# Meaning of r in Schwarzchild coordinates

1. Jul 25, 2015

### PWiz

I'm trying to understand *quote unquote thread title* by performing some simple (heuristic) analysis on my own. Before beginning, I'd like to present what I've been given to understand here at PF:
-r is not the distance from the center of a spherical shell to an arbitrary spatial coordinate on the shell surface (I don't know if one could even properly define the center of a sphere in curved spacetime, but I'm not too sure on this one)
-if two spherical masses have radii $r_1$ and $r_2$ respectively, the distance between the surfaces of these two masses is not equivalent to $|r_1 - r_2|$ (again this seems pretty obvious because space[time] may be curved, but I'd like to derive this fact nonetheless)

So let's look at a feature of the Schwarzchild metric. Since the metric is diagonalized (it describes non-rotating masses, and hence describes static spacetime geometries), it means that there is no need to discriminate between hypersurfaces. (I guess I could do without this statement, but spacetime geometries may not be static/invariant under a time reversal, which I think is the case for the Kerr metric, so I'm putting this in just to be sure) If we investigate hypersurfaces with a constant r coordinate, the metric incidentally reduces to $ds^2 = r^2 (d \theta^2 + sin^2 \theta d\phi^2)$, which exactly describes the surface of a 2-sphere in standard spherical coordinates. The only reasonable deduction that I'm able to make with this is that the set of all 2-spheres surfaces in Schwarzchild coordinates represent the family of hypersurfaces with a fixed r coordinate (or is it the other way around?). I am refraining from changing '2-sphere surfaces' to simply '2-spheres' in the sentence above because I don't think the volumes in a flat and non-flat manifold commute (if memory serves right, I think the amount of deviation is what the contracted Riemann tensor really measures).

Is my reasoning correct? If so, then how do I define r? In terms of a geometric property that associates it with a 2-sphere surface, or maybe even a boundary of a circle?

And now about the difference in radii not being a trivial subtraction:
if I look at the proper distance between two r coordinates $r_1$ and $r_2$ (on an arbitrary submanifold of constant t), the difference takes the form $$s= \int_{r_1}^{r_2} \frac {1} {\sqrt{1- \frac {2GM}{rc^2}}} dr$$. Now I can go ahead and try to show that this does not, in general, equal to $|r_1 - r_2|$, but before I go through the pains of evaluating this integral, I want to know if this is the correct approach to begin with.

P.S. I'm a little tired right now, so please excuse any flagrant mistakes I might've accidently made in this post.

Last edited: Jul 25, 2015
2. Jul 25, 2015

### fzero

I would say that the coordinate $r$ is perfectly defined outside the Schwarschild radius $R_S=2GM/c^2$. So if $r_1,r_2> R_S$, then your formula for the proper distance is correct. If we want to discuss proper distances for $r\leq R_S$, then we would need to use coordinates for which the metric is not singular on the domain of the coordinate.

3. Jul 25, 2015

### PWiz

Yeah, good point. I should've mentioned that I'm restricting the r coordinates as r1,r2>2GM/c^2 (I suppose one could use Eddington-Finkelstein coordinates and remove this condition altogether). I'll work out the integral and post it when my brain feels a bit more rejuvenated (haha).
And I know that r is well-defined, but I don't know what exactly it's defined as. Is my heuristic approach in my OP correct for determining a definiton for r? (Don't give away the defintion to me please. Let me try to reach it myself by giving, ahem, subtle hints :P)

4. Jul 25, 2015

### fzero

Your explanation seems reasonable. If we want to describe points outside the horizon, because of the spherical symmetry, we can put spherical shells at fixed values of $r$ and use the angles on the sphere and the fixed value of $r$ to identify the points. A coordinate system can be thought of as a way to map points like this. Then the notion of coordinate distance $r_2-r_1$ would be the distance between points at the same $(\theta,\phi)$ in our map. However, the proper distance between points is the one that is useful for studying kinematics. For instance if we wanted to know how long it would take to move from point to point at fixed velocity, then we would have to use the proper distance rather than the coordinate distance. The coordinate distance doesn't know anything about the effect of gravitational curvature on the motion of a test particle.

5. Jul 25, 2015

### PWiz

But what's the point of even defining a coordinate distance? The geometry of the manifold is going to change with r, so the idea doesn't seem useful at all. The only use that I can find of this notion is to set it to a constant to describe motion about the spherical surface (by exploiting the symmetry of the situation of course). Even then, since r is constant here, I don't see how the term coordinate distance helps.

