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What happens to the pythagorean theorem in a non-euclidean space?

  1. Apr 4, 2008 #1
    Does the theorem still hold true, but triangle sides are curves?

    pythag_thm.gif
     
  2. jcsd
  3. Apr 4, 2008 #2

    robphy

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    The theorem won't hold as written anymore.
    Working in a non-euclidean space doesn't imply that straight lines (i.e., geodesics) will appear curved. Minkowski space[time] is a non-euclidean space.
     
  4. Apr 4, 2008 #3

    tiny-tim

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    … Pythagoras on a sphere …

    Hi Ulysees! :smile:

    The surface of a sphere is a good example of a non-Euclidean space.

    The equivalent of Pythagoras' theorem is cosC = cosA.cosB.

    If you put cos ~ 1 + ( ^2)/2, you get … ? :smile:
     
  5. Apr 4, 2008 #4
    What's on the left of ^ ? And what is the symbol on the right of cos?
     
  6. Apr 4, 2008 #5

    tiny-tim

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    ooh … sorry … my ~ meant "is approximately", and the whole thing was shorthand for:
    "cosA ~ 1 + (A^2)/2, where A is a very small angle (length, in radians), and the same for B and C".​
     
  7. Apr 4, 2008 #6

    robphy

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    That is, insert into [itex]\cos C =\cos A \cos B[/itex] the "Taylor expansion" of the cosine functions: e.g.,
    [tex]\cos A = 1 - \frac{1}{2!}A^2 + \frac{1}{4!}A^4 \mp \ldots[/tex]
    and keep the lowest nontrivial terms (i.e., let [itex]A[/itex] tend to zero).
     
  8. Apr 4, 2008 #7
    Thats right, on the surface of a sphere you can draw a triangle with three equal length sides that has 3 right angles too! Thats what is meant by curved space to some extent, the ordinary euclidean geometry fails.
     
  9. Apr 4, 2008 #8
    I guess some approximation of the expansion of cos might yield the pythagorean theoreom.

    But I'm pondering, what are we doing here, we're mapping an abstract 2D curved space into ordinary euclidean space. General relativity is doing something analogous?

    Could it be that a 3D or 4D curved space is just a mathematical trick, and in fact we can match experimental measurements with euclidean geometry too and the right formulation of relativity's equations?
     
    Last edited: Apr 4, 2008
  10. Apr 4, 2008 #9

    robphy

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    We're essentially taking limit as the triangle on the sphere gets smaller....
    (more precisely, the triangle-edges are made much smaller than the radius of curvature [of the sphere].)
    ...so that the triangle approaches that of a "triangle on a flat plane (which best approximates the surface at that point)".

    However, this does not mean the "earth is flat"... the earth's surface is certainly curved.
     
  11. Apr 5, 2008 #10
    So the sphere is real, not an abstraction from 4D to 2D?

    The centre of curvature is the sun? And what you're saying is about triangles oriented so the gravity of the sun is at right angles to the triangle?
     
  12. Apr 5, 2008 #11
    Can someone tell me, is it possible to match experimental measurements explained with GR but without including the concept of curved space and instead using relativistic equations in euclidean space?
     
  13. Apr 5, 2008 #12

    tiny-tim

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    Hi Ulysees! :smile:

    Locally, yes … if you're only considering one particle, you can explain away the GR structure of "space-time" by inventing imaginary forces such as "the force of gravity" … which, of course is exactly what we do in ordinary mechanics!

    But, for a region of space (as opposed to just a point), you'll always be able to detect a difference in geometry … the Pythagoras theorem, with which you started this discussion, must work in Euclidean space, but cannot work in GR. :smile:

    To put it succinctly … experimental measurements involving triangles cannot match a Euclidean theory.
     
  14. Apr 6, 2008 #13
    Terminology

    The term Euclidean is, unfortunately, used for two distinct things, one refers to the metric signature and the other to the curvature of a space.

    One can say unambiguously that any space that has an indefinite or a definite negative metric signature is not Euclidean even if that space is flat. Also any space that has a definite positive metric signature and is flat is Euclidean. However a space that has a definite positive metric signature and is not flat is sometimes also called Euclidean.

    A metric that has exactly one single component with a reversed sign is often called Lorentzian.

    The Pythagorean theorem does not hold in a Lorentzian metric space even if that space is flat.
     
    Last edited: Apr 6, 2008
  15. Apr 6, 2008 #14

    robphy

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    "A space that has a definite positive metric signature and is not flat" should be called Riemannian.
     
  16. Apr 6, 2008 #15
    For example, let's take a wire shaped like a triangle, like a hanger for clothes, with a top angle of 90 degrees. We can put this hanger on a nail on the wall in various ways:

    Image3.jpg

    You are saying that the model where there is a constant earth "force" of gravity is false, and the model where there is a varying inverse-square "force" of gravity is false too, and that the reality is that the space of the triangle is curved. I suspect that the GR model does not produce more accurate predictions as to where the triangle will balance in the bottom-right case. Compared to the newtonian model of an inverse-square force of gravity, the prediction should be the same I suspect.

    So let's do this problem correctly then (ie according to GR, and not the false newtonian euclidean-space force mechanics). I think the result will be the same.

    At what point will the triangle balance? (in the bottom-right figure). Given the hypotenuse is 1 metre long, and the other angles are equal. The triangle is on the surface of the earth at the north pole. Make it bigger if you want, with a 1000 km hypotenuse. Let's see if you can get away without using the pythagorean theorem or other euclidean equations...
     
    Last edited: Apr 6, 2008
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