# B How to travel back in time according to General Relativity?

1. Jun 25, 2017

### TheQuestionGuy14

Is there any way to travel back in time in reality according to GR? Let me know!

2. Jun 25, 2017

### Staff: Mentor

What does "in reality" mean?

As a matter of mathematics, the Einstein Field Equation, the central equation of GR, admits solutions with closed timelike curves, which is the technical way of saying "allows travel back in time". But AFAIK nobody thinks those solutions are physically realistic.

3. Jun 25, 2017

### TheQuestionGuy14

In reality I mean, not in the movies, through our understanding of physics.

4. Jun 25, 2017

### Staff: Mentor

Ok, then the answer is what I said in my previous post. (Since this is a forum on physics, not movies, the "in reality" can be left out if all you mean by it is "through our understanding of physics". We assume here that that's what you're asking about.)

5. Jun 26, 2017

### Battlemage!

6. Jun 26, 2017

### Chris Miller

From what I think I've learned here, there is no universal now. Each point in the 4D universe is mapped onto to a unique x,y,z,t coordinate. So traveling backwards in time would be relative. Also, since the universe is a 4D construct, it has no "thickness" in its highest 4th (t) dimension (just as an expanding 3D balloon has no 3rd (z) dimensional thickness. So in the universe, a point's past does not exist. So, to travel back you would have to leave the universe. Really, you can't travel forward in time either. There's only the (relative) present.

7. Jun 26, 2017

### Staff: Mentor

In some senses, yes. But if there are closed timelike curves present in a spacetime, which, as I noted before, is what is usually meant by "traveling back in time", that fact is an invariant, independent of any choice of coordinates.

This is not correct; in fact it doesn't even make sense.

8. Jun 26, 2017

### DrStupid

Yes, in the sense that t2 < t1 for one observer doesn't necessarily imply t2' < t1' for any other.

Going from (x,y,z,t1) to (x,y,z,t2) with t2 < t1 within the same block universe would be a time-travel into the past. Why do I need to leave the universe to do that?

Leaving the universe would be required for different pasts (e.g. in the grandfather paradox with one past where the grandfather survives and another where he dies). But that's another topic.

9. Jun 26, 2017

### Chris Miller

Thanks, Dr. S.

I guess I was assuming that for an observer at any x,y,z there is only one t: the local present/now. Observers in other frames of ref might measure t' at this x,z,y to be < their current t, but that's not time travel. Local time travel is loaded with paradoxes. If it were "ever" possible, seems like time would become irrelevant. The past and the future would merge into a big chaotic blob of a now.

10. Jun 26, 2017

### Staff: Mentor

Why would you assume that? Do you understand that $t$ is a coordinate just like $x, y, z$, and that an observer's worldline is a curve in spacetime, i.e., a continuous set of points, not just a single point?

11. Jun 26, 2017

### Chris Miller

Thanks, PD. Yes, I understand that x,y,z,t are just coordinates in 4D spacetime (the universe). Not sure what you mean by "an observer's worldline" though. Especially "worldline."

12. Jun 26, 2017

### Staff: Mentor

It is the curve in spacetime that describes the observer's history. The usual way of expressing it mathematically is as a set of four functions $x(\tau)$, $y(\tau)$, $z(\tau)$, $t(\tau)$ that parameterize the curve by the observer's proper time $\tau$.

13. Jun 26, 2017

### DrStupid

I don't know, what you mean by "local time travel", but due to the Novikov self-consistency principle paradoxes wouldn't be an issue.

14. Jun 26, 2017

### Paul Colby

Traveling back in time requires the boundary conditions on all fields, not just the metric, must consistent. Having closed time like curves says nothing about these other condition as far as I can see.

15. Jun 27, 2017

### Chris Miller

Thanks. So, according to GR/SR, it's theoretically possible for an observer to return to some x,y,z,t point at a t' that is < t?