About the r definition - I'm having a little trouble at this point. I want to connect the definition to a geometric property of a 2-sphere without referring to the volume. If the situation was being described in 2 (spatial) dimensions, I could just say that r is the coordinate which gives the expression for circumference of a circle at that coordinate as 2πr. How do I do this for 3 dimensions?

6. Jul 25, 2015

### PAllen

The same works. You've pointed out that the geometry (as described by the metric) of the 2 spheres is exactly standard. So they have great circles. So r is taken as circumference / 2 pi, exactly as you suggested. (Equivalently, r = √ (area / 4 pi) ).

7. Jul 25, 2015

### pervect

Staff Emeritus
Right.

Right, but what you may be missing is that the Schwarzschild metric is not applicable if you have two massive bodies. So your need a new metric to start with, trying to work it out in terms of the Schwarzschild metric is not a good approach. Furthermore, even before you start talking about the new metric, you need to figure out what's holding the two massive bodies in equilibrium. Gravity would tend to make them attract each other, if they wasn't something to oppose it. This is important because whatever force is holding them apart will contribut to the gravity via its stress-energy tensor. Apparently there is nowadays a solution for two charged massive bodies which are in equilibrium due to electrostatic repulsion, but I'm not familiar with it. It's a fairly new paper (2007).

http://arxiv.org/abs/0706.1981, originally published in Physical Review D

It's probably more than you want to get into at this point anyway, the short version is that your initial question is developing into a rather complex one, which is probably not what you really want at this point.

I'm not sure what you mean here, but I'll go along for the time being with it not being important. I can't reassure you more definitively on the point because I don't understand exactly what you're worried about.

The Schwarzschild space-time is spherically symmetric, the metric coefficients don't depend on the symmetries of the sphere, which are theta and phi, nor does it depend on time t, since the solution is static.

A general diagonal metric would be

-f1(t,r,theta,phi) dt^2 + f2(t,r,theta,phi) dr^2 + f3(t,r,theta,phi) d phi^2 + f4(t,r,theta,phi )d theta^2

Taking advantage of the natural symmetries , we can simplify this considerably to

-f1(r) dt^2 + f2(r) dr^2 + f3(r) d phi^2 + f4(r) dphi^2

The next simplification we can make is to set f4(r) to r^2 and f3(r) to r^2 sin^2 theta^2, which essentially means using the usual spherical coordinates theta and phi for the spherical hyper-shells, and scaling r such that the surface area of a spherical hypersurface with a constant r coordinate is 4 pi r^2. We don't have to make these choices. The point is that mathematically, there are lots of coordinate choices we could use to describe the surface of a sphere. But to make life easy, we choose familiar ones, theta and phi, the usual angular spherical coordinates.

This leaves us with a metric of the form

-f1(r) dt^2 + f2(r) dr^2 + r^2 d theta^2 + r^2 sin^2 theta dphi^2

Going further requires Einstein's field equations. Solving Einstein's field equations gives us f1(r) and f2(r) up to some integration constants, and the integration constants are set by making the metric far away from the gravitating source the usual Minkowskii metric as expressed in spherical coordinates, a condition that's usually called "asymptotic flatness".

For a more complete and rigorous approach, I'd recommend Wald, who takes a similar approach.

8. Jul 25, 2015

### PWiz

@PAllen Okay, thanks!
Ahhh, right, how could I forget. There isn't any closed form solution for the metric tensor in general for a two body problem right?
What I mean is that you can take a hypersurface at any constant time t without having to worry about the fact that the geometry could depend on your choice of (constant) t.

I remember going through the Schwarzchild metric derivation sometime ago. You just have to calculate the torsion free Christoffel symbols, plug it into the EFE for a vacuum solution, and calculate the final values for the coefficients by selecting those which reduce to the Poisson equation for Newtonian gravity, right?

9. Jul 25, 2015

### Staff: Mentor

What makes the metric static is not just that it's diagonalized, it's that it's diagonalized and none of the metric coefficents depend on $t$. A diagonal metric where the metric coefficients were functions of $t$ would not be static (for example, the FRW metric).

(Also, since $t$ is only timelike outside the horizon, the metric is only static outside the horizon; at and inside the horizon, it's not.)

You've left out the $t$ coordinate. Also, you haven't allowed for the $t$ coordinate being singular at the horizon. Here's a better way of saying it:

Because Schwarzschild spacetime is spherically symmetric, we can view it as an infinite set of 2-spheres, with each 2-sphere being labeled by two parameters. One obvious parameter is $r$, the circumference of the 2-sphere divided by $2 \pi$. The other parameter, in Schwarzschild coordinates, is $t$.

However, in Schwarzschild coordinates, there is a problem at the horizon, $r = 2M$; there are an infinite number of 2-spheres with this $r$, but all of them are labeled by the same $t$ coordinate, $+ \infty$ (or, if you like, their $t$ coordinate is undefined). So to really label all the 2-spheres properly, we need to find a coordinate chart that doesn't have that problem at the horizon. There are several possibilities (Painleve, Eddington-Finkelstein, or Kruskal--note that in the last of these, $r$ is not used as a parameter to label 2-spheres).

10. Jul 26, 2015

### stevebd1

Sample problem 2 on page 2-28 from the following link shows how to calculate the proper distance between 2 radii near a static BH-

http://www.eftaylor.com/pub/chapter2.pdf
'Exploring Black Holes' by E Taylor and J Wheeler

11. Jul 26, 2015

### PWiz

So what exactly does simply diagonalizing the metric do? Suggest that the geometry is invariant under coordinate reversals?
I went through the link, and it would appear that the integral in my OP is correct for determining the proper distance between the surfaces of two such massless spherical shells in the vicinity of a spherically symmetric, non-rotating mass, assuming both the shells have their r coordinates as being greater than 2GM/c^2 . Btw, I didn't really like the rest of the chapter that much - it's too simplistic, almost as if it's giving an introduction to children! (It isn't mathematically rigorous at all, but maybe I'm jumping the gun, as this is only the 2nd chapter of the book)

12. Jul 26, 2015

### Staff: Mentor

It's important to understand that the metric being diagonal is a property of the geometry in a specific coordinate chart, not just the geometry. The metric for Schwarzschild spacetime is not diagonal in every chart; for example, it isn't in Painleve or Eddington-Finkelstein coordinates. But the metric is still static regardless of your choice of coordinates.

A better way to define what it means for a metric to be static is: a static metric has a timelike Killing vector field that is hypersurface orthogonal. These are invariant geometric properties that don't depend on your coordinate choice. However, in any spacetime that has those properties, you will be able to find a coordinate chart with the properties we discussed earlier in this thread:

(1) The metric coefficients will not depend on the time coordinate (because there is a timelike Killing vector field);

(2) The metric will be diagonal (because the timelike KVF is hypersurface orthogonal).

So what diagonalizing the metric (i.e., finding a coordinate chart with the other two properties) does is make the staticity of the metric manifest, i.e., it makes it obvious by looking at the line element. But, again, the staticity is there whether we choose coordinates that make it manifest or not.

13. Jul 26, 2015

### PWiz

So speaking conversely, the metric describing non-static geometries cannot be diagonalized in any coordinate chart (except for the local one where the metric reduces to the Minkowski form)?

14. Jul 26, 2015

### Staff: Mentor

No. I already gave a counterexample: the FRW metric is diagonal in the standard FRW coordinates, but the FRW metric is not static.

15. Jul 26, 2015

### PWiz

Isn't that because the frame describing the FLRW metric itself is non-static? The FLRW metric is a special case because it employs comoving coordinates, which is why it's diagonalizable right? For example, can the Kerr metric be brought into an orthogonal form by transforming to a particular coordinate chart EXCEPT for the co-rotating/free falling one?

16. Jul 26, 2015

### Staff: Mentor

No. Staticity is a property of a metric, not of a coordinate chart. The FRW metric is non-static regardless of which coordinates you use to describe it.

The FRW metric does not employ comoving coordinates; the FRW coordinate chart does. Again, the FRW metric is non-static in any coordinate chart. Please don't confuse coordinate charts with metrics (spacetime geometries).

Which chart on Kerr spacetime are you referring to as "co-rotating/free falling"?

17. Jul 26, 2015

### PWiz

Ok, perhaps I was a little sloppy earlier. I'll try to clarify things.
What I'm trying to say is that the FRW metric appears diagonalized in the FRW metric coordinate chart because the Earth's frame coordinates are comoving right? Let's say I transform to some coordinate system which has a different preferred frame, such as the frame of reference of the Sun. The universe still looks isotropic from the point of view of the Sun, so I will still be able to diagonalize the metric. Virtually any coordinate system used to describe the FRW metric can bring the metric into a diagonalized form because we can use comoving coordinates anywhere as the FRW metric describes the special case of universal expansion (applicable to any point in the universe). Isn't this right?

Onto my main query - can we diagonalize the Kerr metric without using free-fall coordinates (a coordinate system free-falling toward the rotating mass) or coordinate systems which are rotating with the same angular speed as the mass (in the same direction as the mass)?

Last edited: Jul 26, 2015
18. Jul 26, 2015

### PAllen

The ability to diagonalize a metric means you can find a congruence of world lines such that there is a family of hypersurfaces orthogonal to the congruence and these hypersurfaces don't intersect. This is not always possible, but it certainly does not require staticity or stationary character of the metric.

