Bronislaw Maciag
Jan Maciag
THE AETHER
& THE GALILEAN
TRANSFORMATION
Tarnobrzeg 2012
1
Copyright © 2012 by Bronislaw Maciag & Jan Maciag.
Original title ‘ Eter i Transformacja Galileusza’.
Translated by Jadwiga Weglarz-Finnegan.
This version of ‘The Aether and the Galilean Transformation’ incorporates changes and
corrections made by the authors since it was first published in print in 2009
by the Cracow’s Publishing House ‘Tekst’.
All rights reserved, No part of this work may be reproduced or transmitted in any form or by
any means, electronic or mechanical, including photocopying, recording or by any information
storage or retrieval system, without permission in writing from the authors.
Authors:
Bronislaw Maciag
Jan Maciag
Tarnobrzeg, Poland
February 2012
2
CONTENTS
Preface .................................................................................................................................................................5
CHAPTER I:
MATHEMATICAL MODEL ...............................................................................................................7
I.1 ALBERT MICHELSON’S INTERFEROMETER .....................................................................................7
I.1.1 Assumptions and the coordinate systems ..............................................................................7
I.1.2 Rays of light in semi-transparent plate ..................................................................................10
I.1.3 Line equations in the OXY coordinate system .....................................................................10
I.1.4 The coordinates of the A ,...,A points and the lengths of the a ,...,a segments
15 15
in the OXY coordinate system ...................................................................................................11
I.1.5 The coordinates of the B ,...,B points and the lengths of the b ,...,b segments
15 15
in the OXY coordinate system ...................................................................................................15
THE GALILEAN TRANSFORMATION ...................................................................................................19
I.1.6 The coordinates of the A ,...,A points in the O’EQ system ............................................19
15
I.1.7 The coordinates of the B ,...,B points in the O’EQ system .............................................20
15
I.1.8 The lengths of distances traveled by the ray of light after leaving
the S slit at the angle in the O’EQ system ..................................................................21
o
I.1.9 The lengths of distances traveled by the ray of light after leaving
the S slit at the angle in the O’EQ system ..................................................................21
o
I.1.10 The relative difference of distances traveled by the rays of light
reaching one point on the screen M ......................................................................................21
I.1.11 The difference of phases of the light rays reaching one point on the
screen M .............................................................................................................................................22
I.1.12 The interference fringes shift values ......................................................................................23
I.1.13 Interference fringes shift values after changing the mirror-slit distance ...............29
I.2 Why were there no shifts of interference fringes observed in the Michelson–Morley’s
experiments? ....................................................................................................................................................31
I.3 Why was ‘the value of interference fringes shift’ calculated by Albert Michelson not
confirmed during the experiments? ........................................................................................................31
I.4 The velocities at which the centers of the Earth and the Sun travel with respect
to the aether ....................................................................................................................................................32
I.5 The velocity at which the center of our Galaxy travels with respect to the aether .........34
CHAPTER II:
THE VELOCITY OF THE INTERFEROMETER................................................................................35
II.1 The peripheral velocity V of the point U (, ) on the Earth’s surface...............................36
r
II.2 The V velocity at which the Earth’s center revolves around the Sun..................................37
zs
II.2.1 Determining the angle.............................................................................................................38
II.2.2 Determining the angle..............................................................................................................38
II.2.3 Azimuth and the latitude of the Earth’s center velocity V .........................................40
zs
II.2.4 The speed V at which the Earth’s center revolves around the Sun........................42
zs
3
II.3 The velocities V ,V V at which the Sun’s center moves with respect
se se1 se
to the aether ..................................................................................................................................................43
II.3.1 Azimuth and the altitude of the V velocity ........................................................................44
se
II.3.2 Azimuth and the altitude of the V velocity ......................................................................45
se1
II.4 Sum of velocities in the horizontal system .......................................................................................45
II.4.1 Velocity V V .............................................................................................................................46
o o1
II.4.2 Velocity V V .............................................................................................................................47
o o2
II.5 An Example ...................................................................................................................................................48
III.1 CHAPTER III:
NEWTON’S SECOND LAW OF MOTION .................................................................................50
III.2
III.3 Variable mass of a particle in the Newton’s second law of motion ......................................51
III.4 III.1.1 The velocity of the particle ....................................................................................................53
III.5 III.1.2 The energy of the particle ......................................................................................................54
III.1.3 Rest mass of the particle with respect to the aether ..................................................56
III.1.4 The laws of mechanics ........................................................................................................... 56
III.1.5 Determining the F force .......................................................................................................58
1
Times measured by atomic clocks ....................................................................................................63
Decay of particles .....................................................................................................................................64
Determining a sidereal day with atomic clocks ...........................................................................65
Determining the absolute velocities of the Earth and the Sun with atomic clocks ......67
III.5,1 Calculating absolute velocities of the Earth and the Sun (example) ...................75
IV.1 CHAPTER IV:
PROGRAMS ..............................................................................................................................................76
IV.2
IV.3 PROGRAMS: abIntM, baIntM, IntM .................................................................................................76
IV.4 IV.1.1 PROGRAM abIntM .....................................................................................................................77
IV.1.2 PROGRAM baIntM .....................................................................................................................80
IV.1.3 PROGRAM IntM ..........................................................................................................................81
PROGRAM Vo .............................................................................................................................................82
PROGRAM GHA .........................................................................................................................................85
PROGRAM VzeVse ...................................................................................................................................86
RESULTS AND CONCLUSIONS .........................................................................................................87
INDEX OF SYMBOLS ..............................................................................................................................88
LITERATURE .............................................................................................................................................91
4
PREFACE
In the 19th century physicists were convinced that there exists a medium, called the aether,
with respect to which light and all objects are in motion. James Clerk Maxwell believed that
with the use of light, it is possible to determine Earth’s speed in relation to the aether. Under
the Galilean transformation his equations link the speed of light (c) in the inertial frame of
reference with the frame’s velocity with respect to the aether.
Having become familiar with J. C. Maxwell’s deliberations, Albert A. Michelson came up with
an idea for an experiment by which the Earth’s motion with respect to the aether could be
measured with adequate precision and thereby the applicability of the Galilean transformation
to the motion of light could be verified. With an interferometer of his own design he made
calculations from which he obtained the relationship between ‘the shift of interference
fringes’ and the interferometer speed with respect to the aether. After applying the relative
speed of the interferometer against the aether as equal to the orbital speed of the Earth
(approximately 30 km/s) he obtained a specific shift value of about 0.04 of a fringe, and he
expected that the shift he was to observe during the experiment would be no smaller than the
value he had calculated. However, in the experiment which he performed in 1881 – after J. C.
Maxwell had already passed away – he observed no such shift. In 1887 Albert Michelson and
Edward Morley jointly repeated the experiments using a more advanced interferometer with
very much the same result as in 1881 i.e. no shift of interference fringes was observed.
While Albert Michelson’s calculations raised no doubts among physicists though the fact that
Michelson-Morley’s experiments failed to provide the observance of the shift of interference
fringes was weakening their faith in the existence of the aether. Ultimately the aether
concept was abandoned altogaether. In 1905 the Galilean transformation was replaced by
Hendrik A. Lorentz’s transformation after Albert Einstein’s presentation of the Special
Relativity (SR) theory that was based on two key postulates. The first assumes that no
preferred inertial frame of reference exists, which effectively means that the aether does not
exist, and the second assumes that the speed of light in a vacuum is the same in all inertial
frames of reference. The Galilean transformation holds when relative speeds of objects in
inertial frames are negligibly small compared to the speed of light c.
In this work a mathematical model of Albert Michelson’s interferometer was designed
assuming that the aether does exists and that the Galilean transformation is in operation. The
authors have created this model to explain exactly why no shift of interference fringes was
observed with the interferometer used in Michelson and Morley’s experiments.
Based on the data from the Michelson-Morley’s experiments and the values of the
interference fringe shifts resulting from the mathematical model which incorporated a variety
of angles that the interferometer was positioned at and considered its different speeds against
the aether, the speed of the interferometer on the Earth’s surface was determined with
respect to the aether. Then given the interferometer speed on the Earth with respect to the
aether and the speed at which the Sun revolves around the center of our Galaxy as well as
having taken into consideration the aberration of starlight, the relative speeds of the Earth,
the Sun and the Galaxy centers with respect to the aether were determined.
5
For experimental purposes such as investigating particles in linear accelerators, the
coordinates of the absolute velocity of the interferometer, and therefore of any object on the
Earth’s surface, in the horizontal frames of reference were determined. Then, according to
Newton’s second law, the motion of a particle was investigated with its speed-related mass
changes considered.
Finally, the decay of unstable particles was researched and it was shown that the elongation
of the Earth’s sidereal day with respect to the time measured by atomic clocks is merely
apparent. The relationship between the time measured by atomic clocks and the clocks’
speed with respect to the aether was determined. This was applied for calculating the Earth’s
and the Sun’s speeds with respect to the aether with the use of atomic clocks.
Acknowledgements
The authors wish to express their thanks to: Janusz D. Laski, PhD for his valuable comments
and critical remarks regarding this work and Prof. Brian O’Reilly for reading the English
version of the manuscript.
Authors:
Bronislaw Maciag
Jan Maciag
Tarnobrzeg, Poland
February 2012
6
CHAPTER I
MATHEMATICAL MODEL
I.1 ALBERT MICHELSON’S INTERFEROMETER
I.1.1 ASSUMPTIONS AND THE COORDINATE SYSTEMS
Let us assume that a medium, called the aether exists. Light and the interferometer move
with respect to the aether. In our considerations, in order to establish the motion of light and
the interferometer with respect to this medium, we introduce three coordinate systems placed
on one plane (Figs. 2, 3 & 4), namely:
1) A preferred absolute inertial coordinate system OXoYo, motionless with respect to the
aether (a frame of reference).
2) An OXY coordinate system.
Its initial point always corresponds to the OXoYo initial point. The OXY coordinate
system can rotate by any angle with respect to the OXoYo system.
2) An O’EQ coordinate system fixed to the interferometer. The interferometer’s velocity
V is always parallel to the OXo axis. The O’E axis is always parallel to the OX axis.
o
The system’s origin corresponds to the origin of the OXoYo system only at the initial
time t=0 of an interferometer motion under consideration.
The O’EQ is an inertial system which moves togaether with the interferometer along the OXo
axis at a constant velocity V in relation to the OXoYo system. Another inertial system will be
o
obtained when the value of the V velocity modulus is changed and fixed. Thus, if we keep on
o
applying this procedure, any number of O’EQ inertial systems can be obtained. The V
o
velocities are the absolute velocities of the O’EQ systems. The light is an electromagnetic
wave that with respect to the aether travels in a vacuum with the C velocity which modulus
o
(speed) C = const.
o
7
Fig. 1 Diagram of Albert Michelson’s interferometer and the trajectory of light rays in the
interferometer.
