Meaning of r in Schwarzchild coordinates

[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 Schwarzschild 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 Schwarzschild 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.

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

 

[fzero]

I would say that the coordinate ##r## is perfectly defined outside the Schwarzschild 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.

 

[PWiz]

I would say that the coordinate ##r## is perfectly defined outside the Schwarzschild 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.

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)

 

[fzero]

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)

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.

 

[PWiz]

A coordinate system can be thought of as a way to map points like this. Then the notion of coordinate distance r2−r1r_2-r_1 would be the distance between points at the same (θ,ϕ)(\theta,\phi) in our map.

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?

 

[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) ).

 

[pervect]

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

Right.

-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)

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.

So let's look at a feature of the Schwarzschild 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'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 only reasonable deduction that I'm able to make with this is that the set of all 2-spheres surfaces in Schwarzschild coordinates represent the family of hypersurfaces with a fixed r coordinate (or is it the other way around?).

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.

 

[PWiz]

@PAllen Okay, thanks!

Right, but what you may be missing is that the Schwarzschild metric is not applicable if you have two massive bodies.

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?

I'm not sure what you mean here, but I'll go along for the time being with it not being important.

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 Schwarzschild 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?

 

[PeterDonis]

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.

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.)

the set of all 2-spheres surfaces in Schwarzschild coordinates represent the family of hypersurfaces with a fixed r coordinate

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).

 

[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

 

[PWiz]

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 tt.

So what exactly does simply diagonalizing the metric do? Suggest that the geometry is invariant under coordinate reversals?

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

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)

 

[PeterDonis]

hat exactly does simply diagonalizing the metric do?

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.

 

[PWiz]

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.

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)?

 

[PeterDonis]

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)?

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

 

[PWiz]

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

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?

 

[PeterDonis]

Isn't that because the frame describing the FLRW metric itself is non-static?

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 FLRW metric is a special case because it employs comoving coordinates, which is why it's diagonalizable right?

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).

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?

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

 

[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)?

 

[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.

 

[PWiz]

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.

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

 

[PeterDonis]

the FRW metric appears diagonalized in the FRW metric coordinate chart because the Earth's frame coordinates are comoving right?

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).

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.

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.

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?

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).

 

[PeterDonis]

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)?

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.

 

[PAllen]

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

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

 

[pervect]

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.

 

[PeterDonis]

I believe this is the underlying reason the Schwarzschild is metric diagonal, and the Kerr metric non-diagonal.

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.

 

[PWiz]

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.

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

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.

 

[PeterDonis]

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?

It's not enough just to have squared differential elements; all the metric coefficients also need to be functions only of the squares of coordinates. Also, the metric coefficients can't be a function of time (in the FRW metric, they are, so the FRW metric isn't PT symmetric even though it is diagonal).

 

[PWiz]

Also, the metric coefficients can't be a function of time

Yes Peter, I payed attention to it when you first mentioned this fact. Can't believe I forgot to mention that in my post. Sorry.

So a metric which has time independent coefficients and squared differential elements will exhibit PT symmetry, correct?

 

[PeterDonis]

do the meanings of theta and phi just carry over from spherical coordinates?

Sort of. The angular coordinates are coordinates on each 2-sphere (in Schwarzschild spacetime; it's more complicated in Kerr spacetime). See below.

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.

You don't need a center in order to assign coordinates on a 2-sphere; a 2-sphere is a perfectly good manifold in its own right, regardless of how (or whether) it is embedded in a higher-dimensional manifold.

 

[PeterDonis]

a metric which has time independent coefficients and squared differential elements will exhibit PT symmetry, correct?

Actually, a metric with these characteristics will have T symmetry. It may also have P symmetry, depending on the form of the spatial part. If it has both T and P symmetry, it also has PT symmetry.

A metric can have PT symmetry, without having T or P symmetry separately, under somewhat looser conditions (for example, Kerr spacetime does). I'm not sure exactly how one would characterize those conditions.

 

[PWiz]

Actually, a metric with these characteristics will have T symmetry. It may also have P symmetry, depending on the form of the spatial part. If it has both T and P symmetry, it also has PT symmetry.

A metric can have PT symmetry, without having T or P symmetry separately, under somewhat looser conditions (for example, Kerr spacetime does). I'm not sure exactly how one would characterize those conditions.

Okay, let's talk about a diagonal metric with coordinates r, theta, phi and t. If the coefficients are independent of theta, phi and t, then things should look unchanged if we flip the sign of any of these coordinates. So such a metric should exhibits PT symmetry.
If a diagonal metric employs the same coordinates but the coefficients are only independent of t, then the metric should show T symmetry.
Could one explore this chain of reasoning onto non-diagonal metrics? I mean at least establish some basis for these symmetry conditions?

 

[PeterDonis]

If the coefficients are independent of theta, phi and t, then things should look unchanged if we flip the sign of any of these coordinates. So such a metric should exhibits PT symmetry.

And T symmetry, and P symmetry. PT symmetry is a weaker condition than P or T symmetry; PT only means invariance if P and T are both applied, but not necessarily if either one is applied individually. P symmetry means symmetry if just P is applied; T symmetry means symmetry if just T is applied.

If a diagonal metric employs the same coordinates but the coefficients are only independent of t, then the metric should show T symmetry.

Yes, but not P symmetry. It might still show PT symmetry.

Could one explore this chain of reasoning onto non-diagonal metrics?

