I Godel metric in a cylindrical chart

[space-time]

Can someone express the Godel metric line element in cylindrical coordinates? I keep looking for this line element, but no source clearly gives it to me. Can you please express it using the (- + + +) signature and while retaining all c terms?
Thanks.

Here is the line element in Cartesian coordinates:

ds² = (1/2ω²)[-c²dt² - 2ce^xdzdt + dx² + dy² - (1/2)e^2xdz²]

 

[martinbn]

Why can't you do it? Have you tried? How far did you get?

 

[PeterDonis]

I keep looking for this line element, but no source clearly gives it to me.

Where have you looked? Have you Googled "Godel spacetime"?

 

[space-time]

Why can't you do it? Have you tried? How far did you get?

Well, I know that the conversion between Cartesian coordinates and cylindrical coordinates is as such:

x = rcos(θ)
y = rsin(θ)
z = z

However, I don't think that the following is correct:

ds² = (1/2ω²)[-c²dt² - 2ce^rcos(θ)dzdt + dr² - (1/2)e^2rcos(θ)dz²]

I don't quite know where to go from here.

 

[space-time]

Where have you looked? Have you Googled "Godel spacetime"?

I have, but the wiki's section for the cylindrical chart does not actually report the line element in cylindrical coordinates, and other papers that I try to look at either use some strange constants that they don't really define, and it just seems inconsistent from paper to paper.

 

[PeterDonis]

other papers that I try to look at

Can you give any specific examples?

 

[martinbn]

Well, I know that the conversion between Cartesian coordinates and cylindrical coordinates is as such:

x = rcos(θ)
y = rsin(θ)
z = z

However, I don't think that the following is correct:

ds² = (1/2ω²)[-c²dt² - 2ce^rcos(θ)dzdt + dr² - (1/2)e^2rcos(θ)dz²]

I don't quite know where to go from here.

Why do you think it is not correct?

 

[PeterDonis]

I don't think that the following is correct

When you transform the coordinates, you have to transform the differentials as well. So you would have

$$
dx = dr \cos \theta - r \sin \theta d \theta
$$
$$
dy = dr \sin \theta + r \cos \theta d \theta
$$

And you would need to plug those into the Cartesian line element as well.

 

[space-time]

When you transform the coordinates, you have to transform the differentials as well. So you would have

$$
dx = dr \cos \theta - r \sin \theta d \theta
$$
$$
dy = dr \sin \theta + r \cos \theta d \theta
$$

And you would need to plug those into the Cartesian line element as well.

Ok, so here is what I got:

ds² = (1/2ω²)[-c²dt² - 2ce^rcos(θ)dzdt + dr² + r²dθ² - (1/2)e^2rcos(θ)dz²]

Is that right?

 

[PeterDonis]

Is that right?

It looks right, yes.

 

[PeterDonis]

It looks right, yes.

Note, however, that you are using different coordinate names than the Wikipedia article does. The Wikipedia article uses $y$ for what you are calling $z$, and $z$ for what you are calling $y$. So your choice of transformation to cylindrical coordinates is not the same as the one the Wikipedia article's cylindrical chart section is implicitly using.

Another way of seeing the difference is to note that, in your cylindrical chart, $\theta$ is always a spacelike coordinate, since a line element with only $d\theta$ nonzero always has positive $ds^2$. However, the purpose of the cylindrical chart on the Wikipedia page (which, admittedly, the Wikipedia page does not do a good job of explaining) is to show the closed timelike curves in Godel spacetime explicitly by showing under what conditions the angular coordinate $\theta$ in the cylindrical chart is timelike, meaning that a curve with only $d\theta$ nonzero will be a closed timelike curve.

 

[space-time]

Note, however, that you are using different coordinate names than the Wikipedia article does. The Wikipedia article uses $y$ for what you are calling $z$, and $z$ for what you are calling $y$. So your choice of transformation to cylindrical coordinates is not the same as the one the Wikipedia article's cylindrical chart section is implicitly using.

Another way of seeing the difference is to note that, in your cylindrical chart, $\theta$ is always a spacelike coordinate, since a line element with only $d\theta$ nonzero always has positive $ds^2$. However, the purpose of the cylindrical chart on the Wikipedia page (which, admittedly, the Wikipedia page does not do a good job of explaining) is to show the closed timelike curves in Godel spacetime explicitly by showing under what conditions the angular coordinate $\theta$ in the cylindrical chart is timelike, meaning that a curve with only $d\theta$ nonzero will be a closed timelike curve.

