From: KELSEY@SLACVM.SLAC.STANFORD.EDU (KELSEY, Michael H.)
Newsgroups: rec.arts.startrek.tech
Subject: Re: Twin Paradox
Date: 23 Aug 92 21:46:22 GMT
 <7i2wPB7w165w@west.darkside.com>
Organization: Stanford Linear Accelerator Center

In article <7i2wPB7w165w@west.darkside.com>, max@west.darkside.com (Erik Max
Francis) says:
>
>KELSEY@SLACVM.SLAC.STANFORD.EDU (KELSEY, Michael H.) writes:
>
>> Well, that's not entirely true.  Special relativity can certainly handle
>> accelerating objects.  There are Lorentz transformation for the acceleration
>> vector just as there are for the velocity vector.  What SR _can't_ handle
>> (and what GR is for) is curved space-time, that is a metric g_{\mu\nu}
>> which has coordinate dependent terms.
>
>Special relativity _only_ deals with inertial, or non-accelerating,
>reference frames.  If one of the frames accelerates relative to another,
>then special relativity no longer applies -- you have to use general
>relativity.

Acceleration is a time derivative of velocity.  I can perfectly well
take a derivative of velocity with respect to proper time (tau), to get
the acceleration I observe in my frame, and I can apply a Lorentz matrix
transformation on the acceleration four-vector to determine what an
observer in a different, inertial, reference frame will observe.

Special relativity allows you to connect observations made in two different
inertial (non-accelerating) reference frames.  It DOES NOT prevent you
from analyzing observed acclerations of objects, provided you _view_ the
accelerating objects from some other, inertial reference frame.
>
>> If you don't like acceleration, you can analyze the twin paradox in a
>> constant speed system by making the simplifications:  B travels away from
>> home at constant speed.  When he reaches the "midpoint" of his trip, he
>> instantaneously reverses direction, and returns home at the same speed.  The
>> world lines of A and B then form a triangle in a space-time diagram, and you
>> can analyze durations that way.
>
>You've just complicated things, because B has just executed an _infinite_
>acceleration to reverse his velocity like that.  Besides, that doesn't
>resolve the paradox.  When he comes back, he is still travelling faster
>than A.  Thus A should have aged less.  But from A's point of view, B
>has been travelling the whole time.  Thus B should have aged less.
>_That's_ the paradox.  In special relativity, if you pretend no
>acceleration happened, they should both see the other as having aged
>less.  But when they do meet, B aged less, not A.  This is because B
>accelerated, and broke the "symmetry."  B felt something that A did not.,
>and that was the acceleration required to bring him home.
>
>> When I was a first-year grad student, I took a survey course in SR (okay,
>> so I was bored) in which the twin paradox was analyzed and resolved using
>> only special relativity.  So, it _can_ be done.
>
>I'd like to see it, but I suspect that your teacher (or teachers) were
>rather, shall we say, unthorough.  Special relativity applies only to
>inertial reference frames.  If one of them accelerates, then BLAM.
>Special relativity can't apply anymore.  (That's why special relativity
>is "special" -- restricted -- and general relativity is not.)
>
Okay, here it is, in all its mathematical glory.  I just pulled out my
three-year-old notes, so I am copying directly.  The resolution of the
twin paradox is basically a corollary of the relativistic Doppler shift.
I simplify my analysis, as I described above, by letting B undergo
"instantaneous" acceleration (this is valid because the acceleration
occurs at a single event-point, so it is instantaneous for all observers)
at the beginning, end, and midpoint of his trip.  A will be the "rest" or
"stay-at-home" twin.  We direct both A and B to send radio signals (beeps)
to each other at regular, equal, intervals (say 1 second apart) according
to their own clocks.

B travels outward for a time t1B = tB/2 (she turns around at the halfway
point according to her own clock), at a velocity V relative to A.  She
the travels back toward A at the same velocity V for a time t2B = tB/2.
The times tB, t1B, and t2B are measured according to B's clock.

Consider the signal which A sends out, and which will be received by B
_at_the_moment_ B turns around -- call this the "turn around" signal,
since it is like A _telling_ B "turn around now", which B will actually
do _when_ she receives it.  Let the elapsed time in A's frame of
referece be t1A, from when B departed to when A sent out that signal.
Since the signal travels along a null geodesic, the times t1A and t1B
are related by the relativistic Doppler equations, and we obtain:

                  t1B = G(1+V)t1A

where G = 1/sqrt(1-V^2) is the Lorentz gamma factor (I take c=1) of
the relative velocity between A and B.  Now, let t2A be the time that
elapses in A's frame from when she sent out the "turn around" signal
to when B returns.  We can relate that time to the time t2B by

                  t2B = G(1-V)t2A

where the change in sign is because B is now _approaching_ A rather
than leaving.  What is the total trip time in A's frame?

                  tA = t1A + t2A

                     = [ t1B / G(1+V) ] + [ t2B / g(1-V) ]

                     = [ tB/2 / G(1+V) ] + [ tB/2 / G(1-V) ]

                     = G tB

Which is just the Lorentz time dilation factor.  So we have computed in
A's frame that B will be younger than A when she returns, if both A and B
started out the same age.

Now, what of the paradox?  Well, let us compute things from B's point of
view.  B is travelling outward, receives a signal "turn around", and
reverses direction.  At the same time (same event point), B transmits
a signal "I've turned", which will be received by A some time later.
Again, we can use the relativistic Doppler equations to determine what
this time is.  Let t1A' be the time from the start of the trip to when A
_receives_ the "I've turned" message, in A's frame.  This time is related
to t1B (the time when B transmits the message) by

                t1A' = G(1+V)t1B

where the + sign again comes because A and B are moving apart.  Now let
t2A' be the time interval between when A receives the "I've turned"
message, and when B returns.  We can relate t2A' to t2B with

                t2A' = G(1-V)t2B

and again ask what is the elapsed time that B _computes_ for A's clock:

                 tA' = t1A' + t2A'

                     = G(1+V)(tB/2) + G(1-V)(tB/2)

                     = G tB

So B _computes_ that A's clock will have ticked for longer than B's
own clock, and hence that A will be OLDER than B when B returns, if both
A and B started out the same age.

Notice what I did to compute this.  I simplified matters by dividing
the problem into two parts:  the moving apart section which had two
_inertial_ reference frames with a relative velocity +V, and the moving
together section which had two _inertial_ reference frames with a
relative velocity -V.  All of the acceleration occurred at a single
event point, so A, B, and _any_other_ inertial observer could agree on
where-when it occurred, although they might assign different spatial
or temporal coordinate values, depending on their particular frame.

The result is that _both_ twins, and any other inertial observer, can
agree that the travelling twin returns younger than the stay at home
twin.  If you draw a space time diagram, A follows a strictly timelike
worldline, while B follows a worldline with a corner in it.

                                           -- Mike Kelsey

[  My opinions are not endorsed by SLAC, Caltech, or the US government  ]
What is your _name_?    "kelsey@slacvm.slac.stanford.edu"
What is your _quest_?   "To get a Ph.D. in high-energy physics"
When will you _finish_? "I don't know.   Waaaaaaaaahhhhhhhh..."
Newsgroups: rec.arts.startrek.tech