16. Jun 27, 2017

### Chris Miller

Thanks again, Dr. S. By "local time travel" I just meant travelling from x,y,z,t to x,y,z,t' where t' < t.

Very interesting. To me, this translates to "Traveling back in time is impossible," because even just the appearance of an atom changes something. Also, I'd argue that, even were backwards time travel common, it would be impossible to change the past regardless of one's actions. Whatever "new" past then becomes the only past, along with all memory and record of it. It would be no different than how we "change" the future. (E.g., I could assert that a time travel technology was invented, the result of which was so disastrous, someone went back and killed the inventor so that it never happened after all.)

The math and QM, though over my head, seem mis-applied in the way Zeno's arguments mis-apply differential calculus to prove the impossibility of motion and time.

17. Jun 27, 2017

### DrStupid

I'm afraid you didn't got the point. Maybe you realize it when you try to explain what you mean with "change". I think that's the main source for misconceptions regarding time travel.

18. Jun 27, 2017

### Staff: Mentor

No. This doesn't even make sense. A given event in spacetime, in a given coordinate chart, has a unique 4-tuple of coordinate values $x, y, z, t$. As you yourself stated in a previous post:

I strongly suggest that you take some time to work through a textbook. You appear to have some fundamental misunderstandings about how coordinate charts work.

Which doesn't even make sense. See above.

No, that's not what the Novikov self-consistency principle says. It just says that traveling back in time can't change the past; if someone travels back in time, the past they travel back to already includes their traveling back into it. (Note that this doesn't mean time travel must be possible; it is just a proposed constraint on what would happen if time travel is possible.)

If the principle is correct, nothing "appears" in the past when someone time travels; they were always there to begin with. (Any GR solution with closed timelike curves will work like this: to "time travel" in such a spacetime, you don't do anything special, it just so happens that the spacetime is curved in such a way that your worldline ends up traversing the same event multiple times. But your path through spacetime is continuous, so you never "appear" or "disappear" anywhere.)

If it's over your head, how can you tell whether it's mis-applied?

Once again, please take the time to work through a textbook.

19. Jun 27, 2017

### pervect

Staff Emeritus
I haven't read the thread, but I'd like to mention a couple of classic papers on the topic:

"Wormholes, time machines, and the Weak Energy condition" http://authors.library.caltech.edu/9262/1/MORprl88.pdf
"Billiard balls in wormhole space-times" https://docs.google.com/file/d/0B0xb4crOvCgTelBiLXJSRlMwSnM/edit

The second paper discusses some paradoxes involving billiard balls and time machines, and their resolution. Philosophical questions about "free will" cannot be discussed scientifically, but the behavior of billiard balls certainly can. The paper explores the seemingly paradoxical idea of billard balls going back in time and hitting themselves.

The first paper suggests how a time machine might be created from a wormhole, and how a wormhole might be stabilized from a quantum fluctuations.

For a popularized discussion, I'd recommend Thorne's book "Black holes and time warps", https://www.amazon.com/Black-Holes-Time-Warps-Commonwealth/dp/0393312763.

One thing Thorne mentions that the above may not - one can't build a wormhole classically without already having a time machine, so the wormhole route to time machines would require a time machine in a purely classical approach. Quantum mechanics MAY provide a loophole. It also provides obstacles of its own that are not obvious.

In conclusion, there is literature on the topic of time machines in GR. I'd suggest that the thrust is that authors are not especially enthusiastic about the possibility, but haven't been able to totally rule it out yet. This is of course a rather subjective impression, and if you want all the details there's a lot of reading to be done. The two papers above may be a reasonable place to start, along with Thorne's book.

20. Jun 27, 2017

### Arkalius

I think it is accurate to say that in flat spacetime, for any $\tau_1 \lt \tau_2$, $t(\tau_2)^2 - t(\tau_1)^2 \gt x(\tau_2)^2 - x(\tau_1)^2 + y(\tau_2)^2 - y(\tau_1)^2 + z(\tau_2)^2 - z(\tau_1)^2$, correct? Would it also be true in general relativity in curved spacetime geometries with which we are currently familiar? And then would it also be accurate that there exist solutions to the field equations that create geometries wherein this inequality no longer holds true, including closed timelike curves?

