# Michelson Morley experiment is no confirmation of Special Relativity

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3 hours ago, studiot said:

So I would be interested in the mathematical passage that leads you to believe the equation has the status of a postulate or fundament at you call it.﻿﻿

I already tried to explain it in a previous topic. 'What if Einstein's definition of simultaneity is incorrect"

The mathematical passage is in here:

1. Kinematical part § 1Definition of Simultaneity and §2 On the Relativity of Length and Time.

tB - tA = rAB / (c - v )       and      t’A - tB = rAB / (c+v)

These equations are based on the length contraction of rod AB.

To retrieve the first equation I followed the "einsteingenootschap.nl" (a Dutch site dedicated to Einstein)

1- The distance the light beam travels can be expressed as:

rAB + v . ( tB - tA )

Because all clocks on the moving rod and in the stationary system are synchronized, what Einstein assumed is:

tB - tA = { rAB + v . ( tB - tA )} / c

c . ( tB - tA ) = rAB + v . ( tB - tA )

c . ( tB - tA ) - v . ( tB - tA ) = rAB

( c - v ) . ( tB - tA ) = rAB

And there it is:

tB - tA = rAB / (c - v)

So far so good and in line with the einsteingenootschap.

But next I assume that the following is allowed:

tB - tA = { rAB + v . ( tB - tA )} / c

rAB / c = { rAB + v . ( rAB / c )} / c

I replace Rod AB by length  L

L / c =  { L +  v ( L /c )} /c

L  = L +  v ( L /c )

L . v / c = 0

L = 0  ,  c ≠ 0

v = 0  , c ≠ 0

L = rAB  ≠ 0

thus v = 0

Both fundamental equations of Special Relativity eliminate motion?

Edited by vanholten

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56 minutes ago, vanholten said:

I already tried to explain it in a previous topic. 'What if Einstein's definition of simultaneity is incorrect"

?? You want to "assume" that the speed of light is infinite? But we know it isn't. Therefor simultaneity is not an absolute relation between frames and what is simultaneous in one frame of reference is not simultaneous in another. That has been evidenced many times.

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14 hours ago, vanholten said:

These equations are based on the length contraction of rod AB.

But next I assume that the following is allowed:

tB - tA = { rAB + v . ( tB - tA )} / c

You need to justify this assumption

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6 hours ago, swansont said:

You need to justify this assumption

To illustrate the effect of length contraction Einstein sketched an experiment.  Rod AB in motion is equipped with clocks and observers at each end.  The whole thing is observed from a  stationary system with synchrone clocks all over the place.  The clocks on both ends of the moving rod AB are synchronized with the clocks of the stationary non co-moving observers.

Ok this is what I assumed: The two equations are meant for the stationary observers seeing the rod AB in motion. Because those observers will have to deal with the velocity of the rod, as formulated in both equations. And they rely on the timespan tB-tA displayed by their clocks being part of the first equation.

Einstein says that the length of rod AB in motion is measured as being different from the length of rod AB when it is measured as a stationary object. It seems he relies on length contraction, adapted from Lorentz' explanation of the absence of phase shift during the MM experiment. (I think that is strange because the second postulate did us forget about the ether.)

Anyhow, Einstein says : “rAB denotes the length of the moving rod—measured in the stationary system. “

But what does he mean by that? I think there are two explanations possible.

Option 1:  rAB is measured in the stationary system;  the stationary rod is measured to have the length AB.

In that case the stationary observers see the light ray bridge the distance AB according to their clocks in tB-tA seconds.

According to Einstein's definition of simultaneity:  2AB/(t’A-tA) = c       and           tB-tA = t’A-tB

they measure the light to propagate at c so their observation results in

rAB/c = tB-tA  ( my assumption)

Option 2: rAB is measured from the stationary system while the rod AB is in motion.

The speed of light is measured to be c as always for all observers.  The stationary clocks still display the timespan  tB-tA  in which the light at c bridges AB.

