Re: Distance-dependent time contraction
From: mich (mich_at_efni.com)
Date: 06/25/04
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Date: Thu, 24 Jun 2004 22:38:43 -0400
Thanks for your help Dirk; I really appreciate it.If I have some other
problems, I'll let you know.
Andre
"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote
in message news:IIICc.165373$%q1.8157290@phobos.telenet-ops.be...
>
> "mich" <mich@efni.com> wrote in message
news:10dmka6o8pn4215@corp.supernews.com...
> >
> > "Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com>
wrote
> > in message news:rvxCc.4639$9t2.1064@news.cpqcorp.net...
> > >
> > > "mich" <mich@efni.com> wrote in message
> > news:10dkeib1unj4db4@corp.supernews.com...
> > > >
> > > > "Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com>
> > wrote
> >
> > > > in message news:u%gBc.161297$Yb2.8243253@phobos.telenet-ops.be...
> > > > >
> > > > > "mich" <mich@efni.com> wrote in message
> > > > news:10db3ut6mqs0g97@corp.supernews.com...
> > > > > >
> > > > > > "Dirk Van de moortel"
<dirkvandemoortel@ThankS-NO-SperM.hotmail.com>
> > > > wrote
> > > > > > in message
news:0gYAc.160388$BL2.8269913@phobos.telenet-ops.be...
> >
> > I went through our posts and I found something that you wrote:
> >
> > ________________________________________________________
> > You wrote
> > Remark:
> > In order to measure lengths you don't actually need rods.
> > You can measure lengths with light and a clock only,
> > using the assumed fact (postulate) that light speed is
> > always the same. Send a signal, wait for the echo, find
> > the time between sending and receiving, and you know
> > how far away (and when) the signal hit the target.
> > To use this system for measuring a moving rod requires
> > two signals and echoes, in such a way that the front and
> > the back of the moving rod are hit by your signals at the
> > same time (time measured by yourself of course!).
> >
> > _______________________________________________
> >
> > The second explanation I wrote seems to match yours, although I may not
have
> > been as clear.
> > _____________________________________________
> >
> > a quote from you:
> >
> > "To use this system for measuring a moving rod requires
> > two signals and echoes, in such a way that the front and
> > the back of the moving rod are hit by your signals at the
> > same time (time measured by yourself of course!)."
> > _________________________________________
> >
> > The reason why I elaborated a little bit is due to the difficulty of
knowing
> > whether the two signals hit the front and back of the moving rod at the
same
> > time.So here's my second explanation again, and I'll wait for your
response.
>
> Very good! You found exactly what you needed to solve it,
> and there really is no need to repeat that explanation.
> You just found it!
>
> Look:
> "To use this system for measuring a moving rod requires
> two signals and echoes, in such a way that the front and
> the back of the moving rod are hit by your signals at the
> same time (time measured by yourself of course!)."
> That's all there is to it.
> You can measure the distances and the times of the reflection
> events on both sides of the rod, so just make sure that both
> times are the same, and then you take the difference of the
> distances of the reflection events, and hey presto!
>
> Recall the definition of "the time of an event" as
> t = (te+ts) / 2
> and distance as
> x = c*(te-ts)/2.
>
> So for the reflection event on the near side of the rod you
> calculate
> t1 = (te1+ts1)/2 and x1 = c*(te1-ts1)/2
> and likewise for the other reflection event on the far side
> t2 = (te2+ts2)/2 and x2 = c*(te2-ts2)/2 .
> So, if you arrange for things such that
> t1 = t2 [ in other words delta(t) = t2-t1 = 0 ],
> then the length of the rod is given by the difference of the
> distances of the near and far side:
> delta(x) = x2 - x1
>
> It works for a moving rod, and for a non-moving rod it works
> just as well of course. It's *that* simple.
> Nice, isn't it?
>
> Dirk Vdm
>
>
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