SYMBOLS:
ZS source of light,
slit,
S
mirrors,
0
semi-transparent plate,
Z ,Z screen,
points successively reached by a ray of light after leaving the
12
slit S at the angle ,
PP 0
M
points successively reached by a ray of light after leaving the
A ,...,A
slit S at the angle ,
15 0
B ,...,B angles of the light rays refraction in the semi-transparent plate PP.
15
,
12
BASIC DIMENSIONS:
L , L , L, L ,
1 234
g thickness of the semi-transparent plate PP.
The values of basic dimensions and the wavelength of light in a vacuum, can be found on
page 76.
Herein two phenomena i.e. the light diffraction on the slit S and the interference of those
0
rays which after leaving the slit S at , angles reach one point on the screen M were
0
exploited. Points A , B coincide.
55
8
Fig. 2 The trajectory of light rays reaching point A on screen M after leaving the slit S at
50
the angle .
Fig. 3 The trajectory of light rays reaching point B on screen M after leaving the slit S at
50
the angle .
9
I.1.2 RAYS OF LIGHT IN SEMI-TRANSPARENT PLATE
(Figs. 1, 2 & 3)
According to Snell’s law the following equations can be obtained:
sin(45o ) sin(45o ) C n
0
o
sin sin C 2
1 2 pp
Where: , the angles of refraction of the light rays in the PP plate,
12 the index of refraction for the PP plate with respect to a vacuum,
n the speed of light in a vacuum with respect to the aether,
2 the speed of light in the PP plate with respect to the aether,
C the wavelength of light in a vacuum,
o the wavelength of light in the PP plate.
C
p
o
p
From the above equations we have:
(1.1) arcsin sin(45o )
1n
(1.2)
(1.3) 2
(1.4)
arcsin sin(45o )
2n
2
C C /n
p o2
/n
p o2
I.1.3 LINE EQUATIONS IN THE OXY COORDINATE SYSTEM
(Figs. 1, 2, 3 & 4)
The straight line equations of the trajectory of light rays:
y, y , y, y , y
1 234 5
y, y , y , y , y
12 22 32 42 52
The line equation of the mirror Z :
1
(1.5) x L tV cos
1o
The y line equation of the mirror Z :
62
(1.6) y L tV sin
62 o
The y line equation of the screen M:
7
(1.7) y L tV sin
7 4o
10
The y line equation of the PP plate on the side of the S slit.
80
The coordinates of the point A (x , y ) are:
0 a0 a0
(1**) x L tV cos
a0 3 o
(2**) y tV sin
a0 o
The line y passes through point A hence its equation takes the following form:
80
y tg45o x tg45o x y
8 a0 a0
Having considered equations (1**) & (2**) we obtain:
(1.8) y x L tV (cos sin )
8 3o
The y line equation of the other side of the PP plate.
9
y y 2g therefore
98
(1.9) y x L 2g tV (cos sin )
93 o
In equations (1.5) - (1.9) the variable t represents the motion time of the interferometer.
I.1.4 THE COORDINATES OF THE A ,..., A POINTS AND THE LENGTHS OF
15
THE a ,...,a SEGMENTS IN THE OXY COORDINATE SYSTEM
15
The lengths of segments of the distance traveled by a ray of light leaving the slit S at the
0
angle :
a S A , a A A , a A A , a A A , a A A .
1 01 2 12 3 23 4 34 5 45
1. POINT A AND THE LENGTH OF THE a SEGMENT
11
The coordinates of point A (x , y ) are determined by straight line equations:
1 a1 a1
a
y tg x , (1.8) y x L tV (cos sin ) , t t 1 (1*) so
1 8 3o a1 C
o
(1.10) V cos
x [L a o (cos sin )]
a1 3 1 C cos sin
o
(1.11) V sin
y [L a o (cos sin )]
a1 3 1 C cos sin
o
The coordinates of the S A segment are: S A [x 0 , y 0]
01 0 1 a1
a1
We can write an equation: a2 x2 y2
1 a1 a1
which after applying formulae (1.10) and (1.11) takes the following form:
(1.12) a L 3
1V
cos sin (cos sin ) o
C
o
11
2. POINT A AND THE LENGTH OF THE a SEGMENT
22
The equation of the y straight line which passes through the point A is:
21
y tg(90o ) x y tg(90o ) x
2 a1 a1
The coordinates of A (x , y ) point are determined by straight lines equations:
2 a2 a2
a a
y , (1.6) y L tV sin , t t 1 2 (2*) thus
2 62 o a2 C
o
(1.13) x x sin [L y V
(a a ) o sin ]
a2 a1 cos 2 a1 1 2 C
o
(1.14) V
y L (a a ) o sin
a2 2 1 2 C
o
The coordinates of the A A segment are: A A [x x , y y ]
1 2 a2 a1 a2 a1
12
x x x sin a V
o sin
a2 a1 a21 cos 2 C
o
V where:
y y y a o sin
a2 a1 a21 2 C
o
(1.15) x sin (L y V
a o sin )
a21 cos 2 a1 1 C
o
V
(1.16) y L a o sin y
a21 2 1 C a1
o
We can now write the following equation:
a2 (x x )2 (y y )2 which when solved, gives the following:
2 a2 a1 a2 a1
r r
(1.17) a 21 23 where:
(1.18)
(1.20) 2r 22
r V sin ( x sin y ), (1.19) r 1 ( V sin )2
o o
21 C a21 cos a21 22 C cos
o o
r r2 r (x2 y2 )
23 21 22 a21 a21
3. POINT A AND THE LENGTH OF THE a SEGMENT
33
The equation of the straight line y which passes through the point A is given by:
32
y tg(90o ) x y tg(90o ) x
3 a2 a2
The coordinates of the A (x , y ) point are determined by the equations of straight lines:
3 a3 a3
y , (1.8) y x L tV (cos sin ) , a a a (3*) thus
t t 1 2 3
3 8 3o
a3 C
o
(1.21) x sin V cos x
[L y (a a a ) o (cos sin )]
a3 sin cos 3 a2 1 2 3 C sin cos a2
o
(1.22) y sin V cos x L
[L y (a a ) o (cos sin )]
a3 sin cos 3 a2 1 2 C sin cos a2 3
o
V cos V
(a a ) o (cos sin ) a o (cos sin )
1 2C sin cos 3 C
oo
12
The coordinates of the A A segment are: A A [x x , y y ]
23 2 3 a3 a2 a3 a2
x x x sin V
a o (cos sin )
a3 a2 a31 sin cos 3 C
o
y y y cos V where:
a o (cos sin )
a3 a2 a31 sin cos 3 C
o
(1.23) x sin V cos x x
[L y (a a ) o (cos sin )]
a31 sin cos 3 a2 1 2 C sin cos a2 a2
o
(1.24) y sin V cos x L
[L y (a a ) o (cos sin )]
a31 sin cos 3 a2 1 2 C sin cos a2 3
o
V
(a a ) o (cos sin ) y
1 2C a2
o
a2 (x x )2 (y y )2
3 a3 a2 a3 a2
Having solved the above equation, we obtain:
rr
(1.25) a 31 33 where:
(1.26)
(1.27) 3r 32 cos sin
(1.28) sin cos
r (x sin y cos ) V
o
31 a31 a31 C o
r 1 ( V cos sin )2
o
32 C sin cos
o
r r2 r (x2 y2 )
33 31 32 a31 a31
4. POINT A AND THE LENGTH OF THE a SEGMENT
44
The equation of the y straight line which runs through the A point is given by:
43
y tg(45o ) x y tg(45o ) x
4 1 a3 1 a3
Through the plate, light travels with the speed of C C / n (1.3), hence to travel the
p o2
a na
distance a in the plate it requires the following time: 4 2 4
4 CC
po
The coordinates of the A (x , y ) point are determined by the equations of straight lines:
4 a4 a4
y , (1.9) y x L 2g tV (cos sin ) , a a a n a (4*) thus
4 93 o t t 1 2 3 2 4
a4 C
0
cos(45o ) V
(1.29) x 1 [L 2g (a a a n a ) o (cos sin )
a4 sin(45o ) cos(45o ) 3 1 2 3 24C
11
o
tg(45o )x y ]
1 a3 a3
sin(45o ) V
(1.30) y 1 [L 2g (a a a n a ) o (cos sin )
a4 sin(45o ) cos(45o ) 3 1 2 3 24C
11
o
tg(45o )x y ] y tg(45o )x
1 a3 a3 a3 1 a3
The coordinates of the A A segment are: A A [x x , y y ] with
34 3 4 a4 a3 a4 a3
13
cos(45o ) V
x x x 1 n a o (cos sin )
a4 a3 a41 sin(45o ) cos(45o ) 2 4C
11
o
sin(45o ) V
y y y 1 n a o (cos sin ) where:
a4 a3 a41 sin(45o ) cos(45o ) 2 4C
11
o
cos(45o ) V
(1.31) x 1 [L 2g (a a a ) o (cos sin )
a41 sin(45o ) cos(45o ) 3 1 2 3C
11
o
tg(45o )x y ] x
1 a3 a3 a3
sin(45o ) V
(1.32) y 1 [L 2g (a a a ) o (cos sin )
a41 sin(45o ) cos(45o ) 3 1 2 3C
11
o
tg(45o )x y ] tg(45o )x
1 a3 a3 1 a3
a2 (x x )2 (y y )2
4 a4 a3 a4 a3
Having solved the above equation, we obtain:
r r
(1.33) a 41 43 where:
(1.34)
(1.35) 4r 42
(1.35.1)
V cos sin
r [x cos(45o ) y sin(45o )]n o sin(45o ) cos(45o )
41 a41 1 a41 1 2C
11
o
r 1 ( n V cos sin )2
o
42 2C sin(45o ) cos(45o )
11
o
r r2 r (x2 y2 )
43 41 42 a41 a41
5. POINT A AND THE LENGTH OF THE a SEGMENT
55
The equation of the y straight line which passes through the A point is given by:
54
y tg(90o ) x y tg(90o ) x
5 a4 a4
The coordinates of the A (x , y ) point are determined by the equations of straight lines:
5 a5 a5
a a a n a a
y, (1.7) y L tV sin , t t 1 2 3 2 4 5 (5*) thus
5 7 4o a5 C
o
(1.36) x sin [L (a a a n a a V ] x
) o sin y
a5 cos 4 1 2 3 2 4 5 C a4 a4
o
(1.37) V
y L (a a a n a a ) o sin
a5 4 1 2 3 2 4 5 C
o
The coordinates of the A A segment are: A A [x x , y y ] with
45 4 5 a5 a4
a5 a4
x x x sin a V
o sin
a5 a4 a51 cos 5 C
o
V
y y y a o sin where:
a5 a4 a51 5 C
o
(1.38) x sin [L (a a a n a V ]
) o sin y
a51 cos 4 1 2 3 2 4 C a4
o
14
V
(1.39) y L (a a a n a ) o sin y
a51 4 1 2 3 2 4 C a4
o
a2 (x x )2 (y y )2
5 a5 a4 a5 a4
Having solved this equation we obtain:
rr 53
(1.40) a 51 where:
(1.41) 5r
(1.43) 52
r ( y x sin ) V sin , (1.42) r 1 ( V sin )2 ,
o o
51 a51 a51 cos C o 52 C cos
o
r r2 r (x2 y2 )
53 51 52 a51 a51
I.1.5 THE COORDINATES OF THE B ,...,B POINTS AND THE LENGTHS OF
15
THE b ,...,b SEGMENTS IN THE OXY COORDINATE SYSTEM
15
The lengths of distances traveled by the ray of light after leaving the S slit at the angle
0
are:
b S B , b B B , b B B , b B B , b B B .
