Metric, connection, geodesics — free fall is a straight line

Idea. The plane in polar coordinates is the cleanest case of a nonzero \(\Gamma\) that is entirely an artefact of the chart: every symbol is the inhomogeneous term of the transformation law, and nothing else.

1. \(g_{rr} = 1\), \(g_{\theta\theta} = r^2\), \(g^{rr} = 1\), \(g^{\theta\theta} = r^{-2}\), and the only nonzero derivative is \(\partial_rg_{\theta\theta} = 2r\).

• \(\Gamma^r{}_{rr} = \tfrac12g^{rr}\partial_rg_{rr} = 0\); \(\Gamma^r{}_{r\theta} = \tfrac12g^{rr}\partial_\theta g_{rr} = 0\).
• \(\Gamma^r{}_{\theta\theta} = \tfrac12g^{rr}(2\partial_\theta g_{\theta r} - \partial_rg_{\theta\theta}) = -\tfrac12\cdot2r = -r\).
• \(\Gamma^\theta{}_{rr} = \tfrac12g^{\theta\theta}(2\partial_rg_{r\theta} - \partial_\theta g_{rr}) = 0\).
• \(\Gamma^\theta{}_{r\theta} = \tfrac12g^{\theta\theta}(\partial_rg_{\theta\theta} + \partial_\theta g_{r\theta} - \partial_\theta g_{r\theta}) = \tfrac{2r}{2r^2} = \tfrac1r\).
• \(\Gamma^\theta{}_{\theta\theta} = \tfrac12g^{\theta\theta}\partial_\theta g_{\theta\theta} = 0\).

2. The inverse Jacobian entries are \(A^r{}_x = \partial r/\partial x = \cos\theta\), \(A^r{}_y = \sin\theta\), \(A^\theta{}_x = -\sin\theta/r\), \(A^\theta{}_y = \cos\theta/r\). The needed second derivatives of \(x = r\cos\theta\), \(y = r\sin\theta\) are \(\partial_\theta^2x = -r\cos\theta\), \(\partial_\theta^2y = -r\sin\theta\), \(\partial_r\partial_\theta x = -\sin\theta\), \(\partial_r\partial_\theta y = \cos\theta\). So \[ \Gamma^r{}_{\theta\theta} = \cos\theta(-r\cos\theta) + \sin\theta(-r\sin\theta) = -r,\qquad \Gamma^\theta{}_{r\theta} = \Big(-\frac{\sin\theta}{r}\Big)(-\sin\theta) + \frac{\cos\theta}{r}\cos\theta = \frac1r , \] agreeing with part 1. (Also \(\Gamma^r{}_{r\theta} = \cos\theta(-\sin\theta) + \sin\theta\cos\theta = 0\) ✓.)

3. The geodesic equations are \[ \ddot r - r\dot\theta^2 = 0,\qquad \ddot\theta + \frac2r\,\dot r\dot\theta = 0 . \] With the parametrisation of rung 3, \(\lambda\) is arc length along the line (the velocity \((-\sin\theta_0,\cos\theta_0)\) is a unit vector), \(r = \sqrt{b^2+\lambda^2}\) and \(\tan(\theta-\theta_0) = \lambda/b\), which is \(r\cos(\theta-\theta_0) = b\). Then \(\dot r = \lambda/r\), \(\ddot r = 1/r - \lambda^2/r^3 = b^2/r^3\), and \(\dot\theta = \dfrac{b}{b^2+\lambda^2} = b/r^2\). So \(r\dot\theta^2 = b^2/r^3 = \ddot r\) ✓, and \(r^2\dot\theta = b\) is constant, which is the second equation multiplied by \(r^2\) ✓.

4. By symmetry about \(\theta = \pi/4\), the line has \(\theta_0 = \pi/4\) and \(b = 1\cdot\cos(\pi/4) = 1/\sqrt2\). The endpoints are at \(\lambda = b\tan(\mp\pi/4) = \mp b\), so the length is \(2b = \sqrt2 = 1.4142\). The arc \(r = 1\) has length \(\int_0^{\pi/2}\sqrt{r^2}\,\mathrm d\theta = \pi/2 = 1.5708\). The geodesic is shorter, as a chord should be.

5. In the Cartesian chart \(\Gamma\) vanishes on the whole domain, so the symbols computed here are nothing but the inhomogeneous term of the transformation law; and since \(g_{ij}\) is constant in that chart, every derivative of the metric — the second derivatives included — is removed everywhere at once. On the sphere, by contrast, the normal-coordinate expansion left \(\partial_2\partial_2g_{11} = -2/(3R^2)\) at the pole, and P11 proves that no chart can remove it: that is curvature, and here there is none.

Check. Dimensions: \(r\) is a length and \(\theta\) an angle, so \(\Gamma^r{}_{\theta\theta}\) is a length and \(\Gamma^\theta{}_{r\theta}\) an inverse length; \(\Gamma^r{}_{\theta\theta}\dot\theta^2\) and \(\ddot r\) are both lengths per \(\lambda^2\) ✓. Limit \(b\to0\): the line passes through the origin, \(\dot\theta = 0\), \(\ddot r = 0\) — a radial line at constant speed ✓. The chord \(\sqrt2\) is the Euclidean distance from \((1,0)\) to \((0,1)\) ✓.

Idea. A field that is constant in the Cartesian chart has non-constant polar components; the covariant derivative is built so that the correction terms remove exactly that.

1. \(V^r = \cos\theta\), \(V^\theta = -\sin\theta/r\).

• \(\nabla_rV^r = \partial_r\cos\theta + \Gamma^r{}_{rk}V^k = 0 + 0 = 0\).
• \(\nabla_\theta V^r = -\sin\theta + \Gamma^r{}_{\theta\theta}V^\theta = -\sin\theta + \sin\theta = 0\).
• \(\nabla_rV^\theta = \partial_r(-\sin\theta/r) + \Gamma^\theta{}_{r\theta}V^\theta = \sin\theta/r^2 - \sin\theta/r^2 = 0\).
• \(\nabla_\theta V^\theta = -\cos\theta/r + \Gamma^\theta{}_{\theta r}V^r = -\cos\theta/r + \cos\theta/r = 0\).

2. \(\omega_r = \cos\theta\), \(\omega_\theta = -r\sin\theta\).

