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In general relativity, energy bends spacetime. However, this doesn't mean that a fifth dimension for spacetime to "bend into" exists." That is, spacetime isn't embedded in a higher dimensional space, Instead, the curvature is said to be intrinsic.

But what does that mean? One could imagine the sphere on a ball as an example of extrinsic curvature it seems that intrinsic curvature isn't as straight-forward and intuitive. Is there a simple and easy way to understand the differences, and how a space can "curve" without actually be embedded in a higher dimensional space to "bend" in?

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    $\begingroup$ This appears to be a pure differential geometry question. $\endgroup$ Commented Jun 30, 2016 at 17:46

2 Answers 2

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A Riemannian manifold is intrinsically curved if there exists a geodesic triangle bounding a topological disk whose interior angles do not add to $\pi$.

An embedded Riemannian submanifold $M \subset N$ is extrinsically curved if no smooth orthonormal frame of normal vectors of $M$ is parallel (convariantly constant) in $N$.

  • A flat square torus embeds isometrically as a product of circles in four-dimensional space; the image is intrinsically flat (every triangle bounding a disk has total interior angle $\pi$) but extrinsically curved (there is no parallel field of unit normal vectors, much less a parallel frame).

  • A great sphere in a round $3$-sphere is intrinsically curved (isometric to the unit sphere in Euclidean $3$-space, so a triangle bounding a disk has total interior angle $> \pi$) but extrinsically flat (each of the two smooth unit normal fields is parallel). If you don't mind a non-compact example, a circular cylinder in Euclidean $3$-space is intrinsically flat but extrinsically curved for similar reasons.

Gravitational lensing may be viewed as a physical manifestation of intrinsic curvature: In an approximation where the lensing system is static, the path of a light ray is a spatial geodesic. Two light rays seen at distinct points of the sky form a "digon", a geodesic polygon with two sides. Because the sum of the interior angles is not zero, space is intrinsically curved.

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  • $\begingroup$ Could you elaborate on why the great sphere is extrinsically flat? I fail to see the two unit normal fields you mention $\endgroup$
    – Bananach
    Commented Oct 2, 2021 at 0:07
  • $\begingroup$ @Bananach: The essential picture is northward-pointing unit vectors along the equator, or southward-pointing unit vectors. <> Generally, a great sphere on $S^{n} \subset \mathbf{R}^{n+1}$ is the intersection of the sphere with a hyperplane through the center. The two unit normal fields to the hyperplane restrict to the stated unit normal fields to the great sphere. $\endgroup$ Commented Oct 2, 2021 at 1:04
  • $\begingroup$ But in the case of a great 2-sphere on the 3-sphere within $R^4$, don't we need a parallel normal 2-frame, not just vector field, to prove zero extrensic curvature? $\endgroup$
    – Bananach
    Commented Oct 2, 2021 at 10:58
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    $\begingroup$ "Extrinsic" in the sense of the answer means within $N$, here, within the sphere containing the great sphere. A great circle on a sphere, which is geodesic, has no extrinsic curvature in this sense. $\endgroup$ Commented Oct 2, 2021 at 11:14
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    $\begingroup$ That was the missing piece, thanks for your patience $\endgroup$
    – Bananach
    Commented Oct 2, 2021 at 14:33
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Intrinsic curvature comes from the parallel translation of a vector tangent to the path of translation. If a vector is translated around a loop and it fails to come back onto itself that is intrinsic curvature. Extrinsic curvature is computed by the parallel translation of a vector normal to a surface or space. If the translated normal deviates from the normal vector at a point the difference in the two normal vectors $\delta\bf n$ defines the extrinsic curvature $\delta\bf n~=~\bf K\delta e$ for $\delta e$ a unit of translation along the space.

An example of this is the torus. This is easily visualized with the 2-torus. The inner region appears curved in a saddle shape, while the outside is more spherically curved. The sum of the two is zero however. If I cut the torus into a square the geometry is similar to the old video games where Pacman leaves the top and emerges at the bottom, or exists the right and enters the left and visa versa. This square sheet is flat; it has zero intrinsic curvature. The curvature we see in a 2-torus as an inner tube is extrinsic due to its embedding in three dimensions.

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