I feel that this may be a naïve question, but I am struggling with the concept of intrinsic curvature when applied to the surface of a cylinder. In General Relativity: An Introduction for Physicists, the authors argue that the intrinsic curvature of the surface must be zero because it can be constructed by rolling up, without deformation or tearing, a flat surface. This I understand, and I can also visualise what the surface would look like to an observer embedded in that surface (i.e. they would measure the angles of triangles summing to 180° etc.). However, extrinsically we see that the surface is curved. If an observer were to walk along the surface, we (in our 3D world) would see that the observer would eventually return to their starting point (i.e. complete one circuit around the cylinder). Wouldn't the observer embedded in the surface also be able to see that they have returned to their starting point? If so, would that allow the observer to conclude that they inhabit a curved geometry?
2 Answers
An observer embedded in a surface can tell if the surface is curved by walking around what should be a square. That is his path is $4$ segments of equal length with $90^{\circ}$ turns between them.
On earth, such a path might start on the equator. It would go north to the north pole, turn left, go south to the equator, turn left, go west $1/4$ of the way around the world (which is back to the start), turn left, and go north to the north pole.
Since he doesn't wind up back where he started, he can conclude the surface is curved.
Curvature in space time as measured in General Relativity involves parallel transport around such a path. As you walk along the path, carry an arrow and keep it pointing in the same direction. If the arrow points in a different direction when you are done, the surface is curved. The change in direction is a measure of the curvature.
For example, start with the arrow pointing north. As you walk the first leg, it points ahead of you. After you turn left, it point to your right, which is east. After you turn left and walk along the equator, it points behind you, which is east. After you turn left and walk north again, it points to your left, which is east. When you reach the end at the pole, it is pointing to your left. If you turn left again, it is pointing ahead of you. But this is a different direction that when you started.
The example path was large because it is easy to visualize the difference in direction and position. And the differences are large for a large path. But there would still be a small difference if the sides of the square were $1$ mile long.
Also note that instead of walking around the entire square, you can walk halfway around, and see if you wind up at the same place if you walk the other half. In flat space, both halves wind up at the opposite diagonal. In curved space, they wind up in different places.
If you pick a square path on a cylinder, it should be easy to convince yourself that you would come back to the original point, and an arrow you carried would not change direction.
You can see that space time is curved from the fact that time runs slightly slower near the surface of the earth than it does far from the earth.
In space time, you can pick a square where two of the sides are time-like. So start away from earth. In one half path, wait $1$ sec and move $100$ miles closer. In the other half, move $100$ miles closer and wait $1$ sec. Since time moves slower when you wait near the earth, you wind up at the same point, but at different times. These are different points in space time.
- 38,487
- 5
- 49
- 129
-
I understand that on a sphere we can easily deduce that we are on a curved surface. However, it is also true that the surface of a sphere cannot be made into a flat plane without deformation or tearing. My point is, if you can walk along a square on a cylinder, which can be made into a flat plane, does this mean that the observer cannot tell they are on a curved surface? Doesn't ending up where one started, by walking in a straight line indefinitely, prove the surface's curvature? I believe this might be an issue of local curvature (as pointed out by @Chiral Anomaly). – DJTS Dec 13 '20 at 15:37
There are two different understandings of curvature conflicting here, namely intrinsic and extrinsic curvature. Intrinsic curvature is the actual curvature in that is detectable to someone moving inside the space. On the other hand extrinsic curvature can only be defined if the space is embedded in another higher dimensional space, for example the cylinder embedded in $\mathbb{R}^3$. It has an extrinsic curvature but it intrinsic curvature is that of the plane, i.e. it is not curved intrinsically. Note that in the context of general relativity space-time has an intrinsic curvature but it is not assumed that space-time is embedded in any higher dimensional space.
For your question about the observer returning to the start-point. This is not really related to curvature but more about the topology of the space, namely that a cylinder can be viewed an infinite 2-dimensional tape but with its two sides identified.
Look also at https://math.stackexchange.com/questions/2002965/intrinsic-and-extrinsic-curvature.
- 1,119
-
So could an observer travel infinitely in one direction and not return to their start point? – DJTS Dec 12 '20 at 13:32
-
Here I meant a cylinder of infinite height, i.e. the space $[0,1] \times \mathbb{R} \subset \mathbb{R}^2$ with points on ${ 0 } \times \mathbb{R}$ and ${1 } \times \mathbb{R}$ identified. This example was just to show that returning to the starting point after walking in the same direction does not imply that we have intrinsic curvature. Note that in order to show that a space has intrinsic curvate you have to look at the metric of the space. – jkb1603 Dec 12 '20 at 14:08
-
I see. However, if intrinsically a flat plane and a cylinder are the same to an observer embedded in the surface, why is it that an observer can return to the start point on a cylinder but not on a flat plane? I can see that the maths implies that there is no difference (i.e. the metric is the same locally) but overall there is a difference, no? – DJTS Dec 12 '20 at 14:16
-
1@DJTS Curvature (or flatness) is a local property. The existence of straight closed paths is a global (topological) property. For more examples of spaces that are flat but still topologically nontrivial, see https://en.wikipedia.org/wiki/Flat_manifold – Chiral Anomaly Dec 12 '20 at 15:39
-
@DJTS It depends on what you mean by "same". So the plane and the cylinder are not really the same. They are locally homeomorphic, since the cylinder is a manifold, and additionally the intrinsic curvature of the cylinder vanishes (at any point on the cylinder), i.e. it is flat, like the plane. As the previous comment points out they are not the same globally, just like the real interval is not the same as the circle. If you are unsure what local and global means here, consider https://physics.stackexchange.com/questions/595179/does-local-mean-infinitesimally-small/595208#595208 – jkb1603 Dec 12 '20 at 16:16
-
Thank you, I think I understand this now. So the 2D surface of a cylinder and a flat plane are only the same to an observer locally. This is probably what the authors of the book I cited meant. – DJTS Dec 13 '20 at 15:12