The STOE suggests light is photons (matter particles) whose speed depends on the medium through which they travel. There are several experiments which reject the wave models of light. But I am interested in this question.

It is sad that there is no one experiment to clarify the actual drift of lightbeams.

I may be missing the point of your question, but all polarization experiments depend on the wave nature of light and its transverse displacement as it passes through some medium, as indicated by the fact that you can selectively dampen the waves with any specific orientation by an appropriate polarization filter, most generally one that easily conducts electrons in only direction. The fact that you can generate and manipulate both rectilinear polarization and circular polarization adds to the credibility of the conceptual model. It is hard to imagine another form of light that would respond the same way.

Larry L. Baxter ,

please can you clarify better?

NONE. That is if light can be characterized as a wave, it is probably a compression (longitudinal) wave like sound waves.

Dear all,

in the attached file you can find the proposed experiment in a synchrotron and what can be exptected from lorentz invariance: light strikes the point B, and violation of lorentz invariance, light strikes the point A.

SQ: ... the normal component of the source to the light beam direction.

I don't know how you would even define that. The light is moving in some direction so if you then take the normal to that direction, the speed is zero by definition.

Your diagram is also wrong, the radiation is emitted from every point on the circle but beamed along the tangent to the path at the point of emission, see figure 1 here:

https://www.astro.utu.fi/~cflynn/astroII/l4.html

John Hodge, the fact that light can be polarised proves it is a transverse wave in the classical view. That translates to particle spin in the photon approach.

George Dishman ,

The diagram from the source I cited suggests there are two lobes, the stronger being in the forward direction, but no perpendicular lobe.

However, if you avoid that null, the light would travel in a straight line from the point where the source was at the time of emission through the point where the orifice is at the time it passes through, and continue onwards on that path. Of course if the orifice is small then the light will be diffracted by the edges so you'll get a large cone of light coming through or a broad beam if it is large.

The path is attached.

George Dishman ,

no need to stress the fact that at ultrarelativistic speeds the beam is along the instantaneous speed. The structure of the synchrotron itself tells you that.

On the other hand it is clear also that there is always emission in any direction, any speed. At .5c the radial component is still large and there is no need to arrive to .99c to show what happens to the waves when the electron is placed in front of the hole (with no reflections).

Are you aware or not that, in the setup proposed by Einstein for his light clock, the light beams emitted in the moving frame are compelled to go sideways in the stationary frame.

Amrit Šorli

I basically agree on a theory which by changing the energy of the vacuum, it can give account to the known phenomena both in gravitation and electrodynamics.

Although a discriminant experiment has to be performed in order to state if the Lorentz Invariance holds or not.

It is quite strange that an experiment with light rays emitted at high speed and right angles has never been performed in order to determine their actual direction in the reference frame of the lab.

Such experiment were performed to determine the change in frequency of the radiation not their direction.

Stefano Quattrini, there is no emission at low speeds and no emission in the direction of the acceleration, you can see there is a null in picture (A) of your image. The is emission sideways though so you don't have a problem with that but the light will always travel in a straight line. The slower the beam, the shorter the distance the electron moves in the time it takes the light to reach the orifice. That will change the angle a bit. I showed the path for a fairly slow speed, about 30 degrees to the tangent but it is the basic principle that I was illustrating. I've added a couple of green arrows, the upper shows where the electron was when the light was emitted, the second shows where it is when the light exits the orifice. The electron has travelled a slightly smaller distance than the light. The brightest part of the beam is tangential, that to the side is much dimmer but you would get some. For light going to your point B, it would just be emitted a little later than I've shown but for point A it is in the null.

The simple rule is that light travels in straight lines unless you use lenses or consider gravitational bending.

George Dishman ,

let's make a step back then...

From the picture you can see the light clock which can be taken as a reference in order to derive the time dilation effect in the Lorentz Transformations (see also Feynman).

The moving clock configuration in the lab shows a beam departing at (x1,y1),

which would bounce at position (x2,y2). If the beam is not stopped, that would arrive at position (x3,y3) in the lab and absorbed by an absorbe statioanry in the lab.

So the light from the moving emitter to the stationary absorber has gone sideways, although inside the moving frame it has been emitted always in the same way, normal to X.

That means that the light beam leaving the moving emitter took the component of the velocity of the emitter.

Your picture is correct. The line from x1,y1 through x2,y2 to x3,y3 is straight, apply that to the synchrotron.

SQ: That means that the light beam leaving the moving emitter took the component of the velocity of the emitter.

Yes, that is called aberration. In your previous image, the polar diagram (A) shows light being emitted equal in both directions round the loop in the electron frame but in (B) it is concentrated into a narrow beam in the direction of motion in the lab frame. That is just a change of frame.

https://en.wikipedia.org/wiki/Relativistic_beaming

https://en.wikipedia.org/wiki/Relativistic_aberration

George Dishman ,

this is the version of the moving source which, in the synchrotron lab, should generate the radiation and the "aberration" effect... it would be enough to generate visible light at beta=.7 with a short curvature radius of the accelerator, the radiation diagram tells that there is a normal component to the trajectory exiting the plane of rotation.

There it is possible to discriminate if aberration depends on relative speed or not.

That light clock might work for aberration is just an aberration of Physics....

Synchrotron_radiation_energ y_flux.png

This may be interesting to consider:

https://www.emc2-explained.info/The-Light-Clock/

Yes that is the explanation Einstein gave about the light clock and Feynman in his lectures. https://www.feynmanlectures.caltech.edu/I_15.html

look at 15-4 transformations of time...

They bound the aberration of light with the time dilation.... a pure kinematical effect (whose evidence is clear for a moving observer but was never given for a moving source), with a dynamical effect whose evidence has been provided several times in different conditions.

Is the effect at the base of light clocks an actual aberration of light or such aberration does not exist for moving sources and it is just yet another aberration of Physics/nature done by human thoughts...

Stefano Quattrini , «From the picture you can see the light clock which can be taken as a reference in order to derive the time dilation effect in the Lorentz Transformations (see also Feynman).»

