# The very significant and positive result of the MM Exp

1. Jun 19, 2008

### arbol

The following two main principles are the very significant and positive result of the Michelson-Morley Experiment:

1. The laws by which the states of physical systems undergo changes are not affected, whether these changes of states are referred to the one or the other of two systems of coordinates in uniform translatory or accelerated motion.

2. Any period of time t is not affected, whether this period of time be referred to the one or the other of two systems of coordinates in uniform translatory or accelerated motion.

2. Jun 19, 2008

### Staff: Mentor

Neither of these two points are "results" of the Michelson-Morley experiment. The second is not even true, if by "not affected" you mean that the time between two events is independent of reference frame.

3. Jun 20, 2008

### arbol

Yes. By "not affected" I mean independent of reference frame. These two points together is the result of the Michelson-Morley Experiment (MM Exp). And the second one is actually the physical evidence left by the MM Exp by which we can determine that (1) with respect to a stationary system, the time t = L/c a ray of light emitted by a moving system takes to move along a given length L is

(L + D)/(c + v), where v is the given constant velocity the moving system moves along the x-axis of the stationary system, and D = v*L/c is the distance the moving system moves along the x-axis of the stationary system (Note: Actually it is given also that another ray of light, emitted by the stationary system, moves independently along the same given length L. Thus, the time t is also L/c in the stationary system independent of reference frame), and (2) with respect to the stationary system, the time t = L/c the ray of light emitted by the moving system takes to move along the given length L is

v/a, where a is the given constant acceleration by which the moving system achieves velocity v = a*L/c along the x-axis of the stationary system (Note: the initial velocity of the moving system at the time t = 0s is 0m/s).

4. Jun 20, 2008

### Staff: Mentor

As I have pointed out before (in your other threads that say the same thing), this is incorrect. In any case, I fail to see how you deduce this from the M-M experiment. Please show how the null result of the M-M experiment leads one to conclude this.

You seem to be using some thought experiment of your own and mixing it up with the M-M experiment. Please describe the exact scenario you have in mind. A diagram would be helpful.

5. Jun 22, 2008

### arbol

1. Frames of reference having any uniform translatory motion relative to one another are inertial frames of reference.

2. Newton thought the propagation, reflection, and refraction of light were essentially mechanical because light consists of stream of particles traveling with high speed. Thus, inertial frames were equivalent not only in mechanics, but in optics also.

3. However, with the discovery by Young in 1801 of the wave nature of light, it was necessary to imply not only the existence of a medium (the ether) in which the waves travel, but that this ether constitutes a unique and absolute frame of reference to which all motion should properly be referred. The acceptance of this ether as the unique frame of reference required the abandonment of the idea of the complete equivalence of all inertial frames.

4. When Maxwell developed his electromagnetic theory of the propagation of light, states of electric and magnetic stresses in the ether were postulated in place of elastic stresses.

5. The purpose for the Mihelson-Morley Experiment was to determine the velocity v of Earth with respect to the ether.

6. Any wave disturbance must have some definite velocity relative to the medium responsible for its propagation. Thus, argued the physicists of the time, light must have some definite velocity c relative to the ether. Consequently, if an observer on the orbiting Earth moves with velocity v relative to the ether, then relative to Earth the velocity of light might have any value between c + v and c - v, according to the direction from which the light comes to the observer.

7. The Michelson interferometer, which was used earlier with success for the precision determination of optical wave lengths, lend itself to observe an optical interference effect depending on the difference of velocities with which a ray of light emitted by the apparatus moving together with Earth executes equal double journeys to and from a mirror in two directions at right angles. In the theory of the instrument as ordinarily presented, no consideration is given to any possible variation of the velocity of light with its direction of travel. Here, however, this is the very question we which to decide. This decision depends on the difference in the times, with respect to the moving Earth, the said ray of light takes to execute the said equal double journeys.

8. By observing a shift in the system of fringes with a change in the orientation of the apparatus and setting this observable fringe shift equal to the total path difference divided by the wave length of light, the difference in the said times can be calculated.