Again this results in

rAB/c = tB-tA  (my assumption)

Both explanation have the same result. For the stationary observations the velocity  v  in both equations is a meaningless parameter. However Einstein’s experiment suggests that the observers on the rod encounter a different timespan for the light to travel AB.  This means that for those observers the timespan should result in

rAB/c ≠ tB-tA

because their clocks are synchronized with the clocks in the stationary system; the system being not Rod AB.

To make the confusion complete:  Thanks to the equations the observers that are stationary with respect to rod AB are no longer subjected to the definition of simultaneity;

2AB/(t’A-tA) = c

Instead they have to bring the relative motion of the clocks in the non co-moving system into account.  But the equations don't deliver the required parameters to do so. Because by the lack of simultaneity these observers on the rod can't produce a sensible value for velocity v.  In other words: The clocks at both ends of the rod are asynchrone so the observers can't define the relative speed of the other system required to pinpoint v in the equation. Thus also for them  v in the equations is pointless. And if they would adjust their clocks to achieve rAB/c = tB-tA,  they would have only switched sides and end up with the same observations as made by the observers in the stationary system;  velocity is irrelevant.

The intriguing thing is that the second postulate underlines the irrelevance of velocity of the light source during the measurement of the speed of the light emitted by that source.

20 hours ago, beecee said:

?? You want to "assume" that the speed of light is infinite? But we know it isn't. Therefor simultaneity is not an absolute relation between frames and what is simultaneous in one frame of reference is not simultaneous in another. That has been evidenced many times.﻿

I never assumed the speed of light is infinite.

Edited by vanholten
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16 minutes ago, vanholten said:

But what does he mean by that?

As you are having trouble understanding Einstein's description why not find a modern text on the subject, maybe even one in your own language.

I find Einstein's writing to be quite hard to follow, in part because the style of writing 100 years ago was just more wordy and formal than modern English. (The only reason I have tried to read it is because various crackpots insist on using it as their source even though they find it confusing.)

There are quite a few videos that explain these things with all sorts of animations to explain the concepts.

21 hours ago, vanholten said:

What if Einstein's definition of simultaneity is incorrect

It can't be. It follows mathematically from the invariant speed of light.

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54 minutes ago, vanholten said:

I never assumed the speed of light is infinite.

Time dilation and its variable nature arises from the fact that "c" is a finite number.

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On 4/13/2019 at 2:30 AM, vanholten said:

I already tried to explain it in a previous topic. 'What if Einstein's definition of simultaneity is incorrect"

1. Kinematical part § 1Definition of Simultaneity and §2 On the Relativity of Length and Tim﻿e.

Since you are worrying over simultaneity let us examine the passage in more detail.

Perhaps we can make progress that way.

Many authors state without derivation a result from elsewhere that is needed in the paper.

Einstein did exactly this in his 1905 paper to which you refer where he introduces the criterion or condition for simultaneity.

tB - tA = t'A - tB

He does not derive this condition (and perhaps it is an unusual way to ste the condition) so let us do so now.

First let us note that he starts the section on page 2 by establishing a "coordinate system in which the "equations of Newtonian Mechanics hold good".
He does not specifically say he invokes the principle of reversibility of light (which is part of Newtonian Mechanics) but he uses it to state that the transit time for light from fixed point A to fixed point B in this coordinate system is the same as the transit time for light from B to A. I will call this time T, although we don't know it intiially.

So we have.

Light Transit time A to B = Transit time B to A = T

Clock A start reading = tA

Clock B arrival time = Clock B departure time  = tB

Clock A return/arrival time = t'A

I won't go into all the guff about making sure the clocks run at the same rate when side by side, can I take that as read?

When the clocks are side by side at A let the difference in reading be , Δ, maintained a constant over a sufficiently long period.

Then

tB = tA + Δ + T

and

t'A - tA = the time from A to B and back = 2T

Substituting these into Einstein's expression,

For the left hand side       tB - tA = tA + Δ +T - tA = T + Δ

and

For the right hand side     t'A - tB = (t'A - tA) - Δ - T = 2T - Δ - T = T - Δ

The only way these can be equal is if Δ = 0, leading to the identity T = T.

Thus ends the derivation or proof of the condition.

Does this help?

