THE EXPERIMENT GL

 THE EXPERIMENTAL APPARATUS (fig. 3)

D = platform (D).
Φ = light source of wavelength λ.
L = light beam, from the light source Φ.
M_o, M'_o = flat mirrors (or prism T) for seperation of the light beam L in two light beams L₁ and L₂.
S₁, S₂ = closed tubes full water.
S₃, S₄ = closed tubes full benzol.
M₁, M₂, M₃,M'₁,M'₂,M'₃ = flat mirrors.
P = observer on platform (D).
P' = observer standing stationary onto the surface of Earth.
K₀ = central light fringe.
 

THE EXPERIMENT. 

     The experiment GL, is executed as follows:

Phase I. The platform (D), is stationary relatively to an observer P' standing
               stationary, onto the surface of Earth.

     In this case, the light source Φ, emits a light beam L, which then, falling onto the prism T, is divided in two light beams L₁ and L₂.
     The light beam L₁ coming through the water of the tubes S₁, S₂ and the light beam L₂ coming through the benzol of the tubes S₃, S₄ falling onto the mirrors M₁, M₂, M₃ - M'₁, M'₂,M'₃ respectively, are reflected.

     As a result, the above beams L₁ and L₂ interfere forming vertical rectilinear light and dark interference fringes, in the dioptre.
     During this phase of experiment, the vertical thread of the crossthread of dioptre is placed right on middle K_o of the central light fringe.

Phase II.  The platform (D), has uniform motion, with constant velocity υ,
                 relatively to an observer P', standing stationary, onto the surface
                 of the Earth.

     In this case, for the observer P, the total time t₁ for the light, from light source Φ to the middle K_o of central light fringe, through the tubes S₁ and S₂, will be:

   Where:

     In the same way, for the observer P, total time t₂ for the light, from light source Φ to the middle K_o of central light fringe, through the tubes S₃and S₄, will be:

   Where:

From the relations (1) and (2), we get:

     The relation (3) means that, in this case (phase II), the observer P, will observe a shift δ of the interference fringes, which is equal to:

As it is Known, the Theory of Relativity, during the execution of the experiment GL, fig.3 accepts that:
In phase II of the experiment for the observer P the shift δ of the interference fringes, will always be δ=0 fringes.
But is this correct?

In my personal opinion, should the experiment be carried out, observer P will observe in phase II a shift δ of the interference fringes δ≠0.

As a result, if during the execution of the experiment (phase II) the observer P should observe a shift δ of the interference fringes δ≠0, then the whole philosophy and the entire axiomatic foundation of the Special Theory of Relativity is wrong.

EXAMPLE

Let us suppose our experiment is carried out given the following:

Substituting the above values (7), (8), and (9) in the relation (3), we get:

Δt  ≈ 1,333333 . 10^-10  sec     (10)

and from the relation (6), we have a shift δ, which is equal to:

for the observer P.
Conwersely, according to the Theory of Relativity for the observer P the shift δ of the interference fringes, will always be δ=0 fringes.
Consequently, which is right, δ=0 fringes or δ=75.471 fringes?

In my personal view δ= 75.471 fringes is right and not δ=0 fringes.

NOTE: The letters G and L, are the initials of the English words “Greek Light”.

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