19. Jul 26, 2015

### PWiz

So there is no coordinate system for Kerr spacetime which diagonalizes the metric, right (again, excluding the cases I have previously mentioned)?

20. Jul 26, 2015

### Staff: Mentor

The Earth is not a "comoving" object; we on Earth do not see the universe as homogeneous and isotropic. When cosmologists talk about observing the universe to be homogeneous and isotropic, they mean after correcting for the Earth's motion relative to "comoving" observers. The Earth is not at rest in the coordinates in which the FRW metric is usually written (the ones in which it appears diagonal).

No, you won't. The Sun is not at rest in "comoving" FRW coordinates either. An observer at rest in those coordinates would see the universe to be homogeneous and isotropic. We don't actually see that on Earth; for example, we see a dipole anisotropy in the CMB. And since that observed anisotropy allows us to determine Earth's velocity relative to a "comoving" observer, we can determine that the Sun is not "comoving" either.

No. Only one family of observers in the universe will see it as homogeneous and isotropic, and only one coordinate chart will have those observers "at rest" (i.e., with constant spatial coordinates).

21. Jul 26, 2015

### Staff: Mentor

There are no such coordinates as these in Kerr spacetime, at least not globally. There are no global coordinates in which the Kerr metric is diagonal.

22. Jul 26, 2015

### PAllen

There is no way at all to diagonalize the Kerr metric except at a point. This is general feature of 'rotating spacetimes.

23. Jul 26, 2015

### pervect

Staff Emeritus
I don't have a reference, but my current thinking is that a rotating black hole has PT symmetry in place of T symmetry that a non-rotating black hole has. The rotating black hole is symmetrical if you mirror-image the space axes (the P for parity) and also reverse the flow of time (T symmetry), but if you only reverse the flow of time, you change the sense of the rotation (i.e. clockwise vs anticlockwise), meaning it lacks the pure T symmetry.

The non-rotating solution does have pure T symmetry.

I believe this is the underlying reason the Schwarzschild is metric diagonal, and the Kerr metric non-diagonal. The diagonal terms like dt^2 (and also dr^2, dphi^2, d theta^2) have pure T symmetry, the off diagonal terms like dphi dt have only PT symmetry and not T symmetry. So the diagonal metric has the right sort of symmetries (pure T symmetry) for a non-rotating space-time, while the non-diagonal metric has the right sort of symmetries (PT but not pure T) for a rotating one.

Last edited: Jul 26, 2015
24. Jul 26, 2015

### Staff: Mentor

FRW spacetime is neither T symmetric nor PT symmetric, but it has a diagonal metric in appropriate coordinates.

As PAllen said, the condition for being able to find coordinates in which the metric is diagonal is that there is a family of worldlines filling the spacetime and a family of hypersurfaces that are everywhere orthogonal to that family of worldlines. Schwarzschild spacetime and FRW spacetime both meet that requirement, but Kerr spacetime does not (it has a family of worldlines filling it, but they aren't hypersurface orthogonal).

The extra condition that brings in T or PT symmetry is the presence of a timelike Killing vector field. (Note that we are now restricting attention to the region outside the horizon; at and inside the horizon, in both Schwarzschild and Kerr spacetime, the KVF in question is not timelike.) If you have a timelike KVF that is hypersurface orthogonal (i.e., a static spacetime), then you have T symmetry, as in the Schwarzschild case. If you have a timelike KVF but it isn't hypersurface orthogonal (i.e., a stationary spacetime that is not static), you have PT symmetry, as in the Kerr case.

25. Jul 26, 2015

### PWiz

This answers one of my queries neatly. Thanks.
@PeterDonis I guess I'm ignorant about one too many things about the FRW metric and cosmology. Thanks for the correction, I'm going to stop procrastinating and finally go through the cosmology section in GR books! (Seriously though, I didn't know we observe a dipole anisotropy in the CMB from Earth.)

@pervect This was what I was talking about earlier! Since a diagonalized metric only has the squared differential elements, it should be invariant under all spatial/time coordinate reversals, and should therefore exhibit pure PT symmetry, right?
And if we look at the Kerr metric in Boyer-Lindquist coordinates, we are going to have spatial symmetry for the r and theta terms, and no symmetry for the t and phi terms right (since there is a cross term involving these two differenial elements)?

And while we're at it, I might as well ask - do the meanings of theta and phi just carry over from spherical coordinates? If yes, then how do we use this definition with black holes? AFAIK, a BH has no center, and the singularity is a point in time, not in space, so I don't know how one would create an 'origin' at it in that sense.