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Prove: Sum of angles > 180 in curved space

  • Thread starter Niles
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[SOLVED] Prove: Sum of angles > 180 in curved space

Homework Statement


If I have a positively curved space (i.e. a sphere) and I draw a triangle on it, the sum of the angles of the triangle exceed [tex]\pi[/tex], more precisely,

v1 + v2 + v3 = pi + A/R^2

where v1, v2 and v3 are the angles of the triangle, A is the area of the triangle and R is the radius of the sphere.

This is what I have to prove.

The Attempt at a Solution


Ok, first I setup the "environment". I will look at a triangle with equal long sides with one corner at the north-pole. What would be the next step from here?
 

Answers and Replies

  • #2
HallsofIvy
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We are told to verify the angle-relations using a triangle with equally long sides and the other two angles lying on the equator; sorry, I should have written that in post #0.

I really appreciate the links, and I know the identity:

[tex]
ds^2\,=\,dr^2\,+\,R^2 \sin ^2 (r/R)d\theta ^2,
[/tex]
where R is the radius of the sphere and (r, theta) is a point on the sphere (r is the distance from the northpole).

If I stand at the vertex situated at the northpole, then I can write the above identity as

[tex]
ds^2\,=\,R^2 \sin ^2 (r/R)d\theta ^2,
[/tex]

where r = ½*Pi*R, since the side is one quarter of the circumference. Then I integrate, and I get that one side equals 2*Pi*R^2*sin(Pi/2). I have three of those, and since they are equally long, I can multiply by 3 to get the total circumference of my triangle.

The law of sine apparently works as usual (from http://en.wikipedia.org/wiki/Spherical_trigonometry) on the triangle, so I get that the angles must equal eachother.

How am I doing so far?
 
Last edited:
  • #4
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I don't know why, but I'm not able to edit my last post. There is an error: The side equals Pi*R/2.
 
  • #5
HallsofIvy
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Okay, so you don't need to prove the statement, just verify it in that simple case.

Okay, take two points on the equator and the third point as the north pole. It is obvious that the two angles on the equator will be right angles so the angle sum is 180+ [itex]\theta[/itex] where [itex]\theta[/itex] is the angle at the north pole. It's easy to calculate that the triangle takes up [itex]\theta/(4\pi)[/itex] of the entire sphere ([itex]\theta[/itex] is measured in radians). Since the surface area of the entire sphere is [itex]4\pi R^2[/itex], the area of the triangle is [itex]A= \theta R^2[/itex]. Solve that for [itex]\theta[/itex] and you are done.
 

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