Does rotation in Gödel spacetime depend on the frame of reference?

[PeterDonis]

[Moderator's note: Thread spun off to allow discussion of this topic to continue since the previous thread was closed.]

It is actually fairly simple to show that it does not exert the same influence. We can just look up the Christoffel symbols in the Catalog of Spacetimes: https://arxiv.org/abs/0904.4184

Where equation 2.1.30 (rotating bucket in the bucket's frame) has $\Gamma_{tt}^{r}=-\omega^2 r$ but equation 2.10.2 (rotating universe in the bucket's frame) has $\Gamma_{tt}^{r}=0$.

I have had something nagging at me about this for a while, and it finally hit me while looking through this paper about the Godel Universe:

https://elib.uni-stuttgart.de/handle/11682/4945

Equations A.2 in this paper are the same as equations 2.10.2 in the paper you linked to. But as the paper I linked to makes clear, these equations are for a rotating frame! That is, objects at rest in this frame are "rotating with the universe", since the matter in the universe itself is at rest in this frame. The "at rest" frame field associated with this frame, which the paper I linked to calls a "static local tetrad", is geodesic--zero proper acceleration--but it has nonzero vorticity; it's the vorticity you would expect for a family of observers rotating with "the angular velocity of the universe".

So this frame is not the "bucket frame" in the Godel universe case. The "bucket frame" would be non-rotating--that is, we would have to take the worldline at the origin of the above frame and Fermi-Walker transport its spatial basis vectors. This would give us a different frame field than the one described in the paper, one in which every worldline other than the one at the origin would not be at rest in the given coordinates, but would be "rotating" about the origin, relative to the coordinates, in the opposite sense to the "rotation of the universe", i.e., in the negative $\varphi$ direction. And that means we have to look at other Christoffel symbols besides $\Gamma_{tt}^{r}$ to compute the relevant invariants, proper acceleration and vorticity.

I haven't had time yet to do those computations in detail, but just looking at the Christoffel symbols, we have $\Gamma_{t \varphi}^{r} \neq 0$, so we would expect a nonzero proper acceleration in the $r$ direction for an object that is rotating about the origin in the $\varphi$ direction. And that means that water in a "non-rotating" bucket in this universe would not have a flat shape! Whereas water in a rotating bucket--one "rotating with the universe"--would have a flat shape!

I need to look at this more.

 

[Dale]

Equations A.2 in this paper are the same as equations 2.10.2 in the paper you linked to. But as the paper I linked to makes clear,

I see the equations are the same form but not the “makes clear” statement

 

[PeterDonis]

I see the equations are the same form but not the “makes clear” statement

On p. 38: "The mass distribution generating the curvature of Gödel’s universe consists of dust which is rigidly rotating around every point." This dust must be at rest in the given coordinate chart since "dust" means the worldlines of the matter are geodesics, and worldlines at rest in the chart are geodesics, not worldlines with nonzero angular velocity in the chart.

AFAIK this is a well-known property of Godel spacetime and the given chart.

 

[Dale]

worldlines at rest in the chart are geodesics, not worldlines with nonzero angular velocity in the chart.

Hmm, you may be right, but I don't think this statement is correct. Certainly, null geodesics have non-zero angular velocity in the chart. I would suspect that there are timelike geodesics of non-zero angular velocity, the question is if they are constant radius (the null geodesics are not).

My understanding of the metric (admittedly I never bothered too much with it) was that the dust was rotating in these coordinates. However, now that you mention it I am not clear if that is an "orbital" rotation or a "spin" rotation. In other words, if I am a non-spinning observer do I see individual dust particles orbiting around me or do they stay in a constant location spinning about themselves. I had assumed the first, but I think that the second may be correct.

In either case, the fact that $\Gamma_{tt}^r=0$ clearly indicates that these coordinates are not spinning, there is no centrifugal force. That isn't the question. The question is what the matter is doing, and I think that I was misconstruing that.

 

[PeterDonis]

I would suspect that there are timelike geodesics of non-zero angular velocity

I don't think so. As I said before, looking at the Christoffel symbols, it seems like any timelike worldline with nonzero $t$ and $\varphi$ components in the chart, which would include what I think are the worldlines that would describe a non-rotating (i.e., zero vorticity) bucket, will have nonzero proper acceleration.

now that you mention it I am not clear if that is an "orbital" rotation or a "spin" rotation. In other words, if I am a non-spinning observer do I see individual dust particles orbiting around me or do they stay in a constant location spinning about themselves.

My understanding is that you see them orbiting around you. More precisely: if we put you at the origin of these coordinates, and we gyro-stabilize your spatial basis vectors so that the frame field at rest with respect to you has zero vorticity ("non-rotating"), then the worldlines of the individual dust particles, which have nonzero vorticity, will be orbiting around you.

the fact that $\Gamma_{tt}^r=0$ clearly indicates that these coordinates are not spinning, there is no centrifugal force.

