On May 7, 12:57*pm, bz wrote:
Tom Roberts wrote :





bz wrote:
Look at the particle that has just made one trip around the ring.
It is now back at its starting point in space but has moved in time.
The particle that stayed stationary at the center of the ring has NOT
moved in space but it, too has move in time. The integral along both
trajectories must have the same length. Therefore the moving particle
has traveled less distance along 'its time axis' (but the same distance
along the 'Lab' time axis.

This is a fruitful way to look at it, but your statement "The integral
along both trajectories must have the same length." is inadequate --
integral of WHAT? What you mean is that the elapsed time in the lab
frame is the same for both.

Here's a semi-mathematical way of looking at it:

* * *[Notation: \tau is always the proper time of the particle
* * * being discussed. t and x are always lab coordinates, with
* * * x measured tangent to the ring.]

For the particle at rest in the lab, the lab time axis and the
particle's time axis are the same, so dt/d\tau=1. While time T elapses
in the lab frame, the particle experiences T elapsed proper time.

For the particle going around the ring, dx/d\tau is nonzero (this is a
spatial component of its 4-velocity). Because 4-velocity is normalized
to 1, this implies that dt/d\tau1 for this particle [#]. So while the
time T elapses in the lab frame, the particle experiences less than T
elapsed proper time.

* * *[#] U^i = dx^i/d\tau. |U|=g_ij U^i U^j.
* * * * *Remember that {g_ij} = diag(-1,1,1,1).

Yes, you are correct. I should have been a bit more formal in my
explaination. 'Trajectory' seemed to be throwing him a curve and I hoped
to make it a bit clearer. Thanks for cleaning up the loose ends.

--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

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Centrifugal motion slows time.