Reflections on the Michelson-Morley Experiment and the Ineluctable Self-Interview

Reflections on the Michelson & Morley Experiment and the Ineluctable Self-Interview

By Jim Spinosa

ISBN 9781370895526

Published by Jim Spinosa at Smashwords

Copyright 2017 All rights reserved

Dedicated to Steven G. Spinosa The Spinner Cares

Smashwords Edition, License Notes. This e-book is licensed for your personal enjoyment only. This e-book may not be re-sold or given away to other people. If you would like to share this e-book with another person, please purchase an additional copy for each person. If you’re reading this book and did not purchase it, or it was not purchased for your use only then please return toSmashwords.com and purchase your own copy. Thank you for respecting the hard work of this author. Reflections on the Michelson-Morley Experiment and the Ineluctable Self-Interview is a free book.

Table of Contents

Introduction

1. The Michelson-Morley Experiment Probes the Luminiferous Ether

2. A Further Consideration of Certain Aspects of the Michelson-Morley Experiment

3. The Re-conceptualized Michelson-Morley Experiment Probes the Ether

Conclusion

The Ineluctable Self-Interview (Part 1)

Section A. Mold in the Refrigerator

Section B. Watching the Moonrise from the Mindanao Deep

Section C. Pink Sky and Green Stars

Section D. The Cave of the Day

The Ineluctable Self-Interview (Part 2)

Section A. The Stars Are a Sieve

Section B. Myrmecophagous Bridgette

Introduction

This reconsideration of the Michelson–Morley experiment demonstrates that the conceptual theories that underpin the experiment are flawed. Because the conceptual theories are flawed, the mathematics that accompanies the experiment is inappropriate. The conceptual theories have flaws that are serious enough to nullify both the results of the experiment and its conclusions. This reconsideration of the experiment will take into consideration the following important point: There were at least two definitions of the term ether when Albert Michelson and Edward Morley first performed their experiment. Many scientists considered ether a hypothetical, invisible substance that pervaded all of space and served as a medium for the transmission of light waves. That definition of the ether is used in this reconsideration of the experiment in chapters one and two of this e-book. Other scientists considered the ether to be a continuous expanse of empty space, extending in all directions, and chapter three of this e-book examines the experiment employing that definition. The conclusion of the e-book summarizes the experiment’s shortcomings when each of the definitions are used.

1. The Michelson–Morley Experiment Probes the Luminiferous Ether

The Medium for the Propagation of Light Waves

First performed in 1887, the Michelson–Morley experiment used a complex apparatus to probe the nature of the luminiferous ether. The luminiferous ether is often called the aether or simply the ether. Ether wind is a misleading term because the prevailing opinion among scientists in the late 19th century was that the luminiferous ether was stationary under most conditions. These scientists thought the ether wind effect was caused by the motion of the earth through the stationary ether. It is similar to the experience produced by riding a bicycle on a windless day. As the bicycle rider pedals along, he feels as though a wind is blowing on his face. As the speed of the bicyclist increases, the wind increases in its apparent speed.

Another aspect of the Michelson–Morley experiment was that from its results many scientists believed that the absolute velocity of the earth would be determined. To accomplish the task of determining the absolute velocity of the earth, two precise time measurements needed to be made. They are the measurement of the duration of a light beam’s round–trip journey along one arm of the apparatus and the duration of the light beam’s round–trip journey along the other arm of the apparatus. These measurements had to be made when one arm of the apparatus was aligned with the motion of the earth and the other arm was perpendicular to the apparent ether wind that is to say perpendicular to the motion of the earth.

The centerpiece of the Michelson–Morley experiment was a device is called an interferometer; it is a device that divides a light ray into two beams and then brings them together again to cause interference. The recombination of these two beams of light produces interference fringes (bands of more intense color). When these fringes are counted, they give information about the light.

The main part of the apparatus consists of two identical arms fitted together to form a right angle. When one arm is aligned with the motion of the earth, the other arm will be perpendicular to it. Each arm is of equal length. At the arms’ vertex a single ray of light is split into two beams by a half–silvered mirror. One beam travels along one arm, and the other beam travels along the other arm. A fully–silvered mirror is at the far end of each arm. Each half of the split light ray strikes its respective mirror and returns along its respective arm, and the two beams rejoin producing interference fringes.

