Purtill@MIT-MULTICS.ARPA (07/10/85)
From: Mark Purtill <Purtill@MIT-MULTICS.ARPA> <Fnord> 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? Mark ^.-.^ Purtill at MIT-MULTICS.ARPA **Insert favorite disclaimer here** ((")) 2-032 MIT Cambrige MA 02139
rimey@ucbmiro.ARPA (Ken Rimey) (07/12/85)
>From: Mark Purtill <Purtill@MIT-MULTICS.ARPA> >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.)
matt@oddjob.UUCP (Matt Crawford) (07/14/85)
In article <375@sri-arpa.ARPA> Purtill@MIT-MULTICS.ARPA writes: >>[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? I wouldn't say "exactly equivalent", since a single trip at a speed faster than light will have you arriving at a time which to *some* observers is earlier than your departure but which to other observers is later. However ... If I suppose the existence of boxes which can send out signals at a fixed faster-than-light speed (in the reference frame of the sending box) and receive these signals from other boxes, then I can place two of these boxes in motion at speeds slower than light in such a way that the first box can send a signal to the second and receive the reply before it sends the initial signal. Thus any SF writer who wants to invoke FTL signalling will have to either deal with these violations of causality or invoke some preferred frame of reference such as the rest frame of the microwave background. _____________________________________________________ Matt University crawford@anl-mcs.arpa Crawford of Chicago ihnp4!oddjob!matt