Godel metric in a cylindrical chart

[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.