If the earth was moving through ether that was stationary, it was mathematically determined that one light beam would take longer than the other to complete its round–trip journey. This conclusion was decisively drawn although the arms were of equal length. The round–trip journey of the light beam traveling in the arm aligned with the motion of the earth would take longer to complete than the round–trip journey of the light beam traveling in the arm that was perpendicular to the motion of the earth. If the ether surrounding the earth was partially dragged along by the motion of the earth, the overall outcome would still be the same. The only change would be in the size of the difference between the round–trip duration for one arm as compared to the other arm. Albert Michelson and Edward Morley believed that each of these possible outcomes could be detected and distinguished by their interferometer. Each outcome would produce a distinct set of slight changes in the interference fringes of the recombining light beams.

The entire apparatus was mounted on a granite slab. Floating in a basin filled with mercury, the granite slab could be rotated. It was hypothesized that the rotation of the apparatus would produce slight changes in the interference fringes. They would occur as the arm of the interferometer perpendicular to the motion of the earth was rotated until it exchanged position with the arm of the interferometer aligned with the motion of the earth.

Early in chapter 2 of Lillian R. Lieber’s book, The Einstein Theory of Relativity, the formula t1 is introduced. It is the formula for the time it takes the light beam to make a round–trip journey in the arm of the apparatus aligned with the motion of the earth. Lillian R. Lieber writes, "Therefore the time required to travel from A to B would be a/(c–v), where a represents the distance AB, and the time required for the trip from B to A would be a/(c+ v). Consequently, the time for the round trip would be t1 = a/(c–v) +a/(c+ v) or t1 =2ac/ (c²–v²)." This short citation is from Lillian R. Lieber’s book The Einstein Theory of Relativity: A Trip to the Fourth Dimension (New York: Rinehart & Company, Inc., 1936, 1945) page 9. The common denominator for the denominators (c–v) and (c+ v) is obtained by multiplying (c–v) (c+ v) which gives us (c²–v²). The numerator 2ac comes from simplifying the numerator a(c+ v) +a(c–v), which gives us ac +ac or 2ac. The letter c is the velocity of light, and the letter v is the velocity of the apparent ether wind, which has the same magnitude as the earth’s velocity if the ether is stationary.

The formula for the time, t2, it takes the light beam to make a round–trip journey in the arm of the apparatus that is perpendicular to the motion of the earth, is introduced by the author two pages later. Lillian R. Lieber writes, "So the time for the round trip from A to C and back to A, would be

t2 = 2a/ (c²–v²)½." This short citation is from page 11 of Lillian R. Lieber’s book The Einstein Theory of Relativity.

a = the length of the light path or perhaps it would be more accurate to say the length of some sort of hypothetical light path, which is the same in both arms of the apparatus.

c = the speed of light.

v = the velocity of the apparent ether wind.

When the formulas, t1 and t2, are mathematically analyzed, it is revealed that t1 is greater than t2. Therefore, it takes more time for the light beam in the arm of the apparatus aligned with the motion of the earth to make its round–trip journey than it does for the light beam in the arm of the apparatus that is perpendicular to the motion of the earth. Let’s mathematically analyze the formulas t1 and t2. First, we will find a common denominator for the two fractions that appear in the formulas. The common denominator for the two fractions is (c²–v²). This gives us the following: t1 =2ac/ (c²–v²) and t2 = 2a (c²–v²)½/ (c²–v²). Since the denominators are now the same, we can ignore them and concern ourselves with the numerators only. For instance, when comparing the fractions, 3/5 and 4/5, we can ignore the denominators and compare the numerators only. Since 4 is larger than 3, we can determine 4/5 is larger than 3/5. The numerator of t1, is 2ac, and the numerator of t2 is 2a (c²–v²)½. The product 2a is the same in both numerators. We can further determine that if v is greater than zero, the value c is larger than the square root of (c²–v²), which is another way of expressing the term (c²–v²) raised to the ½ power or (c²–v²)½. Thus, t1 is greater than t2. To make this analysis two crucial requirements must be met. The first requirement is that a represents the same length in both equations. The second requirement, is that both equations contain the term (c²–v²), which can be raised to any particular power.

