Insights The Schwarzschild Geometry: Part 3 - Comments

[PeterDonis]

PeterDonis submitted a new PF Insights post

The Schwarzschild Geometry: Part 3

Continue reading the Original PF Insights Post.

 

[fresh_42]

Thanks, even though I might not have understood it all. I'm looking forward to the next part and perhaps some speculations of yours, what a future model would have to accomplish.

There's a little error in the first third "Here will will assume that the GR model holds for all $r>0$."

 

[PeterDonis]

There's a little error in the first third "Here will will assume that the GR model holds for all $r>0$."

Why do you think this is an error?

 

[PAllen]

In exploring the KS chart, I find it useful to look at lines of constant X. Lines of constant T either don't cover the whole chart, or must be treated as split before a certain T, and after another one. Each line of constant X simply goes from a minimum T to a maximum T, giving a smooth connected picture of the S2xR2 manifold. Each of these lines has a nice interpretation: the evolution of a 2-sphere from r > 0, to some maximum r, then down toward 0 again. The smallest maximum r is the horizon radius; for larger X, the maximum radius grows without bound.

 

[fresh_42]

Why do you think this is an error?

I meant it should have been "we will". I guess I should have said mistake instead. I'm not very good at distinguishing between error, failure, mistake and fault. There's only one word for it in my language.

 

[PeterDonis]

I find it useful to look at lines of constant X

The only thing I would clarify here is that these lines are timelike, so the word "evolution" that you use means "evolution in time". We could make that more specific by postulating an observer who follows this line as his worldline. (An interesting exercise is to compute the proper acceleration of this observer; it is not zero.)

Lines of constant X either don't cover the whole chart, or must be treated as split before a certain T, and after another one.

I'm not sure I understand. "The whole chart" is not the entire range of $T$ and $X$ coordinates; it is only those pairs $(T, X)$ that satisfy $T^2 - X^2 < 1$. Lines of constant $X$ have no breaks within this range, and cover it entirely.

 

[PAllen]

The only thing I would clarify here is that these lines are timelike, so the word "evolution" that you use means "evolution in time". We could make that more specific by postulating an observer who follows this line as his worldline. (An interesting exercise is to compute the proper acceleration of this observer; it is not zero.)

I used the word evolution deliberately to capture the timelike nature. Taken together, we have defined a congruence of world lines filling the entire manifold. Yes, I am well aware these world lines are not geodesic. I am going to write a follow on post on some additional physical interpretation for this congruence.

I'm not sure I understand. "The whole chart" is not the entire range of $T$ and $X$ coordinates; it is only those pairs $(T, X)$ that satisfy $T^2 - X^2 < 1$. Lines of constant $X$ have no breaks within this range, and cover it entirely.

Consider a line of constant T with T greater than the minimum T for the future singularity. Then, there are actually two such lines, not one and they are disconnected. This is fine, but the lines of constant X show you more directly how the whole manifold is connected.

 

[PeterDonis]

Consider a line of constant T...

Ah, got it. I had mixed up T and X in that part...

 

[PeterDonis]

I am going to write a follow on post on some additional physical interpretation for this congruence.

Cool!

 

[PAllen]

In exploring the KS chart, I find it useful to look at lines of constant X. Lines of constant T either don't cover the whole chart, or must be treated as split before a certain T, and after another one. Each line of constant X simply goes from a minimum T to a maximum T, giving a smooth connected picture of the S2xR2 manifold. Each of these lines has a nice interpretation: the evolution of a 2-sphere from r > 0, to some maximum r, then down toward 0 again. The smallest maximum r is the horizon radius; for larger X, the maximum radius grows without bound.

Expanding on this, one may make an analogy between this family of sphere evolutions and the exterior of spherical body. Such exterior region can be covered by families of world lines, each such family being a sphere of observers leaving the surface, reaching a maximum, and then returning (no requirement that they they be inertial). They can be set up so their maxima are all at constant T in some chart, and they don't cross, thus forming a valid congruence filling the entire exterior spacetime of the body. An observation is that this shows that the exterior of a spherical body, taken as a manifold by itself, has topology S2xR2.

