Lorentz contraction and simultaneous detection of the ends of the moving rod?

[bernhard.rothenstein]

It is known that with the formula that accounts for the time dilation effect in hand we can derive directly the formula that accounts for the length contraction effect:
L₀/Dt=L/(Dt)₀ (1)
where L₀ and (Dt)₀ are proper length and proper time intervals, L and Dt representing measured length and coodinate time interval. From (1)
L=L₀(1-V²/c²)^1/2 (2)
Is there a special reason (Ockham's razor) for deriving (2) involving the Lorentz transformations and to perform a simultaneous detection of the space coordinates of the moving rod?
I find in the literature of the subject derivations of (2) considering that the two ends are detected at different times. Does the simple derivation (2) involve simultaneous detection of the ends of the rod?
Thanks for your answer.

 

[Meir Achuz]

Your Eq. (2) requires simultaneous detection of the ends of the rod.
The probable motivation for this is that simultaneous detection works in non-relativistic physics. Why should it be expected to work in SR?

 

[bernhard.rothenstein]

Your Eq. (2) requires simultaneous detection of the ends of the rod.
The probable motivation for this is that simultaneous detection works in non-relativistic physics. Why should it be expected to work in SR?

Thanks for your answer. Let K₀ be standard synchronized clocks located along the x-axis of the I inertial reference frame. Let K'₀ a clock of the I' inertial reference frame located at its origin O' moving with velocity V relative to I. Observers from I measure the velocity of the moving clock using a rod of proper length L and the clocks K₀ and K located at the two ends of the rod respectively reading t=0 and t respectively when the moving clock passes in front of them. By definition the speed of the moving clock is
V=L₀/(t-0) (1)
measured as a quotient between a proper length and a coordinate time interval.
In a second experiment an observer R' located at the origin O' of I' uses clock K' as a wrist watch and measures the velocity of the moving rod used in the previous experiment. He detects the presence of the moving rod in front of him during a proper time interval (t'-0) measured as a difference between the readings of his wrist watch when the two ends of the rod cross his location respectively and measures a length L of the rod different from L₀. By definition
V=L/(t'-0) (2)
the detection of the two ends being not simultaneous!
From (1) and (2) and taking into account the formula that accounts for the time dilation effect we receover the formula that accounts for the length contraction effect
L=L₀(1-V²/c²)^1/2.

 

[Meir Achuz]

Sorry, I am better at algebra than lengthy discussion. I can't follow your discussion and don't have time to try. I hope someone else can help you.