Twin paradox, wave-particle paradox and many errors has put Einstein into focus. So, it is about time to ask: who was right? Lorentz ether theory has been suggested - but is in conflict with the twin paradox. Perhaps the problem is in the Michelson-Morley tests (MMX)?
MMX gave no empirical results, but gave a negative result to theoretical physics, due to interpretational error. Michelson said: no effect in transverse arm. However, most of his contemporaries said: half effect in transverse arm and reduced the prediction. Michelson had assumed ray direction c to be defined by mirrors and his contemporaries assumed beam direction c+v to be orthogonal to mirrors. If we respect Michelson's idea we should abolish time dilation and double the FitzGerald contraction. So, we get compensated effect in the longitudinal arm MMX is useless. Using ray direction c means also that stellar aberration is useless (the reason is observer motion).
So, Michelson was right, his contemporaries were wrong and Einstein was a child.
Or - what do you think??
John-Erik
See attachment
Hi everybody,
I'm working on a new theory of fundamental physics (The Daon Theory). I have put the first part, treating the electron, in the corresponding Project.
You have there the beginning to the answer, in the following paper, treating the atom (will be presented in about 2 month time) , you will get an almost complete explanation of SRT.
If you will have some patience, the third part (treating particles) will complete the picture.
OK!
1) The media (ether) exist.
2) It is not pulled by the earth.
3) All relativistic effects comes from the deformation of matter.
Stellan
1) Yes, there is an ether.
2) It is rather the ether that is pulling the Earth. Anyhow, ether winds from distant bodies are zero, as an effect of that. However, Earth is generating its own ether wind in radial direction causing gravity.
3) No, there are no relativistic effects.
Regards from
John-Erik
J.E P
The ether is a also reference system in which the CMBR is equal in all directions. There is no ether wind! It is all matter that are traveling through the ether.
Yes all so called "relativistic effects" are due to the deformation of matter at high velocities relative to the ether.
Stellan
Stellan
If a material body has a velocity in relation to the ether it produces an ether wind, just as a car that is moving in relation to the air.
Regards
John-Erik
Stellan
My opinion is that Michelson was right and Potier (and his contemporaries were wrong. Physics went wrong in 1882, not in 1905.
What is your opinion.
John-Erik
Stellan
Einstein was wrong and before that Lorentz and contemporaries were wrong and before that Potier was wrong. So, Michelson was the last one that was right!
What is your opinion?
With best regards from _______________ John-Erik
Sorry John-Erik
I have been occupied by other things.
Well , I have already stated my principle thoughts about this. Who is wrong or write, is not my principle occupation.
If you have some patience, I will publish the next piece of the daon theory, in about 2 month time; In this you will have a more precise picture of the deformation of mater due to the velocity.
Stellan
Sorry Stellan
I thought that you, as a scientist, would be interested in who is wrong and who is right.
With best regards from ________________ John-Erik
Dissidents have spent lots of time on Einstein but has not observed what happened when Einstein was only 3 years old. At that time Potier wrote a paper, stating that Michelson’s prediction for his experiments, (MMX), with Morley, were interpreted in error by Michelson. Michelson had assumed light to take the fastest – not shortest – way between mirrors, and therefore, wave fronts always are parallel to mirrors in MMX. This means no effect in the transverse arm. However, Potier had a different opinion and stated that transverse ether wind could change the behavior of light and force light to take a longer way.
Since mirrors in MMX have relevance for light only – and not for the ether wind – we can see that Michelson was right. Michelson’s correct prediction was based on wave behavior in both arms of the equipment. Potier reintroduced a particle reasoning in the transverse arm only, and did not see that light behavior is unchanged in the ether’s frame. The increased path length is instead in the frame of the equipment. Potier’s wrong idea got support from most scientists – but not from Michelson. So, Michelson was the last scientist that was right regarding MMX prediction. Poitier’s error supported the Lorentz transform but Michelson’s interpretation is in agreement to the Galilean transform.
So, the wave or particle paradox, and the twin paradox, are both results of Potier’s paper.
John-Erik Persson
All
I have a more detailed description regarding what happened when Einstein wa a child. See attachment!
John-Erik
Two models for light????????
When we detect light based on phase in a coherent system we can only get the ray, or normal to the wave fronts, defined by c. However, if light is focused into a beam we can see real light motion, c+v, as the direction of max amplitude. So, we need 2 models for light motion, beam and ray.
I have explained that on my blog at:
http://www.naturalphilosophy.org/site/johnerikpersson/
See the last post.
The difference between beam and ray is the relevance of ether wind in transverse direction.
What is your opinion????????
Regards ______________ John-Erik
Dear John-Eric,
I try to be a bit more specific on my response.
Let us assume there is an absolute aether. Let us assume that only in this absolute aether the speed of light is absolute c. Then the SRT does predict that in any other reference frame the speed of light appears to be c if we configure there the time and length according to the Lorentz transformations. With this we have a view of flat space with an absolute frame and SRT. I think it is called a Lorentz aether theory.
Now we have in this aether a light source sending light out in all directions. We should agree that in the reference frame of the aether, the ARF, the light wave front should expand out in a circular manner. That is, the light wave front is perpendicular to the moving direction.
An observer at rest in the ARF, OA, will see the light ray. Now this observer OA can observe an optical apparatus moving through the aether. Let us assume it is moving perpendicular to the light ray direction or parallel to the wave front direction. As seen by this OA the path of the light is not perpendicular through the optical apparatus. The optical apparatus has to be rotated a bit to get the light go through correctly. As seen by OA the wave front of the light is not perpendicular to the optical axis of the apparatus. Now the OA can conclude that with respect to the optical apparatus the light moves not only according to the ray direction but also to what you call the beam direction.
I hope until here you agree with me.
An observer at rest with the optical apparatus, OB, is moving with respect to the aether. Thus we could say that he observes an aether wind. You would expect that observer OB will detect this aether wind in that for him the beam has an offset with the ray.
Now we look into this optical apparatus and let us assume it is a big telescope. The mirror of the telescope is configured such that if a wave front traverses through the optical axis perpendicular to that axis, then the focus of that front is in the optical axis. Observer OA will see the light beam move along the optical axis and the ray will point in an offset to that axis. But observer OB will see a perfect focus of the light source in his optical axis. This means that OA and OB disagree about the timing that the wavefront reached the mirror. Only in the focus the light front is in synchronisation. Every where else the front is not focussed simultaneously and thus the interference of different parts will cancel out and there is no image. The disagreement is a physical fact. Optical sensors will detect a focussed image or won't detect a focussed image. But this means that in the reference frame of the moving optical apparatus the simultaneity of reaching of the wave front is interpreted different than it is interpreted by the observer OA. OB only can conclude that in his observations the beam and the ray are congruent. But also OA and OB have a different proper time. OB his time will run slower than the time of OA. This all results in a configuration where the observations, the timing and the length measurements are in accordance with the Lorentz Transformations.
I can have quite some sympathy with your light ray and light beam interpretation. But I think that the factual rotation of the level of simultaneity in a moving reference frame makes that this differentiation is not possible to be detected.
So, in conclusion, I agree with you that the MMX is incapable of detecting any aether. I agree that there is something like a light beam and a light ray. I think that this will result in an effective different wavelength in the two arms of the MMX setup with the effect that we can't see a fringe shift. But I think that you have to include the principles of the SRT to find why we are incapable of detecting the aether.
Regards,
Paul Gradenwitz
Paul
You assume an absolute ether and ether-less SRT cannot predict anything. You can use only Lorentz ether theory LET.
In general the wave front is not perpendicular to motion. Motion is defined by c+v (beam) and perpendicular to wave front is c (ray). I have explained this important distinction in my article.
No, the optical apparatus does not have to adapt to light, since the apparatus itself defines c (ray) (not beam, c+v). This was Potier's mistake.
Reflectors and refractors in the telescope has relevance in relation to ray direction c but not in relation v. You, and Poitier, have missed the important distinction between the dynamic, moving process c and the static, local condition v. Therefore, the distinction between ray and beam was missed. Therefore, stellar aberration is not caused by ether wind.
In my opinion the distinction between c and v can be observed.
If we assume constant frequency, than wavelength is constant in relation to the ether in both arms. In relation to equipment wavelength is constant in transverse arm and ~(1+-v/c) in relation to equipment (with 2-way wavelength ~ (1-v2/c2).
Simultaneity is not a problem since we can use the Galilean transform.
With best regards from
John-Erik
Hi John-Erik,
Your post came up as a personal mail to me but then Paul replied to you and I lost the option of replying? The RG software has some quirks I think.
Anyway, to comment on your post:
JP: So, we need 2 models for light motion, beam and ray.
There seems to be some confusion over names. I understand "beam" to be a low of light with some finite width and a profile of intensity, like the beam from a hand-held torch that may be some cm across. When we think of "ray-tracing" however, we are talking of a mathematical line which is one element of a beam. The surface of the wavefront at any point in the beam is perpendicular to the direction of the ray passing through that point, so for example the beam spreading out from a point source has a spherical wavefront.
Neither a beam nor a ray has any concept of speed related though, they are geometric constructions showing the full path of the light so you mention of c versus c+v in that context means nothing to me.
The modern model of light is based on quantum field theory and has been spectacularly successful. The 'motion' of light is defined by null geodesics both in special and general relativity, which equates to a speed of c in the classical models even though the word motion may be inappropriate in the quantum view. Our resulting understanding of quantum interactions has allowed the development of lasers, highly efficient light emitting devices, quantum entanglement, quantum computing and a great many more phenomena that the modern world relies on.
Dear John-Eric,
John-Eric>: In general the wave front is not perpendicular to motion. Motion is defined by c+v(beam) and perpendicular to wave front is c (ray). I have explained this important distinction in my article.
