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From: rimey@ucbmiro.ARPA (Ken Rimey)
Newsgroups: net.physics
Subject: Re: FTL and time-travel
Message-ID: <9001@ucbvax.ARPA>
Date: Fri, 12-Jul-85 05:42:42 EDT
Article-I.D.: ucbvax.9001
Posted: Fri Jul 12 05:42:42 1985
Date-Received: Sat, 13-Jul-85 11:47:56 EDT
References: <375@sri-arpa.ARPA>
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Reply-To: rimey@ucbmiro.UUCP (Ken rimey)
Organization: U.C. Berkeley
Lines: 38

>From:  Mark Purtill 
>Recently on SF-LOVERS, I read
>>[from: mit-eddie!nessus@topaz.arpa]
>>According to Special Relativity, faster-than-light travel is
>>exactly equivalent to traveling backwards in time: there is no
>>difference.
>
>Is that really true?  If so, could someone please explain how?

Let's say you travel from A to B at a constant velocity of 10 mph.
However if a traveler making the reverse trip regarded himself as
stationary, he would say you were traveling at 20 mph.

Now let's say you travel from A to B at 0.9 times the speed of light.
Does the guy going from B to A at 0.9c see you going at 1.8c?  No.
In special relativity, the rule for combining velocities is not
addition.

NOW, let's say you travel from A to B at twice the speed of light.
The correct rule for combining velocities says that for some
observers you will get to B before you leave A.  If you accept
traveling faster than light, you must accept the concept of
arriving before you leave.

Arriving at ONE PLACE, B, before leaving from ANOTHER PLACE, A, may
not seem like a big deal.  However, if you can do that, then why
can't you travel from B back to A and again arrive before you leave?
Then you meet yourself.

					Ken Rimey

p.s.  The correct velocity addition law is as follows:  If your
speed is u, then your apparent speed to someone going the other
way at speed v is (u + v)/(1 + uv), not the classical u + v.
The units for u and v are such that light travels at speed 1.

For u = v = 0.9, (u+v)/(1+uv) = 0.994  (You don't go faster than light.)
For u = 2 and v = -0.9, (u+v)/(1+uv) = -1.375  (You are going the other way.)