Natural Science Forum / Physics / General Physics / February 2007

According to the stay-at-home twin the trip takes a time T
and the total distance covered is L.
According to the travelling twin the trap takes a time T/gamma
and the total distance covered is L/gamma.
So there is no "lack of symmetry" between space and time.

Mass is an invariant:
  http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

[followup set to sci.physics.relativity]

Dirk Vdm

Create one, then. Here you go, this should get you started:
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img155.gif

"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote
in message news:_kgEh.27738$Ry2.305216@phobos.telenet-ops.be...

news:Y5gEh.1472$8x.19@newssvr14.news.prodigy.net...

Thank you.
I thought that without acceleration, when they are moving away from each
other with constant velocity, the application of gamma applies to each one
viewing the other's inertial frame symmetrically. It is only when one twin
returns to the other's frame (or accelerates at all) that an asymmetric
discrepancy accumulates. That is, without acceleration, if the distance of a
phenomenon is measured in stay-at-home twin's frame (they are both stay at
home and travelling without acceleration, but rather than A and B....) then
the distance of the phenomenon that is measured by the travelling twin is
L/gamma, and vice versa (if the distance of a phenomenon is measure in the
travelling twin's frame is L then the distance of the phenomenon measured in
the stay-at-home twin's frame is L/gamma).
After the travelling twin accelerates, and thereby changes it's inertial
frame, the times and distances measured of phenomena change (with no motion
within the frame itself to account for the changes in position). When the
travelling twin returns home, it's accumulated time (ticks of a clock) is
different than the accumulated time (ticks of a clock) for the stay-at-home
twin, yet no meter stick has an accumulated discrepancy (when the travelling
twin returns to the stay-at-home twin's frame; though there would be a
corresponding discrepancy in the accumulated distances traveled of cyclic
motion within the travelling twin's frame). Could this discrepancy (what I
called asymmetry) be that "local" time is "ordinal" and distance or length
is "cardinal"; that local time is an index of spatial recursion (cyclic
motion)?

Or am I confusing time with clocks, and that
clocks have a
history of time passed, while devices for measuring distance suffer from
none of the same. And while the travelling twin is younger, he/she is not
smaller since no growth while travelling (and staying at home) occurs,
assuming he/she is does not grow during the time/distance travelled.

And the apparent difference between time/distance clocks/meter-sticks is an
artifact of human perception?

It appears that my prior responses did not post.... I hope that this one
does.

Mass is invariant.  So you wouldn't expect it to change.  Length
contraction could only maintained on spacelike paths through
spacetime.  But this would require speeds greater than light, so it
couldn't happen.  All massive bodies can only travel on timelike
paths, so when the twins get back together, one is older, one is
younger, but their meter sticks are still the same size.

Hi  Simple_Simon,  The average Relativistic_Mass ( a.k.a. energy )
of the accelerated twin is higher than that of the inertial twin.

To the extent that these twins could be a real world phenomenon,
General Relativity must be used to model it.