"Harold Ensle" skrev i melding news:arb9p1$n2i$1@slb6.atl.mindspring.net...
>
> PS. I went to the FAQ that you indicated as a source and I couldn't
> even believe how lame it was. First he insults anyone who would
> seriously complain about the paradox, but then when one actually
> reads his solutions, they are a complete bust. What worm hole did
> he pull out the Doppler garbage. All one has to do is flip the graphs
> to put the travelling twin at rest and all the doppler effects are
> reversed. The site does NOT explain the twin paradox. It is just
> the usual (actually, a bit more than usual) hot air.
>
> H.Ellis Ensle 

Does this mean that you do not understand the Doppler explanation?
One would expect that to be easy to understand for anyone.
The travelling twin will see the change in Doppler shift from
red-shift to blue shift the instant he turns around.
So he will see the blue-shift and the red shift for an equal
long time. Isn't that blatantly obvious?
The home twin will however see the Doppler shift turn from
a red shift to a blue shift at the time he _see_ the travelling
twin turn around, and that will of course be after the time
it takes for the travelling twin to reach the turning point
PLUS the time it takes for the light to reach the home twin.
Thus the home twin will see the red shift for a longer time
than he will see the blue shift.

There is no way this scenario can be "flipped" to give the opposite
result, it is inherently unsymmetrical.

A more detailed description:

A is the home twin, B the travelling twin.
They both have emitters flashing with the frequency once per second.

Observations made by A:
When B goes out, A will receive the pulses with the Doppler shifted
frequency f1 = sqrt((c-v)/(c+v))
He will observe this frequency for the time it takes B to reach
the turning point at some distance L, _plus_ the time it takes for
the light to reach him from the turning point: ta1 = L/v + L/c
Thereafter, A will observe the frequency f2 = sqrt((c+v)/(c-v))
for the time ta2 = L/v - L/c (ta1 + ta2 = 2*L/v, obviously)
A will receive the total number of pulses N = f1*ta1 + f2*ta2
N = (2*L/v)*sqrt(1 - v^2/c^2)
A's total time Ta = 2*L/v
Thus N = Ta*sqrt(1 - v^2/c^2)
Since B's clock has emitted one pulse per second, then it must show:
Tb = Ta*sqrt(1 - v^2/c^2)

Observations made by B:
We assume that B uses the time Tb/2 on his way out, and Tb/2 back.
Since B is the one who does the abrupt turn-around, he will see
the frequency change immediately as he does so.
Thus the number of pulses he will count from A is simply:
N = 0.5*Tb*sqrt((c-v)/(c+v)) + 0.5*Tb*sqrt((c+v)/(c-v)) = Tb/sqrt(1-v^2/c^2)
Since clock A has emitted one pulse per second,
Ta = Tb/sqrt(1-v^2/c^2) when B returns.

Why is this garbage?

Paul