# Questions about closed timelike curves

1. Apr 8, 2013

### HomogenousCow

I have read about CTCs from some books and find the explanations confusing.
-Are they simply natural trajectories in the given spacetimes?
-How is energy-momentum conserved if a particle can simply travel back in time? i.e. Observers will observe particles travelling in a CTC to simply disappear from their time slice

2. Apr 8, 2013

### Passionflower

Consider the simplest case of a particle decaying, then the results decaying again and finally forming the original particle again. "Rock, paper scissors", eternally. Nothing goes 'back' in time, a process simply repeats itself.

3. Apr 8, 2013

### Staff: Mentor

No, they won't. Consider carefully what a "time slice" would look like (or whether one can even exist in the sense you mean) in a spacetime with CTCs.

4. Apr 8, 2013

### HomogenousCow

What would an observer travelling in a time like curve observe?

5. Apr 8, 2013

### Passionflower

Nothing particular, he could not even deduce he is in a CTC. Because while it is true that there is no recorded information older than the path length of the curve he could also interpret this as the beginning of his world.

6. Apr 8, 2013

### Staff: Mentor

Meaning a CTC? It would depend on what else was in the spacetime, of course.

But what I think you are getting at is, what would a "time slice" look like for an observer moving along a CTC? The answer is, there can't be a "time slice" in the usual sense that covers the entire spacetime. The whole point of a "time slice" in the usual sense is that any given observer can only pass through a "time slice" in one direction; but if you take any spacelike slice through a spacetime with CTCs in it, observers traveling on one of the CTCs will cross that spacelike slice twice, in opposite directions. It's difficult to make any sense of what such an observer would observe.

7. Apr 8, 2013

### Staff: Mentor

I'm not sure this is correct. See my previous post.

8. Apr 8, 2013

### HomogenousCow

let's say I travel from event A out to somewhere else and then return to event A at a later proper time, thus forming a closed timelike curve, would I "remember" the entire journey? This is essentially what is troubling me because what it really means it that I traverse the same coordinate time twice, but that is unphysical and the only physical time is the proper time I measure.

Actually to be honest I'm not quite sure what any particle on a wordline observes, since we can't talk about relative velocities and pretty much anything tensorial at another point in space time..ugh baffling.

9. Apr 8, 2013

### Passionflower

There would obviously be no memory effect longer than the length of the CTC path.

10. Apr 8, 2013

### bcrowell

Staff Emeritus
Conservation laws in GR are purely local. Except in special cases, there are no globally conserved quantities in GR. Local conservation of matter is expressed by the divergence-free property of the stress-energy tensor, which is equally valid in a spacetime with CTCs (because it follows from the field equations).

11. Apr 8, 2013

### Staff: Mentor

Not unless you already "remembered" it when you were at event A the "first" time. (Actually you will visit event A an infinite number of times, strictly speaking.) Your state must be the same every time you pass through event A, since any physical observables at event A, including those that store your "memories", can only have one state at event A.

No, it means a lot more than that. It means you pass through the *same event* twice. As in, you pass through "the stop light at the corner of 5th Avenue and 59th St. in Manhattan changes from red to green at exactly 12 noon on January 1st, 2001" twice. (Actually, you pass through that event, and all the others on your worldline, an infinite number of times.)

No, but you can talk about what signals (either light signals, traveling on null worldlines, or signals of other kinds traveling on timelike worldlines) are arriving at a particular event on your worldline from other worldlines. Whatever you "observe" has to be expressible in terms of these things. Sometimes there is an intuitive translation from these things to assignments of coordinates to distant events, sometimes there isn't.

12. Apr 8, 2013

### BruceW

When you return to A, the proper time that will have passed since you left is zero. You are returning to the same spacetime point, so you must have the exact same properties as when you started your journey. So if you gained some memories during the journey, then you are going to have to 'forget' them somehow before you return to A.

Edit: I see I have pretty much repeated what PeterDonis posted while I was typing, ah well.

13. Apr 8, 2013

### Staff: Mentor

This isn't quite true (although the rest of your post is, since as you say it's basically what I said ), but explaining why it isn't is instructive in showing the problems with CTCs.

The correct statement is that, if we have an observer traveling along a CTC, and we parametrize events on his worldline by proper time, the proper time itself will run from minus infinity to infinity, but there will be multiple values of proper time that map to the same events. That means that, if we express any observable along the worldline as a function of proper time, it must be a periodic function. The fact that, as I said, the proper time itself ranges from minus infinity to infinity, is why I said the observer will visit each event on his worldline an infinite number of times.

This kind of thing is consistent mathematically, but it's extremely difficult to come up with a physical interpretation of it that seems reasonable, particularly if the observer is supposed to be conscious.

14. Apr 8, 2013

### BruceW

that makes sense. But I don't understand it mathematically. I mean, when we integrate proper time over 1 loop, then it is something like:
$$\oint d \tau$$
Isn't this zero, and so the change in proper time over 1 loop is zero? (p.s. I am new to general relativity, so I apologise in advance if I'm a bit slow to grasp things).

Edit: to explain in more detail my line of reasoning, I would think that the integral of a function between the same end-points is going to be zero, which is why that integral is zero generally. Maybe this is not actually true? Ah, I am also tired right now, so I might not be thinking straight.

15. Apr 8, 2013

### Staff: Mentor

No. There are several ways to see why.

(1) The integral is the length of the loop, not zero. But you could go around the loop multiple times to get a length as large as you want. The object following the loop doesn't just stop when it gets back to its original point; how can it? Moving around the loop, from the object's point of view, is just moving into the future, and how can you stop moving into the future?

(2) You're not really integrating proper time over a loop. As you've written the integral, proper time is the integration variable, so you have to know its range in advance in order to specify the limits of integration; you don't find it out by doing the integral.

(3) If you instead say that proper time in the integral is a function of the coordinates, so we can specify the limits of integration in terms of coordinate intervals, then the problem is that the function is not single-valued; as I said before, multiple values of proper time map to the same event, hence to the same 4-tuple of coordinate values.

16. Apr 8, 2013

### BruceW

ah yeah, of course. Thanks for explaining. I was thinking of the integral of electric potential:
$$\int_{V_1}^{V_2} \ dV$$
And hoping this was analogous to the integral of proper time. But as you say, it is not, because in the integral of proper time, the limits are points in spacetime, which are the same. Not values of the proper time.

edit: or I could guess you could also say that the limits of the integral of proper time are the values of proper time, which are not the same, even though the corresponding points in spacetime are the same.

second edit: can you tell I'm tired? haha. But I do understand now. Even though I might not be able to explain in a very clear way. Thanks for the help :)

Last edited: Apr 8, 2013