Tetrad formalism outside of the equatorial plane

[Vrbic]

I'm interesting in about possibility to express a tetrad of a rotating matter in neutron star (in some approximate known metric, where shape of star is unchanged - Lense-Thirring metric) outside of the equatorial plane.
My idea is: I start from Locally Non Rotating Frame (LNRF) in equatorial plane with know LNRF speed . If I would a rotating frame in equatorial plane I believe it is enough just a boost in $\phi$ (LNRF velocity) direction.
But what about outside of the equatorial plane? There is no LNRF frame. Right? But on the other hand the matter is enforced to move in circle. It is not typical geodesic motion. May I use same proccedure as in equatorial plane in this particular case? I.e. take frame rotating with LNRF speed and boost in $\phi$ direction.
I'm quite sure, If I would interested in frame of test particle in such gravitation field, it si wrong, but the fact that it is enforced circular non geodesic motion, it seems to me it is possible. What do you mean?
Thank you for your comments.
I'm interesting in if is possible to express a tetard of a pieace

 

[PeterDonis]

I start from Locally Non Rotating Frame (LNRF) in equatorial plane with know LNRF speed

By LNRF do you mean a ZAMO (zero angular momentum observer)? It seems like you do, since you talk about "speed" and a boost in the $\phi$ direction.

 

[Vrbic]

By LNRF do you mean a ZAMO (zero angular momentum observer)? It seems like you do, since you talk about "speed" and a boost in the $\phi$ direction.

Yes, I do my LNRF is ZAMO.

 

[PeterDonis]

my LNRF is ZAMO

Ok. (LNRF normally refers to something else.) However, I'm still not sure exactly what kind of "frame" you are thinking about. Are you trying to figure out a rotating coordinate chart for the entire spacetime? By which I mean, coordinates in which ZAMOs (who have nonzero angular velocity) are at rest? If so, that is not possible, since the angular velocity of a ZAMO (relative to infinity) varies with position.

what about outside of the equatorial plane?

There are ZAMO worldlines everywhere, not just in the equatorial plane.

on the other hand the matter is enforced to move in circle. It is not typical geodesic motion.

The trajectories that get "dragged" by the rotation of the central mass are geodesics.

 

[Vrbic]

Ok. (LNRF normally refers to something else.)

Hm, I thought it is the same. What is difference between ZAMO and LNRF. ZAMO is $u_\phi= u_\theta=0$? And LNRF both and $u_\phi= u_\theta=u_r=0$?

 

[Vrbic]

However, I'm still not sure exactly what kind of "frame" you are thinking about.

I would like to transform to a frame rotating with the matter. It is dependent on a position in the star (I guess on $r$ nad $\theta$ only).

 

[PeterDonis]

I would like to transform to a frame rotating with the matter.

In general this won't be possible either, as the angular velocity of the matter won't be constant. For an idealized piece of matter that is all rotating at the same angular velocity, yes, you can do this, but it won't be simple, because no exact solution for the metric inside the matter is known. The Kerr metric only describes a vacuum region with nonzero "rotation", and strictly speaking, it only describes a rotating black hole (i.e., all vacuum); it doesn't describe the vacuum region outside rotating matter like a star except in an approximate sense. So you would have to approach the whole problem numerically.

 

[PeterDonis]

What is difference between ZAMO and LNRF.

A ZAMO is a global concept: it is an observer with zero angular momentum with respect to the global axial Killing vector field.

A LNRF is a local concept; it is a local frame in which there are no inertial forces due to rotation, like centrifugal force or Coriolis force.

ZAMO is $u_\phi = u_\theta = 0$?

No. That should be obvious since a ZAMO must have nonzero angular velocity unless it is at infinity.

And LNRF both and $u_\phi= u_\theta=u_r=0$?

No. A LNRF places no constraints on the 4-velocity. It is a constraint on the spacelike vectors in a tetrad, not on the timelike one. Mathematically, a LNRF means the spacelike vectors in a tetrad are Fermi-Walker transported along the worldline.

