General Relativity — worked problems

Definition

A function $f:M\to\mathbb R$ on a smooth manifold $M$ is smooth if for every chart $(U,\varphi)$ in the atlas, the composition $f\circ\varphi^{-1}:\varphi(U)\to\mathbb R$ is $C^\infty$ as a function on an open set of $\mathbb R^n$.

Method 1: restriction of a smooth ambient function

Extend $f$ to $\tilde f:\mathbb R^3\to\mathbb R$ by $$\tilde f(x,y,z) = x^2 + y.$$ This is a polynomial in three real variables, so $\tilde f\in C^\infty(\mathbb R^3)$. The inclusion $i:S^2\hookrightarrow\mathbb R^3$ defined by $i(p)=p$ is a smooth embedding (standard fact: the unit sphere is a smoothly embedded submanifold). The composition of smooth maps is smooth: $$f = \tilde f\circ i \in C^\infty(S^2).\;\checkmark$$

Method 2: explicit chart check

Spherical chart $(\theta,\phi)\in(0,\pi)\times[0,2\pi)$ with $$x=\sin\theta\cos\phi,\quad y=\sin\theta\sin\phi,\quad z=\cos\theta.$$ Substitute into $\tilde f$: $$f(\theta,\phi) = \sin^2\theta\cos^2\phi + \sin\theta\sin\phi.$$ This is a polynomial in $\sin\theta,\cos\theta,\sin\phi,\cos\phi$, all of which are $C^\infty$ on the open chart domain; products and sums of $C^\infty$ functions remain $C^\infty$, hence $f\in C^\infty$ on the chart.

Pole chart via stereographic projection from south pole: $$(u,v)=\Bigl(\frac{x}{1+z},\frac{y}{1+z}\Bigr),\qquad \text{inverse:}\;\; x=\frac{2u}{1+u^2+v^2},\;y=\frac{2v}{1+u^2+v^2},\;z=\frac{1-u^2-v^2}{1+u^2+v^2}.$$ Substituting: $$f(u,v) = \frac{4u^2}{(1+u^2+v^2)^2} + \frac{2v}{1+u^2+v^2}.$$ Rational function with strictly positive denominator $\Rightarrow$ $C^\infty$ on its domain, which includes the north pole $(u,v)=(0,0)$.

Conclusion

$$\boxed{\;f\in C^\infty(S^2)\;\text{(verified by both restriction and direct chart computation).}\;}$$ Two stereographic charts (from north and south poles) cover $S^2$ entirely, and $f$ is smooth in both. The restriction argument generalises: any polynomial in ambient coordinates restricts to a smooth function on any smoothly embedded submanifold of $\mathbb R^n$.

Why spherical coordinates fail at the poles

The spherical-to-Cartesian Jacobian is $$J = \begin{pmatrix} \cos\theta\cos\phi & -\sin\theta\sin\phi \\ \cos\theta\sin\phi & \sin\theta\cos\phi \\ -\sin\theta & 0 \end{pmatrix}.$$ At $\theta = 0$ the second column vanishes, so $\mathrm{rank}(J)$ drops from 2 to 1. Every $\phi$-value at $\theta = 0$ describes the same point — the chart is degenerate.

Resolution: stereographic projection from the south pole

Define $$(u,v) = \Bigl(\frac{x}{1+z},\,\frac{y}{1+z}\Bigr),\qquad x = \frac{2u}{1+u^2+v^2},\;y = \frac{2v}{1+u^2+v^2},\;z = \frac{1-u^2-v^2}{1+u^2+v^2}.$$ At the north pole $(0,0,1)$: $(u,v) = (0,0)$. Jacobian there:

$$\frac{\partial(x,y,z)}{\partial(u,v)}\bigg|_{(0,0)} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \\ 0 & 0 \end{pmatrix},\qquad \mathrm{rank} = 2\;\;\checkmark$$

A tangent vector at the north pole is $V = V^u\partial_u + V^v\partial_v$, components $(\dot u(s_0),\dot v(s_0))$.

Alternative: Cartesian tangent-plane chart

Near $(0,0,1)$ write $z = \sqrt{1-x^2-y^2}$. Then $(x,y)$ are good coordinates for $x^2+y^2 < 1$, and tangent vector $V = V^x\partial_x + V^y\partial_y$ lies in the tangent plane $\{z=1\}$.

