3\).

1. Worked example 3 used \(\Gamma^i_{00} = -\tfrac12\partial_ih_{00}\) and \(R_{00} = \partial_i\Gamma^i_{00}\) to first order with \(\partial_0 = 0\). Neither step used the range of \(i\): the sum \(\partial_i\partial_i\) is simply the Laplacian in \(n-1\) dimensions. So \(R_{00} = -\tfrac12\nabla^2h_{00} = \nabla^2\Phi/c^2\) for all \(n\).

2. With static dust, \(T_{00} = \rho c^2\), \(T = -\rho c^2\), \(g_{00} = -1\): \[ R_{00} = \kappa\,\frac{n-3}{n-2}\,\rho c^2 . \] Equate with \(\nabla^2\Phi/c^2\): \(\nabla^2\Phi = \kappa c^4\,\frac{n-3}{n-2}\,\rho\). This is Newton's \(4\pi G_n\rho\) iff \[ \kappa = \frac{4\pi G_n}{c^4}\cdot\frac{n-2}{n-3}. \] At \(n = 4\): \(\tfrac21\cdot4\pi G/c^4 = 8\pi G/c^4\) ✓. At \(n = 5\): \(\tfrac32\cdot4\pi G_5/c^4 = 6\pi G_5/c^4\).

3. At \(n = 3\) the source side vanishes: \(R_{00} = 0\) whatever \(\rho\) is, so \(\nabla^2\Phi = 0\) — matter produces no Newtonian attraction, and no choice of \(\kappa\) repairs it. The reason is unit 33's: in three dimensions the Weyl tensor vanishes and Riemann is determined by Ricci, so in vacuum \(R_{\mu\nu} = 0\) forces \(R_{\mu\nu\rho\sigma} = 0\). Outside matter, spacetime is flat; there is no tidal field, hence no \(\partial_i\partial_j\Phi\), hence no Newtonian gravity. (Cited: Deser, Jackiw and 't Hooft, Ann. Phys. 152 (1984) 220, show that a point mass in 2+1 gravity makes space a cone — globally nontrivial, locally flat.)

Check. Dimensions: \(\kappa\) must turn \([T] = \mathrm{J\,m^{-(n-1)}}\) into \(\mathrm{m^{-2}}\), and \(G_n\) has dimensions fixed by \(\nabla^2\Phi = 4\pi G_n\rho\) with \(\rho\) in \(\mathrm{kg\,m^{-(n-1)}}\): \([G_n] = \mathrm{m^{n-1}\,kg^{-1}s^{-2}}\) (at \(n = 4\), the familiar \(\mathrm{m^3kg^{-1}s^{-2}}\)), so \([G_n/c^4] = \mathrm{m^{n-5}kg^{-1}s^{2}}\) and \([\kappa T] = \mathrm{m^{n-5}kg^{-1}s^2\cdot kg\,m^{3-n}s^{-2}} = \mathrm{m^{-2}}\) ✓. Limit \(n\to\infty\): \(\kappa\to4\pi G_n/c^4\), the trace term negligible ✓.

Idea. The trace of the tidal matrix is the rate at which a ball's volume starts to shrink, and the trace of the tidal matrix is a Ricci component, which the field equation sets equal to the local matter. This is the deck's "most compact honest summary of the theory in words", made quantitative.

1. In a local orthonormal frame along the observer's worldline with \(u = c\,e_0\), unit 33 gives \(\mathrm D^2\xi^i/\mathrm d\tau^2 = -R^i{}_{\sigma\mu\nu}u^\sigma\xi^\mu u^\nu = -c^2R^i{}_{0j0}\xi^j\) (the \(\mu = 0\) term vanishes because \(R^i{}_{00 0} = 0\) by antisymmetry in the last pair, and \(\xi^0 = 0\) for deviations measured by the observer). Put \(A_{ij} = c^2R^i{}_{0j0}\). With \(\xi(0) = \xi_0\) and \(\dot\xi(0) = 0\), \(\xi(\tau) = (I - \tfrac12A\tau^2 + O(\tau^3))\xi_0\). The ball is the image of the initial ball under this linear map, so \[ V(\tau) = V(0)\det\big(I - \tfrac12A\tau^2\big) + O(\tau^3) = V(0)\big(1 - \tfrac12\operatorname{tr}A\,\tau^2\big) + O(\tau^3), \] giving \(\ddot V/V = -\operatorname{tr}A = -c^2R^i{}_{0i0}\). Since \(R^0{}_{000} = 0\), \(R^i{}_{0i0} = R^\mu{}_{0\mu0} = R_{00}\), and \(c^2R_{00} = R_{\mu\nu}u^\mu u^\nu\).

