# I Okay, what exactly makes a timelike curve closed?

#### space-time

This question has been bothering me for a long time. It is simple enough to determine whether or not a curve is timelike. You simply
use this formula:

gab(dxa/ds)(dxb/ds)

(where x(s) is our parameterized curve).

Assuming a (- + + +) signature, if the answer to the above summation is negative for all s then the curve is timelike.

That is simple enough. However, how do you know if a curve is closed, or how do you pick out a timelike curve that is closed?

At first, I thought you needed a periodic temporal coordinate and interval in your parameterized curve so that travel along the curve would eventually bring you back to the event from whence you started. An example of such a curve is x(s) = [ sin(s), 0, 0, 0 ] on the interval (0, 2π). However, this is wrong and you guys chewed me out for it in a thread that I made before:

Apparently, a curve has to be one to one which means that for x(s) , no two values of s can yield the same event. (It was also pointed out to me in the above thread that I made a mathematical error and didn't account for the fact that the summation for the above curve in Minkowski space is not negative for all s since it is 0 at certain points).

Even if we ignore that mathematical error though, how can the curve actually loop back on itself to the beginning point if the curve has to be one to one? If it is one to one, I could see the curve being timelike, but I don't see it ever returning back to the same point.

My main reason for asking is because I just want to see a CTC in Godel spacetime already. I once tried to come up with a CTC in Godel spacetime (which is said to have CTC running through every event), and I ended up with a closed spacelike curve rather than a timelike curve. That was in this thread here:

Long story short, can someone please tell me how to establish or notice the "closed" part of a closed timelike curve in something like a Godel spacetime.

If you need the line element for that metric, it is on this wiki:

Thank you

• kent davidge
Related Special and General Relativity News on Phys.org

#### PeterDonis

Mentor
how do you know if a curve is closed
By the fact that after some finite increment of proper time along the curve, it comes back to the same spacetime point.

Apparently, a curve has to be one to one which means that for x(s) , no two values of s can yield the same event.
No, that's not what you were told in that previous thread. What you were told is that no two different events can have the same value of $s$. In other words, you can't have $x(s)$ and $y(s)$ be two different events both mapped to the same value of $s$. But that does not mean you can't have $x(s_1) = x(s_2)$ for two different values of $s$.

#### space-time

By the fact that after some finite increment of proper time along the curve, it comes back to the same spacetime point.

No, that's not what you were told in that previous thread. What you were told is that no two different events can have the same value of $s$. In other words, you can't have $x(s)$ and $y(s)$ be two different events both mapped to the same value of $s$. But that does not mean you can't have $x(s_1) = x(s_2)$ for two different values of $s$.
I see. However, that brings up another question. You say that x(s) and y(s) can't be two different events mapped to the same value of s. Well, in that previous thread, the Minkowski space curve I was using was
x(s) = [ sin(s), 0, 0, 0 ] on the interval (0, 2π). Where exactly on this curve or on this interval am I having a single value of s be mapped to two different events?

After all, sin(0) = 0 (and only 0).
sin(π/6) = 1/2 (and nothing else).

sin(s) for any value of s only maps to one event. For no value of s does sin(s) map to two different values.

Why then, did my curve fall short in regards to this particular issue?

#### PeterDonis

Mentor
in that previous thread, the Minkowski space curve I was using was
x(s) = [ sin(s), 0, 0, 0 ] on the interval (0, 2π)
No, it was $x(s) = [\sin(s), 0, 0, 0]$ without any restriction on the interval. You can't arbitrarily restrict the interval of $s$ because you can't just decree that the curve ends. That would be the equivalent of an object following the curve just disappearing; that can't happen. And as soon as you reach $\sin(s) = \pi$ (not even $2 \pi$), you're back at the same event, $[0, 0, 0, 0]$.

#### space-time

No, it was $x(s) = [\sin(s), 0, 0, 0]$ without any restriction on the interval. You can't arbitrarily restrict the interval of $s$ because you can't just decree that the curve ends. That would be the equivalent of an object following the curve just disappearing; that can't happen. And as soon as you reach $\sin(s) = \pi$ (not even $2 \pi$), you're back at the same event, $[0, 0, 0, 0]$.
So basically, you're saying that when I specify curves, I shouldn't worry about specifying an interval for the curve because I can't just arbitrarily decide that a curve ends somewhere?

If so, that makes sense.

