Solving Einstein Field Equations for Minkowski Space with CTC

[edgepflow]

For Minkowski spacetime, the metric is:

ds^2 = -dt^2 + dx^2 + dy^2 + dz^2

I have read there is a solution when the time dimension is "rolled" into a cylinder forming a closed timelike curve. So the BC is t -> [0,T] with t = 0 identical with t = T.

The Field Equation is:

Rab - 1/2 gab R = 8 Pi G Tab.

With gab = diag [-1,1,1,1]

Could I determine the Ricci Tensor & Scalar and solve for the stress energy tensor?

Please point me in the direction to perform this solution or let me know where I can find it.

 

[bcrowell]

The field equations only depend on the local properties, not the global topology. The curvature is zero, and so is the stress-energy tensor.

 

[edgepflow]

The field equations only depend on the local properties, not the global topology. The curvature is zero, and so is the stress-energy tensor.

I noticed the curvature is zero just after I posted this. If the time dimension of space is rolled into a cylinder forming a closed timelike curve, what would be the metric?

 

[Mentz114]

This is the 'groundhog day' metric where time goes in circles - but it is still flat spacetime.


ds² = -sin(t)²dt² + dx² + dy² + dz²

 

[edgepflow]

This is the 'groundhog day' metric where time goes in circles - but it is still flat spacetime.


ds² = -sin(t)²dt² + dx² + dy² + dz²

Thank you !

 

[K^2]

This is the 'groundhog day' metric where time goes in circles - but it is still flat spacetime.


ds² = -sin(t)²dt² + dx² + dy² + dz²

And what happens to proper time of a massive particle at a turning point? That's right, it diverges. The time might be looped here, but no particle will ever reach the turning point by its own clock, so no Groundhog Day.

Topology and geometry of space-time are two entirely different topics. Topology gives you boundary conditions and is a given when you solve Einstein Field Equations. If you need closed time, just set up a periodic boundary condition in R⁴. The result will be equivalent to TxR³.

 

[George Jones]

This is the 'groundhog day' metric where time goes in circles - but it is still flat spacetime.ds² = -sin(t)²dt² + dx² + dy² + dz²

Interesting, I have never seen the groundhog day metric. I think that t is a poor choice of coordinate label.

And what happens to proper time of a massive particle at a turning point? That's right, it diverges.

I have been playing around a bit, and I don't think so.

The time might be looped here, but no particle will ever reach the turning point by its own clock, so no Groundhog Day.

The t component of of the 4-velocity of a particularly natural family of observers diverges, not proper time for the observers.

Topology and geometry of space-time are two entirely different topics. Topology gives you boundary conditions and is a given when you solve Einstein Field Equations. If you need closed time, just set up a periodic boundary condition in R⁴. The result will be equivalent to TxR³.

By T, do you mean S^1? I think that the metric that Mentz114 gave is intended for the spacetime after (or in combination with) identification, i.e., for S^1 \times \mathbf{R}^3.

More later.

 

[Mentz114]

ds² = -sin(t)²dt² + dx² + dy² + dz²

Solving the equations of motion with the initial condition t=0 \rightarrow \tau=0 gives

<br /> \tau=1-\cos(t)<br />

 

[George Jones]

ds² = -sin(t)²dt² + dx² + dy² + dz²

Solving the equations of motion with the initial condition t=0 \rightarrow \tau=0 gives

<br /> \tau=1-\cos(t)<br />

For and an inertial observer with constant x, y, and z, this is what I get as well. The components of such an observer's 4-velocity are

\left( u_t , u_x , u_y , u_z \right) = \left( \frac{1}{\sin t} , 0 , 0, 0 \right).

 

[PAllen]

For and an inertial observer with constant x, y, and z, this is what I get as well. The components of such an observer's 4-velocity are

\left( u_t , u_x , u_y , u_z \right) = \left( \frac{1}{\sin t} , 0 , 0, 0 \right).

Ah, and the norm of this 4 velocity - using the given metric - is identically 1, as it should be. Tricky to use this metric (for me at least), but all looks valid and consistent.

 

[PAllen]

I see that the cause of the 4 velocity's divergent coordinate representation is that it is actually periodically lightlike. Light paths take the form (in X T plane): x=k-cos(t), so any constant position path is periodically tangent to a light path. This is physically dubious - it should not be possible for a real particle's world line to ever be tangent to null geodesic.

So, I think we have a problem if there are no world lines that are everwhere timelike. My initial take is that, in fact, there are no such world lines.

Reactions?

 

[Altabeh]

Ah, and the norm of this 4 velocity - using the given metric - is identically 1, as it should be. Tricky to use this metric (for me at least), but all looks valid and consistent.