However, there are some key differences from the KS congruence I described. The ordinary body congruence would have spheres evolving from minimum to maximum arbitrarily close to the minimum. There would be no notion of a minimum maxima that is larger by a finite amount than the minimum. Nor would there be a duplicate exterior corresponding to negative X in the KS case. This exterior would be analogous to one KS exterior quadrant, with the SC radius standing in for the body surface.

What GR tells us is that this construction (for the exterior of a body) cannot be shrunken down smoothly to represent a point body. It has a minumum size below which we (if we preserve vacuum and spherical symmetry) we must change the geometry, in the way mandated by the KS interior region(s).

 

[TGlad]

The next thing we notice is that the two boundary hyperbolas are spacelike (because their slope is always less than 45 degrees). That means that they are best viewed, intuitively, as moments of time, not places in space.

Reference https://www.physicsforums.com/insights/schwarzschild-geometry-part-3/

should that be "are timelike (..."?
By two boundary hyperbolas I assume you mean the two thick lined hyperbolas at r=0.

 

[martinbn]

should that be "are timelike (..."?
By two boundary hyperbolas I assume you mean the two thick lined hyperbolas at r=0.

Spacelike, not timelike.

 

[PeterDonis]

should that be "are timelike (..."?

No. A slope of less than 45 degrees (relative to horizontal) is spacelike.

By two boundary hyperbolas I assume you mean the two thick lined hyperbolas at r=0.

Yes.

 

[cianfa72]

Expanding on this, one may make an analogy between this family of sphere evolutions and the exterior of spherical body. Such exterior region can be covered by families of world lines, each such family being a sphere of observers leaving the surface, reaching a maximum, and then returning (no requirement that they they be inertial). They can be set up so their maxima are all at constant T in some chart, and they don't cross, thus forming a valid congruence filling the entire exterior spacetime of the body. An observation is that this shows that the exterior of a spherical body, taken as a manifold by itself, has topology S2xR2.

However, there are some key differences from the KS congruence I described. The ordinary body congruence would have spheres evolving from minimum to maximum arbitrarily close to the minimum. There would be no notion of a minimum maxima that is larger by a finite amount than the minimum. Nor would there be a duplicate exterior corresponding to negative X in the KS case. This exterior would be analogous to one KS exterior quadrant, with the SC radius standing in for the body surface.

What GR tells us is that this construction (for the exterior of a body) cannot be shrunken down smoothly to represent a point body. It has a minumum size below which we (if we preserve vacuum and spherical symmetry) we must change the geometry, in the way mandated by the KS interior region(s).

I was thinking about the above analogy. Since lines of constant $X$ in KS chart are timelike, they actually define a congruence of observers having those as their own worldlines. In their paths through spacetime they 'reach' 2-spheres of minimum and maximum radius.

On the other hand we can image a family of observers starting from each point on the surface of the spherical body. They leave that surface, travel towards points on 2-spheres of maximum radius and then back to the spherical surface they started from.

Why it shows that the exterior of spherical body has topology $S^2 \text {x} R^2$ ?

 

[PeterDonis]

Since lines of constant $X$ in KS chart are timelike, they actually define a congruence of observers having those as their own worldlines.

Yes.

In their paths through spacetime they 'reach' 2-spheres of minimum and maximum radius.

Maximum, yes. Minimum, no; all lines of constant $X$ start on the white hole singularity and end on the black hole singularity, so they all reach 2-spheres of arbitrarily small radius. (The singularities themselves are not part of the manifold, so $r = 0$ is not on any of the lines.)

we can image a family of observers starting from each point on the surface of the spherical body.

There is no spherical body. The spacetime shown in the Kruskal chart is vacuum everywhere.

Why it shows that the exterior of spherical body has topology $S^2 \text {x} R^2$ ?

You don't need a "spherical body" to show that the topology of the spacetime shown in the Kruskal chart is $S^2 \times R^2$. All you need is the fact that the manifold does not contain any points where $r = 0$.

 

[cianfa72]

You don't need a "spherical body" to show that the topology of the spacetime shown in the Kruskal chart is $S^2 \times R^2$. All you need is the fact that the manifold does not contain any points where $r = 0$.

If we take Minkowski spacetime in standard $(t,x)$ global chart each point represents a 2-plane hence its topology is $R^2 \times R^2 = R^4$. On the other hand in spherical polar coordinates each point $(t,r)$ represents a spacelike 2-sphere with $r=0$ as minimum value.

For spacetime shown in Kruskal chart the difference wrt. Minkowski is that the radius $r$ of each spacelike 2-sphere labeled by $(T,X)$ is not allowed to vanish ($r=0$). So its topology cannot be $R^2 \times R^2 = R^4$.

 

[PeterDonis]

If we take Minkowski spacetime in standard $(t,x)$ global chart each point represents a 2-plane hence its topology is $R^2 \times R^2 = R^4$. On the other hand in spherical polar coordinates each point $(t,r)$ represents a spacelike 2-sphere with $r=0$ as minimum value.

For spacetime shown in Kruskal chart the difference wrt. Minkowski is that the radius $r$ of each spacelike 2-sphere labeled by $(T,X)$ is not allowed to vanish ($r=0$). So its topology cannot be $R^2 \times R^2 = R^4$.

Yes, you have just stated in more detail the reason why the statement of mine that you quoted is true.

 

[cianfa72]

Just to clarify myself the point, I tried to draw an analogy.

Consider a flat three-dimensional Minkowski manifold minus a point (an event). We can map it using coordinate time $t$ and polar coordinates $r,\phi$.

In a $(t,r)$ global chart each point represents a $S^1$ circle and each line $t=const$ is defined only for $r>0$ (i.e. points in the chart where $r=0$ do not belong to the manifold). From a topological point of view that means each spacelike hypersurface $t=const$ is basically $R^2 - \{0,0\}$ that is homeomorphic to $S^1 \times R$ -- see for instance here. If we apply it for all $t$ we get that the "three-dimensional Minkowski space minus a point" has topology $S^1 \times R \times R = S^1 \times R^2$.

What if we take a 4D Minkowski manifold minus one point ? The same kind of reasoning above shows that it has topology $S^2 \times R^2$. Hence the spacetime represented in Kruskal chart should have the same topology of "4D Minkowski manifold minus one point".

Does the above make sense ? Thank you.

 

[PeterDonis]

If we apply it for all $t$ we get that the "three-dimensional Minkowski space minus a point" has topology $S^1 \times R \times R = S^1 \times R^2$.

No, we get that 3D Minkowski space minus a line (the line corresponding to the set of points $r = 0$ for each value of $t$) has topology $S^1 \times R^2$.

What if we take a 4D Minkowski manifold minus one point ? The same kind of reasoning above shows that it has topology $S^2 \times R^2$. Hence the spacetime represented in Kruskal chart should have the same topology of "4D Minkowski manifold minus one point".

No, 4D Minkowski space minus a line. See above. You're just replacing circles with 2-spheres in the above construction.

 

[PeterDonis]

@cianfa72 another way of seeing how to get to topology $S^2 \times R^2$ by removing sets of points from $R^4$ (note that since we're talking about topology, the metric is irrelevant) is as follows: if we remove one point from $R^3$, we have topology $S^2 \times R$ (an infinite open set of 2-spheres parameterized by $r > 0$). If we then take an infinite open stack of these, this corresponds to removing a line from $R^4$, and it has topology $S^2 \times R^2$. This may be the construction you were thinking of, but it's not the one you described in post #18. Both constructions, with correct descriptions, are valid.

 

[cianfa72]

No, we get that 3D Minkowski space minus a line (the line corresponding to the set of points $r = 0$ for each value of $t$) has topology $S^1 \times R^2$.

Ok yes, it makes sense.

 

[PeterDonis]

if we remove one point from $R^3$, we have topology $S^2 \times R$ (an infinite open set of 2-spheres parameterized by $r > 0$).

Note, btw, that this is the topology of a surface of constant coordinate time in Kruskal coordinates.