These requirements cause difficulties when analyzing the behavior of the light beam in the arm of the apparatus perpendicular to the motion of the earth that is to say perpendicular to the apparent ether wind. In this arm, the apparent ether wind should sweep the light beam downwind. That would make its light path longer than light path a. This is analogous to a swimmer who tries to swim across a river. To make his journey as short as possible, the swimmer will try to follow a line that runs perpendicularly to the shoreline and intersects his starting point. Nevertheless, inevitably, the current will sweep the swimmer downstream, which increases the distance of his journey. He actually ends up swimming along the hypotenuse of a right triangle, instead of one of its legs.

This difficulty is overcome in the formula t2 by using the term (c²–v²)½. This insures that the light path in formula t2 equals the light path in the formula t1. The term (c²–v²)½ gives us the speed of the light beam. We construct a right triangle in which the speed of light is the hypotenuse of the right triangle, and we denote it by the letter c. It represents the speed of a light beam that is not being influenced by the apparent ether wind. The letter v represents a leg of the right triangle and the speed of the earth. The speed of the earth and the speed of the apparent ether wind are the same; their directions are opposite though. The length of the other leg of the right triangle is obtained using the Pythagorean Theorem. It is represented by the term (c²–v²)½, and it gives us the speed of light under the influence of the apparent ether wind. The equation t2=2a/ (c²–v²)½ is essentially a restatement of the more general equation time = distance/speed. However, the use of the equation t2 =2a/ (c²–v²)½ introduces its own difficulties.

Before we continue any further, an excerpt from pages 11 and 12 of Lillian Lieber’s book will show the confidence she has in the swimmer analogy. She states, "But what has all this [the swimmer analogy] to do with the Michelson–Morley experiment? In that experiment, a ray of light was sent from A to B: At B there was a mirror which reflected the light back to A, so that the ray of light makes the round trip from A to B and back, just as the swimmer did in the problem described above. Now, since the entire apparatus shares the motion of the earth, which is moving through space, supposedly through stationary ether, thus creating an ether wind in the opposite direction, this experiment seems entirely analogous to the problem of the swimmer." This short citation is from Lillian R. Lieber’s book The Einstein Theory of Relativity.

A further examination of the analogy between the light beam and a swimmer will reveal the difficulties with this explanation. The light path a is represented by a perpendicular line running across the river. The line begins at the swimmer’s starting point. If the swimmer knows three facts, he can calculate the precise angle by which he must deviate from the perpendicular in an upstream direction to counteract the effects of the current. These three facts are the following: the velocity of the current, the speed at which he swims and the trigonometric equations known as the Law of Sines. The swimmer must also know that the current flows perpendicularly to a line measuring the width of the river. The swimmer swims upstream at an angle expressly chosen so that, when the current sweeps him downstream, his path forms a line that is perpendicular to the flow of the river and intersects his starting point.

The following is an example of the calculation a swimmer could make, and it should be noted that the velocity of the current is crucial to the calculation. Let us say there is a river that flows east to west. A swimmer on the river’s south bank wants to swim to the north bank along a perpendicular line. The river is 20 miles wide, and we will imagine that the width of the river is side a of a right triangle whose base is on the opposite side of the river. The current flows at a constant rate of 3 mph, and the swimmer swims at a constant speed of 5 mph. The lengths of both side b, the base, and side c, the hypotenuse, of the triangle are unknown. It is known that the ratio b/c = 3 mph/5 mph. The ratio b/c equals this ratio: the speed of the current/the speed of the swimmer. This is true because of the principle of similar triangles. Solving the ratio for b gives: b = c (3/5) = (.6) c.

The Law of Sines is the following: a/Sine A = b/Sine B = c/Sine C. The angles A, B, and C represent the angles directly opposite the respective sides a, b, and c. We know angle C is 90o because the current flows perpendicularly to a line measuring the width of the river. The sine of 90o equals one. If we substitute (.6) c for b and 20 miles for a and one for sine C, we have: 20miles/sine A = (.6) c/sine B = c/1. Solving the equation for sine B we have sine B = .6. The solution is reached by cross multiplying the fractions (.6) c/sine B=c/1, which gives us (.6c) (1) =sine B (c). Next we divide each side of the equation by (c), and we generate .6=sine B. Therefore, angle B equals approximately 36.87o because the sine of 36.87o is approximately .6. A swimmer must swim at an upstream angle of approximately 36.87o from the perpendicular in order for the current to carry him back to the perpendicular.

