A body at rest in Doran and Boyer-Lindquist coordinates

[JimWhoKnew]

Homework Statement

Must a body at rest in Boyer-Lindquist coordinates move (or not) in Doran coordinates?

Relevant Equations

see below

I read this question in the second edition of "Exploring Black Holes" by Wheeler, Taylor & Bertschinger. The book can be freely accessed and downloaded from the author's website.
Chapter 17 deals with Kerr's solution on the equatorial slice $\theta = 0$.
In Doran coordinates, the 2+1 metric is given by $$ d\tau ^2 = dT^2 - [(\frac{r^2}{r^2 + a^2})^{1/2} dr + (\frac{2 M}{r})^{1/2} (dT - a d\Phi)]^2 - (r^2 + a^2) d\Phi ^2 $$ (Eq. 5 p. 17-4)
By the transformation (Eqs. 113-114 p. 17-36) $$ dT = dt + \frac{R \beta}{r H^2} dr $$ $$ d\Phi = d\phi + \frac{\omega R}{r H^2 \beta} dr $$ where $r$ is the same in both coordinate systems, one gets the metric in Boyer-Lindquist coordinates (Eq. 112 p. 17-35) $$ d\tau ^2 = (1-\frac{2 M}{r})dt^2 + \frac{4 M a}{r}dt d\phi - \frac{dr^2}{H^2} - R^2 d\phi ^2 $$ with $$ R^2 \equiv r^2 + a^2 + \frac{ 2 M a^2}{r} $$ $$ H^2 \equiv \frac{1}{r^2} (r^2 - 2 M r + a^2) $$ $$ \omega \equiv \frac{2 M a}{r R^2} $$ $$ \beta \equiv (\frac{2 M}{r})^{1/2} (\frac{r^2 + a^2}{R^2})^{1/2} $$ (p. 17-27)
(the calculation is straight forward but lengthy; I have carried it out and got the desired result).

Now, on page 17-38 the question is asked:

BL-7. Not “at rest” in both global coordinates
Show that a stone at rest in Boyer-Lindquist global coordinates ($dr = d\phi = 0$) is not at rest in Doran global coordinates; in particular, $d\Phi \neq 0$ for that stone.

My problem is that from the transformation relations (presented above), $dr = d\phi = 0$ in B-L, trivially transforms into $dr = d\Phi = 0$ in Doran, and vice versa. What am I overlooking? How can this (apparent?) contradiction be resolved?

 

[JimWhoKnew]

Chapter 17 deals with Kerr's solution on the equatorial slice $\theta = 0$.

I apologize for this error. $\theta$ is the colatitude, of course, and the equatorial slice is at $\theta = \pi /2$ . I'm still looking for a solution to my question.

 

[Ibix]

I agree with you. I also agree with you that it's an odd way to ask the question if it's a typo, so perhaps there's something I've missed.

I'm out of time this evening, but one thought I had was to note that "stationary" observers have non-zero four-acceleration. Can you compute those for the two families of "stationary" observers and transform one into the other system? I don't immediately see that doing so would get the answer they want, but it's at least different to mucking around with the metric (which is what I've been doing).

 

[JimWhoKnew]

Thanks for helping. I appreciate it.

I also agree with you that it's an odd way to ask the question if it's a typo

The question is phrased in a manner that it repeats itself 3 times, as if to eliminate any possibility for misunderstanding. That's why I find it hard to discard as a typo. It looks too intentional.

The book aims to be readable by undergrads, so the questions are not necessarily easy, but are not supposed to be extremely hard either. If we really miss something, I don't expect it to be very elusive.

I'm out of time this evening, but one thought I had was to note that "stationary" observers have non-zero four-acceleration. Can you compute those for the two families of "stationary" observers and transform one into the other system?

That might bring on some advance! Since $a_\mu a^\mu$ is invariant, it can be calculated independently (for static observers) in each coordinate system, and then compared. If the values are different, It supports the book's claim (but as you said, it will not explain why I am wrong). I will do the calculations during the weekend and post the results.

Special thanks to @berkeman who pushes this wagon out of the mud.

 

[PeterDonis]

$r$ is the same in both coordinate systems

But $dr$ is not. To see this, note that if we look at $d\tau^2$ in each chart for an interval with only $dr$ nonzero, we get different quantities.

The above is also true in Painleve vs. Schwarzschild coordinates in Schwarzschild spacetime, which is the analogue of Doran vs. Boyer-Lindquist coordinates in Kerr spacetime. In the Schwarzschild spacetime case, the difference in $dr$ is due to the surfaces of constant coordinate time being "tilted" with respect to one another. But in the Kerr case, in addition to that, the "grid lines" of $r$ in the two charts are rotated with respect to one another (and the angle of rotation varies with $r$). You can see this in Figure 7 where the "rest" vs. "static" frames are shown; the "rest" $y$ direction is the Doran global $r$ direction, but the static $y$ direction is the Boyer-Lindquist global $r$ direction. So saying that $dr = 0$ in Boyer-Lindquist is not the same as saying $dr = 0$ in Doran.

This by itself does not seem to affect the question, but it invites consideration of the fact that the global $\phi$ direction is also not the same in the two charts. Again, if we look at intervals with only $d \phi = 0$ in the two charts, we get different quantities for $d\tau^2$, so the two directions cannot be the same. The question is whether any of the difference is due to the "rotation" of the grid lines described above, in addition to the difference due to the surfaces of constant coordinate time not being the same.

It seems to me that the above might be a possible clue.

 

[JimWhoKnew]

Thanks for joining in. As I wrote to @Ibix above, I really appreciate your help and precious time.

You gave me a lot to think about (well, for me it's a lot), so I'll reply in bits in the coming days.

But $dr$ is not.

As I understand it, given a chart on spacetime and an event $A$, the coordinates of $A$ in this chart are 4 simple real numbers. Transforming into a second chart with $r'=r$, says that the nominal value of $r'_A$ is the same number as $r_A$ (otherwise, what $r'=r$ means?). If we have a second nearby event $B$, then $\Delta r$ is just the simple subtraction of two numbers $r_B-r_A$. If not, how can the coordinates be used to calculate any physical quantity? This leads me to infer $dr'=dr$.

Another way to look at it: the coordinate $r$ is a function on spacetime, and the 1-form $dr$ does not depend on other coordinates or on the existence of a metric. If $r'=r$, then $dr'$ should be the same 1-form as $dr$.

However, $\partial_{r'}\neq \partial_r$ in general, because of the "tilts".

To see this, note that if we look at $d\tau ^2$ in each chart for an interval with only nonzero $dr$, we get different quantities.

Do you mean taking two nearby events for which only $dr$ is nonzero? If so, we can loosely say that these events are "simultaneous" in the first chart. But they will not be simultaneous in the second. So a pair of events with only $dr'$ nonzero, is necessarily different from the first pair, and I wouldn't expect them to have the same $d\tau ^2$.

The above is also true in Painleve vs. Schwarzschild coordinates in Schwarzschild spacetime, which is the analogue of Doran vs. Boyer-Lindquist coordinates in Kerr spacetime. In the Schwarzschild spacetime case, the difference in $dr$ is due to the surfaces of constant coordinate time being "tilted" with respect to one another.

Following this observation, I can think of a simple thought experiment in SR (where things are usually simpler and clearer), that supports (in my opinion) my understanding that $x'=x$ imposes $dx'=dx$. I will describe it if anyone shows interest.

You can see this in Figure 7 where the "rest" vs. "static" frames are shown; the "rest" $y$ direction is the Doran global $r$ direction, but the static $y$ direction is the Boyer-Lindquist global $r$ direction.

I will have to re-read these sections and think more about it. It will take me few days.

