Calculation for angular excess

[Haorong Wu]

TL;DR

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,

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!

 

[mathman]

Try looking at parts of a sphere where you can calculate these numbers. For example two meridians at right angles and the equator. Area is $\frac{\pi}{2}$ while angle sum is $\frac{3\pi}{2}$ for a=1 (I can never trust my arithmetic - check it). It looks like $\pi$ is already there.

 

[Haorong Wu]

Thanks, @mathman . I got the same answer. So I am not sure whether the equation given in the book is wrong, or because it was defined so in the old days since the book is published in the 1970s.