I tried to deduce that I can agree that notion in a certain sense. If I can see an object moving through the aether then the motion of the object is v with respect to the aether and the velocity of the wave front is c. Now I can invert this view and take the object at rest then the light moves with c through the aether and the aether moves with v around the object.
John-Eric>: No, the optical apparatus does not have to adapt to light, since the apparatus itself defines c (ray) (not beam, c+v). This was Potier's mistake.
I don't state that the optical apparatus adapts. I state that we observe perfect focus of the light in the axis of the apparatus. When I explain that from within the reference frame of the apparatus I have to assume that the wave front is perpendicular to the apparatus axis to explain the focus position. When I explain that from the aether frame then I have to assume that the wave front is perpendicular to the ray direction to explain the focus position. There is an apparent rotation in the assumed wave front direction that is in accordance with the change of the level of simultaneity between these two frames.
John-Eric>: Therefore, stellar aberration is not caused by ether wind.
Stelar aberration is caused by relative motion and because of that the different level of simultaneity. In there is active what you distinguish as ray and beam but it only gives a delta and not an absolute value.
John-Eric>: Simultaneity is not a problem since we can use the Galilean transform.
I extract the rotation out of observation and then find that it is also mentioned in SRT. Maybe you show how you get to the same observed effects with Galilean transformation.
John- Eric>: If we assume constant frequency, than wavelength is constant in relation to the ether in both arms. In relation to equipment wavelength is constant in transverse arm and ~(1+-v/c) in relation to equipment (with 2-way wavelength ~ (1-v2/c2).
You have to be more specific for me. I imagine a Minkowski space at rest in the aether. Then I try to understand what happens with an MMX experiment moving through that space. Then I can change observation view and see the same from the observation view of an observer at rest with the MMX. In the aether ray and beam are identical. Now show here how from the moving frame we can observe a difference.
Regards,
Paul Gradenwitz
George
The concepts ray and beam are not related to velocity vectors in your opinion. This seems confusing. Have you read When Physics Went Wrong? Perhaps you should take a look at the attachment below.
Yes, a ray is not a physical reality, but a practical tool to describe the orientation of the wave fronts that are realities.
The important part is the fact that Michelson's first interpretation seems to agree to Galilean transform and no time dilation. This is explained in the complete articles.
Best regards from
John-Erik
Paul
At least we agree upon stellar aberration.
The important part is that Michelson assumed no effect in the transverse arm in MMX. If equipment is moving transverse to light there is no reason for light to change behavior. No wave front bending. We must use the wave model and light is not dependent on source motion. In other words: light takes the fastest way between mirrors, so wave fronts are always parallel to mirrors. No effect in transverse arm. No time dilation. Doubled contraction of matter and Galilean transform.
See the link to my latest article on the earlier post to George.
Best regards from
John-Erik
Dear John-Eric,
I advise you to review your opinion about the MMX experiment. Assume every point a source of a spherical light wave that is summed up. The result is a combined wave front. Now let that wave front travel to a 45° mirror. See how the wave top reaches the mirror points over a time period.
Trace that wave front after reflection and see how the wave length is the same and wave direction is 90°. Now do the same with a moving mirror. Then you will see that the moving mirror will change the time the wave front needs to cover the mirror area and how as a result the angle of reflection is changed from 90° and the wave length also is changed.
The light source of the experiment is normally a laser. The laser will send out light of a wavelength related to the distance of the mirrors. So the moving source will not change the wavelength with resect to the moving frame. From the rest frame the source will have a Doppler shift. The moving 45° mirror will see a Doppler effect. From the rest frame the light wave front will change under an angle that is not 90°.
Regards,
Paul Gradenwitz
Paul
No, you should change opinion.
The motion can cause a change of the angle to deviate from 90 degrees a small amount about 10-6 radians (due to planetary rotation, not motion) and this angle also exists in the reflected ray. However, since relevant wave front is defined by the distant mirror this angle is in the light before the 45 degree mirror. This very small angle is smaller than the beam width of the laser.
Regards from
John-Erik
Paul
No, you should change opinion.
The motion can cause a change of the angle to deviate from 90 degrees a small amount about 10-6 radians (due to planetary rotation, not motion) and this angle also exists in the reflected ray. However, since relevant wave front is defined by the distant mirror this angle is in the light before the 45 degree mirror. This very small angle is smaller than the beam width of the laser.
Regards from
John-Erik
Hi John-Erik,
As you probably know, I'm a strong supporter of the Local-ether model proposed by Prof. Ching-Chuan Su in 2000 and the almost identical concept proposed by Prof. Petr Beckmann in 1987.
The older entrained ether concepts postulated that the "ether wind" at the Earth''s surface was zero. However both Prof. Su and Prof. Beckmann postulate that the Local-ether forms a halo that is aligned with and stationary with respect to the Earth Centered Inertial reference frame (the ECI) but that the Earth rotates within it. This variable density halo is the physical medium that is the unique preferred inertial reference frame for classical EM wave propagation. It extends out to where the Sun's and Earth’s gravitational force vectors are balanced. The Sun’s Local-ether takes over outside this boundary.
This implies that there is an "ether wind" at or near the Earth''s surface due to and equal to the Earth's rotation rate. This is less than 1/1000 the velocity assumed by the universal ether concept . This accounts for and is in accord with the first order Sagnac correction employed by the GPS. It also means that (almost) all MMX type experiments to date have not been sensitive enough to detect this velocity since the MMX interferometer is designed to measure a second order effect in (v/c)². An MMX type experiment would have to be at least 10⁶ more sensitive than the original MMX to detect a ~300 meters /second “ether wind”
Note that Prof. Howard C. Hayden has shown that the Brillet-Hall experiment in 1979 may have detected the effect but assumed it was only a systematic error.
Also note that Prof. Beckmann accounted for the phenomena associated with stellar aberration by showing that the alteration of the angle of a star’s wavefront direction occurs far away in the transition zone between the Sun’s local-ether and the Earth’s local-ether. And therefore it would be expected that experiments like filling a telescope with water (Airy 1871) would not show any affect on the aberration angle. Profs. Su and Beckmann also account for the Fizeau experiment and the Fresnel formula for the index of refraction of a material medium in motion. These are phenomena that are almost always cited to dismiss the entrained ether concept.
Finally note that, Dr. Michelson supported an entrained ether concept which also postulated that it didn’t rotate with the Earth. That’s what led him to perform the successful Michelson-Gale experiment.
I’ve seen some of your older papers that support the entrained ether concept. So I’m curious why you apparently have switched to supporting the universal ether concept.
Best Regards,
Jim Marsen
Dear John-Eric,
Concerning the 90° angle deviation with a 45° beamsplitter. Assume we have a setup of a light source and a 180° angle reflector separated at a distance of about one astronomical distance. Both are moving with the same velocity through our assumed aether and thus share the same IRF. The source beam is in the direction of movement and splits its beam from there to 90°. The full mirror at one AE distance has a size of one meter and the beam deviates from parallel over that distance to a beam of 10 cm width. With a speed of 368 km/sec through the aether will the beam hit the mirror when with a speed of 0 km/sec the setup will have a perfect hit?
All is far from any gravity in deep space.
Regards,
Paul Gradenwitz
Jim Marsen
Thanks for mail.
I also much respect the ideas by Su, Beckman and Michelson. I have always assumed an ether that was entrained in translation but not in rotation. So, we have a horizontal ether wind in the order of 10-6. So, we have only 10-12 to detect, but this effect is compensated in the longitudinal arm and not existent in the transverse arm. Therefore, increasing sensitivity one million wont help.
I do not know how you can think that I believe in an universal ether.
I guess that Airy's test is useless too. If the telescope shall be usable you have adjust to the new focal length and the effect of water is compensated.
Beckman: I think the ether is the cause of gravity and the reference for light motion. I do not think that stellar and pulsar aberrations are caused by ether wind v, but instead by observer motion u. Light moving with c in relation to Sun are moving with c-u in relation to Earth.
Su: I agree to the local ether model, but I think that the ether entrains the Earth to move neutralize effect of distant bodies, so, only effect of Earth is visible. Su has suggested how 1-way speed can be detected by a scaled down version of de Witte's method.
Michelson: I think his first prediction for MMX was correct: no effect in transverse arm.
Best regards from
John-Erik
Paul Gradenwitz
I regard this example more as a joke. MMX is a more important subject where we have different opinions. The question is if a change in the 90 degree angle is localized before or after the reflector. This is much more interesting than your science fiction example.
With best regards from
John-Erik
Hi John-Erik,
I apologize if I misunderstood whether you support Prof. Su's local-ether paradigm.
But I would say that entrained ether models imply that the motion of, for example, a laboratory, with respect to a medium is negligible compared to the sensitivity of the instrument trying to measure it.
For an analogy, think of a cruise ship. You can't use an anemometer located in the ballroom to measure the velocity of the wind outside.
This means the behavior of light in either arm of an interferometer doesn't matter since there's (almost) no motion to detect.
The 19th century mainstream assumed that light and EM waves obeyed classical propagation of waves in a medium analogously to sound and therefore the propagation time is not independent of the motion of the receiver with respect to the medium.
They further assumed that light and EM waves propagate in a medium that filled the Universe that was isotropic, homogeneous, and at rest with the Universe. They further assumed that the Sun centered inertial frame was the preferred inertial reference frame for the propagation of light and EM waves on the Earth's surface, the Solar System, and, in effect, the Universe.
In my opinion, this was the biggest mistake of the 19th and early 20th century physics. It led to Lorentz's length contraction postulate and the Lorentz transform to account for the null results of the MMX. This led to time dilation and, almost inevitably, to Einstein's Relativity.
