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*To*: starship-design@lists.uoregon.edu*Subject*: Re: starship-design: The speed of now*From*: Zenon Kulpa <zkulpa@zmit1.ippt.gov.pl>*Date*: Wed, 27 Aug 1997 18:34:56 +0200 (MET DST)*Cc*: zkulpa@zmit1.ippt.gov.pl*Reply-To*: Zenon Kulpa <zkulpa@zmit1.ippt.gov.pl>*Sender*: owner-starship-design

> From: wharton@physics.ucla.edu (Ken Wharton) > [...] > That value is D = [t ^ 2 - x ^ 2]. The distance between any two > events, measured in terms of D (where t is the time-separation in > seconds and x is the distance separation in light-seconds) will be > identical in all frames. > As far as I remember it should be D = x^2 - t^2 (or rather: D = x^2 + (it)^2, where i = Sqrt(-1)). Or am I wrong? > For light, D=0. Thus my assertion that light "really" travels > infinitely fast. > > For STL, D>0. The time is always larger than the distance. > Rather (using my formula for D) - distance larger than time. It would be intuitively natural: you never have enough time to be in time... ;-) > But for FTL, D<0. You also get D<0 for local time travel; set x=0 (you > don't go anywhere) and set t<0 (you go back in time). > t^2 >= 0 always, also when t < 0. In my version of the formula for D, D < 0 always when x = 0 and t =/= 0. Something is still wrong... > Therefore the D > associated with time travel is equivalent with the D associated with > FTL. It's very simple to take two FTL journeys, one out and one back, > that gets you to return before you left. > > Just another way of thinking about it, I guess. Which, of course, was > probably more confusing than illuminating... > This interpretation wouild be quite illuminating for me, except for the problem with D < 0 for x = 0... -- Zenon

**Follow-Ups**:**Re: starship-design: The speed of now***From:*Steve VanDevender <stevev@efn.org>

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