2. In the fluid's rest frame, which is the observer's frame here, \(R_{00} = \kappa(T_{00} - \tfrac12Tg_{00}) = \kappa\big(\rho c^2 - \tfrac12(-\rho c^2 + 3p)(-1)\big) = \tfrac12\kappa(\rho c^2 + 3p)\). So \[ \frac{\ddot V}{V} = -\frac{c^2}{2}\cdot\frac{8\pi G}{c^4}(\rho c^2 + 3p) = -4\pi G\Big(\rho + \frac{3p}{c^2}\Big). \]

3. Isotropy: \(A = a\,I\), and \(\operatorname{tr}A = 3a = 4\pi G\rho\), so each deviation obeys \(\ddot\xi = -\tfrac{4\pi G\rho}{3}\xi\): simple harmonic motion with \(\omega^2 = 4\pi G\rho/3\). With \(\rho = 5513\): \(4\pi G\rho = 4.624\times10^{-6}\ \mathrm{s^{-2}}\), \(\omega = 1.241\times10^{-3}\ \mathrm{s^{-1}}\), period \(2\pi/\omega = 5061\) s \(= 84.4\) min. Pressure: \(3p/c^2 = 3\times3.6\times10^{11}/8.988\times10^{16} = 1.2\times10^{-5}\ \mathrm{kg\,m^{-3}}\), two parts in \(10^{9}\) of \(\rho\) — negligible, as the slow limit assumes.

Check. 84.4 minutes is the famous period of a frictionless tunnel through a uniform Earth, and also of a grazing orbit, \(2\pi\sqrt{R^3/GM} = \sqrt{3\pi/G\rho}\) — the same number, obtained here from Ricci curvature ✓. Dimensions: \([4\pi G\rho] = \mathrm{s^{-2}}\) ✓. Limit \(\rho\to0\): \(\ddot V = 0\), the vacuum statement that Weyl curvature distorts the ball but does not change its volume ✓.

Idea. The bulk variation gives the field equations; what is left is a boundary integral of normal derivatives of \(\delta g\), which Dirichlet data do not control. The extrinsic curvature is the one boundary scalar whose variation contains exactly those normal derivatives, so a multiple of it cancels them.

1. As in worked example 4, \(\delta(\sqrt{-g}R) = \sqrt{-g}\,G_{\mu\nu}\delta g^{\mu\nu} + \sqrt{-g}\,\nabla_\lambda v^\lambda\) and \(\delta(\sqrt{-g}\mathcal L_{\text m}) = -\tfrac12\sqrt{-g}\,T_{\mu\nu}\delta g^{\mu\nu}\) (unit 29's definition). With P10's identity \(\sqrt{-g}\nabla_\lambda v^\lambda = \partial_\lambda(\sqrt{-g}v^\lambda)\) and the divergence theorem, \[ c\,\delta S = \int_{\mathcal M}\Big[\frac{1}{2\alpha}G_{\mu\nu} - \tfrac12T_{\mu\nu}\Big]\delta g^{\mu\nu}\sqrt{-g}\,\mathrm d^4x + \frac{1}{2\alpha}\oint_{\partial\mathcal M}v^\lambda n_\lambda\sqrt{|h|}\,\mathrm d^3x . \]