 

[Vrbic]

In general this won't be possible either, as the angular velocity of the matter won't be constant. For an idealized piece of matter that is all rotating at the same angular velocity, yes, you can do this, but it won't be simple, because no exact solution for the metric inside the matter is known. The Kerr metric only describes a vacuum region with nonzero "rotation", and strictly speaking, it only describes a rotating black hole (i.e., all vacuum); it doesn't describe the vacuum region outside rotating matter like a star except in an approximate sense. So you would have to approach the whole problem numerically.

I use Lense-Thirring metric - Schwarzschild one with extra $g_{\phi t}$. But yes, "dragging" velocity is calculated up to first order in angular velocity of the star numerically and it is function only $r$. On the other hand I expect constant angular velocity of the matter of the star (rigid rotation). So is it possible to get a tetrad of such frame? For me is enough to express it by metric coefficients.

 

[George Jones]

In this context, I think its is standard to have "ZAMO" and "locally nonrotating observer" mean the same thing. See section 1.3.2 from the notes "An introduction to the theory of rotating relativistic stars" by Eric Gourglhon,

https://arxiv.org/abs/1003.5015

In this context, "locally nonrotating observer" seems to have been introduced around 1970 by James Bardeen, who is the academic son of Feynman and the biological son of Nobel laureate John Bardeen. See (maybe behind a paywall?) the paragraph that includes equation (37) in

http://articles.adsabs.harvard.edu/..._paper=YES&type=PRINTER&filetype=.pdf

 

[PeterDonis]

In this context, I think its is standard to have "ZAMO" and "locally nonrotating observer" mean the same thing. See section 1.3.2 from the notes "An introduction to the theory of rotating relativistic stars" by Eric Gourglhon,

Hm. I find this terminology somewhat misleading, because the definition given only refers to the worldline (or congruence of worldlines--see below), not to a tetrad, and there is no reason why an observer following a ZAMO worldline must carry a tetrad that is Fermi-Walker transported along the worldline.

I believe it is true that if we consider a family of ZAMOs who all carry tetrads set up so that they always point to the same neighboring ZAMO, those tetrads will all be Fermi-Walker transported and will hence be locally non-rotating in the sense I gave before. That would correspond to the statement in that section that the congruence of ZAMO worldlines has zero twist (vorticity). But I still think it's worth keeping those two concepts distinct (angular momentum defined in terms of the axial KVF, and Fermi-Walker transport of a tetrad). As best I remember, MTW, for example, does keep them distinct.

 

[PeterDonis]

I use Lense-Thirring metric - Schwarzschild one with extra $g_{\phi t}$

This is only an approximation, so I would not draw general conclusions from it.

"dragging" velocity is calculated up to first order in angular velocity of the star numerically and it is function only rr.

I don't think that's correct. In the Lense-Thirring metric, the $g_{\phi t}$ term depends on $\theta$ as well as $r$.

 

[Vrbic]

I don't think that's correct. In the Lense-Thirring metric, the $g_{\phi t}$ term depends on $\theta$ as well as $r$.

I agree $g_{\phi t}$ is dependent on $\theta$ but I wrote only velocity of LNRF. See http://adsabs.harvard.edu/cgi-bin/n....150.1005H&link_type=ARTICLE&db_key=AST&high=

 

[Vrbic]

I don't think that's correct. In the Lense-Thirring metric, the $g_{\phi t}$ term depends on $\theta$ as well as $r$.

