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Curvature of space is often intuitively explained as angles of a triangle not adding up to 180 degrees. My questions concerns that.

Suppose you have a perfectly spherical star of uniform density - so that the curvature of space outside the star is described by the Schwarzschild solution.

Let A, B and C be three points on a plane passing through the centre of the star. Assume that the edges of triangle ABC don't intersect the star at all.

Is the angle sum of triangle ABC:

  • Always less than 180 degrees ?

  • Always greater than 180 degrees ?

  • Could be either way ?

  • Does the answer depend on whether the centre of the star is contained inside triangle ABC ?

user12262
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user42761
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    How are you going to define the straight lines that make up the sides of your triangle? Null geodesics i.e. the lines followed by light rays? – John Rennie Jul 30 '15 at 09:45
  • @John Rennie: "How are you going to define the straight lines that make up the sides of your triangle?" -- This is a worthwile question by itself; and it is closely related to the OP question as well as to the (arguably more basic) question "What is the notion of a spatial angle in general relativity?" (PSE/q/108359). (The latter question has been answered already, possibly providing the means of addressing the OP question directly). – user12262 Jul 30 '15 at 19:27
  • The problem with this question is that you are asking about the geometry of space. BUT that depends on how you decompose your space-time in space and time. In one choice you can have flat space, in others non-flat. – MBN Jul 31 '15 at 09:05
  • Hi John, MBN - I believe in cosmology, they decompose spacetime into well-defined spatial slices - so you can talk about "curvature of space". Is that not possible for the Schwarzschild solution ?

    The mental image I had in mind was three satellites orbiting the earth maintaining a fixed position wrt each other (I believe its actually possible to do that). Now the satellites continually fire laser beams at each other and these make up the sides of ABC.

    Is there no way to translate this picture into a meaningful calculation in GR ?

     – user42761 Aug 03 '15 at 03:57
  • Consider a black hole or neutron star. Now put A, B, C in the equatorial plane of the photon sphere. – PM 2Ring Jan 01 '24 at 03:38

1 Answers1

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For my answer, I use geodesics as the lines connecting two points: The path which a photon would travel along. Therefore, I look at three paths of photons from A-B, B-C and A-C.

Let's look at two (extreme) cases:

  1. The photons between A, B, C follow paths equidistant to the surface of the star. There, the situation is the same as on earth, where the angle sum is always larger than 180°.

  2. Two photons, attracted by the star, and meet at point A before they end up in the center. The angle at A is smaller than 60°. Two points B and C on each of the photon paths are connected via a photon path equidistant to the surface of the star. The angles at these two points are smaller than 60°.

Conclusion: The angle sum can be smaller than 180° or larger than 180° - it depends, how exactly the plane of the triangle is oriented within the curved space.

MartyMcFly
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