Fermi-Walker transport of proper acceleration along timelike congruence

[cianfa72]

TL;DR Summary

About Fermi-Walker transport of proper acceleration vector field along worldlines of timelike congruence

Hi,
starting from a recent thread in this section, I decided to start a new thread about the following:

Take a generic irrotational/zero vorticity timelike congruence. Do the 4-velocity and the direction of proper acceleration (i.e. the vector in that direction at each point with norm 1) commute at each point along the congruence's worldlines ? Or is the direction of proper acceleration actually Fermi-Walker transported along such congruence's worldlines ?

 

[PeterDonis]

Do the 4-velocity and the direction of proper acceleration (i.e. the vector in that direction at each point with norm 1) commute at each point along the congruence's worldlines ?

I believe the Rindler congruence is a counterexample if you use the 4-velocity and not the corresponding KVF. The direction of proper acceleration commutes with the KVF (which is $\partial_t$ in Rindler coordinates, and the proper acceleration direction is $\partial_x$), but it does not commute with the 4-velocity (which is $(1 / x) \partial_t$).

is the direction of proper acceleration actually Fermi-Walker transported along such congruence's worldlines ?

The answer to this is yes for the Rindler congruence; this is shown in the Insights article that was referenced in the previous thread.

Another good congruence to examine would be the ZAMO congruence in Kerr spacetime; it is irrotational but it has nonzero expansion and shear and is in a curved spacetime, so it is a more general case. Also it is not a Killing congruence so that confounding factor is not present.

 

[cianfa72]

The answer to this is yes for the Rindler congruence; this is shown in the Insights article that was referenced in the previous thread.

Ok yes, indeed from Insights article the proper acceleration along Rindler congruence's worldlines is $\frac {1} {X} \hat{e}_1$ in Rindler coordinates. The proper acceleration direction is $\hat{e}_1$ everywhere that has norm 1. The Fermi-Walker derivative is $D_{\hat{e}_0} \hat{e}_1 = 0$ hence the direction of proper acceleration is actually Fermi-Walker transported along the Rindler congruence.

Another good congruence to examine would be the ZAMO congruence in Kerr spacetime; it is irrotational but it has nonzero expansion and shear and is in a curved spacetime, so it is a more general case. Also it is not a Killing congruence so that confounding factor is not present.

Have you some reference regarding ZAMO congruence in Kerr spacetime ?

 

[PeterDonis]

Have you some reference regarding ZAMO congruence in Kerr spacetime ?

Most discussions of Kerr spacetime describe it. IIRC Matt Visser's brief paper from a while back on Kerr spacetime gives the math for it.

 

[PeterDonis]

from Insights article the proper acceleration along Rindler congruence's worldlines is $\frac {1} {X} \hat{e}_1$ in Rindler coordinates.

Yes. Or, as the Insights article put it, $\nabla_{\hat{e}_0} \hat{e}_0 = A \hat{e}_1$. And $\nabla_{\hat{e}_0} \hat{e}_1 = A \hat{e}_0$. However, $\nabla_{\hat{e}_1} \hat{e}_0 = 0$, so $\hat{e}_0$ and $\hat{e}_1$ do not commute.

Note, however, that the proper acceleration vector $A \hat{e}_1$ itself, not just its direction, is Fermi-Walker transported along the congruence worldlines. In other words,

$$
D_F (A \hat{e}_1) = \nabla_{\hat{e}_0} (A \hat{e}_1) - (A \hat{e}_1 \cdot A \hat{e}_1) \hat{e}_0 + (A \hat{e}_1 \cdot \hat{e}_0) A \hat{e}_1 = A^2 \hat{e}_0 - A^2 \hat{e}_0 + 0 = 0
$$

where we have used the fact that $\nabla_{\hat{e}_0} A = 0$, i.e., the magnitude of proper acceleration is constant along the congruence worldlines.

 

[cianfa72]

Yes. Or, as the Insights article put it, $\nabla_{\hat{e}_0} \hat{e}_0 = A \hat{e}_1$. And $\nabla_{\hat{e}_0} \hat{e}_1 = A \hat{e}_0$. However, $\nabla_{\hat{e}_1} \hat{e}_0 = 0$, so $\hat{e}_0$ and $\hat{e}_1$ do not commute.

Sorry, in the Insights article the symbol/letter $A$ is defined just for Langevin congruence. In any case for Rindler congruence it should be $A=1/X$ in Rindler coordinates.

Note, however, that the proper acceleration vector $A \hat{e}_1$ itself, not just its direction, is Fermi-Walker transported along the congruence worldlines.

Ok yes. This is a particular case.

Another good congruence to examine would be the ZAMO congruence in Kerr spacetime; it is irrotational but it has nonzero expansion and shear and is in a curved spacetime, so it is a more general case. Also it is not a Killing congruence so that confounding factor is not present.

Kerr spacetime isn't static, however the ZAMO timelike congruence is irrotational/hypersurface orthogonal although it isn't a Killing congruence (otherwise Kerr spacetime would be static while we know it is not).

 

[cianfa72]

Most discussions of Kerr spacetime describe it. IIRC Matt Visser's brief paper from a while back on Kerr spacetime gives the math for it.

I found his paper "The Kerr spacetime: A brief introduction" - 2008 from arxiv.org. However I don't see any reference to ZAMO congruence (in any coordinates introduced there).

 

[PeterDonis]

in the Insights article the symbol/letter $A$ is defined just for Langevin congruence.

That's true, but it should be obvious what it refers to for the Rindler congruence--and evidently it is:

for Rindler congruence it should be $A=1/X$ in Rindler coordinates.

Yes. (Strictly speaking, that requires a particular normalization of the $X$ coordinate.)

 

[PeterDonis]

I found his paper "The Kerr spacetime: A brief introduction" - 2008 from arxiv.org. However I don't see any reference to ZAMO congruence (in any coordinates introduced there).

Hm, that's the paper I was thinking of. I must have misremembered.

 

[cianfa72]

I found this paper from infn. It states that the angular momentum of free-falling particles in Kerr spacetime is conserved along their timelike worldlines (the conserved value is their angular momentum at radial infinity). In B-L coordinates, $\partial_t$ is a timelike KVF. The value $$L=u^{\mu}m_{\mu}=u_{\phi}=0$$ is kept constant along each ZAMO congruence's member. The paper in 21.3 claims that ZAMO observers in the congruence corotate with the rotating black hole described by Kerr spacetime.

What does it mean that Kerr spacetime describes a rotating black hole? Does it simply mean that, for example, the frame field associated to ZAMO congruence is not Fermi-Walker transported along the ZAMO 4-velocity of each ZAMO congruence's member?

In other words the tetrad field's orthonormal spacelike directions actually "rotate" w.r.t. their Fermi-Walker transported "version" along the congruence's worldlines.

 

[PeterDonis]

It states that the angular momentum of free-falling particles in Kerr spacetime is conserved along their timelike worldlines

Yes, this is a well known fact that is true for any axisymmetric spacetime--it is what "axisymmetric" means (that there is a conserved angular momentum, or, to put it another way, there is a spacelike KVF with closed orbits).

The paper in 21.3 claims that ZAMO observers in the congruence corotate with the rotating black hole described by Kerr spacetime.

One has to interpret this very carefully. It does not mean that the angular velocity of a ZAMO is everywhere the same as the angular velocity of the hole's horizon. It only means that the angular velocity of a ZAMO is just the right value to make its angular momentum exactly zero. What this value is varies with $r$ and $\theta$ (in Boyer-Lindquist coordinates).

What does it mean that Kerr spacetime describes a rotating black hole?

It means that there is an event horizon ("black hole") which has nonzero angular velocity (it is the ZAMO angular velocity at the $r$ and $\theta$ coordinates of the horizon). The angular velocity is usually defined relative to a stationary observer at infinity. It turns out that this fact is sufficient to ensure that the timelike KVF in Kerr spacetime is not hypersurface orthogonal.

 

[PeterDonis]

the ZAMO timelike congruence is irrotational/hypersurface orthogonal although it isn't a Killing congruence

This is true, but it's also true that the ZAMO congruence is not Born rigid (it has nonzero expansion and shear, as I have already mentioned).

 

[PeterDonis]

It means that there is an event horizon ("black hole") which has nonzero angular velocity (it is the ZAMO angular velocity at the $r$ and $\theta$ coordinates of the horizon). The angular velocity is usually defined relative to a stationary observer at infinity. It turns out that this fact is sufficient to ensure that the timelike KVF in Kerr spacetime is not hypersurface orthogonal.

For a good discussion of stationary axisymmetric spacetimes (of which Kerr spacetime is one), their definition and implications, see section 7.1 of Wald.

 

[cianfa72]

For a good discussion of stationary axisymmetric spacetimes (of which Kerr spacetime is one), their definition and implications, see section 7.1 of Wald.

I have seen it, however it doesn't help in understanding whether the ZAMO congruence in Kerr spacetime does or does not Fermi-Walker transports the direction of proper acceleration along those congruence's worldlines.

 

[PeterDonis]

it doesn't help in understanding whether the ZAMO congruence in Kerr spacetime does or does not Fermi-Walker transports the direction of proper acceleration along those congruence's worldlines.

For the free-falling ZAMO congruence, the proper acceleration is zero.

For the "orbiting" ZAMO congruence, i.e., ZAMOs in circular orbits in the equatorial plane, the proper acceleration is nonzero and its variation is somewhat complicated. Wald, IIRC, does not solve this problem explicitly, but he does give useful tools.

 

[cianfa72]

For the free-falling ZAMO congruence, the proper acceleration is zero.

Ah ok, hence free-falling ZAMO congruence's worldlines are actually timelike geodesics of Kerr spacetime.

For the "orbiting" ZAMO congruence, i.e., ZAMOs in circular orbits in the equatorial plane, the proper acceleration is nonzero and its variation is somewhat complicated. Wald, IIRC, does not solve this problem explicitly, but he does give useful tools.

You mean ZAMOs in circular orbits in the equatorial plane in Boyer-Lindquist coordinates $\theta =0$.

 

[PeterDonis]

free-falling ZAMO congruence's worldlines are actually timelike geodesics of Kerr spacetime.

Yes.

ZAMOs in circular orbits in the equatorial plane in Boyer-Lindquist coordinates $\theta =0$.

The equatorial plane is $\theta = \pi / 2$.