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roger

Re: On the Motion of the Earth Relative to the Aether

December 28 2017, 7:45 PM 


 
 
roger

Re: On the Motion of the Earth Relative to the Aether

December 28 2017, 7:51 PM 

add on:

it would violate the principle of relativity if the light missed the mirror, because then you would be able to tell if you were moving with respect to an absolute frame i.e. if the light didn't miss the top mirror then you were stationary with respect to the absolute frame, but if the light did miss then you were moving with respect to the absolute frame. AND as per what experiment showed: for inertial motion you can't detect motion with respect to an absolute frame.




 
 
AAF

Re: On the Motion of the Earth Relative to the Aether

December 30 2017, 12:00 AM 











"michelson-morley vertically aimed light beam misses the mirror":
http://www.anti-relativity.com/forum/download/file.php?id=366
http://www.anti-relativity.com/forum/viewtopic.php?f=3&t=7112&start=30









GOOD POINT; ANANYM!






happy.gif






However, it's, absolutely, IMPOSSIBLE, in the Michelson-Morley experiment,
for the vertical light beam to miss the moving mirror, if Earth's velocity
is less than or equal to 30 km/s.


Why?



Well; it's, obviously, because Michelson & Morley used light, from the start,
to align the various parts of their apparatus with respect to each other;
i.e., they made use of electromagnetic radiation, in their experiment,
not once, but two times at once.


It's as simple as that!



wink.gif



And furthermore, according to the Michelson-Morley classical wave theory,
the inverse-square law forces the vertical light beam to expand,
with the square of distance, and to form bigger
& bigger arcs like these:


[linked image]



Accordingly, only the parts of the Michelson-Morley expanding vertical beam
that coincide with the two sides of this isosceles triangle are reflected
by the moving mirror and received by the moving detector:


[linked image]



Nonetheless, if the space motion of Earth is greater than 0.25c,
and if the classical wave theory is correct, THEN
neither the double use of light, nor the inverse-square law
of electromagnetic radiation, would make the Michelson-Morley
transversal beam catch up with the moving mirror.


And that is because:


d' = L/c x 0.25c = 0.25L,


is much GREATER than the actual length
of Michelson-Morley mirror.


And therefore, in that extreme case, the null result of the Michelson-Morley experiment,
is, for sure, entirely consistent with the classical wave theory.


There can be no doubt about that!





wink.gif
















 
 
AAF

Re: On the Motion of the Earth Relative to the Aether

January 3 2018, 12:00 AM 










It's true that Michelson & Morley used light, from the start, to align the various
parts of their apparatus with respect to each other; i.e., they made use of electromagnetic
radiation, in their experiment, not once,
but two times at once.


But how did they do it?


Well . . .



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Let's take, for example how Michelson & Morley, at the beginning of the beginning
- & not just at, perhaps, the end of the beginning – as Winston Churchill
used to say, made sure that the detector and the transverse mirror are, indeed,
at right angles to each other.


Firstly, in the reference frame of the moving detector, Michelson & Morley
received light reflected from the surface of the moving mirror, and used it
to make sure that the mirror image is, indeed, at right angles
to the moving detector.


And secondly, in the reference frame of the moving mirror, Michelson & Morley
received light reflected from the surface of the moving detector, and used it
to make sure that the detector image is, indeed, at right angles
to the moving mirror.


However, because of the finite speed of electromagnetic radiation, neither the observed
image of the moving mirror, nor the observed image of the moving detector, can be,
actually, in real time, or instantaneous.


And so, the BIG question,
now, is this:


Are the real mirror & the real detector, indeed, at right angles
relative to each other?


Of course, they are.


What else can we expect?




wink.gif




And that is the short answer.




















 
 
AAF

Re: On the Motion of the Earth Relative to the Aether

January 5 2018, 12:00 AM 









And, now, this is
the LONG
answer!




happy.gif




Due to the fact that the speed of light finite, from the reference frame of the moving detector,
Michelson & Morley should see the optical image of the moving transversal mirror tilted
and displaced from the vertical line to the backward direction
by a small angle equal to b.


And that is because, during the travel time of light, t, from the moving mirror to the moving
detector, the moving detector, itself, must make a displacement equal to vt; where
v is the orbital velocity of the earth.


And it follows, therefore,
that:


b = arcsin(v/c) ;


where b, as defined above, is the angle by which the moving mirror is tilted
to the backward direction.


But, in reality, Michelson & Morley, from the reference of the moving detector,
did not and could not see any tilted image of the moving mirror,
by any angle at all.


But why?


Well . . .




wink.gif




It's, certainly, because, in the reference frame of the moving detector,
James Bradley's light aberration takes the tilted image of the moving
mirror, as input, shifts it, as output, by an equal angle in the forward
direction, and places it, exactly, on the top of the instantaneous position
of the actual mirror.


