2
$\begingroup$

Consider 2D surfaces. They may be plane or curved. For a 2D surface to satisfy as a plane, it must satisfy that a line joining any two points on the surface must wholly lie on the surface. All other surfaces that do not satisfy this condition are not planes.

Now, extending this analogy to 3D spaces, we could call a space as "straight" (not sure what term I should use) if a plane joining any three points in the space wholly lie within the space. This seems to be true for the 3D space that we can observe (all points on any plane that can be considered in our 3D space lies wholly within the space)

However, is it possible for a space to be not "straight"? In the sense, can we have a situation where a plane connecting 3 points on such a space, such that the plane goes through regions not lying in the space?

$\endgroup$
1
  • 3
    $\begingroup$ It is certainly possible. That's what Riemannian geometry is about: a general theory of curved spaces in any dimension. $\endgroup$
    – Lee Mosher
    Mar 30, 2018 at 14:16

2 Answers 2

1
$\begingroup$

This seems to be true for the 3D space that we can observe

In fact, it isn't—in the theory of general relativity, gravity manifests itself as an "intrinsic curvature" of spacetime. For instance, the triangle determined by three points in space living the same time can contain too much or too little area to be embedded in a Euclidean plane. As the comments note, (pseudo-)Riemannian geometry provides a setting for this physical theory.

$\endgroup$
0
$\begingroup$

A straight line can only leave a 2D surface if that surface is immersed in a higher space and given a curvature with respect to that higher space. Considering the geometry of the 2D surface alone, there will be a definition of "straight line" within that geometry which remains within the surface. For example when a sphere is immersed in 3-space we see its straight lines as equatorial circles known as great circles.

Similarly for a flat plane to cross in and out of a 3-space, the 3-space must be immersed in 4 or higher dimensions and given a curvature with respect to that space.

We know that the spatial shape of the universe is entirely flat, as it is distorted into curves by gravity. But we would only be able to pass a flat plane outside it if there is at least one higher dimension "out there", as predicted by M-theory, to pass it through, and we can define flatness for that higher space.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .