- #1
- 45,370
- 22,523
PeterDonis submitted a new PF Insights post
The Schwarzschild Geometry: Part 3
Continue reading the Original PF Insights Post.
The Schwarzschild Geometry: Part 3
Continue reading the Original PF Insights Post.
fresh_42 said:There's a little error in the first third "Here will will assume that the GR model holds for all ##r>0##."
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 said:Why do you think this is an error?
PAllen said:I find it useful to look at lines of constant X
PAllen said: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 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.PeterDonis said: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.)
PeterDonis said: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 said:Consider a line of constant T...
PAllen said:I am going to write a follow on post on some additional physical interpretation for this congruence.
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.PAllen said: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.
should that be "are timelike (..."?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/
TGlad said:should that be "are timelike (..."?
By two boundary hyperbolas I assume you mean the two thick lined hyperbolas at r=0.
TGlad said:should that be "are timelike (..."?
TGlad said:By two boundary hyperbolas I assume you mean the two thick lined hyperbolas at r=0.
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.PAllen said: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).
Yes.cianfa72 said:Since lines of constant ##X## in KS chart are timelike, they actually define a congruence of observers having those as their own worldlines.
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.)cianfa72 said:In their paths through spacetime they 'reach' 2-spheres of minimum and maximum radius.
There is no spherical body. The spacetime shown in the Kruskal chart is vacuum everywhere.cianfa72 said:we can image a family of observers starting from each point on the surface of the spherical body.
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 said:Why it shows that the exterior of spherical body has topology ##S^2 \text {x} R^2## ?
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.PeterDonis said: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##.
Yes, you have just stated in more detail the reason why the statement of mine that you quoted is true.cianfa72 said: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##.
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##.cianfa72 said: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, 4D Minkowski space minus a line. See above. You're just replacing circles with 2-spheres in the above construction.cianfa72 said: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".
Ok yes, it makes sense.PeterDonis said: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##.
Note, btw, that this is the topology of a surface of constant coordinate time in Kruskal coordinates.PeterDonis said: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##).
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\}##.PeterDonis said:Note, btw, that this is the topology of a surface of constant coordinate time in Kruskal coordinates.
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##.cianfa72 said: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\}##.
See above. No point anywhere in the manifold has ##r = 0##.cianfa72 said:What about the lines of constant Kruskal coordinate time ##T## that do not intersect any boundary hyperbole ?
ok, spacelike hypersurfaces (possibly disconnected) with different geometry but same topology ##S^2 \times R##.PeterDonis said: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 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 said: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 ?
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 said: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.
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 said:my point was map using Kruskal transformations the Minkowski spacetime not the Schwarzschild one
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 said: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##.
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 said: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##.
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).PeterDonis said: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.
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##.cianfa72 said: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).
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.cianfa72 said: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)##.
So ##X## as radial coordiante in spherical coordinate can assume negative values ? (i.e. 2-spheres labeled with negative ##X## hence negative area) ?PeterDonis said: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.
ok got it !PeterDonis said: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).
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.)cianfa72 said:So ##X## as radial coordiante in spherical coordinate can assume negative values ?
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 said:(i.e. 2-spheres labeled with negative ##X## hence negative area) ?
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 said: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.
The Schwarzschild geometry is a mathematical model that describes the curvature of spacetime around a non-rotating massive object, such as a black hole. It is important because it provides a framework for understanding the effects of gravity on the motion of objects and the behavior of light in the presence of massive objects.
The Schwarzschild geometry differs from Newtonian gravity in that it takes into account the curvature of spacetime caused by massive objects, whereas Newtonian gravity assumes a flat spacetime. This means that the predictions of the Schwarzschild geometry are more accurate in extreme gravitational situations, such as near black holes.
Yes, the Schwarzschild geometry can be applied to any non-rotating massive object, not just black holes. This includes planets, stars, and even galaxies. However, the effects of the curvature of spacetime will be more pronounced for objects with larger masses and stronger gravitational fields.
The Schwarzschild geometry has many practical applications, including predicting the motion of objects in the vicinity of massive objects, such as planets orbiting a star or satellites orbiting a planet. It is also used in the study of gravitational lensing, where the curvature of spacetime can bend and distort the path of light from distant objects.
Yes, the Schwarzschild geometry is a simplified model and does not take into account other factors such as the rotation of the object or the effects of other objects in the vicinity. It also does not account for quantum effects, which are necessary to fully understand the behavior of matter and energy in extreme gravitational situations. Therefore, it is not a complete theory of gravity and is often used in conjunction with other theories, such as general relativity, to make more accurate predictions.