A body at rest in Doran and Boyer-Lindquist coordinates

[PeterDonis]

By comparing the two metric forms, we implicitly assume $r'=r$.

This is how the coordinate transformation between the two charts, as I understand it, is defined.

if I don't miss anything, there is possibly a 1-parameter family of transformations that carry the B-L form to Doran's (and maybe more, owing to different roots of the quadratic relations).

As I understand it, the Doran chart is a single chart, with a single transformation from B-L to Doran. There should not be a family of transformations to different possible Doran charts. Everything I read in the book seems to indicate this. Multiple Doran charts would require multiple different forms for the line element, but there is only one.

 

[PeterDonis]

The metric forms provide 5 quadratic equations, relating the differentials of the new coordinates to those of the old

I'm not sure what you mean by this. Equations 113 and 114 in the book chapter define the relationships between the coordinate differentials in the two charts (or more precisely those which are not simple equalities). There are no quadratic equations or free parameters anywhere.

 

[JimWhoKnew]

This is how the coordinate transformation between the two charts, as I understand it, is defined.

Of course it is. But the question in #32 was "How your 3 listed conclusions can be derived from the metric forms ALONE? (i.e. when the coordinates transformation relations are not provided)".

As I understand it, the Doran chart is a single chart, with a single transformation from B-L to Doran. There should not be a family of transformations to different possible Doran charts.

I don't say it is not unique. In the spirit of #32, I'm just considering our ability to deduce the 9 components of the possibly unique transformation (Eq. 113,114 in the book), If we are only given $r'=r$ and the metric forms in the OP. In this case, the counting in #35 is "formally" one constraint short of fully determining the transformation.

There are no quadratic equations or free parameters anywhere.

See the previous paragraphs of this reply.

 

[PeterDonis]

the question in #32 was "How your 3 listed conclusions can be derived from the metric forms ALONE? (i.e. when the coordinates transformation relations are not provided)".

Hm. In flat spacetime, yes, knowing just the line element does not uniquely define a coordinate chart--for example, there are an infinite number of inertial frames in all of which the line element looks the same.

In curved spacetime, though, I'm not sure whether that still holds. We are holding the spacetime geometry constant, and I'm not sure there are multiple distinct charts on Kerr spacetime in which the line element takes either the B-L or the Doran form. If there aren't, then knowing the metric forms alone does uniquely define the coordinate transformation and we don't need to know them separately.

 

[JimWhoKnew]

In curved spacetime, though, I'm not sure whether that still holds. We are holding the spacetime geometry constant, and I'm not sure there are multiple distinct charts on Kerr spacetime in which the line element takes either the B-L or the Doran form. If there aren't, then knowing the metric forms alone does uniquely define the coordinate transformation and we don't need to know them separately.

At any point in curved spacetime we can transform to an inertial frame and then apply an additional arbitrary Lorentz transformation. So there are infinitely many transformations between the line elements. But in general this approach is not integrable, and we can't apply it to a finite "patch" around the point.

If there aren't, then knowing the metric forms alone does uniquely define the coordinate transformation and we don't need to know them separately.

So probably the integrability condition does make the transformation from B-L to Doran unique, and it is fully determined by the metric forms, but we can't deduce all the components of the transformation (solely from the forms), unless we have means to identify which is the integrable one (eqs. 113,114).

Editing: Frobenius theorem?

Editing 2: $\frac{\partial ^2 x'^\mu}{\partial x^\rho \partial x^\nu}=\frac{\partial ^2 x'^\mu}{\partial x^\nu \partial x^\rho}$

 

[PeterDonis]

there are infinitely many transformations between the line elements.

If you allow an arbitrary other coordinate chart in between, sure. But that doesn't mean there are infinitely many such transformations between just the B-L and the Doran chart, with no other charts in between. I don't see the point of talking about any other case.

 

[JimWhoKnew]

So we are left with 6 unknown components in the transformation matrix. The metric forms provide 5 quadratic equations, relating the differentials of the new coordinates to those of the old (the $dr^2$ equation is omitted, since it doesn't bear any additional information).

That's a mistake. No equation should be omitted. So in this particular case, we should be able to fully derive the unique transformation by comparison of the metric forms, provided $r'=r$ (this is a stronger version of @PAllen's claim in #26).

