Find CTCs in Kerr Metric: Visualizing Spacetime Geometry

[space-time]

I have been trying to study some differential geometry and some stuff about manifolds in my efforts to learn about closed timelike curves, but thus far it has been a lot of set theory and I have yet to see the "geometry" aspect. What I really want to know is this:

We know how some spacetimes have "boosts", or in other words, they have spacetime geometries where traveling forward along some curve in spacetime will eventually lead you back to the event from whence you started. If such a curve is timelike, then the curve is a CTC. An example of a spacetime with such a boost is Misner space, which seems to have the boost:

(t, x) -> (tcosh(π) + xsinh(π) , xcosh(π) + tsinh(π))

This is the boost that is reported by the Misner space wiki: https://en.wikipedia.org/wiki/Misner_space

Now in the case of Misner space, that boost was just given to me by the wiki. One can see how a boost like this would indicate the existence of closed timelike curves.

However, let's look at a different metric that is known to have closed timelike curves, such as the Kerr metric:

https://en.wikipedia.org/wiki/Kerr_metric
No such boost was reported on that wiki. I suppose that in order for someone to notice the presence of closed timelike curves within this metric, they would have to be able to somehow visualize or even graph out the spacetime geometry. Somehow, a person would have to recognize that the spacetime geometry is something akin to spherical or circular or just some shape that would allow an object to travel along some curve and then return to the same event (kind of like traveling around a globe until you eventually get back to the same point). Granted, I know that the Kerr metric describes the spacetime geometry around a spherical uncharged rotating mass, but I feel as though knowing the shape of the mass does not guarantee that the spacetime curves into the same shape.

Having said this, are there any tensors or formulas at all that would help me be able to visualize the actual geometry of a spacetime? Is there perhaps anything involving the metric tensor or covariant derivatives that would let me say "hmmm... If I propose some parameterized curve x(s) in this spacetime and traveled forward along it, then the intrinsic curvature of this spacetime would make this curve loop back on itself in a way that I would return to the initial event"?

Thanks for any assistance.

 

[PeterDonis]

No such boost was reported on that wiki.

That's because the "boost" method of constructing a spacetime with CTCs is not the only way to find a spacetime with CTCs.

I know that the Kerr metric describes the spacetime geometry around a spherical uncharged rotating mass

No, it doesn't. It describes the spacetime geometry of an uncharged rotating black hole. The Kerr metric is not spherically symmetric, and it is vacuum; the CTCs occur inside the inner horizon, which is generally believed not to be physically realistic anyway.

Some sources imply that the exterior part of the Kerr metric describes the spacetime around an ordinary rotating object (which will not be spherical--it will be oblate). However, we don't actually know that this is true, and there are plausible arguments to suggest that it can't be. But in any case, the exterior part of the Kerr metric does not contain any CTCs.

Is there perhaps anything involving the metric tensor or covariant derivatives that would let me say "hmmm... If I propose some parameterized curve x(s) in this spacetime and traveled forward along it, then the intrinsic curvature of this spacetime would make this curve loop back on itself in a way that I would return to the initial event"?

I don't know of any general way of doing this for an arbitrary spacetime. In all of the known spacetimes with CTCs, either the CTCs arise by construction, as with Misner space, or the spacetime can be described by a coordinate chart containing angular coordinates (i.e., coordinates that are periodic, going around once by $2 \pi$ takes you back to the same point) which can be shown to be timelike in some region. Kerr spacetime (and also Godel spacetime, another well known example of a spacetime with CTCs) is an example of the latter kind.

 

[space-time]

That's because the "boost" method of constructing a spacetime with CTCs is not the only way to find a spacetime with CTCs.
No, it doesn't. It describes the spacetime geometry of an uncharged rotating black hole. The Kerr metric is not spherically symmetric, and it is vacuum; the CTCs occur inside the inner horizon, which is generally believed not to be physically realistic anyway.