Nevertheless, even though Michelson & Morley, from the reference of
the moving detector, could, only, see the image of the moving mirror at right
angles to the moving detector, the actual path of electromagnetic radiation,
received from the mirror image, is tilted backward by an amount equal to

{arcsin(v/c)};

i.e., along one leg of an isosceles triangle
like this one:



[linked image]

















    
This message has been edited by AAF24 on Jan 5, 2018 12:10 AM


 
 
AAF

Re: On the Motion of the Earth Relative to the Aether

January 7 2018, 12:00 AM 










So, let's, now, carry out the Michelson-Morley calculations, for the longitudinal light beam,
in accordance with the Newtonian assumption, which states that the speed of light is dependent
on the speed of the light source:


(A.) If the light beam is emitted with a muzzle speed, c, in the same direction,
in which the earth is traveling at a speed, v, then it travels at the velocity resultant,
(c + v), and takes an amount of time, T_1, to reach a receding mirror
at a distance, L, away:


T_1 = [L + vT_1] / (c + v) = L / c


And since, according to the above Newtonian assumption, the momentum & the kinetic energy of reflected light
are conserved, the mirror must reflect the incident beam, towards the approaching detector, at the velocity
resultant, (c – v); and hence, the reflected beam takes an amount of time, T_2,
to reach the approaching detector at a distance, L, away:


T_2 = [L - vT_1] / (c - v) = L / c


And so, the total travel time of the light beam, in the whole round trip along
the longitudinal direction, is equal to T:


T = T_1 + T_2 = 2L / c.


And that is how the calculations of Michelson & Morley, in the case of light beams traveling
in the longitudinal direction, are done, on the basis of the ballistic assumption,
in accordance with which, the speed of light is dependent upon
the speed of the light source.



















 
 
AAF

Re: On the Motion of the Earth Relative to the Aether

January 9 2018, 12:00 AM 









And, now, this is how the calculations of Michelson & Morley, in the case of light
beams traveling in the transversal direction, are recalculated, on the basis of
the assumption, in accordance with which, the speed of light is dependent
upon the speed of the light source.


Since the Michelson-Morley transversal light beam is emitted, by the light source,
with the muzzle speed of light, c, at right angles to the direction, in which Earth
is traveling with the orbital speed, v, it follows that, according on the Newtonian
assumption, on the basis of which, the speed of light is dependent on the speed
of the light source, the Michelson-Morley transversal beam travels, towards
the moving transversal mirror, at the velocity resultant, c'; i.e.,


c' = [c2 + v2]0.5.


The total path of the transversal beam, in the Michelson-Morley experiment,
forms an isosceles triangle, like this one:


[linked image]


& whose height is equal to L;
and whose base is equal to vT_3;
& where v is the orbital velocity
of the earth.


And accordingly, T_3 can be obtained
by using the following equation:


T_3 = {2[(0.5vT_3)2 + L2]0.5} / [c2 + v2]0.5 = 2L / c.




Q.E.D.




happy.gif




















 
 
AAF

Re: On the Motion of the Earth Relative to the Aether

January 11 2018, 12:00 AM 










And, now, based on the assumption, in accordance with which, the speed of light
is dependent on the speed of the light source, what is the speed of Michelson-Morley
light beam relative to the receding horizontal mirror?


According to Newton's basic assumption, about the speed of light, the Michelson-Morley beam
is traveling with the velocity resultant, (c + v), towards the horizontal mirror,
which is receding at Earth's orbital speed, v.


And therefore, the speed of the Michelson-Morley light beam, relative to the receding horizontal
mirror, is equal to c.


Now, if the receding horizontal mirror is replaced with a moving observer, will that moving observer
find out that the speed of light, relative to him/her,
is, always, equal to c?


Sure!


happy.gif



If the moving horizontal mirror is replaced with a moving & smart observer, THEN,
on the basis of the assumption, in accordance with which, the speed of light is
dependent upon the speed of the light source, that moving & smart observer will
find out that the speed of light, relative to him/her,
is, always, equal to c.


There is not one shred of a doubt about that.


























 
 
AAF

Re: On the Motion of the Earth Relative to the Aether

January 13 2018, 12:00 AM 










And, once again, based on the assumption, in accordance with which, the speed of light
is dependent upon the speed of the light source, what is the speed of the reflected
Michelson-Morley light beam relative to the approaching detector?


According to Newton's fundamental assumption, about the speed of light, both the momentum
& the kinetic energy of the Michelson-Morley beam are conserved upon reflection
by the receding horizontal mirror.