It indicates that the worldlines that only have a $t$ component in this chart are geodesics. However, that congruence of worldlines also has nonzero vorticity, and that is the invariant that indicates "rotating". A counterintuitive aspect of this spacetime is that this "rotating" congruence of worldlines is geodesic--that would not be the case in flat spacetime. I'm not aware of any other spacetime that has a timelike congruence with this property (geodesic but nonzero vorticity).

It is true that, since worldlines at rest in the chart are geodesics, if you were sitting off axis (away from the origin) at rest in the chart, and you released a rock, it wouldn't move; it would float next to you. However, if you launched a rock with nonzero velocity relative to you, its trajectory would not be straight; it would curve. So even if "centrifugal force" is absent for objects at rest in the chart, "fictitious forces" in general (perhaps "Coriolis force" would be a better term) are still present.

 

[PeterDonis]

the fact that $\Gamma_{tt}^r=0$ clearly indicates that these coordinates are not spinning

If you are implying that a nonzero $\Gamma_{tt}^r$ means the coordinates are spinning, that is clearly false. As a counterexample, $\Gamma_{tt}^r \neq 0$ in Schwarzschild coordinates on Schwarzschild spacetime; those coordinates are clearly not spinning.

 

[Dale]

However, that congruence of worldlines also has nonzero vorticity, and that is the invariant that indicates "rotating".

But is rotating in this sense a spin or an orbital rotation?

If you are implying that a nonzero $\Gamma_{tt}^r$ means the coordinates are spinning, that is clearly false. As a counterexample, $\Gamma_{tt}^r \neq 0$ in Schwarzschild coordinates on Schwarzschild spacetime; those coordinates are clearly not spinning.

No, I was only implying that zero is not rotating. In any case, you are correct, but the sign is reversed. So in the Schwarzschild coordinates it is the inward centripetal force and in the rotating Minkowski coordinates it is the outward centrifugal force.

 

[PeterDonis]

is rotating in this sense a spin or an orbital rotation?

Orbital. That's what nonzero vorticity of a congruence of worldlines means: that the worldlines off center are orbiting around the worldline at the center.

I suppose this could also be considered a "spin" in the sense that, if we imagine sticks pointing out of the object following the center worldline, and connecting it with objects following neighboring worldlines, these sticks would be spinning relative to gyro-stabilized sticks carried along by the object at the center. And if we imagine sticks also pointing out of the off center objects, connecting them with neighboring objects, those sticks would also be spinning relative to gyro-stabilized sticks carried by the off center objects. But it's definitely not a "spin" in the sense of the objects following off center worldlines "spinning about themselves" as opposed to orbiting about the center worldline. The off center objects could be considered to be doing both; but not the first as opposed to the second.

 

[Dale]

Orbital. That's what nonzero vorticity of a congruence of worldlines means: that the worldlines off center are orbiting around the worldline at the center.

I suppose this could also be considered a "spin" in the sense that, if we imagine sticks pointing out of the object following the center worldline, and connecting it with objects following neighboring worldlines, these sticks would be spinning relative to gyro-stabilized sticks carried along by the object at the center. And if we imagine sticks also pointing out of the off center objects, connecting them with neighboring objects, those sticks would also be spinning relative to gyro-stabilized sticks carried by the off center objects. But it's definitely not a "spin" in the sense of the objects following off center worldlines "spinning about themselves" as opposed to orbiting about the center worldline. The off center objects could be considered to be doing both; but not the first as opposed to the second.

Ok, so it is like a rigid tidally locked orbit. The same face of each dust particle faces its neighbor always.

 

[PeterDonis]

so it is like a rigid tidally locked orbit. The same face of each dust particle faces its neighbor always.

Yes.

 

[timmdeeg]

My understanding is that you see them orbiting around you. More precisely: if we put you at the origin of these coordinates, and we gyro-stabilize your spatial basis vectors so that the frame field at rest with respect to you has zero vorticity ("non-rotating"), then the worldlines of the individual dust particles, which have nonzero vorticity, will be orbiting around you.

A and B are a comoving observers with zero proper rotational acceleration. Assuming that the situation is perfectly symmetric should mean that A is orbiting around B and B is orbiting around A.

You said:

(RR) Rotating bucket in rotating universe:

#1: Zero proper acceleration, flat surface.

#2: Nonzero vorticity, nonzero rotation.

#3: Zero angular velocity relative to rest of universe.

#4: Nonzero rotation of rest of universe.

Can I conclude from #1 that A and B see each other all the time in the same distance (because they are rotating themselves with the universe)? But that would mean that they don't recognize that they are in free fall around each other.
If correct how do they verify that their universe is rotating?

 

[PeterDonis]

A and B are a comoving observers with zero proper rotational acceleration.

"Comoving observers", yes. "Zero proper rotational acceleration", no. As I have already said, the congruence of worldlines that are at rest in the given chart on Godel spacetime has nonzero vorticity; that means it is a rotating congruence.