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7 hours ago, studiot said:

Does this help?﻿

Thanks for that I have no trouble understanding it. However I was explaining that Einsteins equations inevitably result in zero velocity. It takes to many words. Explaining my findings in Dutch is already  difficult , and doing so in English is even more complicated. So let me once again try to explain what I mean, again based on Einstein's equations.

The distance the light beam travels can be expressed as

rAB + v . ( tB - tA )﻿

or

c . (tB - tA) + v . (tB - tA)

Because all clocks on the moving rod and in the stationary system are synchronized, what Einstein assumed is:

tB - tA﻿ = { rA﻿B + v . ﻿( tB - tA﻿ )}﻿﻿ ﻿/ c﻿﻿            ...............          leads to the original   "tB - tA = rAB / (c - v)"

or

tB - tA﻿ = { c . (tB - tA) + v . (tB - tA)}﻿﻿ ﻿/ c﻿﻿

tB - tA  is replaced by  x

then

x = (c x + v x)/ c

take for example c = 1

then

x = ( x + v x)

thus

v = 0

WolframAlpha confirms:

"Solutions:

v = o, c ≠ 0

x = 0,  c ≠ 0

v = 0"

Einsteins  "tB - tA = rAB / (c - v)" results in zero velocity.

Edited by vanholten
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17 minutes ago, vanholten said:

Explaining my findings in Dutch is already  difficult , and doing so in English is even more complicated.

I am out tonight, so will not reply again today.

As regards the Dutch, I have not the luxury of being bilingual like my sister-in-law who teaches Maths in Leiden and her husband.
If it is too difficult, I could ask them to translate if you sent me a document by PM.

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26 minutes ago, studiot said:

I am out tonight, so will not reply again today.

Have a nice evening. And many thanks for your kind offer.

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Do a search here for Cahill, he analyses MMXs using refractive index.

Do a search here for Demjanov, he analyses MMXs using permitivity & permeability. And he did a twin media MMX which he called a 1st order MMX (whereas we all call the original MMXs 2nd order).

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Do a search here for Cahill, he analyses MMXs using refractive index.

Crackpot!

Do a search here for Demjanov

Double crackpot!!

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43 minutes ago, Strange said:

Double crackpot!!

I did not know of Demjanov so I searched. Results supports Strange's comment: "Physical interpretation of the fringe shift measured on Michelson interferometer in optical media" paper was retracted by the editors of Physics Letters A.

Quote

the theoretical and experimental claims made by the author cannot be supported and the article should not have been published

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I can’t help feeling that anyone obsessing over the results of a single 100 year old experiment is probably not being entirely rational. At this stage, does it have anything other than historical interest?

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On 4/14/2019 at 12:39 AM, Strange said:

It can't﻿﻿﻿ be. It follows mathematically from the invariant speed of ﻿light﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿.﻿﻿﻿﻿﻿﻿﻿﻿

It relies on the all seeing eye. To establish simultaneity you need at least two events, a rhombus and two observers at opposite edges.

I already showed that you can easily derive the Lorentz-transformation using the pythagorean theorem.

Edited by vanholten
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10 minutes ago, vanholten said:

It relies on the all seeing eye. To establish simultaneity you need at least a rhombus and two observers at opposite edges.

Nonsense. Einstein's simple thought experiment just involves two observers: one on a train and one "stationary" on the platform. No impossible "all seeing eye". No rhombus.

It seems you have given up wasting everyone's time trying to trash the Michelson-Morely experiment. And you are back to your basic inability to understand the simple concept of simultaneity (https://www.scienceforums.net/topic/116999-what-if-einsteins-definition-of-simultaneity-is-incorrect/)

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22 minutes ago, Strange said:

Nonsense. Einstein's simple thought experiment just involves two observers: one on a train and one "stationary" on the platform. No impossible "all seeing eye". No rhombus.