Lol, it is like you read my mind. I was just looking at those differentials and noticed how with the line element that I just used, t and z seem to always be timelike, but r and θ seem to always be spacelike.

In that case, would I need a different line element or something to see the case where the angular coordinate is timelike?

 

[PeterDonis]

would I need a different line element or something to see the case where the angular coordinate is timelike?

You would need a different cylindrical chart, one in which the transformation is $x = r \cos \theta$ and $y = r \sin \theta$ with the definitions of $x$ and $y$ in the Wikipedia article instead of yours (i.e., with $y$ being what you are calling $z$).

 

[space-time]

You would need a different cylindrical chart, one in which the transformation is $x = r \cos \theta$ and $y = r \sin \theta$ with the definitions of $x$ and $y$ in the Wikipedia article instead of yours (i.e., with $y$ being what you are calling $z$).

Ok, I made the necessary switching of the roles of y and z, and I got a new line element. However, I notice a certain problem:

When I set all the differentials except for that of the angular coordinate equal to 0, ds² is negative most of the time, but depending on the value of theta, it is 0 sometimes and it is positive sometimes. In other words, the theta coordinate is not entirely timelike over the entire interval theta = [0, 2π].

Is this problematic? Does this mean I have made some major mistake somewhere, or is this just common?

Can there still be a CTC this way? For reference, the theta coordinate is still timelike at both theta = 0 and theta = 2π. Just not at every angle in between.

 

[PeterDonis]

However, I notice a certain problem

Which we can't possibly help you solve unless we see the new line element you got and how you got it, so we can check your derivation. Please post that information.

Does this mean I have made some major mistake somewhere

Yes, but we can't tell where unless you post the details, as above.

 

[pervect]

To call a metric "cylinderical coordinates", I'd expet it to have some underlying axis symmetry.

This would imply the existence of a Killing vector associated with such axis symmetry.

If we look at https://www.physicsforums.com/threads/schwarzschild-killing-vectors.484181/, we see the sorts of Killing vectors we'd expect for axis symmetric metrics , such as $y \partial_x - x \partial_y$ for symmetry about the z axis. Or more likely $z \partial_x - x \partial_z$ for symmetry about the y axis, wiki mentions something about "rotatation about the y axis".

However, I tested the above candidates in the Godel metric, and none of them were Killing vectors.

So at the moment, I'm not seeing where the cylindrical symmetry in this line element might be, assuming any exists. I haven't tried normalizing $\partial_y$ to be a unit length, though.

Wiki does give the Killing vectors associated with the Godel metric, but I'm not able at the moment to interpret any of them as implying axis symmetry.

 

[PeterDonis]

To call a metric "cylinderical coordinates", I'd expet it to have some underlying axis symmetry.

Normally that's the case when we choose a coordinate chart like this, but it's not in the case of Godel spacetime. Godel spacetime is not axially symmetric [edit--I was wrong here, it is, see post #25]; as you correctly note, none of the five Killing vector fields correspond to axial symmetry. Another way to test for it would be to see if any of the five Killing vector fields have closed orbits, since axial symmetry means there is a KVF with closed orbits that satisfies the SO(2) Lie algebra. Godel spacetime does not have one.

A way of seeing the difference physically is to observe that, in an axially symmetric spacetime [edit--meaning a "normal" one that is only so around one axis--see post #26] (for example, Kerr spacetime), the choice of a cylindrical coordinate chart only works for one particular axis. However, in Godel spacetime, it works for any axis; that is, you can "center" your cylindrical coordinate chart along any axial line you want. The angular $\phi$ coordinate then describes angle about that axis in a plane of constant $z$, and for distances $r$ far enough from the chosen axis, $\phi$ becomes timelike, indicating the presence of closed timelike curves. But since you can do this about any axis you choose, this shows that you can find CTCs anywhere in Godel spacetime. But to describe different CTCs in cylindrical coordinates requires you to choose a different axis. There is no single coordinate chart about one axis that describes all CTCs in this spacetime.