• \(\nabla_r\omega_r = 0 - \Gamma^k{}_{rr}\omega_k = 0\).
• \(\nabla_r\omega_\theta = \partial_r(-r\sin\theta) - \Gamma^\theta{}_{r\theta}\omega_\theta = -\sin\theta + \sin\theta = 0\).
• \(\nabla_\theta\omega_r = -\sin\theta - \Gamma^\theta{}_{\theta r}\omega_\theta = -\sin\theta + \sin\theta = 0\).
• \(\nabla_\theta\omega_\theta = -r\cos\theta - \Gamma^r{}_{\theta\theta}\omega_r = -r\cos\theta + r\cos\theta = 0\).

3. \(\partial_rf = 2r\), \(\partial_\theta f = 0\). Then \(\nabla_r\nabla_rf = \partial_r(2r) - \Gamma^k{}_{rr}\partial_kf = 2\); \(\nabla_r\nabla_\theta f = \partial_r0 - \Gamma^r{}_{r\theta}\cdot2r - \Gamma^\theta{}_{r\theta}\cdot0 = 0\), and likewise \(\nabla_\theta\nabla_rf = 0\) (symmetric, as torsion-freeness promised); \(\nabla_\theta\nabla_\theta f = 0 - \Gamma^r{}_{\theta\theta}\cdot2r = 2r^2\). So \(\nabla_i\nabla_jf = \mathrm{diag}(2,2r^2) = 2g_{ij}\), which is the polar form of the Cartesian Hessian \(2\delta_{ij}\) of \(x^2+y^2\). The trace is \(g^{rr}\cdot2 + g^{\theta\theta}\cdot2r^2 = 2+2 = 4\), and indeed \((\partial_x^2+\partial_y^2)(x^2+y^2) = 4\).

4. \(\nabla_rg_{\theta\theta} = \partial_rg_{\theta\theta} - \Gamma^l{}_{r\theta}g_{l\theta} - \Gamma^l{}_{r\theta}g_{\theta l} = 2r - 2\cdot\tfrac1r\cdot r^2 = 0\) ✓.

Check. Dimensions in part 3: \(f\) is an area; \(\nabla_r\nabla_rf\) is dimensionless and \(\nabla_\theta\nabla_\theta f\) an area, matching \(g_{rr}\) and \(g_{\theta\theta}\) ✓. Part 1 is the statement of the deck's slide "The problem, in one computation", now resolved: the constant field has vanishing covariant derivative although its components vary ✓.

Idea. The Euler–Lagrange operator applied to \(g_{ij}\dot x^i\dot x^j\) produces \(2g_{kl}(\ddot x^l + \Gamma^l{}_{ij}\dot x^i\dot x^j)\) — proved on the slide "Shortest" — so for a diagonal metric each equation, divided by \(2g_{kk}\), displays one row of \(\Gamma\).

1. The \(r\) equation. Equating the two lines of rung 3: \[ \frac{2\ddot r}{1-kr^2} + \frac{2kr\,\dot r^2}{(1-kr^2)^2} - 2r\dot\theta^2 - 2r\sin^2\theta\,\dot\varphi^2 = 0 . \] Multiply by \((1-kr^2)/2\): \[ \ddot r + \frac{kr}{1-kr^2}\,\dot r^2 - r(1-kr^2)\,\dot\theta^2 - r(1-kr^2)\sin^2\theta\,\dot\varphi^2 = 0 , \] so \(\Gamma^r{}_{rr} = \dfrac{kr}{1-kr^2}\), \(\Gamma^r{}_{\theta\theta} = -r(1-kr^2)\), \(\Gamma^r{}_{\varphi\varphi} = -r(1-kr^2)\sin^2\theta\), and \(\Gamma^r{}_{r\theta} = \Gamma^r{}_{r\varphi} = \Gamma^r{}_{\theta\varphi} = 0\).

The \(\theta\) equation. \(\tfrac{\mathrm d}{\mathrm d\lambda}(2r^2\dot\theta) = 2r^2\ddot\theta + 4r\dot r\dot\theta\) and \(\partial_\theta\mathcal L = 2r^2\sin\theta\cos\theta\,\dot\varphi^2\). Dividing by \(2r^2\): \[ \ddot\theta + \frac2r\,\dot r\dot\theta - \sin\theta\cos\theta\,\dot\varphi^2 = 0 , \] so \(\Gamma^\theta{}_{r\theta} = 1/r\) (the \(2/r\) is \(2A\) with \(A = 1/r\)), \(\Gamma^\theta{}_{\varphi\varphi} = -\sin\theta\cos\theta\), the other four zero.

The \(\varphi\) equation. \(\varphi\) is absent, so \(\tfrac{\mathrm d}{\mathrm d\lambda}(2r^2\sin^2\theta\,\dot\varphi) = 0\): \(2r^2\sin^2\theta\,\ddot\varphi + 4r\sin^2\theta\,\dot r\dot\varphi + 4r^2\sin\theta\cos\theta\,\dot\theta\dot\varphi = 0\). Dividing, \[ \ddot\varphi + \frac2r\,\dot r\dot\varphi + 2\cot\theta\,\dot\theta\dot\varphi = 0 , \] so \(\Gamma^\varphi{}_{r\varphi} = 1/r\), \(\Gamma^\varphi{}_{\theta\varphi} = \cot\theta\), the other four zero.

2. \(\Gamma^r{}_{rr} = \tfrac12g^{rr}\partial_rg_{rr} = \tfrac12(1-kr^2)\cdot\dfrac{2kr}{(1-kr^2)^2} = \dfrac{kr}{1-kr^2}\) ✓. \(\Gamma^r{}_{\theta\theta} = -\tfrac12g^{rr}\partial_rg_{\theta\theta} = -\tfrac12(1-kr^2)\cdot2r\) ✓.

3. \((r,r,r)\): left side \(\partial_rg_{rr} = 2kr/(1-kr^2)^2\); right side \(2\Gamma^r{}_{rr}g_{rr} = 2\cdot\dfrac{kr}{1-kr^2}\cdot\dfrac1{1-kr^2}\) ✓. \((r,\theta,\theta)\): left \(2r\); right \(2\Gamma^\theta{}_{r\theta}g_{\theta\theta} = 2\cdot\tfrac1r\cdot r^2 = 2r\) ✓. A different equation from the one used to find \(\Gamma\), and both pass.

4. \(n^2(n+1)/2 = 18\) independent symbols; 7 nonzero (\(\Gamma^r{}_{rr}, \Gamma^r{}_{\theta\theta}, \Gamma^r{}_{\varphi\varphi}, \Gamma^\theta{}_{r\theta}, \Gamma^\theta{}_{\varphi\varphi}, \Gamma^\varphi{}_{r\varphi}, \Gamma^\varphi{}_{\theta\varphi}\)). The coordinate formula, done naively, is 18 symbols \(\times\) 3 values of \(l\) \(\times\) 3 derivatives \(= 162\) derivative evaluations (54 once one notices the metric is diagonal). The trick used 3 equations \(\times\) 2 differentiations \(= 6\).

5. \(k = 1\), \(r = \tfrac12\): \(1 - kr^2 = \tfrac34\), so \(\Gamma^r{}_{rr} = \tfrac{1/2}{3/4} = \tfrac23 = 0.6667\) and \(\Gamma^r{}_{\theta\theta} = -\tfrac12\cdot\tfrac34 = -0.375\).

Check. Dimensions: \(k\) is an inverse area, so \(kr^2\) is a pure number, \(\Gamma^r{}_{rr}\) is an inverse length and \(\Gamma^r{}_{\theta\theta}\) a length ✓. Limit \(k\to0\): the metric is flat space in spherical coordinates, \(\Gamma^r{}_{rr}\to0\) and \(\Gamma^r{}_{\theta\theta}\to-r\), which is P1's symbol ✓. At fixed \(r\), the \(\theta,\varphi\) block is the sphere of radius \(r\) and its two symbols are the deck's ✓.

Idea. The trace of \(\Gamma\) is a logarithmic derivative of the volume element, so a divergence is always "multiply by \(\sqrt{|g|}\), take ordinary partials, divide".

1. In \(\tfrac12g^{il}(\partial_ig_{jl} - \partial_lg_{ij})\) rename \(i\leftrightarrow l\) in the second term: \(g^{li}\partial_ig_{lj} = g^{il}\partial_ig_{jl}\), using symmetry of both \(g^{il}\) and \(g_{jl}\). The two cancel, leaving \(\Gamma^i{}_{ij} = \tfrac12g^{il}\partial_jg_{il}\). \(\square\)

2. Jacobi with \(A = (g_{il})\): \(\partial_jg = g\,g^{li}\partial_jg_{il}\), so \(\partial_j\ln|g| = g^{il}\partial_jg_{il}\) (the sign of \(g\) is constant, so \(\ln|g|\) is differentiable and its derivative is \(\partial_jg/g\)). Hence \(\Gamma^i{}_{ij} = \tfrac12\partial_j\ln|g| = \partial_j\ln\sqrt{|g|}\). Then \[ \nabla_iV^i = \partial_iV^i + \Gamma^i{}_{ij}V^j = \partial_iV^i + V^j\,\frac{\partial_j\sqrt{|g|}}{\sqrt{|g|}} = \frac{1}{\sqrt{|g|}}\,\partial_i\big(\sqrt{|g|}\,V^i\big).\ \square \]

3. \(\nabla_iF^{ij} = \partial_iF^{ij} + \Gamma^i{}_{ik}F^{kj} + \Gamma^j{}_{ik}F^{ik}\). The last term is \(\Gamma^j{}_{ik}\), symmetric in \(i,k\), contracted with \(F^{ik}\), antisymmetric: it equals its own negative under renaming \(i\leftrightarrow k\), so it is zero. The middle term is \((\partial_k\ln\sqrt{|g|})F^{kj}\), and the same product-rule step as in part 2 gives the result. \(\square\)

4. \(\sqrt{g} = r^2\sin\theta\). So \[ \nabla_iV^i = \frac{1}{r^2\sin\theta}\Big[\partial_r(r^2\sin\theta\,V^r) + \partial_\theta(r^2\sin\theta\,V^\theta) + \partial_\varphi(r^2\sin\theta\,V^\varphi)\Big]. \] Substituting the orthonormal components, \[ \nabla_iV^i = \frac{1}{r^2}\partial_r\big(r^2V^{\hat r}\big) + \frac{1}{r\sin\theta}\partial_\theta\big(\sin\theta\,V^{\hat\theta}\big) + \frac{1}{r\sin\theta}\partial_\varphi V^{\hat\varphi}, \] unit 05's divergence in spherical coordinates, now derived rather than remembered. For \(V = \partial_r\): \(r^{-2}\partial_r(r^2) = 2/r\), which is \(1\) at \(r = 2\) (in inverse length units). For \(V = r^{-2}\partial_r\): \(r^{-2}\partial_r(1) = 0\) — unit 05's statement that the inverse-square field has zero divergence away from the source.

Check. Dimensions: \(V^{\hat r}\) and \(\nabla_iV^i\) differ by one inverse length, as a divergence should ✓. Special case: Cartesian coordinates have \(\sqrt g = 1\) and the formula is \(\partial_iV^i\) ✓. The flat plane in polar coordinates has \(\sqrt g = r\), giving \(r^{-1}\partial_r(rV^r) + \partial_\theta V^\theta\), as in the deck's notes ✓.

Idea. The uniqueness argument produced a formula; existence is the check that the formula, used as a definition, has every property it was derived from.

1. By rung 3 the four derivative terms cancel in pairs (\(g\) is symmetric), and the remaining terms are \(f\) times those of \(K(X,Y,Z)\); \(-fZg(X,Y)\) and \(g([X,Y],fZ)\) produce no derivative at all. So for fixed \(X,Y\), \(\omega(Z) := \tfrac12K(X,Y,Z)\) is \(C^\infty\)-linear in \(Z\). By the tensor characterisation lemma (deck, notes of "What a connection is") it is a covector field, and since \(\flat\) is an isomorphism at every point there is exactly one vector field \(W\) with \(W^\flat = \omega\), i.e. \(g(W,Z) = \omega(Z)\) for all \(Z\). Call it \(\nabla_XY\). Smoothness: in a chart \(W^k = g^{kl}\omega(\partial_l)\), smooth.