2. The deck's notes give \(v_\lambda = \nabla^\mu\delta g_{\lambda\mu} - \nabla_\lambda(g^{\mu\nu}\delta g_{\mu\nu})\), with \(\delta g_{\mu\nu}\) the variation of the lower-index metric. On the boundary \(\delta g = 0\), so the \(\Gamma\,\delta g\) terms in \(\nabla\delta g\) vanish and \[ n^\lambda v_\lambda = n^\lambda g^{\mu\alpha}\partial_\alpha\delta g_{\lambda\mu} - n^\lambda g^{\mu\nu}\partial_\lambda\delta g_{\mu\nu}. \] Write \(g^{\mu\alpha} = h^{\mu\alpha} + n^\mu n^\alpha\). In the first term, \(h^{\mu\alpha}\partial_\alpha\) differentiates along the boundary, where \(\delta g \equiv 0\), so it gives zero, leaving \(n^\lambda n^\mu n^\alpha\partial_\alpha\delta g_{\lambda\mu}\). In the second, the \(n^\mu n^\nu\) part gives \(-n^\mu n^\nu n^\lambda\partial_\lambda\delta g_{\mu\nu}\), which cancels it. What survives is \[ n^\lambda v_\lambda = -h^{\mu\nu}n^\lambda\partial_\lambda\delta g_{\mu\nu}. \] This is the normal derivative of the tangential components of \(\delta g\). Fixing \(g\) on the boundary does not fix how it changes as one moves off the boundary, so this is a genuinely free quantity; and fixing it as well would overdetermine a second-order equation.

3. \(K = h^{\alpha\beta}(\partial_\alpha n_\beta - \Gamma^\gamma_{\alpha\beta}n_\gamma)\). On the boundary \(n_\beta\) and \(h^{\alpha\beta}\) are built from \(g\) there, which is not varied, so \(\delta K = -h^{\alpha\beta}n_\gamma\delta\Gamma^\gamma_{\alpha\beta}\). With \(\delta g = 0\) on the boundary, \(\delta\Gamma^\gamma_{\alpha\beta} = \tfrac12g^{\gamma\sigma}(\partial_\alpha\delta g_{\sigma\beta} + \partial_\beta\delta g_{\sigma\alpha} - \partial_\sigma\delta g_{\alpha\beta})\). Contracted with \(h^{\alpha\beta}\), the first two terms are tangential derivatives and vanish; so \[ \delta K = -h^{\alpha\beta}n^\sigma\cdot\big(-\tfrac12\partial_\sigma\delta g_{\alpha\beta}\big) = \tfrac12h^{\alpha\beta}n^\sigma\partial_\sigma\delta g_{\alpha\beta}. \] \(\sqrt{|h|}\) is unchanged. Hence \(\delta\big(\beta\oint K\sqrt{|h|}\big) = \tfrac\beta2\oint h^{\mu\nu}n^\lambda\partial_\lambda\delta g_{\mu\nu}\sqrt{|h|}\), while the boundary term of part 1 is \(-\tfrac{1}{2\alpha}\oint h^{\mu\nu}n^\lambda\partial_\lambda\delta g_{\mu\nu}\sqrt{|h|}\). They cancel iff \(\beta = 1/\alpha\): twice the coefficient of \(R\).

4. With the boundary term cancelled, \(\delta S = 0\) for all \(\delta g^{\mu\nu}\) vanishing on \(\partial\mathcal M\) and, by unit 11's fundamental lemma, \(G_{\mu\nu} = \alpha T_{\mu\nu}\). The deck's section 4 then requires \(\alpha = 8\pi G/c^4\), so the Lagrangian coefficient is \(1/2\alpha = c^4/16\pi G\) and the boundary coefficient is \(\beta = c^4/8\pi G\) — the deck's Gibbons-Hawking-York term.

Check. Dimensions: \([K\sqrt{|h|}\,\mathrm d^3x] = \mathrm{m^{-1}\,m^3} = \mathrm{m^2}\), and \([R\sqrt{-g}\,\mathrm d^4x] = \mathrm{m^{-2}\,m^4} = \mathrm{m^2}\): the same, so \(\beta\) and \(1/\alpha\) have the same dimensions, as a ratio \(\beta = 1/\alpha\) requires ✓. Special case: for flat spacetime and a flat boundary, \(K = 0\) and the term contributes nothing, as it should where there is nothing to cancel ✓.

Idea. A Lagrangian with second derivatives leaves, after integration by parts, a boundary term in \(\delta y'\). Dirichlet data kill \(\delta y\) at the ends, not \(\delta y'\). Subtracting the right total derivative converts the action into a first-order one with the same bulk equation and a clean boundary. This is the Gibbons-Hawking-York term in one dimension.