A swimmer can swim at an angle that precisely deviates from the perpendicular. Swimming at this angle allows the current to carry him back to the perpendicular. The phrase, carry him back to the perpendicular is confusing the reader may think that the current is carrying the swimmer forward to the perpendicular. If we define forward as moving in the direction of the flow of the river then the current is carrying the swimmer forward to the perpendicular. A light beam can also precisely deviate from the perpendicular light path a. This deviation will allow the apparent ether wind to carry the light beam back to the perpendicular light path a. The term (c²–v²)½ is the proper term to use when the light beam is being carried back to the perpendicular by the apparent ether wind. Once the light beam is carried back to the perpendicular, it is mathematically changed from the hypotenuse of a right triangle into the leg of a right triangle. The Pythagorean Theorem determines that the length of the leg a of a right triangle is: a = (c²–b²)½.

The difficulty arises in determining the precise angle the light beam should deviate from the perpendicular. This angle must allow the apparent ether wind to carry it back to the perpendicular. To calculate that angle, the velocity of the apparent ether wind must already be known, but the velocity of the apparent ether wind is unknown. Unless the precise angle of deviation is already known, the formula t2 cannot be employed. It seems clear that we do not know the angle that the light beam (in the arm of the interferometer perpendicular to the motion of the earth) must deviate from the perpendicular in order for the apparent ether wind to blow it back to the perpendicular. Therefore, the equation t2=2a/ (c²–v²)½ is invalid.

However, there is a strategy that might make the equation t2=2a/ (c²–v²)½ valid. That strategy utilizes the rotation of the interferometer. This strategy will be discussed shortly, and its shortcomings will be pointed out.

Instead, a formula similar to the equation t2 must be used. We will call the new equation t2 prime, which we can denote as t2. The new equation is the following: t2=2b/ (c²+v²)½, and it must be used to describe the path of the light beam when it travels perpendicularly to the apparent ether wind. Its use does not require that the light beam travels along a known angle of deviation from the perpendicular.

b = the distance the light beam travels when swept downwind by the apparent ether wind–it is a longer distance than that of light path a.

c = the speed of light.

v = the velocity of the apparent ether wind.

A strategy can be employed that in order to use the formula t2 = 2a/ (c²–v²)½ it is not required that the precise angle of deviation must be known. The strategy is that since the interferometer rotates, it will pass through the precise angle of deviation necessary for formula t2 to be correctly employed. It is true that the interferometer will pass through the precise angle of deviation. It is analogous to a swimmer who swims across a river many times and with each trial he chooses a different angle as measured from the perpendicular. By trial and error he will discover the precise angle that allows the current to carry him back to the perpendicular.

However, when arm A of the rotating interferometer comes to the precise angle of deviation that allows the apparent ether wind to blow the beam of light (traveling along its arm) so that it is perpendicular to the apparent ether wind, the other arm, arm B, of the interferometer will not be perpendicular to that beam of light. The apparent ether wind does blow the beam of light in arm A so that it advances to its correct position. The correct position for the light beam in arm A is perpendicular to the apparent ether wind. The apparent ether wind does not blow the beam of light in arm B so that it advances to its correct position. The correct position for the light beam in arm B is aligned with the apparent ether wind. Therefore, the beam of light in arm B lags behind from its correct position.

Arm B of the interferometer is not perpendicular to the beam of light in arm A because the beam of light in arm A has been blown downwind by the apparent ether wind. It has been blown until it is perpendicular to the apparent ether wind while the beam of light in arm B of the interferometer has lagged behind, and so it will not be aligned with the motion of the earth. Therefore, the formula for the beam of light in arm B, t1 = 2ac/ (c²–v) cannot be correctly employed. Arm B of the interferometer is not perpendicular to the beam of light blown downwind by the apparent ether wind in arm A. Arm B of the interferometer is still perpendicular to arm A because the arms themselves are not affected by the apparent ether wind, only the light beams traveling along the arms are affected by the apparent ether wind.

The light beam traveling along arm A of the interferometer has been blown downwind by the apparent ether wind. Therefore, the light beam is no longer parallel to arm A. Since it is no longer parallel to arm A, it cannot be perpendicular to arm B because the arms are at right angles to one another. Since the light beam traveling along arm A is not perpendicular to arm B, it is not perpendicular to the light beam traveling along arm B.

