The faster-than-light item

Oct 5 16:29 UTC 1995

    Every sci-fi conference needs one.


    Thanks to e-mail, and the PostScript printers at work, I now have the
files miguel94a.ps and fig1.ps in readable form. It's an interesting treatment


of warp theory, but begs some questions and comments:


>     ...Two stars A and B are separated by a distance D in flat
> spacetime. At time t0, a spaceship starts to move away from A at
> a speed v<1 using its rocket engines. The spaceship then stops
> at a distance d away from A.

    Why? Wouldn't it be easier to keep moving at v when activating the warp
drive? It is not stated what that v is relative *to*, though we can presume
that the frame of reference is that in which A and B are stationary. It
should also be noted that stars are not always stationary relative to one
another.

> (3)The two constant-velocity legs at the beginning and end of
> the journey are not crucial for the argument that I wish to
> present here. I only introduce them to guarantee that the two
> stars will remain unaffected by the disturbance of spacetime
> (R<<d), and can therefore be used as unperturbed "clocks" with
> which to compare the proper time on board the spaceship.

    Okay, that answers that question I guess. However, a shuttle coasting
along a vector parallel to B ---> A at half lightspeed can just as well be
used for comparison. At a sufficient distance from the disturbance path,
the crew of this shuttle is going to observe that the arrival at B occurs
*before* the departure from A. This is taking into account the time needed
for information to get from A and B to the shuttle. Again, it has to do with
the constancy of c.
    The problem should be worked out from the points of view in which stars
A and B are moving along the vectors A ---> B and B ---> A; it is very
probable that there will exist values of a for which the trip duration is
negative.


However., in answer to:

> This is the basis of the model for hyper-fast travel that I wish
> to present here: create a local distortion of spacetime that will
> roduce an expansion behind the spaceship and an opposite contraction
> ahead of it. In this way, the spaceship will be pushed away from the
> Earth and pulled towards a distant star by spacetime itself.

    The catch here is that the information that spacetime is to be distorted
in this manner must also travel at lightspeed. Thus, it should still take
a global time of D/c to create the space warp. Once created, however, it
could be possible to send spacecraft through it at an arbitrarily fast rate.
This would be somewhat like building a freeway between two distant cities
separated by rough terrain. It may take years to build the road, and travelers


going the hard way might make several round trips during that time. But once
the road is finished, its length may be traveled in a few hours.

    This is what I have in mind for a FTL travel scheme that deals with
causality, and how I think the jump gates in _Babylon 5_ should have been
handled.

response 2 of 7:

Oct 13 21:19 UTC 1995

How does the strange matter know which end of the ship is the front and which
end is the rear? Also, how does it know how far to extend this effect?

Gravity waves, which are caused by changes in the gravitational field due to
the movement of large masses, would still travel at lightspeed. Or else it
would have to require years to assemble enough strange matter for a space
warping trip - the time required being proportional to the amount of strange
matter needed, which would be proportional to the distance to be traveled.

An observer who is far away from your path, and who is moving at any
significant speed, can still observe the ship arrive before it departs.
(Footnote: by observation, it is assumed that speed-of-light lag is being
taken into account when determining which event occurs "first", but that the
observer is doing his calculations under the assumption that he is stationary
and that the rest of the universe is moving.)

If you say that the space expansion/compression travels instantaniously, you
must answer, instantanious in what reference frame? There is no such thing
as "instantanious in all reference frames"; observers at different velocities
are going to observe different intervals between arrival and departure, both
positive and negative. This is a mathematical consequence of lightspeed being
constant. In which reference frame is this expansion/compression
"instantanious", and why?

If the distortion can be sent instantaniously in any reference frame desired,
then time travel becomes possible. The basic procedure is:

1. Space-warp from A to B in the reference frame of a sublight observer
   moving from A toward B;

2. Space-warp from B to A in the reference frame of a sublight observer
   moving from B toward A.

Doing the math in determining the travel time will show that you end up back
at A *before* the departure time.

response 4 of 7:

Oct 17 04:51 UTC 1995

The basic problem with most science fiction FTL schemes *is* that you can pick
numbers that allow negative travel times, and create what are called closed
timelike loops. The real challenge to science fiction authors is to come up
with a set of rules for FTL drive performance that (a) stay as much as
possible within known physics, and (b) disallow closed timelike loops - that
is, the arrival in ones own past. (It is okay to arrive at planet X at some
theoretical negative delta-t, so long as a return trip is compelled to arrive
*after* departure.)