I assume you are well aware that most researchers who don’t accept the validity of Einstein’s Relativity have continued to assume a variation of the Universal ether paradigm.
I suggest that Prof. Su's local-ehter paradigm is a much more promising paradigm to replace the current paradigm of Einstein’s Relativity. Unfortunately it has not been given serious consideration by orthodox or unorthodox physics.
Finally, I interpreted your discussions of why the MMX didn't detect a 30 km/s motion with respect to the Sun's inertial reference frame seemed like support for the universal ether ocncept.
Regards,
Jim Marsen
Dear John-Eric,
You evade the answer. What I try to get from you is if according to your view the light beam will move with the setup or remain behind and only deviate by 90° in the aether frame. You insist that the small deviation is not visible in the normal MMX size. So I enlarge the size to allow that the deviation could become visible. I still hope you can give me an answer on this question.
Regards,
Paul Gradenwitz
Jim Marsen
I agree to most from Beckmann and Su, but I think like Einstein that Earth is entrained by ether instead. So, gravity from distant bodies is hidden by planetary motion, but gravity from Sun cannot be hidden. The ship is transparent to gravity, but not to air. MMX is transparent to ether, but not to light. Detectable motion is 10-12 for 2-way light.
Yes, if we assume ether entrained by Sun we must also assume entrained by Earth. However, probably not with a GPS satellite due to very low mass.
Yes, I think that Einstein was right when he said that no ether is unthinkable.
Best regards from
John-Erik
Paul
The changed angle due to the motion of 45 degree mirror of is small (10-6) and irrelevant since the change is before the mirror.
What we cannot see in MMX is first order different positions on the mirrors.
We can see second order (12-12), but the effect is compensated in the longitudinal arm and not existent in the transverse arm.
Regards
John-Erik
Paul
In my opinion it is the angle after the 45 degree mirrors that is defined by the mirror. Do you agree?
Regards from
John-Erik
Paul
Due to the feedback it is the distant mirror that decides which wave fronts are relevant. Therefore, if motion causes the effect from the 45 degree mirror to deviate from 90 degrees, this change becomes located between source and 45 degree mirror and not in the long and relevant path.
What do you think? Potier was wrong! Or?
With regards from _____________ John-Erik
Dear John-Erik Persson,
I did let you wait a bit with my answers because I am not all days able to be present here with enough time.
John-Eric>: The changed angle due to the motion of 45 degree mirror of is small (10-6) and irrelevant since the change is before the mirror.
What we cannot see in MMX is first order different positions on the mirrors.
We can see second order (12-12), but the effect is compensated in the longitudinal arm and not existent in the transverse arm.
I don't understand why you come up with a change before the mirror. Imagine the wavefront coming in perpendicular to the light path. One edge will hit the nearest part of the mirror. By the time the other edge of the same wavefront has reached the farthest part of the mirror this one has moved from the 45° position that was valid for the nearest part. The result is that the complete wavefront after the diversion is rotated over less than 90°
John-Eric>: In my opinion it is the angle after the 45 degree mirrors that is defined by the mirror. Do you agree?
Yes I agree. After the mirror the angle is no more 90° because of the movement. John-Eric>: Due to the feedback it is the distant mirror that decides which wave fronts are relevant. Therefore, if motion causes the effect from the 45 degree mirror to deviate from 90 degrees, this change becomes located between source and 45 degree mirror and not in the long and relevant path.
For me this is not clear what you mean. What feedback is there in the deviation? I only see the setup of a moving 90° deviation mirror and a far away 180° mirror (flat).
John-Eric>: What do you think? Potier was wrong! Or?
I see in your :
http://www.naturalphilosophy.org/site/johnerikpersson/2018/08/17/potier-was-wrong/
That you state "So, the distant mirror defines wave fronts to be parallel to the mirror, and therefore, light motion is perpendicular to mirrors in the frame of the ether."
However I see no need for a mirror to define that light is perpendicular to the mirror. It is not defined by the 90° diverting mirror and also not at the far away mirror. For any flat mirror the angle of entry is the angle of exit.
I might need a bit more clarification to give you an answer on the last question. Regards,
Paul Gradenwitz
Paul
Wellcome back. I will try to be more detailed.
The laser has a beam-width much larger than v/c. Therefore containing different wave fronts, and the mirror defines which to reflect. Since we use phase information we cannot see ether wind moving inside the wave fronts. So, we must use the ray direction c (and only longitudinal component of the ether wind). Potier assumed (in error) that total motion c+v was orthogonal to mirrors. You can also see (on my last post on CNPS blog how Armand tried to save Potier's idea by explaining his result as an effect of the 45 degree mirror's motion. So, distant mirror defines wave front (not laser).
Yes, I was unclear regarding the 45 degree mirror. I try to be better. If the effect is changed from 90 degrees, this does not matter, since the distant mirror changes the same wave front from many, since the laser beam width is much larger than v/c.
The idea with the 45 degree mirror is wrong can be seen in another way also. If the ray direction had changed to be larger than 90, than the returning direction would be LESS and the equipment would still be looking in a direction LARGER than 90.
I have explained this in more details on my CNPS blog and in articles to CNPS and to GSJournal.
With best regards from _____________ John-Erik
Hello John-Eric Persson,
John-Eric>: The laser has a beam-width much larger than v/c.
The laser beam width in my understanding has a dimension in length (metre), and the quotient of two velocities v/c is dimensionless. So I don't understand how you compare these two. I don't understand how you come to the conclusion that there are thus different wavefronts. The light has a certain wavelength. We assume with laser something near to optical frequencies and thus in the range of several hundred nanometers. I understand not how you can state that the mirror defines what waves to reflect.
John-Eric>: The idea with the 45 degree mirror is wrong can be seen in another way also. If the ray direction had changed to be larger than 90, than the returning direction would be LESS and the equipment would still be looking in a direction LARGER than 90.
To me it is compete unclear how you come to this conclusion. I repeat my viewpoint. The aether is at rest and the observer is at rest in the aether. The mirrors move with resect to the aether. The light then is diverted over and angle such that the angle source, 45°mirror, diverted beam is greater that 90°. Then the returning light will have the direction less than 90° as you say but that has no influence on the moving system. For the moving system observer the light will travel exactly in the 90° direction forward and backward.
Regards,
Paul Gradenwitz
Paul
The beam width is not a length but an angle around a milliradian. v/c is about a microradian. If the effect deviates from 90 by a microradian this just as unimportant as changing the laser direction by a microradian. In MMX it is the 2 distant mirrors that define the wave fronts to be parallel to mirrors in the 2 arms.
John-Erik
Hello John-Eric Persson,
v/c with the earth speed variation of 60 km/sec is 60*103/3*108=2*10-4 that is larger than micro radials.
You repeat that the 2 distant mirrors define the wavefronts to be parallel. You don't explain why that is the fact. I state I don't understand how you come to that conclusion. Then repeating your statement without elaborating what principle you use to come to that conclusion is for me insufficient.
Regards,
Paul Gradenwitz
v is caused by the rotation of our planet. The translational motion is not available to MMX.
If incoming and reflected wave fronts shall be the same they must be equal to the mirror.
John-Erik
Hello John-Eric Persson,
John-Eric>: v is caused by the rotation of our planet. The translational motion is not available to MMX.
With this statement you assume that the aether is associated with the earth. I started an example without the earth. You evade again my example. I do my best to have an example where we describe what happens based on an observation within the restframe of the aether.
John-Eric>: If incoming and reflected wave fronts shall be the same they must be equal to the mirror.
In the restframe of the aether the light path is without 90° angles. The mirrors have no ability to modify or rotate wavefronts. Thy only reflect. The pat of the light to and from the far mirror is not parallel. The wavefront is perpendicular to the path.
As seen from the frame of the moving setup the light path is with 90° angles. The path to and from the far mirror is parallel. Yet it is the same light and the same mirror. So can you explain why, as seen from this moving frame, the wavefront appears to be rotated such that it is parallel to the mirror orientation?
Regards,
Paul Gradenwitz
Hi John-Eric,
Sorry to take so long to respond.
In the paper in your question above you show two illustrations in Figure 1 and mark one as "correct". In fact both are correct but drawn in different frames. The one on the left is valid in the rest frame of a hypothetical aether in which the apparatus is moving from left to right. The speed measured along each line would be exactly c. The image on the right is valid as drawn in the rest frame of the apparatus such that the aether is moving from right to left. The speed in this frame according to aether theory has to be c-v from left to right, c+v from right to left and have to you use Pythagoras to find the speed on the vertical paths. The time of flight values are given here:
https://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment#Observer_resting_in_the_aether
The correct analysis then predicts a phase shift in aether theory which is not observed.
Paul
Sorry to evade your example. However, i do not see how you can exclude Earth and still do MMX.
The illumination includes many wave fronts. The mirror reflects one of them.
If path to and from are parallel wave fronts are parallel and equal to the mirror. No rotation.
John-Erik
George
Both diagrams are in the frame of the ether. The left is Potier's 'corrected' version and the right one is the right one given earlier by Michelson. So, no effect in the transverse arm. The reference you gave was in error too.
From ____________ John-Erik
Hello John-Eric Persson,
John-Eric>: Sorry to evade your example. However, i do not see how you can exclude Earth and still do MMX.
Light seems to be able to travel all the way from far away stars to us. So an experiment with light at a larger scale would not be impossible. Laser communication over solar system distances is in principle proven to be possible. So laser light on this scale does work. I thus hope you still can go into detail of my example. It is set up to show the consequences of your statements.
