I Spatial geometry of constant Schwarzschild time hypersurface

[MeJennifer]

[Moderator's note: this thread is spun off from another thread since it was a subthread dealing with a separate topic.]

But strictly speaking, the maximally extended spacetime has another exterior region, so the full spacelike surface is actually two paraboloids joined at the $r = 2M$ surface. There is no "hole in spacetime" at $r = 2M$.

There is definitely a maximally extended spacetime but there is no maximally extended spacelike surface of constant Schwarzschild coordinate time t. The spatial curvature (which is coordinate dependent) is infinite at r=2M.

It seems things are conflated here!

 

[PeterDonis]

there is no maximally extended spacelike surface of constant Schwarzschild coordinate time t.

Sure there is. Look at a Kruskal diagram, such as the one shown here:

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

Every straight line from region I through the center point to region III (i.e., with slope more horizontal than 45 degrees and passing through the center point) is a maximally extended line of constant Schwarzschild $t$. When you add back the two other spatial dimensions (each point in the diagram represents a 2-sphere), you get a maximally extended spacelike hypersurface of constant $t$.

The spatial curvature (which is coordinate dependent) is infinite at r=2M.

First, the spatial curvature of a given hypersurface is not coordinate-dependent. What is coordinate-dependent is which hypersurface is a hypersurface of "constant time".

Second, the spatial curvature of a hypersurface of constant Schwarzschild $t$ is finite at the horizon $r = 2M$ (in Schwarzschild coordinates). Do the math. It might help to do it in a chart, such as isotropic coordinates, in which $g_{rr}$ does not have a coordinate singularity at the horizon. (In this chart the horizon is at $\bar{r} = M / 2$, where I have put a bar over the radial coordinate to emphasize that it is a different chart--it just happens to have the same hypersurfaces of constant time as the Schwarzschild chart does.)

 

[MeJennifer]

Sure there is. Look at a Kruskal diagram, such as the one shown here:
https://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

I was talking about the coordinates you brought forward not KS coordinates.

 

[Nugatory]

I was talking about the coordinates you brought forward not KS coordinates.

Ahhh... PeterDonis isn't talking about KS coordinates, he's suggesting that you look at the curves of constant Schwarzschild $t$ in a KS diagram.

 

[MeJennifer]

Ahhh... PeterDonis isn't talking about KS coordinates, he's suggesting that you look at the curves of constant Schwarzschild $t$ in a KS diagram.

Yes there is no problem with KS coordinates.

But eh, we are talking about Flamm's paraboloid right? And in those coordinates the surface of constant t can no longer remain spacelike passed the event horizon.

 

[PeterDonis]

we are talking about Flamm's paraboloid right?

We are talking about the spatial geometry of a spacelike hypersurface of constant Schwarzschild coordinate time in the maximally extended Schwarzschild geometry. The Flamm paraboloid is only a way of helping you to visualize that geometry; it does not define the geometry.

in those coordinates

The Flamm paraboloid has nothing to do with coordinates.

the surface of constant t can no longer remain spacelike passed the event horizon.

Whether or not a hypersurface is spacelike is independent of coordinates. And if you do the math, as I have already asked you to, you will find that the following is true about a spacelike hypersurface of constant Schwarzschild coordinate time in the maximally extended Schwarzschild geometry:

(1) The hypersurface has topology $S^2 \times R$. (Note that this is not $R^3$; this hypersurface does not have the same topology as ordinary Euclidean 3-space. That means you have to be very careful using your spatial intuitions here.)

(2) The hypersurface can be foliated by an infinite series of 2-spheres. If we label each 2-sphere by its Kruskal $X$ coordinate (the "horizontal" coordinate in the Kruskal diagram), we find that the area of the 2-sphere at $X = 0$ is $16 \pi M^2$, where $M$ is the mass of the hole (note that this is the hole's horizon area--this 2-sphere is the "center point" of the Kruskal diagram). The areas of the 2-spheres for $X > 0$ get larger as $X$ gets larger, increasing without bound as $X \rightarrow \infty$ (this is the portion of the hypersurface in region I of the Kruskal diagram); and the areas of the 2-spheres for $X < 0$ get larger as $X$ gets smaller (more negative), increasing without bound as $X \rightarrow - \infty$ (this is the portion of the hypersurface in region III of the Kruskal diagram).

(3) The hypersurface has no portion whatsoever in regions II or IV of the Kruskal diagram; that is, it never goes below the horizon. (There are hypersurfaces of constant Schwarzschild coordinate time that extend through regions II and IV, but they are not spacelike; they are a different family of hypersurfaces altogether from the ones we are considering.) Instead, as item #2 shows, the hypersurface consists of two exterior regions, containing 2-spheres with increasing area, joined by a single 2-sphere with area equal to the hole's horizon area.

