Circular cylinder

A circular cylinder of radius $R$ has principal curvatures $0$ and $\pm 1/R$, according to orientation. Hence $K=0$ and $|H|=1/(2R)$.

This is the circular generating-curve case of a Cylinder; the curvature formulas here refer to its lateral surface.

For $-\pi \le t\le \pi$, this parametric curve closes with a corner on a unit circular cylinder. Its speed is $\cosh t$, so its arc length is $2\sinh \pi$. The cylindrical surface between the curve and $z=0$ has the same surface area, because unrolling gives the integral ${\int }_{-\pi }^{\pi }\cosh tdt$.

The curved surface of a circular cylinder, excluding any flat end caps. A parametrization of a radius-$a$ cylindrical surface is $\mathbf{R}(\theta ,z)=(a\cos \theta ,a\sin \theta ,z)$, with surface area element $ad\theta dz$.

A circular cylinder of radius $\rho$ has induced metric locally equal to $d{u}^2+d{z}^2$ with $u=\rho \phi$. Its Riemann curvature tensor therefore vanishes, despite its nonzero extrinsic curvature. With outward unit normal and convention $K_{ij}=e_i^a{e}_j^b{\nabla }_a{n}_b$, its trace is $1/\rho$ and its principal curvatures are $1/\rho$ and zero. Its averaged mean curvature is $1/(2\rho )$. The angular identification changes global topology, not local flatness.

In coordinates $(\theta ,z)$ on a circular cylinder of radius $a$, the induced metric is $a^{2}d{\theta }^2+d{z}^2$. Its coefficients are constant, so the Christoffel symbols vanish and the affinely parametrized geodesic equation is ${\theta }^{\prime \prime }=z^{\prime \prime }=0$. The nonconstant solutions are generators when $c=0$, circular parallels when $d=0$, and helices when both are nonzero. Unrolling the cylinder with coordinate $u=a\theta$ makes all these curves straight lines in the Euclidean metric. Unit-speed parametrization additionally requires $a^2{c}^2+d^2=1$.

For a worked helix on a circular cylinder of radius $1$, choose $\theta (t)=2\pi t,z(t)=2t$ for $0\le t\le 1$. Unrolling gives the straight segment $(u,z)=(2\pi t,2t)$, so its arc length is $\sqrt{4{\pi }^2+4}$. Its acceleration in space points normally to the surface, consistent with the geodesic equation.

This geodesic is not globally shortest between its endpoints: they share the same angular position, and the vertical generator has length $2$. The helix minimizes length among paths with the same endpoints and one full winding, because their unrolled lifts join $(0,0)$ to $(2\pi ,2)$. In general, geodesics minimize sufficiently short subarcs; changing winding can produce a shorter route. Keeping this local/global distinction makes unrolling useful without mistaking every straight lift for an unrestricted shortest path.

Unroll a circular cylinder of radius $R>0$ using the coordinate $u=R\varphi$. Its line element becomes $d{s}^2=d{u}^2+d{z}^2$, so a geodesic lifts to a straight line in the plane. An endpoint has infinitely many lifts differing by $2\pi Rk$. Minimize the ordinary Euclidean distance over those lifts to obtain the unrestricted shortest path. A fixed lift instead specifies a winding class and need not give the global shortest path.

For angular difference $0\le \delta \le 2\pi$, put $m=\min (\delta ,2\pi -\delta )$. Then $d=\sqrt{R^2{m}^2+(z_1-z_0{)}^2}$. At $\delta =\pi$, two different geodesics have the same shortest length. This is the source of the corner in the distance-versus-angle graph at the opposite generator of the circular cylinder.