 AAF
 Re: On the Motion of the Earth Relative to the Aether  November 2 2017, 12:00 AM 
Of course, Einstein's special relativity & Maxwell's wave theory agree
that the speed of light, relative to its source, is, always,
equal to c, in vacuum.
But that is true, ONLY, in the case, in which the light source
is assumed to be at rest relative to Maxwell's aether!
Is that amazing or what?
Einstein's special theory says that there is no aether.
That is on one hand.
On the other hand, Maxwell's electromagnetic theory says that the speed of light,
relative to its emitting source, will be equal to 299792458 m/s,
if and only if the same emitting source is at absolute rest
relative to the aether.
In short, Maxwell's idea about the constant speed of light is quite different
from Einstein's idea about the constant speed of the same light.

 AAF
 Re: On the Motion of the Earth Relative to the Aether  November 14 2017, 12:00 AM 
As pointed out already, the entire calculations, by Michelson & Morley,
for their experiment, have used nothing else beside the speed of light
relative to the light source, such as (c + v) & (c – v), and the speed
of light relative to the observer such as (c + v) & (c – v), as well.
Let's, now, take a closer look at the MichelsonMorley calculations of the total flight
time of light in the vertical direction of the earth's motion, from the light source
to the transversal mirror, and then from that mirror to the detector.
And once again, the math, here, is quite simple:
If the light beam travels with a speed, c, at right angles to the direction,
in which the earth is traveling at a speed, v, THEN, according to the
Maxwellian assumption, on the basis of which the speed of light is independent
of the speed of the light source, the speed of MichelsonMorley beam relative
to the transversal mirror is equal to
[c^{2}  v^{2}]^{0.5}.
And therefore, the total travel time of the transversal beam
is equal to T_3; i.e.,
T_3 = 2L / [c^{2}  v^{2}]^{0.5}
And that is the final result of the calculations, by Michelson & Morley,
in the case of all light beams traveling in the transversal direction.

 AAF
 Re: On the Motion of the Earth Relative to the Aether  November 18 2017, 12:00 AM 
According to the assumption that, the speed of light is independent of the speed
of the light source, therefore, the MichelsonMorley beam of light takes an interval
of time T_3 to travel a total distance 2L from the light source
to the moving transversal mirror, and from that mirror
to the moving detector; i.e.,
T_3 = 2L / [c^{2}  v^{2}]^{0.5}
Nonetheless, if it's assumed that Earth is at rest, then the same light beam takes
an interval of time T_0 to travel a distance 2L from the light source
to the horizontal mirror, and from that mirror
to the moving detector; i.e.,
T_0 = 2L / c.
And so, the time interval T_3 is longer
than the time interval T_0; i.e.,
T_3 > T_0.
But why is T_3 is greater than T_0?
It's, clearly, because the speed of light relative to the moving
mirror, [c^{2}  v^{2}]^{0.5}
is less than c; i.e.,
[c^{2}  v^{2}]^{0.5} < c.
It's quite clear and simple.
NOTE:
In this equation:
T_3 = 2L / [c^{2}  v^{2}]^{0.5}
the Factor Gamma makes its first appearance, this way:
T_3 = 2L/c x [1  v^{2}/c^{2}]^{0.5}.
And by merely looking at the above equation, George Francis FitzGerald
stumbled upon his idea of length contraction:
https://en.wikipedia.org/wiki/George_Francis_FitzGerald

 AAF
 Re: On the Motion of the Earth Relative to the Aether  November 22 2017, 12:00 AM 
As demonstrated earlier, the amount of time, T_3, is greater than the amount of time,T_0,
because the speed of light relative to the moving mirror,
[c^{2}  v^{2}]^{0.5} is less than c.
Now, if the moving mirror is replaced with a moving observer, will, in this case,
the moving observer measure [c^{2}  v^{2}]^{0.5},
as the speed of light relative to him/her; OR will he/she find out
the relative speed of light is equal to c?
According to Maxwell's theory of electromagnetic radiation, the moving observer,
in this case, must always find out that the speed of light relative to him/her is
always equal to [c^{2}  v^{2}]^{0.5},
and nothing else.
Does that conflict, directly or indirectly, with the Maxwellian assumption, which
states that, the speed of light is independent of the speed of the light source?
The short answer is no.
The speed of light, [c^{2}  v^{2}]^{0.5}, relative to the
moving transversal observer does not conflict, directly or indirectly, with the
Maxwellian assumption, according to which the speed of light is independent
of the speed of the light source.
To the contrary, the relative speed of light, [c^{2}  v^{2}]^{0.5},
is a direct and necessary consequence of the Maxwellian assumption that,
the speed of light is independent of the speed of the light source.
 
   