 

[PeterDonis]

This leads me to infer $dr'=dr$.

If we are considering a given physical displacement, we must hold $d\tau^2$ constant, and if we are considering a displacement in which only $dr$ is nonzero, the two line elements have different coefficients of $dr$, so if $d\tau^2$ is the same, $dr$ must change if it is the only nonzero differential in both charts. Or, alternatively, an interval with only $dr$ nonzero in one chart must have some other differential also nonzero in the other chart. I should have mentioned the latter possibility before.

 

[JimWhoKnew]

Or, alternatively, an interval with only $dr$ nonzero in one chart must have some other differential also nonzero in the other chart.

It seems to me the same as what I wrote above (the paragraph that begins with "Do you mean...").

The invariance of $d\tau ^2$ doesn't forbid $dr'=dr$, unless you insist that only $dr$ and $dr'$ are nonzero - a condition that can't be fulfilled simultaneously in both charts by the same pair of neighboring events.

If we go back to the OP, the assumption $dr'=dr=0$ along with the given $d\phi=0$, leads to $d\Phi=0$ and $dT=dt$, and $d\tau^2$ is the same in both charts. So we can question the assumption, but its consequence is compatible.

 

[PeterDonis]

The invariance of $d\tau ^2$ doesn't forbid $dr'=dr$, unless you insist that only $dr$ and $dr'$ are nonzero - a condition that can't be fulfilled simultaneously in both charts by the same pair of neighboring events.

Yes, that's the "alternative" that I said I should have mentioned in my earlier post. Sorry for the lack of clarity on my part.

 

[Ibix]

I've managed to do it in a different way to what I suggested above.

Spoiler

Deduce the Jacobian matrix from the two line elements and apply that to a unit vector in the "only time coordinate changes" direction in either coordinate system.

Edit: Actually, I think I messed up - see my next post.

 

[PeterDonis]

I've managed to do it in a different way to what I suggested above.

I'm not sure how this helps since if we restrict to $dr = 0$ the two line elements are the same.

 

[JimWhoKnew]

By "the Jacobian matrix" , do you mean $$\frac{\partial x'^\mu}{\partial x^\nu}$$ (or its inverse) of the coordinates transformation?

 

[PeterDonis]

"stationary" observers have non-zero four-acceleration. Can you compute those for the two families of "stationary" observers and transform one into the other system?

I have not done this computation, but the fact that the line elements in both charts are the same for $dr = 0$ indicates to me that the obvious Killing vector fields $\partial_t$ vs. $\partial_T$ and $\partial_\phi$ vs. $\partial_\Phi$ are the same KVFs in both charts. Since "at rest" worldlines are orbits of $\partial_t$ or $\partial_T$, the fact that those KVFs are the same would indicate that the "at rest" worldlines are the same.

It is true that the two 1-form fields $dt$ and $dT$ are not the same, even though $\partial_t$ and $\partial_T$ are the same vector fields. However, the difference between them only shows up for intervals with $dr$ nonzero; if we restrict to a single value of $r$, i.e., to a "cylinder" in spacetime, the difference in the two 1-form fields vanishes.

The more I think about this, the more I wonder if in fact there was a typo in this question in the book.

 

[Ibix]

OK - I'm very confused. If you can see why my four-acceleration isn't perpendicular to my four-velocity when working in Doran coordinates then let me know. The Maxima code below derives the Doran metric tensor from the line element copied from (4) on page 17-3 in the link in the OP, uses ctensor to compute Christoffel symbols (in Maxima $\Gamma^i_{jk}$ is mcs[j, k, i]), defines a function to compute the covariant derivative of a vector, then computes $A^a=U^a\nabla_aU^b$ (which should be the four acceleration) and then $g_{ab}U^aA^b$ which should be zero. It works in Boyer-Lindquist coordinates (there's commented out code to override the metric tensor with the B-L form) but not Doran.


