# I Spatial geometry of constant Schwarzschild time hypersurface

1. Jun 26, 2016

### MeJennifer

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

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!

Last edited by a moderator: Jun 27, 2016
2. Jun 26, 2016

### Staff: Mentor

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 $t$.

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.)

3. Jun 26, 2016

### MeJennifer

4. Jun 26, 2016

### Staff: Mentor

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.

5. Jun 26, 2016

### MeJennifer

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.

6. Jun 26, 2016

### Staff: Mentor

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.

The Flamm paraboloid has nothing to do with coordinates.

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.

7. Jun 26, 2016

### MeJennifer

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)?

8. Jun 26, 2016

### Staff: Mentor

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.

Last edited: Jun 26, 2016
9. Jun 26, 2016

### MeJennifer

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

10. Jun 26, 2016

### Staff: Mentor

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:

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:

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:

And I responded in post #17 with:

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

11. Jun 26, 2016

### MeJennifer

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

12. Jun 26, 2016

### Staff: Mentor

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.

13. Jun 26, 2016

### MeJennifer

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?

14. Jun 26, 2016

### Staff: Mentor

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.

15. Jun 26, 2016

### MeJennifer

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?

Last edited: Jun 26, 2016
16. Jun 26, 2016

### Staff: Mentor

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

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.)

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

17. Jun 26, 2016

### MeJennifer

So it is not one of constant Schwarzschild time?

18. Jun 26, 2016

### Staff: Mentor

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.)

19. Jun 26, 2016