1. \(\delta S = \int_0^1(\delta y\,y'' + y\,\delta y'')\,\mathrm dx\). Twice by parts, \(\int_0^1 y\,\delta y'' = [y\,\delta y']_0^1 - [y'\delta y]_0^1 + \int_0^1 y''\delta y\). With \(\delta y = 0\) at both ends, \[ \delta S = \int_0^1 2y''\,\delta y\,\mathrm dx + y(1)\,\delta y'(1) - y(0)\,\delta y'(0). \] The term \(y\,\delta y'\) at the ends survives.

2. A stationary point needs \(\delta S = 0\) for all admissible \(\delta y\). Taking \(\delta y\) supported in the interior gives \(y'' = 0\) by unit 11's lemma. Then \(\delta S = b\,\delta y'(1) - a\,\delta y'(0)\), and \(\delta y = \epsilon\,x(1-x)\) has \(\delta y'(0) = \epsilon\), \(\delta y'(1) = -\epsilon\), giving \(\delta S = -\epsilon(a + b)\); \(\delta y = \epsilon\,x^2(1-x)\) has \(\delta y'(0) = 0\), \(\delta y'(1) = -\epsilon\), giving \(-\epsilon b\). Both vanish for all \(\epsilon\) only if \(a = b = 0\).

3. The Euler-Lagrange equation \(y'' = 0\) has a two-dimensional solution space, the lines. Prescribing \(y\) and \(y'\) at both ends is four conditions: generically no line satisfies them. Fixing more data is not a repair; it overdetermines, exactly as fixing both \(g\) and its normal derivative on \(\partial\mathcal M\) would.

4. Take \(B = [y\,y']_0^1\). Then \(S - B = \int_0^1 yy'' - [yy']_0^1 = -\int_0^1 y'^2\,\mathrm dx\), and \(\delta(S - B) = -2\int y'\delta y' = -2[y'\delta y]_0^1 + 2\int y''\delta y = 2\int y''\delta y\) when \(\delta y = 0\) at the ends. Well posed, with the same Euler-Lagrange equation. On the solution \(y = x\) (\(a = 0\), \(b = 1\)): \(S = \int_0^1 x\cdot0 = 0\), while \(S - B = -\int_0^1 1 = -1\).

Check. The two actions have the same stationary curves but different values there: the boundary term changes the value of the action and not the equations, which is exactly why it matters for black-hole thermodynamics and not for the field equations. Special case \(a = b = 0\): \(S\) and \(S - B\) agree on every admissible \(y\), since \(B = 0\) ✓.

Idea. A divergence integrates to a boundary term only if it is an ordinary divergence. For a vector it always is; for a symmetric two-index tensor a \(\Gamma T\) term is left over, and in an expanding spacetime that term is the work done by pressure on the expansion — energy leaving with nowhere to go.

1. \(\Gamma^\mu_{\mu\lambda} = \tfrac12g^{\mu\sigma}(\partial_\mu g_{\sigma\lambda} + \partial_\lambda g_{\sigma\mu} - \partial_\sigma g_{\mu\lambda})\); the first and third terms are equal after relabelling \(\mu\leftrightarrow\sigma\) (\(g^{\mu\sigma}\) is symmetric) and cancel. So \(\Gamma^\mu_{\mu\lambda} = \tfrac12g^{\mu\sigma}\partial_\lambda g_{\mu\sigma} = \tfrac12\partial_\lambda\ln|g|\) by Jacobi's formula, which is \(\partial_\lambda\ln\sqrt{-g}\). Then \(\nabla_\mu J^\mu = \partial_\mu J^\mu + \Gamma^\mu_{\mu\lambda}J^\lambda = \partial_\mu J^\mu + J^\lambda\partial_\lambda\ln\sqrt{-g} = \frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}J^\mu)\). For \(T\): \(\nabla_\mu T^{\mu\nu} = \partial_\mu T^{\mu\nu} + \Gamma^\mu_{\mu\lambda}T^{\lambda\nu} + \Gamma^\nu_{\mu\lambda}T^{\mu\lambda}\); the first two combine exactly as for \(J\), and the third is left. \(\square\)

2. With \(x^0 = ct\), \(g_{00} = -1\), \(g_{ij} = a^2\delta_{ij}\), \(\partial_0g_{ij} = 2a\dot a\delta_{ij}/c\). Then \(\Gamma^0_{ij} = -\tfrac12g^{00}\partial_0g_{ij} = a\dot a\delta_{ij}/c\), \(\Gamma^i_{0j} = \tfrac12g^{ik}\partial_0g_{kj} = \dot a\delta^i_j/(ca)\), all others zero; \(\sqrt{-g} = a^3\). At rest, \(T^{00} = \rho c^2\), \(T^{0i} = 0\), \(T^{ij} = pg^{ij} = pa^{-2}\delta^{ij}\). So \[ \nabla_\mu T^{\mu0} = \frac{1}{a^3}\frac{1}{c}\frac{\mathrm d}{\mathrm dt}(a^3\rho c^2) + \frac{a\dot a}{c}\delta_{ij}\,\frac{p}{a^2}\delta^{ij} = \frac{1}{ca^3}\Big[\frac{\mathrm d}{\mathrm dt}(\rho c^2a^3) + 3pa^2\dot a\Big]. \] Setting it to zero, and noting \(3a^2\dot a = \mathrm d(a^3)/\mathrm dt\), gives the result: the first law of thermodynamics, \(\mathrm dE = -p\,\mathrm dV\), for a comoving volume.

3. With \(p = \rho c^2/3\): \(\frac{\mathrm d}{\mathrm dt}(\rho a^3) = -\rho a^2\dot a\). Try \(\rho = Ca^{-4}\): the left side is \(\frac{\mathrm d}{\mathrm dt}(Ca^{-1}) = -Ca^{-2}\dot a\), and the right side is \(-Ca^{-2}\dot a\) ✓, and by unit 01's uniqueness this is the solution. Then \(E = \int a^3\rho c^2\,\mathrm d^3x \propto a^3a^{-4} = 1/a\). For the CMB since emission, the energy in a comoving box has fallen by the factor 1091. No Killing field, no conserved energy (the deck's energy slide), and the loss is exactly the redshift of each photon.

Check. Dust, \(p = 0\): \(\rho a^3\) constant, the number of particles in a comoving box is conserved ✓. Static limit \(\dot a = 0\): all \(\Gamma\) vanish and \(E\) is conserved ✓. Dimensions: with \(a\) dimensionless, \([a\dot a/c] = \mathrm{s^{-1}}/(\mathrm{m\,s^{-1}}) = \mathrm{m^{-1}}\), an inverse length, as a Christoffel symbol in coordinates of length must be ✓.

Idea. Two independent counts, one from the full theory and one from the linearised gauge-fixed wave, should give the same polynomial. They do.

1. Ricci is symmetric: \(n(n+1)/2\).

• \(n = 3\): Riemann 6, Ricci 6, Weyl 0.
• \(n = 4\): Riemann 20, Ricci 10, Weyl 10.
• \(n = 5\): Riemann 50, Ricci 15, Weyl 35.

2. \(\frac{n(n+1)}{2} - n - n = \frac{n(n-3)}{2}\): 0 at \(n = 3\), 2 at \(n = 4\), 5 at \(n = 5\).

3. \(\frac{(n-2)(n-1)}{2} - 1 = \frac{n^2 - 3n + 2 - 2}{2} = \frac{n(n-3)}{2}\) ✓, the same polynomial.

4. \(n - 1 - 1 = n - 2\): two at \(n = 4\), the two polarisations of light; one at \(n = 3\).

Check. \(n = 3\): zero gravitational degrees of freedom and zero Weyl components — no waves and no vacuum tides, which is the deck's statement that \(R_{\mu\nu} = 0\) forces flatness in three dimensions ✓. \(n = 4\) reproduces \(10 - 4 - 4 = 2\) ✓.

Idea. Gauge conditions cut the ten components to six; the residual gauge moves four of those six and leaves two invariant combinations. The invariant two are the physical polarisations, and they match the count by constraints.