In order to get a drawing like you, there is NO need to move anywhere. On the left is shown the clock going «synchronously» in each of the Inertial Systems (IS) (?) And what will happen to such a clock moving in IS(0) as a physical object of IRF?

Regards

SQ: the radiation diagram tells that there is a normal component to the trajectory exiting the plane of rotation.

No, the polar diagram is a torus with a null along the axis but it has emission either side of that.

Stefano, I don't know what you are trying to say here. Synchrotron radiation is very well known, it accounts for the majority of the energy put into accelerators. The beaming effect is also well known as you can see from the diagrams and in nature it is seen in quasar jets so what are you contesting? So far everything you have said is correct but just standard physics.

George Dishman ,

which axis???

The radiation diagram at low speeds is uniform in top and bottom directions, provides zero only in radial direction.

If you noticed from the previous picture, the direction where I want to get the waves from the synchrotron is not anymore the radial direction but the direction normal to the radius.

At .7c the radiation diagram is already deformed along the direction of motion of the electron, but still with a good level of radiation emission in the direction normal to the plane of the cyclotron.

Extend the line marked "Acc", the torus goes to zero along that line, where the shape folds in at the white ellipse.

Here's another view:

https://www.horntorus.com/

George Dishman ,

if you have the patience to read what people write, then we would lose less time maybe.

I do not consider the radial direction anymore!!!!

the rectangle represents the ROOF of the laboratory over the synchrotron.

See also the diagram on page 16 of this article:

https://asd.gsfc.nasa.gov/Volker.Beckmann/school/download/Longair_Radiation2.pdf

There may be a little confusion, I've been saying there is no emission perpendicular to the direction of motion of the particles in the particle frame but as your image shows, that is angled forwards in the lab frame by aberration.

George Dishman ,

in a test on the surface of earth, the direction of the beam I consider is towards the roof of the lab (where the synchrotron stays)...at variance with what I initially proposed (radially emitted).

The smaller lobe is the one considered (at .7c) not the large and long one ( the one peculiar of the synchrotron )..

George Dishman,

so according to your opinion what will be the result of EM waves in that situation,

will they hit the lab in A or B, emitted from the moving electron, (according to the radiation diagram), perpendicularly to the plane of the synchrotron.

Stefano Quattrini: ... perpendicularly to the plane of the synchrotron

That means up the screen in the direction of the axis. I think you meant along the radial line.

There will be light hitting A and there will be a null somewhere around or below B depending on the particle speed, in your final image it is at 46 degrees for beta=0.7 and the null would be where the light path makes that angle to the tangent to the particle path. The light that hits A will be what was launched backwards at about 46 degrees in the rear lobe when plotted in the frame of the particles (where the polar diagram is the torus). If you extend your dotted lines without bending them, you will see where the light was emitted. The brightest light will be along a path from your exit orifice that makes a tangent to the particle path, the rest will be much duller.

You have all the pieces of the puzzle but your sketch looks as though the light originates at the exit instead of from the particle ring, extend the paths and it should fall into place. The reason why the null is moved forward to 46 degrees instead of sideways at 90 is aberration which affects all the paths so I think your "option 1" and "option 2" labels are the wrong way round, but they are referring to the location of the null, not the light.

P.S. I just found a graphic that may help For the synchrotron radiation, you can think of the red arrows showing the location of the null:

https://www.researchgate.net/profile/Jonathan-Granot/publication/48173530/figure/fig1/AS:669530829369358@1536639997546/Aberration-of-light-the-arrows-show-the-directions-of-photons-in-the-lab-frame-for-a_W640.jpg

George Dishman ,

I think now you can understand better.

Some EMwaves emission come out vertically to the roof of the LAB which is above the synchrotron.

There is for sure a normal component pointing up.

Take the gamma*beta =1 of your radiation diagrams where there is a vertical component to the ground.

Oh, OK. From your early diagrams I took you A B plane to be at ground level, sorry.

The emission polar diagram is symmetrical around the direction of the beam so you get the same beaming effect in the direction of motion but the most intense emission will be in the plane so never hit the roof, the orifice will be the limit as the brightest part of the beam will hit at an oblique angle.

Again, all of these paths are shown including aberration. The top panel is relative to the point of emission and shows where the orifice is at the time the light passes through, the bottom panel is drawn relative to the orifice.

George Dishman ,

ok now there is more understanding about the configuration I proposed.

it is clear that at right angles in that case there is emitted radiation.

let's make a step back.. if you state from the beginning that that such behavior is due to aberration it is the dog which bites the tail.

What is interesting is only what happens to the beam exiting normal to the direction of motion which from the radiation diagram is present, since that diagram was experimentally verified I suppose.

There is only one orifice through which the radiation passes and gets that only at right angles. The orifice sufficiently long and with a suitable filter will stop radiation at different angles than 90 deg.

Yes, sorry I missed you configuration earlier but it doesn't make much difference because the polar diagram is symmetrical around the direction of the particle beam.

SQ: let's make a step back.. if you state from the beginning that that such behavior is due to aberration it is the dog which bites the tail.

The attached shows the version without aberration, there is no perpendicular emission so the null would fall at A and the pattern would be symmetrical. Again, you already have the polar pattern in the rest frame of the particles, the torus shown in the papers I referenced 2 days ago.

George Dishman ,

the torus in violet at low speeds is homogeneous and shows that the null is normal to the direction of motion pointing inside and outside the synchrotron, it is not null in the directions up and down.

You're right Stefano, I was still picturing it in the synchrotron frame. What you are asking is even simpler than I thought, if the beam is moving in the x direction and accelerating in y then you are asking about the emission in the x-z plane. In the frame of the beam that is isotropic so this diagram applies. There is no null in this plane so light would exit the orifice in all directions, brightest in the direction of the beam motion and dimmest in the opposite direction.

I also just happened to read your original question again and I found why I've been talking about the demission along the radius, you said:

SQ: The synchrotron light passing through an orifice, radial to the structure of the circle of the synchrotron would show the effect.

You are actually asking about the direction perpendicular to the circle, not radial.

George Dishman ,

as soon as it was clear from your explanations that radially there was no room for any emission, I though that light could instead come out parallel to the axis of rotation of the circle, or parallel to the angular momentum of the circulating electron.

The purple image, I reported two posts ago, represents what is irradiated out at low speeds. Increasing the speed the beam gets concentrated towards the front.

I do not even consider what a co-moving observer would experience if moving with the electron since it is impossible to measure it so it is not falsifiable.

The radiation in that frame is the same as a dipole antenna made from a couple of wires string across your back garden so easily measurable. You then apply aberration and get the lab frame radiation pattern which is again easily measurable.

You can also see the effect in the jets from M87, the one moving towards us is far brighter. I think there's that image on the Wikipedia page about it but I'm on a phone at the moment so can't check.

George Dishman ,

the beam is what is emitted in front of the electron at high speed.

Any observer in front of the electron would see a radiation, it will be more or less doppler shifted according to its relative speed with the electron.

The radiation will be more intense forward irrespective of the observer which will be in front of the electron along its trajectory rather one which stays beside.

That's all correct Stefano though you weren't previously asking about an observer in front of the beam (looking tangentially at the ring), you started by asking about the emission in the radial direction and then switched to asking about the emission normal to the ring plane.

Light will exit the orifice in almost all directions but with a null in the ring plane, aberration causes it to be brighter in the direction of the beam as you say, so what exactly are you asking, or do you now have your answer?

George Dishman ,

you are right!!!

at first, I thought that radiation could come out also radially, but that is not the case, so it was useless to place an exit point radially to the synchrotron to measure the behavior of light from a moving emitter.

It was clear instead that some radiation, beside the beaming, would exit at right angles through z direction (parallel to the axis of rotation of the trajectory of the electron) so there it was the case to let the radiation exit from an orifice, right on the ROOF of the synchrotron.

The point is that I don't find correct to attribute the phenomenon of beaming to aberration. The beaming as "narrowing emission pattern" is always there for any inertial observer which finds itself momentarily in front of the electron in its same instantaneous direction. What changes is just the frequency detected which will be doppler shifted due to the different speeds of inertial observers.

So the beam-forming is a phenomenon observer independent, the beam power depends instead on the relative speed.

Aberration makes sense to be measured as emitted from the moving electron for light which departs as I said towards the ROOF of the synchrotron at right angles.

That would provide a similar configuration to the light clock and show if the phenomenon of aberration occurs also for a moving emitter.

SQ: The point is that I don't find correct to attribute the phenomenon of beaming to aberration.

Aberration means that the angle of the light propagation to the direction of motion of the emitting particles is different in the lab and particle frames Beaming refers to the fact that that change of angle concentrates the emission in the direction of the beam.

SQ: So the beam-forming is a phenomenon observer independent, the beam power depends instead on the relative speed.

That is correct, the observer in this case is the orifice. The speed of the beam relative to the orifice determines the aberration and hence the beam pattern.

SQ: That would provide a similar configuration to the light clock and show if the phenomenon of aberration occurs also for a moving emitter.

Correct. It is very evident in M87 where the second jet moving away from us isn't even visible:

https://en.wikipedia.org/wiki/Relativistic_beaming#/media/File:M87_jet.jpg

I'm still unclear about your question but your comments are again mostly correct. How aberration causes beaming is shown in the attached diagram, as the speed increases, the arrows rotate round to the right causing them to cluster in front and producing the beaming effect.

George Dishman ,

in the case of the orifice in z direction (ROOF of the synchrotron), right angle with the trajectory of the electron, which is the case I'm interesed in, the main "beaming" will never meet the orifice. Although some radiation, according to the radiation diagram, will certainly pass through the orifice at right angles with the trajectory.

In such case one would have emitted radiation from a fast moving charge (non ultrarelativistic) which exists normally from an orifice in the synchrotron towards the roof of the lab.

In the top diagram, you are still using the diagram with a null shown moved forward due to aberration from 90° to 46°. That only applies in the plane of the ring. There is no null in the plane perpendicular to the plane.

In the bottom diagram, the light in the light clock is following a single path, light bouncing between two mirrors is how a laser works, so it exits in a single direction towards B. In the case of the synchrotron, light is emitted in every direction in the x-z plane so exits the orifice in every direction.

In this diagram, I have labelled two light rays a and b and the two panels show the same rays in both the source and lab frames. Ray a will pass through the orifice and hit point A on the roof, similarly ray b will pass through the orifice and hit point B on the roof provided your orifice is essentially just a hole in a thin plate rather than the cylinder you describe. The latter matches the light clock path.

If you put a household bulb at point A in the light clock, the apparatus will behave the same as the synchrotron.

George Dishman ,

those radiation diagrams reveal how many photons per unit angle would be detected for an observer momentarily at rest with the electron or for an observer at rest in the lab.

That tells you also that the two observers with the same detectors (angle of detection along the direction of motion), in the same instantaneous position, *would not* detect the same amount of photons per unit angle, the one at rest in the lab would detect one order of magnitude more or even more... Consider also that length contraction acts parallel to the direction of motion, while the width of the beam is normal, so there is no effect of that on the moving detector in comparison to the LAB one.

The power per unit angle received by the stationary detector is increased by the Doppler effect (frequency) of the single-photon, and the number of photons (intensity) as well that is why the synchrotron light gets so powerful (so the power of 4).

Although the Doppler can be considered observer-dependent (absent at 0 speed with the electron), the number of photons per unit angle is there independently of the instantaneous speed of the observer.

The same instantaneous position and areas of detectors mean the same impinging photons...otherwise, photons would "appear" or "disappear" completely on the sole account of the velocity of bodies, which is a total absurdity... Hence, considering the synchrotron light, a consequence of aberration of light is very wrong!!!!