Can I conclude from #1 that A and B see each other all the time in the same distance

Yes. The congruence has zero expansion and shear so it is rigid--distances between neighboring worldlines are constant.

that would mean that they don't recognize that they are in free fall around each other.

Why not? They can tell they are in free fall, and they can detect the nonzero vorticity (see below).

If correct how do they verify that their universe is rotating?

By detecting the nonzero vorticity: they can carry along gyro-stabilized spatial vectors with them and observe that the neighboring observers are rotating with respect to those gyro-stabilized vectors.

 

[timmdeeg]

"Comoving observers", yes. "Zero proper rotational acceleration", no. As I have already said, the congruence of worldlines that are at rest in the given chart on Godel spacetime has nonzero vorticity; that means it is a rotating congruence.

I don't understand.

I have exchanged here

(RR) Rotating bucket in rotating universe:
#1: Zero proper acceleration, flat surface.

the rotating bucket with an rotating observer. Can you please elaborate a bit the meaning of "rotating congruence", resp. of "nonzero vorticity"?

I understood this quote such that bucket and universe have the same angular speed so that they don't rotate relative to each other. The bucket feels no forces (what I intended to express saying "Zero proper rotational acceleration") and hence the water is flat. Is that wrong?

By detecting the nonzero vorticity: they can carry along gyro-stabilized spatial vectors with them and observe that the neighboring observers are rotating with respect to those gyro-stabilized vectors.

I am not familiar with gyro-stabilized vectors though I guess their function was explained earlier in the thread. Am I correct that they detect "proper rotation" whereby I guess (but am not sure) that this means the same as "proper rotational acceleration"?

 

[PeterDonis]

Can you please elaborate a bit the meaning of "rotating congruence", resp. of "nonzero vorticity"?

Vorticity is part of the kinematic decomposition, which is a very useful (though advanced) tool for analyzing motion in relativity.

See here for a start at the gory details:

https://en.wikipedia.org/wiki/Congr...atical_decomposition_of_a_timelike_congruence

I understood this quote such that bucket and universe have the same angular speed so that they don't rotate relative to each other.

More precisely, the bucket is at rest (no change in spatial coordinates) in the coordinate chart that was specified for Godel spacetime. And all the other matter in the universe is also at rest in this chart. But this chart is a "rotating" chart--the family of worldlines of objects at rest in this chart has nonzero vorticity. The Born chart on Minkowski spacetime, which often comes up in discussions of "rotating disks" in flat spacetime and the Ehrenfest paradox, has the same property.

The bucket feels no forces (what I intended to express saying "Zero proper rotational acceleration")

It has zero proper acceleration, yes. That's why its surface is flat.

However, it does not have zero vorticity. Physically, this means that if you set up three mutually perpendicular rods attached to the bucket, stabilized by gyroscopes, the bucket would be rotating with respect to these gyro-stabilized rods. And this is true even though the bucket's surface is flat and it has zero proper acceleration. Yes, this is highly counterintuitive. Welcome to Godel spacetime.

I am not familiar with gyro-stabilized vectors though I guess their function was explained earlier in the thread. Am I correct that they detect "proper rotation"

Yes. See above.

I guess (but am not sure) that this means the same as "proper rotational acceleration"?

"Proper rotational acceleration" is not a term I'm familiar with; I quoted it from a previous post of yours.

 

[Dale]

But as the paper I linked to makes clear, these equations are for a rotating frame! That is, objects at rest in this frame are "rotating with the universe", since the matter in the universe itself is at rest in this frame. The "at rest" frame field associated with this frame, which the paper I linked to calls a "static local tetrad", is geodesic--zero proper acceleration--but it has nonzero vorticity; it's the vorticity you would expect for a family of observers rotating with "the angular velocity of the universe".

I think that your interpretation of this dissertation is correct, and I see no major flaw with the dissertation either, at least in this topic. I think that the coordinates I had were indeed incorrect for my intended use.

Unfortunately, there seems to be no easy shortcut to find the distortion of a non-rotating "blob" in the Goedel metric. It appears that it will be larger than 0 as I had indicated, but I also am quite certain that it will not be the same as a rotating blob in Minkowski spacetime. In fact, from reading section 4.2.5 I am quite sure that the result will be bizarre. I think I have no intuition about this spacetime.

I think that the easiest that we could do is, instead of my Christoffel-symbol based approach, we could look at the Newtonian potential, which should describe the shape of a blob at rest wrt some coordinates. However, I don't think that will work for the Goedel universe since the assumptions to have a Newtonian potential are questionable. So I think there is simply no shortcut available, and the full calculation is beyond me.

On a broader note, the author of the dissertation makes a great point in regards to Mach's principle (a quote from Rindler). And mathematically the vacuum stress energy in 4.16 is essentially non-Machian. Even if it does not cause flattening of a blob it does produce gyroscopic precession for an object which is at rest with respect to the matter. That is an unambiguous violation of Mach's principle. Of course, Goedel's spacetime is rather contrived, so it does not answer whether or not the actual universe is Machian, but it certainly refutes the idea that GR is Machian.