(4) All curvature invariants on the hypersurface are finite everywhere. (As I noted before, it might be easier to verify this in a chart which does not have a coordinate singularity in $g_{rr}$ at the horizon, such as isotropic coordinates. But invariants are invariants, so if they are finite in one chart, they are finite period.)

All of the statements that I have made above are statements that have probably been verified thousands of times by GR students doing homework problems; there is nothing whatsoever that is in doubt or contentious about any of them. Therefore, I strongly suggest that, before you post again on this topic, you either verify the above statements for yourself, or have an absolutely ironclad mathematical demonstration that they are incorrect. Just asserting that you think they are wrong, even if you do, won't do at this point; all it will get you is a warning.

 

[MeJennifer]

All of the statements that I have made above are statements that have probably been verified thousands of times by GR students doing homework problems; there is nothing whatsoever that is in doubt or contentious about any of them. Therefore, I strongly suggest that, before you post again on this topic, you either verify the above statements for yourself, or have an absolutely ironclad mathematical demonstration that they are incorrect. Just asserting that you think they are wrong, even if you do, won't do at this point; all it will get you is a warning.

Ok, then let me just ask questions!

• Does Flamm's paraboloid demonstrate spatial curvature (no, I am not writing spacetime curvature)? Yes or no?
• What is this spatial curvature at r=2M?
• Could you demonstrate how those, what you call, fully extended Flamm's paraboloids are mathematically constructed (no, I am not talking about the fully extended spacetime I have no issue with that)?

 

[PeterDonis]

let me just ask questions!

Sorry, you've gone beyond that point. Your only option now, at least as far as I'm concerned, is to do the math. Doing that will enable you to answer your questions for yourself.

Just one note: I've already said that the Flamm paraboloid is just a way of visualizing the spacelike hypersurfaces of constant Schwarzschild coordinate time; it is not how those surfaces are defined. If you persist in thinking in terms of the Flamm paraboloid, instead of looking at the actual math of spacelike hypersurfaces of constant Schwarzschild coordinate time in their own right, independent of any visualization, you are going to continue making mistakes.

 

[MeJennifer]

If you persist in thinking in terms of the Flamm paraboloid,...

I am not!
I was merely commenting on a statement you made somehow connecting Flamm's paraboloid with the maximally extended Schwarzschild spacetime.

 

[PeterDonis]

I was merely commenting on a statement you made somehow connecting Flamm's paraboloid with the maximally extended Schwarzschild spacetime.

No, you were making incorrect statements about surfaces of constant Schwarzschild coordinate time in the maximally extended Schwarzschild spacetime. Then you started asking about the Flamm paraboloid even after I had said that that was not a good thing to focus on for the subthread you and I were having.

Here is what I said earlier in the thread (post #7, before the subthread between you and me started) about the Flamm paraboloid and surfaces of constant Schwarzschild coordinate time:

What is correct for a black hole is that the spacelike surfaces of constant Schwarzschild coordinate time $t$ in the exterior of Schwarzschild spacetime (i.e., above the horizon), whose spatial geometry is described by the Flamm paraboloid, do not extend below the horizon. This is usually reflected by cutting off the paraboloid at $r = 2M$. But strictly speaking, the maximally extended spacetime has another exterior region, so the full spacelike surface is actually two paraboloids joined at the $r = 2M$ surface. There is no "hole in spacetime" at $r = 2M$.

This post of mine was responding to Yukterez' incorrect statement that spacetime is somehow "cut off" at $r = 2M$, which you also responded to in post #5 (and I agree with what you said in that post). Yukterez' statement is not true either for a black hole spacetime (what I posted about in post #7) or the spacetime describing an ordinary spherically symmetric object (what you posted about in post #5 and Orodruin posted about in post #6). Also note carefully my description of the relationship between the Flamm paraboloid and the spatial geometry of the full spacelike surface of constant Schwarzschild coordinate time in the maximally extended spacetime.

You responded in post #12 to my post #7, quoted above, with this incorrect statement:

There is definitely a maximally extended spacetime but there is no maximally extended spacelike surface of constant Schwarzschild coordinate time t. The spatial curvature (which is coordinate dependent) is infinite at r=2M.

When I responded in post #13 to correct this, I didn't mention the Flamm paraboloid at all. The next mention of it in this subthread was by you, in post #16:

we are talking about Flamm's paraboloid right?