1 01 2 12 3 23 4 34 5 45
6. POINT B AND THE LENGTH OF THE b SEGMENT
11
The coordinates of the B (x , y ) point are determined by the straight lines equations:
1 b1 b1
y tg x , (1.8) y x L tV (cos sin ) , b (6*) thus
12 tt 1
8 3o
b1 C
o
(1.44) V cos
x [L b o (cos sin )]
b1 3 1 C cos sin
o
(1.45) V sin
y [L b o (cos sin )]
b1 3 1 C cos sin
o
The coordinates of the S B segment are: S B [x 0 , y 0]
01 0 1 b1 b1
b2 x2 y2 .
1 b1 b1
Having solved this equation we obtain:
(1.46) L
b 3
1V
cos sin (cos sin ) o
C
o
7. POINT B AND THE LENGTH THE b SEGMENT
22
The equation of the y straight line which passes through the B point is:
22 1
y tg(45o ) x y tg(45o ) x
22 2 b1 2 b1
b nb
Light travels the distance b within a time interval 2 2 2 .
2 CC
po
The coordinates of the B (x , y ) point are determined by the straight lines equations:
2 b2 b2
y , (1.9) y x L 2g tV (cos sin ) , b n b (7*) thus
22 93 o tt 1 2 2
b2 C
o
15
cos(45o ) V
(1.47) x 2 [L 2g (b n b ) o (cos sin ) y +
b2 sin(45o ) cos(450 ) 3 1 22 C b1
22
o
tg(45o )x ]
2 b1
sin(45o ) V
(1.48) y 2 [L 2g (b n b ) o (cos sin ) y +
b2 sin(45o ) cos(450 ) 3 1 22 C b1
22
o
tg(45o ) x ] y tg(45o )x
2 b1 b1 2 b1
The coordinates of the B B segment are: B B [x x , y y ]
1 2 b2 b1 b2 b1
12
cos(45o ) V
x x x 2 n b o (cos sin )
b2 b1 b21 sin(45o ) cos(45o ) 2 2C
22
o
sin(45o ) V
y y y 2 n b o (cos sin ) where:
b2 b1 b21 sin(45o ) cos(45o ) 2 2C
22
o
cos(45o ) V
(1.49) x 2 [L 2g b o (cos sin ) y
(1.50) b21 sin(45o ) cos(45o ) 3 1C b1
22
o
tg(45o )x ] x
2 b1 b1
sin(45o ) V
y 2 [L 2g b o (cos sin ) y
b21 sin(45o ) cos(45o ) 3 1C b1
22
o
tg(45o )x ] tg(45o )x
2 b1 2 b1
b2 (x x )2 (y y )2
2 b2 b1 b2 b1
Having solved the above equation we obtain:
s s
(1.51) b 21 23 where:
(1.52)
(1.53) 2s 22
(1.54)
V cos sin
s [x cos(45o ) y sin(45o )]n o
21 b21 2 b21 2 2c sin(45o ) cos(45o )
22
o
s 1 ( n V cos sin )2
o
22 2C sin(45o ) cos(45o )
22
o
s s2 s (x2 y2 )
23 21 22 b21 b21
8. POINT B AND THE LENGTH OF THE b SEGMENT
33
The equation of the y straight line which passes through the B point is:
32 2
y tg x y tg x
32 b2 b2
The coordinates of B (x , y ) are determined by the following equations of straight lines:
3 b3 b3
y , (1.5) x L tV cos , b n b b (8*) thus
t t 1 2 2 3
32 1 o
b3 C
o
16
(1.55) V
x L (b n b b ) o cos
b3 1 1 2 2 3 C
o
V
(1.56) y tg[L (b n b b ) o cos ] y tg x
b3 1 1 2 2 3 C b2 b2
o
The coordinates of the B B segment are: B B [x x , y y ]
23 2 3 b3 b2 b3 b2
V
x x x b o cos
b3 b2 b31 3 C
o
y y y sin b V where:
o cos
b3 b2 b31 cos 3 C
o
V
(1.57) x L (b n b ) o cos x
b31 1 1 2 2 C b2
o
V
(1.58) y tg[L (b n b ) o cos ] tg x
b31 1 1 2 2 C b2
o
b2 (x x )2 (y y )2
3 b3 b2 b3 b2
Having solved the above equation we obtain:
s s 33
(1.59) b 31 where:
(1.60) 3s
(1.61) 32
(1.62)
s (x y sin ) V cos
o
31 b31 b31 cos C o
s 1 ( V cos )2
o
32 C cos
o
s s2 s (x2 y2 )
33 31 32 b31 b31
9. POINT B AND THE LENGTH OF THE b SEGMENT
44
The equation of the y straight line which passes through the B point is:
42 3
y tg(180o ) x y tg(180o ) x
42 b3 b3
y tg x y tg x
42 b3 b3
The coordinates of the B (x , y ) point are given by the straight line equations:
4 b4 b4
y , (1.9) y x L 2g tV (cos sin ) , b n b b b (9*)
42 93 o t t 1 2 2 3 4
b4 C
o
thus
x cos [L V
(1.63) b4 sin cos 3 2g (b n b b b ) o (cos sin ) y
1 22 3 4 C b3
+ tg x ]
b3 o
(1.64) y sin V
[L 2g (b n b b b ) o (cos sin ) y
b4 sin cos 3 1 22 3 4 C b3
o
tg x ] y tg x
b3 b3 b3
The coordinates of the B B segment are: B B [x x , y y ]
3 4 b4 b3 b4 b3
34
17
x x x cos V
b o (cos sin )
b4 b3 b41 sin cos 4 C
o
y y y sin V where:
b o (cos sin )
b4 b3 b41 sin cos 4 C
o
x cos [L V
(1.65) b41 sin cos 3 2g (b n b b ) o (cos sin ) y
1 22 3 C b3
tg x ] x
b3 b3 o
sin [L V
(1.66) sin cos 3 2g (b n b b ) o (cos sin ) y
y 1 22 3 C b3
tg x ] tg x
b41 b3 b3
o
b2 (x x )2 (y y )2
4 b4 b3 b4 b3
Having solved the above equation we obtain:
s s 43
(1.67) b 41 where:
(1.68) 4s
(1.69) 42
(1.70)
s (x cos y sin ) V cos sin
sin cos
o
41 b41 b41 C o
s 1 ( V cos sin )2
o
42 C sin cos
o
s s2 s (x2 y2 )
43 41 42 b41 b41
10. POINT B AND THE LENGTH OF THE b SEGMENT
55
The equation of the y straight line which passes through the B point is:
52 4
y tg(90o ) x y tg(90o ) x
52 b4 b4
The coordinates of B (x , y ) are determined by the straight line equations:
5 b5 b5
y, (1.7) y L tV sin , b n b b b b (10*) thus
t t 1 2 2 3 4 5
52 7 4o
b5 C
o
(1.71) x sin [L (b n b b b V ] x
b ) o sin y
b5 cos 4 1 2 2 3 4 5 C b4 b4
o
(1.72) V
y L (b n b b b b ) o sin
b5 4 1 2 2 3 4 5 C
o
The coordinates of the B B segment are: B B [x x , y y ]
45 4 5 b5 b4 b5 b4
x x x sin b V
o sin
b5 b4 b51 cos 5 C
o
V where:
y y y b o sin
b5 b4 b51 5 C
o
(1.73) x sin [L (b n b b b V ]
) o sin y
b51 cos 4 1 2 2 3 4 C b4
o
V
(1.74) y L (b n b b b ) o sin y
b51 4 1 2 2 3 4 C b4
o
18
b2 (x x )2 (y y )2
5 b5 b4 b5 b4
Having solved the above equation we obtain:
s s
(1.75) b 51 53 where:
(1.76)
(1.77) 5s 52
(1.78)
s ( y x sin ) V sin
o
51 b51 b51 cos C o
s 1 ( V sin )2
o
52 C cos
o
s s2 s (x2 y2 )
53 51 52 b51 b51
THE GALILEAN TRANSFORMATION
When recalculating points A ,...,A , B ,...,B from the OXY inertial system into another
15 15
inertial system O’EQ, the Galilean transformation is applied.
