# I I came up with a CTC in Minkowski Space, But To What End?

#### space-time

If you viewed my most recent thread before this one, then you know that I have been studying curves in spacetime (timelike/spacelike/lightlike), and I have especially been looking into the CTCs (closed timelike curves) that the Godel metric is famous for. During my studies I found that I had to take a step back and look into timelike vectors. Upon doing so, I also decided that I should do some mathematical experiments with the simplest spacetime metric, Minkowski space, in order to gain a better understanding of things.

Using the (- + + +) signature where the Minkowski metric ημν = diagonal(-1, 1, 1, 1) and 0 everywhere else, I proposed a very simple yet interesting curve (I denote my curve with the symbol ξ and I parameterize it with the letter s):

ξ(s) = [ sin(s) , 0, 0, 0 ] on the interval 0 < s < 2π

As you can see, the only non-zero coordinate of that curve is the temporal coordinate sin(s).

Now this is definitely a CTC because:

ημνξξ = (-1 * cos(s) * cos(s) ) = -cos2(s)

(You can probably tell this, but ξ is the derivative of component ξμ with respect to s).

The quantity -cos2(s) is less than 0 for all s, so this curve is definitely timelike.

Furthermore, this curve is literally a temporal circle. If you start at s = 0 (which would translate to [0, 0, 0, 0] in the position vector [t, x, y, z], c = 1 ) and moved through the interval from 0 to 2π, then you would return right back to the same event [ct = 0, x = 0, y = 0, z = 0] that you started at once s got to 2π. In fact, you'd actually return to that event at s = π, and then you would shortly thereafter start actually going backwards in time (to t < 0 events) from the point where you initially began. The whole process would be like this: You start at [0, 0, 0, 0] when s = 0. While 0 < s < π/2 , you would keep moving forward in time. Then at s = π/2 you would reach t = 1 (the furthest into the future you go with this CTC). Then while π/2 < s < π you would start moving backwards in time from the t = 1 point where you just were, but you'd still be in the future of the point where you began (or in other words t > 0). Then, when s = π, you would return to the event that you started from. Afterwards, while π < s < 3π/2 , you would move through the past of your original point (t < 0). Then at s = 3π/2, you reach t = -1(the furthest into the past you go with this CTC). Then, while
3π/2 < s < 2π, you start moving towards the future again, but you are still in the past of the point where you originally started. Finally, at s = 2π , you once again return to the event at which you started.

This is definitely a CTC in Minkowski space.

However... I'm pretty sure that such a CTC could not possibly actually occur in Minkowski space, nor could an object in Minkowski space follow such a world line (unlike in a Godel universe where an object apparently could follow such a world line).

This got me thinking:

What about a Godel universe actually theoretically makes the CTCs physically possible in that spacetime (as opposed to a spacetime like Minkowski space where you can mathematically propose a CTC that is consistent with all the math, but there is no way a curve like that would actually physically appear)?

Surely the thing that made the Godel solution famous couldn't have just been the fact that you can simply mathematically propose a CTC, because I have clearly shown here that you can even do that with Minkowski space (and probably just about any metric).

My one hypothesis as to the answer to this question is:

It has something to do with geodesics. Geodesics are the curves that naturally occur in a spacetime when no non-gravitational forces are at play right? So perhaps, the geodesic curves of a Godel universe are CTCs (as opposed to the geodesics in Minkowski space which are just the not so interesting curves that we follow in our every day lives)?

Is that it, or am I way off base?

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#### Dale

Mentor
Now this is definitely a CTC because:
It is not a closed time like curve. It is just the curve [ t , 0, 0, 0 ] on the interval -1 < t < 1 parameterized poorly.

Edit:
The quantity -cos2(s) is less than 0 for all s, so this curve is definitely timelike.
Also, even if you accept this parameterization it is not time like at $s=\pi/2$

Last edited:
• Pencilvester

#### PAllen

Science Advisor
It is not a closed time like curve. It is just the curve [ t , 0, 0, 0 ] on the interval -1 < t < 1 parameterized poorly.
And to make CTC out of in topologically nontrivial flat spacetime, just identify t=1 and t=-1, with t >1 and t < -1 no longer present in the manifold.

#### PeterDonis

Mentor
The quantity -cos2(s) is less than 0 for all s
No, it isn't; it is zero for $s = \pi / 2$.

As @Dale said, what you have actually done is construct a (poor) parametrization of the worldline of an inertial observer at the spatial origin $x = y = z = 0$ for the interval $-1 < t < 1$. This is easily seen when you look at the obvious reparameterization $t = \arcsin s$. Your parameterization cannot be valid outside the range $-1 < t < 1$ because a parameterization of a curve must be one-to-one, and yours is not outside the interval $- \pi / 2 < s < \pi / 2$, which corresponds to $-1 < t < 1$.

#### space-time

No, it isn't; it is zero for $s = \pi / 2$.

As @Dale said, what you have actually done is construct a (poor) parametrization of the worldline of an inertial observer at the spatial origin $x = y = z = 0$ for the interval $-1 < t < 1$. This is easily seen when you look at the obvious reparameterization $t = \arcsin s$. Your parameterization cannot be valid outside the range $-1 < t < 1$ because a parameterization of a curve must be one-to-one, and yours is not outside the interval $- \pi / 2 < s < \pi / 2$, which corresponds to $-1 < t < 1$.
Darn it! I totally forgot about the case of s = pi/2 for -cos2(s). Talk about embarrassing. Well thank you for your reply anyway. I did not know that the parameterization of a curve had to be 1 to 1 so thanks for letting me know this.

#### kent davidge

Two timelike events are causally connected, all inertial observers agree on their temporal order. In particular, the event "I departed from here" will occur before the event "I arrived back here".

So whats the magic in closed timelike curves? Any two pair of events on it would have a definite temporal order.

#### PeterDonis

Mentor
Two timelike events are causally connected
Yes.

all inertial observers agree on their temporal order
In flat spacetime (which has no CTCs), yes.

whats the magic in closed timelike curves?
That in such a curved spacetime, it is no longer possible to say that two timelike separated events have a definite order if they lie on a CTC. All events on a CTC occur both before and after all other events on the same CTC. Just as, for example, every point on the Earth's equator is both due east and due west of every other point on the equator.

• kent davidge

#### Dale

Mentor
Two timelike events are causally connected, all inertial observers agree on their temporal order.
This is always true in SR. In GR it is only always true locally.

So whats the magic in closed timelike curves? Any two pair of events on it would have a definite temporal order.
Closed time like curves are an example where the locally true statement is not globally true.

• kent davidge

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