 

[PAllen]

I've been on a family vacation for a bit. I'll write more later. But for now, on the question of uniqueness of Doran and Boyer-Lindquist metrics, the symmetries tell the story. The 'large' group of transforms leaving the standard SR metric unchanged follow directly from the symmetries flat spacetime has. In the case of a spinning BH, we know that time translations and rotations around the spin axis leave the metrics form unchanged (as long as these symmetries are manifest in the metric form, as is the case for both forms under consideration) . This actually doesn't change some deliberate wording I used - the family of curves being the same in two cases. Adding a rotation to the coordinate transform will mean that the coordinate representation of a member of a family of curves with 2 coordinates constant, will change but the family of such curves will be identical.

 

[JimWhoKnew]

I'll write more later.

I'm looking forward to read

But for now, on the question of uniqueness of Doran and Boyer-Lindquist metrics, the symmetries tell the story. The 'large' group of transforms leaving the standard SR metric unchanged follow directly from the symmetries flat spacetime has. In the case of a spinning BH, we know that time translations and rotations around the spin axis leave the metrics form unchanged (as long as these symmetries are manifest in the metric form, as is the case for both forms under consideration) . This actually doesn't change some deliberate wording I used - the family of curves being the same in two cases.

In post #34 I tried to derive your conclusions on my own (from the metric forms and $r'=r\:$ ALONE). To get there, I used the symmetries as you describe here. At the end of #34 I was wondering: "can it be done without employing the Killing symmetries?". The answer seems to be positive, as I argued in posts #35 & #42. It should be noted however that in the latter procedure the equations actually have the symmetries "built-in".

 

[JimWhoKnew]

I know it's not customary, but I'd like to thank (again) @berkeman, @Ibix, @PeterDonis and @PAllen for helping me. The thought that I might be missing something very elementary and obvious, was really bugging me for almost three months.

 

[PAllen]

I'm not quite sure how it addresses the referenced exercise in the OP, and maybe was pointed out before, but the following can be derived from either the metric forms or the differential coordinate transforms:

1) Holding r and $\phi$ constant represents the same paths as holding r and $\Phi$ constant.
2) Holding t and r constant represents the same paths as holding T and r constant.
3) Holding t and $\phi$ constant represents different paths from holding T and $\Phi$ constant.

Maybe this is what the exercise was meant to bring out, but the wording seems wrong for this??

On deriving this from the metric alone, what I meant, more precisely, in this case is as follows:

We assume the r coordinate is the same. We assume the metrics are related by an unknown coordinate transform (thus representing the same geometry), but we have no idea what the transform is. We then derive that (1) and (2) above are plausible (but not proven); (3) if is fully demonstrated. Note, that all 3 are straightforward to establish from the differential transform, but the question here is what can be derived from the metric forms alone, given the stated assumptions. I do not use any killing vectors or symmetry in this derivation.

For (1), simply consider what each metric says about curve given by $(r_0,\phi_0,t)$ and $(r_0, \Phi_0,T)$. You get, along such curves $d\tau^2=(1-2M/r)dt^2$ and $d\tau^2=(1-2M/r)dT^2$ after a little algebra. This is consistent with them being the same curves.

Similarly, for (2) one gets, after algebra, $d\tau^2=-R^2d\phi^2$ and $d\tau^2=-(2Ma^2/r+r^2+a^2)d\Phi^2$. These are the same, given the definition of R. This is consistent with them being the same curve.

For (3) we get $d\tau^2=-dr^2/H^2$ and $d\tau^2=-\frac {r^2} {r^2+a^2}dr^2$. These are just different (after substituting H definition), so they can't possibly be the same curve, especially since r is assumed the same in both coordinates.

 

[JimWhoKnew]

Thanks for making it clear.

We assume the r coordinate is the same ... Note, that all 3 are straightforward to establish from the differential transform, but the question here is what can be derived from the metric forms alone, given the stated assumptions.

As I wrote in previous posts, I think that the stated assumptions should be sufficient (in principle, at least) for finding all the unknown components of the differential coordinate transform.

I do not use any killing vectors or symmetry in this derivation.

The derivation assumes that $r$ is the same, and also uses the property that all metric components depend only on $r$ (which enables the comparison). The latter means that the symmetries still sneak in.