 

[timmdeeg]

@PeterDonis
Thanks for your enlightening answers in #13 [Moderator's note: was #194 in old thread].

 

[PeterDonis]

there seems to be no easy shortcut to find the distortion of a non-rotating "blob" in the Goedel metric.

I haven't found a shortcut either. I'm working through the brute force computations using the Christoffel symbols and vector components, along the same lines as my Insights articles from a while back on Fermi-Walker transport:

https://www.physicsforums.com/insights/how-to-study-fermi-walker-transport-in-minkowski-spacetime/

https://www.physicsforums.com/insights/fermi-walker-transport-in-schwarzschild-spacetime/

https://www.physicsforums.com/insights/fermi-walker-transport-in-kerr-spacetime/

It appears that it will be larger than 0 as I had indicated, but I also am quite certain that it will not be the same as a rotating blob in Minkowski spacetime.

I think this is correct. More precisely, for the same angular velocity of bucket relative to "rest of universe", I don't think the distortions will be the same. This is based on comparing the Christoffel symbols and local tetrad components for the rotating frame in Minkowski spacetime and the cylindrical chart in Godel spacetime, from the catalog you linked to earlier.

we could look at the Newtonian potential, which should describe the shape of a blob at rest wrt some coordinates

Is this approach workable for the rotating frame in Minkowski spacetime? If so, it could at least be tried for Godel spacetime, since the general form of the metric is similar (although the specific values of the coefficients are different).

 

[PeterDonis]

I'm working through the brute force computations using the Christoffel symbols and vector components, along the same lines as my Insights articles from a while back on Fermi-Walker transport

As a warmup, I'm going to post my computations for the simpler case of a rotating frame in flat Minkowski spacetime. For this case, the metric is (in cylindrical coordinates, which will be the most useful for all the cases we will be discussing):

$$
ds^2 = - \left( 1 - \Omega^2 r^2 \right) dt^2 + 2 \Omega r^2 dt d\varphi + dr^2 + r^2 d\varphi^2 + dz^2
$$

Here $\Omega$ is the "angular velocity of rotation" of the chart (though, as we'll see, we have to be careful in interpreting what that means). I have changed notation slightly from the corresponding equations in the papers linked to previously, because we are going to be considering frame fields describing observers who will also possibly have an "angular velocity of rotation" relative to a given chart, and I want to use lower-case $\omega$ for that, so I'm using upper-case $\Omega$ for the "angular velocity" of the chart itself.

The frame field describing observers at rest in this chart (often called "Langevin observers" in the literature, and the chart itself is often called the "Born chart") is:

$$
\hat{e}_0 = \frac{1}{\sqrt{1 - \Omega^2 r^2}} \partial_t
$$

$$
\hat{e}_1 = \partial_r
$$

$$
\hat{e}_2 = \frac{\Omega r}{\sqrt{1 - \Omega^2 r^2}} \partial_t + \frac{\sqrt{1 - \Omega^2 r^2}}{r} \partial_\varphi
$$

From here on out I will ignore the $z$ coordinate and its corresponding metric coefficient and frame field vector, since they will always be the same. Note that the numbering here is different from what I used in the Insights articles, where the "1" coordinate was $z$.

As shown in my Insights articles, there will be a general pattern in all of the cases we are considering, in the derivatives along an observer's worldline of the frame field vectors. This pattern is as follows:

$$
\nabla_{\hat{e}_0} \hat{e}_0 = A \hat{e}_1
$$

$$
\nabla_{\hat{e}_0} \hat{e}_1 = A \hat{e}_0 + V \hat{e}_2
$$

$$
\nabla_{\hat{e}_0} \hat{e}_2 = - V \hat{e}_1
$$

Here $A$ is the "proper acceleration" and $V$ is the "vorticity" associated with the frame field (and the congruence of worldlines formed by the integral curves of $\hat{e}_0$). (Note that here I am using $V$ for the vorticity, instead of $\Omega$ as I did in the Insights articles, because I have repurposed the symbol $\Omega$ as described above.)