And I responded in post #17 with:

We are talking about the spatial geometry of a spacelike hypersurface of constant Schwarzschild coordinate time in the maximally extended Schwarzschild geometry. The Flamm paraboloid is only a way of helping you to visualize that geometry; it does not define the geometry.

This should make it clear that I'm not talking about the Flamm paraboloid in this subthread. If you aren't either, good.

[Edit: I have spun off this subthread into a separate thread.]

 

[MeJennifer]

I really don't think we have any disagreement regarding KS coordinates.

 

[PeterDonis]

I really don't think we have any disagreement regarding KS coordinates.

The statements I made in posts #2 and #6 of this thread are independent of any coordinates. I only used the Kruskal diagram and the Kruskal $X$ coordinate to help clarify what I was describing.

 

[MeJennifer]

(4) All curvature invariants on the hypersurface are finite everywhere. (As I noted before, it might be easier to verify this in a chart which does not have a coordinate singularity in $g_{rr}$ at the horizon, such as isotropic coordinates. But invariants are invariants, so if they are finite in one chart, they are finite period.)

I ask a question, as statements may generate warnings on this forum, do you identify a difference between the concept of spatial curvature and spacetime curvature?

 

[PeterDonis]

do you identify a difference between the concept of spatial curvature and spacetime curvature?

Of course. The curvature of a spacelike hypersurface is a tensor on a 3-dimensional Riemannian manifold; the curvature of spacetime is a tensor on a 4-dimensional pseudo-Riemannian manifold. Obviously these are not the same thing.

 

[MeJennifer]

Of course. The curvature of a spacelike hypersurface is a tensor on a 3-dimensional Riemannian manifold; the curvature of spacetime is a tensor on a 4-dimensional pseudo-Riemannian manifold. Obviously these are not the same thing.

Ok, so if we consider the spatial curvature in Schwazschild coordinates at r=2M what would you consider this spatial curvature to be?

By the way how do you chose a time coordinate t for r< 2M?
Isn't it true that spacetime is no longer static for r<2M, so don't we have dynamic spacelike hypersurfaces?

 

[PeterDonis]

Ok, so if we consider the spatial curvature in Schwazschild coordinates at r=2M what would you consider this spatial curvature to be?

As I've already said, work it out for yourself.

how do you chose a time coordinate t for r< 2M?

We are talking about a single spacelike hypersurface. There is no "time coordinate". Also, this hypersurface does not include any 2-spheres for which $r < 2M$. (That was already implicit in post #6 of this thread. You aren't reading carefully enough. And don't ask me how this can be true. Work it out for yourself.)

Isn't it true that spacetime is no longer static for r<2M, so don't we have dynamic spacelike hypersurfaces?

This is true, but it is irrelevant to the current discussion. Stop asking irrelevant questions and do the math.

 

[MeJennifer]

We are talking about a single spacelike hypersurface. There is no "time coordinate".

So it is not one of constant Schwarzschild time?

 

[PeterDonis]

don't ask me how this can be true. Work it out for yourself.

I'll give one hint for this, though, along the lines I've already hinted at by suggesting that you work the math in isotropic coordinates. The metric for a single spacelike hypersurface of constant time in these coordinates (these surfaces are the same as for standard Schwarzschild coordinates) is:

$$
ds^2 = \left( 1 + \frac{M}{2\bar{r}} \right)^4 \left( d\bar{r}^2 + \bar{r}^2 d \Omega^2 \right)
$$

where $d \Omega^2$ is the usual line element on a 2-sphere. The horizon in this chart is at $\bar{r} = M / 2$.

Here are some "food for thought" questions about the above metric:

(1) Consider the area $A$ of a 2-sphere as a function of the radial coordinate $\bar{r}$ in this chart. (A 2-sphere is a subspace of constant $\bar{r}$; its area is easy to read off from the above line element, you just set $d\bar{r} = 0$ and integrate over the full range of $d\Omega^2$ to get a factor of $4 \pi$.) What is the derivative of $A$ with respect to $\bar{r}$?

(2) What range of values does $A$ take for the range $0 < \bar{r} < \infty$? (Don't guess; compute from the answer to #1 above.)

 

[PeterDonis]

So it is not one of constant Schwarzschild time?

Read what I've already posted. The answer is there. Stop asking irrelevant questions and work the math out for yourself.

 

[MeJennifer]

My opening post is edited and now taken completely out of context.

 

[PeterDonis]

My opening post is edited and now taken completely out of context.

What is now the opening post in this thread was previously a hijack of another thread. The only context that was needed was the quote from the post of mine that you responded to, which is included in the OP of this thread. I didn't edit your post at all except to add the moderator's note at the top.