And therefore, the Michelson-Morley beam is reflected, by the receding horizontal mirror,
at the velocity resultant, (c – v), towards the approaching detector
with the speed, v.


And so, the speed of the Michelson-Morley light beam, relative to the approaching detector,
is equal to c.


But, now, if the approaching detector is replaced with a moving observer, will that moving
observer find the speed of light, relative to him/her,
is, always, equal to c?


Absolutely!


happy.gif



If the approaching detector is replaced with a skilled & smart observer, THEN,
on the basis of the assumption, in accordance with which, the speed of light
is dependent on the speed of the light source, that moving & skilled observer will
find out that the speed of light, relative to him/her,
is, always, equal to c.


There can be no doubt about that.



























 
 
AAF

Re: On the Motion of the Earth Relative to the Aether

January 15 2018, 12:00 AM 








And, one more time, in accordance with the assumption, on the basis of which, the speed of light
is dependent upon the speed of the light source, what is the speed of Michelson-Morley
light beam relative to the moving transversal mirror?


According to the Newtonian assumption, about the speed of light, the transversal
Michelson-Morley beam is traveling with the velocity resultant, c', towards
the transversal mirror, which is moving at Earth's orbital speed, v; i.e.,


c' = [c2 + v2]0.5.


And therefore, the speed of the Michelson-Morley light beam, relative to
the moving transversal mirror is equal to:


c' = [c'2 - v2]0.5 = [(c2 + v2) - v2]0.5 = c.


And so, now, if the moving transversal mirror is replaced with a moving observer,
will that moving observer find out that the speed of light, relative to him/her,
is, always, equal to c?


Yes, of course . . .



happy.gif



If the moving transversal mirror is replaced with a moving observer, THEN,
on the basis of the assumption, in accordance with which, the speed of light is
dependent on the speed of the light source, that moving observer will
find out that the speed of light, relative to him/her,
is, always, equal to c.

There is no doubt about that.























    
This message has been edited by AAF24 on Jan 17, 2018 12:02 AM


 
 
AAF

Re: On the Motion of the Earth Relative to the Aether

January 17 2018, 12:00 AM 








And, finally, based on the assumption, in accordance with which, the speed of light
is dependent upon the speed of the light source, what is the speed of the reflected
Michelson-Morley transversal light beam relative
to the moving detector?


According to Newton's assumption, about the speed of light, both the momentum
& the kinetic energy of the Michelson-Morley beam are conserved upon reflection
by the moving transversal mirror.


And therefore, the Michelson-Morley beam is reflected, by the moving transversal mirror,
at the velocity resultant, c', towards the moving detector
with the speed, v; i.e.,


c' = [c2 + v2]0.5.


And so, the speed of the Michelson-Morley light beam, relative
to the moving detector:


c' = [c'2 - v2]0.5 = [(c2 + v2) - v2]0.5 = c.


But, now, if the moving detector is replaced with a moving observer, will that moving observer
find the speed of light, relative to him/her,
is, always, equal to c?


Yep!



[linked image]



If the moving detector is replaced with a moving observer, THEN,
on the basis of the assumption, in accordance with which, the speed of
light is dependent on the speed of the light source, that moving observer
will find out that the speed of light, relative to him/her,
is, always, equal to c.


There is, absolutely, no doubt about that.



















 
 
Bill Geist

Re: On the Motion of the Earth Relative to the Aether

January 17 2018, 8:34 AM 


 
 
AAF

Re: On the Motion of the Earth Relative to the Aether

January 19 2018, 12:00 AM 











Oh . . .


I see . . .




happy.gif




Bryan G. Wallace's book:
The Farce of Physics:

http://bryangwallace.dreamhosters.com


What a great book!


However, right now, this Dayton C. Miller's
experimental report:

http://www.anti-relativity.com/Miller1933.pdf

is giving 'MOI' a hard time . . .



wink.gif




Certainly, the quality of Dayton C. Miller's experimental data
is very good.


But, at the same time, his working hypothesis is awful,
to say the least.


And so, for the time being, I just can't choose between these three possible
interpretations of Dayton C. Miller's experimental data:


I. The observed fringe shift, at Mount Willson,
is caused by the multiple-reflection technique, which was used by Dayton C. Miller,
to increase the effective path of his light beam, from about 4 meters
to about 64 meters.


II. The observed fringe shift, at Mount Willson,
is due to the inclination of Dayton C. Miller's experimental apparatus,
by an angle less than 90 degrees, with respect to
the gravitational vector of the earth, g.


III. The observed fringe shift, at Mount Willson,
is the result of the combined effect of multiple reflections as well as by
the inclination of Dayton C. Miller's experimental apparatus,
by an angle less than 90 degrees, with respect to
the gravitational vector of the earth, g.













 
 
 
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