Intuitive understanding of Thomas precession

[PeterDonis]

As has been pointed out several times in this thread, the Lie transported vectors do not remain orthogonal

You left out a key qualifier: the Lie transported vectors of the ZAMO congruence do not remain orthogonal. Bill_K's computation did not use the ZAMO congruence; it used a congruence with constant angular velocity (the Killing congruence that WBN mentioned). The angular velocity of the ZAMO congruence varies with $r$ (and $\theta$ if you're outside the equatorial plane).

 

[PeterDonis]

We don't need to use the ZAMO congruence in order to determine the gyroscopic precession along a single ZAMO orbit.

Strictly speaking, you don't need to use *any* congruence to compute the precession itself; you can just use Fermi-Walker transport along the single worldline. The only role the congruence plays is in defining the Lie transported vectors relative to which the precession is expressed (and strictly speaking, you don't need a congruence for that either, any set of orthonormal basis vectors defined all along the worldline will do).

 

[PeterDonis]

You mentioned earlier that we might be able to represent the Lie transported vectors by a sort of average angular velocity so that we might be able to quantify this quantity.

Yes, I'm still working on that. But one key observation, following on from the discussion of Bill_K's computation, is that changing the congruence we use to define the Lie transported vectors doesn't change the behavior of the gyroscopes (the Fermi-Walker transported vectors) at all; it only changes the Lie transported vectors, which means it changes the *relative* behavior of the F-W vectors with respect to the Lie transported vectors.

In the ZAMO case in Kerr spacetime, I think that the Lie transported $\hat{e}_{\phi}$ will be the same as the $\hat{e}_{\phi}$ of the rigid congruence Bill_K used (since that vector already points towards the neighboring ZAMO in the same circular orbit). So the difference between the two congruences (ZAMO vs. rigid) is in the behavior of the $\hat{e}_r$ Lie transported vectors; in the rigid congruence, they stay orthogonal to $\hat{e}_{\phi}$, whereas in the ZAMO congruence, they don't.

If this is correct, then the results Bill_K and WBN derived will correctly give the precession of gyroscopes relative to $\hat{e}_{\phi}$ in the ZAMO case. What still needs to be computed (and what I'm working on) is the behavior of the $\hat{e}_r$ vectors in the ZAMO congruence. Your paddle wheel analogy is interesting and I'll look at the links you gave to see if it helps in the computation I'm doing.

 

[WannabeNewton]

Strictly speaking, you don't need to use *any* congruence to compute the precession itself; you can just use Fermi-Walker transport along the single worldline.

True, I should have clarified that I was referring to a calculation of the gyroscopic precession that made use of the vorticity. This has the disadvantage of requiring a Killing field and Lie transport of a frame by the Killing field but it has the advantage of being much simpler computationally.

The only role the congruence plays is in defining the Lie transported vectors relative to which the precession is expressed (and strictly speaking, you don't need a congruence for that either, any set of orthonormal basis vectors defined all along the worldline will do).

Well mathematically the Lie derivative is only defined for two vector fields so what we're really Lie transporting is a frame field along a Killing field. Only the Fermi-derivative and the covariant derivative require a single curve (which is another reason for why we should be cautious of the relationship between vorticity and gyroscopic precession in the case of non-Killing fields).

 

[PeterDonis]

mathematically the Lie derivative is only defined for two vector fields

Ah, yes, you're right, you need some specification of at least one other vector field (other than the 4-velocity).

 

[yuiop]

I could use some expert assistance here! With reference to the attached paper n gyroscope precession by Rindler, would someone be kind enough to calculate $e^{\varphi}$, $\omega_{3,1}$, $h^{11}$ and $h^{33}$ from the canonical form of the Kerr metric? Rindler does the calculations on page 12 but he specialises the results to the specific case of a grid rotating at the geodesic angular velocity. I need the more general case for any angular velocity (but still remaining restricted to the equatorial plane).

This would assist greatly in comparing Rindler's results with results I have obtained from other sources.

P.S. If nothing else, just the solution for $\omega_{3,1}$ would be a help. I think i can figure the others out for myself. (maybe).

