Re: Special Relativity

["John S. Denker" <jsd@MONMOUTH.COM>]

John Clement wrote:
...
> As long as the analysis is entirely done from the point of view of the stay
> at home twin there is no paradox.

Philosophical point: I was taught that paradoxes only
come from misunderstanding the problem.  Mother Nature
doesn't see any paradoxes.  We don't see paradoxes
unless we mis-state the problem somehow.

Using paradoxes to teach relativity is abhorrent, although
(alas) common.  By way of analogy:  I could come up with a
dozen "paradoxes" in elementary nonrelativistic mechanics,
but it would be terrible pedagogy to inflict them on
students who are trying to learn elementary mechanics.

Every time we present a "paradox" we are teaching students
to mis-state the problem.

Therefore our goal is to get to the point where we can say
there is no paradox from either twin's point of view.

===============

> Special relativity can not be used to
> analyze the problem from the point of view of the moving twin

I disagree; see below for details on how to do it.

> because part of the time he is not at a constant speed.

I disagree.  There are well-known techniques (typically
using a plurality of instantaneously-comoving observers)
for handling this and many, many other SR problems
involving nonconstant motion.

> General relativity supplies the
> pieces needed to complete the analysis from the point of view of the moving
> twin's acceleration.

That's backwards IMHO.

This situation can be analyzed using nothing but SR.
GR need not be mentioned.  But later, the techniques
used here can be used to provide a running headstart
when learning GR.

> Students will routinely apply equations such as x=X0 + V0 t + 1/2 a t^2 in
> situations where the acceleration changes.  In other words they do not
> recognize that equations are only applicable under certain conditions.

I agree with that.  Note that that's not at all unique to
this problem.  Students are continually overlooking the
breakdown in simultaneity-at-a-distance in particular,
and applying equations outside their domain of applicability
in general.

========================

Here's how I analyze it.  Please refer to
  http://www.monmouth.com/~jsd/physics/gif48/twins.gif

This is a spacetime diagram.  We need three observers.
Their frames are shown in green, red, and blue.

The green observer is comoving with the stay-at-home
twin.  The green lines running up toward 12:00 are
parallel to the green observer's t axis, i.e. they
are lines of constant x_g.  Similarly the green lines
running over toward 3:00 are parallel to the green
observer's x axis, i.e. they are lines of constant
t_g.

The red observer is comoving with the twin travelling
outbound.  The red lines running up toward 1:00 are
parallel to the red observer's t axis, i.e. they are
lines of constant x_r.  Similarly the red lines running
over toward 2:00 are parallel to the red observer's x
axis, i.e. they are lines of constant t_r.

The blue observer is comoving with the twin travelling
inbound.  The blue lines running up toward 11:00 are
parallel to the blue observer's t axis, i.e. they are
lines of constant x_b.  Similarly the blue lines running
over toward 10:00 are parallel to the blue observer's x
axis, i.e. they are lines of constant t_b.

The black dotted line is irrelevant to this problem.
It's just for fun.  It's the world-line of a photon.
It shows that all observers agree that the photon is
lightlike, i.e. it moves right down the diagonal of
the unit squares in all frames.

The path of the travelling twin is marked with a heavy
line, red for outbound and blue for inbound.  The
turnaround point is circled.

The red observer synchronizes his watch to the green
observer at the event where they pass each other (t=0 x=0).
The blue observer synchronizes his watch to the red observer
at the point where they pass each other (the turnaround
point).  Note that synchronization is well defined
if-and-only-if the parties are colocated.

The key point is that when the traveller transfers his
timekeeping from the outbound observer to the inbound
observer, there is a radical disagreement as to
what they think the "current" time is back home.  We
can understand this just by glancing at the spacetime
diagram:  Look at the the red line (toward 8:00) from
the turnaround point and see where it crosses the
green vertical axis (which is the worldline of the
stay-at-home twin).  Then look at the blue line (toward
10:00) from the turnaround point and see where it
crosses the the green axis. These two crossing points
are radically different.

Note that we do not need to worry about what Gee forces
the traveller is subjected to during the turnaround.
That doesn't matter.  No matter how he accomplishes
the turnaround, he must transfer from one timekeeping
system to the other, and must therefore incur the
disgreement as to the "current" time back home.

To complete the analysis, you need to take into account
the time dilatation of the red and blue observers
relative to the green observer, but that's relatively
routine.

This posting is the position of the writer, not that of George Wells,
David Filby, or Phillip Hillyer.

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