On Oct 28, 2:33*am, Pentcho Valev wrote:
Surely there are three conclusive reasons why
acceleration can have nothing to do with the time dilation
calculated:
(i) By taking a sufficiently long journey the effects of acceleration
at the start, turn-round and end could be made negligible compared
with the uniform velocity time dilation which is proportional to the
duration of the journey.
Your first hypothesis is not true. You can't make all the effects
negligible compared to the uniform time dilation.
One problem with your logic is that the less acceleration, the
greater amount of time it takes to reverse the direction. You
apparently think that if you "spread the deceleration" over the length
of a long trip, you eliminate the effects of the acceleration. When
you spread it out, however, you increase the time over which that
"negligible" acceleration acts. So the accumulated effect of the
acceleration remains the same, no matter how you spread it out.
(ii) If there is no uniform time dilation, and the effect, if any, is
due to acceleration, then the use of a formula depending only on the
steady velocity and its duration cannot be justified.
There is a uniform time dilation on the inertial observers only.
The other observers aren't traveling with a uniform velocity.
(iii) There is, in principle, no need for acceleration. Twin A can get
his velocity V before synchronizing his clock with that of twin B as
he passes.
Then A can never prove to B that he didn't start moving before
the actual start of the trip.
He need not turn round: he could be passed by C who has a
velocity V in the opposite direction, and who adjusts his clock to
that of A as he passes. When C later passes B they can compare clock
readings.
However, C can never prove to B that he passed A at the time he
did.
As far as the theoretical experiment is concerned, C's clock
can be considered to be A's clock returning without acceleration
since, by hypothesis, all the clocks have the same rate when at rest
together and change with motion in the same way independently of
direction. [fn. I am indebted to Lord Halsbury for pointing this out
to me.] (...)
Lord Halsbury probably meant something else than what you imply.
If he didn't point out the other issues, then you owe him no favors.
The three examples which have been dealt with above show
clearly that the difficulties are not paradoxes) but genuine
contradictions which follow inevitably from the principle of
relativity and the physical interpretations of the Lorentz
transformations. The special theory of relativity is therefore
untenable as a physical theory."
GIGO. Your initial assumptions are garbage, so your conclusions
are garbage. Garbage in, garbage out.

The following scenario will show that the travelling twin will find
himself OLDER than his brother who remained behind. A long rocket
passes the twin at rest, and the rocket is so long that the twin at
rest will see it passing by all along. According to Einstein's special
relativity, observers in the rocket see their clocks running faster
than the twin at rest's clock, that is, observers in the rocket age
faster than the twin at rest. At some initial moment the travelling
twin, standing so far next to his brother, jumps into the rocket,
joins the observers there and starts, just like them, aging faster
than the twin at rest.
The twin that "jumps aboard" has to undergo a large
acceleration. During the acceleration, the far away universe "appears"
to speed up. The time contraction effect is nonlocal.

Later the rocket stops and immediately starts moving in the opposite
direction. Again, according to Einstein's special relativity,
observers in the rocket, including the travelling twin, age faster
than the twin at rest.
You ignored the acceleration when he "jumped aboard" the craft.
The time contraction effect is nonlocal.

Finally the travelling twin jumps out of the rocket and rejoins his
brother at rest. Who is older?
The brother at rest. The traveling twin has undergone a lot more
acceleration at far distances from the traveling twin.