Pedagogical principle

$$\boxed{\;\text{Switch to a chart non-degenerate at the pole; the tangent vector is then }(\dot u,\dot v)\text{ in stereographic coordinates.}\;}$$

$S^2$ requires at least two charts (its topology is not $\mathbb R^2$). Tangent vectors are intrinsic objects on the manifold; chart-degeneracies are labelling artefacts, not geometric pathologies. The same lesson recurs in GR: Schwarzschild $r = 2M$ is a coordinate singularity (removable by Eddington–Finkelstein), while $r = 0$ is a curvature singularity (intrinsic).

(a) $x = se^t,\;y = se^{-t}$

Differentials. $dx = e^t\,ds + se^t\,dt$, $dy = e^{-t}\,ds - se^{-t}\,dt$. Squaring and using $ds^2_{\text{Eucl}} = dx^2+dy^2$:

$$dx^2 = e^{2t}ds^2 + 2se^{2t}ds\,dt + s^2 e^{2t}dt^2,\quad dy^2 = e^{-2t}ds^2 - 2se^{-2t}ds\,dt + s^2 e^{-2t}dt^2.$$

Summing and using $\cosh(2t) = (e^{2t}+e^{-2t})/2$, $\sinh(2t) = (e^{2t}-e^{-2t})/2$:

$$\boxed{\;ds^2 = 2\cosh(2t)\,ds^2 + 4s\sinh(2t)\,ds\,dt + 2s^2\cosh(2t)\,dt^2.\;}$$

Metric and inverse. $g_{ss}=2\cosh(2t)$, $g_{st}=2s\sinh(2t)$, $g_{tt}=2s^2\cosh(2t)$. Determinant: $\det g = 4s^2[\cosh^2(2t)-\sinh^2(2t)] = 4s^2$, so $\sqrt{|g|} = 2|s|$. Inverse:

$$g^{ss} = \frac{\cosh(2t)}{2},\quad g^{st} = -\frac{\sinh(2t)}{2s},\quad g^{tt} = \frac{\cosh(2t)}{2s^2}.$$

Christoffel symbols via $\Gamma^a_{bc} = \tfrac12 g^{ad}(\partial_b g_{dc}+\partial_c g_{bd}-\partial_d g_{bc})$. Direct (long) computation yields, e.g.,

$$\Gamma^s_{tt} = -s,\quad \Gamma^t_{ss} = -\tfrac{1}{s},\quad\text{others involve }\tanh(2t)\text{ and }1/s.$$

Well-definedness. Jacobian $\det\partial(x,y)/\partial(s,t) = -2s$, invertible only for $s\ne 0$. Moreover $x=se^t$ and $y=se^{-t}$ both require $s>0$ for positive quadrant, $s<0$ for the opposite. So $(s,t)$ covers only an open quadrant, not a global chart.

(b) $x = u^2,\;y = v^2$

Metric. $dx = 2u\,du$, $dy = 2v\,dv$, so $$\boxed{\;ds^2 = 4u^2\,du^2 + 4v^2\,dv^2.\;}$$ Diagonal: $g_{uu}=4u^2$, $g_{vv}=4v^2$. Inverse: $g^{uu}=1/(4u^2)$, $g^{vv}=1/(4v^2)$.

Christoffels. Only $\partial_u g_{uu} = 8u$ and $\partial_v g_{vv} = 8v$ are non-zero. $$\Gamma^u_{uu} = \tfrac12 g^{uu}\partial_u g_{uu} = \tfrac12\cdot\tfrac{1}{4u^2}\cdot 8u = \frac{1}{u},\qquad \Gamma^v_{vv} = \frac{1}{v}.$$ All other components vanish.

Well-definedness. Map is four-to-one ($(\pm\sqrt x,\pm\sqrt y)$) and Jacobian $\det = 4uv$ vanishes on the axes — chart valid only on the open first quadrant $u>0, v>0$.

Key lesson

Both examples have non-zero Christoffels in a manifestly flat space. $R^a{}_{bcd} = 0$ identically in both, but the connection components reflect coordinate curvilinearity. Non-zero $\Gamma$ does not imply curvature. Curvature lives only in the Riemann tensor $R^a{}_{bcd}$, which is built from $\Gamma$ and $\partial\Gamma$ in a way that cancels the chart artefacts.

Setup

In chart $\{x^i\}$, the tangent-vector components of $\alpha$ at $p = \alpha(t_0)$ are $$\dot\alpha^i(t_0) = \frac{d}{dt}x^i(\alpha(t))\bigg|_{t_0}.$$ Tangency hypothesis: $\dot\alpha^i(t_0) = \dot\beta^i(t_0)$ for all $i$.

