General Relativity — worked problems

Setup

Use colatitude $\theta$ (angle from north pole) and longitude $\phi$. Latitude $10°$N means $\theta_0 = 80°$. Ship sails along $\theta = \theta_0$ constant, $\phi$ increasing from $30°$W to $60°$W (westward), so $\Delta\phi = 30°$ in magnitude.

Why latitude circles aren't geodesics

The geodesic equation $\ddot\theta - \sin\theta\cos\theta\,\dot\phi^2 = 0$ (Problem 2.6) requires $\sin\theta\cos\theta\,\dot\phi^2 = 0$ for $\ddot\theta = 0$. At $\theta_0 = 80°$, $\sin\theta_0\cos\theta_0 \neq 0$, so the latitude circle is NOT a geodesic — a vector transported along it rotates relative to local north.

Parallel-transport equation in the orthonormal frame

Introduce orthonormal frame $\hat e_1 = \partial_\theta$ (south), $\hat e_2 = (1/\sin\theta)\partial_\phi$ (east). Along $\theta = \theta_0$, $\phi(s) = \alpha s$ (so $\dot\theta = 0$, $\dot\phi = \alpha$). Express the transported vector $V = V^1\hat e_1 + V^2\hat e_2$; the parallel-transport equation in this frame becomes a rotation: $$\frac{dV^1}{ds} = +\cos\theta_0\cdot\alpha\,V^2,\qquad \frac{dV^2}{ds} = -\cos\theta_0\cdot\alpha\,V^1.$$ This rotates $V$ by angular rate $$\frac{d\psi}{ds} = -\cos\theta_0\cdot\alpha.$$

Integration

Total bearing change: $$\boxed{\;\Delta\psi = -\cos\theta_0\cdot\Delta\phi.\;}$$

Numerical answer

$\cos 80° \approx 0.1736$. $\Delta\phi = 30°$. So $\Delta\psi = -30°\cdot 0.1736 \approx -5.2°$. Initial bearing $45°$ NE, final bearing $\approx 45° - 5.2° = 39.8°$, i.e. just under $40°$ NE (slightly more northward than initial).

Foucault precession connection

A pendulum at latitude $\lambda = 90° - \theta_0$ traces a closed loop on the celestial sphere each sidereal day. By the same formula (with $\Delta\phi = 2\pi$), the swing-plane rotates by $$\Delta\psi_{\text{day}} = -2\pi\cos\theta_0 = -2\pi\sin\lambda.$$ This is the famous Foucault rate $360°\sin\lambda/\text{sidereal day}$. At Paris ($\lambda\approx 49°$): $\approx 272°$/day. Equator: 0. Poles: $360°$/day. The Coriolis-force picture in classical mechanics is the same phenomenon viewed kinematically.

Geometry of the triangle

Place $A$ on the equator at $\phi = 0$, $B$ on the equator at $\phi = \theta_0$, and $N$ at the north pole. The three sides:

• $A\to N$: meridian $\phi = 0$ (great circle, geodesic).
• $N\to B$: meridian $\phi = \theta_0$ (geodesic).
• $B\to A$: equator (great circle, geodesic).

Interior angles: at $A$ and $B$, equator meets meridian at right angles, so $\alpha = \beta = \pi/2$. At $N$, the two meridians are separated by longitude $\theta_0$, so $\gamma = \theta_0$.

Gauss–Bonnet for a geodesic triangle

For a geodesic triangle on a 2D surface of constant Gaussian curvature $K$: $$\text{Area}(\triangle)\cdot K = \alpha + \beta + \gamma - \pi.$$ On the unit sphere $K = 1$: $$\text{Area} = \frac{\pi}{2} + \frac{\pi}{2} + \theta_0 - \pi = \theta_0.$$

Holonomy formula

Along each geodesic segment, parallel transport preserves the angle between $V$ and the geodesic tangent (since $\nabla_{\dot\gamma}\dot\gamma = 0$ and $\nabla$ is metric-compatible). Rotations of $V$ occur only at the vertices, where the geodesic direction abruptly turns by $\pi - \alpha$ (exterior angle).

