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

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On Jul 31, 4:41 am, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
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On Jul 30, 5:14 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
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On Jul 30, 5:06 am, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
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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 to substract the distances and call it the "proper length".

When you understand that, we can continue.
Do you understand that?

Dirk Vdm- Hide quoted text -

- Show quoted text -

Yes, sir!
Regards, LL

Okay, so I guess you understand that for an object at rest
in K (with proper length L), the quantity L' = L * Gamma
is as good as useless? It is the difference between the
distances of the *moving* object's end points marked at
different times :-)

Dirk Vdm- Hide quoted text -

- Show quoted text -

I knew exactly what L' was in the above relation.

Yes, a useless quantity.
Measure the distance to the front of moving train now.
Measure the distance to the back of the train 15 minutes later.
Calculate the difference.
Do go yet a meaningful quantity?

Are you suggesting that you can get the relation between the
*measured* value in frame K and the *measured* value in frame K'
for the rod at rest in frame K?
If yes, how you do it?

If the rod is at rest in frame K (with proper lenght Dx), it does
not matter whether the measuments to front and back are taken
simultaneously (i.o.w. Dt does not matter), but you must make
sure they are simultaneous in frame K', i.o.w. make sure Dt' = 0.
Then you can calculate Dx', the *measured* value in frame K'.

You take the transformation equations:
Dx' = g ( Dx - v Dt ) [1]
Dt' = g ( Dt - v Dx /c^2 ) [2]
with
g = 1/sqrt(1-v^2/c^2)
and of course the inverse
Dx = g ( Dx' + v Dt' ) [3]
Dt = g ( Dt' + v Dx' /c^2 ) [4]
together with the condition
Dt' = 0 [5]

The easiest way to go about is by combining [3] and [5] to get
Dx' = 1/g Dx

Another way is to first combine [2] and [5] to get
Dt = v Dx /c^2 [6]
and combine this [6] with [1] to get
Dx' = g ( Dx - v^2 Dx /c^2 )
= g ( 1 - v^2/c^2 ) Dx
= 1/g Dx

Dx is the mearured length in the rest frame K (aka the proper length).
Dx' is the mearured length in a frame K' (aka the coordinate length).
Coordinate lenght is shorter than proper lenght.

Dirk Vdm- Hide quoted text -

- Show quoted text


-Allow me some remarks regarding your
explanation, which is the standard one.

You can not get the measured value Dx’ from
LT !

If you measure Dx', then indeed "you can not get" Dx'
them from the LT. That's rather obvious.
But you can use the LT in the format
Dx = g ( Dx' + v Dt' ) [3]
Dt = g ( Dt' + v Dx' /c^2 ) [4]
to calculate Dx (and Dt) from Dx' (and Dt')
And then you can also measuse Dx, and compare it to
the calculated value.

The measured value L’ in K’ can only come from your
measurements of X2’ and X1’, the coordinates
of the endpoints through a procedure described
by you many times in this group, so it has nothing
to do with LT.

Indeed, and from X1' and X2' you can calculate X1 and X2
and predict what you would measure for them.

The same applies for frame K and value L.

Yes, from Dx (and Dt) you can use the LT in the format
Dx' = g ( Dx - v Dt ) [1]
Dt' = g ( Dt - v Dx /c^2 ) [2]
to calculate Dx' (and Dt') from Dx (and Dt)

Then, the LT has invariant form in the 2 IRFs.
The invariance of the form of LT, in itself, does not
guarantee that an invariance of the coordinate values
holds true in the 2 IRFs.

I never talked about an "invariance of the coordinate values".
There is no "invariance of the coordinate values".

(Meaning that both observers find the same set L, L’.

This doesn't make sense to me.

This invariance was admitted "by definition"
by Einstein and the whole physics establishment,
as far as I know.)
! At least I have never seen proven such a thing !
If you know differently, I would appreciate some
links to the right sources !
(One last thought… x^2-c^2*t^2 being an invariant
as well as form and value, Dx and Dx’ depend on
only 1 parameter each of them after all,
meaning that the temporal and the spatial coordinates
are not independent variables in the LT. )

There is one thing you seem to fail to understand.

Try to follow this:
If you use a formula like
L' = L / gamma
then this is valid only when L is the proper lenght of a rod at
rest in frame K, and L' is the coordinate length in the 'moving
frame" K'.

If you use a formula like
L = L' / gamma
then this is valid only when L' is the proper lenght of a rod at
rest in frame K', and L is the coordinate length in the 'moving
frame" K.

These are different situations - different rods.
Different equations for different rods.

Dirk Vdm


.

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