Ok but it is not important, tetrad is coordinate independent, right?
My LNRF tetrad in equatorial plane is:$\omega^{(t)}= \Big \{ \sqrt{-\omega(r)g_{\phi t}-g_{tt}},0,0,0 \Big \} \\ \omega^{(r)}=\Big \{0,\sqrt{g_{rr}},0,0 \Big \}\\ \omega^{(\theta)}=\Big \{0,0,\sqrt{g_{\theta\theta}},0 \Big \}\\ \omega^{(\varphi)} = \Big \{-\omega(r) \sqrt{g_{\phi\phi}} \sin\theta,0,0, \sqrt{g_{\phi\phi}} \Big\}$, where $\omega(r)$ is velocity of "dragging".
Than I use lorentz transformation:
$| \gamma; 0; 0; -\gamma \bar{\omega}|\\ | 0 ; 1 ; 0 ; 0| \\ | 0 ; 0 ; 1 ; 0|\\ |-\gamma \bar{\omega}; 0; 0; \gamma|$ it is matrix of transformation, where $\bar[\omega]$ is difference between angular velocity of the star and velocity of dragging (also defined in paper above).
Do you agree with such "jump" between frames?

 

[Vrbic]

In this context, I think its is standard to have "ZAMO" and "locally nonrotating observer" mean the same thing. See section 1.3.2 from the notes "An introduction to the theory of rotating relativistic stars" by Eric Gourglhon,

https://arxiv.org/abs/1003.5015

In this context, "locally nonrotating observer" seems to have been introduced around 1970 by James Bardeen, who is the academic son of Feynman and the biological son of Nobel laureate John Bardeen. See (maybe behind a paywall?) the paragraph that includes equation (37) in

http://articles.adsabs.harvard.edu/..._paper=YES&type=PRINTER&filetype=.pdf

Do you have some link about motion and tetrads outside equatorial plane? Or do you know how to transform from LNRF tetrad in equatorial plane ($\theta=\pi/2$) to frame conected with rotating matter (not to the frame of free particle, but matter forced to rotate in circle as an element of star) in different $\theta$?

 

[PeterDonis]

tetrad is coordinate independent, right?

Tetrad vectors don't change physically when you change coordinates, but their components are certainly different in different coordinates.

However, that's irrelevant to what I actually said. You were claiming that the "frame dragging velocity" only depends on $r$. It doesn't. It depends on both $r$ and $\theta$. In the coordinates you have chosen, that dependence shows up as a dependence on the metric coefficient $g_{\phi t}$, which depends on both $r$ and $\theta$. However, the statement in general is a coordinate-independent statement; $r$ can be defined independently of coordinates, as the "areal radius" of the 2-sphere that contains a chosen event, and your claim then becomes the claim that the frame dragging velocity at a chosen event depends only on the areal radius of the 2-sphere that contains that event, which is false.

 

[Vrbic]

However, that's irrelevant to what I actually said. You were claiming that the "frame dragging velocity" only depends on $r$. It doesn't. It depends on both $r$ and $\theta$. In the coordinates you have chosen, that dependence shows up as a dependence on the metric coefficient $g_{\phi t}$, which depends on both $r$ and $\theta$. However, the statement in general is a coordinate-independent statement; $r$ can be defined independently of coordinates, as the "areal radius" of the 2-sphere that contains a chosen event, and your claim then becomes the claim that the frame dragging velocity at a chosen event depends only on the areal radius of the 2-sphere that contains that event, which is false.

But yes, "dragging" velocity is calculated up to first order in angular velocity of the star numerically and it is function only $r$.

Up to first (and also second) order of magnitude of accuracy, it is true. See http://adsabs.harvard.edu/cgi-bin/n...150.1005H&link_type=ARTICLE&db_key=AST&high=
I know, it isn't right generally.

What about my transformation in #14. Do you agree with it?

 

[Vrbic]

My idea is:
1) Assumptions: a) metric is approximated up to first order of magnitude in the angular velocity of star
b) rigid rotation (angular velocity of the star is constant throughout the star)
c) up to first order (a) the dragging velocity $\omega_L$ is dependent only on $r$
2) In equatorial plane I can define LNRF tetrad (very far observer see such frame as non-rotating)
3) I boost LNRF to RF (Rotating frame with the star) by lorentz transformation (ordinary boost in $\phi$ direction), all in equatorial plane
4) Under assumptions (1) I expect that all frames rotating with the star (in circles) with same distance from the AXIS of rotation are equivalent
Is it right?