Undergrad 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.