I.1.6 THE COORDINATES OF THE A ,...,A POINTS IN THE O’EQ SYSTEM
15 relationship (1*)
POINT A (e , q ) , a relationship (2*)
t 1
1 a1 a1 a1 C relationship (3*)
o relationship (4*)
V
(1.79) e x t V cos x a o cos
a1 a1 a1 o a1 1 C
o
V
(1.80) q y t V sin y a o sin
a1 a1 a1 o a1 1 C
o
POINT A (e , q ) , a a
t 1 2
2 a2 a2 a2 C
o
V
(1.81) e x t V cos x (a a ) o cos
a2 a2 a2 o a2 1 2 C
o
V
(1.82) q y t V sin y (a a ) o sin = L
a2 a2 a2 o a2 1 2 C 2
o
POINT A (e , q ) , a a a
t 1 2 3
3 a3 a3 a3 C
o
V
(1.83) e x t V cos x (a a a ) o cos
a3 a3 a3 o a3 1 2 3 C
o
V
(1.84) q y t V sin y (a a a ) o sin
a3 a3 a3 o a3 1 2 3 C
o
POINT A (e , q ) , a a a n a
t 1 2 3 24
4 a4 a4 a4 C
o
V
(1.85) e x t V cos x (a a a n a ) o cos
a4 a4 a4 o a4 1 2 3 2 4 C
o
V
(1.86) q y t V sin y (a a a n a ) o sin
a4 a4 a4 o a4 1 2 3 2 4 C
o
19
POINT A (e , q ) , a a a n a a relationship (5*)
(1.87) t 1 2 3 24 5
(1.88) 5 a5 a5 a5 C
o
V
e x t V cos x (a a a n a a ) o cos
a5 a5 a5 o a5 1 2 3 2 4 5 C
o
V
q y t V sin y (a a a n a a ) o sin L
a5 a5 a5 o a5 1 2 3 2 4 5 C 4
o
I.1.7 THE COORDINATES OF THE B ,...,B POINTS IN THE O’EQ SYSTEM
POINT 15
(1.89)
(1.90) B (e , q ) , b relationship (6*)
POINT t 1
(1.91) 1 b1 b1 b1 C
(1.92)
POINT o
(1.93)
(1.94) V
POINT e x t V cos x b o cos
(1.95) b1 b1 b1 o b1 1 C
(1.96)
POINT o
(1.97)
(1.98) V
q y t V sin y b o sin
b1 b1 b1 o b1 1 C
o
B (e , q ) , b n b relationship (7*)
t 1 22
2 b2 b2 b2 C
o
V
e x t V cos x (b n b ) o cos
b2 b2 b2 o b2 1 2 2 C
o
V
q y t V sin y (b n b ) o sin
b2 b2 b2 o b2 1 2 2 C
o
B (e , q ) , b n b b relationship (8*)
t 1 22 3
3 b3 b3 b3 C
o
V
e x t V cos x (b n b b ) o cos L
b3 b3 b3 o b3 1 2 2 3 C 1
o
V
q y t V sin y (b n b b ) o sin
b3 b3 b3 o b3 1 2 2 3 C
o
B (e , q ) , b n b b b relationship (9*)
t 1 22 3 4
4 b4 b4 b4 C
o
V
e x t V cos x (b n b b b ) o cos
b4 b4 b4 o b4 1 2 2 3 4 C
o
V
q y t V sin y (b n b b b ) o sin
b4 b4 b4 o b4 1 2 2 3 4 C
o
B (e , q ) , b n b b b b relationship (10*)
t 1 22 3 4 5
5 b5 b5 b5 C
o
V
e x t V cos x (b n b b b b ) o cos
b5 b5 b5 o b5 1 2 2 3 4 5 C
o
V
q y t V sin y (b n b b b b ) o sin L
b5 b5 b5 o b5 1 2 2 3 4 5 C 4
o
20
I.1.8 THE LENGTHS OF DISTANCES TRAVELED BY A RAY OF LIGHT AFTER
LEAVING THE S SLIT AT THE ANGLE IN THE O’EQ SYSTEM
0
(1.99) a (e2 q 2 )1/ 2
1u a1 a1
(1.100) a [(e e )2 (q q )2 ]1/ 2
2u a2 a1 a2 a1
(1.101) a [(e e )2 (q q )2 ]1/ 2
3u a3 a2 a3 a2
(1.102) a [(e e )2 (q q )2 ]1/ 2
4u a4 a3 a4 a3
(1.103) a [(e e )2 (q q )2 ]1/ 2
5u a5 a4 a5 a4
I.1.9 THE LENGTHS OF DISTANCES TRAVELED BY A RAY OF LIGHT AFTER
LEAVING THE S SLIT AT THE ANGLE IN THE O’EQ SYSTEM
0
(1.104) b (e2 q 2 )1/ 2
1u b1 b1
(1.105) b [(e e )2 (q q )2 ]1/ 2
2u b2 b1 b2 b1
(1.106) b [(e e )2 (q q )2 ]1/ 2
3u b3 b2 b3 b2
(1.107) b [(e e )2 (q q )2 ]1/ 2
4u b4 b3 b4 b3
(1.108) b [(e e )2 (q q )2 ]1/ 2
5u b5 b4 b5 b4
I.1.10 THE RELATIVE DIFFERENCE OF THE DISTANCES TRAVELED BY THE
RAYS OF LIGHT REACHING ONE POINT ON THE SCREEN M
Fig. 4
Figure 4 shows points A , B of the screen M, togaether with their coordinates e , e ,
55 a5 b5
which were reached by the rays of light after leaving the slit S at the angles , .
0
The shift of the interference fringes is calculated with respect to point Mo with its coordinate
e on the screen M.
o
21
The coordinates e , e of the points A , B of screen M are dependent upon the variables
a5 b5 55
, , ,V , thus the coordinates e , e take on the form of the following functions:
w a5 b5
e e (, ,V ) relationship (1.87),
a5 a5 w
e e (, ,V ) relationship (1.97),
b5 b5 w
where: V
V o.
wC
o
The interference of the rays of light which have left the slit S at the angles , will only
0
take place on the screen M (fig. 4) when points A and B coincide. This means the coordinates
55
are equal: e e
a5 b5
The relative difference of distances traveled by the rays of light in a vacuum is:
l / [a a a a (b b b b )] / o
oo 1u 2u 3u 5u 1u 3u 4u 5u
The relative difference of distances traveled by the rays of light in the PP plate is:
l / (a b ) / p where:
pp 4u 2u
/n relationship (1.4)
p o2
Thus the total relative difference of distances traveled by the rays of light:
l / l / l /
o oo p p
After transformation of the relationship we obtain:
(1.109) l / [a a a n a a (b n b b b b )] /
o 1u 2u 3u 2 4u 5u 1u 2 2u 3u 4u 5u o
Let us introduce a symbol R :
w
(1.109a) R l /
wo
The relative difference of distances R depends upon the variables , , ,V and therefore
ww
it is defined by the function:
(1.109b) R R (, , ,V )
ww w
We need to calculate the R value at any Mo point with its coordinate e on the screen M,
w0
given the angle and at a fixed value V V / C
n w oo
In order to do this we write the following equations:
(11*) e e ( , ,V ) e
a5 a5 n n w o
(12*) e e ( , ,V ) e
b5 b5 n n w o
Then by applying the appropriate computational software, we can compute such a pair of
angles ( , ) which satisfies the equations (11*) and (12*). Knowing the pair of angles
nn
( , ) at fixed values of ,V we calculate the value of R :
nn nw w
(1.109c) R l / R ( , , ,V )
w o wn n nw
I.1.11 THE DIFFERENCE IN PHASES OF THE LIGHT RAYS REACHING ONE
POINT ON THE SCREEN M
Reaching one point on the screen the light rays may be identical or may vary in their phases.
The phase difference of the light rays equals:
(1.110) 2 frac(Rw )
where : frac(Rw ) is a function denoting the fractional part of the R value.
w
22
I.1.12 THE INTERFERENCE FRINGES SHIFTS’ VALUES
On the screen M let us select a point Mo (Fig. 4) with the e coordinate (a fixed line in the
0
telescope), in relation to which we will calculate the shift of interference fringes.
Corresponding to both the angle 0 and the coordinate e , the pair of angles ( , )
1 0 11
satisfies the following equations:
e e ( , ,V ) e 0
a5 a5 1 1 w
e e ( , ,V ) e so
b5 b5 1 1 w 0
(1.111) R R ( , , ,V )
w1 w 1 1 1 w
Corresponding to both the angle and the coordinate e , the pair of angles ( , )
22
20
satisfies the following equations:
e e ( , ,V ) e
a5 a5 2 2 w 0
e e ( , ,V ) e 0 so
b5 b5 2 2 w
(1.112) R R ( , , ,V )
w2 w 2 2 2 w
Leaving the slit S at angles ( , ), ( , ) the rays of light reach the Mo point of the e
0 11 2 2 0
coordinate.
Calculated with respect to the Mo point, the value k of the interference fringe shift depending
upon the angle and a fixed value V is given by the following:
2w
(1.113) k(2 ,Vw ) Rw2 Rw1
The formula (1.113) can be applied to calculate the values of interference fringe shifts with
respect to any Mo point on the screen M, after rotating the interferometer by any angle
2
and with the V V / C fixed at any value.
w oo
Tables 2 – 7 give the values of the interference fringe shifts with respect to point Mo of the
coordinate
e0 0.1508323849500 m for different values of ,V .
2w
The calculations were carried out using PROGRAM abIntM and PROGRAM baIntM presented
in Chapter IV of this work.
In the calculations the inequality of coordinates e | (e e ) / |<1011 describes the
relative mutual approximation of points A , B . w a5 b5 o
55
In the calculations the approximation of points A , B to the Mo point is described by the
55
following inequalities of coordinates:
| e e | 1011 m, | e e | 1011 m ( Fig. 4 ).
o a5 o b5
The abovementioned approximations of points A , B , Mo are presented in tables 1 to 9.
55
23
e0 0.1508323849500 m V V /C
0 2 w oo
1 R R ( , , ,V )
1 w1 w 1 1 1 w
34
V
R
w1 1 w1
- rad rad -
10 5 2.9222260500 10 4 1.1095088345104 1198.61389486
5 105 3.2967446400 10 4 1.1246435578104 1198.61386441
3.7648875000 10 4 1.1435632972104 1198.61401471
10 4 1.1624845292104 1198.61434710
1.1814072238104 1198.61485214
1.5 10 4 4.2330243900 10 4 1.2949745878104 1198.62194350
2 104 4.7011552800 10 4 1.4843723702104 1198.64892601
5 104 7.5098150500 10 4 4.9188393828104 1202.59429842
103 1.2190436150 103
102 9.6338598000 103
TABLE 1
Relative differences of distances Rw1 Rw (1, 1, 1,Vw ) at 0
1
e0 0.1508323849500 m V V /C
w oo
/4 R R ( , , ,V ) k ( ,V ) R R
2 w2 w 2 2 2 w 2w w2 w1
3
1 2 45
V
Rw2 k ( ,V )
w
2 2 2w
-
rad rad - -
10 5 2.8993247900 10 4 1.1684172723104 1198.61385305 4.181105
3.1822405800 10 4
5 105 3.5358853000 10 4 1.4191841557 104 1198.61358540 2.790104
3.8895300200 10 4
10 4 4.2431747400 10 4 1.7326407963104 1198.61318576 8.289104
1.5 10 4 6.3650428803 10 4
2 104 9.9014893500 10 4 2.0460952664104 1198.61270173 1.645103
7.3557934700 10 3
5 104 2.3595475723104 1198.61214563 2.705103
10 3 4.2402157723104 1198.60709966 1.484102
10 2
7.3744891306104 1198.59227220 5.665102
6.3754765287 103 1196.95554573 5.638
Values of R R ( , , ,V ) are presented in Table 1
w1 w 1 1 1 w
TABLE 2
Values of the interference fringe shifts k( ,V ) at / 4 .
2w 2
24
e0 0.1508323849500 m V V /C
/ 4
w oo
2
R R ( , , ,V ) k ( ,V ) R R
w2 w 2 2 2 w 2w w2 w1
1 2 3 4 5
V R k ( ,V )
2 2
w rad rad w2 2w
- 2.8902800200 10 4 1.0483841619 10 4 - -
10 5 3.1370133100 10 4 8.1902260516 10 5 1198.61393764 4.277 105
5 105 3.4454219502 10 4 5.3232748925 10 5 1198.61420143 3.370 10 4
104 3.7538217200 10 4 2.4563995015 10 5 1198.61488721 8.724 10 4
1.5 10 4 4.0622126680 10 4 4.1039970160 106 1198.61599036 1.643 10 3
2 104 5.9123725167 104 1.7609596901104 1198.61747170 2.619 10 3
5 1 04 8.9952635829 10 4 4.6268835060104 1198.63815940 1.62110 2
103 6.4333659825 10 3 5.6082175560103 1198.69592647 4.700 10 2
102 1206.89437454 4.300
Values of R R ( , , ,V ) are presented in Table 1
w1 w 1 1 1 w
TABLE 3
Values of the interference fringe shifts k( ,V ) at / 4 .
