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Also, does Euclidean Geometry exist naturally in this universe??

Thanks,

Al

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- Thread starter Allojubrious
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- #1

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Also, does Euclidean Geometry exist naturally in this universe??

Thanks,

Al

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- #3

lavinia

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Also, does Euclidean Geometry exist naturally in this universe??

Thanks,

Al

The region in the inside of the circle is curved so radial geodesics are either too long or too short.

not sure what you mean by naturally. In the theory of relativity space is curved near stars.

- #4

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Also, does Euclidean Geometry exist naturally in this universe??

[Philosophy alert!]

Euclidean geometry exists in the minds of humans. The minds of humans exist naturally in the universe. Therefore Euclidean geometry exists naturally in the universe.

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When I say 'naturally' I'm not sure how much simpler to say it, does Euclidean Geometry occur naturally; is it apart of nature anywhere in this universe??

Because when in the presence of a gravitational field (force), space is non-Euclidean. However, the gravitational fields are essentially infinite, so if they are infinite does that mean that Euclidean space does not exist??

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[Philosophy alert!]

Euclidean geometry exists in the minds of humans. The minds of humans exist naturally in the universe. Therefore Euclidean geometry exists naturally in the universe.

Does math exist before or after the human mind or something like that, is that corny philosophy question. They asked the wrong question. Math is manifest by a bunch of people with the same disability. Egads, I'm politically incorrect on the web.

Euclidean geometry is probably a really natural way to represent things, because you can add vectors easily, much more math follows much more straight-forward-ly. But noneuclidean geometry arises quickly as you consider spherical situations for instance. And so forth. It didn't take long before the "modern line of science" realized the earth was round.

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- #7

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When I say 'naturally' I'm not sure how much simpler to say it, does Euclidean Geometry occur naturally; is it apart of nature anywhere in this universe??

Because when in the presence of a gravitational field (force), space is non-Euclidean. However, the gravitational fields are essentially infinite, so if they are infinite does that mean that Euclidean space does not exist??

In physics, you're often picking your battles. It is all based on which level of resolution you need. For some processes, the gravitational effect is negligible. And for instance, the mechanics that an engineer uses would also consider the spherical aspect of gravity negligible, and be able to approximate it as being "rectangular", or "euclidean" to relate it to your wording. Even a non-euclidean geometry like a spherical one is an approximation most likely, so you're not going to get away from this underlying and extremely frequent idea of approximation in physics.

- #8

lavinia

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I Newtonian physics the Universe is Euclidean. In General Relativity it is not.

For most of history,it was believed that Euclidean geometry was intrinsic to space. Simple notions of line and plane were believed to imply Euclidean measurement. In the 18'th century Gauss realized that there is another possible plane geometry - one in which the parallel postulate fails. He tried to measure the sum of the angles of large triangles on the earth in order to see which geometry naturally occurs in space. His measurements were not accurate enough to detect a deviation from Euclidean.

Riemann later came up with an idea of space that did not have any intrinsic geometry. His space could have many geometries, Euclidean, non-Euclidean, and irregular. His space required an externally imposed method of measurement, what is usually called a Riemannian metric although I think Riemann probably thought that other types of metrics might also be possible. In reality, the geometry of space then is determined by the metric that space has. This metric can change and is itself a product of the laws of Physics. From this point of view Euclidean geometry is just wrong. The theory of Relativity applies Riemann's framework to space - time and this theory predicts that the geometry of space is constantly changing.

I do not know if it is possible for an instant to have a small region of space in which all of the gravitational fields exactly cancel out. If so I guess the theory of relativity would predict that the geometry in this region would be Euclidean.

For most of history,it was believed that Euclidean geometry was intrinsic to space. Simple notions of line and plane were believed to imply Euclidean measurement. In the 18'th century Gauss realized that there is another possible plane geometry - one in which the parallel postulate fails. He tried to measure the sum of the angles of large triangles on the earth in order to see which geometry naturally occurs in space. His measurements were not accurate enough to detect a deviation from Euclidean.

Riemann later came up with an idea of space that did not have any intrinsic geometry. His space could have many geometries, Euclidean, non-Euclidean, and irregular. His space required an externally imposed method of measurement, what is usually called a Riemannian metric although I think Riemann probably thought that other types of metrics might also be possible. In reality, the geometry of space then is determined by the metric that space has. This metric can change and is itself a product of the laws of Physics. From this point of view Euclidean geometry is just wrong. The theory of Relativity applies Riemann's framework to space - time and this theory predicts that the geometry of space is constantly changing.

I do not know if it is possible for an instant to have a small region of space in which all of the gravitational fields exactly cancel out. If so I guess the theory of relativity would predict that the geometry in this region would be Euclidean.

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Oh ok, thank you very much lavinia, you answered all of my questions regarding the natural existence of Euclidean geometry! Now, I have one last question, if Euclidean geometry essentially does not exist, then is pi (∏) still practically useful, because all circles are generally spherical, elliptical, or simply curved so the ratio of the circumference to its diameter is no longer pi, **so is pi valid only in Euclidean geometry?** (For in the presence of a gravitational field space is non-euclidean.)

Thanks,

Al

Thanks,

Al

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- #10

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Oh ok, thank you very much lavinia, you answered all of my questions regarding the natural existence of Euclidean geometry! Now, I have one last question, if Euclidean geometry essentially does not exist, then is pi (∏) still practically useful, because all circles are generally spherical, elliptical, or simply curved so the ratio of the circumference to its diameter is no longer pi,so is pi valid only in Euclidean geometry?(For in the presence of a gravitational field space is non-euclidean.)

Thanks,

Al

These days pi is a particular real number that's not defined in terms of circles. So the value of pi is the same regardless of the geometry of the physical universe.

What's true is that the ratio of the circumference of a circle to its diameter depends on the geometry. In Euclidean geometry, the ratio happens to be pi. But in non-Euclidean geometry, the value of that ratio is different than pi.

Regardless of the geometry of the physical universe, the real number pi is very important in math and physics.

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lavinia

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so is pi valid only in Euclidean geometry?(For in the presence of a gravitational field space is non-euclidean.)

Thanks,

Al

The equation C = 2[itex]\pi[/itex]r is valid always only for Euclidean geometry.

But for non-Euclidean it is a good approximation for small r,

The difference, 2[itex]\pi[/itex]r - C where C is the actual length of a circle of geodesic radius ,r, goes to zero as r[itex]^{3}[/itex] and so is eventually a good approximation.

The equation is K[itex]_{r = 0}[/itex] = 3/[itex]\pi[/itex]lim[itex]_{r\rightarrow0}[/itex](2[itex]\pi[/itex]r- C)/r[itex]^{3}[/itex]where K is the Gauss curvature.

So for very small regions of relatively constant curvature - e,g, a corn field or the surface of the ocean on a windless day - the world will appear Euclidean - that is,it will look flat.

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Oh alright, thanks everyone for the great help!

Thanks,

Al

Thanks,

Al

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