Re: SR fundamental contradiction


Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
No, if we choose t2=T, then we get
T=t1-v^2T/c^2+vT/c
t1=T+v^2T/c^2-vT/c

x1'=g(vT-vt1)
=g(vT-vT-v*v^2T/c^2+v*vT/c)
=g((c-v)(v^2T/c^2))
x2'=g(cT-vt2')
=g(cT-vT)
=g(c-v)T
Subtracting, we get
x2'-x1'=g(c-v)T(1-v^2/c^2)
=(c-v)T/g

This is the same result we got when t1=T, so the solution is
still coherent.
Mathematically coherent, in the sense that when t1'=t2', you
always get x2'-x1' = (c-v)T/g when x1=vT and x2=cT, as long as
you use the relation t2-t1 = T(v/c - v^2/c^2).
But notice that with t2=T, you get x1'=g((c-v)(v^2T/c^2))
instead of zero, which is physically false, because when x1=vT
and x2=cT, the origin of S' is at x1=vT and at x1'=0.

The stick is at rest in S, so x1=vT and x2=cT at *all* times in
S. The origin of S' is at x1=vT only when t1=T. When t2=T, and
t2'=t1', then x1<>vT, because t1<>T

In the first scenario, the light arrived at ct and S' arrived at vt
after some
specific time t.
Where does the light come in? The ends of the stick are at vT
and cT at *all* times in S.
The origin of S' is at vT only when t=T (and t'=T/g).


Yes, that was intended in the scenario, reread the first posts of the
sadomaso,
even *he* had no doubt about this. He was the first to specify "at some
specific
time", which means, of course, t = T.


From x' = g(x-vt), x'1 = g(x1-vt), hence x' =0 as x1=vt..
Aftrwards, you complicated the problem by introducing t1 and t2.

No, you complicated the problem by ignoring them, and expecting
S' to mark the ends at two different times.

S and S' measure time differently.


Of course, in the general case, but not in the scenario.


Only setting t1=T leads to a physical coherent result (perhaps
would it be better to say physically 'correct').

No, that would be incorrect, because you are asking S' to mark
ends of the stick at two different times. In between the stick
moves, so you cannot determine the length.


Following the original scenario, the only correct result is obtained
with t1 = T. But I agree that from t2-t1 = T(v/c - v^2/c^2), one gets
an infinity of other solutions.

To measure the stick in S', you must mark the ends at the same
time in S'. You get the correct result for any single value of
t'.

And the correct result is straightforwardly given with my
transformation.



Not at all, the factor (c-v) doesn't mean that light speed
is not c anymore. According to S, light is at ct and S' at vt,
hence in S', after a time t, light has travelled a distance
(c-v)t. To express such distance in S', one has to use t', thus
x' = (c-v)t'.

I don't follow.
You state that:
x'=(c-v)t
and
x'=(c-v)t'

From these two equations, we find that t'=t.


I thought that my sentence was clear enough: According to S, etc...
Thus, I stated that x = (c-v)t and x' = (c-v)t'.
If you look at the text quoted above, you stated > in S', after
a time t, light has travelled a distance
(c-v)t.
Does that not say that x'=(c-v)t?

The answer is given in mine (and your) almost simultaneous post.

Now, a light signal follow the y' axis. Relatively to S,
it travels obliquely, for, while the signal goes
a distance ct, the y'-axis advances a distance x = vt.
Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
But, also, y' = ct' = c * k * t.
I don't see where you obtained y'=ct'. If your are using a
vector sum for the velocity, you obtain y'=sqrt(c^2-v^2)t'
Fistly, I don't use a vector sum (see above), and secondly,
there is no physical reason why y' should be different from y,
at least according to Einstein, with whom I agree here.

Then where do you get y'=ct'?

Einstein got that equation from his light postulate. Because of
that postulate, a light signal travelling along the y' access
must have speed 'c'.

You reject that postulate, so you cannot use it here.

I don't reject the hypothesis that the speed of light is independent
of the motion of its source.

That part of the postulate was not used to derive y'=ct'. The
primary assertion of the postulate is that the speed of light is
a universal constant; it has the same value for every observer,
in every direction.

