Skip to main content
Physics

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

Required fields*

12
  • $\begingroup$ I do not think this is true. Special relativity does not have to happen in one global chart. That is only true for Minkowski spacetime. Special relativity spacetime on a cylinder or a torus cannot (with the usual definitions of charts) be covered by a single global chart. $\endgroup$ Commented Mar 18 at 19:14
  • $\begingroup$ @MarkusKlyver : The tangent space to a 4-manifold at a point looks exactly like the tangent space to any other 4-manifold at any other point. $\endgroup$ Commented Mar 18 at 19:16
  • 1
    $\begingroup$ @MarkusKlyver : They are isomorphic vector spaces, with, a fortiori, isomorphic Stiefel manifolds. Anything one can say about choosing a point in one of those Stiefel manifolds is exactly mirrored by something one can say about choosing a point in the other. As far as its internal structure goes, $T_pM$ knows nothing specific about $M$. It cannot tell the difference between Minkowski space and your cylindrical spacetime. $\endgroup$ Commented Mar 18 at 19:36
  • 1
    $\begingroup$ @CarloC The "curvature" is entirely extrinsic, one can cut a cylinder up and "unfold" it to a planar surface. So there is no meaningful (intrinsic) "curvature" to speak of. Compare that to a sphere, that cannot be "un-curved". $\endgroup$ Commented Mar 19 at 21:41
  • 1
    $\begingroup$ @CarloC The notation $S^n$ refers to the $n$-sphere (in $\mathbb R^n$). It is the set defined by $x_1^2 + \cdots x_{n+1}^2 = 1$. It is equipped with the subset topology, the submanifold smooth structure and the submanifold metric (all from $\mathbb R^n$). In this context, we mean that $M$ is diffeomorphic to $R \times S^1$ with $\mathbb R_t$ signifying that $M$ is time-orientated with the "obvious" time orientation ($t$ increases). $\endgroup$ Commented Mar 19 at 22:09