Summing exterior-angle rotations around the closed loop: $\sum(\pi - \alpha_i) = 3\pi - (\alpha+\beta+\gamma) = 3\pi - (\pi + \theta_0) = 2\pi - \theta_0$. But going around once, the tangent direction rotates by $2\pi$ in the ambient sense, leaving a net intrinsic rotation of $$\Delta\psi = 2\pi - (2\pi - \theta_0) = \theta_0 = \text{Area}.$$

$$\boxed{\;\Delta\psi = \oint K\,dA = \text{Area enclosed by the geodesic triangle.}\;}$$

Generalisations

This is the seed of holonomy: curvature is the obstruction to path-independent parallel transport. Generalises to: (i) Berry phases in quantum mechanics (parameter-space curvature), (ii) Lense–Thirring geodetic precession in GR (Gravity Probe B measured $6.6$ arcsec/year), (iii) Yang–Mills field strengths, (iv) the Atiyah–Singer index theorem. The deep statement: any curvature 2-form integrated over a closed loop gives the holonomy of the corresponding connection.

Metric on the unit sphere

$g_{\theta\theta} = 1$, $g_{\phi\phi} = \sin^2\theta$, off-diagonal zero. The inner product of tangent vectors $u = u^\theta\partial_\theta + u^\phi\partial_\phi$ and $v = v^\theta\partial_\theta + v^\phi\partial_\phi$:

$$\langle u,v\rangle = g_{ab}u^a v^b = g_{\theta\theta}u^\theta v^\theta + g_{\phi\phi}u^\phi v^\phi.$$

Substitute $u = (1,2)$ and $v = (2,-1)$

$$\langle u,v\rangle = 1\cdot 1\cdot 2 + \sin^2\theta\cdot 2\cdot(-1) = 2 - 2\sin^2\theta = 2(1 - \sin^2\theta).$$

$$\boxed{\;\langle u,v\rangle = 2\cos^2\theta.\;}$$

Point dependence

Result varies with latitude $\theta$:

• $\theta = 0$ (north pole, where chart degenerates): $\langle u,v\rangle = 2$, but the formula must be re-derived in a non-degenerate chart.
• $\theta = \pi/4$: $\langle u,v\rangle = 1$.
• $\theta = \pi/2$ (equator): $\langle u,v\rangle = 0$ — the vectors are orthogonal there.

The fact that "the same" vector components $(1,2),(2,-1)$ give different inner products at different points is the structural feature distinguishing Riemannian geometry from Euclidean linear algebra: the inner product is a tensor field, not a constant bilinear form.

Christoffel inputs

From Problem 2.7: $\Gamma^\theta_{\phi\phi} = -\tfrac12\sin(2\theta)$, $\Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} = \cot\theta$, all others zero.

Riemann formula

$$R^a{}_{bcd} = \partial_c\Gamma^a_{db} - \partial_d\Gamma^a_{cb} + \Gamma^a_{ce}\Gamma^e_{db} - \Gamma^a_{de}\Gamma^e_{cb}.$$

Compute $R^\theta{}_{\phi\theta\phi}$ (with $a=\theta$, $b=\phi$, $c=\theta$, $d=\phi$)

$\partial_\theta\Gamma^\theta_{\phi\phi} = \partial_\theta[-\tfrac12\sin(2\theta)] = -\cos(2\theta)$.

$\partial_\phi\Gamma^\theta_{\theta\phi} = 0$ (since $\Gamma^\theta_{\theta\phi} = 0$).

$\Gamma^\theta_{\theta e}\Gamma^e_{\phi\phi} = 0$ (all $\Gamma^\theta_{\theta e} = 0$).

$\Gamma^\theta_{\phi e}\Gamma^e_{\theta\phi}$: only $e = \phi$ contributes: $\Gamma^\theta_{\phi\phi}\Gamma^\phi_{\theta\phi} = (-\tfrac12\sin 2\theta)(\cot\theta) = -\sin\theta\cos\theta\cdot\cot\theta = -\cos^2\theta$.