Maybe you see that your answer style only give snippets of your conclusions without any supporting reasoning. That leaves me with a guess why you cant see how to exclude Earth for an MMX experiment. You might be blind, you might assume that without a fixed support on the Earth it is impossible, you might assume that only over small distances MX can be performed, etc, etc. The MMX setup is used in the LIGO type experiments. There is a plan for a similar GW experiment in space. All this tells me that there is no restriction to only the Earth to do the experiment.
John-Eric>: The illumination includes many wave fronts. The mirror reflects one of them.
So what happens to all the other wave fronts and what do you mean with many wavefronts? What shape do you think these wavefronts have and why do they differ?
John-Eric>: If path to and from are parallel wave fronts are parallel and equal to the mirror. No rotation.
I stated that in the observation of the observer at the rest frame of the aether the wave fronts to and from the mirror have a slight angle while in the observation from the rest frame of the mirror setup the SAME light appears to have a wavefront that is parallel to the mirror. There is no selection by the mirror.
Regards,
Paul Gradenwitz
Hi John-Erik,
JP: Both diagrams are in the frame of the ether. .... The reference you gave was in error too.
No, the suggestion that both are in the same frame is obviously impossible since the splitter changes location in one but remains fixed in the other. The Wikipedia link is correct and you'll find something similar in every valid treatment of the MMX.
Paul
I am not blind and I know about plans for MMX in space. However, those are only plans. There is no need to do that, since we have experiences from 1-way light speed in GPS. We use that speed for navigation.
With best regards from ______________ John-Erik
George
They are in the same frame. You missed the fact that one is correct and the other is wrong.
There are lots of errors in Wiki, especially regarding MMX.
With best regards from ______________ John-Erik
Hello John-Eric Persson,
John-Eric: I am not blind and I know about plans for MMX in space. However, those are only plans.
And you can be pretty sure that they will work. They will find a signal that they will relate to gravity waves.
John-Eric: There is no need to do that, since we have experiences from 1-way light speed in GPS. We use that speed for navigation.
We have in GPS a system that uses the ECI frame. Then in that frame it describes a lot of moving objects. In the ECI frame the timing is adjusted such that the speed of light appears to be c in all directions. Then al other moving objects will have relative to each other speeds that can be the sum or the difference of the own speed and the light speed. All the speeds are described from the viewpoint of one reference frame. This has nothing to do with how a moving observer at earth perceives the speed of light that comes from a satellite in his moving setup.
However, if you have problems to understand the difference between timing in one frame and timing between frames, then you will not be able make the correct judgement.
Because you continue to ignore the thought experiment and only post short statements that have no verifiable reasoning, and are in contradiction to what is generally assumed to be correct I have to discard your message as insufficient to accept. Try to invoke a section on why you think you are correct in your statements.
Regards,
Paul Gradenwitz
JP: They are in the same frame.
No John-Eric, that is not possible. The splitter cannot move in the frame in which it is at rest but it moves in one diagram and doesn't in the other. The answer to this whole question is that you have made an incorrect assumption, both diagrams are valid but you missed the fact that there are drawn in different frames.
PS: I have remembered why Potier corrected the earlier calculation.
JP: The left is Potier's 'corrected' version and the right one is the right one given earlier by Michelson. So, no effect in the transverse arm.
In his 1881 paper, Michelson missed the Pythagoras factor. Potier pointed that out and Michelson corrected it in his 1887 paper with Morley. However, your error remains that the diagrams are in different frames.
JP: The reference you gave was in error too.
No, it is correct.
Paul
You asked me to prove that I am wrong. However, that is not the question. To improve science we have to find the errors to make room for a new theory. Time dilation is one paradox and wave or particle is another paradox (not complementarity). There are lots of errors in modern physics. We cannot prove a theory to be correct in relation to reality. We can only test if it is consistent and without paradoxes. Only correct inside can be tested by logic.
Lots of money has been used in vain to prove SRT and quantum theory, when we instead should try to explain anomalies and there are many.
The reason that we have the idea of light particles is that we do not understand light and ether enough to see how the wave model is all we need for light.
Best regards from ________________ John-Erik
George
Your statement is ridiculous. I made the diagrams and therefore I say that both are in the frame of the ether. I repeat: They represent Potier's and Michelson's different interpretations. I made the diagrams.
Best regards from ________________ John-Erik
John-Erik,
John-Erik>: You asked me to prove that I am wrong. However, that is not the question.
That is my question and you answer my text. If you think that empty statements are sufficient to convince other people then I have to say then for me this means you fail to convince.
John-Erik>: To improve science we have to find the errors to make room for a new theory.
To find errors we have to show where something is inconsistent. That means we have to make sure that we agree about all intermediate steps. You come with end conclusions as declarations without any derivation of your conclusions. That makes these statements empty shells.
John-Erik>: Time dilation is one paradox and wave or particle is another paradox (not complementarity).
You have problems with the wave-particle duality. Maybe it is because you don't understand the subject enough. I have not yet received any justification of your statements.
John-Erik>: There are lots of errors in modern physics.
General statement without evidence. Generally true but it gives to precedence to use it to assume another error exists.
John-Erik>: We cannot prove a theory to be correct in relation to reality.
Again, it is partly true but useless in our discussion
John-Erik>: We can only test if it is consistent and without paradoxes.
We test if there are paradoxes if we invoke in discussion and think something through. We also can compare the theory with observation. So your word only implies that you don't understand the process of truth finding.
John-Erik>: Only correct inside can be tested by logic.
You don't invoke logic but only state empty shells of declarations
John-Erik>: Lots of money has been used in vain to prove SRT and quantum theory, when we instead should try to explain anomalies and there are many.
You have not yet shown to understand SRT and thus I can't accept that you have proven SRT to be wrong.
John-Erik>: The reason that we have the idea of light particles is that we do not understand light and ether enough to see how the wave model is all we need for light.
I might agree with your view here but your description of light makes me conclude that you have heard of the subject but didn't grasp the total. Can you explain why light is quantised?
I have looked into your discussion at:
http://www.naturalphilosophy.org/site/johnerikpersson/2018/08/17/potier-was-wrong/
I find that your answers there are as wrong as you have them here. I really recommend you study some of the basic theory lectures on Youtube that were recommended there. You argue that effects are to small to be detected with 10-12 You missed that effects down to 10-28 are detected in the GW experiments that are effectively MMX setups.
In conclusion I find that the biggest error is that you think you found the truth in this subject. And the biggest problem is that you don't engage in discussion and with that block the possibility to use logic to help you find your errors.
Regards,
Paul Gradenwitz
JEP: Your statement is ridiculous. I made the diagrams and therefore I say that both are in the frame of the ether.
OK, it is your drawings that are wrong. I thought you had offset the lines just to make them visible but it seems you imagine those are the correct paths. The attached correction shows the error you've made.
JEP: They represent Potier's and Michelson's different interpretations.
No they don't. Michelson in 1881 made the error of assuming that the speed would be c in the top right hand diagram, Potier noted that it would be c in the top left hence you apply Pythagoras Theorem to find the speed in the top right. Michelson corrected that and thanked Potier in his 1887 paper with Morley.
Your aether frame diagram shown bottom left is wrong because the returning beam doesn't pass through the splitter on the return path.
Paul
You seems very upset and angry. I find that it is important to allow different opinions without being upset. You also put words in my mouth.
I have never said that I am right. Instead I stated that Einstein was wrong.
Regarding MMX. The mirrors in the cavities are forcing light to move force and back in a right angle to mirrors in both cavities and wave fronts are parallel to mirrors in both cavities. Light behavior is defined by mirror orientations and the state of motion of the ether. So, light does not take part in the equipment's longitudinal motion in one arm and therefore is does not take part in the transverse motion in the other arm either. (Motions inside the planes of the wave fronts does not change the boundary conditions and therefore not light behavior either.) Michelson said so, since mirrors have relevance for light, but not for ether wind, since mirrors are transparent for the ether but not for the light. Mirrors control light, but not ether.
Michelson assumed light to be orthogonal to mirrors and Potier assumed light plus ether wind to be orthogonal to mirrors.
Best regards from ____________ John-Erik
George
See my answer to Paul.
Your statement that light does not pass the beam splitter is wrong. The size of the beam splitter is some centimeters. and the motion of the equipment is only some micrometers, since v/c is 10-6 in the test with cooled cavities.
You did forget to mention that Michelson was trying to convince majority between 1882 and 1887 before he gave up.
Regards from _______________ John-Erik
Hi John-Erik,
JP: You seems very upset and angry.
Not in the slightest, I'm sorry if I gave that impression, I add emphasis to parts that I have said before and you seem to have missed or ignored because I don't think it helps either of us if I have to continually repeat the same points.
JP: I find that it is important to allow different opinions without being upset.
I agree where there is room for opinion in a discussion but in geometrical optics there is only one correct solution.
JP: You also put words in my mouth.
I don't believe so, perhaps your words were not as clear as you intended.
JP: I have never said that I am right. Instead I stated that Einstein was wrong.
You see here again, I have to repeat what I already said. Michelson was wrong in his 1881 paper but he corrected it in his 1887 paper with Morley, and in that he credited Potier with finding the flaw. Einstein didn't say anything until after 1905 (and he later said he wasn't even aware of Michelson and Morley's work at that time) so that was a decade after Michelson fixed the error.
JP: Regarding MMX. The mirrors in the cavities are forcing light to move force and back in a right angle to mirrors in both cavities and wave fronts are parallel to mirrors in both cavities.
First, there were no cavities in Michelson's interferometer, the light made a single pass over each path only.
Second, light reflects from a mirror according the simple law of reflection:
https://www.physicsclassroom.com/class/refln/Lesson-1/The-Law-of-Reflection
In the aether frame, the wavefronts are not parallel to the mirror, that is true only when drawn in the lab frame.
