Re: The real twin paradox.


"colp" <colp@xxxxxxxxxxxxx> wrote in message
news:45e50819-65f6-46a3-a821-5c3698dd146a@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Nov 21, 11:40 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
"colp" <c...@xxxxxxxxxxxxx> wrote in
messagenews:06b84031-18aa-4644-bfb7-43f49f46ae6a@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
This thought experiment is like the classic twin paradox, but in this
expirement both twins leave earth and travel symmetric return trips in
opposite directions.

Since the paths taken by the twins in this experiment are symmetric,
they must be the same age when they meet on their return to earth.

In this experiment the twins maintain constant observation of each
other's clocks, from when they depart until they return and find that
their clocks tell the same time.

Special relativity says that each twin must observe that the other's
clock is running slow, and at no time does special relativity allow
for an observation which shows that the other clock is running fast.

No, special relativity says much more precise than that
"moving clocks" are running slow.

The Lorentz-Fitzgerald transform is more precise that my description,
but that doesn't mean that my description is wrong.


It says something about intertial observers who measure
times between ticks on remote, moving clocks.

When your two clocks fly apart, each clock will measure
this time to be longer and conclude that the other clock
is "running slower".
While clock A is coasting, according to clock A, each
tick on clock A is simultaneous with some tick on clock B
with a smaller time value.
While clock B is coasting, according to clock B, each
tick on clock B is simultaneous with some tick on clock A
with a smaller time value.

Yes, that is the standard theory.


After clock A has made its turnaround, it has shifted to
another inertial frame, in which according to clock A, each
tick on clock A is simultaneous with some tick on clock B
with a larger time value.
After clock B has made its turnaround, it has shifted to
another inertial frame, in which according to clock B, each
tick on clock B is simultaneous with some tick on clock A
with a larger time value.

Wrong. The other clock tick is still observed to have a smaller time
value.
This is because in the Lorentz-Fitzgerald transform the relative
velocity term is squared, making the the issue of the clocks
separating vs the clocks approaching irrelevant to the amount of time
dilation.

http://en.wikipedia.org/wiki/Time_dilation

That is indeed irrelevant but you are still mistaken because time dilation
is ALSO irrelevant at the instant of switching reference frames. Try:
http://en.wikipedia.org/wiki/Relativity_of_simultaneity

Harald


.

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