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*To*: starship-design@lists.uoregon.edu*Subject*: starship-design: Fun With SpaceTime*From*: wharton@physics.ucla.edu (Ken Wharton)*Date*: Tue, 24 Jun 1997 12:37:58 -0700*Reply-To*: wharton@physics.ucla.edu (Ken Wharton)*Sender*: owner-starship-design

Lee writes: >On another vein, some of you already know I am an engineer not a physicis >t and I am having trouble with this violation of causality thing regardin >g FTL. Besides the fact that it just doesn't make any sense, consider thi >s example: > >Example: Let's go from Earth to Alpha Centauri with a young lady named Br >ight (...who's speed is much faster than light. She set out one day. In a > relative way. And returned on the previous night.). > >Distance is 1.3 parsecs, which is 1.3 * 3.26 =3D 4.3 light years. > >Bright travels 500 times light speed. So Bright will take 4.3 / 500 = >.0086 years to travel to Alpha Centauri. The problem here is that this is ill-posed -- all velocity is relative. And that gets you into trouble when you are going faster than light. You can't say how FAST she travels; that depends on your reference frame. The only objective facts are 1) the departure point in space-time, and 2) the arrival point (Alpha Centauri) in space-time. Let's assume Lee means that she travels 500 times c in Earth's reference frame (and let's say Alpha Centauri has the same reference frame as Earth, for sake of argument...). Let's also keep it a one-dimension problem. All the units are in years or light-years. So - she blasts off from Earth at x=0, t=0. She arrives at Alpha Centauri at x=4.3 (light-years), t = 4.3/500 = 0.0086 (years). Using the notation (x, t) the departure is (0, 0) and the arrival is (4.3, 0.0086) -- In Earth's Reference Frame. The actual events are not observer-dependant, but the coordinates are. Now, to answer this question: >Okay, everyone with me so far? Here is the tricky part as I see it: > >She leaves Earth's time frame and travels for 3 days, 3 hours and 20 minu >tes later at Alpha Centauri. So why isn't it just 3 days, 3 hours and 20 >minutes later on Alpha Centauri? Not only did she not time travel, she do >esn't get to Alphan Centauri in time to send any hypothetical message bac >k to Earth that will arrive before she left - it arrives 4.3 years, 3 day >s, 3 hours and 20 minutes later! Anyone care to explain? Feel free to use > math, even engineers know that much <G>. The key issue here is that these two points in space-time -- (0, 0) and (4.3, 0.0087) -- are "space-like points": Delta x (0.43) is greater than Delta t (0.0087). This means that there is no other reference frame where the two points in space-time switch in spatial orientation; no matter where you view this from, the arrival will always be further in the x-direction than the departure. But these two points are NOT "time-like points" (Delta t is not greater than Delta x), which means that there ARE some reference frames where these two points switch in temporal orientation. For some observers, the arrival happens BEFORE the departure. There is a "magical" reference frame -- one travelling from Earth to Alpha Centauri at a sub-light speed (only 1/500th of the speed of light) that sees the two events (arrival and departure) happen at EXACTLY the same time. The perceived velocity is infinite. For a reference frame travelling faster than this (say 1/400th of the speed of light) the travel appears to go in the other direction. Now, despite all the other problems that come along with this (infinite energy, which way is she really going, etc.) This can lead to a causality problem if you have a young man named Dwight travelling back to Earth. If you can travel backwards in time in one reference frame, given the priniciple that all reference frames are equivalent, you can also travel backwards in time in Earth's reference frame. This is how it could work: Dwight can pick up a message from Bright at Alpha Centauri, leisurely get into his FTL ship, and accelerate it AWAY Earth so it's travelling, say 1/10th the speed of light. Now, from his perspective (in his reference frame) Bright's departure from Earth hasn't even happened yet, although her arrival at Alpha Centauri has. From his reference frame (signified by primed coordinates), the (x',t') of Bright's departure is now at (-3.87, 0.38) [the math here is a simple Lorentz Transformation into the new reference frame, available in any relativity book -- his (x', t') location is defined to be (0, 0), equivalent to the (x=4.3, t=0.0087) of Bright's arrival] All he's doing is going into a reference frame where Bright's FTL trip seems to go in the opposite direction. Now, given that he can do all this in less than 0.38 years, and not travel too far in the wrong direction, at this point he can turn the ship around and makes the same FTL jump that Bright did -- travelling 500c from his new reference frame toward Earth. He will arrive back at Earth (in the primed coordinate system) at (x'=-3.87, t'=3.87/500 = 0.008). But Bright doesn't leave until (-3.87, 0.38)! So he gets there before Bright left in the first place, hands her the letter, at which point she decides not to go to Alpha Centauri after all. Nice paradox. If you believe that nature will somehow find a stable variant to this paradox then you can't believe in free will. I'd rather believe that this paradox can't happen in the first place. But who knows... Ken Wharton

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