This is a very poor metric. It certainly has uncountably many coordinate singularities and so it is not a suitable case for Minkowski spacetime! Taking T=\cos(t), we have -dT^2=-sin^2(t)dt^2. But this only works if 0=&lt;t&lt;\pi/2 which means time is bounded from above!

Very bizarre!

 

[JesseM]

This is a very poor metric. It certainly has uncountably many coordinate singularities and so it is not a suitable case for Minkowski spacetime! Taking T=\cos(t), we have -dT^2=-sin^2(t)dt^2. But this only works if 0=&lt;t&lt;\pi/2 which means time is bounded from above!

Very bizarre!

It's not that time is bounded from above, it's that spacetime has an unusual topology such that when you reach what would be t=pi/2 and x=X, y=Y, z=Z, you are actually back at the point in spacetime that was already labeled with coordinates t=0 and x=X, y=Y, z=Z. Think of the analogous case where space is finite but unbounded due to a weird topology (as discussed in this article), here it would also be the case that at least one of the spatial coordinates could only have a finite extent before "resetting" (assuming you want a well-behaved coordinate system that doesn't assign multiple sets of coordinates to the same physical point in spacetime).

 

[Altabeh]

It's not that time is bounded from above, it's that spacetime has an unusual topology such that when you reach what would be t=pi/2 and x=X, y=Y, z=Z, you are actually back at the point in spacetime that was already labeled with coordinates t=0 and x=X, y=Y, z=Z. Think of the analogous case where space is finite but unbounded due to a weird topology (as discussed in this article), here it would also be the case that at least one of the spatial coordinates could only have a finite extent before "resetting" (assuming you want a well-behaved coordinate system that doesn't assign multiple sets of coordinates to the same physical point in spacetime).

In fact I don't think GR accepts such a nonsense! Because if time here is not bounded, you can stand at someplace say (x_0,y_2,z_0) to only observe you getting accelerated to the speed of light as your hand-watch shows t=k\pi.

AB

 

[JesseM]

In fact I don't think GR accepts such a nonsense!

Topology is irrelevant, GR only demands that the field equations are satisfied at every point on the manifold.

Because if time here is not bounded, you can stand at someplace say (x_0,y_2,z_0) to only observe you getting accelerated to the speed of light as your hand-watch shows t=k\pi.

Huh? There is no "acceleration at the speed of light", the point t=0 really is right nearby the point t=pi/2 - epsilon (where epsilon is some very small number). And of course, the choice of where to put t=0 is an arbitrary coordinate choice with no physical significance, you could easily do a coordinate transformation on the same spacetime and put it somewhere different.

Did you read the article about how an unusual topology can allow a flat spacetime to have a finite but unbounded amount of space? If we put a coordinate system on such a spacetime where at least one of the spatial coordinates "resets" in a similar way, do you think you experience an acceleration when you reach the reset point? Just think of a cylinder, if you wanted to put a coordinate system on that then one of the coordinates would have to reset, but nothing special would happen to an ant walking around the cylinder when it reached the reset point, and again it is an arbitrary coordinate choice where the coordinates reset.

 

[edgepflow]

I attempted to compute the Ricci Tensor for this Groundhog Day metric and found all terms equal zero. Is this correct?

I know it is still considered flat spacetime, but I have read that there is a strange energy momentum tensor to bend time like this.

 

[JesseM]

I attempted to compute the Ricci Tensor for this Groundhog Day metric and found all terms equal zero. Is this correct?

I know it is still considered flat spacetime, but I have read that there is a strange energy momentum tensor to bend time like this.

It should be correct, it's just flat spacetime with a weird global topology. My understanding is that the issue of the "strange energy momentum tensor" only arises when you want to have closed timelike curves in a finite region of an infinite space (with a 'Cauchy horizon' as the boundary between the region where CTCs are possible and the region where they're not), using something like a wormhole...see here:

A more fundamental objection to time travel schemes based on rotating cylinders or cosmic strings has been put forward by Stephen Hawking, who proved a theorem showing that according to general relativity it is impossible to build a time machine of a special type (a "time machine with the compactly generated Cauchy horizon") in a region where the weak energy condition is satisfied, meaning that the region contains no matter with negative energy density (exotic matter). Solutions such as Tipler's assume cylinders of infinite length, which are easier to analyze mathematically, and although Tipler suggested that a finite cylinder might produce closed timelike curves if the rotation rate were fast enough,[38] he did not prove this. But Hawking points out that because of his theorem, "it can't be done with positive energy density everywhere! I can prove that to build a finite time machine, you need negative energy."[39] This result comes from Hawking's 1992 paper on the chronology protection conjecture, where he examines "the case that the causality violations appear in a finite region of spacetime without curvature singularities" and proves that "[t]here will be a Cauchy horizon that is compactly generated and that in general contains one or more closed null geodesics which will be incomplete. One can define geometrical quantities that measure the Lorentz boost and area increase on going round these closed null geodesics. If the causality violation developed from a noncompact initial surface, the averaged weak energy condition must be violated on the Cauchy horizon."[40]

 

[PAllen]

Any reactions to my conclusion that there seem to be no extended 'everywhere timelike' world lines with this metric. That seems to make it not physically plausible.

 

[JesseM]

I see that the cause of the 4 velocity's divergent coordinate representation is that it is actually periodically lightlike. Light paths take the form (in X T plane): x=k-cos(t), so any constant position path is periodically tangent to a light path. This is physically dubious - it should not be possible for a real particle's world line to ever be tangent to null geodesic.

Why should a real particle's path be a "constant position path" though? Perhaps in these coordinates timelike paths must have changing position coordinate. Might help to figure out the coordinate transformation between these coordinates and inertial coordinates.

Maybe a naive question, but I don't really get the point in using the type of non-inertial coordinate system suggested by Mentz114 in the first place--why not just use an inertial coordinate system with a t-coordinate that can only range over a certain set of values? Then the metric would just be the regular Minkowski metric.

 

[George Jones]

First, a little more about the construction of the spacetime. Choose a particular inertial coordinate system (T, x) for 2-dimensional Minkowski spacetime. I use T instead of t because Mentzt114 used t for a completely different coordinate, hence

I think that t is a poor choice of coordinate label.

Using this inertial coordinate system, make the indentification (T, x) ~ (T + 2 , x). This turns spacetime into a cylinder with the the T coordinate being circular. S, we can restrict ourselves to the strip of spacetime where x can take on any value, and 0 \leq T &lt; 2.

Now, implicitly define a new coordinate t by T = 1 - \cos t. Consequently, 0 \leq t &lt; \pi. What is the metric in the non-inertial (t, x) coordinate system?

Light paths take the form (in X T plane): x=k-cos(t), so any constant position path is periodically tangent to a light path.

I don't think so. Certainly, if (t, x) = (t, 1 - cos(t)) and (t, 0) (observer at spatial origin) are both plotted, the curves look tangent at t = 0, but here we're dealing with parametrized curves. For parametrized curves, the notion of tangency depends not on the image of the curve (the events in spacetime on the the curve), but also on choice of parameter (how "fast" the events are traversed).

It might be thought that the components of the tangent vectors to the above two curves are (1, sin(t)) = d/dt (t, 1 - cos(t)) and (1, 0) = d/dt (t, 0), which are the same at t = 0. Certainly, (1, sin(t)) is a lightlike vector, but (1, 0) is not the tangent the tangent vector to a worldline parametrized by proper time. What is needed is

\frac{d}{d \tau} \left( t \left( \tau \right), 0 \right) = \frac{dt}{d \tau} \frac{d}{dt} \left( t \left( \tau \right), 0 \right) = \left( \frac{dt}{d \tau} , 0 \right) = \left( \frac{1}{\sin t} , 0 \right).

This is not lightlike "at" t = 0, so the observer worldline is not tangent to a light path.

 

[PAllen]

Why should a real particle's path be a "constant position path" though? Perhaps in these coordinates timelike paths must have changing position coordinate. Might help to figure out the coordinate transformation between these coordinates and inertial coordinates.

I didn't show my work, but some work I did (not complete and rigorous, though) seemed to show there could not be any extended everywhere timelike world lines at all. I wasn't assuming only constant position. Thus I was hoping those more facile with these calculations would contradict or verify my hypothesis.

 

[George Jones]

PAllen, I'm not sure if you saw them (sorry to be a nag) since we moved to a new page, but I posted some thoughts in post #20.

 

[PAllen]

I don't think so. Certainly, if (t, x) = (t, 1 - cos(t)) and (t, 0) (observer at spatial origin) are both plotted, the curves look tangent at t = 0, but here we're dealing with parametrized curves. For parametrized curves, the notion of tangency depends not on the image of the curve (the events in spacetime on the the curve), but also on choice of parameter (how "fast" the events are traversed).

It might be thought that the components of the tangent vectors to the above two curves are (1, sin(t)) = d/dt (t, 1 - cos(t)) and (1, 0) = d/dt (t, 0), which are the same at t = 0. Certainly, (1, sin(t)) is a lightlike vector, but (1, 0) is not the tangent the tangent vector to a worldline parametrized by proper time. What is needed is

\frac{d}{d \tau} \left( t \left( \tau \right), 0 \right) = \frac{dt}{d \tau} \frac{d}{dt} \left( t \left( \tau \right), 0 \right) = \left( \frac{dt}{d \tau} , 0 \right) = \left( \frac{1}{\sin t} , 0 \right).

This is not lightlike "at" t = 0, so the observer worldline is not tangent to a light path.

