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*To*: starship-design@lists.uoregon.edu*Subject*: starship-design: More Fun with Spacetime*From*: wharton@physics.ucla.edu (Ken Wharton)*Date*: Tue, 24 Jun 1997 17:48:25 -0700*Reply-To*: wharton@physics.ucla.edu (Ken Wharton)*Sender*: owner-starship-design

Lee, Quick caveat: I made a slight typo (or one that I noticed, anyway). When Dwight zooms off in the wrong direction, he needs a GAMMA [(1-v^2/c^2)^(- 0.5)] = 1.1, not 1/10th light speed as I said before. I think this turns out to be 0.4c. Anyway... >Okay here is where you lost me. WHY did you change frames of reference? I needed to prove that Adding a normal sub-light vector (Dwight's wrong-way trip) with two identical FTL vectors (Bright and Dwight's FTL trip) can wind you up with a net spatial movement of zero, and a net time movement back in time. This can also be done without the extra sub-light vector, but then the two FTL vectors will be different, and the one that takes you "back in time" will of course seem counter-intuitive. If I tried to go that route, you would say that only one "type" of FTL travel was possible, when in fact the two journeys were identical from different frames. In order to spell this out more clearly, I only used one "type" of FTL journey -- 500c from the current frame of the ship. In both of the FTL legs, from Dwight's perspective, the time at Earth advances in a positive direction. This way you can't say that the FTL is taking Dwight back in time and disallow it on those grounds alone. The back-in-time leg was the sub-light journey that Dwight starts out with! His journey away from Earth at 0.4c is the only vector in which time seems to "go backwards" at Earth. Now, this "back-in-time" behavior will happen with or without FTL. This happens in the twin paradox when one twin turns around; the stationary twin "sees" the other twin travel backwards in time. But this is not a true paradox, because by the time they catch up to each other, time has still advanced for both of them; you can never get a net-negative time without FTL. But combined with FTL, this strange but well-established aspect of time -- where a far away event can seem to go backwards in time as you accelerate away from it -- can turn itself into a paradox. With FTL you CAN create a net-negative time travel vector. >It seems obvious from normal physics that if you have a vector A>B with = >time t1 and a vector B>C with time t2 the sum of the vectors is A>C = >where T =3D t1 + t2. Now if you turn around (reverse vectors) the sum of = >the vectors is A>C + C>A =3D 0 for x, y and z and T =3D 2(t1 + t2) if = >you maintain ANY constant frame of reference. I'm afraid this doesn't quite apply with FTL time-vectors because the only way to measure the magnitude of the vector is to find something that is frame-independant. You can't say A>B when in some other frame B>A. The frame-independant number here is (t^2 - x^2). Because (for FTL travel) x>t, this "vector" is negative. The sub-light vectors are all positive. I'm combining two "negative" vectors and one "positive" vector to get a net negative vector; delta t<0, delta x=0. This is true for ALL frames. >Every description of a causality paradox I have ever read seems to have = >included a change of frame of reference for no logical reason. It seems = >to me that the mistake is in the change of frames! Any set of events must be explainable in ANY set of frames. That's the fundamental principle of relativity. If the same thing looks different in two different frames then there's a logical error involved. >If you look at this my way and maintain an Earth frame of reference: > >Bright leaves at t=3D0y and arrives at t=3D0.0082y, Dwight leaves at = >t=3D0.0082y travels on same vector for some arbitrary time, let's pick = >0.0003y, then reverses his vector and travels back to Earth. He will = >arrive at Earth at t=3D(0.0082+0.0003)+(0.0082+0.0003). All this assumes = >an Earth observer's frame of reference which is the ONLY ONE THAT = >MATTERS. > >So where did I goof? I still don't get it... We're talking apples and oranges here. The FTL trip that Dwight takes in your example ( let's call it 'L') is not the same one he takes in my example (let's call it 'K'). From Earth's perspective, Dwight's K trip takes him backwards in time. From Earth's perspective, Dwight's L trip doesn't. Therefore, these are different trips we're describing. Now,you may be thinking, ah hah - if you stay on Earth, 'K' takes Dwight backwards in time! Therefore his FTL trip must not be allowable. I agree with you perfectly. But as I pointed out, in Dwight's frame of reference, HIS FTL trip is IDENTICAL to Bright's FTL trip. Therefore, if you're going to disallow HIS trip, you also have to disallow Bright's trip, and therefore disallow FTL. This, I think, is the correct conclusion to draw. Although certain FTL journeys may seem fine in some frames, in others it will inevitably take you back in time. If you want to stay in the same frame the whole time, it's possible to think you're sheltering yourself from paradoxes by not allowing any backwards-in-time travel in Your frame. But that's a frame-dependant physics you're describing, where the rules in your frame (no backwards-in-time travel) don't apply to any other frame. In other words, that's saying that there IS an ether, a preferred frame, in which you can avoid paradoxes. That's saying you can make paradoxes in other frames, but that's okay, because you can't ever translate it to a paradox on Earth. But this doesn't solve the problem: If you have a colony in a spaceship, travelling away from Earth at a sub-light velocity, what happens when a Paradox happens to them? Once you allow FTL, the philosophical dilemmas just keep on coming... Hope that was more illuminating than confusing! Ken

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