Natural Science Forum / Physics / Research / February 2005

Suppose we have a spacetime that has a wormhole, and that spacetime is
(nearly) flat outside of the wormhole. Suppose further that the throat
of the wormhole is infinitesimally short.

The wormhole is a time machine, i.e., spacetime has a closed timelike
curve (CTC), if Han Solo can do the following: jump into mouth 2, move
through the wormhole's throat, emerge mouth 1, move from mouth 1 to
mouth 2 through spacetime external to the thoat, and bump into himself
as he jumps into mouth 2.

Call jumping into 2 event C' and emerging from 1 event C. The above is
possible if event C' is in the future of C in the the external
spacetime. In other words, there has to be a (future-directed) timelike
worldline joining C to C' for Han to traverse. The CTC C'CC' consists of
a segment from C' to C through the throat of the wormhole together with
a timelike segment from C to C' external to the wormhole.

Because the throat of the wormhole is short (and thin), Every event on
the worldline of mouth 1 is identified with a corresponding event on the
worldline of mouth 2. A and A' are one such pair of identified points.

Suppose the mouths of the wormhole are initially close together and at
rest with respect to a particular inertial reference frame for the
external flat spacetime. Now let the mouths play the roles of the twins
in the twin paradox, i.e., mouth 1 stays at rest with respect to the
original reference frame, while mouth 2 moves out and back.

Which pairs of events are identified in this scenario? At the instant
that the mouths start to move apart, place a (zeroed) clock in mouth 1.
The clock is also in mouth 2, again because of the infinitesimal throat
length. Therefore, the clock measures proper time for both "twins", and
an event A on the worldline of mouth 1 is identified with an event A' on
the worldline of mouth 2 iff the proper time elapsed for event A
according to mouth 1 equals the proper time elapsed for A' according to
mouth 2.

The following spacetime diagram shows the worldlines of the 2 mouths and
3 pairs of identified events. If Han goes into the wormhole at one event
in the pair, he comes out of the wormhole at the other event of the
pair. As explained above, there has to be a timelike relationship
between events in a pair in order to have a CTC.

     t

     |
     |\
     | \
     |  \
     |   \
     |    \C'
     |     \
     |      \
     |       \
    C|        \
     |         \B'
     |         /
    B|        /
     |       /
     |      /
     |     /A'
     |    /
    A|   /
     |  /
     | /
     |/
    O----------------------------- x

The proper time of any event on the worldline of mouth 1 is just the t
coordinate of that event. The proper time tau of any event on the
worldline of mouth 2 is related to the t coordinate of that event by

tau = t*sqrt(1 - v^2).                           (1)

Consider the A and A' events. The identification condition is

t_A = tau_A'.                                    (2)

To find the causal relationship between A and A', consider

(x_A' - x_A)/(t_A' - t_A)

 = v*t_A'/(t_A' - t_A'*sqrt(1 - v^2))    (from (1) and (2))

 = v/(1 - sqrt(1 - v^2)).                      (3)

Note:
1) (3) is independent of A and A' , and so is valid for any part of the
   outgoing part of mouth 2's worldline;

2) (3) > 1, so the relationship between A and A' is spacelike, and thus
   is part of any worldline for Han.

Like you said, no time machine forms as mouth 2 moves away from mouth 1.

Consider C and C'.

x_C' = x_B' - v*(t_C' - t_B')

     = 2*x_B' - v*t_C'

gives

t_C' = (2*x_B' - x_C')/v                          (4)

Suppose a time machine forms at C', i.e., C and C' lightlike related.
Then,

1 = (x_C' - x_C)/(t_C' - t_C)

  = x_C'/(t_C' - t_C'*sqrt(1 - v^2))

  = v*x_C'/[ (2*x_B' - x_C')*(1 - sqrt(1 - v^2))].  (5)

After playing around with (5), I get (maybe incorrectly)

x_C' = x_B'*[1 - sqrt((1 - v)/(1 + v))].            (6)

In your example, v = 7/10 and x_B' = lightyears.

I get x_C' = 58 lightyears. A time machine forms when mouth 2 get within
58 lightyears of mouth 1 on the return trip.

I haven't checked the above carefully, so there may be mistakes.

Note the Doppler shift factor in (6). In terms of of twins, a time
machine forms when twin 2 sees via light, maybe  through a telescope,
the time on twin 1's watch to be the same as the time on his watch.

Regards,
George