2w 2
e0 0.1508323849500 m V V /C
/2
w oo
2
R R ( , , ,V ) k ( ,V ) R R
w2 w 2 2 2 w 2w w2 w1
1 2 3 4 5
V R k ( ,V )
2 2
w rad rad w2 2w
- 2.8349912400 10 4 1.1906017454 10 4 - -
105 2.8605717800 10 4 1.5301090535 10 4 1198.61386731 2.755105
5 105 2.8925448400 10 4 1.9544967058 10 4 1198.61371836 1.460 10 4
104 2.9245150500 10 4 2.3788883244 10 4 1198.61366595 3.487 104
1.5 10 4 2.9564823600 10 4 2.8032838545 10 4 1198.61372009 6.270104
2 104 3.1482259889 10 4 5.3497398800 10 4 1198.61393648 9.156104
5 1 04 3.4675687794 10 3 9.5941484000 10 4 1198.61794096 4.002103
103 9.1664807771104 8.6060727920 10 3 1198.63526760 1.365 10 2
102 1201.28038669 1.313
Values of R R ( , , ,V ) are presented in Table 1
w1 w 1 1 1 w
TABLE 4
Values of the interference fringe shifts k( ,V ) at / 2 .
2w 2
25
e0 0.1508323849500 m V V /C
/ 2
w oo
2
R R ( , , ,V ) k ( ,V ) R R
w2 w 2 2 2 w 2w w2 w1
12 3 45
V R k ( ,V )
w2 2 w2 2 w
- rad rad - -
105 2.8222003000 10 4 1.0208490571104 1198.61396654 7.168 10 5
5 105 2.7966170000 10 4 6.813455335110 5 1198.61423265 3.682 10 4
2.7646353000 10 4 2.5696968167 105 1198.61470493
104 2.7326507024 10 4 1.6740222578 10 5 1198.61530207 6.902104
1,5 104
9.549 10 4
2 104 2.7006632500104 5.9177019877 105 1198.61602312 1.170103
5 1 04
2.5086783385104 3.1378951600104 1198.62313876 1.195 10 3
103
102 2.1884737400104 7.3811214228104 1198.64554335 3.382103
3.6241999271104 8.3691589093103 1201.24009280 1.354
Values of R R ( , , ,V ) are presented in Table 1
w1 w 1 1 1 w
TABLE 5
Values of the interference fringe shifts k( ,V ) at / 2 .
2w 2
e0 0.1508323849500 m V V /C
w oo
2
R R ( , , ,V ) k ( ,V ) R R
w2 w 2 2 2 w 2w w2 w1
1 23 4 5
V R k ( ,V )
22
w rad rad w2 2w
- 2.7349653500104 1.1019418554104 - -
105 2.3604410300104 1.0868085674104 1198.61395537 6.050 10 5
5 105 1.8922802500104 1.0678932908104 1198.61414555 2.81110 4
104 1.4241135000104 1.0489795001104 1198.61456080 5.460 10 4
1.5 10 4 9.5594077664105 1.0300672000104 1198.61517355 8.264 10 4
2 104 1.8532210108104 9.9166245470105 1198.61596454 1.112 10 3
5 1 04 6.5356348070104 7.7672443000105 1198.62463692 2.693 10 3
103 1198.65393517 5.009 10 3
10 2 No light interference
Values of R R ( , , ,V ) are presented in Table 1
w1 w 1 1 1 w
TABLE 6
Values of the interference fringe shifts k( ,V ) at .
2w 2
26
Fig. 5
Fig. 6
Tables 2 to 6 present the values of the interference fringe shifts k( ,V ) ; Figures 5 and 6
2w
provide their graphic representation.
When the interferometer’s relative speed reaches the value of V 1.5104 , the shift of
w
interference fringes takes its maximum value of | k | 1.645103 . At any lower relative speed
values V 1.5104 the shifts are not observable.
w
The value of the interferometer’s relative speed cannot be lower than the value of the Earth’s
relative rotation speed, which is about 104 . Hence the relative speed of the interferometer
located on the Earth’s surface takes values within the following range:
(1.114) 104 ≤ V 1.5104 (Fig. 5).
w
27
e0 0.1508323849500 m V V /C
0 w oo
1 R R ( , , ,V )
w1 w 1 1 1 w
12
34
V
R
w1 1 w1
- rad rad -
0.1 9.2579809820102
0.3 2.6004654118101 4.1812603372 10 3 1909.681149
0.5 3.2170261308101 1.3836889860 10 2 24249.102056
2.4834178376 10 2 212815.140048
e0 0.1508323849500 m V V /C
/4
w oo
2
R R ( , , ,V ) , k ( ,V ) R R
w2 w 2 2 2 w 2w w2 w1
12 3 4 5
R
V k (2 ,Vw )
w2 -
w2 2
- 873.670
- rad rad 1036.010585 24592.198
0.1 7.1068798638102 343.096524
0.3 2.1407339919101 6.2422111812 10 2 216368.335
0.5 3.6167734252101 1.8572583970 10 1 3553.195447
3.0959467087 101
TABLE 7
Values of the interference fringe shifts k( ,V ) at / 4 .
2w 2
Fig. 7
Table 7 provides values of the interference fringe shifts k( ,V ) , which are graphically
2w
presented in Figure 7.
28
I.1.13 VALUES OF THE INTERFERENCE FRINGE SHIFTS AFTER CHANGING
THE MIRROR-SLIT DISTANCE
We will calculate the values of the interference fringe shifts with respect to the Mo point at a
given angle after the distance between the mirror Z and the slit S has been changed.
n 20
The distance L is replaced by the distance L L .
2 22
A pair of angles ( , ) , which corresponds to: the angle , the coordinate e and the
22 n0
distance L2 , satisfies the following equations:
e e ( , ,V ) e 0
a5 a5 2 n w
e e ( , ,V ) e
b5 b5 2 n w 0
The relative difference of distances traveled by rays of light equals:
(1.115) R R ( , , ,V )
w2 w 2 2 n w
A pair of angles ( , ) , which corresponds to: the angle , the coordinate e
2L2 2L2
n 0
and the distance L L , satisfies the following equations:
22
e e ( , ,V , L ) e
a5 a5 2L2 n w 2 0
e e ( , ,V , L ) e
b5 b5 2L2 n w 2 o
The relative difference of distances traveled by rays of light equals:
(1.116) R R ( , , ,V , L )
w2L2 w 2L2 2L2 n w 2
The rays of light leaving the slit S at angles ( , ) and ( , ) reach the Mo point on
0 22 2L2 2L2
the screen M.
Depending on the distance increment L the k value of the interference fringe shift with
2
respect to the Mo point equals:
(1.117) k k( ,V , L ) R R
nw 2 w2L2 w2
In Tables 8 and 9 the values of interference fringe shifts were given with respect to the Mo
point of the coordinate e0 0.1508323849500 m at the distance L 1.25 and at the angles
2o
/ 4 and / 2 .
nn
MEASURING LENGTH WITH THE MICHELSON’S INTERFEROMETER
The evaluation of the measured length.
We are going to evaluate how accurately the length was calculated by the means of the
mathematical model of the Michelson’s interferometer.
W (k / 2) the length calculated with the mathematical model,
mo the length determined with the physical model,
W (k / 2)
ff o
where: k the read value of the interference fringes shift.
f
W L , W L
m2 f2
With a mathematical model the accuracy of the calculated length W is specified by the
m
following formula:
| L W |
2m
L
2
In the mathematical model the length L is known by assumption, whereas in the physical
2
model the length L is the length that is measured.
2
29
e0 0.1508323849500 m L 1.05 m L 1.25 V V /C
/4 2o
2 w oo
n2
R R ( , , ,V , L )
w2L2 w 2L2 2L2 2 w 2
k k( ,V , L ) R R
2w 2 w2L2 w2
1 2 3 4 5
V R k
2L2 2L2 -
w rad rad w2L2 2.500002274
2.500002465
- 2.8993229999 10 4 1.1684172614 10 4 -
105 3.1822388000 10 4 1.4191841515 10 4 1201.1138553246 2.500000353
5 105 3.5358835200 10 4 1.7326407898 10 4 1201.1135878711 2.500007475
104 3.8895282499 10 4 2.0460952718 10 4 1201.1131861135 2.500006704
1.5 10 4 4.2431729800 10 4 2.3595475866 10 4 1201.1127092074 2.500003427
2 104 6.3650410900 10 4 4.2402157582 10 4 1201.1121523343 2.499994957
5 1 04 9.9014875798 10 4 7.3744891363 10 4 1201.1071030963 2.500004779
103 7.3557932900 10 3 6.3754765270 10 3 1201.0922671575
10 2 1199.4555505123
Values of R R ( , , ,V ) are presented in Table 2
w2 w 2 2 2 w
TABLE 8
Values of the interference fringe shifts k k( ,V , L ) R R at / 4 .
2w 2 w2L2 w2 n2
e0 0.1508323849500 m L 1.05 m L 1.25 V V /C
/2 2o
2 w oo
n2
R R ( , , ,V , L )
w2L2 w 2L2 2L2 2 w 2
k k( ,V , L ) R R
2w 2 w2L2 w2
1 2 34 5
V R k
2L2 2L2 -
w rad rad w2L2
- 2.8349894800 10 4 -
105 2.8605699600 10 4
5 105 2.8925430500 10 4 1.1906017666104 1201.1138711255 2.500003815
104 2.9245132000 10 4
1.5 10 4 2.9564805000 10 4 1.5301090309104 1201.1137162018 2.499997839
2 104 3.1482240330 10 4
5 1 04 3.4675666035 10 4 1.9544967294104 1201.1136558912 2.49998994
103 9.1664750538 10 4
10 2 2.3788883096104 1201.1137241032 2.500004007
2.8032838497 104 1201.1139420345 2.500005549
5.3497398992104 1201.1179393790 2.499998419
9.5941484000104 1201.1352681443 2.500000583
8.606072795103 1203.7803835719 2.499996878
Values of R R ( , , ,V ) are presented in Table 4
w2 w 2 2 2 w
TABLE 9
Values of the interference fringe shifts k k( ,V , L ) R R at / 2 .
2w 2 w2L2 w2 n2
.
30
An Example.
The accuracy of the calculated length is the lowest at / 2 and V 104
2w
(the lowest k value in Table 9).
Table 9: / 2 , V 104 , L 1.25 , k k( ,V , L ) 2.49998994 .
