New Improved Twin Paradox Explanation

Good morning.

Some time ago I made a rather stuttering attempt at this explanation
on YouTube, (math free version)

http://www.youtube.com/watch?v=nmAVkj9fTnU
http://www.youtube.com/watch?v=76BB-qZXEho
http://www.youtube.com/watch?v=eT-3R8ju9V0

but I did not feel like doing the mathematics at that time.

This morning, I'd like to try to put a FEW equations down. Don't
expect here any profound discovery. Typically an explanation has a
quality in inverse proportion of the explainer's pride. I'm pretty
proud of this explanation, so that's probably not a good sign.

This morning I wanted to describe in my own words and at least give
what little bit of the math that I know how to do. I apologize in
advance for the math and grammatical errors you'll probably find.

In any case, let us consider two parties, A and B, A stays at home,
while B moves away into the distance to point C, 1 "light" unit away
and then returns home.

The speed of light in this model is 1 and the distance is 1 and the
velocity which B chooses to travel is v. (where, of course, v=1
represents the speed of light)


----------------
Part 1, What do things look like from the Home Twin, A?

So A watches B travel toward point C at velocity v. How long does
this take? B arrives at C at time t = distance/velocity = 1/v.
However when B arrives at C, there will be a time delay before A sees
the event of arrival because of the time the light takes to return
from that point. So A observes B's arrival at C at time t = 1+1/v

An interesting thing to consider here is that the observed velocity is
explicitly calculable from the two observed events (departure
observation and arrival observation). A will have observed B's
journey at a constant speed. Observed Velocity of Outward trip =
distance/time = 1/(1+1/v) = v/(1+v)

A similar consideration of the inbound trip yields an Observed Time of
the inbound trip of (1/v)-1, and observed velocity v/(1-v)

Thus if v=.5, then the observed velocity outbound =.5/1.5 = .3333c,
and the obseved velocity inbound = .5/.5 = 1c.

If v=.9 then the observed velocity outbound = .9/1.9 = 0.47c and the
observed velocity inbound = .9/.1 = 9c.

So the maximum observable outbound velocity is .5c but there is NO
maximum observable inbound velocity.

---------
Part 2. What do things look like from the perspective of the
traveling twin, B?

http://www.youtube.com/watch?v=eT-3R8ju9V0

From B's perspective there must be some things which are the same, and
some things that are different. If all of the same physics apply,
then we must assume that the same chemical, nuclear, or matter/anti-
matter reactions that produced the velocity v (Between A and B) for
observer A must also have produced the same velocity v (Between A and
B) for observer B.

The real velocities are the same. If the real velocities are the same
then should the observed outbound velocity v/(1+v) and observed
inbound velocity v/(1-v) be the same? Yes, these calculations are
general. The observed inbound and observed outbound velocities are
based ONLY on the actual velocity, and nothing else. Though I did the
calculation based on a specific pair of events, departure and arrival
of observer B, any two events could be used, and they cancel out of
the final calculation.)

A thing that will be different for the two observers, though is the
apparent time that it takes B to travel from A to C and back. Whereas
A observed the return trip to happen much faster than outbound trip, B
will observe the outbound trip and inbound trip to take the same
amount of time. (Other famous gedanken experiments determine that
this time would be (1/v)*(1/sqrt(1-v^2)) whereas 1/v would be the
expected time it takes to travel distance 1, and the other factor is
the time dilation factor)

The Observed velocities are the same, but the amount of time traveled
at those velocities is different. This means that the observed
distance traveled by B is different for B than it is for A.
(Technically, B remains stationary relative to observer B, just to be
troublesome.)

But the greatest point of interest is what happens to the Observed
distance to A from the perspective of B during the turn-around. The
apparent relative velocity of A on the way out was v/(1+v). The
apparent relative velocity of A on the way in will be v/(1-v).

If the time is the same, but the apparent velocity is different, then
there is only one possibility. The distance CHANGED. You cannot move
away from an object at one apparent velocity and then return at a
faster apparent velocity for the same period of time unless that
object has moved away from you.

In effect, the traveling twin at his point of turning-around is
transitioning himself into a new reference frame where his home-world
is much further away than it was before. Also as his home goes from
slow-motion [v/(1+v) .5] to fast forward [v/(1-v) = unlimited.] For
the moving twin it may be a surprise that the earth lurches away from
him at the point of turnaround, but he will be able to watch the rise
and fall of civilizations in fast-forward on his return trip. It
won't be a surprise to him when he gets home and everyone he knew is
dead unless he has no windows on his space-ship.