I Schwarzschild spacetime in Kruskal coordinates

[cianfa72]

TL;DR

Line element for Schwarzschild spacetime in Kruskal coordinate chart

As explained here in Kruskal coordinates the line element for Schwarzschild spacetime is:

$$ds^2 = \frac{32 M^3}{r} \left( – dT^2 + dX^2 \right) + r^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)$$
My simple question is: why in the above line element are involved 5 coordinates and not just 4 ? Thanks.

 

[Dale]

In this expression $r$ is not a coordinate (notice that there is no $dr$). Instead, $r$ is just shorthand for
$$r=2GM\left( 1+W_0\left(\frac{X^2-T^2}{e}\right)\right)$$
See

https://en.m.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

 

[cianfa72]

Ah ok, but in the Kruskal diagram ($X$ as horizontal axis and $T$ as vertical one) loci of constant $r$ are those of constant Schwarzschild radius $r$, right ?

 

[Orodruin]

Ah ok, but in the Kruskal diagram ($X$ as horizontal axis and $T$ as vertical one) loci of constant $r$ are those of constant Schwarzschild radius $r$, right ?

The function $r(X,T)$ is the Schwarzschild $r$ coordinate. (Not to be confused with the Schwarzschild radius!)

 

[Dale]

Ah ok, but in the Kruskal diagram ($X$ as horizontal axis and $T$ as vertical one) loci of constant $r$ are those of constant Schwarzschild radius $r$, right ?

Yes, the $r$ in the KS expression at any event is the same value as the Schwarzschild $r$ at that same event.

Edit: @Orodruin for the win!

 

[cianfa72]

Another point: I have not a clear understanding why hypersurfaces of constant $X$ in the KS diagram start as disconnected pieces then connected through the wormhole $T=0$.

 

[Orodruin]

Do you mean surfaces of constant $T$? That’s just what the spacetime looks like in those coordinates.

 

[cianfa72]

Do you mean surfaces of constant $T$? That’s just what the spacetime looks like in those coordinates.

From that insight, hypersurface $T=0$ i.e. Schwarzschild coordinate $t=0$ should be spacelike with topology $S^2 \times R$ (the wormhole).

This is sometimes described as the wormhole “pinching off” before anything can travel through it; this interpretation comes from looking at surfaces of constant $X$ from the bottom to the top of the diagram. These surfaces start out as two disconnected pieces, which then get connected by the wormhole, which increases from zero radii (at the bottom hyperbola) to radius $2M$ (at the origin), then decreases back to zero radii (at the top hyperbola), and the surfaces then split back into two disconnected pieces.

What does it mean?

 

[Ibix]

From that insight, hypersurface $T=0$ i.e. Schwarzschild $t=0$ should be spacelike with topology $S^2 \times R$ (the wormhole).What does it mean?

Each point on the Kruskal diagram is one point on a sphere with surface area $4\pi r^2$, right? Draw a horizontal line somewhere on a Kruskal diagram. Everywhere it crosses a hyperbola of constant $r$, draw a little circle of that radius - that's the equator of the sphere.

If you drew a line that intersects one of the singularities then you'll have drawn two sets of circles that go to zero at the singularity. You could nest each circle inside the next larger one, and you get two surfaces each covered by a polar coordinate system (they're not Euclidean planes, since the radii don't work out).

If you drew the line so it didn't intersect a singularity then you'll have drawn one set of circles that get smaller towards the middle of the diagram, reach a minimum size, then start to grow again. As the circles get smaller you can nest them inside each other, but when the next circle is larger than the smallest one, where are you going to put it? The solution is to think of it as two funnels connected at their narrowest points - i.e., a wormhole.

 

[cianfa72]

Each point on the Kruskal diagram is one point on a sphere with surface area $4\pi r^2$

Yes, each point on KS diagram represents a 2-sphere.

if you drew a line that intersects one of the singularities then you'll have drawn two sets of circles that go to zero at the singularity. You could nest each circle inside the next larger one, and you get two surfaces each covered by a polar coordinate system (they're not Euclidean planes, since the radii don't work out).

Why two set of circles? One set for each line that intersect the top or the bottom of singularity?

 

[Ibix]

Why two set of circles? One set for each line that intersect the top or the bottom of singularity?

Once your line crosses the singularity it's left spacetime. The radius isn't defined there - so you've got a set of circles on the left of the singularity,a set of circles on the right and a gap between, so they're not connected.

 

[cianfa72]

Once your line crosses the singularity it's left spacetime. The radius isn't defined there - so you've got a set of circles on the left of the singularity, a set of circles on the right and a gap between, so they're not connected.

Ah ok, but the line you're talking about is an horizontal line as in the first part of your post #9 or is it just a generic (not straight) line ?

 

[Ibix]

Ah ok, but the line you're talking about is an horizontal line as in the first part of your post #9 or is it just a generic (not straight) line ?

I was talking about a horizontal line (constant Kruskal-Szekeres $T$) yes.

 

[cianfa72]

I was talking about a horizontal line (constant Kruskal-Szekeres $T$) yes.