However, as for the part about getting back to the event [0, 0, 0, 0] before even reaching 2π:

Isn't that kind of the goal (at least as far as CTCs go)? Don't we want to reach the event from which we started? Or... Does the problem emerge from the fact that once you go beyond s = π, you start heading towards t = -1 ?

#### PeterDonis

Mentor
So basically, you're saying that when I specify curves, I shouldn't worry about specifying an interval for the curve because I can't just arbitrarily decide that a curve ends somewhere?
When you specify a curve, the interval of the parameter will always be $- \infty$ to $\infty$ unless the spacetime itself has a singularity somewhere along the curve (for example, at $r = 0$ in Schwarzschild spacetime).

Isn't that kind of the goal (at least as far as CTCs go)? Don't we want to reach the event from which we started
Yes, but you have to actually reach the same event while continuing to go forward on the curve, not because you stopped going forward and started going backward. See below.

Does the problem emerge from the fact that once you go beyond s = π, you start heading towards t = -1 ?
More precisely, once you go beyond $s = \pi / 2$, you are not going forward along the curve, you are going backward along the curve. The derivative of $x^0(s)$ with respect to $s$, which is $\cos(s)$, becomes negative after $s = \pi / 2$. A valid parameterization would not do that.

• space-time

#### space-time

Yes, but you have to actually reach the same event while continuing to go forward on the curve, not because you stopped going forward and started going backward. See below.

More precisely, once you go beyond $s = \pi / 2$, you are not going forward along the curve, you are going backward along the curve. The derivative of $x^0(s)$ with respect to $s$, which is $\cos(s)$, becomes negative after $s = \pi / 2$. A valid parameterization would not do that.
Okay. This basically means that I want a curve whose component functions are never decreasing (meaning that their derivative is never negative)? Well, this brings up three questions:

1. Would it be correct to say that this means that we also don't want a function whose derivative would be 0 at some point (since that would imply that the object at some point simply stopped moving along the curve)?

2. What kind of function actually loops back to the same point (to make a closed curve) without ever decreasing (or being 0 if my first question is valid)? I mean, I can think of functions that would not decrease or be 0 such as a linear function with a positive slope. That function would always increase, but it would never loop back to the same point, so we couldn't meet the goal of a CTC with that. On the other hand, something like a quadratic function or a trig function would have at least one value of s where it gets back to the same original point (as seen with sin(s) ), but those functions involve the negative derivatives that you speak of (as well as derivatives of 0 at some values of s). I honestly can't think of any type of function that is always increasing, while still being able to loop back to the same event at some point (unless I specify some kind of arbitrary piecewise function or something like that).

3. Does this rule for valid parameterizations only apply to the temporal part of a curve, or the spatial components too? I ask this because when it comes to space, you literally could have something like a circle in space. For example, if an object is literally just moving around in a circle of radius 2 in space, I could see the parameterization being something like:

x(s) = [ 10s, 2cos(s), 2sin(s), 0 ]
(Let's just say we're using Minkowski space - + + + signature)

In this case, the temporal coordinate is always increasing, but the x and y coordinates are not. Nevertheless, as far as the spatial part goes, this does seem to describe a circle of radius 2.

Is this invalid?

Basically, with this question, I just want to know what changes (if anything) if our curve is not just simply a temporal coordinate with trivial spatial coordinates (such as [sin(s), 0, 0, 0] ).

#### PeterDonis

Mentor
Would it be correct to say that this means that we also don't want a function whose derivative would be 0 at some point (since that would imply that the object at some point simply stopped moving along the curve)?
Yes.

What kind of function actually loops back to the same point (to make a closed curve) without ever decreasing (or being 0 if my first question is valid)?
This is not a property of the function, it's a property of the underlying topology of the spacetime. For example, consider ordinary angular coordinates, say longitude on the Earth. As I go East around the Earth, say along the equator for definiteness, I eventually reach 180 degrees longitude; assigning + signs to East longitude and - signs to West longitude (which is the normal convention), as I cross that meridian I go from +180 to -179 in terms of longitude. But a curve parameter along the curve I am following would go from, say, 180 to 181. In other words, the curve parameter "181" maps to the same point on the Earth as the curve parameter "-179". But nothing about the curve parameter itself, if that's all I know, tells me that. I have to know by some other means that I'm on a sphere and the curve parameter maps to longitude, in order to know that curve parameters "181" and "-179" map to the same point.

