Intuitive understanding of Thomas precession

[PeterDonis]

From what I can glean from the Rindler paper, $h^{33}$ should be inverted

Hm, yes, it looks like what I wrote down is $h_{33}$ (and $h_{11}$ as well), so if you want the inverse it would be $h^{33} = 1 / h_{33}$ (since the metric is diagonal).

 

[yuiop]

Hm, yes, it looks like what I wrote down is $h_{33}$ (and $h_{11}$ as well), so if you want the inverse it would be $h^{33} = 1 / h_{33}$ (since the metric is diagonal).

I think your $h^{11}$ is OK. It is already the inverse of the factor in the canonical form of the Kerr metric.

I put it all into maple software (worksheet attached) and this is the end result with a little manual simplification:

$$
\Omega = \frac{2}{r^3}\frac{ma\omega^2(a^2+3r^2)+ma(1-2a\omega)+\omega (1-3m/r)r^3 }{1 - \omega^2 \left( a^2 + r^2 \right) - \frac{2M}{r} \left( 1 - a \omega \right)^2}
$$

Other than the factor of 2, the result agrees with other sources for $\omega=0$ case and the ZAMO angular velocity. The result also collapses correctly to the Schwarzschild and Minkowski forms for $a=0$ and $m=o$ respectively.

Rindler consistently leaves out the factor of 2 that appears in the derivative of $w_3$ in all his derivations. I notice that a factor of 2 is also left out when reading off $g_{t\phi}$ from the Kerr metric in most references so it seems standard practice, but I am not sure of the reason why. Can anyone enlighten me?

 

[PeterDonis]

Rindler consistently leaves out the factor of 2 that appears in the derivative of $w_3$ in all his derivations. I notice that a factor of 2 is also left out when reading off $g_{t\phi}$ from the Kerr metric in most references so it seems standard practice, but I am not sure of the reason why. Can anyone enlighten me?

If you're reading $g_{t \phi}$ off the line element, you have to divide by 2 because the line element counts both $g_{t \phi}$ and $g_{\phi t}$, which are equal since the metric is symmetric, but both have to be counted in the line element because the summation $ds^2 = g_{ab} dx^a dx^b$ includes "cross terms" twice.

 

[yuiop]

If you're reading $g_{t \phi}$ off the line element, you have to divide by 2 because the line element counts both $g_{t \phi}$ and $g_{\phi t}$, which are equal since the metric is symmetric, but both have to be counted in the line element because the summation $ds^2 = g_{ab} dx^a dx^b$ includes "cross terms" twice.

Is it a similar argument for why Rindler leaves out the factor of 2 for the $w_{3,1}$ term?

 

[PeterDonis]

Is it a similar argument for why Rindler leaves out the factor of 2 for the $w_{3,1}$ term?

I don't think so, because if I'm understanding the formulas right, $w_{i, j}$ is not a symmetric tensor whose off-diagonal terms appear twice in a "line element" type of expression; $w_i$ appears in the canonical line element, but only once, not "double counted", since it only has one index and it's contracted with $dx^i$.

 

[yuiop]

I don't think so, because if I'm understanding the formulas right, $w_{i, j}$ is not a symmetric tensor whose off-diagonal terms appear twice in a "line element" type of expression; $w_i$ appears in the canonical line element, but only once, not "double counted", since it only has one index and it's contracted with $dx^i$.

I think I have an explanation. I suspect Rindler's equation $\Omega = \epsilon^{\varphi}(h^{11}h^{33}\omega^2{[3,1]})^{1/2}$ is actually the curl (or vorticity?) and the rotational velocity is 1/2 that value. See line (11) of this paper. (It mentions my favourite object, the paddle wheel ).

 

[yuiop]

Section 3 of the paper linked in the last post mentions that a paddle wheel with a single blade would rotate unevenly which is logical in the fluid context. However, it takes me back to an issue that was possible unresolved in these threads. If we have two independent gyroscopes in at the same location in the Kerr metric, one with its spin axis parallel to $e_{\phi}$ (tangential to a circular orbit) and the other with its spin axis parallel to $e_r$, (so pointing outward from the gravitational body centre), would the $e_r$ gyroscope precess faster relative to $e_r$ than the other gyroscope precesses relative $e_{\phi}$? I ask this, because it seems the gyroscope parallel to $e_{\phi}$ is subject to much less sheer or frame dragging than the other gyroscope. Ihis is pretty much the picture painted by Ohanian. If any of the above is true, it would of course imply the gyroscopes would not remain orthogonal to each other. Is there any law that says untorqued gyroscopes that are initially orthogonal must remain orthogonal in any kind of gravitational field, if not subjected to torque forces?

 

[WannabeNewton]

If we have two independent gyroscopes in at the same location in the Kerr metric, one with its spin axis parallel to $e_{\phi}$ (tangential to a circular orbit) and the other with its spin axis parallel to $e_r$, (so pointing outward from the gravitational body centre), would the $e_r$ gyroscope precess faster relative to $e_r$ than the other gyroscope precesses relative $e_{\phi}$?

