The Schwarzschild Geometry: Part 3 - Comments

[PeterDonis]

so why is Kruskal $X$ coordinate called 'a radial coordinate?

Because, as noted, the points in the Kruskal diagram label 2-spheres, not 2-planes. Since $X$ is spacelike everywhere, and does not point along any of the 2-spheres, it must point in a radial direction.

 

[cianfa72]

Since $X$ is spacelike everywhere, and does not point along any of the 2-spheres, it must point in a radial direction.

In other words since the metric in Kruskal coordinates $(T,X,\theta,\phi)$ is diagonal then $\partial_X$ is orthogonal both to $\partial_{\theta}$ and $\partial_{\phi}$ hence it does not point along any of the 2-spheres.

 

[PeterDonis]

since the metric in Kruskal coordinates $(T,X,\theta,\phi)$ is diagonal then $\partial_X$ is orthogonal both to $\partial_{\theta}$ and $\partial_{\phi}$ hence it does not point along any of the 2-spheres.

Yes. Although you can see that the 2-spheres in any spherically symmetric spacetime are orthogonal to the other two spacetime dimensions by purely invariant reasoning. The argument is simple: suppose that there were some vector field in the spacetime not lying within the 2-spheres (i.e., not expressible as a linear combination of the 2-sphere basis vectors) but also not orthogonal to the 2-spheres (i.e., nonzero inner product with some vector on each of the 2-spheres). This would be mathematically equivalent to having a vector field on a 2-sphere that is nonzero everywhere, which is impossible by the "hairy ball" theorem. So the remaining 2 dimensions of the spacetime must be everywhere orthogonal to the 2-spheres. (I first encountered this argument in MTW; I don't know what other GR textbooks describe it.)

 

[cianfa72]

suppose that there were some vector field in the spacetime not lying within the 2-spheres (i.e., not expressible as a linear combination of the 2-sphere basis vectors) but also not orthogonal to the 2-spheres (i.e., nonzero inner product with some vector on each of the 2-spheres).

ok, that would mean that such a vector field would always have a nonzero component lying within the 2-spheres.

This would be mathematically equivalent to having a vector field on a 2-sphere that is nonzero everywhere, which is impossible by the "hairy ball" theorem.

By the same argument at some point on each 2-sphere the basis vector fields (coordinate basis) $\partial_{\theta}$ and $\partial_{\phi}$ must vanish (not at the same point). Is it related to the reason why it is impossible to map a complete 2-sphere with only one chart ?

 

[PeterDonis]

that would mean that such a vector field would always have a nonzero component lying within the 2-spheres.

Yes, exactly. And that is impossible by the hairy ball theorem: it is impossible to have a vector field on a 2-sphere that is everywhere nonzero.

By the same argument at some point on each 2-sphere the basis vector fields (coordinate basis) $\partial_{\theta}$ and $\partial_{\phi}$ must vanish (not at the same point). Is it related to the reason why it is impossible to map a complete 2-sphere with only one chart ?

Yes, it is the reason why it is impossible to map a complete 2-sphere with only one chart.