vanholten

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

But if these were the first two objects in the universe, they were not in motion relative to something else. By the absence of any reference they were at an absolute distance.Their geometrical relation in relativity still is a definite line segment. I just don’t see the theoretical need for stating that both endpoints of a segment are at rest with respect to each other. And what do you think of the following? If two objects are in uniform motion, a point of rest exists from which these objects move away in opposite direction at half their relative velocity.

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

Apparently you are unable to comprehend the simple concept of relative motion. You seem to have missed that the relations between the clock and their observers essential to define motion can't be relative, since nobody can be in relative rest compared to a stationary object. Rest defines a definite or absolute relation. SR totally relies on that definite relationship. Imagine empty space. What is to be defined first, rest or motion, or relative motion? How many positions do you need to define rest? And how many positions do you need to create motion? Once you know, you might even have the answer to how many positions you need to experience relative motion.

It relies on the all seeing eye. To establish simultaneity you need at least two events, a rhombus and two observers at opposite edges. I already showed that you can easily derive the Lorentz-transformation using the pythagorean theorem.

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

Thanks for that I have no trouble understanding it. However I was explaining that Einsteins equations inevitably result in zero velocity. It takes to many words. Explaining my findings in Dutch is already difficult , and doing so in English is even more complicated. So let me once again try to explain what I mean, again based on Einstein's equations. The distance the light beam travels can be expressed as rAB + v . ( tB - tA ) or c . (tB - tA) + v . (tB - tA) Because all clocks on the moving rod and in the stationary system are synchronized, what Einstein assumed is: tB - tA = { rAB + v . ( tB - tA )} / c ............... leads to the original "tB - tA = rAB / (c - v)" or tB - tA = { c . (tB - tA) + v . (tB - tA)} / c tB - tA is replaced by x then x = (c x + v x)/ c take for example c = 1 then x = ( x + v x) thus v = 0 WolframAlpha confirms: "Solutions: v = o, c ≠ 0 x = 0, c ≠ 0 v = 0" Einsteins "tB - tA = rAB / (c - v)" results in zero velocity.

To illustrate the effect of length contraction Einstein sketched an experiment. Rod AB in motion is equipped with clocks and observers at each end. The whole thing is observed from a stationary system with synchrone clocks all over the place. The clocks on both ends of the moving rod AB are synchronized with the clocks of the stationary non co-moving observers. Ok this is what I assumed: The two equations are meant for the stationary observers seeing the rod AB in motion. Because those observers will have to deal with the velocity of the rod, as formulated in both equations. And they rely on the timespan tB-tA displayed by their clocks being part of the first equation. Einstein says that the length of rod AB in motion is measured as being different from the length of rod AB when it is measured as a stationary object. It seems he relies on length contraction, adapted from Lorentz' explanation of the absence of phase shift during the MM experiment. (I think that is strange because the second postulate did us forget about the ether.) Anyhow, Einstein says : “rAB denotes the length of the moving rod—measured in the stationary system. “ But what does he mean by that? I think there are two explanations possible. Option 1: rAB is measured in the stationary system; the stationary rod is measured to have the length AB. In that case the stationary observers see the light ray bridge the distance AB according to their clocks in tB-tA seconds. According to Einstein's definition of simultaneity: 2AB/(t’A-tA) = c and tB-tA = t’A-tB they measure the light to propagate at c so their observation results in rAB/c = tB-tA ( my assumption) Option 2: rAB is measured from the stationary system while the rod AB is in motion. The speed of light is measured to be c as always for all observers. The stationary clocks still display the timespan tB-tA in which the light at c bridges AB. Again this results in rAB/c = tB-tA (my assumption) Both explanation have the same result. For the stationary observations the velocity v in both equations is a meaningless parameter. However Einstein’s experiment suggests that the observers on the rod encounter a different timespan for the light to travel AB. This means that for those observers the timespan should result in rAB/c ≠ tB-tA because their clocks are synchronized with the clocks in the stationary system; the system being not Rod AB. To make the confusion complete: Thanks to the equations the observers that are stationary with respect to rod AB are no longer subjected to the definition of simultaneity; 2AB/(t’A-tA) = c Instead they have to bring the relative motion of the clocks in the non co-moving system into account. But the equations don't deliver the required parameters to do so. Because by the lack of simultaneity these observers on the rod can't produce a sensible value for velocity v. In other words: The clocks at both ends of the rod are asynchrone so the observers can't define the relative speed of the other system required to pinpoint v in the equation. Thus also for them v in the equations is pointless. And if they would adjust their clocks to achieve rAB/c = tB-tA, they would have only switched sides and end up with the same observations as made by the observers in the stationary system; velocity is irrelevant. The intriguing thing is that the second postulate underlines the irrelevance of velocity of the light source during the measurement of the speed of the light emitted by that source. I never assumed the speed of light is infinite.

