Ok, folks, sorry this took me so long, but I’m just finishing up finals week here. WARNING! Physics follows
Ok, the first thing we need to address is the concept of simultaneity. Basically, there ain’t none such. In relativity, if in one reference frame two events are simultaneous, then in another frame moving relative to the first, they won’t be.
Consider three observers, Alice, Bob, and Charlie. Charlie is moving to the left relative to Bob with a speed v, and Alice is moving to the right with a speed v, again with respect to (WRT) Bob. They all pass each other at the same time, and agree that they’ll call that time t=0, and that position x=0. An important note here is that each observer considers himself to stationary, and the other two to be moving-- so, for instance, Alice says that her position is always x=0, and likewise for Bob and Charlie.
Now, somewhere, somewhen, two flashbulbs go off. Both bulbs are stationary with respect to Bob, and according to him, the first flash is a distance x=X_b to the left of him, and the second is the same distance right of him. Let’s also assume that the light from both bulbs reaches Bob at the same time, which happens to be t=0. Since the speed of light is constant, Bob concludes that T_1b = T_2b = -X_b/c , so in Bob’s reference frame, the two flashes were simultaneous.
Now let’s look at the same situation from another frame of reference, say, Charlie’s. Since Charlie is at the same position as Bob at t=o, he also sees both flashes at the same time. However, since Charlie is moving left at v WRT Bob, that means that Bob and the bulbs are moving to the right at v WRT Charlie, so that the flash 1 was farther away when it went off than flash 2. Now, since he sees them at the same time, he interprets this to mean that flash 1 went off at an earler time than flash 2-- a correct interpretation, in his reference frame. Specifically, (X_2a - X_1c)/(T_2a - T_1c) = c²/v .
The same reasoning can be applied to Alice, except for her, flash 2 went off before flash 1, and (X_2a - X_1a)/(T_1a - T_2a) = c²/v .
In other words, one observer sees one event as happening first, one observer sees theother as happening first, and another sees them as happening at the same time. Now we’re ready to tackle the original question, that of FTL (Faster than light) travel. Here, the two events we’re interested in are the ship leaving one point, and it arriving at another. It doesn’t matter what happens in between, it could be a hyperjump, or a wormhole, or a magic carpet, or anything else you feel comfortable calling it-- The point is that the ship is at one place at one time, and another place at another time. Even though the ship might not be actually “moving”, let’s define an effective velocity: V = (X2 - X1)/(T2 - T1) . Now, since we’re talking about an FTL ship here, V > c .
All right, let’s define some more reference frames again. Alice is on the Earth, Bob is moving left at speed v = c²/V with respect to Alice, and Charlie is moving left at c²/V with respect to Bob. Similarly, we’ve got Diane moving right at c²/V relative to Alice, and Edgar moving at the same speed relative to Diane. A few points to note here: First, since we said that V > c, c²/V < c , so all of our five observers are going at ordinary speeds below the speed of light. Secondly, relativistic velocities don’t add like normal velocities, so we can’t just say that Edgar is moving at a speed 4c²/V relative to Charlie. In actuality, the relative speed is less than that, and guaranteed to be less than the speed of light. The way we’re going to approach this, though, we don’t need the exact formula for velocity addition, so it’s just something to keep in mind.
Ok, now, ellis555 gets into his nifty keen-o Ellismobile on Earth, and kicks on the (normal, sub-light) engines to match velocity with Charlie. Having done so, he then activates the flux capacitors, or tells his helmsman “Engage”, or whatever it is he does to travel FTL, and jumps to the right at an effective speed V, from Charlie’s point of view. Now, this means that for Charlie, (X_2c - X_1c)/(T_2c - T_1c) = V = c²/v . Now, remember our discussion of simultanaity above? When, according to Charlie, event 2 happens after event 1, subject to that formula, they’ll be simultaneous according to Bob, and according to Alice, event 2 happened before event 1.
Now, ellis555 turns around and matches velocity with Edgar, and jumps the same distance (according to him) to the left. The same thing happens: Edgar claims that the ship moved at FTL speed, Diane claims that it jumped instantaneously from one point to the other, since its departure and arrival were simultaneous, and Alice claims that it went back in time again, and all three claims are correct, for their respective frames of reference. If ellis555 then kicks in the normal engines again and brings himself to a stop relative to the Earth, then the net result is that according to Alice on the Earth, he departed a given place at a given time, and arrived at that same place, at an earlier time. Time travel, in other words.
Q.E.D.