 

[space-time]

Which we can't possibly help you solve unless we see the new line element you got and how you got it, so we can check your derivation. Please post that information.
Yes, but we can't tell where unless you post the details, as above.

The new line element that I got is:

ds² = (1/2ω²)[-c²dt² - 2ce^rcos(θ)sin(θ)drdt - 2ce^rcos(θ)rcos(θ)dθdt - (1/2)e^2rcos(θ)sin²(θ)dr² - re^2rcos(θ)sin(θ)cos(θ)drdθ - (1/2)e^2rcos(θ)r²cos²(θ)dθ² + cos²(θ)dr² - 2rsin(θ)cos(θ)drdθ + r²sin²(θ)dθ² + dz²]

This was the transformation from the cartesian line element:
ds² = (1/2ω²)[-c²dt² - 2ce^xdydt - (1/2)e^2xdy² + dx² + dz²]

My coordinate transformations were:

x = rcos(θ)
y = rsin(θ)
dx = cos(θ)dr - rsin(θ)dθ
dy = sin(θ)dr + rcos(θ)dθ

 

[PeterDonis]

The new line element that I got is

Please use the PF LaTeX feature. Your post is unreadable without it.

 

[PeterDonis]

The new line element that I got is

Also, you need to tell us what coordinate transformation you used.

 

[space-time]

Also, you need to tell us what coordinate transformation you used.

I hope this one is readable

The new line element that I got is:

ds² = (1/2ω²)[-c²dt² - 2ce^rcos(θ)sin(θ)drdt - 2ce^rcos(θ)rcos(θ)dθdt - (1/2)e^2rcos(θ)sin²(θ)dr² - re^2rcos(θ)sin(θ)cos(θ)drdθ - (1/2)e^2rcos(θ)r²cos²(θ)dθ² + cos²(θ)dr² - 2rsin(θ)cos(θ)drdθ + r²sin²(θ)dθ² + dz²]

This was the transformation from the cartesian line element:
ds² = (1/2ω²)[-c²dt² - 2ce^xdydt - (1/2)e^2xdy² + dx² + dz²]

My coordinate transformations were:

x = rcos(θ)
y = rsin(θ)
dx = cos(θ)dr - rsin(θ)dθ
dy = sin(θ)dr + rcos(θ)dθ

 

[PeterDonis]

I hope this one is readable

Again, please use the PF LaTeX feature:

https://www.physicsforums.com/help/latexhelp/
It makes equations much more readable.

My coordinate transformations were

Note that, while these are the obvious coordinate transformations from a Cartesian into a cylindrical chart, they are not the only possible ones; one can also try rescaling the $r$ coordinate in various ways, which is a tactic suggested by the appearance of the exponential in the original line element. I suspect that the transformation implicitly being used in the Wikipedia article is doing something like that.

 

[PeterDonis]

Godel spacetime is not axially symmetric; as you correctly note, none of the five Killing vector fields correspond to axial symmetry.

Given this...

one can also try rescaling the $r$ coordinate in various ways, which is a tactic suggested by the appearance of the exponential in the original line element. I suspect that the transformation implicitly being used in the Wikipedia article is doing something like that.

...I am not sure that I was correct to say that Godel spacetime is not axially symmetric [edit: I was wrong--see post #25]. I suspect that in the cylindrical chart that is implicitly being used in the Wikipedia article, the line element is independent of $\phi$, which would mean that $\partial_\phi$ is a Killing vector field in this chart, which makes the spacetime manifestly axially symmetric. I think this KVF would have to correspond to the fifth one listed in the article, the one that looks very messy in the Cartesian chart.

However, I have not checked the math in detail on this so I might be mistaken. I will need to try to work through the computations in more detail when I get a chance.

 

[PeterDonis]

one can also try rescaling the $r$ coordinate in various ways

Another thing to keep in mind is that the worldlines of different particles of the "fluid" in Godel spacetime are twisting around each other, which means that the coordinate transformation to the cylindrical chart might have to mix together $t$ and $\phi$, similar to the way they are mixed in the transformation to Born coordinates for a rotating frame in flat spacetime.

