In article ,
Stephen Bint wrote:

The problem is not "deciding" which frame is the one running slow, by
clutching at an asymmetry. The problem is that time dilation fails to
prevent witnesses seeing the bouncing beam going faster than light, unless
it works both ways and the Earth-bound observer has also aged less than the
twins.

If you say either frame is the slow one, you deviate from what the formula
asserts, which is plainly that both are slow by the same factor.

In your previous example, all factors (time dilation and kinematic) are
symmetrical between the two twins, so they experience the same elapsed
time. In the classic "twin paradox", the time dilation is symmetric, but
the kinematic factors aren't. Therefore the twins experience different
elapsed times.

Here's a description of a classic "twin paradox" situation, in the same
style as my description of your previous example:

The scenario: You stay behind on Earth while your twin goes on a space
journey. From your point of view he travels away from the Earth at a
speed of 0.8c for 5 years, covering a distance of 4 light-years in the
process, then immediately turns around and returns, again at a speed of
0.8c, taking another 5 years for the return trip. The total trip duration
is 10 years by your reckoning.

Relativity predicts that your twin experiences less elapsed time
because of time dilation:

(5 years) * sqrt (1 - 0.8^2) = 3 years

for each leg of the trip, and from his point of view the round trip lasts
only 6 years.

The relativistic time dilation equation predicts that each twin's clocks
"run slower" in the other twin's reference frame. So why can't your twin
conclude that the trip must be shorter for you, than it is for him? The
answer lies in the fact that your experiences are not symmetrical. Your
twin is at rest in two different inertial reference frames, one during the
outbound trip and another one during the inbound trip. You remain at rest
in a single inertial reference frame during the entire journey. Your twin
has to fire his spaceship's engines at the turnaround point. You do
nothing.

By examining what both of you actually *see* by watching each other's
clocks/calendars through telescopes, we can show that both of you must
come to the same conclusion: the trip lasts 10 years for you, and 6 years
for him.

First let's look at what's happening from your point of view.

To be specific, assume that your twin starts out on his journey at the
very beginning of the year 2004. You watch his clock (calendar?) through
your telescope as he recedes. The rate at which you receive the images of
each of his "new years" depends not only on his time dilation, but also on
the fact that he is moving away from you, so we have to use the
relativistic Doppler-effect equation. You see his calendar turn over to a
new year at intervals of

(1 year) * sqrt ((1 + 0.8) / (1 - 0.8)) = 3 years.

Therefore,

At the beginning of 2007, you see his calendar turn over to 2005.

At the beginning of 2010, you see his calendar turn over to 2006.

At the beginning of 2013, you see his calendar turn over to 2007.

This is the point at which you see him turn around and begin his return
trip. In your reference frame, he actually began the return trip at the
beginning of 2009 (2004 + 5), but you don't see it until 4 years
later because it happened 4 light-years away from you.

During his return trip, we have to switch the signs in the Doppler-shift
equation because now he's approaching, not receding. You now see his
calendar turn over to a new year at intervals of

(1 year) * sqrt ((1 - 0.8) / (1 + 0.8)) = 1/3 year.

Therefore,

At the beginning of May 2013, you see his calendar turn over to 2008.

At the beginning of September 2013, you see his calendar turn over to
2009.

At the beginning of January 2014, you see his calendar turn over to 2010,
and he arrives home. Ten years have elapsed on your calendar, and 6 years
have elapsed on his.

What does this look like from your twin's point of view, as he watches
your clock/calendar through his telescope?

As he is traveling away, he sees your calendar turn over a new year at
intervals of three years, just like you do his. Therefore,

At the beginning of 2007 (2004 + 3), he sees your calendar turn over to
2005.

But three years is the length of the outbound trip, according to him,
because of time dliation. So at this point he turns around and begins his
return trip. Now he sees your calendar turn over a new year at intervals
of 1/3 year, just like you do his. Therefore,

At the beginning of May 2007, he sees your calendar turn over to 2006.

At the beginning of September 2007, he sees your calendar turn over to
2007.

At the beginning of January 2008, he sees your calendar turn over to 2008.

At the beginning of May 2008, he sees your calendar turn over to 2009.

At the beginning of September 2008, he sees your calendar turn over to
2010.

At the beginning of January 2009, he sees your calendar turn over to 2011.

At the beginning of May 2009, he sees your calendar turn over to 2012.

At the beginning of September 2009, he sees your calendar turn over to
2013.

At the beginning of January 2010, he sees your calendar turn over to 2014.

But now three years have elapsed (on his calendar) since he turned around,
so he has now returned home. Six years have elapsed on his calendar, and
10 years have elapsed on yours, in agreement with what you observe.

Note that although we did not use the time dilation equation directly in
predicting what your twin sees your clock/calendar doing, it is implicit
in the relativistic Doppler-shift equation. In deriving the relativistic
Doppler shift equation, one must use the time-dilation equation.

--
Jon Bell Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA