- #1
Haorong Wu
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- How to calculate the angular excess on a sphere?
Hello. I am not familiar with spherical trigonometry while I am reading a solution in a GR problem book. It reads,
I study spherical trigonometry on Wikipedia and some other sites, but I am still not sure how to calculate the angular excess.
First, is angular excess equivalent to spherical excess? I have not found a clear definition for angular excess. But the definition for spherical excess makes me think that they are the same concept. Maybe angular excess is just an old-fashioned name?
Second, Girard's theorem states that the area of a spherical triangle is equal to its spherical excess.
Then for a sphere with radius ##a##, Girard's theorem gives that ##Area=a^2 \times E## where ##E## is the spherical excess.
So the spherical excess is given by ##E=Area/a^2##.
Now I am not sure where the ##\pi## comes from. Maybe angular excess differs from spherical excess by a factor ##\pi##?
Thanks!
If we construct a coordinate patch from geodesics we can then bisect that coordinate box with a geodesic diagonal, forming two geodesic triangles. The angular excess of a triangle made from great circles is ##\pi [Area/a^2]## where a is the radius of the sphere.
I study spherical trigonometry on Wikipedia and some other sites, but I am still not sure how to calculate the angular excess.
First, is angular excess equivalent to spherical excess? I have not found a clear definition for angular excess. But the definition for spherical excess makes me think that they are the same concept. Maybe angular excess is just an old-fashioned name?
Second, Girard's theorem states that the area of a spherical triangle is equal to its spherical excess.
Then for a sphere with radius ##a##, Girard's theorem gives that ##Area=a^2 \times E## where ##E## is the spherical excess.
So the spherical excess is given by ##E=Area/a^2##.
Now I am not sure where the ##\pi## comes from. Maybe angular excess differs from spherical excess by a factor ##\pi##?
Thanks!