1. By the deck's plane-wave lemma, \(\Box\bar h_{\mu\nu} = -\eta^{\alpha\beta}k_\alpha k_\beta A_{\mu\nu}\cos(k\cdot x)\), zero for all \(x\) iff \(k\) is null or \(A = 0\). Also \(\partial^\mu\bar h_{\mu\nu} = -k^\mu A_{\mu\nu}\sin(k\cdot x)\), so \(k^\mu A_{\mu\nu} = 0\). With \(k^\mu = w(1,0,0,1)\): \(A_{0\nu} + A_{3\nu} = 0\), i.e. \(A_{3\nu} = -A_{0\nu}\) for \(\nu = 0,1,2,3\). This gives \(A_{30} = -A_{00}\), \(A_{31} = -A_{01}\), \(A_{32} = -A_{02}\), \(A_{33} = -A_{03} = A_{00}\). Free: \(A_{00}, A_{01}, A_{02}, A_{11}, A_{12}, A_{22}\) — six, which is \(10 - 4\).

2. The trace of \(\partial_\mu\xi_\nu + \partial_\nu\xi_\mu\) is \(2\partial^\lambda\xi_\lambda\), so \(h\to h - 2\partial^\lambda\xi_\lambda\) and \(\bar h_{\mu\nu}\) changes by \(-(\partial_\mu\xi_\nu + \partial_\nu\xi_\mu) + \eta_{\mu\nu}\partial^\lambda\xi_\lambda\). With \(\partial_\mu\xi_\nu = k_\mu B_\nu\cos(k\cdot x)\), the change in \(A\) is as stated. (It preserves the Lorenz condition since \(\Box\xi = 0\), by the deck's gauge computation.) With \(k_\mu = w(-1,0,0,1)\) and \(k^\lambda B_\lambda = w(B_0 + B_3)\):

• \(\delta A_{00} = 2wB_0 - w(B_0 + B_3) = w(B_0 - B_3)\);
• \(\delta A_{01} = wB_1\), \(\delta A_{02} = wB_2\);
• \(\delta A_{11} = \delta A_{22} = w(B_0 + B_3)\);
• \(\delta A_{12} = 0\).

Choose \(B_1 = -A_{01}/w\), \(B_2 = -A_{02}/w\), \(B_0 - B_3 = -A_{00}/w\) and \(B_0 + B_3 = -(A_{11} + A_{22})/2w\). Then \(A_{00} = A_{01} = A_{02} = 0\), hence all \(A_{3\nu} = 0\), and \(A_{11} + A_{22} = 0\). What is left is \(A_{11} = -A_{22} = h_+\), \(A_{12} = h_\times\): the deck's matrix. Its trace is zero, so \(\bar h = h\).

3. \(A_{12}\) and \(A_{11} - A_{22}\) are untouched by every \(B\). The other four directions are pure gauge. So: \(10\) components, \(-4\) from the Lorenz condition, \(-4\) from the residual gauge, \(= 2\); and independently, the deck's \(10 - 4 - 4\) from gauge and constraints. The two fours are different fours with the same count — Lorenz gauge is where the constraints went.

4. The deviation equation gives \(\ddot\xi^i = -c^2R^i{}_{0j0}\xi^j = \tfrac12\ddot h_{ij}\xi^j\) (dots are \(t\)-derivatives, \(\partial_0 = c^{-1}\partial_t\)). To first order in \(h\), with \(\xi^j\) on the right replaced by its unperturbed value, integrate twice: \(\delta\xi^i = \tfrac12h_{ij}\xi^j\). A plus wave: the \(x\)-arm changes by \(+\tfrac12h_+L\), the \(y\)-arm by \(-\tfrac12h_+L\), and their difference by \(h_+L = 1.0\times10^{-21}\times4000 = 4.0\times10^{-18}\) m, which is \(4.0\times10^{-18}/0.84\times10^{-15} = 1/210\) of a proton radius.

Check. Dimensions: \(h\) is dimensionless, so \(\delta\xi\) is a length ✓. Limit \(h\to0\): no displacement ✓. The ring is squeezed along \(y\) as it stretches along \(x\), with area unchanged to first order since \(h\) is traceless — the vacuum statement of P7 that Weyl curvature preserves volume ✓.

Idea. An infinitesimal diffeomorphism changes \(h\) by a symmetrised gradient, and the linearised Riemann tensor is built so that symmetrised gradients drop out, just as \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\) is blind to \(A\to A + \partial\chi\). What survives is physical, and in the Newtonian limit it is the tidal tensor.