Edit: figured it out - the indices are inconsistent in the calculation of the covariant derivative. See next post. I've deleted the buggy code from this post.

 

[JimWhoKnew]

OK - I'm very confused. If you can see why my four-acceleration isn't perpendicular to my four-velocity when working in Doran coordinates then let me know. The Maxima code below...

I am not familiar with Maxima, so I can't help here. I have calculated the acceleration too (as promised) and got the same $a^\mu a_\mu$ for observers "at rest" in both metrics.

In either metric I get $$u^\mu=(1-\frac{2M}{r})^{-1/2}\: \delta^\mu_{\: 0}$$ and from $$a^\mu = \frac{du^\mu}{d\tau}+\Gamma^\mu_{\: \nu \rho}u^\nu u^\rho$$ I get $$a^\mu=(1-\frac{2M}{r})^{-1} \: \Gamma^\mu _{\: 00}$$ So it's all about calculating $\Gamma^\mu _{\: 00}$.
The results I got do satisfy $a^\mu u_\mu=0$, and have the desired property that $a^\mu a_\mu$ diverges at the static limit. However, in Boyer-Lindquist I got that only $a^r$ and $a_r$ are nonzero. I expected a $\phi$ component to oppose the frame dragging.

(In 3+1 dimensions, $g_{0 \theta}=0$, so the results for $a^\mu$ ($\mu\neq\theta$) should be the same whether calculated in 3+1 or 2+1. I tried both, just to verify).

 

[Ibix]

Fixed the code and deleted buggy version (turned out to be using indices inconsistently) from above. Corrected and expanded version is below.

In B-L coordinates I get $A^t=A^\phi=0$ and $$A^r=\frac{M(r^2-2Mr+a^2)^2}{r^5(r-2M)}$$In Doran coordinates I get $$\begin{eqnarray*}
A^T&=&\frac{\sqrt{2}M^{3/2}\sqrt{r^2+a^2}}{r^{5/2}\left(r-2M\right)}\\
A^r&=&\frac{M(r^2-2Mr+a^2)}{r^3(r-2M)}\\
A^\Phi&=&\frac{\sqrt{2}M^{3/2}a}{r^{5/2}(r-2M)\sqrt{r^2+a^2}}
\end{eqnarray*}$$
This matches the analysis in #15.

I think these two are actually the same vector - if you use $(J^{-1})^T$ to convert the BL $A$ into Doran coordinates their inner product is equal to the square root of the products of the magnitudes, so they appear parallel.

/* https://www.physicsforums.com/threads/a-body-at-rest-in-doran-and-boyer-lindquist-coordinates.1063139/ */

/* Some useful definitions, from Taylor et al Exploring Black Holes */
assume(r>0);
H:(r^2-2*M*r+a^2)/r^2;
R:sqrt(r^2+a^2+2*M*a^2/r);
omega:2*M*a/(r*R^2);
beta:sqrt(2*M/r)*sqrt((r^2+a^2)/(R^2));

/* Convert Doran coordinates (Kerr spacetime) line element into metric tensor */
dT^2-(sqrt(r^2/(r^2+a^2))*dr+sqrt(2*M/r)*(dT-a*dPhi))^2-(r^2+a^2)*dPhi^2;
doranMetric: collectterms(expand(%), dT, dr, dPhi);
doranTr: ratsimp(part(doranMetric, 1) / (dT*dr));
doranPhir: ratsimp(part(doranMetric, 2) / (dPhi*dr));
doranrr: ratsimp(part(doranMetric, 3) / (dr*dr));
doranPhiPhi: ratsimp(part(doranMetric, 4) / (dPhi*dPhi));
doranTT: ratsimp(part(doranMetric, 5) / (dT*dT));
doranTPhi: ratsimp(part(doranMetric, 6) / (dT*dPhi));

doranLG:matrix([doranTT, doranTr/2, doranTPhi/2],
		[doranTr/2, doranrr, doranPhir/2],
		[doranTPhi/2, doranPhir/2, doranPhiPhi]);