 

[PeterDonis]

Ok, I have looked at Godel's original paper, which is the first reference in the Wikipedia article, and can be found here:

https://journals.aps.org/rmp/pdf/10.1103/RevModPhys.21.447
He exhibits a transformation to a cylindrical chart which is somewhat complicated and I won't write it down here, but here is the line element that results (where I have written the factor in front as $2 / \omega^2$ instead of his $4 a^2$ and have used $z$ here for what he calls $y$ in his chart, to be consistent with the notation we have been using in this thread, and I have also switched metric signature conventions from +--- to -+++):

$$
ds^2 = \frac{2}{\omega^2} \left[ - dt^2 - 2 \sqrt{2} \sinh^2 r \ d t d \varphi - \left( \sinh^4 r - \sinh^2 r \right) d \varphi^2 + dz^2 + dr^2 \right]
$$

This line element makes it evident that $\partial_\varphi$ is indeed a Killing vector field, so the spacetime is in fact axially symmetric (so I was wrong earlier when I said it wasn't). It also shows that if we have $\sinh^4 r - \sinh^2 r > 0$, then $\partial_\varphi$ is timelike, so its orbits are closed timelike curves. This will be true whenever $\sinh r > 1$.

 

[PeterDonis]

A further note on what I said in post #17, given that I was wrong there about Godel spacetime not being axially symmetric (it is, see post #25 just now): the comparison with Kerr spacetime that I gave there is still correct. However, the fact that you can choose any axis you like for the cylindrical coordinates in Godel spacetime does not mean it isn't axially symmetric; it just means it's axially symmetric about any axis, instead of just about one particular one.

 

[pervect]

Am I correct in calculating and reading from Godel's paper above that the Godel metric does NOT have exotic matter - that the matter density and pressure are both positive? Usually I associate CTC's with exotic matter, but it doesn't seem to be the case here. I forget the exact loopholes that permits CTC's without exotic matter, though.

 

[PeterDonis]

Am I correct in calculating and reading from Godel's paper above that the Godel metric does NOT have exotic matter

The stress-energy tensor satisfies all the energy conditions, yes.

I forget the exact loopholes that permits CTC's without exotic matter, though.

CTCs are more generally a failure of the causal structure of the spacetime to meet certain requirements. The details of how this works are in Hawking & Ellis. The strongest condition on causal structure is global hyperbolicity; this is the one that is usually considered to be necessary for a physically reasonable spacetime, since it is the one that ensures that the initial value problem is well posed (assuming the matter fields meet certain requirements as well). But there are weaker conditions that any spacetime with CTCs also violates. I don't know if any conditions are known on the stress-energy tensor that are either guaranteed to prevent CTCs or guaranteed to give rise to them.

 

[space-time]

Ok, I have looked at Godel's original paper, which is the first reference in the Wikipedia article, and can be found here:

https://journals.aps.org/rmp/pdf/10.1103/RevModPhys.21.447
He exhibits a transformation to a cylindrical chart which is somewhat complicated and I won't write it down here, but here is the line element that results (where I have written the factor in front as $2 / \omega^2$ instead of his $4 a^2$ and have used $z$ here for what he calls $y$ in his chart, to be consistent with the notation we have been using in this thread, and I have also switched metric signature conventions from +--- to -+++):

$$
ds^2 = \frac{2}{\omega^2} \left[ - dt^2 - 2 \sqrt{2} \sinh^2 r \ d t d \varphi - \left( \sinh^4 r - \sinh^2 r \right) d \varphi^2 + dz^2 + dr^2 \right]
$$

This line element makes it evident that $\partial_\varphi$ is indeed a Killing vector field, so the spacetime is in fact axially symmetric (so I was wrong earlier when I said it wasn't). It also shows that if we have $\sinh^4 r - \sinh^2 r > 0$, then $\partial_\varphi$ is timelike, so its orbits are closed timelike curves. This will be true whenever $\sinh r > 1$.

Cool. I recently decided to use this line element that you have found (I added the missing c terms back in)

$$
ds^2 = \frac{2}{\omega^2} \left[ - c^2dt^2 - 2c\sqrt{2} \sinh^2 r \ d t d \varphi - \left( \sinh^4 r - \sinh^2 r \right) d \varphi^2 + dz^2 + dr^2 \right]
$$

I was able to verify that $$ds^2$$ is timelike when all differentials except for $$d\varphi$$ are 0 and $$\left( \sinh^4 r - \sinh^2 r \right)$$ is greater than 0.

