Triangle on a sphere (Schutz, 6.10)

In summary, the exercise shows that the amount of rotation experienced by a vector parallel transported around a triangle on a sphere with sides as arcs of great circles is equal to the excess of the sum of the triangle's interior angles over 180°. This is proven by recognizing that the rotation only occurs at the vertices of the triangle and is equal to 180° minus the angle between the two sides at that vertex. Thus, the total rotation after returning to its starting point is 360° minus the excess angle.
  • #1
hellfire
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A first course in general relativity. B.F. Schutz. Exercise 6.10:
A straight line in a sphere is a great circle, and it is well known that the sum of interior angles of any triangle on a sphere whose sides are arcs of great circles exceeds 180°. Show that the amount by which a vector is rotated by parallel transport around such a triangle equals the excess of the sum of the angles over 180°.
How to proceed to prove this?
 
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  • #2
hellfire said:
A first course in general relativity. B.F. Schutz. Exercise 6.10:
A straight line in a sphere is a great circle, and it is well known that the sum of interior angles of any triangle on a sphere whose sides are arcs of great circles exceeds 180°. Show that the amount by which a vector is rotated by parallel transport around such a triangle equals the excess of the sum of the angles over 180°.
How to proceed to prove this?
The first step is to recognize that since each side of the triangle is a great circle (geodesic), a vector parallel transported along that geodesic experiences no rotation. The rotation occurs only at each vertex where the vector is rotated by an angle equal to 180 less the angle made by the two sides of the triangle. So the total angle of rotation after returning to its starting point is (180-a1 + 180 - a2 + 180 - a3). But since we know that a1 + a2 + a3 = 180 + gamma (where gamma is the difference you are looking for):
[tex]\theta = 540 - (180 + \gamma)) = 360 - \gamma[/tex]

AM
 
  • #3
Thank you Andrew. This was quite easier than I had supposed.
 

FAQ: Triangle on a sphere (Schutz, 6.10)

What is a triangle on a sphere?

A triangle on a sphere is a geometric shape formed by connecting three points on the surface of a sphere with straight lines. It is similar to a triangle on a flat surface, but takes into account the curvature of the sphere.

How do you find the area of a triangle on a sphere?

To find the area of a triangle on a sphere, you can use the formula A = r²(θ1 + θ2 + θ3 - π), where r is the radius of the sphere and θ1, θ2, and θ3 are the angles of the triangle.

What is the sum of the angles of a triangle on a sphere?

The sum of the angles of a triangle on a sphere is always greater than 180 degrees due to the curvature of the sphere. The exact sum can be calculated using the formula θ1 + θ2 + θ3 = π + A/r², where A is the area of the triangle and r is the radius of the sphere.

Can a triangle on a sphere be equilateral?

Yes, a triangle on a sphere can be equilateral if all three angles are equal. This would mean that the three points on the sphere are equally spaced from each other, forming an equilateral triangle. However, the sides of the triangle on a sphere will not be equal in length due to the curvature of the sphere.

How is a triangle on a sphere different from a triangle on a flat surface?

A triangle on a sphere takes into account the curvature of the surface, while a triangle on a flat surface does not. This means that the angles and sides of a triangle on a sphere will be different from those on a flat surface, and the formulas for finding the area and angles will also be different.

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