So to find $A$ and $V$ we simply compute $\nabla_{\hat{e}_0} \hat{e}_0$ and $\nabla_{\hat{e}_0} \hat{e}_2$. (As a check, we can also compute $\nabla_{\hat{e}_0} \hat{e}_1$ to make sure things match up, but we won't do that here; it is left as an exercise for the reader.) For the frame field given above, these computations are simple:

$$
\nabla_{\hat{e}_0} \hat{e}_0 = \left[ \Gamma^r_{tt} \left( \hat{e}_0 \right)^t \left( \hat{e}_0 \right)^t \right] \partial_r
$$

$$
\nabla_{\hat{e}_0} \hat{e}_2 = \left[ \Gamma^r_{tt} \left( \hat{e}_0 \right)^t \left( \hat{e}_2 \right)^t + \Gamma^r_{t \varphi} \left( \hat{e}_0 \right)^t \left( \hat{e}_2 \right)^ \varphi \right] \partial_r
$$

Since $\hat{e}_1 = \partial_r$, we easily find that (remembering that $V$ is proportional to minus $\hat{e}_1$):

$$
A = - \frac{\Omega^2 r}{1 - \Omega^2 r^2}
$$

$$
V = \frac{\Omega}{1 - \Omega^2 r^2}
$$

The physical meaning of the proper acceleration is straightforward: it is the reading on an accelerometer carried by an observer following the given worldline. The physical meaning of vorticity, however, is a little more complicated. The spatial unit vectors $\hat{e}_1$ and $\hat{e}_2$ of the frame field "point" at fixed neighboring observers moving along integral curves of $\hat{e}_0$; that is, the vector $\hat{e}_1$ always points at the same neighboring observer, the one in the "radial outward" direction, and the vector $\hat{e}_2$ always points at the same neighboring observer in the "tangential forward" direction. (Think of being at a specific point on a merry-go-round with a radial outward and a tangential forward arrow painted on the floor of the merry-go-round where you are standing; those arrows will always point at the same neighboring object on the merry-go-round.)

Now, suppose that in addition to those frame field vectors, our observers are also carrying along a pair of perpendicular spatial vectors, in the same plane as $\hat{e}_1$ and $\hat{e}_2$, but whose directions are determined and stabilized by gyroscopes. Then the vorticity $V$ is the angular velocity at which the frame field vectors $\hat{e}_1$ and $\hat{e}_2$ are rotating, with respect to the gyroscope-stabilized vectors. A positive vorticity, as we have in this example, means that the frame field vectors are rotating in the positive $\phi$ direction, relative to the gyro-stabilized vectors.

So when we say that nonzero vorticity means "rotation", more precisely, it means "rotation relative to gyro-stabilized vectors". (The more technical way to say "gyro-stabilized" is "Fermi-Walker transported", and when we compute $V$ above we are computing by how much the frame field vectors $\hat{e}_1$ and $\hat{e}_2$ fail to be Fermi-Walker transported. My Insights articles go into much more detail about this.) "Rotation" here does not mean that neighboring observers are rotating "relative to each other", if we use the frame field vectors as our standard; the frame field vectors remain rigidly pointing at the same neighboring observers. It only means they are rotating relative to each other if we use the gyro-stabilized vectors as our standard for "not rotating". (In merry-go-round language, we can view everything riding along on the merry-go-round as being "at rest" and not rotating relative to the merry-go-round, or we can view it as being rotating relative to gyro-stabilized vectors.)

There are further wrinkles here as well. First, notice the extra factor of $1 / \left( 1 - \Omega^2 r^2 \right)$ in $A$ and $V$, compared to what the Newtonian values would be. In the case of $A$, this factor is entirely due to time dilation; heuristically, since "acceleration" is a second time derivative, it will contain the square of the time dilation factor $\sqrt{1 - \Omega^2 r^2}$ that appears in the denominator of the 4-velocity $\hat{e}_0$. In the case of $V$, however, since it is an angular velocity, only one time dilation factor should appear. Yet there are two (the square of the factor appears, just as in $A$). As I discussed in my Insights article, this second factor is due to Thomas precession: relative to an object fixed at infinity, the gyro-stabilized vectors carried by observers following integral curves of our frame field in this example will appear to rotate backwards.

The other wrinkle worth noting here concerns "fictitious forces". If we compute the geodesic equation in the rotating chart given above, we will find that it contains terms that go by names like "centrifugal force" and "Coriolis force". In this particular case, those forces look like we would intuitively expect based on the Newtonian case (but with some corrections in magnitudes due to time dilation, just as for the proper acceleration above). However, since ultimately the geodesic equation is determined by the spacetime geometry, we will find that a "rotating" frame field (i.e., one with nonzero vorticity) will not always give rise to the same set of "fictitious forces" if we look at different spacetime geometries.

More to come.

 

[PeterDonis]

Okay, second warmup post. This time we're going to look at the same rotating chart in Minkowski spacetime as in the previous post, but with a frame field that is not at rest in the chart; now we're going to allow the frame field to have an angular velocity of $\omega$ relative to the coordinate chart. This will only change the frame field vectors $\hat{e}_0$ and $\hat{e}_2$; the new vectors are:

$$
\hat{e}_0 = \frac{1}{\sqrt{1 - \left( \Omega + \omega \right)^2 r^2}} \partial_t + \frac{\omega}{\sqrt{1 - \left( \Omega + \omega \right)^2 r^2}} \partial_\varphi
$$

$$
\hat{e}_2 = \frac{\left( \Omega + \omega \right) r}{\sqrt{1 - \left( \Omega + \omega \right)^2 r^2}} \partial_t + \frac{1 - \Omega \left( \Omega + \omega \right) r^2}{\sqrt{1 - \left( \Omega + \omega \right)^2 r^2}} \frac{1}{r} \partial_\varphi
$$