/* Boyer-Lindquist coordinates metric */
blLG:matrix([1-2*M/r, 0, 2*M*a/r],
		[0, -1/H^2, 0],
		[2*M*a/r,0,-R^2]);

/* Write down B-L -> Doran Jacobian matrix */
J:matrix(	[1,             0,       0],
		[-R*beta/(r*H), sqrt(H), -omega*R/(r*beta*H)],
		[0,             0,       1]);
radcan(J.blLG.transpose(J)-doranLG);

/* Transform "stationary in B-L coordinates" to Doran coordinates */
invert(transpose(J)).[1,0,0];

/* Setup to check proper accelerations - load ctensor package and provide */
/* functions for covariant differentiation of a vector.                   */
load(ctensor);
covdif(U):=block(
	[i,j,k,ans],
	ans:zeromatrix(dim,dim),
	for i:1 thru dim do block (
		for j:1 thru dim do block (
			ans[i,j]: diff(U[j],ct_coords[i]),
			for k:1 thru dim do block (
				ans[i,j]: ans[i,j] + mcs[i,k,j] * U[k]
			)
		)
	),
	ans
);

/* Compute 4-acceleration of hovering observer in B-L coordinates */
ct_coords:[t,r,phi];
dim:3;
lg:blLG;
cmetric(false);
christof(mcs);
blU:[1/sqrt(lg[1,1]),0,0];
blA:ratsimp(blU.covdif(blU));
ratsimp(blU.blLG.blA);  /* Check U orthogonal to A */

/* Ditto in Doran coordinates */
ct_coords:[T,r,Phi];
dim:3;
lg:doranLG;
cmetric(false);
christof(mcs);
doranU:[1/sqrt(lg[1,1]),0,0];
doranA:ratsimp(doranU.covdif(doranU));
ratsimp(doranU.doranLG.doranA);  /* Check U orthogonal to A */

 

[JimWhoKnew]

I have not done this computation, but the fact that the line elements in both charts are the same for $dr = 0$ indicates to me that the obvious Killing vector fields $\partial_t$ vs. $\partial_T$ and $\partial_\phi$ vs. $\partial_\Phi$ are the same KVFs in both charts. Since "at rest" worldlines are orbits of $\partial_t$ or $\partial_T$, the fact that those KVFs are the same would indicate that the "at rest" worldlines are the same.

I am not sure about it, so please point out where I'm wrong:
$\partial _t$ in B-L must be a linear superposition of the form $A\,\partial _T+B\,\partial _\Phi$, $A$ and $B$ constant (because these 2 vector fields span the Killing subspace). So motion in B-L along a worldline which is an orbit of $\partial_t$, should be viewed in Doran as motion along a worldline which is an orbit of $A\,\partial _T+B\,\partial _\Phi$, and therefore, in this case, $dr=0$ in both charts.

 

[PeterDonis]

I expected a $\phi$ component to oppose the frame dragging.

There isn't one for observers "hovering" at constant $r$; the $\phi$ component only appears once $dr / d\tau$ is nonzero.

 

[PeterDonis]

I am not sure about it, so please point out where I'm wrong:
$\partial _t$ in B-L must be a linear superposition of the form $A\,\partial _T+B\,\partial _\Phi$, $A$ and $B$ constant (because these 2 vector fields span the Killing subspace).

Generally speaking, yes, but if you work it out for this particular case I think you will find that $B$ must be zero. Otherwise the KVF does not approach a Minkowski time translation at infinity and is therefore not a possible KVF for describing an "at rest" observer.

 

[JimWhoKnew]

This matches the analysis in #15.

In Doran, I got the same results as you. My $a^r$ in B-L is exactly the same as in Doran, and different from yours. I got the same $a^\mu a_\mu$ in both charts. Do you get it too?

