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From: carroll@uiucdcsb.CS.UIUC.EDU
Newsgroups: net.space
Subject: Re: Time Retardation
Message-ID: <15700032@uiucdcsb>
Date: Sun, 17-Nov-85 16:34:00 EST
Article-I.D.: uiucdcsb.15700032
Posted: Sun Nov 17 16:34:00 1985
Date-Received: Tue, 19-Nov-85 04:03:03 EST
References: <8511111807.AA01976@s1-b.arpa>
Lines: 16
Nf-ID: #R:<8511111807.AA01976@s1-b.arpa>:-30:uiucdcsb:15700032:000:994
Nf-From: uiucdcsb.CS.UIUC.EDU!carroll    Nov 17 15:34:00 1985


It is possible to see two things receeding from each other at 2c from
YOUR point of view. The people on the ships, however, see something
different. To transform their speed in YOUR reference frame, you have to use
the transform (a+b)/(1+(ab/c*c)) where a,b are the speeds observed in
YOUR reference frame. Note that if a,b = c (speed of light), then each
ship sees the other receeding at (c+c)/(1+(c*c/c*c)) = c. So, no one
ever sees something moving away from him at a speed greater than c, i.e.
the speed of an object in an inertial frame is always <= c.
	This is in the same vein as the question "What happens if some
ship is going at .9c and fires something out the front at .9c? Doesn't
it go at 1.8c?". No, it doesn't. The ship sees it go at .9c, but the
"stationary" observer sees it move at something less than c. (For an
exact answer, switch to the ship reference frame, and compute the above
as if the "stationary" observer and the object are moving at .9c in
opposite directions).