1. To first order, \(\Gamma\Gamma\) is second order and \(R_{\rho\sigma\mu\nu} = \eta_{\rho\lambda}R^\lambda{}_{\sigma\mu\nu} = \partial_\mu\Gamma_{\rho\nu\sigma} - \partial_\nu\Gamma_{\rho\mu\sigma}\). Then \(\partial_\mu\Gamma_{\rho\nu\sigma} = \tfrac12(\partial_\mu\partial_\nu h_{\rho\sigma} + \partial_\mu\partial_\sigma h_{\rho\nu} - \partial_\mu\partial_\rho h_{\nu\sigma})\) and \(\partial_\nu\Gamma_{\rho\mu\sigma} = \tfrac12(\partial_\nu\partial_\mu h_{\rho\sigma} + \partial_\nu\partial_\sigma h_{\rho\mu} - \partial_\nu\partial_\rho h_{\mu\sigma})\). The first terms cancel (Clairaut), leaving the stated formula.

2. Substitute \(h_{\alpha\beta} = -\partial_\alpha\xi_\beta - \partial_\beta\xi_\alpha\). Then \(2R_{\rho\sigma\mu\nu}\) is minus \[ \partial_\mu\partial_\sigma(\partial_\rho\xi_\nu + \partial_\nu\xi_\rho) + \partial_\nu\partial_\rho(\partial_\mu\xi_\sigma + \partial_\sigma\xi_\mu) - \partial_\mu\partial_\rho(\partial_\nu\xi_\sigma + \partial_\sigma\xi_\nu) - \partial_\nu\partial_\sigma(\partial_\rho\xi_\mu + \partial_\mu\xi_\rho). \] Sort by which component of \(\xi\) appears. \(\xi_\nu\): \(\partial_\mu\partial_\sigma\partial_\rho\xi_\nu - \partial_\mu\partial_\rho\partial_\sigma\xi_\nu = 0\). \(\xi_\rho\): \(\partial_\mu\partial_\sigma\partial_\nu\xi_\rho - \partial_\nu\partial_\sigma\partial_\mu\xi_\rho = 0\). \(\xi_\sigma\): \(\partial_\nu\partial_\rho\partial_\mu\xi_\sigma - \partial_\mu\partial_\rho\partial_\nu\xi_\sigma = 0\). \(\xi_\mu\): \(\partial_\nu\partial_\rho\partial_\sigma\xi_\mu - \partial_\nu\partial_\sigma\partial_\rho\xi_\mu = 0\). All eight terms cancel in pairs, since partial derivatives commute. \(\square\)

3. Put \(\rho = 0, \sigma = i, \mu = 0, \nu = j\), static: \(R_{0i0j} = \tfrac12(\partial_0\partial_ih_{0j} + \partial_j\partial_0h_{0i} - \partial_0^2h_{ij} - \partial_j\partial_ih_{00}) = -\tfrac12\partial_i\partial_jh_{00} = \partial_i\partial_j\Phi/c^2\). Raising the first index with \(\eta^{ii} = 1\) and using pair symmetry, \(R^i{}_{0j0} = \partial_i\partial_j\Phi/c^2\), and geodesic deviation with \(u = ce_0\) gives \(\ddot\xi^i = -c^2R^i{}_{0j0}\xi^j = -(\partial_i\partial_j\Phi)\xi^j\): unit 05's tidal equation, term for term.

4. In transverse-traceless gauge \(h_{0\mu} = 0\), so the same component is \(R_{0i0j} = -\tfrac12\partial_0^2h_{ij}\). For a wave \(h_{ij} = A_{ij}\cos(k\cdot x)\) this is \(\tfrac12(\omega/c)^2A_{ij}\cos(k\cdot x)\), nonzero whenever \(h_+\) or \(h_\times\) is. By part 2 no coordinate change can remove it, so the polarisations are curvature, not coordinates.

Check. Dimensions: \([\partial\partial h] = \mathrm{m^{-2}}\) and \([\partial_i\partial_j\Phi/c^2] = \mathrm{m^{-2}}\) ✓. Special case: the point mass \(\Phi = -GM/r\) gives tidal eigenvalues \((-2, 1, 1)GM/r^3\) along the radial and two transverse directions, unit 33's result, and their sum \(\nabla^2\Phi = 0\) is \(R_{00} = 0\) in vacuum ✓.