 

[Ibix]

In Doran, I got the same results as you. My $a^r$ in B-L is exactly the same as in Doran, and different from yours. I got the same $a^\mu a_\mu$ in both charts. Do you get it too?

I've double checked, and what's written is what I've got. The vectors are parallel, but B-L $A^aA_a$ is $H$ times Doran $A^aA_a$. I'll have another look just in case there's another bug lurking there.

 

[Ibix]

I note that my B-L $A^r=\Gamma^r_{tt}(1-2M/r)^{-1}$, where$$\Gamma^r_{tt}=\frac{M(r^2-2Mr+a^2)^2}{r^6}$$That's consistent with your #15. The B-L metric tensor I'm using is$$\begin{pmatrix} 1-{{2\,M
}\over{r}}&0&{{2\,M\,a}\over{r}}\cr 0&-{{r^4}\over{\left(r^2-2\,M\,r
+a^2\right)^2}}&0\cr {{2\,M\,a}\over{r}}&0&-r^2-{{2\,M\,a^2}\over{r
}}-a^2\cr \end{pmatrix}$$and the Christoffel symbol seems consistent with that.

 

[JimWhoKnew]

I note that my B-L $A^r=\Gamma^r_{tt}(1-2M/r)^{-1}$, where$$\Gamma^r_{tt}=\frac{M(r^2-2Mr+a^2)^2}{r^6}$$That's consistent with your #15. The B-L metric tensor I'm using is$$\begin{pmatrix} 1-{{2\,M
}\over{r}}&0&{{2\,M\,a}\over{r}}\cr 0&-{{r^4}\over{\left(r^2-2\,M\,r
+a^2\right)^2}}&0\cr {{2\,M\,a}\over{r}}&0&-r^2-{{2\,M\,a^2}\over{r
}}-a^2\cr \end{pmatrix}$$and the Christoffel symbol seems consistent with that.

Note that $H^2$ is defined as $$H^2 \equiv \frac{1}{r^2} (r^2 - 2 M r + a^2)$$ so actually you are using $$-\frac{dr^2}{H^4}$$ instead of $$-\frac{dr^2}{H^2}$$

 

[JimWhoKnew]

Generally speaking, yes, but if you work it out for this particular case I think you will find that $B$ must be zero. Otherwise the KVF does not approach a Minkowski time translation at infinity and is therefore not a possible KVF for describing an "at rest" observer.

Thanks. I did the calculation (which turns out to be very easy in this particular case), and got that indeed $\partial_T=\partial_t$ , $\partial_\Phi=\partial_\phi$ , and $\partial_{r'} \neq \partial_r$ , as expected.

 

[PeterDonis]

Thanks. I did the calculation (which turns out to be very easy in this particular case), and got that indeed $\partial_T=\partial_t$ , $\partial_\Phi=\partial_\phi$ , and $\partial_{r'} \neq \partial_r$ , as expected.

To summarize what I think this means, in conjunction with Figure 7 in the book that I referred to earlier:

In both charts $\partial_\phi = \partial_\Phi$ points in the direction of the global $\Phi$ in the figure, which is also the "static" $x$ direction.

The Boyer-Lindquist $\partial_r$ points in the "static" $y$ direction in the figure. This direction is orthogonal to $\partial_\phi$.

The Doran $\partial_r$ points in the "rest" $y$ direction in the figure. This direction is not orthogonal to $\partial_\phi$.

$\partial_t = \partial_T$ points out of the page of the figure, "tilted" relative to (i.e., not orthogonal to) $\partial_\phi$, but orthogonal to the Boyer-Lindquist $\partial_r$. (But not orthogonal to the Doran $\partial_r$.)

 

[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??

 

[Ibix]

Note that $H^2$ is defined as $$H^2 \equiv \frac{1}{r^2} (r^2 - 2 M r + a^2)$$ so actually you are using $$-\frac{dr^2}{H^4}$$ instead of $$-\frac{dr^2}{H^2}$$

Right. Way too many typos from me in this - sorry. With that correction, I agree with you.