Or, to put it another way, angular coordinates like longitude are inherently multi-valued; longitude -179 can also be expressed as longitude +181 (or longitude +541, or longitude -539, or...). So the function that maps curve parameter to longitude will just be $\phi = s$ (assuming I use the same units for both). The knowledge that $\phi = -179$ and $\phi = 181$ represent the same point comes from somewhere else, not from the curve parameter or the function that maps it to longitude.

In the case of a spacetime with closed timelike curves, a timelike coordinate will have to have the same multi-valued property as $\phi$ above.

Does this rule for valid parameterizations only apply to the temporal part of a curve, or the spatial components too?
It applies to all components.

#### space-time

Yes.

This is not a property of the function, it's a property of the underlying topology of the spacetime. For example, consider ordinary angular coordinates, say longitude on the Earth. As I go East around the Earth, say along the equator for definiteness, I eventually reach 180 degrees longitude; assigning + signs to East longitude and - signs to West longitude (which is the normal convention), as I cross that meridian I go from +180 to -179 in terms of longitude. But a curve parameter along the curve I am following would go from, say, 180 to 181. In other words, the curve parameter "181" maps to the same point on the Earth as the curve parameter "-179". But nothing about the curve parameter itself, if that's all I know, tells me that. I have to know by some other means that I'm on a sphere and the curve parameter maps to longitude, in order to know that curve parameters "181" and "-179" map to the same point.

Or, to put it another way, angular coordinates like longitude are inherently multi-valued; longitude -179 can also be expressed as longitude +181 (or longitude +541, or longitude -539, or...). So the function that maps curve parameter to longitude will just be $\phi = s$ (assuming I use the same units for both). The knowledge that $\phi = -179$ and $\phi = 181$ represent the same point comes from somewhere else, not from the curve parameter or the function that maps it to longitude.

In the case of a spacetime with closed timelike curves, a timelike coordinate will have to have the same multi-valued property as $\phi$ above.

It applies to all components.
I see. This makes a lot of sense. Basically, you're saying that spacetimes that contain closed timelike curves have topologies that are basically "periodic" in the sense that if you keep going forward along a curve you will eventually come back to the same point even if it numerically looks like a different point (kind of like a globe or the unit circle).

In that case, what exactly is the "period" (if you will) of the Godel metric? By that I mean, if I start at event
[0, 0, 0, 0] in a Godel spacetime and go forward along some curve, what later event will be equivalent to the event [0, 0, 0, 0] (in the same way that the point θ = 2π is the same point on the unit circle as θ = 0)?

How can I tell or determine what this "period" is?

Here is the wiki for the Godel metric (which contains the line element and other info that you may need):

I certainly can't tell what the period is from that line element. That's for sure.

• kent davidge

#### PeterDonis

Mentor
what exactly is the "period" (if you will) of the Godel metric?
As I understand it, it depends on the particular CTC; the "period" is not the same for all CTCs in the Godel metric. In this respect the Godel metric is much harder to visualize regarding how the CTCs work.

#### Dale

Mentor
When you specify a curve, the interval of the parameter will always be $- \infty$ to $\infty$ unless the spacetime itself has a singularity somewhere along the curve (for example, at $r = 0$ in Schwarzschild spacetime).
I don’t think this is generally correct. I believe that it is perfectly OK to have a curve that is parameterized over a finite interval. For example, the curve representing the worldline of an unstable particle is clearly a finite curve. It makes no sense to require such a curve to be extended to infinity.

I think that the problem with his previous curve is not the finiteness of the parameterization, but rather that the parameterization was non affine.

#### PeterDonis

Mentor
I don’t think this is generally correct.
I agree there are cases where a particular object whose lifetime is finite for reasons having nothing to do with the spacetime geometry would be modeled by a curve with a finite parameter range. However, any such curve will be extendible in the spacetime geometry sense. CTCs are a feature of the spacetime geometry, so I wanted to focus on the geometry, and therefore limit attention to inextendible curves.

I think that the problem with his previous curve is not the finiteness of the parameterization, but rather that the parameterization was non affine.
Not just non-affine, but non-monotonic (if that's the word). There are plenty of possible non-affine parameterizations that still have positive $dx^\mu / ds$ everywhere; those won't show the problem that his did.

• Dale

#### Dale

Mentor
However, any such curve will be extendible in the spacetime geometry sense. CTCs are a feature of the spacetime geometry, so I wanted to focus on the geometry, and therefore limit attention to inextendible curves.
Got it, I agree.

#### space-time

As I understand it, it depends on the particular CTC; the "period" is not the same for all CTCs in the Godel metric. In this respect the Godel metric is much harder to visualize regarding how the CTCs work.
The period depends on the CTC huh? Well in that case, would you by any chance be able to tell the period of the following timelike curve (if it even is a CTC, I can't tell since I don't know how I would find the period)? Here is the metric tensor for Godel spacetime (using c = 1 convention):

g00 = -1/(2ω2)
g11 = g22 = 1/(2ω2)
g33 = -e2x/(4ω2)
g03 = g30 = -ex/(2ω2)

Now here is the curve:

x(s) = [10s, s, s, s]

This curve is definitely timelike since
gab(dxa/ds)(dxb/ds) = (-196 - 40ex - e2x)/(4ω2)

This is less than 0 for all s, which means that this curve is definitely timelike. Plus, this time I made sure that the curve is always increasing.

Now this is a timelike curve. However, is there any way that I can tell whether or not this curve is closed (since I don't know how to find any kind of period in this case)? If it is closed, what is the period of this CTC?

#### PeterDonis

Mentor
This curve is definitely timelike
I'm not sure if your formula is exactly right, but I agree the curve will be timelike since you've set things up so the negative metric coefficients far outweigh the positive ones.

is there any way that I can tell whether or not this curve is closed
Sure: it's not, because Godel spacetime is simply connected, so the single coordinate chart you give covers it entirely. So any curve that never returns to the same coordinate values twice for different values of $s$ is not closed.

The more challenging question is how to find a curve that is closed but always timelike and express it in these coordinates. You might find the the fact that, in cylindrical coordinates (given later on in the Wikipedia article), the coordinate basis vector $\partial_{\phi}$ is timelike for large enough values of $r$, helpful. (Note that there is no preferred choice of origin/axis in cylindrical coordinates in Godel spacetime, so it's always possible to choose an origin/axis so that $\partial_{\phi}$ is timelike at any point in the spacetime you choose.)

#### space-time

I'm not sure if your formula is exactly right, but I agree the curve will be timelike since you've set things up so the negative metric coefficients far outweigh the positive ones.

Sure: it's not, because Godel spacetime is simply connected, so the single coordinate chart you give covers it entirely. So any curve that never returns to the same coordinate values twice for different values of $s$ is not closed.

The more challenging question is how to find a curve that is closed but always timelike and express it in these coordinates. You might find the the fact that, in cylindrical coordinates (given later on in the Wikipedia article), the coordinate basis vector $\partial_{\phi}$ is timelike for large enough values of $r$, helpful. (Note that there is no preferred choice of origin/axis in cylindrical coordinates in Godel spacetime, so it's always possible to choose an origin/axis so that $\partial_{\phi}$ is timelike at any point in the spacetime you choose.)
What do you mean you're not sure that my formula is exactly right? Did I make a math error somewhere?

As far as I understand, you can tell whether or not a curve x(s) is timelike, spacelike or light like by using the formula:

gab(dxa/ds)(dxb/ds)

If it is negative for all s then the curve is timelike.

Positive is spacelike

Zero is light like.

This is assuming a (- + + +) signature.

What part of this is wrong?

#### pervect

Staff Emeritus
I believe there is some topological definition of a closed curve, but I am not sure of the details. Wiki gives us the topological definition of a closed manifold, which might be overkill.

Wiki mentions that the only 1-dimensional closed manifold is a circle. There's probably some way to leverage this for a better/simpler definition of a closed curve.

#### PeterDonis

Mentor
What do you mean you're not sure that my formula is exactly right?
Just that I hadn't had a chance to check it. I've checked it now and I agree with it.

What part of this is wrong?
None of it. You correctly identified your curve as a timelike curve. But it's not a closed timelike curve.

• space-time

#### PeterDonis

Mentor
Wiki mentions that the only 1-dimensional closed manifold is a circle. There's probably some way to leverage this for a better/simpler definition of a closed curve.
Topologically, that is the definition of a closed curve. The problem is that this is a topological property and has nothing to do in general with either the geometry of the manifold or the parameterization of the curve.

### Want to reply to this thread?

"Okay, what exactly makes a timelike curve closed?"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving