[QUOTE=Chronos]
Assuming that special relativity is correct, you could, in fact, go into the past.
The short version of the explanation relies on the fact that the concept of “simultaneity” does not exist in relativity; or rather, it depends on which reference frame one chooses (all of which are equally valid). As an example, suppose we have two stars which are dying, and which are at rest relative to each other. Further suppose that, to an observer at rest relative to the two stars, they both go supernova at the same time. If that’s the case, then to an observer flying from star A to star B, star B exploded before star A. Meanwhile, to an observer flying in the other direction, star A exploded before star B. If you take an observer, and that observer measures the positions and times of both supernovae, he can calculate the value u = (x[sub]B[/sub] - x[sub]A[/sub]) / (t[sub]B[/sub] - t[sub]A[/sub]). You’ll notice that this value has units of a speed, and if you do the math, this speed will always be greater than the speed of light (if you’re in the frame at rest relative to the two stars, it’ll be infinite, since in that frame, t[sub]B[/sub] = t[sub]A[/sub]). Note, by the way, that this isn’t actually the speed of anything; it’s just the result of a calculation. But even though this speed u is always greater than c, it can be as close to c as you like. As your observer’s speed relative to the stars approaches c from below, this calculated speed approaches c from above. In fact, it turns out that u = c[sup]2[/sup]/v , where v is the speed you’re going relative to the stars.
So, now suppose you’ve developed a sufficiently advanced technology which lets you travel at an effective speed of (say) 1.002c. It doesn’t matter how this technology works, whether it be a wormhole, a warp drive, or a magic carpet: All that matters is that it can get you from point A to point B, and arrive there before a photon travelling through vacuum could get there. Let’s now bring this FTL device on board an ordinary rocket. Since all reference frames are equally valid, we can, if we want, start off in a reference frame travelling at .999c, in the direction from star A to star B (we can get into this reference frame using our non-warp drive rocket engine, if we’re not in it already). So we fly at this velocity right past star A, and we time our trip so that, just at the moment we pass the star, it goes supernova. We’ll call this position t = 0, x = 0 for simplicity. We know, then, that after a time t = t[sub]B[/sub], and at a distance x = x[sub]B[/sub] from us, star B explodes, and from our previous calculation, we know that u = x[sub]B[/sub] / t[sub]B[/sub] = 1.001c.
Well, now suppose that right at (0,0), as we’re passing the supernova of A, we aim our warp drive at B, and turn it on. Our warp drive has an effective speed of 1.002c, which is faster than u. So our warp drive actually arrives at star B a little bit before B explodes.
But wait a moment, here: We already said that, in the reference frame where the stars are at rest, both supernovae happen at the same time. The problem is completely symmetrical. So we can have a traveller going the opposite direction at high speed, who catches our warp drive at star B, and sends it back to us as soon as they catch it. By the same reasoning above, if it leaves star B a little before B explodes, then it’ll come back to A before A explodes.
But the last time the warp drive was at star A, it was right at the moment of explosion. Now here it is, having travelled some ways, and returning to A before A explodes. In short, it’s gone back in time.
[/QUOTE]
Man, what if, like, “C-A-T” spelled “dog” ?
[sub]That’s the only line from that movie I remember[/sub]
