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Reading John Rennie's answer, it made it very clear as to why it is justified to talk of time as a coordinate. So, for now we put time and space on the same footing i.e. consider them dynamical coordinates (they can bend, stretch etc).

Let us talk of only simple closed curves (A connected curve that does not cross itself and ends at the same point where it begins). A planetary orbit around a star can be said to be a closed spacelike curve (loop) (excluding parabolas and hyperbolas) because the curve does not intersect itself and can be said to begin and end at the same point.

I highlighted the last word because that word holds some ambiguity in itself. Point in space? Or in time? Or both? From what we can judge, it's the second one, because for example, we don't go back to a former time but rather to a former space (coordinate) after one revolution.

We also have taken that time and space are atleast on the same dimensional footing.

Then what restricts closed timelike curves (or loops) in General Relativity? Or framed in another way Why don't closed lime loops exist if space loops do?

Some might say that CTCs exist in some cases like Gödel universe and others. But why do we require so exotic situations for a time loop? Why don't a normal metric provide us with it?

Yuzuriha Inori
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Space and time are only "sort of" on the same dimensional footing. Space and time are physically inseparable, but within four-dimensional spacetime, some directions are "spacelike", and some directions are "timelike". ("Lightlike" a.k.a. "null" is a third type of direction in spacetime.) And if you choose four orthogonal directions in spacetime, with no "lightlike" directions, only one of those directions will be "timelike", and the three other directions will be "spacelike".

The reason closed timelike curves are harder to form than closed spacelike curves is because although there's only one timelike dimension, there's more than one spacelike dimension, which makes it possible to form a closed spacelike curve without even needing spacetime to be curved or have any topological weirdness.

For example, in flat spacetime with no topological weirdness, choose a timelike direction to call "time", identified using the coordinate $t$, and three orthogonal spacelike directions identified using coordinates $x$, $y$ and $z$. One closed spacelike curve is the set of events given parametrically by

$$r_s(\lambda) = (0,\ \cos\lambda,\ \sin\lambda,\ 0)$$

for $0 \le \lambda \le 2 \pi$, where the event coordinates are given in the order $(t,x,y,z)$. $r_s(2\pi)=r_s(0)$, so the curve is closed, and the tangent four-vector to that curve,

$$\vec{T}_s(\lambda)=\frac{d}{d\lambda}r_s(\lambda)=(0,\ -\sin\lambda,\ \cos\lambda,\ 0)$$

clearly always points in a spacelike direction, so the curve is spacelike.

Now try to do something similar (still in regular flat spacetime) to form a closed timelike curve. Define the curve parametrically as being the set of events

$$r_t(\lambda) = (f_t(\lambda),\ f_x(\lambda),\ f_y(\lambda),\ f_z(\lambda))$$

for $0 \le \lambda \le \lambda_{max}$.

For the curve to be closed, we require $f_t(\lambda_{max})=f_t(0)$, $f_x(\lambda_{max})=f_x(0)$, etc. The tangent four-vector to that curve is

$$\vec{T}_t(\lambda)=\frac{d}{d\lambda}r_t(\lambda)=(f'_t(\lambda),\ f'_x(\lambda),\ f'_y(\lambda),\ f'_z(\lambda))\ \ ,$$

where $'$ denotes differentiation with respect to $\lambda$.

Define $\lambda_{oops}$ such that $f_t(\lambda_{oops}) \ge f_t(\lambda)$, for all $\lambda$ in the range $0 \le \lambda \le \lambda_{max}$. For the curve to be smooth, it must be the case that $f'_t(\lambda_{oops})=0$. But that means that at $\lambda=\lambda_{oops}$, the tangent four-vector to the curve has a zero component in the timelike direction $t$, so it must point in a spacelike direction, and we've failed to construct a closed timelike curve.

If spacetime had a second timelike dimension, call the extra timelike coordinate $\tau$, then we could solve the above problem by arranging for $\vec{T}_t(\lambda_{oops})$ to point in the $\tau$ direction. It's the lack of a second timelike dimension that keeps us from successfully constructing a closed timelike curve without relying on curvature or topological weirdness.

Red Act
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A closed, space-like curve (CSC) would have to return to its original point in (t,x,y,z) to be considered truly closed in GR. Of course, you can draw such a curve, but a massive particle (or even photon) could never traverse it, as the definition of space-like means that it cannot be the trajectory of a physical particle. So really, even open space-like curves are forbidden. That is why the focus of time travel is based on time-like curves that could actually be the trajectories of real particles.

Jeremy
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The path that a planet takes around a star is a closed curve in space, but not a closed curve in spacetime. Like you said, it would be a bit ambiguous to say that the curve ends at the same spatial point, because the splitting between spatial coordinates vs. time depends on what coordinate system you're using. Closed curves make sense when the curve returns to the same point in spacetime.

As for your other question, I don't have a good answer because there isn't a strict definition of "exotic". But one thing to note is that closed spacelike curves exist in Minkowski space, whereas closed timelike curves don't. That's one indication that it takes some work to construct situations where closed timelike curves exist.

pianyon
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