The classical example of the conversion of a wormhole into a time machine is to consider two wormholes, one stationary and another one accelerating for some period before going back to its original position. From the point of view of an observer it is fairly simple to see : for a thin shell wormhole in flat space, take two observers arbitrarily close to the boundary, one is accelerating and one isn't, hence the proper time of the accelerated observer will be shorter despite being just next to another observer at a different time.

From the point of view of purely the metric and topology though, I'm having some trouble showing it (this question is probably somewhat related to this one).

Consider the spacetime $(\mathbb{R}^2 \# T^2) \times \mathbb R$, with the usual thin-shell construction : remove the disks $D_1$ and $D_2$, such that at a time $t$, we have the identification of $D_1(t)$ with $D_2(t + \Delta t)$, with initially $D_1(0)$ identified with $D_2(0)$. For every slice of constant $t$, we have

\begin{eqnarray} D_1 &=& \{ (x,y) | (x - x_a)^2 + (y - y_a)^2 \leq R^2 \}\\ D_2 &=& \{ (x,y) | (x - x_b(t))^2 + (y - y_b(t))^2 \leq R^2 \}\\ \end{eqnarray}

With $\gamma(t)$ some accelerated timelike curve (between $t_1$ and $t_2$) in the original $\Bbb R^2$ , which would then have a proper time inferior to the difference between $t_1$ and $t_2$, and we have $\gamma$ static outside of that acceleration phase.

How to show that, given a long enough acceleration period, this induces closed timelike curves without resorting to handwavy argument about the time shift between observers on both sides? I'm having trouble, possibly because there isn't really a good coordinate patch for an observer crossing the wormhole on which to define the metric all throughout the trip, and everything else is flat outside of this portion.


Your Answer

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

Browse other questions tagged or ask your own question.