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Geometry on a sphere

  1. Sep 27, 2015 #1
    Hope I am on the right forum (and that my question makes some sense-so here goes.

    Imagine we are a race of people living on a sphere (not hard because we are)

    However , rather than buying into the idea that lines are ideally straight we are and have always been well aware of how "parallel " lines meet ,effectively being "great circles".

    It has never occurred to this race that parallel lines could be imagined as lines that never meet and so all their geometrical measurements are made on the basis of lengths along this curved surface.

    To make this "more realistic" (well ,maybe less) this race lives on the event horizon of a black hole so that even light seems /does follow the surface around.

    So how could their geometry have evolved from its first uses?

    Obviously we might start with the Pythagoras formula. What would have been their alternative formula?

    If you draw 2 lines from an origin how do you calculate the distance between their end points?

    Everything must be done in 2D. So it will have to be something along the lines of f(a) G(fb)=h(c)

    where "G "indicates some process like adding ,multiplying or somehow combining in another way.
    and a,b and c are the first ,second and (hopefully ) 3rd calculated measurements. (f and h are different , appropriate functions)

    In other words ,is this race which lives on a curved surface condemned to see the idealised Euclidean view of the world in order to invent mathematics/geometry or can they (in a simple way) plough their own furrow?
     
    Last edited: Sep 27, 2015
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  3. Sep 27, 2015 #2

    mfb

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    A physical detail:
    You have to be at 1.5 times the Schwarzschild radius for that, well away from the event horizon. Below that, light won't follow the surface, it will fall inwards.

    About the mathematics:
    You can set up a general formula but that has to involve the size of the triangle and the curvature, and it gets really messy. It would be obvious to consider the limit of small triangles or large curvature radius, and then you are back to Euclidean geometry.
     
  4. Sep 27, 2015 #3
    thanks. So ,from your "messy" comment can I deduce that (because of its simplicity" the Euclidean geometry is in a sense "correct" even though it is idealised?

    I mean I understand that the universe is not really "euclidean " and yet it seems to be the only assumption that allows us to do mathematics about other kind of geometries (you cannot start from a non -Euclidean geometry and try to describe a Euclidean geometry in its terms.

    I do realise that my "logic" may be very garbled ,ignorant and erroneous so I do apologize in advance for that -but perhaps you can see what (misguidedly ,probably) I am trying to say.
     
  5. Sep 27, 2015 #4

    mfb

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    Sure you can. The Euclidean geometry is just a special case of non-Euclidean geometry. A very useful special case.
     
  6. Sep 27, 2015 #5
    I don't want to advertize my ignorance too much but if you went to a school (eg I started learning geometry at the age of 11 I think) where non Euclidian geometry was the "default" learning system wouldn't the curriculum be incredibly curtailed ?

    If we take the curvature as a constant wouldn't the lessons still be "starting at the deep end" There would be nothing simple at all would there.? You would need a supercomputer to work out the simplest calculations .

    Nothing would be apparent , nothing elementary....

    I do understand what you are saying about Euclidean geometry being a special case of non Euclidean geometry but all the formulae that are useful in that domain seem (owing to my ignorance I anticipate) to have no corresponding formulae in the general domain .
     
  7. Sep 27, 2015 #6

    mfb

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    I think it would be very hard to learn non-Euclidean geometry without understanding Euclidean geometry first. It's like starting with General Relativity instead of classical mechanics.
     
  8. Sep 27, 2015 #7
    thanks again. Maybe I need to learn a little about the actual subject now ; )
     
  9. Sep 27, 2015 #8

    WWGD

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    OP: You have a good point, and it is a tricky issue. You dont want to start with something too complicated, e.g., bringing up curvature, multiple-connectedness, non-convexity etc. (unless you have a particularly clear way of explaining it at an intro. level), but presenting an extremely simple method may be misleading and may not prepare students for more advanced work. Like when working with Real numbers and ## \mathbb R^n ## , which has all sorts of nice properties: it is a convex vector space, complete metric space*, 0 curvature, Lie group, simply-connected, a Hilbert Space, etc. and then having to deal with all sorts of weird spaces. There may be a beneficial in-between .

    * Though in a sense, using metric completion, every metric space is complete.
     
    Last edited: Sep 27, 2015
  10. Sep 28, 2015 #9
    Suppose this region can be accepted as existing, would all beams of light emitted on it complete a "grand circle "? In that case it would be impossible to "see" anything except yourself.

    Are there any circumstances that a beam of light might connect two points on this sphere that were not both on a grand circle ?

    Could an incidental beam of light that came in from outside the sphere accomplish this?

    (I know I said I should learn the subject a bit before coming back so I will not object to being lectured to at proposing ludicrous suggestions .But I hope this suggestion "holds up" , at least in theory)
     
  11. Sep 28, 2015 #10

    mfb

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    Only those emitted orthogonal to the radial direction.
    If the black hole is rotating, things get complicated. Otherwise: no.
    That won't stay at this radius, it will fall into the black hole.
     
  12. Sep 28, 2015 #11
    Can we have a beam of light that meets the surface tangentially(arriving from beyond the surface)? Would those beams of light get trapped (ie not fall into the black hole)and have trajectories along the surface that are not great circles ?
     
  13. Sep 28, 2015 #12

    mfb

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    No. At this radius, photons either orbit at a constant distance along great circles forever (or until they hit something), fall into the black hole, or escape. Photons emitted somewhere else can never achieve that orbit. To make it worse, the orbit is unstable - even a very careful alignment is not perfect so actual light will never exactly stay in that orbit.
     
  14. Sep 28, 2015 #13
    thanks, appreciated.
     
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