 

[yuiop]

This paper http://www.ias.ac.in/jarch/jaa/20/103-120.pdf might be of interest. it gives a general equation (35) for the precession of a gyroscope in the Kerr metric. there appears to be a typo in the denominator of the equation as it does not reduce to the Schwarzschild case (36) when a=0. They give the denominator as:

$\left[1-(r^2+a^2)\omega^2 - \frac{2M(a\omega)^2}{r}\right]$

I think it is intended to be the Kerr time dilation factor for an object in a circular orbit and should actually be:

$\left[1-2M/r -(r^2+a^2)\omega^2 - \frac{2M(a\omega)^2}{r}+\frac{4Ma\omega}{r}\right]$

 

[PeterDonis]

would someone be kind enough to calculate $e^{\varphi}$, $\omega_{3,1}$, $h^{11}$ and $h^{33}$ from the canonical form of the Kerr metric?

You can read them off from equation 14, which is the Kerr metric in canonical form for arbitrary $\omega$. I get

$$
e^{2 \psi} = 1 - \omega^2 \left( a^2 + r^2 \right) - \frac{2M}{r} \left( 1 - a \omega \right)^2
$$

(the above is actually given as equation 15)

$$
h^{11} = h^{rr} = 1 - \frac{2M}{r} + \frac{a^2}{r^2}
$$

$$
h^{33} = h^{\phi \phi} = \frac{r^2 - 2 m r + a^2}{1 - \omega^2 \left( a^2 + r^2 \right) - \left( 2 M / r \right) \left( 1 - a \omega \right)^2}
$$

$$
w_3 = w_{\phi} = \frac{\omega \left( a^2 + r^2 \right) - \left( 2 a M / r \right) \left( 1 - a \omega \right)}{1 - \omega^2 \left( a^2 + r^2 \right) - \left( 2 M / r \right) \left( 1 - a \omega \right)^2}
$$

Then we need to differentiate the last equation with respect to $r$ to get

$$
w_{3, 1} = w_{\phi , r} = \frac{2 \omega r + \left( 2 a M / r^2 \right) \left( 1 - a \omega \right)}{1 - \omega^2 \left( a^2 + r^2 \right) - \left( 2 M / r \right) \left( 1 - a \omega \right)^2} - \frac{\omega \left( a^2 + r^2 \right) - \left( 2 a M / r \right) \left( 1 - a \omega \right)}{\left[ 1 - \omega^2 \left( a^2 + r^2 \right) - \left( 2 M / r \right) \left( 1 - a \omega \right)^2 \right]^2} \left[ - 2 \omega^2 r + \frac{2M}{r^2} \left( 1 - a \omega \right)^2 \right]
$$

If I've done the algebra right, this simplifies to

$$
w_{3, 1} = w_{\phi , r} = \frac{2 \omega r - 2 \omega M \left[ 2 + \left( a^2 + r^2 \right) / r^2 \right] \left( 1 - a \omega \right) \left( 1 - 2 a \omega \right) + \left( 2 a M / r^2 \right) \left( 1 - a \omega \right)}{\left[ 1 - \omega^2 \left( a^2 + r^2 \right) - \left( 2 M / r \right) \left( 1 - a \omega \right)^2 \right]^2}
$$

which doesn't look very enlightening, so it's quite possible that I've made an algebra error.

 

[yuiop]

... which doesn't look very enlightening, so it's quite possible that I've made an algebra error.

Thanks Peter! That is really awesome! :thumbs:

The first easy check is for the $\omega=0$ case and I get $\frac{ma}{r^3(1-2m/r)}\frac{2(r^2-2mr-a^2)}{(1-2m/r)}$ so there might be a problem there. I will check it all out thoroughly. You have given me a solid start. Thanks again!

 

[yuiop]

$$
h^{33} = h^{\phi \phi} = \frac{r^2 - 2 m r + a^2}{1 - \omega^2 \left( a^2 + r^2 \right) - \left( 2 M / r \right) \left( 1 - a \omega \right)^2}
$$

From what I can glean from the Rindler paper, $h^{33}$ should be inverted, which simplifies things a lot due to cancellations.

With that change, I now get the correct result for the w=0 case. I will check the rest later. Thanks again for your help. Thanks to you, I think I understand the paper a whole lot better now!

 

[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).