Some sources imply that the exterior part of the Kerr metric describes the spacetime around an ordinary rotating object (which will not be spherical--it will be oblate). However, we don't actually know that this is true, and there are plausible arguments to suggest that it can't be. But in any case, the exterior part of the Kerr metric does not contain any CTCs.
I don't know of any general way of doing this for an arbitrary spacetime. In all of the known spacetimes with CTCs, either the CTCs arise by construction, as with Misner space, or the spacetime can be described by a coordinate chart containing angular coordinates (i.e., coordinates that are periodic, going around once by $2 \pi$ takes you back to the same point) which can be shown to be timelike in some region. Kerr spacetime (and also Godel spacetime, another well known example of a spacetime with CTCs) is an example of the latter kind.

So in the end, it comes down to the coordinate chart huh? Well, I think I understand, but that brings up one question. You say that the Kerr metric is an example of a metric that has a coordinate chart with periodic coordinates. Well, going based off of the wiki, I'm pretty sure the coordinate chart for the Kerr metric is:

x⁰ = t
x¹ = r
x² = θ
x³ = ∅

The latter two coordinates are periodic, but the first two (most importantly the temporal coordinate) are not. How then, is any curve supposed to ever loop back around to the same event? I don't think t = 2π is the same as t = 0 unlike with the angular coordinates. The same goes for r.

At most, I could see a curve returning back to the same point in space, but at a different time. Even then, I question even that possibility since you said that when you have a parameterized curve x^μ(s), the derivatives (dx^μ/ds) have to be positive for all s in order for the curve to be valid. This would mean that the r component of the curve could not just be a constant, which would in turn mean that r would have to change as you move along the curve.

This would mean that instead of ever returning to the exact same event, you would keep circling around the angular coordinates while continually increasing in the t and r coordinates.

 

[Pencilvester]

At most, I could see a curve returning back to the same point in space, but at a different time. Even then, I question even that possibility since you [Peter] said that when you have a parameterized curve xμ(s), the derivatives (dxμ/ds) have to be positive for all s in order for the curve to be valid. ...
This would mean that instead of ever returning to the exact same event, you would keep circling around the angular coordinates while continually increasing in the t and r coordinates.

You are misremembering. This is what he said:

The derivative of $x^0(s)$ with respect to s... becomes negative after s=π/2. A valid parameterization would not do that.

Meaning you can’t change direction in time, but of course you’re free to change direction in space.

 

[PeterDonis]

So in the end, it comes down to the coordinate chart huh?

Not as far as the spacetime geometry is concerned. The spacetime geometry is independent of any choice of coordinates.

However, in some spacetimes with particular symmetries, such as Kerr spacetime, a well chosen coordinate chart can make certain features of the spacetime geometry easier to see. See below.

The latter two coordinates are periodic, but the first two (most importantly the temporal coordinate) are not.

The fact that a coordinate is called $t$ does not mean it is "the temporal coordinate" everywhere. You need to actually look at the metric to see what coordinates are spacelike, timelike, or null in various regions. If you look at the Kerr metric--not just the coordinates, but the line element--you will see that there is a region inside the inner horizon where the coordinate $\phi$ is timelike. That means the worldline of an observer can point purely in the $\phi$ direction in this region. But $\phi$ is still periodic, so that means the worldline of an observer in this region can be a closed timelike curve.

you said that the derivatives (dxμ/ds) have to be positive for all s in order for the curve to be valid

Not for all the coordinates on all curves. I'm not sure where you think you are quoting me from, but the condition as you state it is much too strong.

For the $\phi$ coordinate, which is timelike in the region of Kerr spacetime I described, it is true that the derivative $d \phi / d \tau$ along such a timelike worldline is positive everywhere. But when you get to $\phi = 2 \pi$, that is the same event as $\phi = 0$, because the $\phi$ coordinate is periodic and no other coordinates are changing ($d x^\mu / d \tau = 0$ for every coordinate except $\phi$). So you have a closed timelike curve.

 

[space-time]

You are misremembering. This is what he said:

Meaning you can’t change direction in time, but of course you’re free to change direction in space.

