Re: An interesting SR puzzle

From: Dirk Van de moortel (dirkvandemoortel_at_ThankS-NO-SperM.hotmail.com)
Date: 02/07/05 

Date: Mon, 07 Feb 2005 14:04:46 GMT

"Jesse Mazer" <vze2ztqw@mail.verizon.net> wrote in message news:4207605E.4090903@mail.verizon.net...
>
>
> RP wrote:
>
> >
> >
> > glbrad01 wrote:
> >
> >> "Paul B. Andersen" <paul.b.andersen@deletethishia.no> wrote in
> >> message news:cu62db$oiq$1@dolly.uninett.no...

[snip]

> >>> The "whirling clock" is accelerated.
> >>> That makes all the difference.
> >>> The correct answer is they will both agree
> >>> that the wrist watch runs faster.
> >>>
> >>> Paul
> >>
> >>
> >>
> >> That is not the correct answer.
> >>
> >> Brad
> >
> >
> > I agree.
> > The correct answer is that in this system GR must be applied.
>
>
> No, SR can deal with the question of the proper time recorded by an
> accelerating clock by considering its path as seen in an inertial frame.
> Just integrate [squareroot(1 - v(t)^2/c^2) dt] between two times t1 and
> t2 in your frame, with v(t) being the clock's instantaneous velocity at
> time t in your frame; this will give the correct answer for the time
> elapsed on the accelerating clock during the interval (t1, t2) in your
> frame

You can see this in action on
  http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html

And SR is even sufficient in a non-inertial frame:
  http://www.geocities.com/slithytove5/AccelClocks.htm
  http://groups-beta.google.com/group/sci.physics.relativity/msg/806a08d082376014
  http://groups.google.co.uk/groups?&threadm=buphbm$m65$1@dolly.uninett.no
In that case a factor (1+a(t) x(t) / c^2 ) must be appended
to the square root.
It's a bit messy, but it works and it's still SR.

Dirk Vdm

> (you also have to take into account the finite speed of light to
> figure out how much time you'd actually *see* elapse on the accelerating
> clock during that time-interval, but in this problem the answer would be
> the same). Then imagine that your own clock is sending signals to the
> accelerating clock at the speed of light, and figure out what
> time-signal the accelerating clock will be receiving at time t1 and what
> time-signal it will be receiving at time t2; this tells you what amount
> of time an observer sitting on the accelerating clock would see elapsed
> on your clock during the same interval. If the accelerating clock is
> maintaining a constant distance from your clock, and a constant velocity
> in your frame, it's pretty easy to show that an observer on the
> accelerating clock will see your clock sped up by the same amount that
> you see the accelarating clock slowed down.
>
> Jesse
>

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