Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.3 4.3bsd-beta 6/6/85; site ucbvax.ARPA Path: utzoo!watmath!clyde!cbosgd!ihnp4!ucbvax!rimey 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> Sender: nobody@ucbvax.ARPA 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.)