It seems you have given up wasting everyone's time trying to trash the Michelson-Morely experiment. And you are back to your basic inability to understand the simple concept of simultaneity (https://www.scienceforums.net/topic/116999-what-if-einsteins-definition-of-simultaneity-is-incorrect/)

Apparently you are unable to comprehend the simple concept of relative motion. You seem to have missed that the relations between the clock and their observers essential to define motion can't be relative,  since nobody can be in relative rest compared to a stationary object. Rest defines a definite or absolute relation. SR totally relies on that definite relationship.

Imagine empty space. What is to be defined first, rest or motion, or relative motion?  How many positions do you need to define rest? And how many positions do you need to create motion? Once you know, you might even have the answer to how many positions you need to experience relative motion.

Edited by vanholten
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3 minutes ago, vanholten said:

Rest defines a definite or absolute relation.﻿ SR totally relies on that definite relationship.

With this level of ignorance, it is not surprising that you have a hard time understanding things.

There is no definite and absolute rest. That is fundamental to relativity.

Motion (or rest) can only be defined relative to something else.

Looks like you need to start from the beginning and understand the basic principles first.

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Too busy crackpotting everyone with and deviant view you apparently forgot to prove that my calculations resulting in the zero velocity are plain wrong. I was hoping for that actually. Because if my childish calculations are right there is no fundament for special relativity whatsoever. Perhaps reading that wordy and formal 100 years old text with a bit more attention will help you understand why.

Edited by vanholten
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1 minute ago, vanholten said:

Too busy crackpotting everyone with and deviant view you apparently forgot to prove that my calculations resulting in the zero velocity are plain wrong. I was hoping for that actually. Because if my childish calculations are right there is no fundament for special relativity whatsoever. Perhaps reading that wordy and formal 100 years old text with a bit more attention will help you understand why.

You can't falsify a theory with calculations. You need evidence. And, so far, all evidence is consistent with relativity.

Maybe you should read some ore modern texts (and look into more modern experiments) instead of being stuck in the past. Quantum field theory is based on special relativity and, as well as mountains of experimental confirmation, it has been used to develop many modern technologies.

You have a lot of work to do if you want to show that more than 100 years of science and technology doesn't work.

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52 minutes ago, vanholten said:

Apparently you are unable to comprehend the simple concept of relative motion. You seem to have missed that the relations between the clock and their observers essential to define motion can't be relative,  since nobody can be in relative rest compared to a stationary object. Rest defines a definite or absolute relation. SR totally relies on that definite relationship.

SR does nothing of the sort. If v=0, two objects are at rest with respect to each other. (v is always a speed with respect to something) But they can both be in motion relative to something else.

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1 hour ago, swansont said:

SR does nothing of the sort. If v=0, two objects are at rest with respect to each other. But they can both be in motion relative to something else.﻿

But if these were the first two objects in the universe, they were not in motion relative to something else. By the absence of any reference they were at an absolute distance.Their geometrical relation in relativity still is a definite line segment. I just don’t see the theoretical need for stating that both endpoints of a segment are at rest with respect to each other.

And what do you think of the following? If two objects are in uniform motion, a point of rest exists from which these objects move away in opposite direction at half their relative velocity.

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7 hours ago, vanholten said:

I﻿ j﻿ust don’t see the theoretical need for stating that both endpoints of a segment are at rest with respect to each other.﻿

The “theoretical need” is that you have specified that as the initial condition.

7 hours ago, vanholten said:

And﻿ what do you think of the following? If two objects are in uniform moti﻿on, a point of rest exists from which these objects move away in opposite direction at half their relative velocity.﻿﻿﻿﻿

So what. You can choose a frame of reference where either is stationary and the other is moving. Or where one is moving slowly and the other fast. Or both are moving in the same direction.

All of these are equally valid descriptions. None of them is more “true” than any other.

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9 hours ago, vanholten said:

But if these were the first two objects in the universe, they were not in motion relative to something else. By the absence of any reference they were at an absolute distance.Their geometrical relation in relativity still is a definite line segment. I just don’t see the theoretical need for stating that both endpoints of a segment are at rest with respect to each other.

There are an infinite number of frames, but in only one of them will they be at rest.

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7 hours ago, swansont said:

There are an infinite number of frames, but in only one of them will they be at rest.

I am afraid one frame is enough to undo the relativity of all others.

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