2. \(X\to fX\). Derivative terms: \((Yf)g(Z,X)\) from \(Yg(Z,fX)\); \(-(Zf)g(X,Y)\) from \(-Zg(fX,Y)\); \(-(Yf)g(X,Z)\) from \(g([fX,Y],Z)\); \(+(Zf)g(X,Y)\) from \(-g([fX,Z],Y)\). They cancel in pairs, so \(\nabla_{fX}Y = f\nabla_XY\).

\(Y\to fY\). Derivative terms: \((Xf)g(Y,Z)\) from \(Xg(fY,Z)\); \(-(Zf)g(X,Y)\) from \(-Zg(X,fY)\); \(+(Xf)g(Y,Z)\) from \(g([X,fY],Z)\); \(+(Zf)g(Y,X)\) from \(-g([fY,Z],X)\). The \(Zf\) terms cancel, leaving \(K(X,fY,Z) = fK(X,Y,Z) + 2(Xf)g(Y,Z)\). Halving: \(g(\nabla_X(fY),Z) = g(f\nabla_XY + (Xf)Y,\,Z)\) for all \(Z\), so by nondegeneracy \(\nabla_X(fY) = f\nabla_XY + (Xf)Y\). Additivity in each slot is plain because \(K\) is additive. So \(\nabla\) is an affine connection. \(\square\)

3. Torsion. \(K(Y,X,Z) = Yg(X,Z) + Xg(Z,Y) - Zg(Y,X) + g([Y,X],Z) - g([Y,Z],X) - g([X,Z],Y)\). Subtracting from \(K(X,Y,Z)\): the three derivative terms cancel by symmetry of \(g\), the two brackets involving \(Z\) cancel, and \(g([X,Y],Z) - g([Y,X],Z) = 2g([X,Y],Z)\) remains. So \(g(\nabla_XY - \nabla_YX,Z) = g([X,Y],Z)\) for all \(Z\), and \(T = 0\).

Metric. \(K(X,Z,Y) = Xg(Z,Y) + Zg(Y,X) - Yg(X,Z) + g([X,Z],Y) - g([X,Y],Z) - g([Z,Y],X)\). Adding to \(K(X,Y,Z)\): \(Yg(Z,X)\) and \(Zg(X,Y)\) cancel against their negatives, \(g([X,Y],Z)\) and \(g([X,Z],Y)\) likewise, and \(-g([Y,Z],X) - g([Z,Y],X) = 0\) by antisymmetry of the bracket. What remains is \(2Xg(Y,Z)\). Halving, \(g(\nabla_XY,Z) + g(Y,\nabla_XZ) = Xg(Y,Z)\). \(\square\)

Check. On \(\mathbb R^n\) with coordinate fields \(X = \partial_i\), \(Y = \partial_j\), \(Z = \partial_l\) all brackets vanish and \(K = \partial_ig_{jl} + \partial_jg_{il} - \partial_lg_{ij}\), which is \(2\Gamma^k{}_{ij}g_{kl}\) — the boxed formula, so the connection constructed here is the one the rest of the course computes with ✓. For the Euclidean metric in Cartesian coordinates \(K = 0\) and \(\nabla_{\partial_i}\partial_j = 0\) ✓.

Idea. A geodesic in the wrong parametrisation has an acceleration along its own tangent and nothing else; a non-geodesic has a transverse acceleration that no change of clock removes.

1. By rung 3, \(\nabla_{\dot\gamma}\dot\gamma = 0\cdot\partial_\theta + 6\lambda\,\partial_\varphi\), and \(\dot\gamma = 3\lambda^2\partial_\varphi\). So \(\nabla_{\dot\gamma}\dot\gamma = f\dot\gamma\) with \(f = 6\lambda/3\lambda^2 = 2/\lambda\).

3 (done before 2, since 2 uses it). Let \(y(s)\) be the same curve in a new parameter, \(x(\lambda) = y(s(\lambda))\) with \(s' \gt 0\). Then \(\dot x = s'y'\) and \(\ddot x = s'^2y'' + s''y'\), so \[ \ddot x^k + \Gamma^k{}_{ij}\dot x^i\dot x^j = s'^2\big(y''^k + \Gamma^k{}_{ij}y'^iy'^j\big) + \frac{s''}{s'}\,\dot x^k . \] The left side is \(f\dot x^k\) by hypothesis, so \(y'' + \Gamma y'y' = (f - s''/s')\dot x/s'^2\), which vanishes iff \(s''/s' = f\), i.e. \((\ln s')' = f\), i.e. \(s' = \exp\int f\). That \(s'\) is positive and continuous, so \(s\) is a strictly increasing \(C^1\) reparametrisation. \(\square\)

2. \(s' = \exp\int(2/\lambda) = \lambda^2\) (constant of integration chosen 1), so \(s = \lambda^3/3\). In terms of \(s\), \(\varphi = 3s\): \(\mathrm d^2\varphi/\mathrm ds^2 = 0\) and \(\theta = \pi/2\), which satisfies both geodesic equations ✓ — the equator at constant rate.

4. At \(\theta = \pi/3\) the \(\theta\)-component is \(-\sin\tfrac\pi3\cos\tfrac\pi3\,\dot\varphi^2 = -\tfrac{\sqrt3}{4}\cdot9\lambda^4\ne0\), while \(\dot\sigma\) has no \(\theta\)-component. So \(\nabla_{\dot\sigma}\dot\sigma\) is not a multiple of \(\dot\sigma\), and by the formula in part 3 any reparametrisation only adds multiples of the tangent: the \(\theta\)-component of the geodesic equation stays \(-\tfrac{\sqrt3}{4}(\mathrm d\varphi/\mathrm ds)^2 \ne 0\) in every parameter.

Check. \(f\) has dimension \(1/\lambda\), as \(s''/s'\) must ✓. Special case: for \(\gamma = (\pi/2,\lambda)\) the same computation gives \(f = 0\), already affine ✓. The number \(-\sqrt3/4 = -0.43301\) is the deck's "fails by" figure for the latitude \(\pi/3\) ✓.

Idea. A cyclic coordinate gives a conserved momentum, and a conserved momentum plus unit speed reduces the geodesic problem to one quadrature — the same two first integrals that solved the sphere.

