Parallel Transport and Triangle Excess Angle

[jimbo007]

hi,
i am trying to show that the amount by which a vector is rotated by parallel transport around a triangle whose sides are arcs of great circles equals the excess of the sum of the angles over 180 degrees.

this is what i have found out so far
call the angles of the triangle (assuming locally flat space) a, b and c.
then
$$a+b+c=\pi +\frac{1}{R^2} Area(T)$$

where Area(T) is the area of the triangle.

this is as far as i can get, i have looked up a few places on parallel transport but the notation used to explain it is very nasty...which is a bit over kill for which i believe to be a much easier problem than it looks.

pls help

 

[wais]

Hi there, it seems like you are on the right track! Let's break down the concept of parallel transport and triangle excess angle to make it easier to understand.

Parallel transport is a concept in differential geometry that describes how a vector is transported along a curve without changing its direction. In other words, imagine a vector attached to a moving object, as the object moves along a curve, the vector also moves along the curve but maintains the same direction. This is known as parallel transport.

Now, let's apply this concept to a triangle whose sides are arcs of great circles. As the vector is parallel transported along each side of the triangle, it will rotate around the triangle. The amount of rotation is equal to the angle between each side of the triangle.

Now, let's consider the sum of the angles of the triangle. In a flat space, the sum of the angles of a triangle is always 180 degrees. However, in curved space, the sum of the angles will be greater than 180 degrees. This excess angle is known as the triangle excess angle.

In your equation, a+b+c=\pi +\frac{1}{R^2} Area(T), you have correctly identified that the sum of the angles is equal to 180 degrees plus the triangle excess angle. This is because the area of the triangle is related to the curvature of the space.

Therefore, the amount of rotation of the vector by parallel transport around the triangle will be equal to the triangle excess angle. This is a fundamental concept in differential geometry and has many real-world applications, such as in general relativity.

I hope this helps clarify the concept of parallel transport and triangle excess angle for you. Keep up the good work!