JP: So, light does not take part in the equipment's longitudinal motion in one arm and therefore is does not take part in the transverse motion in the other arm either.
The light spreads out from the source lamp and the portion we are interested is the part that takes the path it needs to reach the point on the screen where it combines to interfere with the light traversing the other arm.
JP: Michelson assumed light to be orthogonal to mirrors
When drawn in the lab frame, that is correct but he forgot to take account of the effect of the ether wind on the speed in that frame. That's why I thought your second drawing was in the lab frame.
JP: Your statement that light does not pass the beam splitter is wrong. The size of the beam splitter is some centimeters. and the motion of the equipment is only some micrometers, ...
Yes, that is true, my mistake. What I meant to say was that to interfere the two beams have to pass through the same point on the splitter to reach the same point on the viewing screen and that point has moved so your drawing is wrong.
John-Erik,
John-Erik>: You seems very upset and angry. I find that it is important to allow different opinions without being upset. You also put words in my mouth.
I did my best to show my disappointment with your style to answer. So I tried to show how all your sentences are like postulates without supporting evidence. Then you only can end up with a yes/no fight that leads to nothing.
John-Erik>: I have never said that I am right. Instead I stated that Einstein was wrong.
You have made a picture with two versions of light path and stated that one is wrong and the other is correct. For me that is synonym with a declaration that you think you are right with that description. you stated to George Dishman: "They are in the same frame. You missed the fact that one is correct and the other is wrong."
John-Erik>: Regarding MMX. The mirrors in the cavities are forcing light to move force and back in a right angle to mirrors in both cavities and wave fronts are parallel to mirrors in both cavities.
This view is correct when observed from the reference frame of the setup. I started to describe the whole from the reference frame of the aether.
John-Erik>: Light behaviour is defined by mirror orientations and the state of motion of the ether. So, light does not take part in the equipment's longitudinal motion in one arm and therefore is does not take part in the transverse motion in the other arm either.
You assume that the aether has an influence. I can agree with that. However you miss that because of that assumption you have to conclude that light will not divert under 90° in the aether frame. The mirror does not define the angle of the wavefront to be only parallel. It reverts the wavefront movement in one direction and keeps the wavefront movement in the other directions.
John-Erik>: (Motions inside the planes of the wave fronts does not change the boundary conditions and therefore not light behaviour either.)
The motion has an influence on the way the beamsplitter diverts the light. There the splitter surface is at 45° angle to the wavefront.
John-Erik>: Michelson said so, since mirrors have relevance for light, but not for ether wind, since mirrors are transparent for the ether but not for the light. Mirrors control light, but not ether.
They reflect light. Any point of the mirror emits the momentary EM field back.
Regards,
Paul Gradenwitz
Dear George Dishman ,
About the MMX experiment I have my restrictions on what is written in Wiki. So I sat down to try to make a picture of what happens. I have drawn all in the rest frame of the aether. I assume two wave fronts that move from left to right from the source. I let the wave fronts move with c through the aether. The fist wave front is in black and the second is in red. Each next line then is one time unit later for the same wave front. When the black has advanced one step the red happens to be at the old position of the black wave front. So I have a lot of moments stacked on top of each other in one picture.
The moment that the first wavefront (black) meets the splitter I set the time at 0 for counting. The concentric circle sectors are a method to show what happens with the light of that meeting point of splitter and wavefront. At t=1 the black wavefront has advanced one step. Now it meets the same splitter that has moved by a smaller step related to its velocity v
Hi Paul,
PG: When the black has advanced one step the red happens to be at the old position of the black wave front. So I have a lot of moments stacked on top of each other in one picture.
I'm just starting to read this but there is an obvious error right at the start, the reflected wave will be Doppler shifted so if you show one wave at the moment it hits the mirror, the nearest red line should be farther way than the nearest black line because it was reflected when the mirror was to the left of its current position.
PG: But with this picture I hope that I can convince you that as seen from the aether frame the wavelength changes for different parts of the light path.
It does but you haven't shown it.
PG: Thus their derivation why there should be a signal seems not correct for me.
I haven't checked the Wikipedia article but wavelength changes do occur and they are what causes the phase shift. However, the wavelength and frequency depend on the frame so you have to be careful. It is better to calculate the path travel time and work out the difference in arrival at the screen, that's a proper time and as I have said in several places, it is always proper quantities that drive the physics.
I'll need to go through the rest later, it a somewhat messy diagram. The wavefronts in the vertical beam look OK but you also need to reflect the returning wavefronts of the horizontal beam from the moving splitter and then down towards the bottom right. The second reflection cancels the Doppler frequency difference but freezes the accumulated phase difference, that's the source of the fringe shift.
P.S. Note also the splitter moves between waves from the source hitting it which increases the wavelength on the near-vertical paths but by a different amount from that on the horizontal return path.
If I were trying this, I'd work out the values in a spreadsheet to make sure my maths was correct and then plot the results or you can end up doing a lot of nugatory sketching.
Hi George Dishman ,
George>: I'm just starting to read this but there is an obvious error right at the start, the reflected wave will be Doppler shifted so if you show one wave at the moment it hits the mirror, the nearest red line should be farther way than the nearest black line because it was reflected when the mirror was to the left of its current position.
The upward and downward path has a Doppler shift visible in red. The horizontal arm has a Doppler shift with change of colour. So the red wavefront changed to orange in the return path. The black is set to mirror on top of each other, so these lines count double. Because both waves meet at t=15 the continued path is with the same wavelength for both beams. That is why I didn't draw that. Yes it is a bit messy picture. Sorry.
Regards,
Paul Gradenwitz
Gentlemen,
I suggest that you take a loke at:
R. de Sangro, G. Finocchio, P. Patteri, M. Piccolo, G. Pizzella "Measuring Propagation speed of Coulomb Fields". arxiv:1211.2913 [gr-qc] 10/11-2014
Which is an experimental demonstration that the electromagnetic field is not produced by photons!
So, the Feynman diagrams!?
Hi Paul,
PG: The horizontal arm has a Doppler shift with change of colour. So the red wavefront changed to orange in the return path. The black is set to mirror on top of each other, so these lines count double. Because both waves meet at t=15 the continued path is with the same wavelength for both beams.
The reflected wave doesn't have the same wavelength. I've done a much simpler version showing just the horizontal. In the top half you see three black wavefronts approaching the blue mirror, the first labelled 'a' has just reached it. The wavelength is λ.
The lower set show the setup when the second wave 'b' reaches the mirror. In the time between, the mirror moved a distance d so the wavefront had to move distance λ+d. That means wavefront 'a' also moved distance λ+d back from the point of its reflection (shown by the vertical red line) so it is now λ+2d from the mirror. All the reflected wavefronts should therefore have that greater spacing than the incident wavefronts, they shouldn't overly and "count double".
The same principle will apply to the near-vertical beam but you have the added effect of the change of propagation direction.
Overall, this is just applying Huygens' method.
George
...seems very upset...
Was addressed to Paul and not to you.
In order to increase precision the arms were realized with 2 cooled down cavities instead. Many short reflektions instead of one long.
No, Michelson did not forgot to account for transverse ether wind. Instead he realized that transverse ether wind is irrelevant. This follows from the fact that light moves with speed c in a right angle to mirrors in the frame of the ether in the same way as it moves in relation to the ether in the longitudinal arm. Consistency demands that, as Michelson said for 5 years, until he gave up. Therefore, no effect in the transverse arm. Speed becomes sqrt(c2+v2) in equipment frame.
Since light does not always move in a right angle to wave front we must make distinction between beam as c+v and ray c.
Regards from ___________ John-Erik
Paul
See my answer to George.
If the effect of the beam splitter deviates from 90 is without relevance, since the distant mirror defines which wave front is of interest, as George said. You are wrong regarding the beam splitter.
Regards from John-Erik
Hi John-Erik,
JP: ...seems very upset... Was addressed to Paul and not to you.
I must have missed that, no problem then.
JP: In order to increase precision the arms were realized with 2 cooled down cavities instead.
JP: Many short reflections instead of one long.
Yes but it was still a single zig-zag path, they didn't use a cavity as far as I know.
JP: No, Michelson did not forgot to account for transverse ether wind. Instead he realized that transverse ether wind is irrelevant. This follows from the fact that light moves with speed c in a right angle to mirrors in the frame of the ether in the same way as it moves in relation to the ether in the longitudinal arm.
No, it can't do that or it doesn't reach the same point on the screen as the longitudinal beam. You are calculating the interference between the two beams from source to some point on the screen so both paths have to have exactly the same destination and that means the transverse beam is perpendicular to the mirror in the lab frame so it cannot be in the aether frame, it has to hit at an angle. That is basic geometry.
Paul said:
PG: The motion has an influence on the way the beamsplitter diverts the light. There the splitter surface is at 45° angle to the wavefront.
That is correct, the mirrors do not define the light path, we are interested only in rays that both reach the same point on the screen so that they can cause interference. In the aether frame, the light deflection needed to do that is not exactly 90 degrees, Paul is perfectly correct about the geometry. All he missed is the resulting change of wavelength which we call Doppler shift.
Hi George Dishman ,
I didn't miss the Doppler shift in my drawing. Let me start with your drawing. You have in the upper row 3 stages of one wave front. At the time that the wavefront is on the 3rd position the mirror is also there. Your a is at the mirror. So this wavefront will now move back with the same speed and overlap each time step the position of the earlier stages. The second wavefront reaches one time step later the position where the mirror was at the time step before. But at that moment the mirror has moved by one velocity step. Now the light front has to move a bit further while also the mirror moves. Hence the mirror will be somewhere in between the time step positions when the wavefront catches up with the mirror. I calculated that position with two slanted lines. The rest of the distance for the wavefront is then in reversed direction. But the sum of the forward and return distance still has to be c x t. I indicate that with the curved arrow.