1. The \(x\) equation (rung 3), divided by \(2/y^2\): \(\ddot x - \tfrac2y\,\dot x\dot y = 0\), so \(\Gamma^x{}_{xy} = -1/y\), \(\Gamma^x{}_{xx} = \Gamma^x{}_{yy} = 0\). The \(y\) equation: \(\tfrac{\mathrm d}{\mathrm d\lambda}(2\dot y/y^2) = 2\ddot y/y^2 - 4\dot y^2/y^3\) and \(\partial_y\mathcal L = -2(\dot x^2+\dot y^2)/y^3\); equating and multiplying by \(y^2/2\): \[ \ddot y + \frac{\dot x^2}{y} - \frac{\dot y^2}{y} = 0 , \] so \(\Gamma^y{}_{xx} = 1/y\), \(\Gamma^y{}_{yy} = -1/y\), \(\Gamma^y{}_{xy} = 0\). Check: \(\Gamma^y{}_{xx} = \tfrac12g^{yy}(-\partial_yg_{xx}) = \tfrac12y^2\cdot2y^{-3} = 1/y\) ✓; \(\Gamma^x{}_{xy} = \tfrac12g^{xx}\partial_yg_{xx} = \tfrac12y^2\cdot(-2y^{-3}) = -1/y\) ✓.

2. On \(x = x_0\), \(\dot x = 0\) and the \(x\) equation holds. The \(y\) equation is \(\ddot y = \dot y^2/y\), solved by \(y = e^s\) (then \(\ddot y = e^s = \dot y^2/y\)). So \(s = \ln y\) is an affine parameter, and it is unit speed: \(\dot y^2/y^2 = 1\).

3. \(\dot x = py^2\) with \(p\ne0\), and \(\dot y^2 = y^2 - p^2y^4\). Then \[ \Big(\frac{\mathrm dy}{\mathrm dx}\Big)^2 = \frac{y^2 - p^2y^4}{p^2y^4} = \frac{1/p^2 - y^2}{y^2}. \] The circle \((x-x_0)^2 + y^2 = a^2\) with \(a = 1/|p|\) has \(y\,y' = -(x-x_0)\), so \(y'^2 = (x-x_0)^2/y^2 = (a^2 - y^2)/y^2\): the same equation. That is how the answer is found, but it is not yet a proof, because two first integrals do not give back the second-order equations. The horizontal line \(y \equiv a\), \(\dot x = a\) satisfies \(\dot x = py^2\) with \(p = 1/a\), has unit speed, and solves the first-order equation above — yet the \(y\) equation gives \(\ddot y + \dot x^2/y - \dot y^2/y = a \ne 0\). It is the same trap as the latitudes in the deck's route B.

So verify the circles directly. Parametrise the semicircle by \(x = x_0 + a\tanh s\), \(y = a\,\mathrm{sech}\,s\), which satisfies \((x-x_0)^2 + y^2 = a^2\). Then \(\dot x = a\,\mathrm{sech}^2s\), \(\dot y = -a\,\mathrm{sech}\,s\tanh s\), and \((\dot x^2+\dot y^2)/y^2 = \mathrm{sech}^2s + \tanh^2s = 1\): unit speed. The \(x\) equation: \(\ddot x - 2\dot x\dot y/y = -2a\,\mathrm{sech}^2s\tanh s + 2a\,\mathrm{sech}^2s\tanh s = 0\) ✓. The \(y\) equation: \(\ddot y = a\,\mathrm{sech}\,s\tanh^2s - a\,\mathrm{sech}^3s\) and \((\dot x^2 - \dot y^2)/y = a\,\mathrm{sech}^3s - a\,\mathrm{sech}\,s\tanh^2s\), which sum to \(0\) ✓. Through any point of \(H\), in any non-vertical direction, there is exactly one circle centred on the \(x\)-axis with that tangent (the centre is where the normal line meets the axis); shifting \(s\), and reversing it if needed, makes this parametrisation match any unit initial velocity. By uniqueness of geodesics given \((p,v)\), every geodesic that is not vertical is one of these semicircles.

4. On \(x = \sqrt2\cos t\), \(y = \sqrt2\sin t\): \(\sqrt{\dot x^2+\dot y^2}/y = \sqrt2/(\sqrt2\sin t) = 1/\sin t\). The points \((\pm1,1)\) are at \(t = 3\pi/4\) and \(\pi/4\), so \[ L = \int_{\pi/4}^{3\pi/4}\frac{\mathrm dt}{\sin t} = \Big[\ln\tan\tfrac t2\Big]_{\pi/4}^{3\pi/4} = \ln\frac{\tan(3\pi/8)}{\tan(\pi/8)} = \ln(1+\sqrt2)^2 = 2\ln(1+\sqrt2) = 1.7627 , \] using \(\tan(3\pi/8) = 1+\sqrt2\) and \(\tan(\pi/8) = \sqrt2-1 = 1/(1+\sqrt2)\). The horizontal segment has \(y = 1\) throughout and length \(\int_{-1}^1\mathrm dx/1 = 2\). The geodesic, bulging up to where \(y\) is larger and lengths are cheaper, is shorter by \(0.2373\). From \((0,1)\) to \((0,2)\): \(\int_1^2\mathrm dy/y = \ln2 = 0.6931\).

Check. The metric is dimensionless here (lengths are counted in units in which the curvature radius is 1), so all three answers are pure numbers ✓. Limit: for two points close together at height \(y\), the distance tends to (Euclidean distance)\(/y\), and the semicircle through them is nearly the straight segment ✓. Special case \(p\to0\): \(a\to\infty\) and the semicircles straighten into the vertical lines of part 2 ✓.

Idea. In Lorentzian signature the spatial part of the velocity is subtracted under the root, so every wiggle costs proper time; in Euclidean signature it is added, so every wiggle costs length. Same computation, opposite extremum.

1. In an inertial chart \(\eta\) is constant, so \(\Gamma = 0\), and \(x^\mu(\tau) = (c\tau, 0)\) has \(\ddot x^\mu = 0\) ✓.

2. With \(u = \dot x^2/c^2\in[0,1)\), rung 3 gives \(\sqrt{1-\dot x^2/c^2} \le 1 - \dot x^2/(2c^2)\) pointwise. Integrating, \(\tau[x] \le T - \frac{1}{2c^2}\int_0^T\dot x^2\,\mathrm dt\). If \(x\not\equiv0\) then, since \(x(0) = x(T) = 0\), \(\dot x\) is not identically zero, so the integral is strictly positive and \(\tau[x]\lt T = \tau[0]\). \(\square\) This is a global statement, not only a second-variation one: no worldline between the two events, however large the excursion, beats the geodesic.