Hi Stefano Quattrini

SQ: those radiation diagrams reveal how many photons per unit angle would be detected for an observer momentarily at rest with the electron or for an observer at rest in the lab.

Exactly, you get the diagram on the right from the one on the left by adding the source velocity vector to the light vector following the rules for relativistic composition of velocities. That is what is called aberration.

SQ: ... two observers with the same detectors (angle of detection along the direction of motion), in the same instantaneous position, *would not* detect the same amount of photons per unit angle, the one at rest in the lab would detect one order of magnitude more or even more.

Yes.

SQ: Although the Doppler can be considered observer-dependent (absent at 0 speed with the electron), the number of photons per unit angle is there independently of the instantaneous speed of the observer.

No, both depend on the speed of the source relative to the observer.

SQ: Hence, considering the synchrotron light, a consequence of aberration of light is very wrong!

The concentration of photons in the beam direction is specifically aberration. The classical geometric component of Doppler also contributes to the increase in beam power in that direction while it decreases the power in the opposite direction. Time dilation decreases the frequency of the source regardless of direction and that plus the geometric part combine to give the relativistic Doppler equation.

Hi George, George Dishman

What you just affirm means that in the same instantaneous position and orientation an observer at rest with the lab detects Q photons, while another observer comoving with the electron detects a number Q/k, where k>>1 is a reduction factor due to aberration.

So photons in such configuration disappear or appear only on accounting of relative speed....

To make the story very short let's make the moving source emit just 360 photons at once and then nothing.

Since the source emits in all directions for a comoving observer, two comoving detectors would detect in a solid angle of 10 degrees 10+10= 20 photons.

We could in principle have a speed in the lab such that 350 photons (strong beaming) are all detected by the detector at rest at a solid angle of 10 degrees.

So basically out of 360 photons produced in one shot, 370 have been detected....

Conclusion 10 photons were created from nowhere....

Hi Stefano Quattrini ,

I have a better suggestion, let's make the moving source emit just 16 photons at once and then nothing.

The path of the 16 photons is then shown by the 16 arrows in the two diagrams, none are gained or lost.

Hi George Dishman ,

very good... 16 photons are generated at once and nothing else.

Since the source emits in all directions for a comoving observer,

two comoving detector with the electron, place in the back of it, would detect, in a solid angle of 22.5 degrees, at least 2 photons, from back the moving emitter, according to the radiation diagram.

The electron was accelerated at the event of emission at a speed in the lab such that 15 photons (strong beaming) are all detected by the detector at rest at a solid angle of 10 degrees.

Out of 16 photons produced in one shot, 15 have been detected by the detector at rest, while 2 photons detected from the moving detectors....

1 photon not emitted has been absorbed...aberration created a photon.

There are only 16 photons so that is the most that can be detected, but if you are mixing up moving and stationary detectors then some photons might escape between the detectors, not hitting either, and some photons may hit one detector when it is in front of the other. If you place a detector circling the source moving with the source, the locations where the photons will be detected will be uniformly distributed. If you fix it in the lab, they will be concentrated in the direction of the beam but in both cases 16 photons should be detected (ignoring detector efficiency).

It's really very simple Stefano, there is no problem here, you're just trying to invent non-existent problems.

George Dishman ,

"the test of the transverse Fresnel–Fizeau light drag effect"

a light beam in z direction

http://muj.optol.cz/richterek/data/media/ref_str/gjurchinovski2004a.pdf

the beams will result deflected in the direction of motion. In the case described with the rotating glass (or flowing water), there is actually a material medium which drags light and that effect is experimentally verified.

For inertial frames in relative motion in empty space, there is no medium in between, but according to the prediction of LT, light would behave as if something drags it. The two relative moving observers will see incoming light in the same way as if something dragged the emitted light, but as a matter of fact there is nothing which drags light with it....

Yes, where you have a refractive index, the light is interacting with the charges in the material so gets dragged along a bit. In a vacuum that doesn't happen and you just get aberration. Fizeau showed that the amount of drag was as predicted by SR and different from what simple aether theories would suggest.

Hi George, George Dishman

ok..I agree it is aberration, that depends on v/c as well. Then it should not at all different than the aberration of light between inertial frames of references, which allegedly explains the beaming in synchrotron..

In the experiment with rotating glass the aberration has a phyiscal reason to occur which is the dragging effect of glass...but with two inertial frames in vacuo (as predicted by LT and showed in Einsetin's clcok) there are not such substances in between: only by assuming that each IRF is by itself a sort of a "aether" we can match the experiment...

http://muj.optol.cz/richterek/data/media/ref_str/gjurchinovski2004a.pdf

the author says that "there exists no reference frame in which Snell’s law of refraction takes place, so there is nothing to be Lorentz transformed".

Snell's Law refers to refraction which requires a material. I have no idea why he would imagine any transform was required. If the material is moving then you get the same situation as Fizeau.

The key evidence for light waves is the Young’s double-slit interference experiment. After the discovery of the double-slit interference experiment of particles (electrons, etc.), the conclusion of light waves can no longer be established. "Wave-particle duality" denies light as particles because of the wave phenomenon, but recognizes the wave nature of particles. It is better understood that light particles have wave phenomena. Of course, light particle behavior and wave mechanism need new understanding.

Here is a new understanding of particle diffraction and interference，

https://book4you.org/book/18269199/bf2351

There is not an experiment to-date that observably demonstrates time-dilation. The double slit is right as Yuan states. The photoelectric effect can be easily interpreted as UV waves going in and waves of lower energy (discharge i.e. 'electrons') coming out of the piece of sodium. The one thing the corpuscular theory had going for it can be easily explained if electrons are also dielectric field waves-forms. Relativity exists only psychologically. Photons have never been observed in isolation. A later is an example of a point-source electromagnetic geometry. Double slit and gold foil experiments also prove is that electromagnetic wave-forms can take many geometries. I invite you to read my short discussion on relativity Preprint Aberrations in Spatio-Temporal Physical Theory

Dear Stefano Quattrini

I don't know if it is possible to measure the influence in a lab. There is the "zitterbewegung" of the energy of every moving phenomenon - even light - and there is the non-locality (mutual influence at exactly the same moment) of all the phenomena - even light - in our universe.