In my drawing I show in the longitudinal direction two waves. One wave is for the direction left to right and one is for the direction right to left. The first wave front, your a, has advanced 2 steps when the second wavefront has just finished the mirroring step. So the right to left wave length is nearly 1.6 times the left to right length.
You state that the wavelength is λ+2d. That is wrong. Take your position of the mirror on the top. A does hit the mirror. When B is at the position of A let the mirror have moved 1/4 of the wavelength. Now the wavefront moves with c and the mirror with 1/4th c. So we have for the wavefront y=x and for the mirror position y=4x-1. They are equal when x=1/3. So the wavefront has to move 1/3 in the direction left right, 1/3 back to position A and then again 1/3 to reach to position of the next time step of the picture. At that moment the wavefront A has moved again one unit to the left. Your d is 1/4. My d is 1/3.
I hope you now agree that I incorporated the Doppler shift in my drawing.
Regards,
Paul Gradenwitz
Hi Paul,
PG: Let me start with your drawing. You have in the upper row 3 stages of one wave front.
No, on the top row I have a single snapshot of three wavefronts which are separated by wavelength λ.
On the bottom row I have the same three wavefronts one time step later. The mirror has moved by a distance d and wavefront b has moved by distance λ+d so the time step shown is slightly greater than the period of the wave.
PG: A does hit the mirror. When B is at the position of A let the mirror have moved 1/4 of the wavelength.
In that case, wavefront 'b' hasn't reached the mirror yet. Try instead saying that when 'b' reaches the mirror (that's what I've shown), the mirror has moved λ/4 so the wavefront has moved 5λ/4. The reflected wave 'a' has also moved at the same speed c but in the opposite direction so it too has moved 5λ/4 from where it reflected so it is now 6λ/4 from the mirror. Wavefront 'b' is at that moment also at the mirror being reflected. Thereafter wavefronts 'a' and 'b' both move right to left at speed c so they maintain a constant separation of 6λ/4.
PG: My d is 1/3. I hope you now agree that I incorporated the Doppler shift in my drawing.
No, it looks to me as though you have the separations equal at λ in both directions. The black wavefronts going left to right and the red ones going right to left in the attached detail edited from your drawing have identical separation unless I'm misunderstanding what is shown.
George
No, they do not have to reach the same point. The deviation is about a few micrometers and the mirrors are a few centimeters and the fringes are about a few millimeters.
However, the original MMX has relevance only in relation to ether wind from planetary translation, so forget that.
2 orthogonal cavities are more interesting since they can detect planetary rotation. No beam splitter. If cavity mirrors move inside their own planes boundary conditions are unchanged and light moves perpendicular in the frame of the ether. In relation to ether in both arms.
John-Erik
JP: No, they do not have to reach the same point.
They must if you want to get fringes. The intensity at any point on the screen depends on the difference in paths lengths from the source to that point.
JP: However, the original MMX has relevance only in relation to ether wind from planetary translation, so forget that.
The MMX is what we are discussing at the moment. There's no point moving on to other subjects as long as you don't undertsand the MMX so please don't change the subject.
Dear George Dishman ,
You should careful read my initial description. I only trace 2 wave fronts over time. With the steps before the t=0 location I have 26 times the red/orange position of the same wavefront and more than 26 times the black wave front.
Paul>: "I assume two wave fronts that move from left to right from the source. I let the wave fronts move with c through the aether. The fist wave front is in black and the second is in red. Each next line then is one time unit later for the same wave front."
You draw two time steps of 3 wavefronts. I draw many time steps of two wavefronts. So each black line is the same wave front one time unit later. Each red/orange line is the same wavefront in many time steps. So the distance between each red/orange line is (c x ∆t) the distance between black and red/orange is one wavelength. You got of track because initial the wavelength also is (c x ∆t).
So, you interpret one colour for one direction. I intended one colour for one wavefront. Then I notice that the forward red and backward red would be not so clear so then I changed the backward red to orange. I talk about the 15th version of the wavefront. That means counting from the t=0 place you have 15 time steps stacked on top of each other.
Your reasoning with the increased wavelength is correct. You do it again from a different viewpoint. I start with a velocity of the mirror and calculate the point where the second wave will meet the mirror. You just assume a point where the second wavefront meets the mirror and leave open the speed of the mirror. I could not do that because I also need that same speed to show the 90° deflection problem. In the end we get both λ+2d. However d ≠ v x ∆t
Can you now agree with my drawing?
Regards,
Paul Gradenwitz
Hi Paul,
PG: I only trace 2 wave fronts over time. .. You draw two time steps of 3 wavefronts.
I showed three but I didn't use 'c' so you can think about just 'a' and 'b'.
PG: So each black line is the same wave front one time unit later.
That's what I missed, I thought the black were going one way and the red returning.
PG: So, you interpret one colour for one direction. I intended one colour for one wavefront.
Right, that was the confusion.
PG: Your reasoning with the increased wavelength is correct.
Thank you. Previously you had said " You state that the wavelength is λ+2d. That is wrong." so I'm glad we can now agree there is a change.
PG: You just assume a point where the second wavefront meets the mirror and leave open the speed of the mirror.
Yes, it is easy enough to find the speed v/c from λ and d.
PG: Can you now agree with my drawing?
Not without going into great detail counting lines, I can't look at an orange line and see that it is farther from it's companion black line than the red was from its matching black and there are no red/black lines going down the page after reflecting from the beam splitter which is where the comparison must be made.
I really think you need to draw just a single snapshot showing many wavefronts with the Doppler-shifted spacing to get an illustration that tells you anything.
Dear George Dishman ,
My drawing needs the counting of lines. I agree that is cumbersome. I draw no downward beam because I show that the first and the second front meet at the 15th position of the beam splitter. That means that from there on they will go "on top of each other" downward. To see if this will give a signal I would need to draw the same picture with different mirror speed.
The drawing has the mirrors placed such that a certain phase relation results at the downward final path. Without changing to a different orientation of the movement through the aether or changing the velocity there would be no fringe shift.
You can see the difference of the red and orange lines. The red vertical lines are nearly on top of the black lines. The orange lines are more than one unit trailing the black lines. At the right hand side near the end mirror I have drawn two sine waves. One has its peaks at black to red and one has its peaks from black to orange.
What interests me is how this picture would be Lorentz transformed to the view in the restframe of the mirrors with the aether as a wind. Now my drawing is pure handwork in MS-Word. I have no software where I could make such pictures with dynamic changing of parameters.
Once we can agree on the picture, even if you think another would give better information, then we might discuss what happens if some parameter changes.
Regards,
Paul Gradenwitz
Hi Paul,
PG: My drawing needs the counting of lines. I agree that is cumbersome.
It is but it's not the fundamental problem.
PG: I draw no downward beam because I show that the first and the second front meet at the 15th position of the beam splitter.
That would be a phase shift of 360 degrees. For a lesser mirror movement hence a smaller value of v, you would get a smaller phase shift, so it is your choices in the way you've illustrated the scenario that are hiding that shift by making it equal to a complete cycle. The shift Michelson expected was never as much as a whole cycle.
Paul and George
If the effect of the beam splitter deviates from 90 by a small amount is irrelevant, since the distant mirror defines which wave front is relevant. The effect of beam splitter in the other arm is also irrelevant since it is very near the detector.
The old MMX is relevant only in relation to planetary motion. So, forget about that, since the later test with cooled resonators is relevant also in relation to planetary rotation. In these tests there are cavities, but not any beam splitter.
In the longitudinal arm light moves with speed c transverse to mirrors with the ether as reference. Therefore, in the transverse arm light also moves with the speed c transverse to mirrors with the ether as reference. The direction of v is the only thing that changes. Therefore, since we get c+-v in one arm we must have sqrt(c2+v2) in the other arm. Michelson was right, not Potier. Wave fronts are always parallel to mirrors. There is no effect of ether wind in the transverse arm.
Regards from John-Erik
Paul and George
No, the 2 returned signals must not hit the same point. The difference is only some micrometers and the size of the mirrors is some centimeters and the size of a fringe is some millimeters.
Regards fr om _____________ John-Erik
JP: If the effect of the beam splitter deviates from 90 by a small amount is irrelevant,
That is correct, the source Michelson used was a lamp (lasers hadn't been invented) so it was sending out light as a portion of a spherical wavefront. From that, you choose the paths that obey the laws of optics to reach the final point of interest. As you say, the mirrors are wide enough that the light won't fall off the ends.
JP: No, the 2 returned signals must not hit the same point.
If you want to know the intensity at any point, you need to know the phase difference between the two rays that hit that point, one from each path. You cannot add the waves from two different points to calculate the amplitude at a single location. Think about it before replying again, this is obvious but you've missed it several times now.
JP: Wave fronts are always parallel to mirrors
No they aren't, haven't you learned the law of refraction yet? Wavefronts are always perpendicular to the direction of propagation. Look at figure 2 here:
https://www.ck12.org/book/CBSE_Physics_Book_Class_XII/section/9.1/
Dear George Dishman ,
George>: That would be a phase shift of 360 degrees. For a lesser mirror movement hence a smaller value of v, you would get a smaller phase shift, so it is your choices in the way you've illustrated the scenario that are hiding that shift by making it equal to a complete cycle. The shift Michelson expected was never as much as a whole cycle.
The experiment sets a certain phase relation. The question is if that phase relation changes when the orientation is changed. I could have opted for 180° or another phase shift. I need two drawings to prove a change of that phase. I can't hide any thing with just one picture. Michelson was working with decent speeds. I have already a relativistic speed in the drawing. If I have about 360° then it is also about 0°. There is no need for both arms to be equal long down to sub light wavelength dimensions. But both light waves of the same wavefront need to meet again.
