# Schwarzchild radial coordinate

1. Jun 3, 2013

### maxverywell

The Schwarzchild spacetime can be foliated by 2-sphere, which are spacelike hypersurfaces of constant t and r (Schwarzchild coordinates) with a normal vector $\partial_t$ (outside the horizon). Because a 2-sphere has no center, the coordinate r is not the radius of the sphere and we consider it as a function of the total area of each such 2-sphere.

But why we can't consider these 2-spheres as eccentric spheres with center the r=0 singularity?
Isn't r the distance from the r=0 singularity? So the coordinate r can make a perfect sense as the radial coordinate, i.e. distance from the singularity.

And what happens inside the horizon? Did the 2-spheres become timelike?

2. Jun 3, 2013

### WannabeNewton

In $(\mathbb{R}^{3},d)$, where $d$ is the euclidean metric, we define the 2-sphere $S^{2}\subseteq \mathbb{R}^{3}$ as the set $S^{2} = \{x\in \mathbb{R}^{n}:d(x,0) = 1\}$. We call the distance $d(x,0) = 1$ the radius from the origin, which in this case is just the unit radius, and $0\in \mathbb{R}^{3}$ the center of the 2-sphere. Any other sphere is simply a translation and rescaling of the 2-sphere.

If the manifold is not $\mathbb{R}^{3}$ (or any euclidean space for that matter), then how will you make sense of things in the same way? A sphere embedded in a non-euclidean manifold need not even have a notion of a center and even if it did, what canonical metric would you endow on the manifold that would allow you to interpret the radial coordinate as a straight line distance from the center to points in the surface?

For any connected Riemannian manifold, there is always a metric we can endow on the manifold given by $d(x,y) = \inf \{L(\gamma):\gamma \text{ is a } C^{\infty} \text{ curve joining } x \text{ and } y\}$ where $L(\gamma)$ is the arc length. However there is absolutely nothing that says a priori this must give us the same notion of the radius of a sphere as the distance from some notion of a center in the same way the euclidean metric on $\mathbb{R}^{3}$ does.

Last edited: Jun 3, 2013
3. Jun 3, 2013

### maxverywell

How do we foliate the spacetime by 2-spheres?
Isn't the r=0 singularity the center of these spheres?

4. Jun 3, 2013

### WannabeNewton

No, how are you coming to that conclusion exactly?

See the section starting at the bottom of page 120 in Wald or section 5.2 of page 197 in Carroll to see how the foliation is achieved, and how the Schwarzschild coordinate system is constructed for points where $\xi^{a}$ and $\nabla^{a} r$ are not colinear (which fails in the strong field region).

5. Jun 3, 2013

### maxverywell

A falling observer will move from one 2-spheres of constant t and r to another 2-sphere in such a way that r decreases to r=0 (i.e. its area A decreases to 0). The distance from one 2-sphere to another is independent of $\theta$ and $\phi$, so the singularity is at the center of the black hole.

6. Jun 3, 2013

### WannabeNewton

1. There can be bound orbits in the Schwarzchild space-time so freely falling observers don't have to fall towards $r = 0$. Furthermore there can be hyperbolic trajectories in which observers free fall from spatial infinity, reach a turning point, and head back towards spatial infinity. There is no reason to think all time-like geodesic trajectories must terminate at $r = 0$. The existence of a singularity at $r = 0$ simply requires there exist some time-like geodesic that terminates at $r = 0$ i.e. there is geodesic incompleteness.

2. What notion of "distance" are you using? Again, the most natural metric on an arbitrary connected Riemannian manifold is the Riemannian distance function I gave in post #2 but I highly suspect this is what you have in mind when you say "distance".

3. What does the fact that $r = 0$ represents a physical singularity have anything to do with the question of why the radial coordinate cannot in general be interpreted as the straight line distance from some "center" to points on the surface of an orbit sphere generated by the action of the $SO(3)$ isomorphic subgroup of the space-time's isometry group on the space-time?

7. Jun 3, 2013

### pervect

Staff Emeritus
No, inside the horizon, $\partial_\theta$ and $\partial_\phi$ are spacelike. But $\partial_r$ becomes timelike inside the horizon, and $\partial_t$ becomes spacelike. Physicists will often say, informally, that "r and t switch roles", mathemeticians are usually more careful.

This picture leads to the idea of a black hole as being dynamic, since the timelike symmetry (represented by a Killing vector) that existed outside the horizon becomes spacelike inside the horizon. On further analysis, the geometry becomes a dynamic wormhole, the so-called Einstein-Rosen bridge. See MTW for details.

And dr does not measure distance. Inside the horizon, as we've seen, it's a spatial separation, not a time separation.

But even outside the horizon, you need to adjust the value of dr by the metric coefficient to get distances.

