Re: SR fundamental contradiction


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.
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.


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.




Thank you for your pertinent analysis.
Contrary to those critics, who transformed this NG into a hornet's
nest,
you adopted a positive attitude, the only one that makes scientific
progress possible.
Anyhow, I probably had a big headache when I wrote my derivation.
Here is the revised version:

Derivation of the correct transformation:
----------------------------------------

Let's consider two frames of reference, S and S', each in uniform
translatory motion relative to the other, the velocity of S' relative
to S being v.

Basis relations:

x' = ax + bt
t' = ex + kt

At the origin of S', x' = 0 and x = vt.
Hence, 0 = (av+b)t, whence b = -av

The basis relations are now

(1) x' = a(x - vt)
(2) t' = ex + kt

Now, let's suppose that a light signal, starting from the coincident

origins of frames S and S' at t = t' = 0, travels toward positive x.
After a time t, it will be at
x = ct, and also at
x' = (c-v)t'
(Notice that Einstein postulated here that x' = ct')
It appears that, again, you are assuming a vector sum for the
velocity vector.

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'.

I would like to make clearer why, after a time t, the light signal will
be at
x' = (c-v)t' in S'.
Indeed, according to S, the distance between the endpoint of the signal
and the origin of S' is ct - vt after a time t. In S', the
corresponding distance
is (c-v)t', hence the S'-coordinate of the endpoint is x' = (c-v)t'.
In S, the endpoint is of course at x = ct.
But I am convinced that you already grasped what I meant.

Marcel Luttgens





Substituting these values of x and x' in relations
(1) and (2), one gets:

(c-v)t' = a(c-v)t, thus t' = at
t' = (ec+k)t, hence

(3) ec + k -a = 0

If the signal travels toward negative x,
x = -ct and x' = -(c+v)t'
(Here, Einstein postulated that x' = -ct')

Substituting these values of x and x' in relations
(1) and (2), one gets

-(c+v)t' = a(-ct-vt) = -a(c+v)t, thus t' = at
t' = (-ec+k)t, hence

(4) -ec + k - a = 0

From (3) and (4), one gets e = 0 and a = k

Hence, relations (1) and (2) become

(1) x' = k (x - vt)
(2) t' = k t

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.

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

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),

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

As the MMX has shown that the moving stick is in fact *contracted*,
one has to conclude that the Einsteinian LT is false.

Marcel Luttgens

.

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