Regards,
Paul Gradenwitz
John-Erik Persson ,
You might have studied my picture. I discussed it with George Dishman . In that picture I show a deviation of 90° of the transversal arm of the setup. The upward and downward wavefront are clearly not parallel to the mirror. can you show where in your opinion is the error of that picture?
You try to jump from one subject to another without proving that your basic assumptions are correct. I started with a setup that invoked a solar system scale. You ignored it. Now I set a smaller scale. It shows how light is diverted by the moving beam splitter with respect to the aether at rest.
To convince us you need to join the discussion. Until now you have made declarations and received the answer that you are wrong.
I work in chip design. There the geometries are well below the wavelength of light. To focus there they use mirrors of large size with short focus length. They can get an image in focus over more than an inch in diameter. For that all the light of the mask has to travel over the complete surface of the mirror with an area of tens of inches and focus back in an area less than on wavelength of light. If there would be an aether deviation in the way you state we could not get that focus. But the fact that you communicate here with me on RG means that they can sell chips that work.
I look forward to your response.
Regards,
Paul Gradenwitz
Hi Paul,
PG: The experiment sets a certain phase relation. The question is if that phase relation changes when the orientation is changed. I could have opted for 180° or another phase shift.
No, he was using a non-coherent source so the contrast degraded for whole fringe shifts. The experiment was set up carefully to have zero shift in the first orientation before rotation so the path lengths were the same to much better than one wavelength.
It's easier to consider a single orientation of the apparatus, for example pointing the "longitudinal" arm at a distant star, then comparing the fringes 6 months apart and looking for the change due to the orbital velocity.
For the illustration, all you need is to show how the fringes move as a function of speed through the aether. At zero speed, obviously the wavefronts arrive in sync at the screen, all you need to show is why there is any phase shift at all when the apparatus is moving through the aether. The details of longitudinal versus transverse speed components and so on are then just another layer of detail on top of the basic operation.
What you've done isn't wrong Paul, I'm just making suggestions to make your life easier.
Dear George Dishman ,
George>: No, he was using a non-coherent source so the contrast degraded for whole fringe shifts. The experiment was set up carefully to have zero shift in the first orientation before rotation so the path lengths were the same to much better than one wavelength.
Ok, the setup uses two arms with equal length and the tuning is such that the phase shift is 180°. Then all light is cancelled. He was using one wavelength?
George>: For the illustration, all you need is to show how the fringes move as a function of speed through the aether.
How do I create a fringe?
George>: At zero speed, obviously the wavefronts arrive in sync at the screen, all you need to show is why there is any phase shift at all when the apparatus is moving through the aether.
I made one picture with one speed. To see a change I have to repeat that picture with another speed.
George>: The details of longitudinal versus transverse speed components and so on are then just another layer of detail on top of the basic operation.
If in the longitudinal arm something changes and the same change is in the transversal arm then the phase shift of one path with respect to the other has not yet changed.
George>: What you've done isn't wrong Paul, I'm just making suggestions to make your life easier.
I make my life easier when I can imagine the whole setup. I like to see all parts in one picture. Then I understand it better than when I only take one detail.
Maybe you can explain why in the view of the aether the wavefront is not perpendicular to the transversal arm but in the view of the setup the wavefront is perpendicular. I only can explain that with relativity of simultaneity. But then the speed of the light in the transversal arm is in the setup view not c. This then has to be compensated with the time dilation. Is that correct?
Regards,
Paul Gradenwitz
Hi Paul,
Michelson's 1887 paper is here:
https://history.aip.org/history/exhibits/gap/PDF/michelson.pdf
The discussion of the error in the 1881 paper starts on page 334, note the acknowledgement to Potier at that bottom of that page.
The light source is described on page 339 (part attached). He focussed the optics using a sodium source then did the measurements using white light.
PG: How do I create a fringe?
That would need a model of the effect of the optics over the screen like this:
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/michel.html
You'll find applets online that do that but I think its beside the point, you only need to illustrate the phase shift at the centre of the rings.
PG: I make my life easier when I can imagine the whole setup
See the perspective view on page 337 and the diagram of the light paths in the original paper on page 338.
PG: I made one picture with one speed. To see a change I have to repeat that picture with another speed.
No, you assume the setup is in phase for zero speed then your existing drawing is valid for any non-zero value.
PG: Maybe you can explain why in the view of the aether the wavefront is not perpendicular to the transversal arm but in the view of the setup the wavefront is perpendicular. I only can explain that with relativity of simultaneity.
RoS only happens in SR, the MMX was based on assuming Galilean relativity. The explanation is that the beam splitter is moving (like the mirror) in the aether frame but because it is at 45 degrees in the apparatus frame, the wavefronts hit it at different times over the surface, it hits earlier in the centre than at the top edge, and the motion then creates an apparent rotation of the mirror.
Dear George Dishman ,
Thank you for these references. With some tricks I was able to see the pdf and that other link even here in China.
The fringes are circular and relate to the difference in path length for a light path through the centre and off centre. There is no way that I can make them in the picture and I don't need that. All I need is to see phase shift.
About RoS I maybe have to be more clear. You agree that in my drawing with a velocity trough the aether the wavefront is not parallel to the mirror. But when I would like to draw the same events from the frame of the setup then I would get an even more slanted wavefront. But in the frame of the setup the wave should be parallel to the mirror and deviate 90°. Because I have a stack of pictures with equal time difference I should be able to move each time instance back by the same distance and then get the setup frame. Only this does not take into account the rotation of the wave front.
I did select each line of the same time stamp with the mirror position of the same time stamp and moved al mirrors on top of each other. That gave the new picture. I get again the same 15th wave instance meet on the mirror. However the transversal arm has slanted wavefront. When you look at the level of simultaneity then you see that it would rotate the wave front to parallel. Only I am puzzled where I would set the pivot point. Also that would alter the wave positions of the longitudinal arm. New in this picture the longitudinal first wavefront is in the return path with a dark blue colour.
Regards,
Paul Gradenwitz
OK, this is just a very simple answer Paul, I'm sure you can work out the details. In the attached aether frame diagram, the mirror in blue is at exactly 45 degrees. Again I've shown a single snapshot of three consecutive wavefronts but the angles may be exaggerated, it's just a crude sketch of the principle. Wavefront 'a' is being reflected and 'b' is just reaching the point where the bottom of 'a' was reflected ('b' hasn't reached the mirror yet). The left edge of the reflected 'a' has travelled a distance shown by the orange line, the top edge has travelled the same distance horizontally. That will let you find the angle of rotation of the wavefronts.
George and Paul
No, they must not hit the same point. I said that the fringes was some millimeter over-lapping with about the same phase. Therefore, it does not matter if one is shifted some micrometers in relation to the other. I thought that you understood that.
Michelson's older method can only detect planetary motion and is later substituted by tests with cavities that can detect planetary rotation. They contain no beam splitter. So, the older test is without interest.
Wave fronts in standing waves are always parallel to mirrors in a cavity, since it is the mirrors that define direction of light to be transverse to mirrors in the frame of the ether. Light takes the fastest way, as Michelson said in contrast to Potier, who deviated from the wave model in only one arm.
Now let us see if you have answers to these important questions?
Regards from ___________ John-Erik
Dear George Dishman ,
I am puzzled by your answer. It is not that I can't understand that answer but it is that I used exactly that for the generation of the slanted wavefront.
Have a detailed look in my initial picture. I have one quarter of circular wavefronts. That is used to show that for each point of the mirror the wavefront generates a point source of light. The sum of all these point sources forms the slanted wavefront of the reflected upward beam. I give also an effective mirror angle. So your whole story is already in the picture.
I am not arguing that the wavefront is not perpendicular in the aether frame where the setup is moving. I am arguing that when I make a simple transformation from that frame to the setup frame then in that transformation the wavefront remains slanted. But because that frame moves with respect to the aether frame there should be a different LoS. That will make that in that frame the wavefront rotates to perpendicular.
I think this is a completely different issue than what you talk about in your answer.
Regards,
Paul Gradenwitz
Hi John-Erik,
JP: No, they must not hit the same point. I said that the fringes was some millimeter over-lapping with about the same phase. Therefore, it does not matter if one is shifted some micrometers in relation to the other. I thought that you understood that.
I understand what you are saying but it is wrong. Did you look at the illustration I gave Paul?
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/michel.html
You can see that the fringes form concentric circles on the screen so the intensity varies only with the radius. If you want to find the intensity at some radius r, you have to combine the amplitude and phase of the waves that move down the two paths to that point, at radius r, not one wave to r and the other wave falling at r+dr. If the amplitudes are equal, the intensity of the light at any radius can only depend on the phases arriving at that radius.
JP: Therefore, it does not matter if one is shifted some micrometers in relation to the other.
If you truly think that, you should approach the problem as Michelson did, you'll get the right answer regardless. His figures 1 and 2 on page 335 are both correct, as is the mathematical analysis on page 336.
JP: Wave fronts in standing waves are always parallel to mirrors in a cavity,
In the rest frame of the apparatus, that is true, but as seen from the aether frame, the whole cavity is moving so the beam takes a zig-zag path and the wavefronts are no longer parallel to the mirror in that frame, they must obey the law of reflection.
JP: Light takes the fastest way ..
Yes, and the fastest way between the two locations where the light bounces off the mirror at one end of the moving cavity is if it obeys the law of reflection at the other end.
Hi Paul,
PG: I am puzzled by your answer.
In that case I probably don't understand the question.
PG: So your whole story is already in the picture.
OK.
PG: I am arguing that when I make a simple transformation from that frame to the setup frame then in that transformation the wavefront remains slanted.