8. Jun 3, 2013

### yenchin

One way to see how things can become misleading if you are not careful with coordinate vs distance, is to look at metric which is not Schwarzschild. Consider the Einstein-Maxwell-Dilaton theory, one of its [charged] black hole solution is the Garﬁnkle-Horowitz-Strominger solution [Charged Black Holes in String Theory, Phys. Rev. D, 43(10): 31403143 (1991).]:
$ds^2 = - \left[1-\frac{2M}{r}\right]dt^2 + \left[1-\frac{2M}{r}\right]^{-1} + r\left[r-\frac{Q^2}{M}\right]d\Omega^2$.​
This black hole has the property that in the $t-r$ plane it looks exactly the same as that of Schwarzschild, but with smaller 2-sphere radius at fixed $r$ and fixed $t$.

As you increase charge $Q$, the radius of the 2-sphere appears to shrink and becomes degenerate at the extremal limit $r=M=Q$. Yet at the same time we see that the horizon remains always at $r=2M$, independent of charge. So what is going on? how can $r$ remain unchanged yet the 2-sphere is shrinking? The answer of course is that $r$ does not have geometrical meaning of distance, but only coordinate. By defining a new coordinate $R^2=r^2-\frac{Q^2}{M}r$, one could re-write the metric in the form
$ds^2 = - \sigma(R)^2 f(R)dt^2 + f(R)^{-1} dR^2 + R^2 d\Omega^2$. Then you will see that in the extremal limit, $R$ does shrink to zero, consistent with the area shrinkage.

9. Jun 3, 2013

### Staff: Mentor

This is an unusual use of the term "foliated". A foliation is normally taken to be a one-parameter family of spacelike hypersurfaces; but in order to fill the entire Schwarzschild spacetime with 2-spheres, you need a two-parameter family of them. (The Schwarzschild $t$ and $r$, btw, will *not* work as the two parameters, since Schwarzschild coordinates are singular at the horizon--there are an infinite set of 2-spheres at the horizon that all have the same $t$ and $r$. You need to pick coordinates that aren't singular at the horizon, such as Painleve, Eddington-Finkelstein, or Kruskal. In any such coordinates, each 2-sphere has *two* unique coordinate labels, not one.)

They aren't "eccentric" if they all have the same center, are they? However, this still won't work because the r=0 singularity is a spacelike line, not a point (or a timelike line). A spacelike line can't be the center of a family of spheres, at least not in any usual sense of the word "center".

No, they're still spacelike; but curves of constant $r$ inside the horizon are *also* spacelike, and curves of constant Schwarzschild $t$ inside the horizon are timelike. So if you try to view a 2-sphere inside the horizon as being embedded in a surface of "constant time", it won't be a surface of constant $t$ (it could be a surface of constant $r$, though there are also other ways of defining surfaces of constant time inside the horizon, as in the other coordinate charts I mentioned).

10. Jun 3, 2013

### Staff: Mentor

Note that this is only true in Schwarzschild coordinates; in other charts (such as Painleve or Eddington-Finkelstein), $\partial_r$ remains spacelike inside the horizon.

11. Jun 3, 2013

### maxverywell

I wanted to say concentric spheres. The R^3 can be foliated in this way.
However spacetime is 4 dimensional and I was trying to understand/visualize where these 2-spheres (of constant r and t) are located. I thought that they would be concentric, somehow... But now I have realized that this doesn't make sense in 4d.

12. Jun 3, 2013

### Staff: Mentor

Yes, but a spacelike slice through Schwarzschild spacetime does not have the topology of R^3. It has the topology of R x S^2, because the singularity at r = 0, strictly speaking, is not part of the spacetime at all; it is only definable as a limit point of ingoing geodesics.

I believe there is still a useful definition of "concentric" that allows a spacelike slice taken out of Schwarzschild spacetime to be foliated by concentric 2-spheres, but the foliation will have the central point (the "2-sphere" at r = 0, which degenerates to a point) missing--see above.

However, as you note, all this applies only to a single spacelike slice; the full 4-D spacetime is different.

13. Jun 3, 2013

### WannabeNewton

Peter makes another good point. The topology of the spacelike hypersurfaces is that of $\mathbb{R}\times S^{2}$ (i.e. the product topology of the 2-sphere and the real line), not $\mathbb{R}^{3}$. This goes back to what I said in post #2. The manifold structure of the spacelike hypersurfaces is not euclidean; the notion of the radius of a sphere as being the straight line distance from the center of the sphere to points on its surface relies heavily on the fact that we are using the euclidean metric on $\mathbb{R}^{3}$. Yenchin's example is a brilliant example of why we cannot interpret the radial coordinate in such a geometric way.

14. Jun 3, 2013

### pervect

Staff Emeritus
Yes. The second fundamental confusion of calculus comes into play here. Suppose we have coordinates (t1,r1) and (t2,r2). Then $\partial_{t1}$ is not necessarily equal to $\partial_{t2}$ even if the coordinate transformation between t1 and t2 specifies that t1=t2.

My remarks wrere specifically about the Schwarzschild coordinates.