Furthermore, I have found that t is always timelike. The r coordinate is always spacelike with this line element. The z coordinate is always spacelike.

Now, it just so happens that $$\left( \sinh^4 r - \sinh^2 r \right)$$ is greater than 0 at many points including r = 1 (this is not the only point, I just chose this r value for the example I am about to present to you).

I have chosen the region r = 1 for what should finally be a successful example of a closed timelike curve proposed by me.

Firstly, here are the non-zero metric tensor elements:
g₀₀ = $$\frac{-2c^2}{\omega^2}$$

g₀₂ = g₂₀ = $$\frac{-2c\sqrt{2}\sinh^2 r}{\omega^2}$$

g₁₁ = g₃₃ =
$$\frac{2}{\omega^2}$$

g₂₂ = $$\frac{-2\left( \sinh^4 r - \sinh^2 r \right)}{\omega^2}$$

You can probably tell this, but:
x⁰ is the t coordinate
x¹ is the r coordinate
x² is the angular coordinate
x³ is the z coordinate

Anyway, moving along to my parameterized curve:

x(s) = [s, 1, s, 0]

Evaluating the formula:
g_ab(dx^a/ds)(dx^b/ds) yields:

$$\frac{-2c^2}{\omega^2} + \frac{-4c\sqrt{2}\sinh^2 r}{\omega^2} + \frac{-2\left( \sinh^4 r - \sinh^2 r \right)}{\omega^2} $$

At r = 1, this result is negative, meaning that this curve is certainly timelike.

Since this curve is timelike, and since the angular coordinate (which is periodic) is timelike at r = 1, this curve is a closed timelike curve.

Am I correct? Have I finally proposed a valid CTC?

 

[PeterDonis]

the missing c terms

They're not missing; Godel was using "natural" units in which $c = G = 1$ to avoid cluttering up the equations. I do the same thing. If you're not used to that convention, you might consider trying it when dealing with relativity problems; it makes things a lot easier once you get used to it.

 

[PeterDonis]

Since this curve is timelike, and since the angular coordinate (which is periodic) is timelike at r = 1, this curve is a closed timelike curve.

No, it isn't, because the angular coordinate isn't the only one that's changing.

Go back and read what you quoted from my post again. I specifically described a closed timelike curve in it.

 

[pervect]

If I might suggest, consider the curve

$$t(s)=0 \quad r(s)=1 \quad \varphi(s) = s \quad z(s)=0 \quad s=0..2\pi$$

Introducing t(s)=s is unnecessary and confusing, just set t(s)=0. Compare the result you from the Godel metric you've just quoted with the standard flat space-time cylindrical line element, i.e.

$$ds^2 = -c^2\,dt^2 + dr^2 + r^2 \,d\varphi^2 + dz^2$$

You'll find you can ignore all the metric coefficients except $g_{\varphi\varphi}$ or in your notation $g_{22}$, because dt=dr=dz=0. The only thing that is varying with s is $\varphi$.

Then in words - in flat space-time, we calculate that the circumference of a unit circle as being spacelike and having a value of $2 \pi$. In the Godel spacetime, we find that the circumference of a unit circle is timelike (!).

 

[space-time]

No, it isn't, because the angular coordinate isn't the only one that's changing.

Go back and read what you quoted from my post again. I specifically described a closed timelike curve in it.

But I thought you said that timelike coordinates had to have their derivatives with respect to s be greater than 0. That is why I set t to being s. Besides, how would that even work with t being held constant? The t coordinate is the time measured in the "lab frame". If t is held constant then that means that observers in the lab frame would experience no time passing while the observer in the Godel spacetime would be moving around in a circle. Are you saying that an observer in the lab frame simply would not perceive the movement of the observer in the Godel spacetime at all since the lab frame observer would perceive no passage of time? If not, then I just don't understand how the lab frame observer could perceive the Godel spacetime observer as traversing any space in 0 time.

 

[PeterDonis]

I thought you said that timelike coordinates had to have their derivatives with respect to s be greater than 0.

I have said no such thing. I have said things that are somewhat similar, but with qualifiers that do not apply to this case.

The t coordinate is the time measured in the "lab frame".

No it isn't. Coordinates are just labels. They don't have to match anyone's lab time.