This in turn changes the proper acceleration and vorticity calculations as follows:

$$
\nabla_{\hat{e}_0} \hat{e}_0 = \left[ \Gamma^r_{tt} \left( \hat{e}_0 \right)^t \left( \hat{e}_0 \right)^t + 2 \Gamma^r_{t \varphi} \left( \hat{e}_0 \right)^t \left( \hat{e}_0 \right)^\varphi + \Gamma^r_{\varphi \varphi} \left( \hat{e}_0 \right)^\varphi \left( \hat{e}_0 \right)^\varphi \right] \partial_r
$$

$$
\nabla_{\hat{e}_0} \hat{e}_2 = \left[ \Gamma^r_{tt} \left( \hat{e}_0 \right)^t \left( \hat{e}_2 \right)^t + \Gamma^r_{t \varphi} \left( \hat{e}_0 \right)^t \left( \hat{e}_2 \right)^\varphi + \Gamma^r_{t \varphi} \left( \hat{e}_2 \right)^t \left( \hat{e}_0 \right)^\varphi + \Gamma^r_{\varphi \varphi} \left( \hat{e}_0 \right)^\varphi \left( \hat{e}_2 \right)^\varphi \right] \partial_r
$$

Which leads to:

$$
A = - \frac{\left( \Omega + \omega \right)^2 r}{1 - \left( \Omega + \omega \right)^2 r^2}
$$

$$
V = \frac{\Omega + \omega}{1 - \left( \Omega + \omega \right)^2 r^2}
$$

This is nice because we can immediately see that, if $\omega = - \Omega$, both $A$ and $V$ vanish; and of course $\omega = - \Omega$ is just the condition for the congruence of ordinary inertial observers! In other words, we have laboriously shown in the rotating chart what we already knew was obviously true, that a family of inertial observers in Minkowski spacetime, all at rest relative to each other, have zero proper acceleration and zero vorticity. However, the labor was still not wasted, because, first, we have seen how to find the frame field of "non-rotating" observers in a rotating chart, and second, the form of the calculations we have done will remain the same in other spacetime geometries, although the results will no longer be quite so simple.

More still to come!

 

[PeterDonis]

An interlude with a question: is there an error in both of the papers we've linked to? Meaning, the "Catalog of Spacetimes" that @Dale linked to and the dissertation I linked to?

The possible error is in the Christoffel symbol $\Gamma^r{}_{t \varphi}$. I'll use the form from the dissertation; it is given as (leaving out factors of $c$ since I'm using units where $c = 1$ for simplicity and to reduce clutter):

$$
\Gamma^r{}_{t \varphi} = \frac{\sqrt{2} r}{r_G} \left[ 1 + \left( \frac{r}{r_G} \right) \right]^2
$$

However, I think the last factor should have the square inside the square brackets instead of outside. I have two reasons for thinking this, the second of which I'll save for my next post when I start showing the computations for Godel spacetime. The first reason, though, is simple; when I compute this Christoffel symbol directly, I get this:

$$
\Gamma^r{}_{t \varphi} = - \frac{1}{2} g^{rr} \frac{\partial g_{t \phi}}{\partial r} = - \frac{1}{2} \left[ 1 + \left( \frac{r}{r_G} \right)^2 \right] \frac{\partial}{\partial r} \left( - \frac{\sqrt{2} r^2}{r_G} \right) = \frac{\sqrt{2} r}{r_G} \left[ 1 + \left( \frac{r}{r_G} \right)^2 \right]
$$

In other words, the factor in the square brackets is $g^{rr}$, which has the square only on the $r / r_G$ term.

Thoughts?

 

[Ibix]

Using the metric $$g_{ab}=\pmatrix{-1&0&{{r^2}\over{\sqrt{2}\,a}}&0\cr 0&{{1}\over{{{r^2 }\over{4\,a^2}}+1}}&0&0\cr {{r^2}\over{\sqrt{2}\,a}}&0&r^2\,\left(1- {{r^2}\over{4\,a^2}}\right)&0\cr 0&0&0&1\cr }$$(copied from 2.10.1 in Dale's link in the OP edit: and copied incorrectly - the off-diagonal elements should have a negative sign) Maxima tells me that $$\Gamma^r_{t\phi}=-\frac r{a\sqrt 2}{{r^2+4a^2}\over{4a^2}}$$which agrees with PeterDonis (give or take a sign - possibly metric signature? Edit: no, it's the mistake with the off-diagonal elements. Correcting them corrects the sign here.), assuming $2a=r_G$ (which it is, comparing $g_{t\phi}$ components).

 

[PeterDonis]

which agrees with PeterDonis

Ok, good. Various computations work out quite a bit more neatly that way.

assuming $2a=r_G$

Yes.