 

[JimWhoKnew]

To summarize what I think this means, in conjunction with Figure 7 in the book that I referred to earlier:

In both charts $\partial_\phi = \partial_\Phi$ points in the direction of the global $\Phi$ in the figure, which is also the "static" $x$ direction.

The Boyer-Lindquist $\partial_r$ points in the "static" $y$ direction in the figure. This direction is orthogonal to $\partial_\phi$.

The Doran $\partial_r$ points in the "rest" $y$ direction in the figure. This direction is not orthogonal to $\partial_\phi$.

$\partial_t = \partial_T$ points out of the page of the figure, "tilted" relative to (i.e., not orthogonal to) $\partial_\phi$, but orthogonal to the Boyer-Lindquist $\partial_r$. (But not orthogonal to the Doran $\partial_r$.)

I am confused, so I'll start from the bottom line: do you think the question in the book (from the OP) is correct?

I still have to study Figure 7 carefully. But even now, it seems to me that the relations you describe are reasonable consequences of $g_{\mu \nu}=\partial_\mu \cdot \partial_\nu\:$ .

 

[JimWhoKnew]

1) Holding r and $\phi$ constant represents the same paths as holding r and $\Phi$ constant.

This is what the whole thread is about
If you are right, the book is wrong.

Thanks for joining.

 

[PeterDonis]

do you think the question in the book (from the OP) is correct?

As I said in post #13, the more I think about it, the more I think the question as it appears in the book is not correct.

 

[JimWhoKnew]

As I said in post #13, the more I think about it, the more I think the question as it appears in the book is not correct.

Thanks for making that clear. The confusion is my fault. Sorry.

 

[JimWhoKnew]

but the following can be derived from either the metric forms or the differential coordinate transforms:

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

 

[PAllen]

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

Strictly speaking, you can't show the families of curves holding two coordinates constant are the same across the two metrics. All you can show is that metric interval along the curves is the same function the third coordinate. This is a necessary, but not sufficient condition. So, I was imprecise in my claim.

 

[JimWhoKnew]

Strictly speaking, you can't show the families of curves holding two coordinates constant are the same across the two metrics. All you can show is that metric interval along the curves is the same function the third coordinate. This is a necessary, but not sufficient condition. So, I was imprecise in my claim.

Showing that the conclusions hold in an infinitesimal vicinity of the starting point, is good enough for me.

Let's try:
If I use the property that each metric form has 2 explicit Killing directions, one timelike and one spacelike, I can use @PeterDonis' argument of the asymptotic form at infinity (post #19), and get that $\partial_t$ and $\partial_T$ are proportional. This is your first conclusion above. Moreover, by substitution in the metric forms, we can get $\partial_t=\partial_T$ .
By a similar procedure we get $\partial_\phi=\partial_\Phi$ . This is your second conclusion. By now we have determined 6 out of 9 components of the inverse transformation matrix (or it's transposed...).
Now we can see that in B-L, $g_{r \phi}=\partial_r\cdot\partial_\phi=0$ , but in Doran $g_{r' \Phi}\neq 0$ . This leads us to your third conclusion. ( $g_{T \Phi}=g_{t \phi}$ so we're good here).

Can it be done without employing the Killing symmetries?

 

[JimWhoKnew]

Strictly speaking, you can't show the families of curves holding two coordinates constant are the same across the two metrics. All you can show is that metric interval along the curves is the same function the third coordinate. This is a necessary, but not sufficient condition. So, I was imprecise in my claim.

I thought about it today. By comparing the two metric forms, we implicitly assume $r'=r$. Otherwise it's a mess. 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). We know that a solution exists. So, 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).
Must they all satisfy the 3 conclusions listed in post #26 ?

 

[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.