The Schwarzschild Geometry: Part 3 - Comments

[cianfa72]

The fact that the full Kruskal chart for Schwarzschild spacetime looks a lot like a spacetime diagram of an inertial chart in 1+1 Minkowski spacetime is misleading in this respect.

I take it as follows. An inertial chart in $T,X$ coordinates for 1+1 Minkowski spacetime (1 timelike + 1 spacelike dimension) has a diagram that actually extends on all over the $T,X$ plane (i.e. there is not any boundary like the two hyperbolas of Kruskal chart for Schwarzschild spacetime).

As you said for 1+1 Minkowski spacetime the "Kruskal" $X$ coordinate is basically the spherical polar $r$ . Since in this case there is 1 spacelike dimension only, the 'spherical polar $r$' (i.e. $X$) takes actually all values in the range $(- \infty, + \infty)$.

 

[PeterDonis]

An inertial chart in $T,X$ coordinates for 1+1 Minkowski spacetime (1 timelike + 1 spacelike dimension) has a diagram that actually extends on all over the $T,X$ plane (i.e. there is not any boundary like the two hyperbolas of Kruskal chart for Schwarzschild spacetime).

Yes, that's correct. But here $X$ is not a radial coordinate; these coordinates are Cartesian, not spherical. In spherical coordinates the $T, R$ "plane" of Minkowski spacetime is only a half-plane, with $0 \le R < \infty$.

As you said for 1+1 Minkowski spacetime the "Kruskal" $X$ coordinate is basically the spherical polar $r$ . Since in this case there is 1 spacelike dimension only, the 'spherical polar $r$' (i.e. $X$) takes actually all values in the range $(- \infty, + \infty)$.

No, it doesn't. Minkowski spacetime only goes from $0 \le r < \infty$. But the maximally extended Schwarzschild spacetime in Kruskal coordinates goes from $- \infty < X < \infty$ even though $X$ is a radial coordinate in spherical coordinates, not a Cartesian coordinate.

So the Kruskal diagram for Schwarzschild spacetime in $T, X$ coordinates, with $X$ a "radial" coordinate in spherical coordinates (i.e., every point in the diagram represents a 2-sphere, and there are hyperbolic boundaries) looks like a standard spacetime diagram for Minkowski spacetime in Cartesian coordinates, i.e., $X$ is a Cartesian coordinate, not a radial coordinate (and if we take the diagram to represent 1+3 Minkowski spacetime, then every point in the diagram represents a 2-plane, not a 2-sphere, and there are no hyperbolic boundaries).

If we take the $M = 0$ case of Schwarzschild spacetime, and keep spherical coordinates, then the whole left half-plane of the Kruskal diagram for Schwarzschild spacetime disappears and only the right half-plane remains (with no hyperbolic boundaries).

 

[cianfa72]

No, it doesn't. Minkowski spacetime only goes from $0 \le r < \infty$. But the maximally extended Schwarzschild spacetime in Kruskal coordinates goes from $- \infty < X < \infty$ even though $X$ is a radial coordinate in spherical coordinates, not a Cartesian coordinate.

So $X$ as radial coordiante in spherical coordinate can assume negative values ? (i.e. 2-spheres labeled with negative $X$ hence negative area) ?

So the Kruskal diagram for Schwarzschild spacetime in $T, X$ coordinates, with $X$ a "radial" coordinate in spherical coordinates (i.e., every point in the diagram represents a 2-sphere, and there are hyperbolic boundaries) looks like a standard spacetime diagram for Minkowski spacetime in Cartesian coordinates, i.e., $X$ is a Cartesian coordinate, not a radial coordinate (and if we take the diagram to represent 1+3 Minkowski spacetime, then every point in the diagram represents a 2-plane, not a 2-sphere, and there are no hyperbolic boundaries).

If we take the $M = 0$ case of Schwarzschild spacetime, and keep spherical coordinates, then the whole left half-plane of the Kruskal diagram for Schwarzschild spacetime disappears and only the right half-plane remains (with no hyperbolic boundaries).

ok got it !

 

[PeterDonis]

So $X$ as radial coordiante in spherical coordinate can assume negative values ?

The $X$ coordinate in Kruskal coordinates in Schwarzschild spacetime can, yes. (Questions about coordinates are meaningless unless you specify which particular chart and which spacetime.) But while the Kruskal $X$ is a radial coordinate, it's not the same radial coordinate as $r$. (In Kruskal coordinates $r$ is not even a coordinate, it's a function of the coordinates. See below.)

(i.e. 2-spheres labeled with negative $X$ hence negative area) ?

The 2-spheres labeled with negative $X$ do not have negative area. $X$ is not an "areal radius"; a radial coordinate does not have to have a direct relationship with area of 2-spheres. The areal radius $r$ in Kruskal coordinates is a function of both the Kruskal $X$ and the Kruskal $T$. I believe I give the function in the Insights article.

 

[cianfa72]

The 2-spheres labeled with negative $X$ do not have negative area. $X$ is not an "areal radius"; a radial coordinate does not have to have a direct relationship with area of 2-spheres. The areal radius $r$ in Kruskal coordinates is a function of both the Kruskal $X$ and the Kruskal $T$. I believe I give the function in the Insights article.

Ah ok, so why is Kruskal $X$ coordinate called 'a radial coordinate? Just because -- as pointed out in the Insight -- it is given as transformation from $(t,r)$ coordinates of Schwarzschild chart ?

 

[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.