 

[PeterDonis]

give or take a sign

You have your off-diagonal metric coefficients as positive. They should be negative.

 

[Ibix]

You have your off-diagonal metric coefficients as positive. They should be negative.

Ah yes. Corrected, and now we agree.

 

[Dale]

Is this approach workable for the rotating frame in Minkowski spacetime? If so, it could at least be tried for Godel spacetime, since the general form of the metric is similar (although the specific values of the coefficients are different).

I think it is workable. This is the Newtonian approach I was thinking of:

https://farside.ph.utexas.edu/teaching/celestial/Celestial/node52.html

It may not be as much of a short cut as I had remembered.

The key is just adding the centrifugal and the gravitational potentials. This is automatically done in the metric. But then the other steps were not as trivial as I thought.

 

[PeterDonis]

Ok, having done some warmup with rotating coordinates in flat spacetime, we will now start in on Godel spacetime. It will be helpful to define some convenience functions of $r$; some of these are defined in the references, but I have added a few more:

$$
K^2 = 1 + \left( \frac{r}{r_G} \right)^2
$$

$$
\bar{K}^2 = 1 - \left( \frac{r}{r_G} \right)^2
$$

$$
\Omega = \frac{\sqrt{2}}{r_G}
$$

$$
\Gamma = \frac{1}{\sqrt{1 + \sqrt{2} \Omega \omega r^2 - \omega^2 r^2 \bar{K}^2}}
$$

$$
\Delta = \frac{1}{r K}
$$

$$
a = - \Omega r^2 + \omega r^2 \bar{K}^2
$$

$$
b = 1 + \Omega \omega r^2
$$

Note that I am using units in which $c = 1$ and that I am using $\omega$ instead of $\zeta$ for the angular velocity of the frame field with respect to the coordinate chart.

We can now write the metric as follows (as before, leaving out the $z$ coordinate and its corresponding metric coefficient since those will play no role in the analysis):

$$
ds^2 = - dt^2 - 2 \Omega r^2 dt d \varphi + \frac{dr^2}{K^2} + r^2 \bar{K}^2 d\varphi ^2
$$

Our frame field vectors are:

$$
\hat{e}_0 = \Gamma \partial_t + \Gamma \omega \partial_\varphi
$$

$$
\hat{e}_1 = K \partial_r
$$

$$
\hat{e}_2 = \Gamma \Delta a \partial_t + \Gamma \Delta b \partial_\varphi
$$

And the only two connection coefficients we will need are:

$$
\Gamma^r{}_{t \varphi} = \Omega r K^2
$$

$$
\Gamma^r{}_{\varphi \varphi} = r K^2 \left( 1 - \Omega^2 r^2 \right)
$$

We can now write down the covariant derivatives we will need:

$$
\nabla_{\hat{e}_0} \hat{e}_0 = \left[ 2 \Gamma^r_{t \varphi} \left( \hat{e}_0 \right)^t \left( \hat{e}_0 \right)^\varphi + \Gamma^r_{\varphi \varphi} \left( \hat{e}_0 \right)^\varphi \left( \hat{e}_0 \right)^\varphi \right] \partial_r
$$

$$
\nabla_{\hat{e}_0} \hat{e}_2 = \left[ \Gamma^r_{t \varphi} \left( \hat{e}_0 \right)^t \left( \hat{e}_2 \right)^\varphi + \Gamma^r_{t \varphi} \left( \hat{e}_2 \right)^t \left( \hat{e}_0 \right)^\varphi + \Gamma^r_{\varphi \varphi} \left( \hat{e}_0 \right)^\varphi \left( \hat{e}_2 \right)^\varphi \right] \partial_r
$$

Since now we have $\hat{e}_1 = K \partial_r$, we can divide each of the above formulas by $K$ (which is easy because it just means canceling one factor of $K$ in each connection coefficient--this is why I said in an earlier post that things work out much more neatly with the correction to $\Gamma^r{}_{t \varphi}$) to obtain formulas for the proper acceleration $A$ and vorticity $V$ (again remembering that the vorticity is minus the above formula divided by $K$):

$$
A = \omega r K \Gamma^2 \left[ 2 \Omega + \omega \left( 1 - \Omega^2 r^2 \right) \right]
$$

$$
V = - \Gamma^2 \left[ \left( \Omega + \omega \right) b + \omega \left( \Omega a - \Omega^2 r^2 b \right) \right]
$$

Before going into this in the general case (which I'll save for my next post), we can gain some initial insight by considering the special case $\omega = 0$, i.e., the case of observers at rest in our coordinate chart, which means they are at rest "relative to the universe"--the matter in the universe is moving along worldlines at rest in this chart. For $\omega = 0$, we have $\Gamma = b = 1$ (we also have $a = - \Omega r^2$, but we will not need this since the only term in $a$ in the vorticity is multiplied by $\omega$ so it vanishes for this special case), and this gives us:

$$
A = 0
$$

$$
V = - \Omega
$$

Of course you can now see why I defined the symbol $\Omega$ as I did: the magnitude of the vorticity for this special case is expected to be the "angular velocity of rotation of the universe". (Note that we do not expect any time dilation factor, since the metric coefficient $g_{tt}$ has no such factor in it; coordinate time is the same as proper time along worldlines at rest in the chart.) And we see that these worldlines are geodesics--zero proper acceleration--which again is as we expected.