2w 2o 2w 2
So W (2.49998994 / 2) 1.24999497
m oo
| L W | (1.25 1.24999497) 4.024 106 .
o
2 m
L 1.25
2o
The calculated accuracy applies also to the length W determined with the physical model i.e.
f
the real model.
I.2 WHY WERE THERE NO SHIFTS OF INTERFERENCE FRINGES OBSERVED
IN THE MICHELSON-MORLEY’S EXPERIMENTS?
The relative speed V of the interferometer located on the Earth is specified by the
w
relationship (1.114): 104 ≤ V 1.5104
w
Within this range of relative speeds V , the shift values are very small | k | 1.645103 (see
w
Tab.1-6), hence non-observable (Fig. 5).
I.3 WHY WAS ‘THE VALUE OF THE INTERFERENCE FRINGES SHIFT’ CALCULATED BY
ALBERT MICHELSON NOT CONFIRMED DURING THE EXPERIMENTS?
With the aim of calculating the values of the interference fringe shifts, Albert Michelson
considered the mutually perpendicular rays of light that were reaching the Z , Z mirrors.
12
This happens when the rays of light leave the slit S at the angles 0 , 0 .
0
Table 10 contains calculations which indicate that the rays of light that leave the slit S at
0
the angles 0 , 0 reach distant points A , B of the screen M. The distance between
55
the two points is equal to several hundreds wavelengths of light, therefore no interference of
the light waves occurs.
Let us introduce the following symbols:
(1.118) R R (,V ) l / the relative difference of distances traveled by the
rw rw w o
rays of light, reaching distant points A , B of the
55
screen M in the O’EQ system,
(1.119) K R ( ,V ) R ( ,V ) the difference of relative differences of distances R .
r rw 2 w rw 1 w rw
31
In accordance with the results of calculations contained in Table 10 at / 2 and
2
V 104 , the K takes the value:
wr
K R ( ,V ) R ( ,V ) 1198.4682686 1198.5038720 3.56102 .
r rw 2 w rw 1 w
The calculated value of K 3.56102 is not the shift value k . The distance | e e |
r a5 b5
between the points A5 and B5 on the screen M which were reached by the rays of light
equals: 273.974180 when 2 /2 and 985.786396 o at 1 0 .
o
It is evident that by assuming perpendicularity between the light rays and the Z , Z mirrors,
12
Albert Michelson actually calculated the value of │ K r │ 0.04 (1.119) and not the shift
value k (1.113).
V V / C 104
w oo
0, 0 K R ( ,V ) R ( ,V )
r rw 2 w rw 1 w
12 34 5
e ( 0, ,V ) e ( 0, ,V ) |e e |/ R ( ,V )
1 a5 1w b5 1 w a5 b5 o
rw 1 w
rad m m - -
985.786396 1198.5038720
0 0.1499477388 0.1505293528
e ( 0, ,V ) e ( 0, ,V ) |e e |/ o R ( ,V )
a5 2 w b5 2 w a5 b5
2 m m rw 2 w
rad -
0.1501527315 0.1503143762 -
/2 273.974180
1198.4682686
TABLE 10
The table presents the values of: R ( ,V ) , R ( ,V ) and | e e | / together with
rw 1 w rw 2 w a5 b5 o
the coordinates e , e of the A , B points reached by the light rays that have left the slit S
a5 b5 55 0
at the angles 0 , 0 and V V / C 104 . These calculations were carried out with
w oo
the computational program PROGRAM IntM (see Chapter IV).
I.4 THE VELOCITIES AT WHICH THE CENTERS OF THE EARTH AND THE SUN TRAVEL
WITH RESPECT TO THE AETHER
(in relation to a specific absolute OXoYoZo system)
In relation to the aether, the interferometer velocity V on the Earth’s surface is the sum of
o
three vectors:
(1.120) V V V V
o r zs se
The vector V is the peripheral velocity of a point i.e. a place on the Earth’s surface where
r
the interferometer is located. The plane of this vector is parallel to the one at the equator.
Its modulus value equals: V 0.464 cos km / s , where:
r
is the latitude of the interferometer’s position.
32
The vector V is the velocity of the Earth’s center around the Sun. This vector is located on
zs
the Earth’s ecliptic plane.
V 29.29 km / s , V 30.28 km / s
zs min zs max
In our considerations an approximate modulus value of the vector V will be adopted, namely
zs
V 30 km / s
zs
(1.121) V / C 104
zs o
The vector V is the velocity of the Sun’s center with respect to the aether. This vector is
se
perpendicular to the ecliptic plane, which is conclusive from starlight aberration.
The vector V can be omitted due to its small modulus value compared to that of the V vector.
r zs
Consequently the equation (1.120) takes the following form:
(1.122)
V V V
o zs se
Since the vectors V ,V are mutually perpendicular, the following equation can be written:
zs se
(1.123) V 2 V 2 V 2
o zs se
According to (1.114): 104 ≤ V 1.5104 , V V / C and therefore
w w oo
(1.124) 104 ≤ V / C 1.5104
oo
The interferometer is located on the Earth’s surface so its velocity V is equal to the
o
velocity of the point on the Earth’s surface (a laboratory) with respect to the aether, which is
approximately the velocity V of the Earth’s center with respect to the aether:
ze
(1.125) V V
o ze
After considering the inequality (1.124) we obtain:
(1.126) 104 ≤ V / C 1.5104 , Vze Vo
ze o
This inequality (1.126) specifies the speed of the Earth’s center relative to the aether,
expressed with respect to the speed of light Co .
The speed Vse of the Sun center with respect to the aether can be determined from the four
relations i.e. (1.121), ((1.123), (1.125) and (1.126).
From the equation (1.123): V 2 V 2 V 2
se o zs
and the equation (1.125) we obtain:
V 2 V 2 V 2
se ze zs
consequently after applying (1.121) and (1.126), we further obtain:
(1.127) 0 ≤ V / C 1.12104
se o
The inequality (1.127) specifies the speed of the Sun’s center relative to the aether,
expressed with respect to the speed of light Co .
33
I.5 THE VELOCITY AT WHICH THE CENTER OF OUR GALAXY TRAVELS
WITH RESPECT TO THE AETHER
(with respect to a specific absolute OXoYoZo system)
With respect to the aether, the center of the Sun travels at the velocity V which is the sum
se
of the following vectors:
(1.128) V V V
se sg ge
The vector V is the velocity with which the Sun center rotates around the center of our
sg
Galaxy. It takes an approximate modulus value of: V 250 km / s
sg
(1.129) V / C 8.33104
sg o
The vector V is the velocity at which the center of our Galaxy moves with respect to the
ge
aether.
From the equation (1.128) we ob tain:
(1.130) V V V
ge sg se
Then from (1.127) , (1.129) and (1.130) we can determine the speed V of the Galaxy
ge
center with respect to the aether:
(1.131) (8.33 1.12) 104 Vge / Co (8.33 1.12) 104
The inequality (1.131) specifies the speed of the Galaxy center relative to the aether,
expressed with respect to the speed of light Co .
Knowing the apex A ( , ) of solar motion around the Galaxy center, we can estimate
sg sg sg
approximately the apex A ( , ) of the Galaxy center’s motion with respect to the aether:
ge ge ge
,
ge sg
180o ,
ge sg
where: , declination of apexes,
sg ge
, right ascension of apexes.
sg ge
34
CHAPTER II
THE VELOCITY OF THE INTERFEROMETER
The interferometer absolute velocity V is the sum of three vectors:
o
V V V V as in relation (1.120)
o r zs se
where: V peripheral velocity of the point U on the Earth’s surface where
r
the interferometer is located,
V the velocity at which the Earth’s center revolves around the Sun,
zs
V the velocity at which the Sun’s center travels relative to the aether.
se
The aether-relative velocity V of the Sun center is perpendicular to the plane of the ecliptic.
se
However, the direction of that velocity (a vector) is not known. Hence in our deliberations, we
will consider two vectors perpendicular to the ecliptic plane, namely:
vector V and vector V V .
se se1 se
Thus two vectors are obtained:
(2.1) V V V V
01 r zs se
(2.2) V V V V
02 r zs se1
Therefore the interferometer absolute velocity V is:
o
either the vector V V or the vector V0 V02
0 01
In this chapter the coordinates of the vectors V and V were established in the
01 02
horizontal coordinate system.
35
II.1 THE PERIPHERAL VELOCITY Vr OF THE U (, ) POINT ON THE EARTH’S SURFACE
Fig. 8
Peripheral velocity V and its azimuth A
rr
SYMBOLS:
U (, ) a location (point) with geographical coordinates , ,
at which the interferometer has been located,
lp a vertical line which runs through the point U (, ) and the center of the globe 0,
pha the plane of astronomical horizon i.e. its projection, which runs through the globe
center 0 and is perpendicular to the vertical line lp ,
ph the plane of the horizon i.e. its projection, which runs through the point
U (, ) and is perpendicular to the vertical line lp ,
N the northern point of the horizon,
S the southern point of the horizon,
Nu Su line the line of intersection between the horizon plane and the meridian plane, both of
which run through the U (, ) point,
R the radius of the globe,
Ar the azimuth of the Earth’s peripheral velocity Vr .
The peripheral speed Vr of the point U (, ) :
(2.3) Vr R cos
where:
the angular speed of the Earth’s rotation.
The peripheral velocity V is located on the horizon plane which runs through
r
the point U (, ) .
36
II.2 THE VELOCITY V AT WHICH THE EARTH’S CENTER REVOLVES AROUND THE SUN
zs
.
Fig. 9 The Earth’s motion on its orbit around the Sun.
SYMBOLS: an average Earth-Sun distance,
a small semi-axis of the Earth’s orbit,
a a radius vector,
b true anomaly,
r
an angle ∡ (r,Vzs ) ,
p annual precession within ecliptic (in longitude),
e the eccentricity of the Earth’s orbit,
Sn the Sun,
P z a point on the orbit in which the center of the Earth is located,
V
the velocity at which the center of the Earth revolves around the Sun,
zs
Winter’s position (Earth’s location when astronomical winter starts),
A1 Summer’s position (Earth’s location when astronomical summer starts),
A2 Earth’s location at the time of spring equinox,
B1 Earth’s location at the time of spring equinox of previous tropical year,
B1’ Earth’s location at the time of autumn equinox,
B2 Earth’s location at the beginning of a new calendar year of the UT time,
No Universal Time.
UT
B1’A2 spring, A2 B2 summer,
B2 A1 autumn, A1 B1 winter.
