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From: @S1-A.ARPA:host.MIT-MC.ARPA
Newsgroups: net.space
Subject: Re:  Speed Of Light (explain this too)
Message-ID: <1768@mordor.UUCP>
Date: Sun, 12-May-85 17:03:45 EDT
Article-I.D.: mordor.1768
Posted: Sun May 12 17:03:45 1985
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From: Rick McGeer <mcgeer%ucbkim@Berkeley>

	Well, in fact you are observing the objects A and B from a third
reference frame with observer, C, and we presume that observers in A,
B, and C all have meter sticks, clocks and what have you.  From
the question we say that in C's reference frame, both A and B are
moving in one dimension on opposite vectors with velocity .6 c.  The question
is, in A's reference frame, what is B's velocity?  From Tipler, pg 680,
we obtain:

u(x') = (u(x) - v)/(1 - v u(x) / c^2)

where v is the velocity of A in the frame of C, u(x) is the velocity of B
in the frame of C, and u(x') is the velocity of B in the frame of A.  Solving
for u(x) = -v = -.6c, we get:

u(x') = -1.2c/1.36

or almost exactly -.97c.

					Rick.

ps -- the formula I gave above can be obtained by differentiating the
Lorentz Transform:

x' = gamma (x - vt)

where v is the velocity of the frame S' in terms of the frame S.

					R.