Line element on a sphere 2026-10-18
In spherical coordinates on a sphere of radius , the position is . Its coordinate tangent vectors satisfy , , and . Expanding proves the displayed line element, the first fundamental form of the sphere. Along a parametrized path it gives the arc length by integrating . The vanishing azimuthal coefficient at a pole is a coordinate singularity, not a singularity of the sphere.
For an oriented piecewise smooth surface with boundary curve , and a continuously differentiable vector field near , Stokes theorem states
Here is the chosen unit normal, is positive surface area, and is the directed tangent line element. The positive traversal of is fixed by the right-hand rule relative to . Thus the flux of the curl through the surface equals the circulation along its oriented boundary.
Differentiate the position vector, using the angular factor from the scale factors of orthogonal coordinates:
The basis is an orthonormal basis, so all cross terms in its squared length vanish. The Euclidean line element is therefore
In particular, an angular change at fixed radius traverses a physical distance .
For a finite separation use the Cartesian distance, rather than substituting finite coordinate differences into the local line element. Expanding the two horizontal components gives
The vertical contribution is , independently. Thus
and
Equivalently, the horizontal contribution is , which also makes its nonnegativity explicit.
The downward drop from the starting rim to angle is . With no dissipative forces, conservation of energy gives , so . Dividing the line element by this speed yields
The endpoint singularities are integrable for the minimizing path. The integrand is independent of ; its Euler-Lagrange equation gives a constant , after absorbing the overall prefactor:
Solving for and taking the positive branch required by gives
The constant is fixed implicitly by
This integral is strictly increasing from zero to infinity as increases from zero to one, so every finite positive separation selects a unique . The time integrand has second derivative with respect to . Its supporting-line inequality, integrated between fixed endpoints with constant first derivative with respect to , proves this stationary path is the unique global minimum within the stated single-valued class.
As , . Therefore the skater should initially head directly down the curved ramp, perpendicular to the cylinder axis, and gain axial motion later. The path differs from the constant-pitch shortest route because extra distance is cheaper in time near the faster, lower part of the ramp. This is the brachistochrone on a horizontal half-cylinder.
Figure 1.
Shortest and fastest routes on the unrolled half-cylinder for
. Details.
On the cylindrical surface , the line element is . For a single-valued path parametrized by increasing ,
The integrand has no explicit , so its Euler-Lagrange equation gives the constant conjugate quantity . Its derivative with respect to is strictly positive, hence itself is constant. The two endpoints determine
This is a constant-pitch helix on the cylinder. To verify global minimality, unroll the cylinder with planar coordinates . The path becomes a straight line between points separated by and , so the triangle inequality for arc length gives
This is the standard description of geodesics on a circular cylinder.