3. \(\sqrt{1+\dot x^2}\ge1\) pointwise, so the length is \(\ge T\), with equality only for \(\dot x\equiv0\). The flip is the sign in front of \(\dot x^2\) under the root, which is the sign of \(g_{xx}\) relative to \(g_{tt}\): the straight line minimises in Riemannian signature and maximises in Lorentzian.

4. \(\dot x = \beta c\cos(\pi t/T)\), so \(\int_0^T\dot x^2 = \beta^2c^2T/2\) and the bound is \(\tau\le T(1-\beta^2/4) = 10\times0.91 = 9.1\) years. Simpson with 8 intervals on \(\tfrac1\pi\int_0^\pi\sqrt{1-0.36\cos^2u}\,\mathrm du\) gives \(0.90278\), so \(\tau = 9.028\) years (four intervals give \(9.037\); the converged value is \(9.0278\)). The travelling twin ages \(0.972\) years less.

Check. Dimensions: \(\dot x^2/c^2\) is a pure number and \(\tau\) a time ✓. Limit \(\beta\to0\): \(\tau\to T\) ✓. The exact value lies below the bound, as part 2 requires, and the bound is close because \(\beta^2 = 0.36\) is modest: the next term in \(\sqrt{1-u}\) is \(-u^2/8\) ✓.

Idea. On a geodesic polygon, transport is bookkeeping of angles, and the rotation on return equals the enclosed area over \(R^2\). This is the deck's latitude example with the non-geodesic side replaced by geodesics, and unit 33 checks it against the integral of curvature.

1. Leg \(N\to A\): \(v\) is the unit tangent, which is parallel, so \(v\) arrives at \(A\) as \(\hat e_\theta\), pointing south. Leg \(A\to B\) along the equator: the tangent is \(\hat e_\varphi\); \(v\) starts perpendicular to it and stays so, with the same orientation, so \(v = \hat e_\theta\) all along (worked example 3 at \(\theta_0 = \pi/2\) says the same: nothing rotates on the equator). Leg \(B\to N\) along \(\varphi = \alpha\): the tangent is \(-\hat e_\theta\), and \(v = \hat e_\theta\) is minus the tangent, so it stays minus the tangent and arrives at \(N\) pointing away from \(N\) along the meridian \(\varphi = \alpha\).

At \(N\), the directions of the meridians \(\varphi = 0\) and \(\varphi = \alpha\) differ by the angle \(\alpha\): in normal coordinates at \(N\) (the deck's example) the meridian \(\varphi\) leaves the origin along \((\cos\varphi,\sin\varphi)\). So \(v\) has rotated by \(\alpha\), in the sense of increasing \(\varphi\) — the same sense in which the loop was traversed about \(N\).

2. \(\text{Area} = \int_0^\alpha\!\!\int_0^{\pi/2}R^2\sin\theta\,\mathrm d\theta\,\mathrm d\varphi = \alpha R^2\). So the rotation is \(\text{Area}/R^2\), the same law as for the latitude cap.

3. \(\alpha = \pi/2\): a rotation of \(90^\circ\); the triangle is an octant of the sphere with area \(\pi R^2/2 = 6.3758\times10^7\ \mathrm{km}^2\). For area \(10^6\ \mathrm{km}^2\): \(\alpha = 10^6/6371^2 = 0.024637\) rad \(= 1.4116^\circ\), and the vector rotates by the same \(1.4116^\circ\).

Check. Dimensions: area over \(R^2\) is an angle ✓. Limit \(\alpha\to0\): the triangle collapses, area \(\to0\) and no rotation ✓. The angle sum of the triangle is \(\pi/2+\pi/2+\alpha = \pi + \alpha\): the excess over \(\pi\) is again \(\alpha\) — Girard's theorem, that a spherical triangle's area is \(R^2\) times its angle excess (cited: do Carmo, Differential Geometry of Curves and Surfaces, §4-5), a third route to the same number ✓.

Idea. A cone is flat everywhere it is smooth, yet transport round the apex rotates vectors: the curvature is all at the apex, which is not part of the manifold.

1. As in P1 with \(g_{\theta\theta} = \beta^2r^2\): \(\Gamma^r{}_{\theta\theta} = -\tfrac12\partial_r(\beta^2r^2) = -\beta^2r\), \(\Gamma^\theta{}_{r\theta} = \tfrac12(\beta^2r^2)^{-1}\cdot2\beta^2r = 1/r\), the rest zero.

2. On a region not encircling the apex, \(\theta\) can be taken single-valued in an interval of length less than \(2\pi\), and \(\vartheta = \beta\theta\) turns the metric into \(\mathrm dr^2 + r^2\mathrm d\vartheta^2\), the flat plane in polar coordinates on a sector of angle less than \(2\pi\beta\le2\pi\). Then \(x = r\cos\vartheta\), \(y = r\sin\vartheta\) is a Cartesian chart there, with \(\Gamma = 0\) identically (P1, part 5).

3. With \(a = V^r\), \(b = \beta r_0V^\theta\) (components in the orthonormal frame \(\hat e_r = \partial_r\), \(\hat e_\theta = (\beta r_0)^{-1}\partial_\theta\)): \(\dot a = \beta^2r_0V^\theta = \beta b\) and \(\dot b = \beta r_0\dot V^\theta = -\beta a\). So \((a+\mathrm ib)^{\boldsymbol\cdot} = -\mathrm i\beta(a+\mathrm ib)\) and \(a + \mathrm ib = e^{-\mathrm i\beta\lambda}(a_0+\mathrm ib_0)\). After \(\lambda = 2\pi\) the vector has turned by \(-2\pi\beta\) relative to the frame, which has itself returned to its starting position. As a rotation this is the same as \(2\pi(1-\beta)\). It does not depend on \(r_0\).

4. \(\beta = 3/4\): rotation \(-3\pi/2\equiv\pi/2\), that is \(90^\circ\) — exactly the angle of the wedge removed. The curvature of the smooth part is zero (part 2), so the holonomy must come from the apex, where the removed wedge concentrates a total "enclosed curvature" equal to the deficit angle \(2\pi(1-\beta)\). Every loop around the apex encloses all of it, which is why \(r_0\) does not matter.