I already tried to explain it in a previous topic. 'What if Einstein's definition of simultaneity is incorrect" The mathematical passage is in here: ‎hermes.ffn.ub.es-luisnavarro-nuevo_maletin-Einstein_1905_relativity.pdf.webloc 1. Kinematical part § 1Definition of Simultaneity and §2 On the Relativity of Length and Time. tB - tA = rAB / (c - v ) and t’A - tB = rAB / (c+v) These equations are based on the length contraction of rod AB. To retrieve the first equation I followed the "einsteingenootschap.nl" (a Dutch site dedicated to Einstein) 1- The distance the light beam travels can be expressed as: rAB + v . ( tB - tA ) Because all clocks on the moving rod and in the stationary system are synchronized, what Einstein assumed is: tB - tA = { rAB + v . ( tB - tA )} / c c . ( tB - tA ) = rAB + v . ( tB - tA ) c . ( tB - tA ) - v . ( tB - tA ) = rAB ( c - v ) . ( tB - tA ) = rAB And there it is: tB - tA = rAB / (c - v) So far so good and in line with the einsteingenootschap. But next I assume that the following is allowed: tB - tA = { rAB + v . ( tB - tA )} / c rAB / c = { rAB + v . ( rAB / c )} / c I replace Rod AB by length L L / c = { L + v ( L /c )} /c L = L + v ( L /c ) L . v / c = 0 L = 0 , c ≠ 0 v = 0 , c ≠ 0 L = rAB ≠ 0 thus v = 0 Both fundamental equations of Special Relativity eliminate motion?

Sorry You wrote that Lorentz and Fitzgerald explained the absence of interference due to length contraction. Then they must have supported the idea of the ether I presume. So why did Einstein at one hand adapt the Lorentz/Fitzgerald contraction as the fundament of simultaneity according to tB - tA = rAB / (c - v ) and t’A - tB = rAB / (c+v) and at the other made the ether superfluous with his second postulate?

The idea that there is not even an actual reflection is the whole point. It is supposed to be remittance of energy in the shape of a new photon; a fresh emission creating a fresh wave propagating at c. If that is going on the mirrors act like stationary light sources with respect to the telescope. From a relativistic perspective there is only a single stationary frame of reference, which as such can't confirm the second postulate. The thing you might expect caused by the rotation of the interferometer is the Sagnac effect. But that effect is predicted based on classic physics and adapted by relativity, so neither the Sagnac effect can be a confirmation of SR. I agree it makes you wonder why the remittance of light by electrons is forced in the reflective angle. It needs an explanation. Probably it is imposed by the collaboration of surrounding electrons in the reflective surface?

Sure it wasn’t discovered but it was acting anyhow. So what is your point? Earth wasn’t flat before it was discovered to be sphere. The fact that at the current state of science according to you is impossible to describe the speed of individual photons doesn’t imply it’s meaningless. It might even be essential to get a grip on the behaviour of light.

https://en.wikipedia.org/wiki/Reflection_(physics) "mechanism" -"In the case of dielectrics such as glass, the electric field of the light acts on the electrons in the material, and the moving electrons generate fields and become new radiators. The refracted light in the glass is the combination of the forward radiation of the electrons and the incident light. The reflected light is the combination of the backward radiation of all of the electrons."- This says the mirrors become radiators that emit light at c independent of their relative motion as usual. The telescope in the MM experiment is the stationary observer of the mirrors so it should measure the light within the instrument to propagate at c. It is not even a relativistic observation. That is why I don't t see why this experiment confirms the second postulate. It only comments the ether or on the absence of it.

My point is that the MM experiment is frequently brought forward as confirmation of the second postulate. If your assumption would be correct it would undermine the theory of SR and its underlying calculations all by itself.

Hello. Excuse me, but I do question the relevance of the Michelson-Morley experiment regarding Special Relativity. At the time the MM interferometer didn’t show any phase shift at the receiver within the instrument. The outcome of the MM experiment suggests that the velocity of Earth relative to the Sun being the light source, had no effect on the measurement of the propagation velocity of sunlight in neither directions. This absence of phase shift is considered to be a confirmation of the second postulate of Special Relativity: “2. Second postulate (invariance of c) As measured in any inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body. Or: the speed of light in free space has the same value c in all inertial frames of reference.” But is the absence of phase shift indeed a solid confirmation of second postulate postulate? Not quite I think. To measure the propagation velocity of light, you should measure the velocity of one and the same photon. (@swanson) In the decennia after the MM experiment and SR, Quantum mechanics evolved mead thanks to Einstein explanation of photo-electric effects. The interferometer makes use of mirrors. However when a photon strikes a mirror and it would be absorbed by an electron, that electron will be excited, gaining energy and momentum. Later, it will drop back to the previous state, emitting a photon. When indeed the electrons in the mirrors of the interferometer act like light sources, they will emit other photons than those being intercepted and absorbed from the sunlight. If that is true the MM experiment compares the velocity of different photons coming from different light sources. Of course photons emitted by the MM mirrors would have propagated at the velocity of light but independently from the photons emitted by the Sun. The fact that no phase shift occurred seems to be exactly in line with quantum mechanical effects and as such is no confirmation of Special Relativity at all. Perhaps to do a proper test to confirm the irrelevance of motion of the light source, one should make use of the refraction index of light, because that alters when the speed of light deviates.