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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.

tiny-tim

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Hi Ulysees!

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 … ?

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What's on the left of ^ ? And what is the symbol on the right of cos?If you put cos ~ 1 + ( ^2)/2, you get … ?

tiny-tim

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"cosA ~ 1 + (A^2)/2, where A is a very small angle (length, in radians), and the same for B and C".

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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.,

"cosA ~ 1 + (A^2)/2, where A is a very small angle (length, in radians), and the same for B and C".

[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).

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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?

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?

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We're essentially taking limit as the triangle on the sphere gets smaller....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?

(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.

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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?

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tiny-tim

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Hi Ulysees!without including the concept of curved spaceand instead using relativistic equations in euclidean space?

Locally, yes … if you're only considering

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,

To put it succinctly … experimental measurements involving

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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.

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"A space that has a definite positive metric signature and is not flat" should be called Riemannian.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.

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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:experimental measurements involving triangles cannot match a Euclidean theory

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

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...

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