Parallel Transport on a Sphere

Draw a loop on a sphere and carry an arrow round it, keeping it pointing the same way all the time. The page shows where it ends up, how far that is from where it started, and how the turn compares with the area inside the loop.

Worked example — Parallel Transport on a Sphere

Sides, as angles at the centre	90.00°, 90.00°, 90.00°
Angle at A	90.00°
Angle at B	90.00°
Angle at C	90.00°
The three angles together	270.00°
Angle excess	90.00°
Share of the sphere inside	0.1250
Area inside	1.5708 R²

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Parallel Transport on a Sphere

Parallel transport carries an arrow along a path on a surface without turning it: at every step the arrow stays in the surface and changes only as much as the surface forces it to. On a flat sheet an arrow carried round any closed loop comes back as it left. On a curved surface it can come back turned, and the turn equals the curvature added up over the area inside the loop, less any whole turns. On a sphere of radius R that is area/R² in radians, which for a triangle of great-circle arcs is its angle excess A + B + C − 180°.

What each symbol means

E: a triangle's angle excess, A + B + C − 180°; in radians it is the area inside in units of R²
R: the sphere's radius; for the Earth here 6,371.0088 km
φ: the latitude of the circle the arrow is carried round

Never turned, the arrow still comes home 90° off 🖖

Press Three right angles. The triangle runs from the North Pole down to the equator, a quarter of the way round it, and back up to the pole, and all three of its angles are 90°. Now watch the orange arrow. Along each side it keeps the same angle to the side, which is all that “not turning” can mean on a curved surface, and at each corner it does not move at all. It reaches the pole again turned 90° anticlockwise from the dashed arrow it set out as. The dial says where the 90° went. Someone walking the triangle turns 90° at each corner, 270° in all, and a closed walk on paper turns through exactly 360°. The 90° that is missing is the arrow's turn, and it is also the triangle's area: π/2 in units of R², an eighth of the sphere.

Only the area counts, and Foucault's Paris is almost an eighth of the sphere 🖖

Press Same area, round: a circle of latitude at 48.59° N, where sin φ = 3/4, so the cap above it is an eighth of the sphere, like the octant. A circle has no corners and nothing else in common with that triangle, and the arrow comes home turned by the same 90°. The turn depends on nothing but the area enclosed: 720° times the share of the sphere inside, less any whole turns, and the second chart puts every loop on one line. Press Paris, 48.836° N, the latitude of Foucault's pendulum in Coriolis Effect and Foucault's Pendulum: the arrow comes home 88.98° anticlockwise. The pendulum is this arrow. In one sidereal day the Earth carries its swing once round the circle of latitude, nothing twists it, and against the turning ground it falls behind by 360° − 88.98° = 271.02°, clockwise, which is the 11.32° an hour that page prints for Paris. At 30° N the cap is a quarter of the sphere and the arrow comes home reversed.

An ant can find the radius of its world without leaving it 🖖

The turn divided by the area is the same for every loop on one sphere that holds less than half of it, and it is 1/R²: that ratio is the curvature. Press London, Tokyo, Los Angeles. On the Earth the triangle covers 47,696,300 km² and turns the arrow 67.33°, which is 1.17508 radians, and √(47,696,300 / 1.17508) is 6,371.0 km, the Earth's radius, found from two measurements made on the surface. It takes a big loop to see it. Halve an equilateral triangle's sides and once it is small the turn falls to a quarter (the table: 4.059, 4.015, 4.004), so a triangle with 100 km sides turns the arrow by 22.0 seconds of arc. Surveyors have allowed for exactly that since Legendre: each angle of a small spherical triangle is bigger than the flat triangle's with the same sides by very nearly a third of the excess.

Problem solved in full

A triangle whose angles add to 181° 4 steps
On the Earth, an equilateral triangle of great-circle arcs has angles adding to exactly 181°. How much ground does it cover, and how long are its sides? Press Three right angles first: there the excess is 90° and the page gives the area on the Earth.

Open this problem in the tool
1. On the octant the excess is a quarter turn, and the area is that excess, in radians, times R²:
  
  $E = 27 0^{\circ} - 18 0^{\circ} = 90.0 0^{\circ} = \frac{π}{2}, A = \frac{π}{2} R^{2} = 63, 758, 235 km^{2}$
2. An excess of 1° is π/180 of a radian, so the area follows at once, whatever the triangle's shape:
  
  $A = \frac{π}{180} \times 6371.008 8^{2} = 708, 424.8 km^{2}$
3. For the sides, use the triangle being equilateral. Each angle is 181°/3, and the law of cosines for a spherical triangle with three equal sides s reduces to cos A = cos s / (1 + cos s), which turns round to give cos s:
  
  $A = \frac{18 1 ^{\circ}}{3} = 60.333 3^{\circ}, cos s = \frac{cos A}{1 - cos A} = \frac{0 . 49495324}{0 . 50504676} = 0.9800147 \Rightarrow s = 11.474 1^{\circ}$
4. and the side, in kilometres, is that angle in radians times R:
  
  $s = 11.474 1^{\circ} \times \frac{π}{18 0 ^{\circ}} \times 6371.0088 km = 1, 275.9 km$
Answer

708,425 km², with sides of 1,275.9 km. A flat triangle with the same three sides would have angles of exactly 60° and an area of about 704,900 km², 0.50% less. A whole degree of excess needs a triangle the size of a large country: halve its sides and the excess falls to 0.249°, about a quarter, which is why a triangle paced out in a field adds to 180° as far as any protractor can tell.

$A = 708, 425 km^{2}, s = 1, 275.9 km$

Learning path

Geometry on a curved surface

References (2)

The turn of a parallel-transported vector round a circle of latitude, 2π(1 − sin φ), equal to the area of the cap it bounds on the unit sphere; the Foucault pendulum as that transport, with Paris turning about 271° against the ground in a sidereal day; and, for a geodesic triangle, the curvature inside equal to the angle sum minus π: O. Stoytchev, "From Foucault's Pendulum to the Gauss–Bonnet Theorem", arXiv:1701.01666 [math.HO] (2017).
The area of a spherical triangle, (A + B + C − π)r², proved from lunes; the law of cosines the results table uses for the angles; and Legendre's theorem, that each angle of a small spherical triangle exceeds the flat triangle's with the same sides by a third of the excess, with its use in measuring the Earth: I. Todhunter, Spherical Trigonometry for the Use of Colleges and Schools, 5th edition, Macmillan, 1886, Arts. 96–97 and 106–108 (Project Gutenberg ebook 19770).

By the 42Math team · Reviewed against published sources. Corrections welcome.