37
II.2.1 DETERMINING THE ANGLE
In the OX1Y1 system the coordinates of the Earth’s center on the orbit are defined as follows:
(13*) x ea r cos ea a cos (1 e2 ) , r a(1 e2 )
z 1 e cos 1 e cos
(14*) y r sin a sin (1 e2 )
z 1 e cos
The equation of the line tangent to the Earth’s orbit in the P (x , y ) point is:
zz z
xx yy b a 2 (ea)2
1 z 1 z 1,
a2 b2
After transformation we obtain: y b2x x b2
z
1 a2y 1 y
z z
Thus the angular coefficient of the line tangent to the orbit in point P (x , y ) equals:
zz z
(15*) tg b 2 xz (b / a) 2 xz
3 a2 yz yz
Applying equations (13*), (14*) we obtain a quotient:
x e(1 e cos ) 1
z
y sin (1 e2 ) tg
z
From the equation (15*) we obtain:
(2.4) arctg [ (b / a)2 ( e(1 e cos ) 1 )], 0 , 180o , 360o
3 sin (1 e2 ) tg
(2.5) | | (Fig. 9)
23
(2.6) arctg b (Fig. 9) so
0 ea
(2.7)
(2.8) when 0 ≤180o
(2.9) 2 0
(2.10)
when 180o 180o
2 0
180o when 180o ≤ 180o
2 0
180o when 180o 360o
2 0
where: true anomaly (Fig. 9).
II.2.2 DETERMINING THE ANGLE
The true anomaly is the angle between the radius vector r and the direction from the Sun
center towards the point on the orbit nearest to the Sun i.e. the perihelion.
Corresponding to a specific time, the angle can be determined from the Kepler second law:
r 2 d C const
dt 1
r a(1 e2 ) (the radius vector)
1 e cos
From the above the following integral is obtained:
t( ) [a(1 e2 )]2 d
C (1 e cos ) 2
1
38
After integration we have:
(2.11) t( ) [a(1 e2 )]2 [ e sin 1 ( 2 arctg 1 e2 tg( / 2) )]+ C
C (1 e2 )(1 e cos ) 1 e2 1 e2 1 e 2
1
Let us adopt an initial condition:
0 t 0, hence the integration constant C 0 , so
2
t( ) a 2 (1 e2 )e ( 2 arctg 1 e2 tg( / 2) sin )
C 1 e 1 e cos
e 1e2
1
T where: T is the stellar year, we can
From the condition that 180o t( ) rg
rg
2
determine the C constant
1
T a 2e(1 e2 ) 2 hence:
rg
2 C e 1e2 2
1
(2.12) C 2 a 2 (1 e2 ) then
1 T 1e2
rg
(2.13) e 1 e2 T ( 2 arctg 1 e2 tg( / 2) sin )
t( )
rg
2 e 1 e2 1 e 1 e cos
The t( ) function is of negative value when 180o .
In order to avoid negative time values we introduce two functions:
(2.14) t ( ) t( ) when 0 ≤ ≤ 180o
(2.15) when 180o 360o
1
t ( ) T t( )
2 rg
Then we define the following symbols:
T tropical year
rz
T the duration of astronomical winter.
z
T t (90o ) T t (360o ) (Fig.9) which after transformation
z1 4 rg 2 1
T t(90o ) t(360o )
z4 1
Angle p (Fig. 9), where:
41
(2.16) p (Tz / Trz ) p precession in the ecliptic (in longitude) during the time of
astronomical winter.
Therefore
(2.17) Tz t(90o p 1) t(360o 1) .
If the astronomical winter duration time T is known, the angle can be determined from
z1
the equation (2.17) by the method of successive approximations.
Let us say that Ta means the time which has elapsed from the moment the astronomical
winter of the UT time started (point A , Fig. 9) up to the moment the Earth is nearest to the
1
Sun (the perihelion).
Ta can be determined from the relationship:
Ta Trg t2 (360o 1) t(360o 1) t(1)
(2.18) Ta t(1)
Then let us say that Tb means the time which has passed from the start of astronomical
winter up to the end of a calendar year of the UT time (point No , Fig. 9).
Tb Ta
39
The difference of the Ta ,Tb times equals:
Ta Tb Trg t2 (360o 5 ) .
After transforming the equation, the following is obtained:
Ta Tb t(360o 5 ) t(5 ) so
(16*) Ta Tb t(5 )
Referring to equation (2.11) and adopting an initial condition:
t( ) 0
5
with a constant value C specified by the relationship (2.12),
1
an integration constant C can be calculated:
2
C e 1 e2 T ( 2 arctg 1 e2 tg( / 2) sin ) t( ) ,
5 5
rg 1 e
2 2 e 1 e 2 1 e cos 5
5
C t( ) .
25
Having considered the equation (16*) we obtain:
(2.19) C2 Ta Tb
Now we can specify the relationship between the UT time and the angle i.e. true anomaly:
(2.20) t3( ) t( ) (Ta Tb ) when 0 ≤ ≤ 180o
(2.21) t4 ( ) Trg t( ) (Ta Tb ) when 180o 360o
From equations (2.20) and (2.21) the value of the angle for any given time UT can be
calculated with the use of the method of successive approximations.
II.2.3 AZIMUTH AND THE ALTITUDE OF THE EARTH’S CENTER VELOCITY V
zs
The definitions that follow refer to to the following vectors: V , V and V V .
zs se se1 se
The V and V vectors are also the velocities of the Earth’s center. The declination of a
se se1
vector is the angle between the vector and the plane of the celestial equator.
The Greenwich hour angle GHA of a vector is a dihedral angle between the semi-circle of the
celestial meridian in Greenwich and the hour semi-circle which runs through the vector. The
GHA angle counting starts at the semi-circle of the celestial meridian in Greenwich and up
towards the West.
The local hour angle LHA of a vector is a dihedral angle between the celestial meridian
semi-circle of the observer and the hour semi-circle which runs through the vector.
The right ascension of a vector is a dihedral angle between the hour semi-circle which
runs through the spring equinoctial point i.e. the Aries point and the hour semi-circle which
runs through the vector. The right ascension counting starts at the Aries point up towards the
East.
The altitude H of a vector is an angle between the vector and the horizon plane.
Starting from the northern point of the horizon, the azimuth A of a vector is a dihedral angle
between the celestial meridian of the observer and the semi-circle which runs through the
vertical line and the vector; whereas starting from the northern direction ( Nu ,, Fig.12) the
azimuth A of a vector is an angle between the Nu Su line and the projection of the vector on
the horizon plane that runs through the point U (, ) .
The observer is located in the same place as the interferometer.The above definitions
correspond to the definitions which refer to celestial bodies.
40
Fig. 10 Coordinates of the equatorial system: declinations, right ascensions.
SYMBOLS: O the center of the globe,
1 celestial equator,
2 ecliptic,
3 celestial zone,
4, 5 hour semi-circles,
Sn the Sun,
PB the Aries point,
inclination of the ecliptic to the equator,
right ascension of the Sun,
s
s declination of the Sun,
right ascension of the Vzs velocity,
zs
declination of the Vzs velocity,
zs
an angle ∡ (r,Vzs ) relationships (2.7 – 2.10)
From two perpendicular spherical triangles, shown in Fig.10, the right ascension as well
zs
as the declination of the Earth’s center velocity V will be determined.
zs zs
(2.22) tg tgk cos , hence 0 ≤ 90o
(2.23) s1 when s
(2.24) when
tg when 90o 270o
k arctg s s
1 cos
270o 360o
tg s
k 180o arctg s
1 cos
tg
k 360o arctg s
1 cos
41
(2.25) k k (Fig. 10).
21
(2.26)
(2.27) tg tgk cos , hence
zs 2
arctg(tgk cos ) when 90o k 90o
zs 2
2
zs 180o arctg(tgk2 cos ) when
90o k 270o
2
sin sin k sin , hence
zs 2
(2.28) arcsin (sin k sin )
zs 2
The angles in the equatorial system which are necessary to determine the coordinates of the
vector V in the horizontal system are:
zs
(2.29) GHA GHAaries
zs zs
LHA GHA so
zs zs
(2.30) LHA GHAaries
zs zs
where:
GHAaries Greenwich Hour Angle of the Aries Point,
GHA Greenwich Hour Angle of the V velocity,
zs zs
Local Hour Angle of the V velocity,
LHA
zs
zs right ascension of the V velocity,
ZS zs
longitude of a place (point U) where the interferometer is located.
The altitude H of velocity V in the horizontal system:
zs zs
sin H cos cos cos LHA sin sin , where:
zs zs zs zs
the latitude of a place (point U, Fig. 8) where the interferometer is located.
Hence:
(2.31) H arcsin(cos cos cos LHA sin sin)
zs zs zs zs
The azimuth A of velocity V , calculated within the range from 0 to 360 o starting from
zs zs
the northern point of the horizon, is expressed as:
sin A cos zs sin LHA
zs cos H zs
zs
sin sin H sin
cos A zs zs
zs cos H cos
zs
Let us introduce the following symbols:
sin sin H sin
(2.32) d zs zs , ( = )
(2.33) d cos A
zs cos H cos zs
zs
zs
z d /|d | , Azs 900 , Azs 2700 . Therefore
zs zs zs
(2.34) A 90o (3 z )z arcsin ( cos sin LHA )
zs
zs zs zs cos H zs
zs
II.2.4 THE SPEED V AT WHICH THE EARTH’S CENTER REVOLVES AROUND THE SUN
zs
The speed V at which the center of the Earth revolves around the Sun can be calculated
zs
from the Kepler second law: r 2 d C
dt 1
42
We can write r ( r d ) m C m where: m the mass of the planet.
dt 1
r d V cos( 90o ) V sin(180o ) so
dt zs zs
(17*) rV m sin(180o ) C m
zs 1
The left-hand side of the equation (17*) expresses the modulus of the planet’s angular
momentum (Fig. 9). From the equation ( 17*) we obtain:
C where: C 2 a 2 (1 e2 ) relationship (2.12),
V 1 , 1 T 1e2
zs r sin rg
r a(1 e2 ) the value of the radius vector,
1 e cos relationships (2.7 - 2.10),
angle
Hence T stellar year.
rg
(2.35) V 2 a(1 e cos )
zs T 1 e2 sin
rg
II.3 THE VELOCITIES V AND V V AT WHICH THE SUN CENTER MOVES WITH
se se1 se
RESPECT TO THE AETHER
Fig. 11
The coordinates of the vectors V and V V (declinations, right ascensions).
