The Lorentz transformation (LT) are false 

Proof:
_______

INTRODUCTION
____________

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. 

Let's count time from the instant at which the origins of coordinates,
O and O', momentarily coincide. At that instant, let's assume that
t = t' = 0.

Let an object start from the coincident origin at t = t' = 0, with a 
constant velocity V, the object's clock then reading t0 = 0. 
 
Let's note that there is absolutely no physical relation between 
the motion of the frame S' and that of the object.

Lets call x the position, according to S, of the origin of S' after 
a time interval t, and X, also according to S, the position of the
object after the same interval t. Then, of course, x = vt and X = Vt.
After the time interval t, the clock of S' will read t', whereas the 
object's clock will read to.
According to S', the object will be at X' = V't'. 

In order to obtain the relation between t and t', according to S,
let's consider that a light signal, starting from the coincident origins
at time t = t' = 0, follows 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 = ct* sqrt(1 - v^2/c^2).
As y = y' = ct',
 
t' = t * sqrt(1 - v^2/c^2).
  
As the motion of S' is physically independent from that of the 
object, the above relation between t and t' is always true.

However, it is possible to mathematically link the times t and t'
to the time to of the object:

Indeed, after a time interval t, 

to(S) = t * sqrt(1 - V^2/c^2), according to S, and 
to(S') = t' * sqrt(1 - V'^2/c^2), according to S'.

As V' = (V - v) / (1 - Vv/c^2),
this last relation can be written

to(S') = t' * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2).

By assuming that to(S) = to(S'), one gets
t' = t * (1 - Vv/c^2) / sqrt(1 - v^2/c^2), or, as X = Vt,

t' = g(t - Xv/c^2), where g = 1 / sqrt(1 - v^2/c^2), the famous
Einstein's gamma.

Stupendously, we got Einstein's (Lorentz) time transformation!

But to(S) is generally different from to(S'). 

Only when V = v are they identical !
Then, to(S) = t * sqrt(1 - v^2/c^2)
      to(S') = t', and, as expected 

t' = t * sqrt(1 - v^2/c^2).

On the other hand, as V = v, V' = 0, and from X' = V't', one gets

X' = 0.

This is the first proof of the falseness of the Lorentz transformation.

Einstein's derivation of the LT:
___________________________________________

After supposing that two frames of reference, S and S', are each
in uniform translatory motion relative to the other, the velocity
of S' relative to S being v,

1) Einstein began his derivation with the relations 
   (1) x' = ax + bt, and 
   (2) t' = ex + gt

2) Then he claimed that at the origin of S', x' = 0 and x = vt.
   Hence, 0 = (av+b)t, whence b = -av

3) Now, he supposed 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' = ct', since the speed of light is the same in
   all frames.
   Substituting these values of x and x' in relations
   (1) and (2), and eliminating t and t', he found
   0 = ac + b - ec^2 - gc.
   If the signal travels toward negative x, x = -ct and x' = -ct', thus
   0 = -ac + b -ec^2 + gc.
   Hence, a = g and b = ec^2 (or e = b/c^2).

Comments:

Einstein has mixed up two different positions reached after the time
interval t, that of the frame S' (x) and that (X) of some object moving 
at V, in this case the light signal travelling at c. 

In fact, his basis relations should have been written

    (1) X' = aX + bt
    (2) t' = eX +gt

Let's remember that 

X (the position of the object wrt S) = Vt and 
X' (the position of the object wrt S') = V't'.

Assuming that X = vt,

X' = 0, because the frame S' is also at the distance vt relatively to S.
Hence, 0 = (avt +bt), whence b = -av

In the third step, X = ct, and X' = ct' on one hand, and X = -ct
and X' = -ct' on the other hand.
The same results are obtained, i.e.
a = g and b = ec^2 (or e = b/c^2).

After the third step, one gets a = g and b = -gv = ec^2.
Hence, e = -gv/c^2.
Substituting these constants in

(1) X' = aX + bt, and 
(2) t' = eX + gt, one obtains

(1') X' = gX - gvt = g(X-vt), and 
(2') t' = gt - gvX/c^2 = g(t-Xv/c^2) 

Let's note that since X = Vt, 

V' = X'/t' = gt(V-v) / gt(1 - Vv/c^2) = (V-v) / (1 - Vv/c^2) 

4) 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(ev + g) * t. 
  
   Equating y' to y, he found

   c(ev + g) = sqrt(c^2 - v^2) = c * sqrt(1 - v^2/c^2), and claimed:

   "Since, by prior results, e = b/c^2 = -av/c^2 = -gv/c^2,
   it follows that cg(1 - v^2/c^2) = c * sqrt(1 - v^2/c^2)
   and g = 1/sqrt(1 - v^2/c^2).

   All constants thus being determined, the relations (1) and (2)
   can be written 
   x' = g(x - vt) and t' = g(t - vx/c^2)."

Comments:

As stated in the above introduction, the motion of S' is physically 
independent from that any object moving at V, so that the relation 
t' = t * sqrt(1 - v^2/c^2) is always true.

Claiming that the y'-axis advances a distance vt is equivalent
to saying that after a time t, the frame S' arrives at the position
x = vt. This is of course correct, but simultaneously, the object
reaches the position X = Vt, what Einstein simply overlooked.

By equating
y' = ct' = c(ex + gt) to y = sqrt(c^2 - v^2) * t, 
Einstein made the big logical mistake of mixing up the two positions x and X.
Instead of t' = ex + gt, he should have used the relation t' = eX + gt,
thus getting c(eX + gt) = ct(eV + g) = ct * sqrt(1 - v^2/c^2), and,
after simplification, eV + g = sqrt(1 - v^2/c^2).
By replacing e by -gv/c^2, he would have obtained

g(1 - Vv/v^2) = sqrt(1 - v^2/c^2), and

g = sqrt(1 - v^2/c^2) / (1 - Vv/v^2).

As shown above, t' = g(t-Xv/c^2). Replacing X by Vt, and g by
its correct value sqrt(1 - v^2/c^2) / (1 - Vv/v^2), one gets
the correct result

t' = gt(1 - Vv/c^2) = t * sqrt(1 - v^2/c^2).

As X' = V't', the correct position relation is given by
X' = [(V - v) / (1 - Vv/c^2)] * [t * sqrt(1 - v^2/c^2)], which is
equivalent to

X' = [(X - vt) / (t-Xv/c^2)] *  [t * sqrt(1 - v^2/c^2)]
 

Einstein's g = 1 / sqrt(1 - v^2/c^2) corresponds to the special
case, where V = v. Then,

X' = 0
t' = t * sqrt(1 - v^2/c^2).

Conclusively, Einstein's LT, that he called the Lorentz transformation,
are, at best, the result of a logical error, and at worst, a HOAX. 
(Cf. also "There is no length contraction" ).


Marcel Luttgens

May 23, 2003.

muttgens@spamorange.fr

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