Re: Twin paradox revisited ll

On 29 jul, 20:07, bill <cosmo...@xxxxxxxxxxxxxxx> wrote:
On Jul 30, 4:03 am, Mike Fontenot <mlf...@xxxxxxxxxxx> wrote:



bill wrote:

There seems to be a discrepancy in your article athttp://home.comcast.net/~mlfasf.

In paragraph 4 you have the traveler accelerating away from the earth
at 1g for a period of two years his time then coasting for 9 years.

In paragraph 5 he accelerates at -1g for two years [...]

You apparently misread the 5th paragraph...it actually says that
he accelerates at -1g for 3 years, not 2. When he starts that
acceleration, his speed is 0.968c (moving away from his twin).
When he finishes that acceleration, his speed is -0.774c (moving
toward his twin).

Mike Fontenot

Thank you, I did make that mistake - reading 2 years ilo the 3 years
you specified.

Bill

I think the main interpretation problem in this topic, is the fact
that every observer (both twins and possibly an external observer)
will have knowledge of what is going on the travel, only by his local
information (clocks) and by the outside information he is receiving at
Earth or at the ship (like radio messages or position of guiding
stars). This received information can only travel from its source at
c, since c is the maximum interaction propagation speed.

So the total trip look like this. Let us assume Twin A is staying
home, and Twin B is traveling to a star located 6 light years away, at
a speed v=0.6c. Every birthday, each twin sends a message to his
brother. The only information each has is his local time for the
sending of the message and his local time for the reception of his
twin message.

a) Twin B will travel, according to twin A, for 10 years before
reaching the star. In his local clock, this travel time will only
measure 8 years (due to time dilation).
b) Acceleration need not to be considered, since the acceleration
period is short with respect to the distance to be covered (at 2g, 4
months will make the ship to reach 0.6c).
c) At the 1st, 2nd, 3rd, 4th anniversary, twin A messages will have to
catch the traveling ship, the information will reach twin B at 2, 4, 6
and 8 years of his local clock. The last message will reach twin B
just as he is arriving to the destination star.
d) At his own 1st, 2nd, 3rd and 4th anniversaries, twin B will send
back to Earth his messages. These messages will reach twin A in his
own local time at 2, 4, 6 and 8 years. Twin B 5th to 8th messages will
be received by twin A at 10 to 16 years of his local clock.

So we see up to this point that the situation is quite symmetric, with
both twins seeing the other as "younger", according to received
messages.

e) Direction of travel changes and ship returns to Earth. Trip will
last 10 years, according to twin A, but only 8 years, according to
twin B.
f) Messages from twin A now arrive more frequently to the ship. So
messages from 5th up to 20th anniversaries will be received at twin B
location at 8.5, 9, 9.5 and so on to 16 years of his local clock.
g) Messages from twin B, on his way back to Earth, also arrive more
frequently to twin A. Thus twin B messages 9th, 10th up to 16th
anniversaries will be received at twin A location at 16.5, 17, and up
to 20 years of his local clock.

Again the situation is quite symmetric, but at the end twin B's clock
is showing 16 years has passed, while twin A's clock is showing 20
years has passed.

Miguel Rios

.

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