They must precess at the same rate because they always remain orthogonal. See below.

Is there any law that says untorqued gyroscopes that are initially orthogonal must remain orthogonal in any kind of gravitational field, if not subjected to torque forces?

Yes. Fermi-transport preserves orthogonality. This is why we often just speak of a compass of inertia instead of referring to individual gyroscopes.

I ask this, because it seems the gyroscope parallel to $e_{\phi}$ is subject to much less sheer or frame dragging than the other gyroscope.

Well frame dragging is not exactly the same thing as shear but they're related in this case so no harm there. However the gyroscopes aren't the things subjected to the shear but rather the connecting vectors are the objects subjected to the shear. The connecting vectors are by definition the vectors Lie transported by the congruence whereas the gyroscopes are Fermi-transported by the congruence. You can think of the connecting vectors as representing the paddles of the wheel as I already explained in the other thread.

 

[yuiop]

They must precess at the same rate because they always remain orthogonal. See below.

I have started a new thread just to satisfy myself there are no counter examples in a relatively easily analysed situation (hopefully).

The connecting vectors are by definition the vectors Lie transported by the congruence whereas the gyroscopes are Fermi-transported by the congruence. You can think of the connecting vectors as representing the paddles of the wheel as I already explained in the other thread.

Earlier in these threads it was stated that $\Omega$ was the precession of the Lie Transported Basis Vectors relative to a set of gyroscopes and that the negative of this was the precession on the gyroscopes relative to the LTBVs. Has this now been retracted? Some sources state that $-\Omega$ is the precession of the gyroscopes relative to the orthonormal basis vectors $e_r$ and $e_{phi}$ which is a different kettle of fish. This latter viewpoint means all the talk of LTBVs was irrelevant to the issue of gyroscope precession. The former viewpoint means we need to quantify the rotation of the LTBVs relative to a fixed point at infinity or to the the orthonormal basis vectors, if we want to quantify the precession of the gyroscopes relative to a fixed point at infinity.

Finally just for clarity, consider this thought experiment in Kerr spacetime. We have an orbiting lab. To measure rotation, the lab operators has a device in the centre with lightly sprung weights that essentially measure centrifugal force. The weights are thrown outwards if the device is rotating. They also have some large weights that they can position at the out limits of the lab. If the lab is rotating, when they bring the weights to the centre of the lab, the intrinsic spin of the lab will increase. The lab has small orientation thrusters that can correct its rotation. They adjust the rate of rotation of the lab until it has no rotation according to the two lab devices. They can place the large weights wherever they like in the lab and there will be no detectable change. This to me, would be a sensible way of defining locally non rotating. Now if we introduce some gyroscopes into this lab, will they precess relative to the the lab? Intuition would say no, but I just want to be sure.

 

[WannabeNewton]

Earlier in these threads it was stated that $\Omega$ was the precession of the Lie Transported Basis Vectors relative to a set of gyroscopes and that the negative of this was the precession on the gyroscopes relative to the LTBVs. Has this now been retracted?

No that's perfectly fine but you cannot have a Lie transported set of spatial basis vectors if there is shear. Shear by definition is the deformation of connecting vectors, which are Lie transported vectors that are not spatial basis vectors (if they were then the only way they could be Lie transported is if there is no shear as already mentioned).

Some sources state that $-Omega$ is the precession of the gyroscopes relative to the orthonormal basis vectors $e_r$ and $e_{phi}$ which is a different kettle of fish. This latter viewpoint means all the talk of LTBVs was irrelevant to the issue of gyroscope precession.

We've already addressed this multiple times in the thread. See posts #38-41 as well as posts #47 and #49. Is the math posing a hurdle?

Now if we introduce some gyroscopes into this lab, will they precess relative to the the lab? Intuition would say no, but I just want to be sure.

What you've basically done is define a Fermi-transported laboratory (so no they won't). Making the centrifugal forces vanish in a given coordinate system is equivalent to having the coordinate system be Fermi-transported. Another way to do this is to have the observer in the laboratory place a mirror up against any of the walls such that the plane of the mirror is orthogonal to the line joining the mirror to the center of the laboratory-if the observer emits a beam of light from the center towards the mirror then it will be reflected back to the center if and only if the laboratory is non-rotating. However this definition of "locally non-rotating" is applicable to any observer whatsoever in any space-time. It's not the same sense in which "locally non-rotating" is being used when referring to zero angular momentum observers. In the latter case "locally non-rotating" is referring to the vanishing angular momentum which consequently makes the Sagnac effect vanish for these observers.

 

[WannabeNewton]

Hey guys! I just finished reading an interesting and rather instructive paper on gyroscopic precession. I may, in the immediate future, have some questions regarding specific parts of it but for now I just thought you guys would like to read it. Here's the link: http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19890018085.pdf (if the pdf stops loading partway through then just download it).