Another way

The circle of latitude \(\theta_0\) on the sphere touches a cone along it (the cone tangent to the sphere there), and the two surfaces have the same metric to first order along that circle, hence the same Christoffel symbols on it and the same transport. That cone has \(\beta = \cos\theta_0\): unrolled, its slant length to the circle is \(R\tan\theta_0\) and the circle's length is \(2\pi R\sin\theta_0\), so \(\beta = \sin\theta_0/\tan\theta_0 = \cos\theta_0\). Then part 3 gives the deck's \(-2\pi\cos\theta_0\) ✓.

Check. \(\beta = 1\): the plane, rotation \(2\pi\equiv0\) ✓. Dimensions: \(a,b\) are components in an orthonormal frame, \(\beta\) is a pure number ✓. For physics: a straight cosmic string of mass per length \(\mu\) produces exactly this geometry in the plane perpendicular to it, with deficit angle \(8\pi G\mu/c^2\) (cited: A. Vilenkin, Phys. Rev. D 23 (1981) 852); for \(G\mu/c^2 = 10^{-7}\) that is \(2.513\times10^{-6}\) rad \(= 0.518''\).

Idea. Geodesic polar coordinates are already normal coordinates in disguise; converting them to Cartesian form shows the metric flat to first order, with one number at second order that is the curvature.

1. With \(\dot\phi = 0\) the \(\phi\) equation holds and the \(\rho\) equation reads \(\ddot\rho = 0\), solved by \(\rho = s\). The speed is \(\sqrt{\dot\rho^2} = 1\).

2. Substituting rung 2's expressions into \(g = \mathrm d\rho^2 + f^2\mathrm d\phi^2\): \[ g_{ab} = \frac{y_ay_b}{\rho^2} + \frac{f^2}{\rho^4}\big(\rho^2\delta_{ab} - y_ay_b\big) = \delta_{ab} + \Big(\frac{f^2}{\rho^2} - 1\Big)\Big(\delta_{ab} - \frac{y_ay_b}{\rho^2}\Big). \] Since \(f\) is odd, \(f^2/\rho^2\) is a smooth even function of \(\rho\) equal to \(1\) at \(0\), and \((f^2/\rho^2-1)\) is \(\rho^2\) times a smooth function of \(\rho^2\), so \(g_{ab}\) is smooth at the origin with \(g_{ab}(0) = \delta_{ab}\). The vectors \(\partial_{y^1},\partial_{y^2}\) are therefore orthonormal at the origin, and by part 1 the geodesic with initial velocity \(v = s(\cos\phi_0,\sin\phi_0)\) reaches, at parameter 1, the point at \(\rho = s\) in direction \(\phi_0\) — whose \(y\)-coordinates are \(v\). So \(\mathbf y(\exp_0(v)) = v\): normal coordinates.

3. From rung 3, \(f^2/\rho^2 - 1 = -K\rho^2/3 + O(\rho^4)\), so \(g_{ab} = \delta_{ab} - \tfrac K3(\rho^2\delta_{ab} - y_ay_b) + O(\rho^4)\) with \(\rho^2 = |\mathbf y|^2\). Explicitly \(g_{11} = 1 - Ky_2^2/3\), \(g_{22} = 1 - Ky_1^2/3\), \(g_{12} = Ky_1y_2/3\) to this order. So at the origin: \(g_{ab} - \delta_{ab} = 0\); every \(\partial_cg_{ab} = 0\), since the correction is quadratic; hence \(\Gamma = 0\) by the coordinate formula. But \(\partial_2\partial_2g_{11}(0) = -2K/3\), nonzero unless \(K = 0\).

Why no chart removes it. The counting on the local-flatness slide says a surface keeps at least one number; here is the number, and a proof it is kept. Put \(Q := \epsilon^{ac}\epsilon^{bd}\,\partial_c\partial_dg_{ab}(0) = \partial_2\partial_2g_{11} + \partial_1\partial_1g_{22} - 2\partial_1\partial_2g_{12}\), which for the expansion above is \(-\tfrac{2K}3 - \tfrac{2K}3 - \tfrac{2K}3 = -2K\). Any other chart with \(g = \delta\) and \(\partial g = 0\) at the origin differs from this one by \(\tilde y^a = O^a{}_by^b + \tfrac16P^a{}_{bcd}y^by^cy^d + \dots\) with \(O\) orthogonal: \(\Gamma(0) = 0\) in both charts, so by the transformation law of \(\Gamma\) the quadratic term is zero. The orthogonal part leaves \(Q\) alone, since \(\epsilon^{ac}\epsilon^{bd}\) is multiplied by \((\det O)^2 = 1\). For the cubic part, pulling the metric back gives \(\partial_c\partial_d\tilde g_{ab}(0) = \partial_c\partial_dg_{ab}(0) - P_{acdb} - P_{bcda}\) with \(P_{abcd} = P^a{}_{bcd}\) symmetric in its last three indices (indices moved with \(\delta\)). In \(Q\), \(\epsilon^{bd}P_{acdb}\) vanishes because \(P_{acdb}\) is symmetric in \(d,b\), and \(\epsilon^{ac}P_{bcda}\) because it is symmetric in \(c,a\). So \(Q = -2K\) in every such chart, and when \(K\ne0\) the second derivatives cannot all vanish. Unit 33 names \(K\) the Gaussian curvature.

4. \(R\sin(\rho/R) = \rho - \rho^3/(6R^2) + \dots\): \(K = 1/R^2\), the deck's sphere. \(f = \rho\): \(K = 0\), the plane, where normal coordinates are Cartesian and \(g = \delta\) exactly. \(a\sinh(\rho/a) = \rho + \rho^3/(6a^2) + \dots\): \(K = -1/a^2\), negative, and the correction has the opposite sign. For the Earth, the largest deviation of \(g_{11}\) at \(|\mathbf y| = 1\) km is \(\tfrac13(1/6371)^2 = 8.21\times10^{-9}\).

Check. Dimensions: \(K\) is an inverse area and \(K|\mathbf y|^2\) a pure number ✓. With \(K = 1/R^2\) the expansion is the deck's slide "Normal coordinates on the sphere" ✓. Limit \(|\mathbf y|\to0\): the deviation falls quadratically, the precise sense of "locally flat" ✓.

Idea. The whole of Newtonian gravity sits in one symbol, \(\Gamma^i{}_{00}\), and the factor of two in \(g_{00}\) is there to cancel the \(\tfrac12\) of the Christoffel formula.

