Re: Special Relativity

[Ken Caviness <caviness@SOUTHERN.EDU>]

Nice comparison!  I particularly like this because it highlights exactly in
what way the two twins views are not symmetric (or "equally good", as my
students phrase it).  If you truly have a symmetric situation, the twins will
age the same amount, and their calculations will agree on the amount.

I also appreciated John S. Denker's very clear explanation from a special
relativistic point of view, and the pains he went to in explaining why two
inertial reference frames are necessary (and as an approximation, sufficient)
to describe the world-line of the travelling twin, while only one is needed
for the stay-at-home twin.

Incidentally, I prefer to refer to the problem as "the so-called twin
paradox".  Several people have commented in this thread on the use of the word
paradox.  My take is that a paradox is a statement that is self-contradictory,
it can neither be true nor false.  The statement "I am lying" is a
(self-referrent) paradox.

To many people, special relativity appears paradoxical at first glance.  Take
length contraction:  John thinks Jane's meter sticks are too short, Jane
thinks John's are too short.  Likewise time dilation:  each thinks the other's
clocks are running too slow.  Yet there is no paradox, because the observers
always agree on the things that matter: the laws of physics and value of the
speed of light in vacuum (c).  We just need to give up our insistence that
simultaneous events for one observer are necessarily simultaneous for another,
and that one observer's length and time measurements will necessarily be the
same as another's, etc..  Notice that the space-time interval between events
will be measured as the same for all observers -- "some things never change".

I see the so-called twin paradox as a very natural attempt to pin down the
feeling that one has when first faced with relativity that the observers'
views are in conflict -- something "must" be wrong.  If both John and Jane are
right in saying that the other's clocks are running slow, what happens when
they later place them side by side?  Won't we be able to see then who was
right?  And then how can special relativity be correct, if it were to give an
answer that we could verify to be wrong?  If the twins' view are symmetrical,
any answer other than that they measure the same result would result in a
paradox.

Of course, there is no paradox, since the symmetry between the two observers
is not complete.  Both see the other move away and then return, but one twin
feels the acceleration of her rocket at the turn around point (although she
may choose to interpret it as a gravitational field!)  The stay-at-home twin
feels no such effect.  The beautiful thing about the problem is that from
either point of view we get the same answer:  when the twins meet again, there
calculations will agree that the travelling twin aged less than her brother.
This is in spite of the fact that during the entire trip out and back (except
for the turnaround point) each says that the other's clocks were running
slow.  The "radical change" (JSD's words) in what the travelling twin thinks
the "current" time is back home is precisely what is needed to make the twins
agree on how their clocks disagree at the end of the trip!

Note:  The problem has been done in many other ways, too.  Here's a sampling:

Robert H. Romer, AJP, 27, 131-135 (1959)
Geoffrey Builder, AJP 27, 656-658 (1959)
Robert Perrin, AJP 47, 317-319 (1979)

(This last uses the following interesting scenario:  one twin accelerates away
from the earth with constant acceleration g until reaching final velocity v at
some distance d from the earth.  Then the acceleration is cut off for the main
portion of the trip.  Constant deceleration with a=-g is begun a distance d
from the target destination.  The process is precisely reversed for the trip
home.)

Great fun,

Ken Caviness
Physics Department
Southern Adventist University
http://physics.southern.edu/


"Edmiston, Mike" wrote:

> A way I have found effective for explaining this to students is to
> imagine two infinite conveyor belts moving rapidly in opposite
> directions.  The two belts are side-by-side and pass right next to
> the twins' house here on earth.  The first twin steps onto one of
> the belts and is whisked into space.  The second twin stays at home.
> At some point the first twin needs to come home.  To get home she
> has to step off the belt she originally stepped on, and onto the
> other belt thatwill take her home.
>
> We have three obvious frames of reference: earth, belt 1, belt 2.
> The twin who stayed home stayed in one frame the whole time.  The
> twin who traveled experienced all three frames, and had to make two
> frame jumps.  Clearly the experiment is not symmetric for the two.
>
> We could make it symmetric.  The twins could simultaneously step onto
> opposite belts.  They would stay on their respective belts for an
> agreed-upon time duration that each would measure with his/her own
> clock, then each would jump to the other belt and come home.
>
> In the first case (one twin travels and changes reference frames)
> there would be an age difference when they are reunited.  In the
> second case there would not be an age difference when they are
> reunited.
>
> Michael D. Edmiston, Ph.D.
> Professor of Chemistry and Physics
> Bluffton College
> Bluffton, OH 45817
> (419)-358-3270
> edmiston@bluffton.edu
>
> This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.