What is the case is that the $t$ coordinate labels proper time for a particular class of observers. These observers do not follow CTCs; their timelike worldlines are "normal" ones that just go from past infinity to future infinity. But those observers are not the only possible observers. There are other observers whose worldlines are CTCs--such as the ones that go around circles where all the coordinates are constant except $\phi$, with $r$ large enough for those closed circles to be timelike curves.

If t is held constant then that means that observers in the lab frame would experience no time passing

It means no such thing. What it does mean is that the worldlines of those observers, the ones for whom $t$ matches their proper time, will only cross a given CTC once, at a single event (a single point in spacetime). They fly by the observer who is following the CTC and are gone. In other words, you simply have two different sets of curves in spacetime with different properties. It's just geometry.

All the rest of your post is just incorrect inferences from your incorrect mental picture of what is going on.

Basically, you seem to me to be trying to take shortcuts instead of doing the full math. That doesn't work in a general curved spacetime. There are shortcuts that happen to work in the particular case of flat Minkowski spacetime, but if you are going to look at general curved spacetimes, you need to stop using them, because they don't work. Godel spacetime is actually an excellent example of a spacetime where none of your intuitive shortcuts work. You simply have to do the full math.

 

[space-time]

I have said no such thing. I have said things that are somewhat similar, but with qualifiers that do not apply to this case.

I suppose what is confusing me then is this. Here is a quote from you from another thread:

More precisely, along a future-directed timelike curve, $d x^\mu / d \tau > 0$ for any timelike coordinate $x^\mu$, assuming the usual convention that $\tau$ increases towards the future along the curve and the timelike coordinate increases in the future direction.

That is from the following thread:

https://www.physicsforums.com/threa...tice-a-boost-or-a-loop-on-a-spacetime.977237/
Now I just took notice of this as I was typing this reply, but I notice that you did specifically say that the rule for having the derivatives of timelike coordinates be greater than 0 applies to future directed timelike curves (assuming that tau increases towards the future along the curve and the timelike coordinates increase towards the future).

Now, this leaves me with a few questions to check up on:

1. Is it that a CTC is not future directed due to it's closed nature? Is that why the t coordinate can be held constant even though it is a timelike coordinate?

2. Is it the case that tau is not increasing towards the future along the curve in a CTC? Is that why t can be held constant instead of having it's derivative be positive?

3. You also mentioned the timelike coordinates themselves increasing towards the future. Of course if t is held constant, then surely it is not increasing towards the future, right? Why is this allowed for t in the case of a CTC (or at least a CTC in this metric)?

Basically I just need an outright clarification of what circumstances necessitate that (dx^a/ds] be greater than 0.

In the CTC x(s) = [0, 1, s, 0] , why is the t coordinate allowed to be constant even though it is a timelike coordinate?

 

[PeterDonis]

Here is a quote from you from another thread

Yes, but, as I noted before, this was assuming qualifiers that are not applicable in Godel spacetime.

Is it that a CTC is not future directed due to it's closed nature?

Sort of. In a spacetime with CTCs everywhere, like Godel spacetime, there is no single way to distinguish the "future" and "past" halves of the light cones that works everywhere. You can make that choice along a particular timelike curve, but you can't continuously extend that choice to all timelike curves the way you can in flat Minkowski spacetime.

I just need an outright clarification of what circumstances necessitate that (dxa/ds] be greater than 0.

There isn't one that works for all spacetimes. That's why I said in an earlier post that you can't take shortcuts. There simply is no way to state such a criterion in terms of coordinates. Coordinates are not physically meaningful. You need to look at invariants. Sometimes, if you have a well chosen coordinate chart, you can figure out meaningful invariants from it, but this is never guaranteed and there is no general rule for how you do it. You have to look at each individual case.

 

[pervect]

A useful tool for getting some physical insight into a metric or line element is the orthonormal basis. This can most easily be done with an algebraic manipulation of the line element. For the Godel metric, if we write

$$ \textbf{d}\omega0= \textbf{d}t + \sqrt{2}\sinh^2r \,\textbf{d}\varphi \quad \textbf{d}\omega1 = \textbf{d}r \quad \textbf{d}\omega2 = \sqrt{\sinh^4 r + \sinh^2 r} \, \textbf{d}\varphi \quad \textbf{d}\omega3 = \textbf{d}z$$

[add]. I've omitted a multiplicative factor from the line element, out of laziness.