1. \(\Gamma^i{}_{00} = \tfrac12\delta^{ij}(2\partial_0g_{j0} - \partial_jg_{00}) = \partial_i\Phi/c^2\), exactly, since \(g^{ij} = \delta^{ij}\). \(\Gamma^0{}_{0i}\) as in rung 3. \(\Gamma^0{}_{00} = \tfrac12g^{00}\partial_0g_{00} = 0\) (static). \(\Gamma^i{}_{0j} = \tfrac12(\partial_0g_{ji} + \partial_jg_{0i} - \partial_ig_{0j}) = 0\). \(\Gamma^i{}_{jk} = 0\) since \(g_{ij}\) is constant. The remaining ones, \(\Gamma^0{}_{ij} = \tfrac12g^{00}(\partial_ig_{j0}+\partial_jg_{i0}-\partial_0g_{ij}) = 0\).

2. The spatial equation is \(\ddot x^i = -\Gamma^i{}_{00}(\dot x^0)^2 = -\partial_i\Phi\,\dot t^2\). The time equation is \(c\,\ddot t = -2\Gamma^0{}_{0j}\,c\dot t\,\dot x^j\), so \(\ddot t = -2\Gamma^0{}_{0j}v^j\dot t^2\). Then \[ \frac{\mathrm d^2x^i}{\mathrm dt^2} = \frac{\ddot x^i - v^i\ddot t}{\dot t^2} = -\partial_i\Phi + 2v^i\,\Gamma^0{}_{0j}v^j , \] which is the stated result. \(\square\) The first term is Newton's law with no approximation at all for this metric; the factor \(2\) in \(g_{00}\) and the \(\tfrac12\) in \(\Gamma\) cancelled exactly.

3. At rest, \(|\mathrm d^2\vec x/\mathrm dt^2| = GM_\oplus/R_\oplus^2 = 9.7983\ \mathrm{m\,s^{-2}}\). With \(g_{00} = -(1+\Phi/c^2)\), \(\Gamma^i{}_{00}\) halves and so does the answer: \(4.8991\ \mathrm{m\,s^{-2}}\).

4. Relative to \(|\nabla\Phi|\), the second term is at most \(2v^2/c^2\). The stone: \(v = \sqrt{2gh} = 44.29\ \mathrm{m\,s^{-1}}\), and \(v\parallel\nabla\Phi\), so the ratio is \(2v^2/c^2 = 4.37\times10^{-14}\). A circular orbit: \(\vec v\perp\nabla\Phi\), so \(\vec v\cdot\nabla\Phi = 0\) and the term vanishes identically. Terms of this order (\(v^2/c^2\) relative) are exactly what the metric's spatial part \(g_{ij} = (1-2\Phi/c^2)\delta_{ij}\), dropped here, also contributes to; the deck's notes say why slow bodies cannot see it and light can.

Check. Dimensions: \(\Phi\) is \(\mathrm{m^2s^{-2}}\), \(\partial_i\Phi\) an acceleration, and \((\vec v\cdot\nabla\Phi)v/c^2\) an acceleration ✓. Limit \(c\to\infty\): the second term vanishes, Newton exactly ✓. The halved value is \(g/2\), as the deck's warning box says ✓.

Idea. The divergence of \(T\) splits into a part along \(u\) (energy) and a part orthogonal to it (momentum). The orthogonal part is "density times acceleration equals minus pressure gradient", so the only thing that can push a fluid element off its geodesic is a pressure gradient.

1. \(u_\nu a^\nu = u_\nu u^\mu\nabla_\mu u^\nu = \tfrac12u^\mu\nabla_\mu(u_\nu u^\nu) = \tfrac12u^\mu\partial_\mu(-c^2) = 0\), using metric compatibility to move \(\nabla\) through the contraction. Contracting rung 3's line with \(u_\nu\): \(-c^2\nabla_\mu[(\rho+p/c^2)u^\mu] + 0 + u^\mu\partial_\mu p = 0\), which rearranges to the stated equation.

2. Apply \(h^\lambda{}_\nu\) to rung 3's line. \(h^\lambda{}_\nu u^\nu = u^\lambda + u^\lambda(u\cdot u)/c^2 = 0\) kills the first term; \(h^\lambda{}_\nu a^\nu = a^\lambda + u^\lambda(u\cdot a)/c^2 = a^\lambda\) by part 1; and \(h^\lambda{}_\nu g^{\mu\nu}\partial_\mu p = h^{\lambda\mu}\partial_\mu p\). So \((\rho+p/c^2)a^\lambda = -h^{\lambda\mu}\partial_\mu p\). \(\square\) If \(\rho + p/c^2 \gt 0\), then \(a = 0\) — geodesic motion — if and only if \(h^{\lambda\mu}\partial_\mu p = 0\). For dust \(p = 0\) and the right side vanishes, recovering the deck's theorem.

3. For air at rest, \(u^i = 0\) and \(u^0 = c\,\mathrm dt/\mathrm d\tau\). Then \(a^i = \mathrm du^i/\mathrm d\tau + \Gamma^i{}_{00}(u^0)^2 = 0 + (\partial_i\Phi/c^2)\,c^2(\mathrm dt/\mathrm d\tau)^2 = \partial_i\Phi\,(1+O(\Phi/c^2))\): an acceleration of magnitude \(g\), pointing up (\(\Phi\) increases upwards). Air at rest is not freely falling; it is being held up. Since \(u\) has no spatial part, \(h^{zz} = g^{zz} = 1\), and part 2 gives \[ \frac{\mathrm dp}{\mathrm dz} = -(\rho + p/c^2)\,g = -1.2\times9.81 = -11.77\ \mathrm{Pa\,m^{-1}} , \] the hydrostatic equation: pressure falls by about 11.8 Pa per metre of height. The correction \(p/c^2 = 101\,325/c^2 = 1.13\times10^{-12}\ \mathrm{kg\,m^{-3}}\) is twelve orders of magnitude below \(\rho\): irrelevant here, though it matters inside a neutron star.

Check. Dimensions: \(\rho + p/c^2\) is a density, \(a\) an acceleration, and their product is a force per volume, as is \(\partial p\) ✓. Limit \(p\to0\): dust, geodesic motion ✓. Sign: pressure decreases upwards, as in every atmosphere ✓.