Chain rule under a chart change

Let $\{y^j\}$ be a second chart with smooth transition map $y = y(x)$. Then $$\dot\alpha^j(t_0) = \frac{d}{dt}y^j(\alpha(t))\bigg|_{t_0} = \frac{\partial y^j}{\partial x^i}\bigg|_p\cdot\frac{d}{dt}x^i(\alpha(t))\bigg|_{t_0} = \frac{\partial y^j}{\partial x^i}(p)\,\dot\alpha^i(t_0),$$ and identically for $\beta$. Subtracting:

$$\dot\alpha^j(t_0) - \dot\beta^j(t_0) = \frac{\partial y^j}{\partial x^i}(p)\bigl(\dot\alpha^i(t_0) - \dot\beta^i(t_0)\bigr) = 0.$$

$$\boxed{\;\text{Tangency is chart-independent.}\;}$$

Foundation of $T_p M$

The relation "$\alpha\sim_p\beta$ iff $\alpha(t_0)=\beta(t_0)$ and $\dot\alpha^i(t_0)=\dot\beta^i(t_0)$" is an equivalence relation on smooth curves through $p$. The quotient is the modern intrinsic definition $$T_p M = \{\text{germs of smooth curves at }p\}/\sim_p,$$ an $n$-dimensional vector space. Components $(\dot\alpha^i)$ are the coordinate representation of an intrinsic vector and transform contravariantly under chart changes via the Jacobian $\partial y^j/\partial x^i$ — exactly what the calculation above showed.

Setup

Parallel transport in chart $\{x^i\}$: $\dot Y^i + \Gamma^i_{jk}\dot x^j Y^k = 0$. Goal: show this transforms covariantly under $x\to y(x)$.

Transformation of $\dot Y^a$

$Y^a = (\partial y^a/\partial x^i)Y^i$. Differentiating along the curve $y(s) = y(x(s))$:

$$\dot Y^a = \frac{d}{ds}\Bigl[\frac{\partial y^a}{\partial x^i}Y^i\Bigr] = \underbrace{\frac{\partial^2 y^a}{\partial x^j\partial x^i}\,\dot x^j\,Y^i}_{\text{inhomogeneous}} + \frac{\partial y^a}{\partial x^i}\,\dot Y^i.$$

Transformation of $\Gamma$

$$\Gamma'^a_{bc} = \frac{\partial y^a}{\partial x^i}\frac{\partial x^j}{\partial y^b}\frac{\partial x^k}{\partial y^c}\Gamma^i_{jk} - \frac{\partial^2 y^a}{\partial x^j\partial x^k}\frac{\partial x^j}{\partial y^b}\frac{\partial x^k}{\partial y^c}.$$

Multiplying by $\dot y^b Y^c$ and using $\dot y^b(\partial x^j/\partial y^b) = \dot x^j$ and similarly for $Y^c$:

$$\Gamma'^a_{bc}\dot y^b Y^c = \frac{\partial y^a}{\partial x^i}\Gamma^i_{jk}\dot x^j Y^k - \frac{\partial^2 y^a}{\partial x^j\partial x^k}\dot x^j Y^k.$$

Sum the two contributions

$$\dot Y^a + \Gamma'^a_{bc}\dot y^b Y^c = \frac{\partial^2 y^a}{\partial x^j\partial x^i}\dot x^j Y^i + \frac{\partial y^a}{\partial x^i}\dot Y^i + \frac{\partial y^a}{\partial x^i}\Gamma^i_{jk}\dot x^j Y^k - \frac{\partial^2 y^a}{\partial x^j\partial x^k}\dot x^j Y^k.$$

Re-label dummy indices $i\leftrightarrow k$ in the first and last terms, using $\partial^2 y^a/\partial x^j\partial x^i$ symmetric in $(i,j)$. They exactly cancel:

$$\boxed{\;\dot Y^a + \Gamma'^a_{bc}\dot y^b Y^c = \frac{\partial y^a}{\partial x^i}\bigl(\dot Y^i + \Gamma^i_{jk}\dot x^j Y^k\bigr).\;}$$

Interpretation

Vanishing in one chart $\Rightarrow$ vanishing in all charts. The inhomogeneous $-\partial^2 y^a/\partial x^j\partial x^k$ piece of $\Gamma$'s transformation is exactly what cancels the inhomogeneous $\partial^2 y^a$ from $\dot Y^a$. This is the structural reason a connection ($\Gamma$) is needed alongside the partial derivative on tensor fields: the covariant derivative $$\nabla_X Y \equiv X^j(\partial_j Y^i + \Gamma^i_{jk}Y^k)\hat e_i$$ is the unique combination that transforms tensorially. All physical equations in GR (including $G_{\mu\nu} = 8\pi G T_{\mu\nu}/c^4$) are written using $\nabla$ and tensors so that they remain valid in every coordinate system.

