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The Michelson-Morley experiment [1] is a well-known experiment, but not well understood in details. The paper “A new derivation of the Michelson-Morley experiment [2] presents in details the first two positions of Michelson’s interferometer. This

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Page 1 of 9
A new derivation of the Michelson-Morley experiment Four positions of the interferometer
Filip Dambi E-mail: filipdambi1@gmail.com
1.
Introduction
The Michelson-Morley experiment [1] is a well-known experiment, but not well understood in details.
The paper “A new derivation of the Michelson
-Morley experiment [2] presents in details the first two positions
of Michelson’s
interferometer. This paper completes the derivation of the light paths and the fringe shift in Michelson-Morley experiment for four positions of the interferometer.
2.
Derivation of the light paths with the initial position of the interferometer
Figure 1 illustrates the initial position of the interferometer. The interferometer is
moving with the Earth’s inertial frame in
a fixed frame. Throughout this paper, points marked by a letter without an index are points as seen by an observer in the inertial frame. Points marked by a letter with an index are instances of inertial frame points in the fixed frame. For example, point A is a physical point on the beam splitter seen in the inertial frame and points A0, A1, A2, A3, A4, and A5 are different instances of point A in the fixed frame. The veloci
ty of the Earth’s orbit around the Sun
=3×10
/
, the velocity of light
=3×10
/
, the wavelength of light
=550×10
−
,
aberration angle
=arcsin/
, and length of the interferometer arms
=11
. In figure 1, each ray of light from the source is split into two light rays by the beam splitter. The light rays that travel through the beam splitter
M
toward mirror
M
1
are called transmitted rays and are drawn in red. The light rays that are reflected by the beam splitter
M
toward mirror
M
2
at the aberration angle are called reflected rays and are drawn in blue. The interfering rays are drawn in green. The light path geometry is depicted so that the reflected rays travel with the aberration angle, and the transmitted and reflected rays that interfere coincide. The derivation of the light paths starts when the wavefront of light from the source is at line
A
1
B
1
. Points
A
,
B
,
D,
and
G
are aligned vertically. Point
B
is a fixed point in the interferometer space and does not belong to the beam splitter as point A. The point under observation is point
A
. There is only one transmitted ray that starts at point
B
that continuously intercepts point
A
. The light from
B
1
intercepts point
A
at
A
4
after traveling the path length
. The light at
A
4
is reflected toward the screen, making the angle
C
1
A
4
G
2
, which is equal to
90°
, and
Page 2 of 9 therefore coincides with the reflected ray at
A
4
. The instance when the light from
B
1
arrives at
E
1
includes points
E
1
,
F
1
,
C
1
, and
A
3
. In time
, the light travels from
B
1
to
E
1
and point
A
travels from
A
1
to
A
3
. In time
, the light travels from
E
1
to
A
4
and point
A
travels from
A
3
to
A
4
. The time
=
/=
tan/
, and the time
=
tan/cos2+
. The distance
=
2
and then
=
2tan/cos2+
⇒
=/ cos2+c2tan+
. The distance
can be calculated and then the time
=
tan/
. The time
=
+
=7.333,333,333,333,33008
. In time
, the light travels from point
B
1
to point
A
4
and the beam splitter travels the distance
A
1
A
4
. The path length
and the distance
A
1
A
4
are:
=
=2.200,000,000E+01=4.000,000,000E+07 and
=
=2.200,000,000E03=4.000,000,000E+03
.
M
2
M
1
A
0
A
1
A
2
A
3
A
4
A
5
D
1
D
2
M I
1
H
1
B
1
B
2
SourceScreen
E
1
E
2
F
1
aaaaa
2/2
a
2
v C
1
45°
G
1
G
2
G
3
Figure 1.
Light path geometry with the initial position of the interferometer. The reflected light rays travel at an angle
a
equal to the aberration angle; thus, the reflected ray that starts at point A intercepts the same point
A
continuously. The light from
A
1
intercepts point
A
5
after traveling the path length
. In triangle
A
1
D
1
D
2
, the speed along
A
1
D
1
is
×
. In time
, the light travels from point
A
1
to point
D
1
then back to point
A
5
, and the beam splitter travels the distance A
1
A
5
.
The time
is defined as:
= 2/ =7.333,333,370,000,000E08s.
The path length
and distance
A
1
A
5
are:
=
=2.200,000,011E+01=4.000,000,020E+07 and
Page 3 of 9
=
=2.200,000,011E03=4.000,000,020E+03
. The difference
∆
of the path lengths
and
and the distance
A
4
A
5
are:
∆
=
=1.100E07=2.000E01 and
=
=1.100E11=2.000E05
.
3.
Derivation of the light paths with the interferometer rotated 90° from its initial position
Figure 2 depicts the light paths of the interferometer rotated 90° clockwise from its initial position shown in figure 1. The distances
AP
and
AH
are the arm length equals to L. The derivation of the light paths starts when point
A
of the beam splitter is at instance
A
1
and the wavefront of light coming from the source is at line
B
1
C
1
. Points
A
1
,
E
1
,
B
1
,
C
1
,
D
1
,
J
1
,
U
1
, and
V
1
belong to the same instance, and points
B
1
, and
C
1
belong to the same wavefront. Points
B
, and
E
are fixed points in the interferometer space and do not belong to the beam splitter as points
A
and
C
. Points
A
,
D
, and
P
are vertically aligned. The point under observation is point
A
. There is only one reflected ray that starts at point
C
that continuously intercepts point
A
. The light from
C
1
intercepts point A at A
4
after traveling the path length
. In triangle
C
1
F
1
G
1
, the speed along
C
1
G
1
is
×
and the speed along
G
1
F
1
is
×a=
. The time
, in which the light travels from point
C
1
to point
F
1
, is given by:
=
+ =
⇒
=
+ .