9. But Michelson and Morley were unable to observe any effect when they rotated their apparatus.

10. The absence of any fringe shift means that the times, with respect to the moving Earth, the ray of light emitted by the apparatus moving together with Earth, takes to execute the said equal double journeys are equal, and equal times means that the velocities with which the said ray of light executes the said equal double journeys are also equal.

11. This result of the Michelson-Morley Experiment is described as the negative Michelson-Morley result.

12. This negative Michelson-Morley result means that the motion of Earth through the ether was not detected.

13. Let us trace back what happened up to this point:

a. The motion of Earth with respect to the ether was not detected.

b. Failure to detect the motion of Earth through the ether presupposes equal velocities with which the ray of light in question executes the said equal double journeys.

c. Equal velocities presupposes equal times, with respect to the moving Earth, the ray of light emitted by the apparatus moving together with Earth, takes to execute the said double journeys.

d. Equal times presupposes failure to observe any fringe shift with a change in the orientation of the apparatus.

e. Failure to observe any fringe shift presupposes what? It presupposes the fact that the ray of light in question was emitted by the apparatus moving together with Earth. Nothing but failure to observe any fringe shift could be expected. The Michelson-Morley result is meaningless, and we ought to just postulate the propagation of light in empty space and go back to Newton's complete equivalence for all inertial frames.

14. Let S' be an x'-coordinate system, and let S be an x-coordinate system. Let the x'-axis of S' coincide with the x-axis of S. Let S' move with constant velocity v relative to S in the direction of increasing x and the origin of S' coincide with the origin of S at the time t = t' = 0s.

15. Let a ray of light depart from x' = 0m at the time t' = 0s and reach x' = x'1 at the time t' = t'1, and let the ray of light at x'1 be reflected back to x' = 0s, reaching x' = 0m at the time t'2.

16. Let the length of the path of the ray of light from x' = 0m to x' = x'1 be the length L' of a rigid rod, and let us say that we are able to determine that the length L' of the path of the ray of light emitted by the moving S' is the length L (= x1- 0m) in the stationary system S.

17. Out of the meaningless result of the Michelson-Morley Experiment, Einstein postulated the constancy of c independent of reference frame. He also developed his theory of relativity of times (which implies his theory of relativity of lengths) based on the following argument:

If t'1 = L/(c - v), and

t'2 - t'1 = L/(c + v), then

t'1 = t'2 - t'1 with respect to the stationary system S, but

t'1 ~(=) t'2 - t'1 with respect to the moving system S'.

The folly in this argument is the arbitrary introduction of the irrelevant path of length L to the ray of light emitted by the moving system S'. This argument is just as meaningless as the result of the Michelson-Morley Experiment.

18. Let us say that we are able to design an experiment by which we are able to determine the time, with respect to the moving system S', a ray of light emitted by the stationary system S takes to execute a complete journey to and from x1. Let us say also that we are able to compare this time with the time, also with respect to the moving system S', the ray of light emitted by the moving system S' takes to execute a complete journey to and from x'1.

If by the result of this physical experiment

L/(c - v) = L/(c + v) = L'/c, then we ought to embrace (1) the constancy of any given period of time independent of reference frame and (2) the relativity of c.

But if by the result of this experiment

L/(c - v) = L/(c + v) ~(=) L'/c, then

we ought to embrace (1) Einstein's postulate of the constancy of c independent of reference frame and (2) Einstein's Relativity of times.

19. In my original post I said, "The following two main principles are the very significant and positive result of the Michelson-Morley Experiment:

1. The laws by which the states of physical systems undergo changes are not affected, whether these changes of states are referred to the one or the other of two systems of coordinates in uniform translatory or accelerated motion.

2. Any period of time t is not affected, whether this period of time is referred to the one or the other of two systems of coordinates in uniform translatory or accelerated motion."

20. In light of the argument I set above, I have instead the disposition now to agree to postulate (1) Newton's complete equivalence for all inertial frames and (2) the propagation of light in empty space.