Then the line element would be written as
$$-ds^2 = (d\omega0)^2 )+ (d\omega1)^2 + (d\omega2)^2 + (d\omega3)^2 $$

The metric would be written in this notation as

$$g = -\textbf{d}\omega0 \otimes \textbf{d}\omega0 + \textbf{d}\omega1 \otimes \textbf{d}\omega1 + \textbf{d}\omega2 \otimes \textbf{d}\omega2 + \textbf{d}\omega3 \otimes \textbf{d}\omega3$$

Hopefully , it's clear that this is a locally Lorentizn metric, similar to the line element $-dt^2 + dx^2 + dy^2 + dz^2$. The similarity to the flat space metric allows for easy physical interpretation.

Note that $\textbf{d}\omega0$ is a linear functional of a vector, i.e. a scalar-valued function of a vector. It's a rank 1 tensor. Without the boldface, $d\omega0$ is a differental.

$\otimes$ represents the tensor product. The expression for the metric gives the metric g, a rank 2 tensor, as a sum-of-tensor-products of rank 1 tensors.

Much as vectors form a basis, dual vectors form a cobasis. So we describe what we're doing as creating an orthonormal basis of one-forms, or an orthonormal cobasis.

Notationally , this has so far all been written in index free notatoin. $\textbf{d}\omega0$ might also be written in index notation as some one-form $v_a$, with a lower index. Raising the index by multiplying $v_a$ with the inverse metric, i.e. $g^{ab} v_a$ would create the associated dual vector $v^a$.

 

[space-time]

Yes, but, as I noted before, this was assuming qualifiers that are not applicable in Godel spacetime.
Sort of. In a spacetime with CTCs everywhere, like Godel spacetime, there is no single way to distinguish the "future" and "past" halves of the light cones that works everywhere. You can make that choice along a particular timelike curve, but you can't continuously extend that choice to all timelike curves the way you can in flat Minkowski spacetime.
There isn't one that works for all spacetimes. That's why I said in an earlier post that you can't take shortcuts. There simply is no way to state such a criterion in terms of coordinates. Coordinates are not physically meaningful. You need to look at invariants. Sometimes, if you have a well chosen coordinate chart, you can figure out meaningful invariants from it, but this is never guaranteed and there is no general rule for how you do it. You have to look at each individual case.

I see. Well, just one more question

Let's say that I do use the CTC
x(s) = [0, 1, s, 0]

Now, we know that the proper time experienced when traveling along a timelike curve between two events can be calculated by evaluating the following integral from s₁ to s₂:

$$\int \sqrt{-g_{ab}\dot x^a \dot x^b} \, ds$$

Now, in the case of this CTC, from s = 0 to s = 2π, this integral evaluates to:

(2π/ω)$\sqrt{2sinh^4 r - 2sinh^2 r}$

At r = 1, this evaluates to about 6.45/ω

Would this in turn mean that the amount of time covered by the CTC would be
(6.45/ω) seconds (or whatever unit of time you are using)? By "covered", I mean to say: Does the fact that an observer moving along this curve would experience a proper time of 6.45/ω seconds (from s = 0 to 2pi) mean that from this observer's perspective, time would progress "normally" for 6.45/ω seconds before reverting back to the original point in time that the observer began at, and that the observer would just keep reliving cycles of the same 6.45/ω seconds over and over again?

In short, I am basically asking if the proper time along this CTC from s = 0 to s = 2pi would dictate how much time would pass "normally" before the loop got reset.

If not, then how would you figure out how much time passes before the loop resets itself?

 

[PeterDonis]

Does the fact that an observer moving along this curve would experience a proper time of 6.45/ω seconds (from s = 0 to 2pi) mean that from this observer's perspective, time would progress "normally" for 6.45/ω seconds before reverting back to the original point in time that the observer began at, and that the observer would just keep reliving cycles of the same 6.45/ω seconds over and over again?

Basically, yes, but the proper time doesn't "revert" or "reset"; there is no sharp discontinuity anywhere along the curve. It is just that the events along the curve are such that they repeat in a cycle that is 6.45/ω seconds long.