Ah ok, I was confused since the insight talked of hypersurfaces of constant $X$ and not constant $T$.

 

[Ibix]

Ah ok, I was confused since the insight talked of hypersurfaces of constant $X$ and not constant $T$.

Indeed, it says "this interpretation comes from looking at surfaces of constant $X$ from the bottom to the top of the diagram". The $X$ is a typo, I think @PeterDonis? It should be a set of surfaces of constant $T$ at different vertical positions moving from bottom to top.

 

[PeterDonis]

Indeed, it says "this interpretation comes from looking at surfaces of constant $X$ from the bottom to the top of the diagram". The $X$ is a typo, I think @PeterDonis?

Yes, it should say constant $T$. I will edit the Insight to fix. Thanks for the catch!

 

[PeterDonis]

I will edit the Insight to fix.

Now edited.

 

[cianfa72]

Ok, the spacelike hypersurface $T=0$ has topology $S^2 \times R$. Now if we slice up the spacetime diagram with horizontal straight lines (i.e. constant $T$), each spacelike hypersurface still has got $S^2 \times R$ topology ?

 

[Ibix]

Ok, the spacelike hypersurface $T=0$ has topology $S^2 \times R$. Now if we slice up the spacetime diagram with horizontal straight lines (i.e. constant $T$), each spacelike hypersurface still has got $S^2 \times R$ topology ?

The problem is that once you get above $T=1$ a line of constant $T$ actually exits spacetime. So it is spacelike at each end and nothing at all in the middle - not spacelike, not null, not timelike. The surface represented by each spacelike part is $S^2\times R$, yes, but you have two disjoint surfaces of that topology.

 

[cianfa72]

The surface represented by each spacelike part is $S^2\times R$, yes, but you have two disjoint surfaces of that topology.

Ok, so how do we come to the conclusion that this spacetime has topology $S^2 \times R^2$ ?

 

[Ibix]

Ok, so how do we come to the conclusion that this spacetime has topology $S^2 \times R^2$ ?

I would say the obvious thing to do is switch to a conformal diagram (aka Penrose diagram), which turns the singularities into horizontal lines, then slice along horizontal lines in that chart.

 

[PeterDonis]

Ok, so how do we come to the conclusion that this spacetime has topology $S^2 \times R^2$ ?

By not looking at any particular coordinate surfaces, which are irrelevant to the topology of the spacetime. The spacetime itself, as should be obvious from any diagram of the Kruskal or Penrose type, is an $R^2$ worth of 2-spheres, since each point within the region of the diagram that is occupied by the spacetime, which is an $R^2$ region, represents a 2-sphere, i.e., an $S^2$. So the overall topology is obviously $S^2 \times R^2$.

 

[cianfa72]

since each point within the region of the diagram that is occupied by the spacetime, which is an $R^2$ region

This region is an open region of $R^2$, hence it is homeomorphic to the entire plane $R^2$. From this we get the $S^2 \times R^2$ topology.

 

[PeterDonis]

This region is an open region of $R^2$, hence it is homeomorphic to the entire plane $R^2$. From this we get the $S^2 \times R^2$ topology.

Yes.

 

[cianfa72]

Another general question: take a timelike path starting from point/event A to event B. B is of course inside the future light cone centered in A. If we consider all the events along the timelike path their future light cones will always be "inside" the light cone in A, I believe.

 

[PeterDonis]

Another general question: take a timelike path starting from point/event A to event B. B is of course inside the future light cone centered in A. If we consider all the events along the timelike path their future light cones will always be "inside" the light cone in A, I believe.

Yes. In fact this is one way of defining the ordering relation of the events along the timelike path: inclusion of future light cones.

 

[cianfa72]

In fact this is one way of defining the ordering relation of the events along the timelike path: inclusion of future light cones.

So along the timelike path, the light cones centered on the events along it in a such way "restrict" the set of events possibly reachable from event A.

 

[PeterDonis]

So along the timelike path, the light cones centered on the events along it in a such way "restrict" the set of events possibly reachable from event A.

Reachable from the event at which you are evaluating the light cones. If you are somewhere to the future of event A along some particular timelike path, the set of events reachable from the event where you are is a proper subset of the events reachable from event A. But that is not a restriction on the events reachable from event A. It's a restriction on the events reachable from the event where you are.

 

[cianfa72]

If you are somewhere to the future of event A along some particular timelike path, the set of events reachable from the event where you are is a proper subset of the events reachable from event A. But that is not a restriction on the events reachable from event A

Yes, my term "restriction" meant to say a proper subset of all events reachable from event A.

 

[cianfa72]

From the Insight part 3

But what if we trace its worldline backward, to negative values of $T$ ? What we will find is that, no matter how negative $T$ gets, the worldline never reaches $r = 2 M$ ; it just approaches it asymptotically, in much the same way that $r$ approached $2M$ asymptotically in Schwarzschild coordinates for $t → ∞$.

$T$ should be the Gullstrand-Painleve coordinate time. From which region is the free-falling Painleve observer traced backward ? I believe its path should be in the past light cone centered at the event its worldline is traced backward.