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starship-design: Re: Fun With SpaceTime



Ken explains to Lee:

>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.

Do I understand correctly:
Is this because the light takes a finite time to reach the observer?
And if you would reckon with the time it took for the light to reach the
observer, would you be able to configure the right order of the events?

If so, I don't understand the fuzz about it.
If not, I don't understand at all (and can you explain in a different way?).

>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.  

You don't really travel back in time, some observers just think they see you
travel back in time. But if they would do their homework (reverse calculate
while including distances and time intervals), they would see what happened
first and what happened last.

>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...

I'd think the paradox isn't there, because it has several false observations:

- When Dwight meets Bright on Alpha Centauri, he'll also see an image of
Earth where Bright has not left (that image has been traveling for 3.0087
years).
- With this info he already can figure out that something is strange here.
- When he still decides to travel FTL to Earth to stop Bright from leaving,
he will meet her light-image (that she reflected during her trip to AC) all
the way, until some small distance from Earth where the image suddenly
changes from "Bright going into FTL" to "Bright suddenly being disappeared".

Timothy