Re: SR fundamental contradiction

mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
>From x1 = vT and x2 = cT, values of the endpoints of the stick in S,
that are admittedly known, you now calculate x1' and x2' by using
the 'direct' LT

x1'=g(vT-vt')
x2'=g(cT-vt'), thus obtaining
Those equations are incorrect. I used 'inverse' equations.
x1 = vT = g(x1'+vt')
x2 = cT = g(x2'+vt')

To obtain:
x1'=vT/g-vt'
x2'=cT/g-vt'
E.g.
Start with vT=g(x1'+vt')
Divide by g: vT/g=x1'+vt'
Subtract vt' from both sides: vT/g-vt'=x1'
Swap sides to obtain: x1'=vT/g-vt'

Any questions?

(Note that these equations give <x2'-x1'=(c-v)T/g> at all times t'.)
No, only a couple of remarks:

>From the direct transform, one gets x1'=0, so, in
your relation x1'=vT/g-vt', vT/g = vt', thus t' = T/g.

And indeed, from the 'direct' time transform
t' = g(T-vx/c^2), one gets
t' = g(T-vT*v/c^2)
= gT(1-v^2/c^2)
= T/g

>From the 'direct' transform, one also gets
x2' = g(x2-vT) = g(cT-vT) = g(c-v)T
x2'-x1' = g(c-v)T
L' = gL
No, that is not correct. You get
x2'=g(cT-vt)

You got cT=g(x2'+vt')

According to the 'direct' time transform,
t' = g(T-cT*v/c^2),
= gT(1-v/c), hence, by substitution in your formula
No, the direct transform (when x2=cT) yields
t'=g(t-cT*v/c^2)
=g(t-vT/c)

From this equation, and t'=T/g (with which you agreed above),
you get
T/g=g(t-vT/c)
T/g^2=t-vT/c
T(1-v^2/c^2)=t-vT/c
t=T(1-v^2/c^2)+vT/c

Substituting this value into x2'=g(cT-vt) gives
x2'=g(cT-vT(1-v^2/c^2)-v^2T/c)
=g(cT(1-v^2/c^2)-vT(1-v^2/c^2))
=g*(1/g^2)*(c-v)T
=(c-v)T/g

(You can see why I chose to work with the inverse transforms.
With them, we don't need to calculate the values of t).

Your fundamental error seems to be that you assume that "the
same time" in S' is the same as "the same time" in S. It is
important to use the values of the endpoints "at the same time"
in S', if you want to measure the length according to S'

If x1=vT, and x2=cT, for all times in S, then x1'=0 and
x2'=g(c-v)T occur at different times in S'.

Those times are given by the direct transform as

t' = T/g (from x1 = vT)
t' = g(t-cT*v/c^2)
= g(t-vT/c) (from x2 = cT)

Let's call those times in S'

t1' = T/g
t2' = g(t-vT/c)

Boldly equating t1' to t2', you got

t = T(1-v^2/c^2)+vT/c
= T/g^2 + vT/c,
t2' = g(T/g^2 + vT/c - vT/c)
= T/g

You had to assume that t1' = t2' = T/g to get the correct solution
L'=L/g.
Yes! You finally got it.


Notice that from the 'inverse' transform t = g(t' + vx'/c^2), one gets
- from x1' = 0,
t1 = gt'
- from x2' = (c-v)T/g,
t2 = gt' + gvx'/c^2
= gt' + v(c-v)T/c^2,
hence t1<>t2 for a same value of t'.
Correct.

So, if x1'=0 and x2'=(c-v)T/g, for all times in S', then x1=vT
and x2=cT occur at different times in S

With the Einsteinian LT, there is no coherent solution.
Can you explain that statement? We have obtained a coherent solution, and you followed the derivation.


The correct LT is
x' = (x-vt)/g
t' = t/g
What is your derivation?
Have you noticed that if you apply your equations, first with v, then with -v (which should get you back to the original system), your time equation results in
t''=t/g^2
It should be
t''=t

More generally, your equations only represent a group when g=1.
.

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