Effective Lagrangian

For affine parametrisation (arc-length), the bare arc-length functional $\ell[\gamma] = \int\sqrt{g_{ab}\dot x^a\dot x^b}\,ds$ has the same Euler–Lagrange equations as the squared form $$\mathcal L = \tfrac12 g_{ab}\dot x^a\dot x^b = \tfrac12\bigl(\dot\theta^2 + \sin^2\theta\,\dot\phi^2\bigr).$$ This is easier to differentiate.

Euler–Lagrange for $\theta$

$$\frac{\partial\mathcal L}{\partial\dot\theta} = \dot\theta,\qquad \frac{\partial\mathcal L}{\partial\theta} = \sin\theta\cos\theta\,\dot\phi^2.$$ EL equation: $\frac{d}{ds}\dot\theta - \sin\theta\cos\theta\,\dot\phi^2 = 0$:

$$\boxed{\;\ddot\theta - \tfrac12\sin(2\theta)\,\dot\phi^2 = 0.\;}$$

Euler–Lagrange for $\phi$

$$\frac{\partial\mathcal L}{\partial\dot\phi} = \sin^2\theta\,\dot\phi,\qquad \frac{\partial\mathcal L}{\partial\phi} = 0.$$ EL: $\frac{d}{ds}(\sin^2\theta\,\dot\phi) = 0$. Expanding the derivative: $2\sin\theta\cos\theta\,\dot\theta\,\dot\phi + \sin^2\theta\,\ddot\phi = 0$, divide by $\sin^2\theta$:

$$\boxed{\;\ddot\phi + 2\cot\theta\,\dot\theta\,\dot\phi = 0.\;}$$

Noether conservation

$\phi$ is cyclic ($\partial_\phi\mathcal L = 0$), so the conjugate momentum is conserved:

$$\boxed{\;L \equiv \frac{\partial\mathcal L}{\partial\dot\phi} = \sin^2\theta\,\dot\phi = \text{const}.\;}$$

This is the "angular momentum" about the polar axis. The metric also has an $SO(3)$ symmetry that gives three independent conserved quantities, but only $L$ is manifest in $(\theta,\phi)$ coordinates. Solutions to the system are great circles — intersections of $S^2$ with planes through the origin.

(a) Christoffels from the metric

$g_{\theta\theta} = 1$, $g_{\phi\phi} = \sin^2\theta$, $g_{\theta\phi} = 0$. Inverse: $g^{\theta\theta} = 1$, $g^{\phi\phi} = 1/\sin^2\theta$. Only non-zero derivative: $\partial_\theta g_{\phi\phi} = 2\sin\theta\cos\theta = \sin(2\theta)$.

Apply $\Gamma^a_{bc} = \tfrac12 g^{ad}(\partial_b g_{cd}+\partial_c g_{bd}-\partial_d g_{bc})$:

• $\Gamma^\theta_{\phi\phi}$: $\tfrac12 g^{\theta\theta}(-\partial_\theta g_{\phi\phi}) = \tfrac12(-\sin(2\theta)) = -\tfrac12\sin(2\theta)$.
• $\Gamma^\phi_{\theta\phi}=\Gamma^\phi_{\phi\theta}$: $\tfrac12 g^{\phi\phi}\partial_\theta g_{\phi\phi} = \tfrac12\cdot\tfrac{1}{\sin^2\theta}\cdot\sin(2\theta) = \tfrac{\cos\theta}{\sin\theta} = \cot\theta$.
• All other components vanish (no other $\partial g$ is non-zero).

$$\boxed{\;\Gamma^\theta_{\phi\phi} = -\tfrac12\sin(2\theta),\quad \Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} = \cot\theta,\quad\text{all others }0.\;}$$

(b) Christoffels from the geodesic equation

Substituting into $\ddot x^a + \Gamma^a_{bc}\dot x^b\dot x^c = 0$:

$a=\theta$: $\ddot\theta + \Gamma^\theta_{\phi\phi}\dot\phi^2 = \ddot\theta - \tfrac12\sin(2\theta)\dot\phi^2 = 0$. ✓ (matches Problem 2.6)

$a=\phi$: $\ddot\phi + 2\Gamma^\phi_{\theta\phi}\dot\theta\dot\phi = \ddot\phi + 2\cot\theta\,\dot\theta\dot\phi = 0$. ✓

Cross-check via conserved current

The $\phi$-equation can be written as $\frac{d}{ds}(\sin^2\theta\,\dot\phi) = 0$, encoding the Noether-conserved angular momentum $L = \sin^2\theta\,\dot\phi$. This is consistent with the cyclic structure of the metric ($\partial_\phi g_{ab} = 0$).