I agree that you accurately quoted the fanciful Einstein's postulate.
I disagree with your first sentence.


You have not explained how you derived y'=ct'.


y' = ct' is a direct consequence of the hypothesis that c is
independent
of the motion of S'.


But I keep claiming that the Einsteinian LT are false:

Here again is a classical explanation why a moving stick appears
contracted to an observer at rest.
Remember that you considered it as correct.

BEGIN

Consider a stick which, when at rest in S, has a length Lo in the
direction of the x-axis.

Let the stick be set moving relative to S at such velocity that it is
at rest in S'. Its length as measured in S' will still be Lo,
because it must have a certain fixed value in any frame in which
the stick is at rest.

Let us see how the length now measure in S, relative to which the
stick is moving with a velocity v.
It seems reasonable to define the length as the distance between
two points fixed in S, which are occupied by the ends of the stick
simultaneously, i.e., at the same time t.
If the coordinates of these points are x1 and x2, the length is
then L = x2 - x1.

Since the stick is at rest in S', its ends have fixed coordinates
x1', x2' such as Lo = x2' - x1'.

If one substitutes in this last equation values of x2' and x1'
calculated from the LT x' = g(x - vt), one obtains, for a given
value of t,

x2' = g(x2 - vt)
x1' = g(x1 - vt)

Lo = g(x2 - x1) = gL, or L = Lo/g

END

Such demonstration amounts to sophistry.

Indeed, it began by calling Lo the length of a stick at rest in S,
whose end coordinates are x1 and x2. Now, it calls deceitfully
that length L.

No, it states that the length of the stick, in the system in
which it is at rest, is Lo.

Yes, it states that its length is Lo when it is at rest in S, and Lo
when it is at rest in S'.


To avoid confusion, such length should be called Lo(S).

Only if the stick is at rest in S.

And Lo(S') when it is at rest in S'. Notice that Lo(S') = Lo(S),
according
to the premise.



Moreover, it calls Lo the length x2'-x1', i.e. the length of the
stick as measured in S'. Again, to avoid confusion, let's call
such length Lo(S').

So, the end formula should be written

Lo(S') = gLo(S),

No, you stated that the stick was at rest in S', so Lo(S')
represents the rest length of the stick: Lo. You did not define
Lo(S) when the stick is in motion, but we will simply call it L.
Lo=gL
L=Lo/g

You should closely follow the reasoning:

"Let us see how the length now measure in S, relative to which the
stick is moving with a velocity v.
It seems reasonable to define the length as the distance between
two points fixed in S, which are occupied by the ends of the stick
simultaneously, i.e., at the same time t.
If the coordinates of these points are x1 and x2, the length is
then L = x2 - x1."

We agreed to call the length of the stick at rest in S, Lo(S).
Now, it is called L. Iow, L = Lo(S).

"Since the stick is at rest in S', its ends have fixed coordinates
x1', x2' such as Lo = x2' - x1'."

We also agreed to call Lo(S') the length of the stick at rest in S'.
Hence, Lo = x2' - x1' sould be written Lo(S') = x2' - x1'.

"If one substitutes in this last equation values of x2' and x1'
calculated from the LT x' = g(x - vt), one obtains, for a given
value of t,

x2' = g(x2 - vt)
x1' = g(x1 - vt)

Lo = g(x2 - x1) = gL, or L = Lo/g"

Logically, as Lo represents in fact Lo(S') and (x2-x1) represents
Lo(S),
Lo = g(x2 - x1) = gL should be written Lo(S') = gLo(S), meaning, as I
said,
that the length of a moving stick is *dilated* relative to the length
of the
same stick at rest.

I realize that it is no easy to detect the logical 'hocus pocus' that
Einstein had to use in order to sell his postulate that "that the speed

of light is a universal constant; it has the same value for every
observer,
in every direction."



meaning that, according to the Einsteinian LT x' = g(x - vt), a moving
stick is *dilated* relative to a stick at rest.

No, the stick is at rest in S', so it is moving in S. The
moving stick is shorter that at rest.

Are you an Einstein's integrist? I hope not.

Marcel Luttgens

.

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