It should be noted that the effect that the apparent ether wind has on the light beam traveling along arm B is very slight, but it actually increases the amount by which the beam of light in arm B lags behind from its correct position.

Though the rotating interferometer passes through the precise angle of deviation required for formula t2 to be correctly employed, it does so in a way that makes it incorrect to employ formula t1.

There is another problem with the argument. The mirror at the far end of arm A of the interferometer would need to be tilted at a precise angle so that once the light beam has struck the mirror, the returning light beam will be traveling at the proper angle for it to be blown back to the perpendicular by the apparent ether wind. The precise angle the mirror must be tilted to fulfill this requirement is unknown. Therefore, for all the reasons listed above, t2 is the incorrect equation to employ, and a formula similar to t2 is the correct equation to employ.

The equation t2’ =2b/ (c² +v²)½ as with the other equations already mentioned, is an expression of the general equation: time = distance/speed. The distance is 2b because the light beam makes a round–trip journey, and it is assumed that each leg of the journey has a length which equals b. (We will soon see this assumption is incorrect.) The speed of the light beam is (c²+v²)½. The speed is determined by the addition of vectors.

A vector is a mathematical expression denoting a combination of both magnitude and direction. In these equations, the vectors used express the speed and the direction of one of the three following quantities: a light beam, the apparent ether wind, or a light beam under the influence of the apparent ether wind. Assuming the earth does not drag or even partially drag the ether along with it, then the vector for the velocity of the apparent ether wind will have the same speed as the vector for the absolute velocity of the earth. However, the direction for the vector of the absolute velocity of the earth is the reverse of the direction of the apparent ether wind’s vector. When the vectors, which represent the apparent ether wind and the velocity of a beam of light, are added they form a third vector. It results from the combination of the speeds and directions of these two vectors.

The following is an example of vector addition. Let us say there is a river that flows east to west at a rate of 3 mph. A swimmer is positioned on the river’s south bank who is ready to swim to the north bank, at a speed of 4 mph. The river’s current can be represented by an arrow 3 units long, pointing toward the west, and beginning at the swimmer’s starting point. The swimmer can be represented by an arrow 4 units long, pointing toward the north. This arrow also begins at the swimmer’s starting point. These two vectors form two sides of a parallelogram. The other two sides of the parallelogram can be constructed because they mirror the two sides already formed by these two vectors. Once we construct the parallelogram, we can draw the third vector. It represents the addition of the current’s vector and the swimmer’s vector. We begin the third vector at the swimmer’s starting point. We end the third vector at the diagonally opposite corner of the parallelogram. The length of the third vector represents the speed of the swimmer under the influence of the river’s current. The position of the third vector represents the direction of the swimmer under the influence of the river’s current.

In this example, the third vector is the hypotenuse of a right triangle. One leg of the triangle is three units, and the other leg is four units. Using the Pythagorean Theorem we find the length of the hypotenuse/vector is (3²+4²)½ or 5 units. The length of the vector equals the speed of the swimmer under the influence of the river’s current. It is 5 mph.

The same reasoning is used to learn the speed of the light beam under the influence of the apparent ether wind. The speed of the light beam is (c² +v²)½ when the light beam begins its journey perpendicular to the apparent ether wind. Likewise, the speed of the light beam under the influence of the apparent ether wind is (c²–v²)½ when the light beam begins its journey at that one specific upwind angle from a line drawn perpendicularly to the apparent ether wind that we have designated as the known angle of deviation from the perpendicular.

For the reasons given before, we are compelled to use the term (c²+v²)½. When that term is used, deciding mathematically which of the two possible round–trip journeys will have the greatest duration is impossible. We cannot mathematically determine which will take longer: the journey of the light beam in the arm of the interferometer perpendicular to the motion of the apparent ether wind or the journey of the light beam in the arm of the interferometer aligned with the motion of the apparent ether wind.

It cannot be mathematically determined whether t1 = 2ac/ (c² –v²)½ is greater than, less than, or equal to t2=2b/ (c²+v²)½. Since this is the case, the experimental result that the light beams in each arm of the interferometer take the same amount of time to complete their round–trip journeys is not in conflict with this new mathematical analysis. The original mathematical analysis of equations t1 and t2 conflicted with the empirical results of the Michelson–Morley