The space expansion-compression set of rules fails the causality challenge.

response 6 of 7:

Oct 19 16:33 UTC 1995

    Okay, time to put up, I guess.


    For this exercise, we will assume that a ship travels
between two planets, A and B, which are at rest relative to
each other and a distance L apart, at an arbitrary FTL speed
w. The transit time in the AB reference frame, of course, will
be L/w.


    0                           x                  L
    |                           |                  |
   (_) A.                      |::> C.            (_) B.


Now we will have the FTL ship transmit a laser signal toward
B as he is entering jump space, and then transmit another
laser signal toward A upon re-emergence into normal space. The
two beams will meet at some point C, which is a distance x
from A and L-x from B. The beams reach point C at time

     t = x/c = L/w + (L-x)/c.

from which we find that  x = L/2 ( c/w + 1 ). (At instantanious
speed, c/w would approach zero and point C would be the
midpoint.) An observer at point C who is at rest relative to
A and B would detect both beams at the same time, and would
conclude that the jump took time L/w.

    Now consider an observer who is moving at some arbitrary
sublight speed v toward B. He, too, encounters both beams at
once, and predictably, measures their speeds to be the
constant c. At speed v, the universe is shrunken by a factor
of b = sqrt( 1 - (v/c)^2 ), and x becomes  b x  while L-x
becomes  b (L-x).
    Observer C must conclude that planet A must be distance
b x  behind him, and planet B is distance  b (L-x)  ahead, in
HIS OWN reference frame when the beams meet. Also note that
planets A and B are moving toward his rear at speed v in his
reference frame.

    Now let's work in C's reference frame and trace back
signal A. It traveled a distance  b x  from where A is "now",
minus a certain distance from this "now" point to where it
was when signal A was transmitted, and this took time

   t[A]  =  b x / c  -  dt[A].

While signal A was travelling, planets A and B were moving
back through a distance  v t[A]. So if planet A is distance
b x  "now", it was at  b x  -  v t[A] when signal A was sent,
the travel time being t[A] and the speed being, of course, c.

                b x  -  v t[A]
         c  =  ----------------
                    t[A]       .

From this, we conclude that t[A] = b x / (c+v). Likewise, planet
B, being b (L-x) away "now", was at b (L-x) + v t[B] when
signal B was sent.

               b (L-x)  +  v t[B]
        c  =  --------------------
                     t[B]        ,

leading to  t[B] = b (L-x) / (c-v). The transit time of FTL
ship W is the difference between the signal transit times:

                        b x           b (L-x)
  t  =  t[A] - t[B]  = ------   -    ---------
                       (c+v)           (c-v).

Plugging in the formula for x and reducing yields

               / c (c/w + 1)           1     \
   t  =  b L  {   ---------    -   --------   }
               \  c^2 - v^2         c - v    /.


    Time travel occurs at the reference-frame speed at which
t=0. There are two solutions for t=0: the speed of light itself
(for b=0), and the speed that satisfies the equation

        c (c/w + 1)            1
       -------------    =   -------
        c^2 - v^2            c - v,

or v = c^2/w. For a cargo ship in an episode of ST:TNG, w is
"Warp Three", or 39 times the speed of light. For such a
ship to experience backward time travel, it must change the
reference frame of its jump by a speed of
7687 kilometers per second. (Sorry, I am unclear as to what
"Warp Eight" means, so cannot calculate the critical reference
frame speed for the _Enterprise_.)
    This speed of 7687 km/sec would be achieved by accelerating
at 1 gravity for just over 9 days. The mass ratio for this
speed for total conversion (which the _Enterprise_, at least,
has) is 1.026. For catalytic C-N-O cycle hydrogen fusion, the
mass ratio would be in the area of 1.25.

    On conclusion: It would take some expense, but a private
individual or group with such a cargo ship *could* strip out
its cargo hold, fill it with hydrogen or deuterium, strap on
a fusion thruster, and accelerate to the normal-space speed
required to go a time-travelin' --- AND slow down once it's
there. A businessman might find it unprofitable, but a terrorist
group would certainly, pardon the pun, jump, at the chance.


    The details of the faster-than-light drive don't have much of
an effect on the basic problem: *someone* is going to observe a
negative travel time. And since no one is special, it might as
well be the people at the point of departure.