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From: mcgeer@KIM (Rick McGeer)
Newsgroups: net.space
Subject: Re: Unified Field Theory and space trav
Message-ID: <8511172327.AA28832@kim>
Date: Sun, 17-Nov-85 18:27:48 EST
Article-I.D.: kim.8511172327.AA28832
Posted: Sun Nov 17 18:27:48 1985
Date-Received: Sat, 23-Nov-85 10:06:07 EST
Sender: daemon@ucbvax.BERKELEY.EDU
Organization: The ARPA Internet
Lines: 71

>> 
>> For math types, the important quantity is sqrt(1-(v/c)**2). If we call that
>> R, then time, mass, and length change as follows, from the viewpoint of
>> someone on EARTH! Please note, everything depends on who is the observer.
>> T' = T * R
>> M' = M / R
>> L' = L * R
>> 
>> 	Where T',M',L' are what you see while the voyagers are traveling
>> rapidly, and T,M,L are what they were just before they left. So, when v
>> is close to c, one hour on Earth (T) becomes a very small amount of time
>> on the ship (T'). The inevitable question is: "I thought it was all relative,
>> why don't the voyagers see Earth as running slow in time?", or the Twins
>> Paradox. The answer is that someone (the voyagers) accelerated, and the other
>> people didn't, and that distinguishes them. Accleration and gravity slow
>> down clocks too, just like high speeds.
>OK, lets rephrase the question.  The two twins get into identical space ships
>and accelerate at the same amount but in opposite directions.  What do they
>observe about the other?

	If their roles are symmetric, then clearly neither twin is preferred:
when they meet again, *so long as their roles have been prefectly symmetric*
(identical accelerations, identical velocity vectors (in magnitude) wrt to
a third observer, identical trip times, again wrt a third observer) they will
be the same age.

	Let's consider the following: let's suppose that the twins (call them
Harold & Maude) leave from Earth and head for planets P & Q, each of which
are 8 light years from Earth in radially opposite directions.  Harold &
Maude travel at .8 c, and acceleration and deceleration times are negligible
and can be ignored.  They return immediately.

OK.  For each, 1-v^2/c^2 = .36, hence tau = .6.  Hence the distance for each
of them are foreshortened by an identical amount, namely to 4.8 light years.
At a velocity of .8c, they reach the planets in six years, by their own clocks,
return in an identical time, and have each (according to their own clocks)
aged 12 years.  According to an observer of earth, 20 years have passed.

But while their roles have been symmetric, each believes the other has been
aging very strangely.  Suppose H & M, before they leave, agree to send a signal
to the other at the end of each year.  Since their roles are symmetric, we need
only consider Harold.  Maude's velocity relative to him is

1.6c/1.64 ~= .975c

Hence, by the Doppler effect, Harold will receive .113 flashes/year on the
outbound trip.  Hence (according to Harold) Maude has aged only .678 years
on his outbound trip...

Then Harold turns around.  He still hasn't received 5.322 of Maude's outbound
flashes.  Now, he's moving with speed  zero relative to Maude *when she fired
the flashes*.  Hence, 0 doppler effect.  Hence, for the first 5.322 years of
his return voyage he gets a flash a year.

Now Harold starts to pick up the flashes Maude sent on the inbound journey.
Her velocity relative to him is -.975c, which means that the frequency of
the flashes is now 1/.113 flashes * year, or 8.84 flashes/year, which means
that over the last .678 years of his trip Harold picks up six flashes from
Maude.  Of course, Maude perceives exactly the same phenomenon...

To summarize:

Harold thinks Maude aged .678 years though the first six years of his trip.
Harold thinks Maude aged 5.322 yrs. through the next 5.322 years
Harold thinks Maude aged 6 years during the last .678 years of his trip.

and Maude thinks the reverse.  Who's right and who's wrong?  Well, neither
and both: the key thing is that events separated by space can't be temporally
compared.

					-- Rick.