21. I will postpone any judgment with respect to Einstein's constancy of c independent of reference frame and his Theory of Relativity of times until an adequate physical experiment shows one way or another.

6. Jun 22, 2008

### JesseM

Are you suggesting that we can explain the results by postulating that light always moves at c relative to the emitter, but that Newtonian velocity addition still works, so that if the emitter is moving at speed v in our frame and the light is emitted in the same direction, in our frame the light is moving at v+c? If so, the problem here is that there have been plenty of experiments to show that we cannot change the measured speed of light in our frame by looking at light from an emitter which is moving in our frame--see this page for some examples. And there's also the fact that Maxwell's laws of electromagnetism explicitly predict that the speed of all electromagnetic waves is the same regardless of the velocity of the emitter.
In which frame are you imagining the rod to be at rest, the primed frame or the unprimed frame? In whichever frame the rod is moving, the distance the light travels in that frame will not be the same as the length of the rod in that frame, since if the light is emitted at the back end of the rod, the front end will be moving away from it and so in time t the front end will have moved an extra distance vt beyond where it was at the moment the light started its journey from the back end.

Last edited: Jun 22, 2008
7. Jun 22, 2008

### Staff: Mentor

OK.

OK, you imagine a rigid rod in the S' frame of length L' that extends from x' = 0 to x' = x'1. And then you define L as the distance x1 in frame S (I assume that x1 coincides with x'1 when the light reaches x'1). Thus L is the path length that the light travels as measured in frame S. Note that L is not the length of the rigid rod (L') as measured in frame S.

Please don't attribute this "argument" to Einstein.

Don't confuse these equations, which seem to be partly lifted from the classical analysis of the M-M experiment, with a proper analysis of your thought experiment--which has little to do with the M-M experiment.

To properly describe your thought experiment, you'll have to correct these equations.

Your first equation t'1 = L/(c - v), should be replaced by t'1 = D'(c + v), where D' is the length of a rigid rod of length L in frame S as measured in frame S'. (D' does not equal L'.)

Perhaps you meant to view things from frame S, in which case your first equation should be:
t1 = E/(c - v), where E is the length of the rigid rod L' as measured in frame S. (E does not equal L.)

I see no point in continuing until you correct these errors.

8. Jun 22, 2008

### arbol

There is no science here.

This science and mathematics is not right.

For what purpose you say that, not only the two perpendicular paths of light are equal, but that they execute the two equal paths at the same time, and then say that the times the ray of lights takes to execute these two equal paths ought not to be equal? For what purpose you are confusing?

I drew on a piece of paper a right angle with the two adjacent sides equal. Then I asked my daughter of 6 years old the following:

You see this figure? The distance from this point to this point to the right is the same as the distance from this point to this point up.

Suppose a ray of light moves from this point to this one to the right and bounces back to this point again at the same time a ray of light moves from this same point to this one up and bounces back to this one again. What would you say about the time the two ray of lights takes to do this at the same time?

She said, "Mommy I don't want to eat that" (her mother was serving breakfast). And her mother said, "Don't ask her those questions. That doesn't make any sense."

I later asked my son of 14 years old the same question, and he said, "It's the same." I asked, "What is the same'? He said, "The time. You just said that the lights do that at the same time." Then I turn the paper, and asked him again, "And now, what would you say about the time the light would take to do that?" He said, "It's still the same." I asked, "Why?" He said, "Because is the same from here to here and from here to here." Then I turned the paper even more. I asked the same question again. He said, "From here to here now looks shorter than from here to here."

He was referring to the adjacent side that originally from his point of view was directed upward. Nevertheless this is only an optical illusion.

Just before sending this message to pf, my wife said to me, "You are waging psychological war against those people."

9. Jun 22, 2008

### MeJennifer

Arbol is the point of your postings to challenge the theory of relativity? If so, you are wasting your time, this forum is to discuss and improve understanding of the theory not a platform to challenge it.