 

[cianfa72]

Note, btw, that this is the topology of a surface of constant coordinate time in Kruskal coordinates.

Ok yes, since in Kruskal chart any line of constant coordinate time $T$ that intersects one of the two boundary hyperboles does not include the '$r=0$ 2-sphere'. Therefore those spacelike 'slices' have topology $R^3 - \{ 0,0,0\}$.

What about the lines of constant Kruskal coordinate time $T$ that do not intersect any boundary hyperbole ?

 

[PeterDonis]

Ok yes, since in Kruskal chart any line of constant coordinate time $T$ that intersects one of the two boundary hyperboles does not include the '$r=0$ 2-sphere'. Therefore those spacelike 'slices' have topology $R^3 - \{ 0,0,0\}$.

Actually, those are not single regions, they're two disconnected regions (because the boundary hyperbolas and the regions above/below them are not part of the manifold at all). Each individual disconnected region does have topology $S^2 \times R$.

Strictly speaking, what I said as I said it applies to surfaces of constant Kruskal coordinate time that don't intersect the boundary hyperbolas and so are in fact single connected regions. (In fact what I said applies to any spacelike hypersurface whatever that lies entirely within the manifold, i.e., does not intersect either boundary hyperbola.)

Note, btw, that the latter kind of hypersurface (i.e., one that doesn't intersect either boundary hyperbola) does differ geometrically from the former kind (one that does intersect one of the boundary hyperbolas): the latter kind of hypersurface has a minimum value of $r$, and extends from that minimum value to $r \to infty$ in both directions from that minimum value. The former kind of hypersurface just goes from $0 < r < \infty$ in one direction.

What about the lines of constant Kruskal coordinate time $T$ that do not intersect any boundary hyperbole ?

See above. No point anywhere in the manifold has $r = 0$.

 

[cianfa72]

Note, btw, that the latter kind of hypersurface (i.e., one that doesn't intersect either boundary hyperbola) does differ geometrically from the former kind (one that does intersect one of the boundary hyperbolas).

ok, spacelike hypersurfaces (possibly disconnected) with different geometry but same topology $S^2 \times R$.

I do not know if it might make sense apply Kruskal transformations as in the insight to Minkowski spacetime in spherical polar coordinates. Which kind of chart would we get ?

 

[PeterDonis]

I do not know if it might make sense apply Kruskal transformations as in the insight to Minkowski spacetime in spherical polar coordinates. Which kind of chart would we get ?

The standard Kruskal chart is in spherical polar coordinates. Each point in the standard Kruskal diagram, which represents the $T-X$ plane of the chart, represents a 2-sphere.

 

[cianfa72]

The standard Kruskal chart is in spherical polar coordinates. Each point in the standard Kruskal diagram, which represents the $T-X$ plane of the chart, represents a 2-sphere.

Yes, my point was map using Kruskal transformations the Minkowski spacetime not the Schwarzschild one (as explained in the insight what we get applying Kruskal transformations is the Schwarzschild spacetime in Kruskal chart).

 

[PeterDonis]

my point was map using Kruskal transformations the Minkowski spacetime not the Schwarzschild one

You can't; the Kruskal transformations only make sense for $M > 0$. For the $M = 0$ case, i.e., Minkowski spacetime, spherical polar coordinates are identical to "Kruskal" coordinates, with the "Kruskal" $X$ being the spherical polar $r$.

 

[cianfa72]

You can't; the Kruskal transformations only make sense for $M > 0$. For the $M = 0$ case, i.e., Minkowski spacetime, spherical polar coordinates are identical to "Kruskal" coordinates, with the "Kruskal" $X$ being the spherical polar $r$.

So for Minkowski spacetime (i.e. $M=0$ case) the "Kruskal" chart in $X,T$ diagram is defined only for the half plane $X \geq 0$.

 

[PeterDonis]

So for Minkowski spacetime (i.e. $M=0$) the "Kruskal" chart in $X,T$ diagram is defined only for the half plane $X \geq 0$.

Yes. That follows from the fact that the topology of Minkowsi spacetime is $R^4$, not $S^2 \times R^2$. 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.

 

[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.