Paul B. Andersen wrote:
> mluttgens@wanadoo.fr wrote:
> > *According to SR*, the speed of light is independent from
> > the motions of its source and of the observer. Iow, for the S observer,
> >
> > it is independent from the velocity of the interferometer, hence
> > the transit time of light along the arm which is parallel to
> > the x,x'-axis is L0*sqrt(1-v^2/c^2)/c, if one believes in the
> > SR length contraction. But for the arm, which is perpendicular to the
> > velocity vector, the transit time is L0/c, because no length
> > contraction
> > occurs, again *according to SR*.
> > Conclusively, the transit times of the light waves being different
> > along the two arms, a fringe shift should be observed, which is not
> > the case. Hence, the MMX proves the non-existence of length contraction
> > and moreover falsifies SR.
> >
> > Remember that the SR length contraction has been put forward by SRists
> > to explain the null result of the MMX, but those SRists forgot their
> > own postulates.
>
> Since you numerous times have been shown the correct
> SR calculation of the transit times in a moving MMX-interferometer,
> I find it remarkable that you can write something as naive as this.
>
> You have got the transit time along both arms wrong.
> The round trip time along both arms is: (2*L0/c)/sqrt(1 - v^2/c^2),
> and the derivation is simple - "if one believes in the SR length contraction".
>
> Paul

Paul B. Andersen wrote:

"You have got the transit time along both arms wrong.
The round trip time along both arms is: (2*L0/c)/sqrt(1 - v^2/c^2),
and the derivation is simple - "if one believes in the SR length
contraction"."

Lets first remember how Einstein derived 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).

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)."



Let's apply Einstein's LT derivation to the MMX:

Let S be the frame relatively to which the interferometer
is moving at v, and let's call S' the interferometer frame.

According to step 3 of the derivation, the speed of light
is the same in all frames, thus the round-trip transit time
of the light wave moving along the "horizontal" arm is
t(h) = 2*(L0/c) for both frames S and S'.

According to step 4, relatively to S, the "vertical" light wave
travels obliquely, for, while the wave goes a distance ct,
the y'-axis advances a distance x = vt.
Thus c^2t^2 = v^2t^2 + L0^2, whence
L0 = ct(v) * sqrt(1 - v^2/c^2), and the round-trip time is
t(v) = 2*(L0/c) * 1/sqrt(1 - v^2/c^2)

As the MMX showed no fringe shift, t(v) must be equal to t(h).
This is only possible if the length of the "horizontal" arm
is "dilated" by 1/sqrt(1 - v^2/c^2).
Indeed, if L0 becomes L0 * 1/sqrt(1 - v^2/c^2),
t(h) = 2*(L0/c) * 1/sqrt(1 - v^2/c^2)
     = t(v)

Or, according to SR, the "horizontal" arm measures shorter
in terms of the frame S relatively to which it is moving with
velocity v than it does in the frame S' of the interferometer
in which it is at rest, the ratio of shortening being
sqrt(1-v^/c^2).
As its length is L0 in the interferometer frame, it becomes
L = L0 * sqrt(1-v^/c^2) according to frame S.

We have just seen that the length "horizontal" arm must be
"dilated", not "contracted", in order to explain the null
result of the MMX. Thus, SR is falsified by the MMX.

Of course, "true crackpots" will not recognize this.

Marcel Luttgens