Yes, by looking at which non-diagonal terms are present. For example, in Kerr spacetime (in appropriate coordinates), there is only one non-diagonal term, and it is a term in ##dt d \phi##. So reversing the sign of both ##t## and ##\phi## (which a PT transformation does) leaves the metric invariant. But reversing ##t## or ##\phi## individually (i.e., P alone or T alone) does not. (You also have to look at the metric coefficients, of course; in Kerr spacetime those are all functions of ##r## and ##\theta## only, not ##t## or ##\phi##.)

 

[PWiz]

And T symmetry, and P symmetry. PT symmetry is a weaker condition than P or T symmetry

Hmmm, interesting, I did not previously know that.

Yes, but not P symmetry. It might still show PT symmetry

OK. I think I finally wrapped my head around this.

Thanks @fzero @PAllen @stevebd1 @pervect @PeterDonis for helping me out! I appreciate it

 

[pervect]

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

I thought we were talking Schwarzaschild vs Kerr? At least that's what I was talking about.

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).

I'm not quite following - which family of worldlines are we talking about? If we're talking about a static or stationary metric, I'd assume we were talking about the worldlines whose tangents were timelike Killing vectors, but if I'm following, you're trying to extend the analysis to include more general cases than I was initially considering. But if we are considering a more general case, then it's not clear to me what worldlines we are talking about here.

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.) '

Right, I agree with this.

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.

That's pretty much the condition I was assuming, analyzing only the case where a timelke KVF was present. Rather than use the language of Killing vectors, though, I tried to make the same point used different language, The Killing vectors are basically about the symmetries of the space-time, I thought that the argument in terms of the symmetries would be more accessible to a wider audience than an argument based on the Killing vector fields. To my mind, though, the arguments are basically the same, only the manner in which it is presented changes, as long as one agrees that Killing vector fields represent underlying symmetries.

A few more little nits. I think that I've really only shown that for our example, we can eliminate the time-space terms like dt*dr, dt*phi, dt*dtheta by my time reversal argument. We need a different argument than I presented (probably involving the other KVF's and/or their associated radial symmetries) to eliminate terms like dphi*dtheta., space-space terms that are not diagonal.

Additionally I glossed over the whole issue of equivalence classes , i.e. metrics related by diffeomorphisms. As an example, one could use the non-diagonal Painleve metric to represent the same space-time as the Schwarzschild metric. So we don't really have to use a diagonal metric to represent the Schwarzschild space-time, it's just that it's a coordinate choice that respects the underlying coordinate-independent symmetries.

 

[PeterDonis]

thought we were talking Schwarzaschild vs Kerr? At least that's what I was talking about.

If we're restricting discussion to just those spacetimes, then yes, what you say about symmetries is correct. In the thread as a whole we have been considering more general conditions for a diagonal metric (see the recent posts between me and PWiz).

which family of worldlines are we talking about?

The family of worldlines that have some particular symmetry property. In the case of spacetimes with a timelike KVF, the worldlines are the orbits of the KVF, as you say. In the case of spacetimes which are homogeneous and isotropic, such as FRW spacetime, the worldlines are those of observers for whom those properties are manifest, i.e., they actually see the spacetime as homogeneous and isotropic. We could also express this in terms of spacelike KVFs; see below.

The Killing vectors are basically about the symmetries of the space-time

Yes, and these can be spacelike as well as timelike. For example, spherical symmetry means that there is a 3-parameter family of spacelike KVFs whose orbits generate 2-spheres. Homogeneity and isotropy means that there is a 6-parameter family of spacelike KVFs (3 parameters for spherical symmetry, plus 3 for spatial translation symmetry) whose orbits generate spacelike hypersurfaces; the "comoving" worldlines in FRW spacetime are the ones that are everywhere orthogonal to those hypersurfaces.

We need a different argument than I presented (probably involving the other KVF's and/or their associated radial symmetries) to eliminate terms like dphi*dtheta., space-space terms that are not diagonal.

Yes, to show that those terms can always be eliminated, you would look at the 3-parameter family of spacelike KVFs corresponding to spherical symmetry.

one could use the non-diagonal Painleve metric to represent the same space-time as the Schwarzschild metric. So we don't really have to use a diagonal metric to represent the Schwarzschild space-time, it's just that it's a coordinate choice that respects the underlying coordinate-independent symmetries.

More precisely, that respects all of them (including the spacelike as well as the timelike ones) and their relationships. Painleve coordinates still respect the timelike KVF: the metric is independent of ##t## and the integral curves of the timelike KVF are curves with constant spatial coordinates. And Painleve coordinates still respect spherical symmetry; the angular part of the metric is the same as for Schwarzschild. The only difference is that the surfaces of constant time are not orthogonal to the integral curves of the timelike KVF; in other words, the coordinates do not "respect" the hypersurface orthogonality of the timelike KVF.

 

[PAllen]

It seems we have explored this topic before, in a thread I forgot all about - finding more definitive answers:

https://www.physicsforums.com/threads/when-can-a-metric-be-put-in-diagonal-form.798380/

In particular, see Ben Niehoff's comment #6, and Bcrowell's #12, and the unusual case Peter mentioned in #5. This means the condition I gave in this thread is necessary but not sufficient for diagonalizability. This ZAMO exception seem really mysterious to me since (as covered in http://arxiv.org/abs/0809.3327, linked to by bcrowell), any 3-manifold is diagonalizable. Thus, if you have a congruence orthogonal to a foliation, why can't you then diagonalize the 3-manifolds of the foliation, and then build the 4th basis as the surface orthogonal congruence tangent, and end up with coordinates everywhere orthgonal? Of further relevance is that ZAMO congruence is twist free, so that is not a problem. All quite mysterious to me, but it is well established that there is no diagonalization of Kerr in a finite region.