It shouldn't. I've shown a mirror at 45 degrees so in the apparatus frame, the light should reflect through 90 degrees and the wavefronts must be parallel to the mirror. If you don't get that result, you must have made a mistake in your transform. Remember in this frame there is no Doppler whereas in the aether frame, the moving mirror will change the wavelength as well as the reflection angle.
PG: But because that frame moves with respect to the aether frame there should be a different LoS.
No, simultaneity is absolute in the Galilean Transforms.
PG: I think this is a completely different issue than what you talk about in your answer.
It's the other side of the same coin I think, if you set the speed of the 45 degree splitter to zero, you should get 90 degree reflection. It's so trivial I hadn't thought you were asking about that. What am I missing?
George
As long as the difference between the 2 images is much smaller than the size of the fringes that difference is irrelevant. This is self evident. If you do not see this very simpel fact I cannot help you.
Michelson assumed light to move in relation to the ether in ether in both arms according to the wave model. Potier shifted to particle model in only one arm.
John-Erik
Dear George Dishman ,
George>: It's the other side of the same coin I think, if you set the speed of the 45 degree splitter to zero, you should get 90 degree reflection. It's so trivial I hadn't thought you were asking about that. What am I missing?
You are missing that in the aether frame the relative speed of the light to the mirror is less than c. So with the same absolute LoS also in the frame of the apparatus the speed in longitudinal direction is less than c, (c-v), but in transversal direction it is faster √(c2-v2). So again the wave front will be slanted and not 90°. Only after applying an LT and not a GT you get that the wavefront is rotated by 90°.
That is why I showed you the second picture. There I only translate each time stamp set of wavefronts, splitter and mirror back to the t=0 time stamp location. Then you get the relative position of all of them to the location of the apparatus but in the view of the aether. That is no LT but GT x'=x-vt.
Can you agree?
Paul Gradenwitz
Hello John-Erik Persson ,
John-Erik>: As long as the difference between the 2 images is much smaller than the size of the fringes that difference is irrelevant. This is self evident. If you do not see this very simpel fact I cannot help you.
I think we see that simple fact that with small enough speed and small enough apparatus the difference is less than the width of the fringe, but that does not mean that light will move in the way you assume. The evidence is not that small difference but that you miss that light deviates from 90°.
You can take a derivation using waves, as I did, or you can take a derivation using photons and elastic bouncing. In both cases you get that with the moving mirror the deviation is not 90°.
So, when you look to the last picture that George Dishman gave to me with the tree wavefronts and the two mirror positions, then you first have to show why that picture is wrong before you can state that it is self evident that light moves like you suppose.
Regards,
Paul Gradenwitz
Dear John-Erik,
JP: As long as the difference between the 2 images is much smaller than the size of the fringes that difference is irrelevant.
No, the shift alters the angle of the path and Pythagoras Theorem means that alters the effective length to the root of the sum of the squares. As Michelson said, it halved the amount of fringe shift they expected, the majority of the signal would come from the longitudinal arm. It is a minor point in that no shift was observed so Galilean aether theory failed either way, but it meant is 1881 test which was already marginal could not be taken as convincing. That is why he repeated the experiment with Morley and achieved much higher resolution by improving their technique.
JP: Michelson assumed light to move in relation to the ether in ether in both arms according to the wave model. Potier shifted to particle model in only one arm.
No, Michelson thought the speed in the transverse arm would be c while that in the longitudinal arm would be c-v from the splitter to the mirror then c+v on the return. In the 1881 experiment he anticipated a shift of 0.08 fringes, Potier's correction reduced that to 0.04 which was comparable to the noise.
In the 1887 experiment, the multiple reflections increased that by a factor of 10 to an expected value of 0.40, with your incorrect approach (as Michelson used in 1881), that would have been doubled to 0.80 of a fringe. The test results given in the table on page 340 and shown in the graphs beneath are around 0.02 of a fringe. Your argument, if it were correct, would rule out the aether even more convincingly.
As you mention, modern cavity tests now give zero effect to one part in 1018 compared with Michelson's achievement of 1 in 20, a remarkable illustration of the power of modern technology.
Hi Paul,
PG: You are missing that in the aether frame the relative speed of the light to the mirror is less than c.
In the aether frame, the light always moves at c. The mirror and splitter move at v and you have to account for both motions.
When you say "the relative speed of the light to the mirror", you are thinking in the apparatus frame.
PG: the speed in longitudinal direction is less than c, (c-v), but in transversal direction it is faster √(c2-v2).
In the apparatus frame, the longitudinal speed of the light is c-v one way and c+v returning so higher or lower than c depending on the direction. The transverse speed is √(c2-v2) as you say but that is slower than c, not faster. That slight difference does mean that the mirror would need a small alteration from the optimum position but it is not significant in the experiment.
Paul
I repeat again: If the beam splitter does not produce 90 than the distant mirror just finds another wave front. So that is irrelevant. If there is a deviation in the other arm it is also irrelevant, since it is after the long path and very near the detector.
Another reason is that there is no beam splitter in the tests that have relevance for planetary rotation. Namely tests with resonators.
Regards from ___________ John-Erik
George
No light must not hit the same point. Fringes of some millimeters means that phase difference is constant over an area much larger than the small difference of a few micrometers between the 2 images.
Light behavior is defined by ether wind and the orientation of the mirrors that put boundary conditions on light waves. When light is moving over 10 meters light is completely ignorant whether the mirrors have moved a few micrometers or not. Light is defined by mirror orientation - not by mirror motion.
You are siting me wrong. I also said that Michelson's first (and correct) interpretation mean speed c in relation to the equipment in the transverse arm. When light is moving inside the cavity it can feel the ether, but not what the mirrors are doing. Since light takes the fastest way light moves transverse to mirrors in the ether frame with speed c. Wave fronts become parallel to mirrors and speed is sqrt(c2+v2) in equipment frame.
See my article Einstein was wrong - who was right. You will see that we need 2 models for light.
Regards from _____________ John-Erik
Hi John-Erik,
JP: No light must not hit the same point.
I repeat again, if you want to calculate the intensity at any point, then you can only sum the two waves that arrive at that point. I don't care how often you repeat your stupidity, it is obviously incorrect.
JP: You are siting me wrong. I also said that Michelson's first (and correct) interpretation ...
I was not citing you, I was pointing out that not only was the 1881 paper wrong but your understanding of the nature of the error was also incorrect. The 1887 paper is correct, the error in the 1881 paper had the effect of doubling the predicted fringe shift, not cancelling it as you seem to imagine.
Hi John-Erik Persson ,
John-Eric>: I repeat again: If the beam splitter does not produce 90 than the distant mirror just finds another wave front. So that is irrelevant. If there is a deviation in the other arm it is also irrelevant, since it is after the long path and very near the detector.
You repeat again nonsense. If you are convinced that light behaves as you state then you should come up with any documentation not written by you that proves that light behaves as you say. There is only one wavefront and the mirror can't chose. So the product of beam splitter is the only relevant wavefront.
John-Eric>: Another reason is that there is no beam splitter in the tests that have relevance for planetary rotation. Namely tests with resonators.
Again you try to deviate to another experiment. I invite you to show where the statements of George and me are wrong. There are detailed drawings posted here. Show that you have a basic understanding of light. Don't evade discussions.
Regards,
Paul Gradenwitz
Dear George Dishman,
George>: In the aether frame, the light always moves at c. The mirror and splitter move at v and you have to account for both motions. Yes. I take that into account George>: When you say "the relative speed of the light to the mirror", you are thinking in the apparatus frame.
Yes, but not in the sense of SRT with LT but in the sense of GT. PG: the speed in longitudinal direction is less than c, (c-v), but in transversal direction it is faster √(c2-v2). George>: In the apparatus frame, the longitudinal speed of the light is c-v one way and c+v returning so higher or lower than c depending on the direction. The transverse speed is √(c2-v2) as you say but that is slower than c, not faster. That slight difference does mean that the mirror would need a small alteration from the optimum position but it is not significant in the experiment.
I was talking about the slanted wavefront in the forward path. Of cause in the return path the speed of light is c+v. The two speeds: c-v < √(c2-v2). That is why I said what you quoted of me above. This speed difference makes that the wave is not 90° rotated. The GT does not rotate the wave front. The LT does rotate the wave front. Because c+v > √(c2-v2) in the return path the wave front is rotated more than 90°.
Regards,
Paul Gradenwitz
Hi Paul,
I understand what you are saying and I don't disagree with any of it, but I'm not sure I see the point. Michelson constructed the equipment with adjusting screws on the mirrors so that the beams could be carefully aligned onto the telescope so whether the angle was 45 degrees or 44.9999999 to get the light to where it had to go doesn't seem relevant, the position of the various parts could not have been constructed to that accuracy. That's why adjustments were needed in the first place.
George
You call it stupidity. I say that the fact that the sizes of the fringes are some millimeters wide indicates the phases are varying very slow and are constant over regions much, much larger than the few micrometers that differ between the 2 images means that this difference is irrelevant. I call it stupidity when you earlier said that few micrometers would cause the return signal to miss the beam splitter.
I have explained to you why 1881 paper is correct by assuming wave behavior in both arms and motion in relation to the ether. Potier introduced a particle behavior in one arm only. Remember that Michelson resisted majority for 5 years before he gave up.
Regards from ___________________ John-Erik
Paul
Your statement that there are only one wave front is absurd. The light source emits light over an angle that is many, many times the few microradians that the beam splitter can change.
The older MMX can only detect effects around 10-8 and we need at least 10-12. So we have to go to the modern type with cooled resonators. There is no beam splitter and I have also explained how it is irrelevant also in the older MMX.
Regards from _____________ John-Erik
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