Re: SR fundamental contradiction

Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
Of course, the measuring instruments are also moving with S', so
they will be contracted as well, and S' will not measure any
change of length.
You misunderstood, I didn't refer to the constancy of the length of
the stick at rest in S' and measured in S'. I clearly said " in S, its
length is L, in S', it is L'=L/g", meaning that the length of the stick
measured in S' is contracted wrt its length measured in S.
Using your equations, we conclude that the length of the stick
is independent of is speed for any observer. We do find that
its length depends on the speed of the observer, but that the
observer will not be able to measure the change. Of what use
are your equations?

Quibbling again, unless you didn't understand what I said.
Apparently, you don't understand your equations.

It is clear that I meant that the length L' of a moving stick is
contracted by g
relative to a stick of length L at rest. In S', the stick is at rest,
so its length is L.
OK. In S', its length is L, so we have for the endpoints:
0=(x1-vt)/g
:: x1=vt

L=(x2-vt)/g
:: x2=gL+vt

x2-x1=gL.
Conclusion: The moving stick is measured to be longer in the S
system than in its rest system.


You really don't understand what you are speaking about.
Till now, one considered that S' -and the stick- are moving away
at v from the S-frame, hence L'=L/g. Knowing L', the length of the
stick
in S', one conclude that the corresponding length in S is L=gL'.
Now L' is called L by you ("In S', the stick is at rest, so its length
is L")
so L must of course be called L'. Hence, its length in S is L'=gL,
according to your new terminology.
I am becoming rather annoyed by your sophistry.
..

What is the original frame? I presume that you mean the rest frame.
Yes, in S, the velocity ot the light signal following the y'-axis is
sqrt(c^2+v^2), but it remains c in S'.
Previously, you stated that the speed of the light signal in the
rest system was 'c'. Which is it?
Then I was inattentive, sorry.
Are you now saying that, in S, you have <y=ct>?

Can't you read? I just said that in S, the velocity ot the light signal

following the y'-axis is sqrt(c^2+v^2), but it remains c in S'.
Hence (cf. my derivation), y = sqrt(c^2 - v^2) * t and y'=ct'.

Your math fails. If y=sqrt(c^2-v^2)t and x=vt, then the speed
of the signal is c, not sqrt(c^2+v^2). OTOH, that is the
correct speed, under your assumptions, if y=ct and x=vt.

This has nothing to do with math, it's a matter of geometry.
According to S, the signal travels *obliquely*. Apply the Pythagorean
theorem, and you will get y=sqrt(c^2-v^2)t.
I am now more than annoyed by your unceasing quibbling.
Unless quibbling is not the right word, it could be incompetence
instead.

OK. I will do it again slowly.
The transformation from S' to S", with speed <u-v> is
x"=a(u-v)(x'-(u-v))t
t"=a(u-v)t'

You don't seem to understand the meaning of the transformation
x'=(x-vt)/g
t' = t/g,
where x is the position of some object in frame S,
v is the velocity of a frame S' relative to S,
x-vt is the difference in S between x and the origin of frame S'
after a time t,
x'=(x-vt)/g is the coordinate of the object in S'.
Correct, but I will continue to use the multiplier a(v), because
using 1/g implies something that you have not demonstrated.


In the former discussions, we used x=ct, but you can of course
choose x=ut. Then x'=(u-v)t/g. But notice that x' is the coordinate
of the object in S'.
I demonstrated before that, if x=ct implies x'=(c-v)t', x=ut
implies x'=(u-v)t' for any u. And yes, it is understood that
x' is the coordinate in S', and x is the coordinate in S, and x"
is the coordinate in S".



Now you consider the frame S" of the object, and claim that
the transformation from S' to S", with speed <u-v> is
x"=a(u-v)(x'-(u-v))t"
t"=a(u-v)t'

This is plainly wrong, because the coordinate of the object
in S" is *zero*, thus x"=0.
What is wrong? I used your equations.
Why would the be wrong when x"=0, if your equations are correct
when x'=0?

If you had understood my equations, you would have concluded
that the way you use them is irrelevant, as your S" is nothing more
than another name for the object. When you use sophistry, you should
at least show some intelligence.

Marcel Luttgens



Your demonstration must be a hoax!
Where is the error?


Substituting the equations we already have for the
transformation from S to S':

x"=a(u-v)(a(v)(x-vt)-(u-v)a(v)t)
=a(u-v)a(v)(x-vt-(u-v)t)
=a(u-v)a(v)(x-vt-ut+vt)
x"=a(u-v)a(v)(x-ut)

t"=a(u-v)a(v)t


The direct equations give:
x"=a(u)(x-ut)
t"=a(u)t
????????
These are simply your equations for the case v=u, giving the
transformation from S to S".


Consistency requires that a(u-v)*a(v)=a(u).
This is a functional equation, that must be satisfied by a, for
all values of u and v.

Setting u=0, we find that a(v)*a(-v)=1. This gives a(v)^2=1, or
a(v)=1 (the negative root is excluded, because a(0)=1).
Setting u=0, we have
x'= (u-v)t'= -vt'
I don't know if you were trying to advance your objection, or
just posting an equation. Was there a point?

t'= t/g
You ignored the functional equation, and substituted a=1/g,
without any justification. (Of course, the functional equation
can be expressed in terms of g, and, after inverting, gives
g(u-v)g(v)=g(u), leading to g=1.)


We do not need to examine the y direction to conclude that your
transformation must reduce to:
x'=x-vt
t'=t .


.

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