The V se se1 se
and V vectors are also the velocities of the Earth’s centre.
se
se1
43
SYMBOLS IN FIG. 11:
0 the center of the globe,
1 plane of the celestial equator (its projection),
1 a the celestial equator,
2 plane of the ecliptic (its projection),
the inclination of the ecliptic to the equator,
PB the Aries point,
Vse
Vse1 Vse the velocity at which the Sun center moves with respect to the aether,
se the velocity at which the Sun center moves with respect to the aether,
se1
right ascension of the Vse velocity,
right ascension of the V V velocity.
se1 se
Vector Modulus of the Right ascension of Declination of the
vector the vector vector
Vse
Vse 270o 90o
Vse
Vse1 se se
90o (90o )
se1 se1
TABLE 11 (refers to Fig. 11)
The following relationship specifies the speed of the Sun’s center relative to the aether,
expressed with respect to the speed of light C0 :
0 ≤ V / C 1.12104 (1.127).
se o
II.3.1 VELOCITY
AZIMUTH AND THE ALTITUDE OF THE V
se
The Local Hour Angle LHA of the V velocity:
se se
(2.36) LHA GHAaries
se se
The altitude H of the V velocity:
se se
(2.37) H arcsin(cos cos cos LHA sin sin)
se se se se
The azimuth of the V velocity, calculated within the range from 0 to 360o starting from the
se
northern point of the horizon is:
sin A cos se sin LHA
se cos H se
se
sin sin H sin
cos A se se
se cos H cos
se
Let us introduce the following symbols:
sin sin H sin
(2.38) d se se , ( = )
(2.39) d cos A
se cos H cos se
se
se
z d /|d | , Ase 900 , Ase 2700 . Therefore
se se se
(2.40) A 90o (3 z )z arcsin ( cos sin LHA ).
se
se se se cos H se
se
44
II.3.2 THE AZIMUTH AND THE ALTITUDE OF THE V VELOCITY
se1
The Local Hour Angle LHA of the velocity V V :
se1 se1 se
(2.41) LHA GHAaries
se1 se1
The altitude H of the V velocity:
se1 se1
(2.42) H arcsin(cos cos cos LHA sin sin)
se1 se1 se1 se1
The azimuth of the V velocity is calculated within the range from 0 to 360o starting from the
se1
northern point of the horizon as follows:
sin A cos se1 sin LHA
se1 cos H se1
se1
sin sin H sin
cos A se1 se1
se1 cos H cos
se1
We introduce the following notations:
sin sin H sin
(2.43) d se1 se1 , ( = )
(2.44) d cos A
se1 cos H cos se1
se1
se1
z d /|d | , Ase1 900 , Ase1 2700 . so
se1 se1 se1
(2.45) A 90o (3 z )z arcsin ( cos sin LHA )
se1
se1 se1 se1 cos H se1
se1
The angles , are the geographical coordinates of the U point (Fig. 8) in which the
interferometer is located.
The previously introduced relationships (2.31), (2.34), (2.37), (2.40), (2.42) and (2.45) for
calculating the altitudes and the azimuths of velocities relate to the astronomical horizon
plane which runs through the globe center.
This plane is perpendicular to the vertical line running through the U (, ) point (Fig. 8).
The abovementioned relationships apply as well as to the horizon plane which runs through
the U (, ) point and is also perpendicular to the vertical line.
II.4 SUM OF VELOCITIES IN THE HORIZONTAL SYSTEM
Let us introduce a rectangular system of coordinates O’EQW (Fig. 12) with the two axes O’E
and O’Q on the horizontal plane which runs through the point U (, ) . The O’E axis coincides
with the Nu Su line. The O’W axis coincides with the vertical line which runs through the point
U (, ) (Fig. 8).
45
Fig.12 The rectangular system of coordinates O’EQW
The V ' ,V ' ,V ' vectors represent the projections of these vectors on the horizon plane
zs se se1
which runs through the point U (, ) .
The coordinates of the velocities: 0]
V [ 0, V,
r rq V]
V [ V , V ,
zs zse zsq zsw
V [ V , V , V]
se see seq
sew
V [ V , V ,
se1 se1e se1q V]
se1w
(2.46) Vrq Vr R cos (2.3)
(2.47) V V cos H cos A (2.31), (2.34)
(2.48) zse zs zs zs
V V cos H sin A
zsq zs zs zs
(2.49) V V sin H
(2.50) zsw zs zs
(2.51)
V V cos H cos A (2.37), (2.40)
see se se se
V V cos H sin A
seq se se se
(2.52) V V sin H se
(2.53) sew se
(2.54)
V V cos H cos A (2.42), (2.45)
se1e se se1 se1
V V cos H sin A
se1q se se1 se1
(2.55) V V sin H se1
se1w se
II.4.1
VELOCITY V V
(2.1)
0 01 46
V V V V
01 r zs se
V, V, V]
The coordinates of the velocity V [
01e 01q 01w
01
(2.56) V V V
01e zse see
(2.57) V V V V
01q rq zsq seq
(2.58) V V V
01w zsw sew
The modulus of the velocity V :
01
(2.59) V V 2 V 2 V 2
01 01e 01q 01w
The altitude H and the azimuth A of the V velocity:
01 01 01
(18*) V V cos H cos A
01e 01 01 01
(19*) V V cos H sin A
01q 01 01 01
(20*) V V sin H
01w 01 01
From the equation (20*) the altitude H of the velocity V can be determined:
01 01
(2.60) V
H arcsin 01w
01 V
01
V 01q
From the equation (19*) we obtain: sin A
01 V cos H
01 01
Let us introduce the following notation:
(2.61) z V / |V | , A01 900 , A01 2700 .
01 01e 01e
The azimuth A of the velocity V calculated within the range from 0 to 360 o starting
01 01
from the northern point of the horizon is:
A 90o (3 z ) z arcsin V 01q
(2.62) 01 01 01 V cos H
01 01
(2.2)
II.4.2 VELOCITY V V V]
0 02 02w
V V V V
02 r zs se1
The coordinates of velocity V :
02
V [ V , V ,
02 02e 02q
(2.63) V V V
02e zse se1e
(2.64) V V V V
02q rq zsq se1q
(2.65) V V V
02w zsw se1w
The modulus of the velocity V :
02
(2.66) V V 2 V 2 V 2
02 02e 02q 02w
The altitude H and the azimuth A of the velocity V .
02 02 02
(21*) V02e V02 cos H 02 cos A02
(22*) V V cos H sin A
02q 02 02 02
(23*) V V sin H 02
02w 02
From the equation (23*) the altitude H of the vector V can be determined:
02 02
47
(2.67) V
H arcsin 02w
02 V
02
V 02q
From the equation (22*) we obtain: sin A
02 V cos H
02 02
Let us introduce the following notation:
(2.68) z V / |V | , A02 900 , A02 2700 .
02 02e 02e
The azimuth A of the velocity V calculated within the range from 0 to 360o starting from
02 02
the northern point of the horizon is:
A 90o (3 z ) z arcsin V 02q
(2.69) 02 02 02 V cos H
02 02
Parameter The value of the parameter
a 149597103 km
e 0.01671
23o.439 0.4090877 rad
p 50".292
T 365 d .256366
rg
T 365 d .242199
rz
R 6378.1 km
7.292115105 rad/s
TABLE 12
Table 12 gives the values of astronomical parameters, used in a computation program,
referred to as PROGR AM Vo in Chapter IV, to calculate the coordinates of velocities:
V, V01 (2.1), V02 (2.2).
zs
II.5 AN EXAMPLE
We are to calculate the coordinates of the V , V (2.1) and V velocities (2.2) at the U
zs 01 02
point (Fig.8) with its geographical coordinates 50o34' , 21o41' on 15th December 2009
at 10.30 UT. The coordinates of the vectors should be determined in a horizontal system.
In order to solve the problem we will use the previously mentioned PROGRAM Vo (see
Chapter IV). In addition to the astronomical quantities, contained in Table 12 and introduced
into the program, we also need to introduce the values of the angles corresponding to the
case-specific time, namely:
- right ascension of the Sun,
s
- Greenwich Hour Angle of the Aries point GHAaries ,
- angle (true anomaly).
The values of both i.e. the sun right ascension and the Greenwich Hour Angle of the Aries
point can be found in the astronomical annual and they read as follows:
s 263o0'.5 263o.008333
GHAaries 241o42'.9 241o.715000
48
The value of the angle can be calculated from relationships (2.13) - (2.21).
Astronomical winter duration time T .
z
Astronomical winter started on 21st December 2008 at 12.04 UT.
Astronomical spring started on 20th March 2009 at 11.44 UT.
Hence the astronomical winter duration time T in the years 2008 - 2009 equals:
z
Tz 88d 23h40m 88.986111 days .
Precession (in longitude) during astronomical winter:
from the relationship (2.16) p (Tz / Trz )50''.292 12''.252 0o.003403 .
From the equation (2.17) 88.986111 t(90o 0o.003403 1) t(3600 1) and with the use of
the method of successive approximation, the value of the angle can be calculated:
1
1 13o.212402
From the relationship (2.18): Ta t(1) 12.966631 days .
Tb is the time that elapsed from the start of the 2008 astronomical winter until the end of the
2008 calendar year i.e. Tb 10d11h56m 10.497222 days
hence Ta Tb 2.469409 days .
180o 360o
The time t ( ) that elapses from the start of the 2009 calendar year until 10.30 UT on 15th
4
December 2009 will amount to:
t4 ( ) 348d10h.5 348.437500 days .
From the equation (2.21) we have:
348.437500 Trg t( ) 2.469409 and with the use of the method of
successive approximations, the value of the angle can be calculated:
340o.353014
Having introduced to PROGRAM Vo the values of the following angles:
50o.566666 s 263o.008333
21o.683333 GHAaries 241o.715000
and 340o.353014
we obtain the coordinates of velocities V , V and V in the horizontal system.
zs 01 02
THE RESULTS OF CALCULATIONS:
Vzs 30.257974 km / s
Vector V H zs 2o.506181
zs Azs 270o.768878
V0 V01 44.793772 km / s V0 V02 44.773691 km / s
Vector V H 01 47o.951082 Vector V H 43o.141687
01 A01 288o.751884 02 02
(2.1) (2.2) A02 254o.313146
49
CHAPTER III
NEWTON’S SECOND LAW OF MOTION
Michelson-Morley experiments and the values of the interference fringe shifts, calculated
from the mathematical model, confirm the premise of the existence of the aether and the
applicability of the Galilean transformation.
Therefore let us apply the Galilean transformation.
Fig. 13
Then let us introduce two rectangular coordinate systems (Fig. 13).
1) Preferred absolute inertial rectangular coordinate system 1, named
OXoYoZo, motionless with respect to the aether.
2) An inertial system 2 i.e. the O’EQW system that is in motion relative to
the system 1 with constant velocity V .
o
Axis O’E is parallel to axis OXo.
Axis O’Q is parallel to axis OYo.
The times in both inertial systems 1 and 2 are equal: t2 t1 t .
The velocity V of particle P relative to the inertial system 1 (Fig. 13) equals:
1
(3.0) V V V ,
1o2
where: V the velocity of particle P in the inertial system 2.
2
The accelerations of particle P in inertial system 1 and 2 respectively:
dV dV d(V V ) dV
, 1 o 2 , then .
1 a 2 a a2 a1
dt 1 dt dt dt 2
50
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1 Bronislaw Maciag Jan Maciag THE AETHER & THE GALILEAN TRANSFORMATION Tarnobrzeg 2012
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