I'm still not convinced. Obviously(?) d/d tau is undefined along a light path. Your (1/sin(t),0) is undefined at precisely the points where I think the world line is lightlike. This seems consistent with my claim.

 

[George Jones]

Sorry, I have to run for my bus, and I don't know if I'll have a chance to get to this tonight or tomorrow. Try transforming the curves (T, x) = (T, 0) and (T, x) = (T, T) to t-x coordinates. Also transform the tangent vectors.

 

[Mentz114]

Maybe a naive question, but I don't really get the point in using the type of non-inertial coordinate system suggested by Mentz114 in the first place--why not just use an inertial coordinate system with a t-coordinate that can only range over a certain set of values? Then the metric would just be the regular Minkowski metric.

I never intended that metric to be taken seriously - it's rubbish.
I'm sure a much better 'ground-hog' metric can be found.

 

[K^2]

ds² = -sin(t)²dt² + dx² + dy² + dz²

Solving the equations of motion with the initial condition t=0 \rightarrow \tau=0 gives

<br /> \tau=1-\cos(t)<br />

Except that what you want is the t(\tau), which does diverge. It does end up periodic, though, so yeah, I made a mistake. But I still don't like the divergence where the sign flips.

 

[edgepflow]

It should be correct, it's just flat spacetime with a weird global topology. My understanding is that the issue of the "strange energy momentum tensor" only arises when you want to have closed timelike curves in a finite region of an infinite space (with a 'Cauchy horizon' as the boundary between the region where CTCs are possible and the region where they're not), using something like a wormhole...see here:

Yes, I read that the energy momentum tensor would need exotic matter (maybe an unphysical tensor).

I wanted to try to use my pea brain and calculate this energy momentum tensor with exotic matter. Anyone have any suggestions how to do this?

 

[JesseM]

Yes, I read that the energy momentum tensor would need exotic matter (maybe an unphysical tensor).

I wanted to try to use my pea brain and calculate this energy momentum tensor with exotic matter. Anyone have any suggestions how to do this?

In what spacetime? In Minkowski spacetime (even one with a weird topology) there is no exotic matter, since if there was it would make the curvature nonzero.

 

[Altabeh]

Topology is irrelevant, GR only demands that the field equations are satisfied at every point on the manifold.

Topology is irrelevant that allows you to have Deutsch-Politzer spacetime in GR with CTCs. The fact is that here the choice of coordinates is messy! By "messy" I mean if det g = 0 at some point(s) in spacetime, then either you have chosen a bad coordinate system, or you have to bond coordinates in such a way that spacetime gets bounded to avoid having singularities!

Huh? There is no "acceleration at the speed of light", the point t=0 really is right nearby the point t=pi/2 - epsilon (where epsilon is some very small number). And of course, the choice of where to put t=0 is an arbitrary coordinate choice with no physical significance, you could easily do a coordinate transformation on the same spacetime and put it somewhere different.

You better read carefully! You're at rest and suddenly you're accelerated to the speed of light, so as to have ds^2=0 hold. I guess there is no misunderstanding on my side in this case!

Did you read the article about how an unusual topology can allow a flat spacetime to have a finite but unbounded amount of space? If we put a coordinate system on such a spacetime where at least one of the spatial coordinates "resets" in a similar way, do you think you experience an acceleration when you reach the reset point? Just think of a cylinder, if you wanted to put a coordinate system on that then one of the coordinates would have to reset, but nothing special would happen to an ant walking around the cylinder when it reached the reset point, and again it is an arbitrary coordinate choice where the coordinates reset.

Oh wait a sec! This is a very poor definition of the Deutsch-Politzer spacetime where time resets after you walk around the cylinder-like spacetime. Yet, this doesn't solve anything about the problem we have here! The problem is that staying at rest (i.e. spatial coordinates do not change) in this spacetime is equivalent to starting to move at the speed of light whenever the observer's clock ticks any multiple of pi seconds! Even starting at t=0 is nonsense because there you'll have either a speed greater-than the speed of light or the same feeling that a photon has when traveling at speed c! Indeed, the null geodesic generators of Deutsch-Politzer spacetime are (not all of them) those discussed here! Most important thing here is that this is not even because of a poor choice of metric, but irremovablity of its singularities!

See, for example, http://arxiv.org/abs/gr-qc/9803020v1

AB

 

[edgepflow]

In what spacetime? In Minkowski spacetime (even one with a weird topology) there is no exotic matter, since if there was it would make the curvature nonzero.

I must admit I am an engineering graduate trying to teach myself some GR. So when you ask "what spacetime" I would ask: what are my choices? Are there other non-Minkowski metrics that describe the time dimension folded into a cylinder I could use to find a stress-momentum tensor with exotic matter?