However, note the sign of the vorticity above. It is negative! This is the opposite of what we might "naively" expect. Let's walk through again what the vorticity means. We imagine that each of the observers at rest in our chart, who are moving along with the matter in the universe, carry a set of gyro-stabilized vectors with them, and watch how the vectors that connect them to neighboring observers (the ones that are painted on the floor of the merry-go-round) rotate wtih respect to the gyro-stabilized vectors. The negative vorticity tells us that the merry-go-round vectors rotate with angular velocity $\Omega$ in the minus $\varphi$ direction with respect to the gyro-stabilized vectors--i.e., in the opposite direction from the direction we might have expected to be the "direction of rotation of the universe".

(Comparing the metric above with the metric of the rotating chart in Minkowski spacetime, we can see that the reversal of sign of the vorticity comes from the reversal of sign of the $g_{t \varphi}$ metric coefficient, which is positive in the rotating chart in Minkowski spacetime but is negative in Godel spacetime.)

I'll leave all this as food for thought until my next post.

 

[Dale]

Ok, I officially have no intuition about this spacetime. Bizarre

 

[PeterDonis]

Bizarre

The negative vorticity is mentioned in passing on the Wikipedia page, of all places; but I have not seen any real discussion of it.

 

[vanhees71]

Well, after all it's Gödel who found this solution. It's amazing that one of the greatest logicians ever was so weird in everyday-life matters ;-).

 

[MathematicalPhysicist]

Well, after all it's Gödel who found this solution. It's amazing that one of the greatest logicians ever was so weird in everyday-life matters ;-).

It's logical being illogical...
That's why I am so fascinated by this subject of paraconsistent logics.

 

[pervect]

Some general random comments at a broad, non-detailed level that I hope may be helpful.

Tensor quantities, like proper acceleration can (I beleive) be regarded as absolute, because tensors are coordinate independent. Note that this implies that 4-velocity, being a tensor quantity, should be regarded as absolute, even though 3-velocities are famously relative. It's possible this point could be argued, I suppose, I do not have a hard reference that defines the English terms "absolute and relative" mathematically. But I believe that this approach makes sense.

Vorticity is another example (besides proper acceleration) of a tensor quantity. As such, it should be regarded as absolute, because of the coordinate independent property of tensors.

However, there is no guarantee that when someone talks about an object "not rotating", that they are talking about the tensor quantity of vorticity. So some work needs to be done to figure out what a person may be talking about when they talk about non-rotating.

To take a specific example, an object in a Kerr spacetime that is fixed relative to the absolute stars, i.e. a telescope pointed at a distant "guide star", will in general have a non-zero vorticity.

The tensor concept of vorticity applies directly to timelike congruences of worldlines. So some general discussion of what a time-like congruence of worldlines means would seem helpful. If one has a physical object (a bucket full of water, for example), the timelike congruence associated with the physical object can be regarded as the set of 4-velocities of all points in the object at every moment in time. This is just a unit vector field (a timelike unit vector field for a timelike congruence) at every event, since the 4-velocity has unit length.

Worldlines enter into this picture as the integral curves of the above vector field. The integral curves are worldlines of "points on the object", and the tangent vectors of these worldline generate the unit timelike vector field that mathematically defines the timelike congruence.

It'd get too far afield to discuss worldines and tangent vectors, apologies if the terms are not familiar.

 

[PeterDonis]

Tensor quantities, like proper acceleration can (I beleive) be regarded as absolute, because tensors are coordinate independent.

Yes.

this implies that 4-velocity, being a tensor quantity, should be regarded as absolute

Yes. A way of describing it that makes the absolute nature more apparent is that it is the tangent vector to a worldline at a particular point. (Proper acceleration is the path curvature of the worldline at a particular point.)

Vorticity is another example (besides proper acceleration) of a tensor quantity.

Yes.

there is no guarantee that when someone talks about an object "not rotating", that they are talking about the tensor quantity of vorticity.

Yes. If one wants to be more precise, one can describe zero vorticity as "not rotating with respect to local gyroscopes".

an object in a Kerr spacetime that is fixed relative to the absolute stars, i.e. a telescope pointed at a distant "guide star", will in general have a non-zero vorticity.

Yes. Note, however, that this comparison requires the spacetime to be asymptotically flat, and Godel spacetime is not. In other words, in Godel spacetime, there is no analogue to "the distant stars" that can be used to define an alternative sense of "not rotating".