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:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
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.
It was your condition. If you go back and read your statement
of the conditions, you will find, "In S', the stick is at rest,
so its length is L."

No matter what you call the length, your equations predict that
a moving stick will be measured to be longer than in its rest
system.

If you can, use my equations correctly.
x' = (x-vt)/g
t' = t/g
For instance, x=cT, thus x'=(c-v)T/g.
Consider a stick of length (c-v)t in S, such as x1=vT and x2=cT
x1' = (vT-vT)/g = 0
x2' = (cT-vT)/g
x2'-x1' = (cT-vT)/g = (x2-x1)/g
L' = L/g, L' being the length of the stick in S'. It is obviously
contracted.
I will not repeat such elementary demonstration anymore.



.
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.
I am trying to find a rational reason to accept your <y'=ct'>.
It makes perfect sense if <y=ct>, or if we accept Einstein's
light postulate, but I don't understand where you get it.
Sometimes you assert that the light speed in S is c, sometimes
you assert that it is sqrt(c^2+v^2).

Your argument that the direction is perpendicular to the motion
does not work, because, in S', the velocity is zero. All
directions are perpendicular to a zero vector. In S, where the
velocity is directed along the x axis, the light signal is *not*
perpendicular to the motion.


I can't help you. How many time did I tell you and explained to you
that in S', y'=ct', and in S, y=sqrt(c^2-v^2)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.
S" is the frame moving with the speed u relative to S, just as
S' is the frame moving with speed v relative to S. It is also
true that, with your equations, S" is moving with speed u-v
relative to S'.

How did I misuse the equations?

You case is desperate. Apply my equations to the object, which has
been done more than once, then call the object S" if you want.
This change nothing to the result.

To conclude (for good) those discussions about the so-called relativity
of simultaneity, an Einsteinian hoax, I invite you to try to understand
the following point of view from

Ben Rudiak-Gould, Oct 26 2006, in the thread "question (length
contraction)"

"In Newtonian physics, there's no difference between you alone
accelerating
and the whole rest of the universe accelerating in the opposite
direction.
The only thing that matters is the relative distances and angles
between
different objects at each instant of absolute time, and those will be
the
same either way. It's easy to define a reference frame which gives all
distances and all angles at all times in terms of some rigid reference
body,
and it's just like an inertial frame except that there are extra
"fictitious
forces" whose sole purpose is to make sure that the rest of the
universe
accelerates in the proper way. Someone once suggested that Newton's
second
law should be changed to

F = (Mm/(M-m)) a

where m is the mass of the object that's apparently being acted upon,
and M
is the mass of the rest of the universe. This is indistinguishable from
F =
ma when m << M, and it has the nice property that the force needed to
accelerate an object is the same as the force needed to accelerate
everything else in the opposite direction.

In special relativity this doesn't work, because of the way that space
and
time are mixed together. If you naively try to define a global
reference
frame around an arbitrarily moving object, it has all kinds of
pathological
properties: in general there are surfaces where time stops, and regions

where time runs backwards, and branch cuts where the time coordinate of
a
point is discontinuous with the time coordinate of neighboring points,
and
regions of spacetime that don't have coordinates at all. The laws of
physics
become a mess, and you can't just patch up the situation by adding
fictitious forces.

So the answer to your question is that the contraction of the universe
when
you accelerate is an artifact of the pathological coordinate system
that
people have to use in order to claim that the universe contracts when
you
accelerate. The universe couldn't really accelerate like that. In fact
the
universe can't accelerate at all in special relativity, at least not
rigidly, because the rigid acceleration of an object of length L can
never
exceed c^2 / L."

Thank you nevertheless,

Marcel Luttgens




Your equations give directly:
x'=a(v)(x-vt)
t'=a(v)t

and

x"=a(u)(x-ut)
t"=a(u)t

I wanted the transformation from S' to S". Assuming that the
transformations will have the same form, I got:
x"=a(w)(x'-wt')
t"=a(w)t'

Where is the error? What did I misunderstand?

.

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