The instance when the light from
C
1
arrives at
F
1
includes points
F
1
,
G
1
, and
H
1
. In triangle
A
4
F
1
H
1
, the speed along
H
1
A
4
is
×
, and the speed along
F
1
H
1
is
×=
. The time
, in which the light travels from point
F
1
to point
A
4
, is given by:
=+ =
⇒ F
H
=+ .
Triangle
gives: /4/2 =
⇒
=
/4/2=
+
/4/2.
Substituting the formula for
F
1
G
1
and
F
1
H
1
in the formula for
C
1
D
1
:
=
+/4/2 +/4/2+ ⇒
(1 /4/2 )= /4/2(1 +1+ ).
The distance
C
1
D
1
can be calculated, and then distance
A
1
D
1
and time
can be calculated as follows:
=
/4/2 and
=
+
=
+ ++ =7.334,066,850,025,67008.
The light path length
and distance
A
1
A
4
are:
=
=2.200,220,055E+01=4.000,400,100E+07 and
=
=2.200,220,055E03=4.000,400,100E+03
.
Page 4 of 9
v C
0
C
1
D
1
B
1
B
2
A
0
A
1
A
2
A
3
A
4
E
1
F
1
H
1
G
1
J
1
J
2
P
1
M
2
M
1
M 2aaa J
2
SourceScreenDetail
1
S
1
RV
1
U
1
P
1
U
1
V
1
45°-a/2 J
1
Figure 2.
Light path geometry when the interferometer is rotated by 90° from its initial position. There is only one transmitted ray that starts at point
B
that continuously intercepts point
A
. The light from
B
1
intercepts point
A
at
A
3
after traveling a path length of
. In time
, the light from point
B
1
travels to point
J
2
, point
A
travels from
A
1
to
A
2
, and point
J
travels from
J
1
to
J
2
. In triangle
J
1
J
2
U
1
, point
J
travels along
J
1
J
2
with the speed
, and the speed
along
U
1
J
2
is
′
=
. A ray from the source sees the mirror
M
1
coming toward it at the speed
′
. In triangle
P
1
U
1
V
1
,
=
tan
. The time
=
++
/+
, then
=
tan
, and
=
. The time
, in which the light travels from point
J
2
to point
A
3
, is given as:
=
+
cos2,then
=
,and
=
sin2.
The distance
=
=
+
⇒
=
tan=
tan+
tan
tan=
tan+
tan
sin2tan
. Substituting the formula for J
2
U
1
and U
1
V
1
in the formula of time
⇒
=
tan+
tan+
tan
sin2tancos2 ⇒
=cos2+sin2tantan.
The time
can be calculated, then
=
++
tan+
tanc
sin2tan+ ⇒
Page 5 of 9
=
++
tanc
sin2tan.
The time
=
+
=7.334,066,740,018,34008
, and the path length
and distance A
1
A
3
are:
=
=2.200,220,022E+01=4.000,400,040E+07 and
=
=2.200,220,022E03=4.000,400,040E+03
.
The difference ∆
L
2
of the path lengths and distance
A
3
A
4
, for this position, are:
∆
=
=3.30007=6.000E01 and
=
=3.300E11=6.000E05
. Rotating the interferometer from the initial position to the 90° position, the difference
∆
=∆
∆
=2.20007=4.00001
. The fringe image is then expected to shift by
=∆
⁄=0.400
fringes in the positive direction from the initial position. The positive direction is chosen arbitrarily.
4.
Derivation of the light paths with the interferometer rotated by 180° from its initial position
Figure 3 depicts the light paths of the interferometer rotated 180° clockwise from its initial position shown in figure 1. The distances
AG
and
AD
are the arm length, equals to L. The derivation of the light paths starts when point
A
of the beam splitter is at instance
A
1
and the front of light coming from the source is at the line
C
1
B
1
. The point under observation is point
A
. There is only one reflected ray that starts at point
C
that continuously intercepts point
A
. The light from
C
1
travels to
M
2
then back to beam splitter, and intercepts point
A
at
A
4
after traveling the path length
. In time
, the light travels from point
C
1
to point
A
4
, and the beam splitter travels the distance A
1
A
4
. The speed of light along the vertical is
×
, and the speed along the horizontal is
×=
. Thus, the distance
H
1
A
4
is equal to the distance
A
1
A
4
and
=2
=2
. In triangle
C
1
A
1
H
1
, angle
C
1
A
1
H
1
is equal to
/4/2
. Thus,
tan/4/2=
⇒
=
tan/4/2.
The time
is given as:
=
+2 =
⇒
tan/4/2 +2 =
2 ⇒
(12tan/4/2 )=2 .
The distance
A
1
H
1
can be calculated and then the time
=
/2=7.334,800,183,355,34008
. The path length
and the distance
A
1
A
4
can be calculated.
=
=2.200,440,055+01=4.000,800,100+07
, and
=
=2.200,440,05503=4.000,800,100+03
. There is only one transmitted ray that starts at point
B
that continuously intercepts point
A
. The light from
B
1
intercepts point
A
at
A
3
after traveling the path length
. The light at
A
3
is reflected towards the screen, making angle
G
1
A
3
K
1
, which is equal to
90°
, and therefore coincides with the reflected ray at
A
3
.

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