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

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

  cfdeda6f-9219-4338-9a08-13086221f...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx





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.
The rest frame of the rod must not make exception.
He has no other way to get the length, thought to be valid for
the other frame, than using the LT.
The fact, that the events of marking the endpoints are not
simultaneous in the other frame, does not matter.
They can not be! (According to SRT. Did you change your opinion
about that?) Based on his measurement and the LT the observer from
frame K
calculates an L' value HE attributes to the moving frame.

You don't measure the length on a moving train by subtracting
the distance to the front and the end of the train is these are
measured at different times.
If you say that the rod is at rest in K, then you must make sure
that T'1 = T'2. And *that* gives you L' = L / Gamma..

This tells you that measured length L' of a moving rod is shorter
than the "proper length"



In frame K’ :
   Measured values are : X’1, X’2  in moments T’1=T’2
  Calculated values, from Lorentz transformations are
   X1 and X2
    which give L=L’* Gamma
 where  L is the calculated length for K  (supposed to be the proper
length)
 and     L’ is the measured  length in K’.
  As Gamma>1
      we get L>L’ ,  that is : the length attributed to be valid
 for frame K  IS BIGGER than the measured length in K’ !!!!!!!!!!!!!!!

This ALSO tells you that measured length L' of a moving rod is shorter
than the "proper length"



Comments:

Do you understand your error?

No!
It seems that, by your opinion, no matter which observer
does the measurements, he has to make sure that in BOTH
frames the same time coincidence is considered, that is either
T1=T2 or T'1=T'2 .
Is that so?

Dirk Vdm- Hide quoted text -

- Show quoted text -

Regards, LL
.

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