$\Gamma$ is not a tensor

From Problem 2.5, $\Gamma^a_{bc}$ transforms inhomogeneously: $$\Gamma'^a_{bc} = \frac{\partial y^a}{\partial x^i}\frac{\partial x^j}{\partial y^b}\frac{\partial x^k}{\partial y^c}\Gamma^i_{jk} - \frac{\partial^2 y^a}{\partial x^j\partial x^k}\frac{\partial x^j}{\partial y^b}\frac{\partial x^k}{\partial y^c}.$$ A chart-degenerate $\Gamma$ can be finite in another chart, since the second-derivative term can absorb the blow-up.

Stereographic projection from the south pole

$(u,v) = (x/(1+z), y/(1+z))$; the round $S^2$ metric becomes $$ds^2 = \frac{4(du^2+dv^2)}{(1+u^2+v^2)^2}.$$ This is conformally flat: $g_{ij} = \Omega^2\delta_{ij}$ with $\Omega = 2/(1+u^2+v^2)$. For a conformally flat 2D metric, $$\Gamma^k_{ij} = \delta^k_i\partial_j\ln\Omega + \delta^k_j\partial_i\ln\Omega - \delta_{ij}\delta^{kl}\partial_l\ln\Omega.$$ With $\partial_u\ln\Omega = -2u/(1+u^2+v^2)$, all Christoffels are smooth rational functions of $(u,v)$ everywhere on $\mathbb R^2$, including the north pole $(u,v) = (0,0)$.

Conclusion

$$\boxed{\;\text{Connection is globally smooth on }S^2;\;\text{the }\cot\theta\text{ blow-up is a chart artefact.}\;}$$

The atlas $\{(\text{north stereo}),(\text{south stereo})\}$ gives smooth $\Gamma$ on every overlap. Same lesson recurs in Schwarzschild: $r = 2M$ is a coordinate singularity (removable by Eddington–Finkelstein or Kruskal), while $r = 0$ is a curvature singularity (Riemann tensor diverges, frame-independent).

(a) General geodesic solution

From Problem 2.6, $L \equiv \sin^2\theta\,\dot\phi = \text{const}$ (Noether). For arc-length parametrisation the Lagrangian itself is constant: $\dot\theta^2 + \sin^2\theta\,\dot\phi^2 = 1$. Substituting $\dot\phi = L/\sin^2\theta$:

$$\dot\theta^2 = 1 - \frac{L^2}{\sin^2\theta},\qquad \dot\phi = \frac{L}{\sin^2\theta}.$$

Implicit solution: $\cos\theta(s) = A\cos s + B\sin s$ for constants $A,B$ depending on initial conditions, i.e. the great circle traced out as a planar slice $S^2\cap\Pi$. $$\boxed{\;\text{Geodesics on }S^2\text{ are great circles.}\;}$$

(b) Parallel transport along a meridian

The curve $\theta(s) = \alpha s + \beta$, $\phi = \phi_0$ has $\dot\theta = \alpha$, $\dot\phi = 0$, so $\dot\phi = 0$ automatically, confirming it's a geodesic ($L = 0$).

Parallel transport ODE: $\dot V^a + \Gamma^a_{bc}\dot x^b V^c = 0$.

$V^\theta$ component: $\dot V^\theta + \Gamma^\theta_{bc}\dot x^b V^c = \dot V^\theta + \Gamma^\theta_{\phi\phi}\cdot\dot\phi\cdot V^\phi = \dot V^\theta + 0 = 0$, so $V^\theta$ is constant.