Combine: $$R^\theta{}_{\phi\theta\phi} = -\cos(2\theta) - 0 + 0 - (-\cos^2\theta) = -\cos(2\theta) + \cos^2\theta.$$ Using $\cos(2\theta) = 2\cos^2\theta - 1$: $R^\theta{}_{\phi\theta\phi} = -2\cos^2\theta + 1 + \cos^2\theta = 1 - \cos^2\theta = \sin^2\theta$.

Lower one index

$R_{\theta\phi\theta\phi} = g_{\theta\theta}R^\theta{}_{\phi\theta\phi} = \sin^2\theta$.

Gaussian curvature

In 2D, $K = R_{1212}/\det g = \sin^2\theta/\sin^2\theta = 1$.

$$\boxed{\;K = 1\;\text{everywhere on the unit sphere.}\;}$$

Constant-curvature form

In 2D the entire Riemann tensor is determined by the single component $R_{1212}$, and equivalently by $K$: $$R_{abcd} = K(g_{ac}g_{bd} - g_{ad}g_{bc}).$$ The sphere is the simplest example of a maximally symmetric 2D manifold with positive curvature.

(a) Commutator $[X,Y]$

$[X,Y]^i = X^j\partial_j Y^i - Y^j\partial_j X^i$. With $X = (-y, x)$, $Y = (x, y)$:

$[X,Y]^1 = X^j\partial_j Y^1 - Y^j\partial_j X^1 = (-y)\partial_x x + x\partial_y x - x\partial_x(-y) - y\partial_y(-y) = -y + 0 - 0 + y = 0$.

$[X,Y]^2 = (-y)\partial_x y + x\partial_y y - x\partial_x x - y\partial_y x = 0 + x - x - 0 = 0$.

$$\boxed{\;[X,Y] = 0.\;}$$ Rotation and dilation generators commute on $\mathbb R^2\setminus\{0\}$.

(b) Riemann tensor

Torsion-free condition: $\nabla_X Y - \nabla_Y X = [X,Y] = 0$, so $\nabla_Y X = \nabla_X Y = X$.

$$R(X,Y)X = \nabla_X\nabla_Y X - \nabla_Y\nabla_X X - \nabla_{[X,Y]}X = \nabla_X(X) - \nabla_Y(-Y) - 0 = -Y - (-Y) = 0.$$

$$R(X,Y)Y = \nabla_X\nabla_Y Y - \nabla_Y\nabla_X Y - 0 = \nabla_X(Y) - \nabla_Y(X) = X - X = 0.$$

Since $\{X,Y\}$ span $T_p(\mathbb R^2\setminus\{0\})$ at every point, $R(\cdot,\cdot)\cdot \equiv 0$ on the whole open set.

$$\boxed{\;R \equiv 0\;\text{(flat connection).}\;}$$

Lesson

The Christoffels $\Gamma^i_{jk}$ in the $(x,y)$ basis are non-zero (e.g. $\nabla_x\partial_y = ?$ would have entries encoding the rotation/dilation). Yet Riemann vanishes — the curvature combination cancels. Non-zero $\Gamma$ ≠ curvature. Curvature is an intrinsic property captured only by $R^a{}_{bcd}$.

Compute $[X,Y]$

$X = (x^2, -x^1)$, $Y = (x^1, x^2)$. $[X,Y]^1 = X^j\partial_j Y^1 - Y^j\partial_j X^1 = x^2\cdot 1 + (-x^1)\cdot 0 - x^1\cdot 0 - x^2\cdot 1 = 0$. Similarly $[X,Y]^2 = 0$. So $[X,Y] = 0$.

Torsion check

$\nabla_X Y - \nabla_Y X = (X+Y) - (X-Y) = 2Y \ne 0$, while $[X,Y] = 0$. The torsion tensor $T(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y] = 2Y$ is non-zero. So this connection has torsion.

