This question is an extension of my a question that I have recently asked: Why doesn't a global frame of reference exist for GR?, where it was recommended that I post another question (so I am sorry if this question is inappropriate). The question above was suggested to me, but below, I will offer my own thoughts on the matter just to provide some context.

I have originally heard that GR generally doesn't allow for the concept of a global frame of reference, or really, a global coordinate system. Now when I say "global frame" I DON'T mean an "absolute frame" like an ether or anything. I just mean a frame that could be applied to the entire universe. Now moving on, it appears that GR does generally allow them, but under certain conditions they are not possible (such as ones that involve wormholes). I have also read that GR has solutions which allow for closed timelike curves, which involve time travel. Now, my question is whether or not both of these concepts are related. When we say that GR doesn't allow for a global frame, are we referring to the fact that there are solutions to GR that allow for CTCs?

  • $\begingroup$ Such a coordinate system doesn't even exist for a static sphere. You need at least two maps to capture every point on a sphere (if you try to use just one continuous map, then at least one of the poles gets lost). This isn't even a property of general relativity, it's a much more common phenomenon of manifolds. $\endgroup$ – CuriousOne Jun 12 '15 at 18:29
  • 3
    $\begingroup$ The question in your title doesn't make sense without context and it doesn't seem to be the same as what you ask in the body text. $\endgroup$ – Jim Jun 12 '15 at 19:02
  • $\begingroup$ @Jim the Enchanter: That exact question was recommended to me to ask, and apparently is supposed to help address the concerns that I have laid out. I am not entirely sure about the connection myself, but I just wanted to know if the existence of models with CTCs would lead to no global coordinate systems, so if you can answer that latter question, then at least I would be happy. $\endgroup$ – Mike H Jun 12 '15 at 19:08
  • $\begingroup$ I'm no mathematician, but I think existence of global coordinate systems is model dependent itself. A spacetime with CTCs wouldn't have a globally valid manifold that describes the spacetime (again, not a mathematician so I could be wrong about that), but there certainly are other models for which you could find a spacetime manifold that is completely specified by one coordinate system. $\endgroup$ – Jim Jun 12 '15 at 19:13
  • $\begingroup$ If you don't know what you are asking and if that becomes obvious, as in this case, we tend to vote to close. Are solutions to Einstein's equations around a point unique up to coordinate transformations? I believe so. The physical difficulty is in the "up to coordinate transformations" part. $\endgroup$ – CuriousOne Jun 12 '15 at 19:30

Here is one solution to Einstein's field equations:

$$ds^2=dt^2-dx^2-dy^2-dz^2 \text{ on } \{(x,y,z,t):x,y,z,t\in\mathbb R\}.$$

It has a global coordinate system, a global reference frame, no closed time like curves, and a famous name, Minkowski space.

Our second solution is a different manifold $\mathbb R^3\times \mathbb S$, which can (but doesn't have to be) though of as $\{(v,w,x,y,z)\in\mathbb R^5:x,y,z\in\mathbb R, v^2+w^2=1\}.$ Intuitively you can just treat it as a submanifold with the metric $ds^2=dv^2+dw^2-dx^2-dy^2-dz^2$ (I tell you that so the following steps will be easy to follow, but you shouldn't imagine that spacetime is a surface is a larger space, there are multiple ways to do so and they make the same predictions so you have to be careful not to read too much into things that don't affect your predictions). But if you treat it as a manifold in its own right then because it is topologically a cylinder there isn't a global coordinate system. As a manifold in its own right it is 4 dimensional, and it is flat.

So for the second manifold we need two coordinate patches, for instance:

$$ds^2=dT^2-dX^2-dY^2-dZ^2 \text{ on } \{(X,Y,Z,T):X,Y,Z\in\mathbb R, T\in(-2\pi/6,2\pi/6)\}, \text {and}$$

$$ds^2=d\tau^2-dA^2-dB^2-dC^2 \text{ on } \{(A,B,C,\tau):A,B,C\in\mathbb R, \tau\in(\pi/6,11\pi/6)\}.$$

To make an actual manifold we need transition maps between the charts. Here is why I introduced the surface $\{(v,w,x,y,z)\in\mathbb R^5:x,y,z\in\mathbb R, v^2+w^2=1\},$ just to make these transitions maps easy to see. Think of time ($T$ or $\tau$) as like angle, we can get $v=\cos (\tau)$ and $w=\sin (\tau)$ (and $x=A$, $y=B$, $z=C$). Similarly $v=\cos (T)$ and $w=\sin (T)$ (and $x=X$, $y=Y$, $z=Z$). In both cases, $w/v=\tan (\tau)=\tan (T)$ when $v\neq 0$. But the whole region of overlap is where $v>0$ so $w/v=\tan(T)$. So the map that sends $(\tau,A,B,C)$ to $(T,X,Y,Z)=(\arctan \left(\tan (\tau)\right),A,B,C)$ is our transition map.

Now you can dispense with $\{(v,w,x,y,z)\in\mathbb R^5:x,y,z\in\mathbb R, v^2+w^2=1\}$ entirely and just say that when in $$ds^2=d\tau^2-dA^2-dB^2-dC^2 \text{ on } \{(A,B,C,\tau):A,B,C\in\mathbb R, \tau\in(\pi/6,11\pi/6)\}$$ if you are in $\tau < 2\pi/6$ and want to switch to the other coordinate just change set $T=\tau$, $X=A$, $Y=B$, and $Z=C$. Whereas if you are in $\tau > 10\pi/6$ and want to switch to the other coordinate just change set $T=\tau -2\pi$, $X=A$, $Y=B$, and $Z=C$. And that is what $\arctan \left(\tan (\tau)\right)$ does after all, so this isn't different, its just that we don't need an embedding space we just need to transition from one coordinate system to the other before we leave its domain of applicability.

And going the other way, when in $$ds^2=dT^2-dX^2-dY^2-dZ^2 \text{ on } \{(X,Y,Z,T):X,Y,Z\in\mathbb R, T\in(-2\pi/6,2\pi/6)\}$$ if you are in $T>\pi/6$ and want to switch to the other coordinate just change set $\tau=T$, $A=X$, $B=Y$, and $C=Z$. Whereas if you are in $T < -\pi/6$ and want to switch to the other coordinate just change set $\tau=T+2\pi$, $A=X$, $B=Y$, and $C=Z$.

OK, so that is a groundhog day type universe, the whole universe just repeats after $2\pi$ units of time. For no reason. It doesn't even have any matter, let alone any curvature (no Riemann curvature) anywhere at any time. And you are forced to have two coordinate charts (unless you want to embed it into a larger set). But not because of any curvature or anything, just because it is a cylinder and cylinders require more than one coordinate system unless you want to have jumps, for instance you could have a regular four dimensional flat space and just identify $t=0$ and $t=2\pi$ and get your coordinate system that way, and most physicists are fine with that.

So if you look at the region near $t=0$ from our first manifold and $T=0$ from our second manifold they look exactly the same from $T=-\pi/6$ all the way to $T=\pi/6$ everything flat, everything empty, everything Minkowski even. So they agree in a neighborhood of the spacelike slice $t=0$.

But the second solution has closed time like curves and hence time travel.

But we could have done the same thing with a spatial direction. Had $z$ be an angle in the $v,w$ plane like $\{(v,w,x,y,t)\in\mathbb R^5:x,y,z\in\mathbb R, v^2+w^2=1\}$ with metric $ds^2=dt^2-dx^2-dy^2-dv^2-dw^2$ in which case going in the $z$ direction effectively jumps back $2\pi$ so the universe repeats itself spatially and is still uncurved and does so for no reason and technically requires at least two coordinate systems if you don't want to allow discontinuous jumps of coordinate values.

To connect this back to your question about global coordinate systems, these examples can't locally be distinguished, so it also doesn't make much sense to make a big deal about there differences. For instance if you walk $2\pi $ units in the z direction and everything looks the same, maybe the universe looped back, or maybe the universe just looks similar every $2\pi$ units with many copies of you so you each moved to the next region which is different but looks the same. Both options look the same. So how big a deal can it be?

For time travel it might sound like a bigger deal whether you are in your own past or just in a region of space and time that looks like your own past looked. But again, how are you going to distinguish between the two options? If you don't know how to tell the difference between the two then it might be too early to get excited about the differences (if any) between the two.

| cite | improve this answer | |
  • $\begingroup$ Ha, I can see why you wanted me to make another question. The math is unfortunately beyond me, but your description makes your previous response clearer (specifically, why the existence of groundhog and pac-man universes would be indistinguishable from the ones without them). Thanks again. $\endgroup$ – Mike H Jun 12 '15 at 20:38
  • $\begingroup$ @MikeH If you imagine longitude going from -180 to -90 to 0 to +90 all the way to +180 then jumping back down to -180 if you go a tiny bit more then just keep in mind that GR can allow the same thing to happen with space or with time, without even having any curvature. If you want to force there to be CTC you might need singularities or some exotic matter or both and even then it might just allow it without requiring it. $\endgroup$ – Timaeus Jun 12 '15 at 21:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.