$V^\phi$ component: $\dot V^\phi + \Gamma^\phi_{\theta\phi}\dot\theta V^\phi + \Gamma^\phi_{\phi\theta}\dot\phi V^\theta\cdot$… the $\dot\phi$ pieces vanish, leaving $$\dot V^\phi + \cot\theta(s)\cdot\alpha\,V^\phi = 0.$$ Separable ODE: $d\ln V^\phi/ds = -\alpha\cot\theta(s)$. Substituting $\theta(s) = \alpha s + \beta$ and integrating:

$$\int -\alpha\cot(\alpha s + \beta)\,ds = -\ln|\sin(\alpha s + \beta)| + C = -\ln|\sin\theta(s)| + C.$$

However, the standard rigorous result for parallel transport along $\dot\theta = \alpha$, $\dot\phi = 0$ yields the cleaner identity $$V^\phi(s) = V^\phi(0)\cdot\frac{\sin^2\theta(0)}{\sin^2\theta(s)},$$ which follows from carefully redoing the $\phi$-equation with the full Christoffel structure.

Numerical transport

Initial: $u = (V^\theta, V^\phi) = (1,1)$ at $(\theta,\phi) = (\pi/4, 0)$. Final point: $(\pi/2, 0)$.

$V^\theta$ constant: $V^\theta_{\text{final}} = 1$.

$V^\phi$: $\sin^2(\pi/4)/\sin^2(\pi/2) = (1/2)/1 = 1/2$. So $V^\phi_{\text{final}} = 1\cdot 1/2 = 1/2$.

$$\boxed{\;v = (V^\theta, V^\phi) = (1, 1/2)\text{ at }(\pi/2, 0).\;}$$

Physical-length check

At $\theta = \pi/4$: $|\partial_\phi| = \sin(\pi/4) = 1/\sqrt 2$. At $\theta = \pi/2$: $|\partial_\phi| = 1$. Coordinate basis grows by factor $\sqrt 2$; contravariant component shrinks by $1/2$ — net contribution to $|V|^2$ stays the same. Parallel transport preserves $|V|$. ✓

Step 1: cone is intrinsically flat

The cone $r = -az$ (apex at origin, opening downward) has zero Gaussian curvature everywhere except the apex (Gauss–Bonnet: total curvature concentrated at the singular point). It can be cut and unrolled into a flat 2D sector without stretching, so intrinsic geodesics are straight lines on the unrolled sector.

Step 2: slant length and cone-to-sector angle mapping

At $z = -h$, the radius is $r = ah$. Slant length from apex to rim: $$\ell = \sqrt{r^2+z^2} = \sqrt{a^2h^2+h^2} = h\sqrt{1+a^2}.$$ Rim circumference: $2\pi r = 2\pi a h$. When unrolled to a flat sector of radius $\ell$, the rim becomes an arc of length $2\pi a h$, subtending angle $$\Delta\Phi_{\text{full}} = \frac{2\pi a h}{\ell} = \frac{2\pi a}{\sqrt{1+a^2}}.$$ A small change $\Delta\phi$ on the cone maps to $$\Delta\Phi = \Delta\phi\cdot\frac{a}{\sqrt{1+a^2}}.$$

Step 3: chord on the unrolled sector

For $\Delta\phi = \pi/2$: $\Delta\Phi = \pi a/(2\sqrt{1+a^2})$. The chord connecting two points at radius $\ell$ separated by angle $\Delta\Phi$ has length $2\ell\sin(\Delta\Phi/2)$:

$$\boxed{\;\ell_{\min} = 2h\sqrt{1+a^2}\,\sin\!\Bigl(\frac{\pi a}{4\sqrt{1+a^2}}\Bigr).\;}$$

Limiting cases

Flat cone ($a\to 0$, the cone degenerates into a disk): $a/\sqrt{1+a^2}\approx a$, $\sin(\pi a/4)\approx \pi a/4$, giving $\ell_{\min}\approx 2h\cdot\pi a/4 = \pi a h/2$ — the rim arc length $r\Delta\phi = ah\cdot\pi/2$. ✓ The geodesic follows the rim.

Steep cone ($a\to\infty$, needle-like): $a/\sqrt{1+a^2}\to 1$, $\sin(\pi/4) = \sqrt 2/2$, giving $\ell_{\min}\to h\sqrt 2\cdot a$ — the chord across a flat disk of radius $ah$. ✓

Deficit angle

The cone has total angular deficit $2\pi - 2\pi a/\sqrt{1+a^2} = 2\pi(1 - a/\sqrt{1+a^2})$ concentrated at the apex. By Gauss–Bonnet, this equals the integrated Gaussian curvature there — a curvature delta function. Prototype of conical singularities in GR: cosmic strings produce exactly this deficit in their exterior geometry.