Noticing CTC's / time travel in a Godel space-time

[space-time]

I have been working with the Godel solution to the Einstein field equations which is known to contain closed time-like curves.
The metric I am using is the following:
ds² = dt²/(2ω²) + (e^xdzdt)/ω² + (e^2xdz²)/(4ω²) - dx²/(2ω²) - dy²/(2ω²)
This is sign convention (+ - - -)

Now from what I've researched, the term ω is the angular velocity of the dust grains in this space-time around the y-axis. Now let's get to an example scenario that I came up with:

Some dust grains in a Godel space-time, travel around the y-axis at an angular velocity of ω (I won't specify what ω is just yet for a certain reason). The observer named Boruto decides to stand in this space-time at the spatial coordinates (x,y,z) = (0,0,0). He stays there and doesn't move. This of course means that x(t) = 0, y(t) = 0, and z(t) = 0. Meanwhile, another guy named Kawaki decides to relax by simply sitting on one of the dust grains. Kawaki's mass is negligible, so let's just say he doesn't slow the dust grain down. Now, while riding the dust grain, Kawaki counts some arbitrary coordinate time t. In this case, how much proper time τ would pass from Boruto's perspective?

Now, here is my process (and yes I purposely didn't give a value for t):
τ = ∫sqrt(ds²) = ∫sqrt( dt²/(2ω²)) = [1/(ω√2)] * ∫₀^t 1 dt = [t/(ω√2)]

Just in case it is not clear, the bounds on that previous integral are 0 to t. Anyway, as you can see here, if an observer is standing still in a Godel universe (and not riding one of the dust grains), then the relationship between the proper time τ that they count, and the time t seen by an observer who just rides the grains is:

τ = t / (ω√2) Note: This is why I didn't specify a value for ω or t earlier. I just wanted to derive this particular formula.

Now this leads into my questions about closed time-like curves and time travel:

1. As can be seen in the above formula, if a negative value is plugged in for ω, then τ would turn out negative and it would seem as if the observer standing still saw time moving in reverse. However, is it actually physically possible for ω to be negative? I know that sometimes when taking direction into consideration, one particular direction of rotation is considered positive while the other is considered negative. However, it is a matter of convention regarding which direction is chosen as positive and which is negative. That is why I am questioning whether or not the dust grains could actually move in a way so that ω is negative, and if this would actually cause the sought out time reversal effect. Here's a thought: Based on what I've researched, a Godel universe has to be rotating, so maybe the dust grains could achieve a negative ω if their direction of motion is opposite to the direction of the universe's rotation. What is you guys' response to this?

2. If a negative ω is not what would be responsible for backwards time-travel in a space-time like this, then there is another part of this metric that I have noticed that may have something to do with backwards time-travel or closed time-like curves. According to this wiki on the Godel metric https://en.wikipedia.org/wiki/Gödel_metric, the variables t, x, y, and z can range from -∞ to ∞ .How can t be negative though? This certainly would lead to the possibility of τ coming out negative after all the integration and other calculations are done, but I don't know how it is physically possible for the external observer (the guy riding the grains) to count a negative coordinate time t in the first place.

3. If neither of these two things have any effect on the presence of closed time-like curves or backwards time travel in a Godel space-time, then how can you tell that the Godel metric contains closed time-like curves? Can you tell that because of some feature within the line element? Can you tell it because of any other relativistic tensors such as the Riemann or Ricci curvature tensors or some other tensor? Can you tell by looking at the velocity 4 - vector? I would like to see the mathematical representation of a closed time-like curve. Please show/tell me what on paper screams closed time-like curve or backwards time travel.

 

[PeterDonis]

As can be seen in the above formula, if a negative value is plugged in for ω, then τ would turn out negative

You can't deduce this from what you've done. You took a square root in deriving your formula. Taking the square root is where you impose a sign convention: either increasing $t$ corresponds to increasing $\tau$ (positive square root), or increasing $t$ corresponds to decreasing $\tau$ (negative square root). The sign of $\omega$ itself has nothing to do with it, because the interval has $\omega^2$ in it, not $\omega$, so the sign of $\omega$ drops out of the calculation before you even take the square root.

As for which sign you choose when you take the square root, that's just a matter of convention. Virtually everybody uses the convention where increasing $t$ corresponds to increasing $\tau$, which means taking the positive square root. Note that this has nothing to do with the presence of $\omega$ in the formula; you would encounter the same problem for an observer at rest in Minkowski spacetime, for whom $d\tau^2$ = $dt^2$. To get his $\tau$ by integrating you have to take a square root, and which sign do you use?

If a negative ω is not what would be responsible for backwards time-travel in a space-time like this

Which it isn't. See above.

How can t be negative though?

Because the spacetime is infinite in both directions of time. This is true of many spacetimes, including, again, simple Minkowski spacetime; the range of the $t$ coordinate in inertial coordinates in Minkowski spacetime is $- \infty < t < \infty$.

I don't know how it is physically possible for the external observer (the guy riding the grains) to count a negative coordinate time t in the first place.

The increment of coordinate time he counts will be positive, but there's no reason he can't start counting at a negative value of $t$. That just depends on where we arbitrarily choose to define $t = 0$. This is true of any spacetime. Again, consider Minkowski spacetime, and an inertial observer at rest in a particular frame. Suppose we arbitrarily choose noon today as $t = 0$ in this frame. Then if the observer starts counting off time at 9 am today, he started at $t = - 3 \text{hours}$, i.e., a negative value of $t$.

You should be seeing a pattern here: before even tackling a spacetime like the Godel spacetime, you need to be sure you fully understand how coordinates work in the simplest possible spacetime, Minkowski spacetime. That way you won't confuse yourself by thinking about issues that arise in Minkowski spacetime too, and therefore tell you nothing about the particular properties of Godel spacetime (or whichever other spacetime you are interested in).

how can you tell that the Godel metric contains closed time-like curves?

It's easier to see if you use a cylindrical chart, as described on the Wikipedia page:

https://en.wikipedia.org/wiki/Gödel_metric#A_cylindrical_chart

The CTCs are described later in that same section; they are just circles of constant radius $r$ from a chosen symmetry axis, for large enough $r$.

 

[MeJennifer]

This solution is interesting because it proves that topological features of a spacetime can have physical consequences. It also shows that numerical approaches like using ADM equations can never completely cover all (theoretical) solutions of general relativity.

 

[PeterDonis]

This solution is interesting because it proves that topological features of a spacetime can have physical consequences.

How so? The topology of the Godel spacetime is $R^4$, just like Minkowski spacetime.

 

[MeJennifer]

How so? The topology of the Godel spacetime is $R^4$, just like Minkowski spacetime.

True, and that implies there cannot be spacelike hypersurfaces in a Godel solution.

 

[PeterDonis]

that implies there cannot be spacelike hypersurfaces in a Godel solution.

That's not because of the topology, it's because of the geometry. It can't be because of the topology since there is another spacetime with the same topology that does not share this property.