Mostly we try to understand "isolated" phenomena. But more and more there is a need for an overall understanding of what is really happening from the smallest changes in the microcosmos to the largest observable structures in the macrocosmos. But that is an enormous "excercise". I am trying but it is really difficult to get a realistic (mathematical) "big-picture".

The problem is that all those equations we are used too don't describe what is really going on in our universe (like Newtonian gravity and mechanics, QM or Einstein's theory of Special and General relativity). Nevertheless these equations predict the "evolution" of distinct properties in space and time in a convincing way. Sometimes so precise that it is hard to believe that it is "physically" possible because of (e.g.) all the quantum fluctuations in vacuum space.

By the way, the proposed motion of our solar system (direction and velocity) in relation to the CMB - actually absolute space - is verified and shows to be correct:

"The Universe is Brighter in the Direction of Our Motion: Galaxy Counts and Fluxes are Consistent with the CMB Dipole", by Jeremy Darling (The Astrophysical Journal Letters, 931:L14 (8pp), 2022 June 1) [Open Access]

So we have to rethink the meaning of the concepts that underly the theory of Special relativity, maybe even Einstein's clock.

With kind regards, Sydney

dear Vladimir Onoochin ,

In a synchrotron, electrons are forced to travel in a closed path by strong magnetic fields. The radiation pattern is distorted from the isotropic dipole pattern, expected from non-relativistic theory, into an extremely forward-pointing cone of radiation.

Lorentz Transformations predict that:

a) an observer comoving with a charge at high speed in a circular accelerator sees an equally distributed radiation,

b) an observer stationary with the accelerator/laboratory sees a narrow beam of radiation, a very sound experimental fact that gives the synchrotron light.

At the event of deviation of the charges by one magnet, more than 99% of the radiation/energy is in a quite narrow solid angle, as detected by the laboratory. How could a comoving observer (with the charges), around the same event, see a uniform radiation when almost all the sychrotron light photons are detected in a quite narrow solid angle?

Consider an example where the interaction magnet-travelling charge (Travelling quite close to the speed of light) is a source of 10^5 photons. At the event of emission 10^5 - 10 photons are detected by the stationary observer in a narrow solid angle, only 10 are not absorbed by that detector.

Suppose the comoving observer is located just behind the charged particles (for example), along the direction of their motion. In that case, since he is supposed to see uniform radiation, it should be able to detect a suitable fraction of 10^5 photons, 10% if its solid angle is 10% of 360 deg, meaning 10^4 photons (with lower energy).

That is a source of a non-trivial paradox:

according to what can occur, 10^5-10 photons are detected by the Lab. That represents a real energy detection corresponding quite closely to the energy taken from the electrons which decelerate.

On the other hand, 10^4 photons should be detected by the comoving observer according to LT.

The total emission was 10^5 photons, but the absorbed number is predicted as 10^5*(1+1/10).

The only solution which takes into account the number of photons generated is that a fraction of the 10 photons not belonging to the beam are detected by the comoving observer.

But that is at variance with the predictions of the Lorentz Transformations.

In other words the reason according to LT of the beam-forming in the synchrotron radiation is due to aberration of light...!!!!!!!!!!!!

Dear George Dishman ,

Aberration of starlight:

We observe stars in our Galaxy that have relative speeds from a few km/s up to over 400 km/s depending on their position relative to us.

If aberration of starlight depended on the Relative speed between the earth and stars, then the tilting should be quite larger in order to observe them.

so it does not depend on the relative speed but it seems that it depends on the continous change of frame of reference of the observer. That is not what happens in the light clock where only one reference frame is present. That is why it comes out the speed of earth around the Sun.

At the same time earth rotation should have a bigger effect on a observer on earth since it position varies considerably, which doesn't....

So the light clock has nothing to do with the aberration of light... but simply on the funny effect that light takes the inertia of the source thanks to the term vx/c2

Hi Stefano Quattrini,

Aberration means we see the light coming from where it was at some time in the past based on its distance from us. The offset due to the speed of the star through the galaxy is constant because stars (including binary systems) are far apart so don't interact to any significant degree. When we look through a telescope, we can only see relative changes and that is dominated by the motion of the Earth.

SQ: So the light clock has nothing to do with the aberration of light ...

No, I have no idea why you would think that they were connected.

Dear all,

in the LT the term gamma*vx/c2 , responsible also of the light bending in the light clock configuration, is just a re-synchronization term. That depends on the failure of the Einstein Poincarè synchronization procedure in moving frames. Such feature is not difficult to be shown in Sagnac type experiments which also bring the Sagnac corrections made on earth base stations of GPS.

The term gamma*vx/c2 is derived reliably and found experimentally only from the Sagnac configurations it measures the gap which is found from simultaneity due to the difference between absolute synchronization with central light beams and local synchronization of Einstein Poincare'.

Since it is the absolute sync of the beams (departing same place same time) which determines afterwards their absolute desync, same place different times, and the absolute sync gives the same result as the "central synchrnonization", it is evident that the EP procedure fails in synchronizing clocks at a distance in a circle... hence light measured in a inertial co-moving frame is different from c.

Here it is the demonstration of what I said above

https://arxiv.org/pdf/2106.09537.pdf

Light-clock cannot explain anything, since it is Lorentz Electrodynamics is the theory which can account for the effect at higher speeds/energies...

Preprint Einstein train-embankment thought experiment revisited

Dear Stefano Quattrini

1) "Light-clock cannot explain anything, since it is Lorentz Electrodynamics is the theory which can account for the effect at higher speeds/energies..."

Light-clock is a mathematical-physical model and the light signal in a light-clock propagate according to the laws of physics. It has nothing to do with Lorentz's transformations!

If we are talking about SR, the conditions that exist there apply. We cannot constantly mix different aspects in the debate.

Dear Jan Slowak ,

Dear Stefano Quattrini

2) "The motion is not relative anymore but it is locally absolute in the isotropic frame of light"

This is interesting, it fits with how I think about the motion of light.

Possibly see my book Light - The Absolute Reference in the Universe.

I wonder if one can read more about "the isotropic frame of light"?

Dear Jan Slowak ,

it is basically Lorentz Electrodynamics

Hi Stefano Quattrini,

SQ: "The motion is not relative anymore but it is locally absolute in the isotropic frame of light"

Light does not have a frame, the axes are degenerate.

The speed of light is isotropic in every inertial frame according to Maxwell's Equations.

Hi George Dishman ,

1) "Light does not have a frame"

I agree with you

2) "the axes are degenerate"

What do you mean by this?

3) "The speed of light is isotropic in every inertial frame according to Maxwell's Equations."

What do you mean by "the speed of light is isotropic"?

George Dishman ,

in Special Relativity light does not have a preferred frame, or rather no specific frame in a physical problem..

. On the contrary, In Lorentz Electrodynamics light has a preferred frame (local) like the ECIF if you are on earth.

Dear Jan Slowak ,

an introduction you can find here about the Lorentz-Larmor interpretation vs Einstein-Poincare' interpretation

Preprint LORENTZ'S WORK OF 1904 AND THE LORENTZ TRANSFORMATIONS

from Vladimir Onoochin

while a comparison between the two you can find it here

Preprint Einstein train-embankment thought experiment revisited

page 6 you can find some reasons why the Lorentzian one must be adopted.

Basically the light clock cannot work as Einstein conceived but it must work according to Lorentz as I have already said..

Einstein's description is based on the concept that that light takes the transverse inertia of the source, which composes with its own direction if it is seen by another frame.

This "side effect" is not in agreement with the well known experimental evidence of light speed independent on the emitter. That means that in the longitudinal direction there is no vector composition, while in the transversal direction there is vector composition.

Hi Stefano Quattrini,

GD: Light does not have a frame, the axes are degenerate.

SQ: in Special Relativity light does not have a preferred frame, or rather no specific frame in a physical problem.. . On the contrary, In Lorentz Electrodynamics light has a preferred frame (local) like the ECIF if you are on earth.

No, there is a little confusion here. What I said is correct because in both time dilation and length contraction go to zero so you cannot assign coordinates as a multiple of zero. Light doesn't have a frame in either.

In Lorentz's interpretation, the 'preferred' frame is that in which the aether is at rest and through which the Earth is moving, not the ECIF.

GD: The speed of light is isotropic in every inertial frame according to Maxwell's Equations.

SQ: it was postulated by Einstein, it is not a direct consequence of Maxwell equations.

In Maxwell's Equations, the speed is c=1 / √(ε₀ μ₀) Here is how to derive it:

https://www.wikihow.com/Derive-the-Speed-of-Light-from-Maxwell%27s-Equations

Since ε₀ and μ₀ are both scalar quantities, the value of c must also be scalar and hence isotropic. That was Einstein's starting point for his 1905 paper.

Hi Jan Slowak,

GD: 1) "Light does not have a frame"

JS: I agree with you

GD: 2) "the axes are degenerate"

JS: What do you mean by this?

The easiest way to see it is by drawing a Minkowski Diagram. As you rotate the moving frame axes, as the vertical axis turns clockwise, the horizontal axis rotates anticlockwise. You can assign coordinates to any point in the chart by finding where lines drawn through it and parallel to the two axes intercept the axes as usual. At the speed of light however, both axes have rotated 45 degrees so become a single line. That means you can no longer obtain 2 separate coordinates even though the graph is 2 dimensional.

Second, if you look at the spacing of the tick marks along the axes, they get farther apart and go to infinity at the speed of light, all points have a single coordinate of exactly zero on the single axis.

GD: 3) "The speed of light is isotropic in every inertial frame according to Maxwell's Equations."

JS: What do you mean by "the speed of light is isotropic"?

It has the same value in all directions, see my last reply to Stefano for the details.

George Dishman ,

>

this is well known, it is the part in common with Lorentz Electrodynamics (1904) and is related to the fact that light is a wave with its characteristic speed hence its speed is independent on the speed of the source.

Dear George Dishman

1) "Minkowski Diagram"

Minkowski Diagram is a mathematical construct but it does not verify our reality!

2) "at the speed of light"

What is moving at the speed of light? It is absurd to mix with models and believe that reality behaves according to them in every way!

3) "the speed of light is isotropic"

Yes, it is true, as Maxwell said:

the speed of light, c = 1/(μ0ε0)1/2;

the permittivity of free space, ε0

the permeability of free space, μ0.

Note that there are no reference systems here!

Dear Jan Slowak ,

this has to be interpreted. The ECIF is isotropic in both theories SR an LE, so being stationary there one can only measure

c. SR would just say that for every moving observer in ECIF, speed of light is isotropic, since it is isotropic in every inertial frame.

LE would say that all the moving non accelerated observers in ECIF will detect the speed of light as not isotropic, c/(1-v/c) or c/(1+v/c)

Dear Stefano Quattrini

"all the moving non accelerated observers in ECIF will detect the speed of light as not isotropic, c/(1-v/c) or c/(1+v/c)"

How can you write that the speed of light is a function of the speed of light? The speed of light is c, then the speed of light cannot be c/(1-v/c) or c/(1+v/c)!

Dear Jan Slowak ,

the speed of light is c as measured by a stationary observer in the isotropic domain or frame (ECIF for example). The same is not c but something more or something less according to the direction of motion and the speed in the isotropic domain.

Dear Stefano Quattrini

1) "ECIF" = Earth-Centered Inertial Frame

A reference system connected to the Earth is not inertial!

2) "The same is not c but something more or something less"

If you mean by "the same" speed of light, then it is wrong to reason the way you do. I think.

According to me, the speed of light is equal to c (in vacuum). Other speeds had to be given a different name and a different definition otherwise it gets confusing.

So, according to me: The speed of light in vacuum is c and the relative speed between a frame of reference and the wavefront of the light signal is c±v, where v is the absolute speed of the frame of reference in space.

Hi Stefano Quattrini,

SQ: Time dilation and length contraction are represented by the gamma factor in the transformations which can be written also in the following form

t' = ɣ-1 t - vx' /c2

x' = ɣ (x-vt)

You have an x' on the right hand side of the first of those, it should be x. The correct equations to convert (x, t) to (x', t') are

t' = ɣ (t - vx/c2)

x' = ɣ (x-vt)

where ɣ = 1 / √(1 - (v/c)2)

At v=c, ɣ becomes infinite. However, it is easier to see in a Minowski diagram. I've knocked this up quickly but it is interactive, move the speed indicator at the top left or right from zero and the red grid shows the (x',t') coordinates while the black is (x,t). The point Pt is at (0.5, 1.5)

Set v=-0.33 and you see that x' will be close to 0. Now move the slider to either -1 or +1 and you will see that the red grid turns into a single line, the axes are degenerate and that is why it is impossible to construct a frame referenced from a photon.

SQ: I don't know what you are talking about,

I hope the interactive graphic helps.

SQ: The original ether idea was evidently wrong

Indeed we know that now but it is what it is, the concept of the "preferred frame" refers to the aether, not the Earth.

SQ: Lorentz Electrodynamics though, contains the reference to a preferred frame since the term vx'/c2 is not a coordinate time for Lorentz but just the additional light time to cover the distance x'.

In Lorentz's interpretation, clocks at rest in the aether would measure time correctly but that frame is undetectable, both t and t' refer to measurements from clocks where only their relative motion can be measured.

The term vx/c2 is the offset due to relativity of simultaneity, it has nothing to do with the preferred frame.

GD: Since ε₀ and μ₀ are both scalar quantities, the value of c must also be scalar and hence isotropic.

SQ: this is well known

That is the key.

GD: That was Einstein's starting point for his 1905 paper.

SQ: Einstein's starting point making a difference with what was previously affirmed was : the light time to connect A and B is the same as connect B to A, that is the requirement of the constancy of the speed of light for every inertial observer.

No, the time from A to B is the same as the time from B to A for the same observer follows from the fact that it is isotropic. The principle of relativity requires that any other observer must also measure the speed to be isotropic though not necessarily of the same value, but if all observers measure μ0 and ε0 to have the same values, they will also agree the speed.

Dear Jan Slowak,

JS: Minkowski Diagram is a mathematical construct but it does not verify our reality!

It models it accurately, verification comes from experimental testing.

GD: "Light does not have a frame"

JS: I agree with you

GD: At the speed of light however, both axes have rotated 45 degrees so become a single line.

JS: What is moving at the speed of light?

The point I answered was about trying to construct a frame relative to light so it would be the light that is moving at the speed of light.

Dear George Dishman

Yes, a model of reality must be able to be verified and confirmed experimentally. For example SR does NOT verify our reality! That is my opinion and it appears from my research, which is of a theoretical nature. People can talk all they want, but my research is of a mathematical/physical/logical nature and then people should respond to it with the same type of argument.

SR is primarily a mathematical/theoretical construction, therefore it fits well in all discussions with a mathematical model and figures. That's how I work. In these models and figures it is not possible to talk in passing, but you have to talk about concrete concepts.

Dear Jan Slowak ,

JS: Yes, a model of reality must be able to be verified and confirmed experimentally. For example SR does NOT verify our reality! That is my opinion and it appears from my research, which is of a theoretical nature.

A study of a theoretical nature is not experimental. The Michelson-Morley experiment refuted the Galilean expectation and all subsequent experiments have been consistent with SR, that is the concrete evidence.

George Dishman ,

>

sure but this is not the classic approximation, the classic approximation is for v

Dear George Dishman

1) "A study of a theoretical nature is not experimental."

And an experiment is not a theoretical study. But that's no argument.

2) "The Michelson-Morley experiment refuted the Galilean expectation and all subsequent experiments have been consistent with SR, that is the concrete evidence."

MME is used as an argument for SR. But MME was carried out about 20 years before SR was created. Basing a theory on a failed experiment is wrong.

3) SR is primarily a mathematical/theoretical construction, therefore it fits well in all discussions with a mathematical model and figures. That's how I work. In these models and figures it is not possible to talk in passing, but you have to talk about concrete concepts.

Dear Jan Slowak,

JS: But MME was carried out about 20 years before SR was created. Basing a theory on a failed experiment is wrong.

Einstein did not know of the M-M experiment at the time, he based SR on Maxwell's Equations which show the speed of light is isotropic. Modern experiments continue to show that SR is accurate to better than one in 1018 but more importantly that the Galilean transforms are ruled out to a huge confidence, far beyond the usual 5 sigma.

https://phys.org/news/2015-09-precise-lorentz-symmetry-photon-constant.html

Dear Stefano Quattrini,

SQ: is mathematically equivalent, it is enough to make some algebraic passages.

It is not wrong but it assumes you know the answer for (t', x') before you calculate it so results in a circular argument if you aren't careful. Try to learn the theory properly first.

GD: At v=c, ɣ becomes infinite. However, it is easier to see in a Minowski diagram. I've knocked this up quickly but it is interactive, move the speed indicator at the top left or right from zero and the red grid shows the (x',t') coordinates while the black is (x,t).

SQ: sure but ...

If you know I'm right, why are you wasting everyone's time by arguing about it? You seem to just want to create the appearance of a disagreement where there is none so I have to question your attitude.

SQ: ... this is not the classic approximation, the classic approximation is for v

George Dishman .

>

it is not very understandable why one should consider the case where gamma goes to infinite, and to which purpose.

Dear George Dishman , you wrote:

>> SQ: sure but ...

If you know I'm right, why are you wasting everyone's time by arguing about it? You seem to just want to create the appearance of a disagreement where there is none so I have to question your attitude. //

Do you really not know the answer to your question or are you pretending to be too incoherent? The answer to your question is very simple: the text that you advertise with the help of a link is ordinary relativistic garbage, not worthy of any critical analysis.

As for your personal contribution to the discussion, I have always been interested in people who refer to the achievements of such a magnificent representative of the degenerative way of thinking as Herr Herman Minkowski. Am I correct in understanding that you are not in a position to recount/confirm the logical and/or mathematical validity of the admirable linguistic revelations of the aforementioned gentleman? Or could you actually add something to the completed discussion on this topic https://www.researchgate.net/post/Who_can_explain_H_Minkowskis_theory_to_me?

With kind regards

Dear Jan Slowak , 5 days ago you wrote:

>> Minkowski Diagram is a mathematical construct but it does not verify our reality! //

Do you agree that this mathematical construction is created and used by people who are too superficially familiar with both logic and mathematics?

>> the permittivity of free space, ε0 ; the permeability of free space, μ0. //

Curious, what do you mean by sets of letters «free space»? What is free of such a mysterious space, from dipole matter, which can only be characterized by such phenomenological parameters as «permittivity» and «permeability»? Or is this space free of anything else?

With kind regards

Dear Sergey Sheludko ,

mathematically I cannot say anything, it is certainly good, there are several things which are possible in math but they do not correspond to any Physics. It is the physical interpretation of that which is at stake and let's say the rational part which is experiments and logic..

Dear George Dishman

4) "Einstein did not know of the M-M experiment at the time, he based SR on Maxwell's Equations which show the speed of light is isotropic."

How can anyone know if Einstein knew about the Michelson-Morley experiment or not? It is extremely unlikely that he did not!

5) "Modern experiments continue to show that SR is accurate"

If you mean similar experiments as MME, then it goes without saying that they reach the same results as they reached in 1887, but with better precision. If you use the same method/model, it goes without saying that you will reach the same result. I have shown in my book Special Relativity is Nonsense that Michelson's interferometer was not suitable for use in making measurements of the Earth's velocity.

6) The link you provide shows a device of similar construction to the Michelson interferometer: two right-angled arms. That is the problem!

Dear Jan Slowak ,

experiments tested the accuracy of the Lorentz Transformations basically. The point is which version??

For sure the Lorentz Electrodynamics (LE)

a) twin effect with muons

b) transverse Doppler blue-shift

c) Rectilinear Doppler (and RADAR)

etc.

d) Sagnac effect

these are quite easily accounted by LE

Dear Sergey Sheludko

1) "Do you agree that this mathematical construction is created and used by people who are too superficially familiar with both logic and mathematics?"

I never discuss people behind an article or book. I refer only to the content of their work. So your question is irrelevant to me.

2) "Curious, what do you mean by sets of letters «free space»?"

These are not my words but it is in other people's articles and books. You probably mean vacuum. I refer to Maxwell's work:

the speed of light, c = 1/(μ0ε0)1/2;

the permittivity of free space, ε0

the permeability of free space, μ0.

Dear Stefano Quattrini

Can you refer to a figure/mathematical model for the "twin effect with muons"? Then I can comment.

Dear Sergey Sheludko,

SQ: sure but ...

GD: If you know I'm right, why are you wasting everyone's time by arguing about it?

SS: Do you really not know the answer to your question ...

Since the words I quoted were written by Stefano Quattrini, obviously I cannot know what motivated him to write them. He said "sure" which means he agreed with the mathematical fact I stated but then went on to talk about a different situation which was unrelated to my answer to Jan Slowak. I can only enquire why he did that.

SS: are you pretending to be too incoherent?

I will give you the benefit of the doubt and assume you just hadn't followed the previous series of posts, otherwise perhaps your command of English isn't as good, I might think it is not your native language.

SS: ... garbage, not worthy of any critical analysis.

Coincidentally I take the same view of your misguided reply.

Dear Jan Slowak,

GD: Einstein did not know of the M-M experiment at the time, he based SR on Maxwell's Equations which show the speed of light is isotropic."

JS: How can anyone know if Einstein knew about the Michelson-Morley experiment or not? It is extremely unlikely that he did not!

When interviewed he stated that he did not know of it and nobody at the time or since has found any evidence to contradict his statement. That fact remains either way that he only used the well-known result from Maxwell's Equations that the speed of light should be isotropic in deriving SR so whether he knew or not is academic.

GD: Modern experiments continue to show that SR is accurate

JS: If you mean similar experiments as MME ... The link you provide shows a device of similar construction to the Michelson interferometer: two right-angled arms.

Although somewhat similar to the interferometric method, there no arms in the method used in the article, it uses two cross-coupled microwave cavities excited by sapphire crystals. It still confirms Lorentz Invariance to the limit of the best current test accuracy and falsifies Galilean invariance to an extreme level.

Dear Stefano Quattrini,

SQ: try to learn where the theory comes from first ...

The theory comes from Maxwell's Equations as I have pointed out repeatedly.

GD: At v=c, ɣ becomes infinite. However, it is easier to see in a Minowski diagram. I've knocked this up quickly but it is interactive, move the speed indicator at the top left or right from zero and the red grid shows the (x',t') coordinates while the black is (x,t).

SQ: it is not very understandable why one should consider the case where gamma goes to infinite, and to which purpose.

The purpose is very simple, I had previously said:

GD: Light does not have a frame, the axes are degenerate.

then Jan Slowak asked

JS: What do you mean by this?

so the purpose of my reply was to answer his question by explaining why the axes are degenerate when v is exactly equal to c as it is for light in vacuo.

SQ: time dilation is determined by

SQ: t' = ɣ-1 t (this term)- vx' /c2

SQ: length contraction by this term

SQ: x' = ɣ (x-vt)

SQ: one is the reciprocal of the other in gamma.

I did warn you about using muddled equations, your error is what is likely to happen. The correct equations for converting the (t,x) pair into (t',x') are:

x' = ɣ (x-vt)

t' = ɣ (t-vx/c2)

Both are affected by gamma in the same sense, not the inverse.