🚀 go-pugleaf

On 30 Oct, 16:14, Tom Roberts <tjroberts...@sbcglobal.net> wrote in
sci.physics.relativity:
> Alen wrote:
> > On Oct 31, 12:02 am, stevendaryl3...@yahoo.com (Daryl McCullough)
> > wrote:
> >> Alen says...
> >>> You are right, and they have to resort to houdini-like
> >>> contortions to avoid it.  The accumulated time dilations of
> >>> A in B's view, and B in A's view, are created when A and
> >>> B are inertial frames. Yet, all this is magically invalidated
> >>> because A and B accelerate and decelerate, but it is not
> >>> invalidated in a third frame C, into which they decelerate.
> >> Your problem, Alen, is that you rely on fuzzy concepts
> >> without attempting to make them clear. What do you mean
> >> by saying that "the accumulated time dilations...[are]
> >> magically invalidated"?
> >> The meaningful quantity is not "accumulated time dilations"
> >> but the proper time recorded by each twin. How can time dilations
> >> accumulate? What does that even mean?
> >> In Special Relativity, the time shown on any ideal clock
> >> that travels from event P1 to event P2 is given by:
> >>     Tau = Integral from t1 to t2 of square-root(1 - (v/c)^2) dt
>
> > At constant velocity, integration is not necessary.
>
> But the twin scanario does not have "constant velocity".
>
> > You need
> > only write Dt' = (1/g)Dt, where D means a macroscopic time
> > interval
> > A sees B's Dt' to be less than his Dt
> > B sees A's Dt' to be less than his Dt
>
> Yes. But remember this ONLY applies to an inertial frame. And remember
> how these are measured: A and B each need assistants comoving with
> themselves to measure an interval of the other's clock, because the
> other's clock is MOVING relative to themselves. A marks the time on his
> own clock and the time on B's clock ONLY when B is right next to A; A's
> assistant marks the time of his (the assistant's) clock and the time on
> B's clock ONLY when B is right next to him (the assistant). So there are
> no "propagating photons". (interchange A and B for the other case; the
> assistants' clocks are synchronized with their respective observers'
> clocks -- do this via slow clock transport so there still are no
> "propagating photons"). Note this is not at all the twin scenario; while
> it can be applied to the inertial twin, it cannot be applied to the
> accelerated (traveling) twin, because it requires an inertial frame
> THROUGHOUT.
>
>         If you want to apply this to the traveling twin, assume
>         he moves with constant velocity during the outbound and
>         return legs, with an instantaneous turn-around. This analysis
>         can then be applied during the outbound leg, and SEPARATELY
>         during the return leg. But the traveling twin's assistant
>         at the location of the inertial twin much CHANGE CLOCK
>         SYNCHRONIZATION WITH THE TRAVELING TWIN when the traveling
>         twin turns around. This change in clock synchronization is
>         essential for this type of analysis, and fully explains the
>         effect.
>
> > A's view of B, and B's view of A appear to
> > magically vanish in the orthodox solution.
>
> Not true. There's no "magic", just basic physics. For an actual twin
> scenario, A and B are distinguishable by the acceleration of one and the
> non-acceleration of the other. When one does the ACTUAL COMPUTATION one
> always finds that the accelerated (traveling) twin has less elapsed
> proper time than the inertial (non-traveling) twin.

Roberts Roberts are you really a distinguished hypnotist in Einstein
criminal cult or just a hyperactive zombie? Countless times people
explained to you that acceleration is immaterial in the quantitative
interpretation of twin paradox; countless times you repeated the above
idiocy. At least you should learn what your cleverer brothers
hypnotists teach:

http://isites.harvard.edu/fs/docs/icb.topic193296.files/chap11.pdf
p. 14: "For the entire outward and return parts of the trip, B does
observe A's clock running slow, but enough strangeness occurs during
the turning-around period to make A end up older. Note, however, that
a discussion of acceleration is not required to quantitatively
understand the paradox, as Problem 11.2 shows."

Pentcho Valev