21. Jun 27, 2017

### Staff: Mentor

If $\tau_1$ and $\tau_2$ are proper times of events on the same observer's worldline, yes, your statement is just another way of saying that the worldline is timelike.

No, because the metric won't be the same, and the metric is what determines the "interval formula" that you used to derive your expression. You will be able to write down some expression involving the coordinates as functions of $\tau$ that corresponds to an observer's worldline being timelike, but it won't be the same one in general that you wrote down for flat spacetime.

As you state it, this question doesn't have an answer, because it assumes that the answer to your previous question was "yes", and it isn't--see above.

Rather than speak in generalities, it might be helpful to look at some specific examples. One obvious one is Schwarzschild spacetime, more specifically the region outside the horizon (to keep things simple). If we look at the parameterization of any worldline in Schwarzschild coordinates, we will find that we can write down an expression something like the one you wrote down, but note carefully the differences:

$$\left( 1 - \frac{2M}{r} \right) dt^2 > \frac{1}{1 - 2M / r} dr^2 + r^2 \left( d\theta^2 + \sin2 \theta d\phi^2 \right)$$

As you can see, first of all, there are extra factors in the terms, corresponding to the extra factors in the metric that would not be there in flat spacetime. But notice also that the expression involves differentials of coordinates, i.e., infinitesimally small differences--whereas your original expression involved finite differences in coordinates. You were able to get away with finite differences for two reasons: (1) you were in flat spacetime, and (2) you chose standard Minkowski coordinates, which have certain special properties that do not generalize.

In the general case, if we want to derive an expression involving finite coordinate differences, we have to integrate the above expression over some range of coordinates, and since the extra factors in the expression depend on the coordinates, such an integral gets complicated. (The special property of Minkowski coordinates in flat spacetime that lets you avoid this is that all of the metric coefficients are constants--none of them are functions of any of the coordinates. You should be able to convince yourself that that specific case is the only case that has that property.) But we can still investigate some special cases, in spacetimes which have some symmetries. For example, we can see that, since the metric coefficients only depend on $r$ and $\theta$, any worldline of constant $r$ and $\theta$ will have the nice property that the metric coefficients are constant along that worldline. Which means that, for such worldlines, the integrals are simple just the way they are in Minkowski coordinates in flat spacetime, and we can write an expression involving finite differences:

$$\left( 1 - \frac{2M}{r} \right) \left( t(\tau_2)^2 - t(\tau_1)^2 \right) > r^2 \sin^2 \theta \left( \phi(\tau_2)^2 - \phi(\tau_1)^2 \right)$$

If you think about it, you will see that this places a limit (which depends on $r$ and $\theta$) on the angular velocity at which an object can orbit the central mass.

I'll follow up with a similar discussion of the Godel spacetime in another post.

22. Jun 27, 2017

### stoomart

Sure, if you have access to a wormhole and don't mind the risk of getting stuck in an infinite time loop. From a nature.com article:

23. Jun 27, 2017

### Staff: Mentor

This requirement is highly likely to mean you can't do this "in reality" in the usual sense--that is, while there are solutions of the Einstein Field Equations that describe wormholes like this, they require "exotic matter" (i.e., stress-energy that violates the energy conditions), and it's highly likely that exotic matter of this kind cannot exist.

24. Jun 27, 2017

### stoomart

I assume there aren't any realistic conditions where CTCs are possible (maybe inside black holes rotating at relativistic speeds?).

25. Jun 27, 2017

### Staff: Mentor

Not that I'm aware of.

The deep interior of Kerr spacetime (inside the inner horizon) has CTCs, yes, but it is not considered "realistic" (because the geometry inside the outer horizon is unstable to small perturbations).