Riemann tensor in $\{X,Y\}$ basis

$$R(X,Y)X = \nabla_X(\nabla_Y X) - \nabla_Y(\nabla_X X) - \nabla_{[X,Y]}X = \nabla_X(X-Y) - \nabla_Y(0) - 0 = (X+Y) - (X+Y) \cdot\ldots$$ Carefully: $\nabla_X(X-Y) = \nabla_X X - \nabla_X Y = 0 - (X+Y) = -(X+Y)$.

$$\boxed{\;R(X,Y)X = -(X+Y).\;}$$

Similarly $R(X,Y)Y = \nabla_X(0) - \nabla_Y(X+Y) - 0 = -(\nabla_Y X + \nabla_Y Y) = -(X-Y+0) = -X+Y$.

Components in $\{\partial_1,\partial_2\}$ basis

Express $X = x^2\partial_1 - x^1\partial_2$, $Y = x^1\partial_1 + x^2\partial_2$, transformation matrix $A$ with $\det A = (x^1)^2+(x^2)^2 = r^2$. Solving for $\partial_1,\partial_2$ in terms of $X,Y$ and substituting into $R(\partial_1,\partial_2)\partial_1$ yields non-zero components.

$$\boxed{\;R^1{}_{1ij}\ne 0\text{ in general — genuinely curved connection (with torsion).}\;}$$

Contrast with Problem 2.15

Problem 2.15 had a flat connection with $[X,Y] = 0$ and balanced $\nabla$ relations. Here the imbalance $\nabla_X Y \ne \nabla_Y X$ (with $[X,Y]=0$) means torsion, and the iterated derivatives don't commute — producing genuine Riemann curvature.

Koszul formula

The Levi-Civita connection on a Lie group with bi-invariant metric, applied to an orthonormal basis with $\langle L_i, L_j\rangle = \delta_{ij}$ constant on the basis vectors:

$$2\langle\nabla_{L_i}L_j, L_k\rangle = L_i\langle L_j,L_k\rangle + L_j\langle L_k,L_i\rangle - L_k\langle L_i,L_j\rangle + \langle[L_i,L_j],L_k\rangle - \langle[L_j,L_k],L_i\rangle + \langle[L_k,L_i],L_j\rangle.$$

The first three terms vanish (constant inner products). The bracket terms use $[L_i,L_j] = \epsilon_{ijk}L_k$:

$$\langle[L_i,L_j],L_k\rangle = \epsilon_{ijm}\delta_{mk} = \epsilon_{ijk},\quad \langle[L_j,L_k],L_i\rangle = \epsilon_{jki} = \epsilon_{ijk},\quad \langle[L_k,L_i],L_j\rangle = \epsilon_{kij} = \epsilon_{ijk}.$$

Sum: $2\langle\nabla_{L_i}L_j, L_k\rangle = \epsilon_{ijk} - \epsilon_{ijk} + \epsilon_{ijk} = \epsilon_{ijk}$.

$$\boxed{\;\nabla_{L_i}L_j = \tfrac12\epsilon_{ijk}L_k.\;}$$

Riemann tensor

$$R(L_i,L_j)L_k = \nabla_{L_i}\nabla_{L_j}L_k - \nabla_{L_j}\nabla_{L_i}L_k - \nabla_{[L_i,L_j]}L_k.$$

Compute: $\nabla_{L_i}\nabla_{L_j}L_k = \tfrac12\epsilon_{jkm}\nabla_{L_i}L_m = \tfrac14\epsilon_{jkm}\epsilon_{imn}L_n$. By symmetry, $\nabla_{L_j}\nabla_{L_i}L_k = \tfrac14\epsilon_{ikm}\epsilon_{jmn}L_n$. And $\nabla_{[L_i,L_j]}L_k = \epsilon_{ijm}\nabla_{L_m}L_k = \tfrac12\epsilon_{ijm}\epsilon_{mkn}L_n$.

Use $\epsilon_{abc}\epsilon_{ade} = \delta_{bd}\delta_{ce} - \delta_{be}\delta_{cd}$. After careful algebra: $$R(L_i,L_j)L_k = \tfrac14(\delta_{ik}L_j - \delta_{jk}L_i).$$

Equivalently: $$\boxed{\;R_{abcd} = \tfrac14(g_{ac}g_{bd} - g_{ad}g_{bc}),\quad K = \tfrac14.\;}$$

Geometric identification

This is the bi-invariant metric on $SU(2) \cong S^3$, the round 3-sphere of radius 2 (since $K = 1/R^2 = 1/4$ gives $R = 2$). On any compact Lie group with bi-invariant metric, the exponential map is the Riemannian exponential, and 1-parameter subgroups $e^{tA}$ are geodesics from the identity (cf. Problem 2.24). The relation $\mathfrak{su}(2) \cong \mathbb R^3$ with cross-product structure makes $S^3$ a natural setting for studying constant-curvature geometry, gauge theory, and the universal cover of $SO(3)$.

Metric compatibility

The Levi-Civita (or any metric-compatible) connection satisfies $\nabla_X g = 0$ for every $X$. Consequence: along any curve $\gamma$ with tangent $\dot\gamma$, $$\frac{d}{ds}g(V,W) = g(\nabla_{\dot\gamma}V, W) + g(V, \nabla_{\dot\gamma}W).$$ For parallel-transported vectors $\nabla_{\dot\gamma}V = \nabla_{\dot\gamma}W = 0$, so $g(V,W)$ is constant along $\gamma$.

Consequence for angles

The angle $\psi$ between $V$ and $W$ satisfies $\cos\psi = g(V,W)/(|V||W|)$. Since both numerator and denominators are conserved, $\psi$ is conserved along parallel transport. Even when traversing a closed loop with non-trivial holonomy, individual vectors rotate but their pairwise angles do not change.

$$\boxed{\;\text{Final angle between vectors} = \pi/3\;\text{(unchanged).}\;}$$

Holonomy of each individual vector

Each vector rotates by $\Delta\psi = \int_D K\,dA$ (the integrated Gaussian curvature over the disk bounded by the loop). On the specific surface in this problem, this rotation can be computed explicitly, but both $V$ and $W$ rotate by the same angle, so their relative orientation is preserved.

Physical realisation

Gravity Probe B (2004–2011) measured the geodetic precession of four gyroscopes orbiting Earth in a polar orbit. All four gyroscopes precessed by the same angle ($\approx 6.6$ arcsec/year), confirming GR's prediction. Had the precessions differed by gyroscope, that would have signalled a violation of metric compatibility — a fundamental departure from Riemannian geometry.

Induced metric

Compute differentials: $dx = \cos\phi\,dr - r\sin\phi\,d\phi$, $dy = \sin\phi\,dr + r\cos\phi\,d\phi$, $dz = r^{1/2}\,dr$ (since $z = \tfrac23 r^{3/2}$ gives $dz/dr = r^{1/2}$). Then:

$$dx^2 + dy^2 = dr^2 + r^2\,d\phi^2,\qquad dz^2 = r\,dr^2,$$

so the induced metric is

$$\boxed{\;ds^2 = (1 + r)\,dr^2 + r^2\,d\phi^2.\;}$$

Geodesic equations

From the Lagrangian $\mathcal L = \tfrac12[(1+r)\dot r^2 + r^2\dot\phi^2]$:

$\phi$-equation (cyclic): $\frac{d}{ds}(r^2\dot\phi) = 0$, so $L \equiv r^2\dot\phi$ is conserved.

$r$-equation: $\frac{d}{ds}[(1+r)\dot r] - \tfrac12\dot r^2 - r\dot\phi^2 = 0$.

Test each ant

Ant 1 ($r = \lambda$, $\phi = 0$): $\dot\phi = 0$ so $L = 0$ ✓. $\dot r = 1$, $\ddot r = 0$. Check the $r$-equation: $\frac{d}{ds}[(1+r)\cdot 1] - \tfrac12\cdot 1 - 0 = \dot r - \tfrac12 = 1 - \tfrac12 = \tfrac12 \ne 0$. Strictly speaking $\lambda$ is not an affine parameter, but the path $\phi = 0$ is a meridian on a rotation-symmetric surface and is geodesic in shape.

Ant 2 ($r = \lambda^{2/3} - 1$, $\phi = \pi/2$): also meridian. Geodesic.

Ant 3 ($r = \lambda^{1/2}$, $\phi = \ln\lambda$): this is a logarithmic spiral $\phi = 2\ln r$ on the surface. $L = r^2\dot\phi = \lambda\cdot(1/\lambda) = 1$, conserved ✓. But substituting into the $r$-equation produces a non-zero residual, so the spiral is not a geodesic for any reparametrisation.

$$\boxed{\;\text{Ants 1 and 2 walk on (meridian) geodesics; Ant 3's log spiral is not a geodesic.}\;}$$

Geometric reason

Any rotation-symmetric surface in $\mathbb R^3$ has its meridians as geodesics (the normal acceleration is purely centripetal, normal to the surface). Latitude circles and spirals are not geodesics unless they coincide with great circles — which spirals never do.

Induced metric

Constraint $x^2 + y^2 - z^2 = -a^2$ on the upper sheet $z > 0$: $z = \sqrt{r^2 + a^2}$ where $r^2 = x^2 + y^2$. Differentiate: $dz = (r/\sqrt{r^2+a^2})dr$, so $dz^2 = r^2\,dr^2/(r^2+a^2)$.

Ambient Euclidean metric $ds^2_{\text{amb}} = dx^2 + dy^2 + dz^2 = dr^2 + r^2 d\phi^2 + dz^2$:

$$ds^2 = dr^2 + r^2 d\phi^2 + \frac{r^2}{r^2+a^2}dr^2 = \frac{(r^2+a^2)+r^2}{r^2+a^2}dr^2 + r^2 d\phi^2.$$

$$\boxed{\;ds^2 = \frac{2r^2 + a^2}{r^2 + a^2}\,dr^2 + r^2\,d\phi^2.\;}$$

Christoffels

$g_{rr} = (2r^2+a^2)/(r^2+a^2)$, $g_{\phi\phi} = r^2$. $\partial_r g_{rr} = 2r\cdot a^2/(r^2+a^2)^2$ (after applying quotient rule). $\partial_r g_{\phi\phi} = 2r$. Inverse: $g^{rr} = (r^2+a^2)/(2r^2+a^2)$, $g^{\phi\phi} = 1/r^2$.

$$\Gamma^r_{rr} = \tfrac12 g^{rr}\partial_r g_{rr} = \frac{r^2+a^2}{2(2r^2+a^2)}\cdot\frac{2ra^2}{(r^2+a^2)^2} = \frac{ra^2}{(2r^2+a^2)(r^2+a^2)}.$$

$$\Gamma^r_{\phi\phi} = -\tfrac12 g^{rr}\partial_r g_{\phi\phi} = -\frac{r^2+a^2}{2(2r^2+a^2)}\cdot 2r = -\frac{r(r^2+a^2)}{2r^2+a^2}.$$

$$\Gamma^\phi_{r\phi} = \tfrac12 g^{\phi\phi}\partial_r g_{\phi\phi} = \frac{1}{2r^2}\cdot 2r = \frac{1}{r}.$$

All others vanish.

Geodesic equations

$$\ddot r + \Gamma^r_{rr}\dot r^2 + \Gamma^r_{\phi\phi}\dot\phi^2 = 0,\qquad \ddot\phi + 2\Gamma^\phi_{r\phi}\dot r\dot\phi = 0.$$ The second integrates to $L = r^2\dot\phi = \text{const}$ (Noether/$\phi$-cyclic).

Geometric interpretation

This hyperboloid is the Bolyai–Lobachevsky model of 3D hyperbolic geometry — constant negative Gaussian curvature $K = -1/a^2$. Geodesics are intersections with planes through the origin. The model satisfies all Euclidean axioms except the parallel postulate (through any point not on a given geodesic, infinitely many non-intersecting geodesics exist).