That makes sense I suppose. If I'm understanding correctly then, the derivatives of the spatial components of the curve are allowed to be positive, negative, or even 0. However, the derivative of the temporal component must always be positive. Is that correct?Also, even if that is correct, that still doesn't settle the issue about the temporal coordinate not being periodic. The lack of a periodic temporal coordinate would make it so that you could never return to the same point in time, even if you did return to the same point in space.

 

[PeterDonis]

the derivative of the temporal component must always be positive

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.

However, that in itself does not tell you which coordinates, if any, are timelike. You have to look at the metric for that. Note that it is perfectly possible to have a coordinate chart where no coordinates are timelike in a region of spacetime. You certainly, as I said in my previous post just now, can't assume which coordinates are timelike or spacelike based on their names.

 

[space-time]

Not as far as the spacetime geometry is concerned. The spacetime geometry is independent of any choice of coordinates.

However, in some spacetimes with particular symmetries, such as Kerr spacetime, a well chosen coordinate chart can make certain features of the spacetime geometry easier to see. See below.
The fact that a coordinate is called $t$ does not mean it is "the temporal coordinate" everywhere. You need to actually look at the metric to see what coordinates are spacelike, timelike, or null in various regions. If you look at the Kerr metric--not just the coordinates, but the line element--you will see that there is a region inside the inner horizon where the coordinate $\phi$ is timelike. That means the worldline of an observer can point purely in the $\phi$ direction in this region. But $\phi$ is still periodic, so that means the worldline of an observer in this region can be a closed timelike curve.
Not for all the coordinates on all curves. I'm not sure where you think you are quoting me from, but the condition as you state it is much too strong.

For the $\phi$ coordinate, which is timelike in the region of Kerr spacetime I described, it is true that the derivative $d \phi / d \tau$ along such a timelike worldline is positive everywhere. But when you get to $\phi = 2 \pi$, that is the same event as $\phi = 0$, because the $\phi$ coordinate is periodic and no other coordinates are changing ($d x^\mu / d \tau = 0$ for every coordinate except $\phi$). So you have a closed timelike curve.

Ok, so I previously didn't know that you could check to see if an individual coordinate was timelike/spacelike/null. I only knew how to check to see if a parameterized curve or a 4-vector was timelike/spacelike/null.

I just found this cool thread however, that seems to tell me how to check for an individual coordinate. Can you verify for me if what I am about to say is correct?:

Here is the thread by the way: https://www.physicsforums.com/threads/time-coordinate-timelike.398341/

In order to find out whether or not a coordinate is timelike/spacelike/null, you set all of the dx^μ terms in the line element (except for the ones for the coordinate that you are working with) equal to 0. Then, based on what remains (and on what sign convention you're using), you can classify that coordinate as timelike/spacelike/null depending on the sign of what remains.

Using Minkowski space as an easy example:

ds²= -c²dt² + dx² + dy² + dz²

If I set dx, dy, and dz equal to 0, then I am left with:

ds²= -c²dt²

This is always negative, so I suppose the t coordinate is always timelike in Minkowski space.

Note: The thread said that you set the (dx^μ)² terms equal to 0, but I assume that he really just meant the dx^μ terms (even the ones that aren't squared), and he only put the squared there because a lot of metrics like Minkowski space have most of the (dx^μ) terms squared.

Is my understanding correct?

 

[PeterDonis]

In order to find out whether or not a coordinate is timelike/spacelike/null, you set all of the dxμ terms in the line element (except for the ones for the coordinate that you are working with) equal to 0. Then, based on what remains (and on what sign convention you're using), you can classify that coordinate as timelike/spacelike/null depending on the sign of what remains.

Yes.

I suppose the t coordinate is always timelike in Minkowski space

In that coordinate chart, yes. Note that the chart you are using is not the only possible coordinate chart on Minkowski spacetime. Also, you are using the convention that a negative $ds^2$ means a timelike interval, but as @DrGreg pointed out in the thread you linked to, not all of the literature uses that convention.

The thread said that you set the (dxμ)2 terms equal to 0, but I assume that he really just meant the dxμ terms (even the ones that aren't squared), and he only put the squared there because a lot of metrics like Minkowski space have most of the (dxμ) terms squared.

Yes.