Re: Is there length contraction in SRT, uncle Ben?

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On Jul 30, 1:00 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
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On Jul 30, 5:06 am, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
xray4abc <lemhen...@xxxxxxxx> wrote in message

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What SRT does say about length contraction?
Consider a rod along OX axis be resting in IRF K and
moving in frame K’.
Consider 2 observers in the 2 frames, measuring the
length of the rod, setting up measurements at their will,
where they localize the endpoints in a simultaneous
manner, each in his frame.

In frame K :
Measured values are: X1, X2 in moments T1=T2
Calculated values, from Lorentz transformations are
X’1 and X’2
which give L’=L* Gamma
where L’ is the calculated length for K’
and L is the measured (and at the same time the proper) length
of the object in frame K.
As Gamma>1
We get L’>L that is : the length attributed to be valid
for frame K’ (the moving frame) IS BIGGER than the measured length
(that is the proper length)
( So far nothing new!)

Alas, you are completely wrong here.

If the rod is at rest in K (T1=T2), then it is not at rest in K',
so the value L’=L* Gamma is not the calculated length for
K’ since the times of measurement not the same and the
rod is *moving* in K'.

Then, what is it L'=L*Gamma by your opinion?
One uses simultaneous marking of the endpoints of the rod
no matter in which frame he is in.

When the thing is moving, you must use simultaneous measurements
to be able to substract the distances and call it the "measured length".
When it is not moving, it doesn't matter whether the measurements
are simultaneous

It does, when you need the measured X1 and X2
coordinates
in the Lorentz transforms, to calculate the length applicable for the
moving frame ! This is the scope of the LT after all, to let you
calculate coordinates valid in one inertial frame when they are known
in
another one.

See my other reply.

Dirk Vdm

.

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