Lammaite2

Every annoyance is welcome. When providing the predicted time of transverse journey Feynman says: …the light travels a distance ct3 along hypotenuse of a triangle… If this was possible there must have been assumed some counter-intuitive relativistic effect i.e. what is seen in aether - the transverse light just bouncing up and down, is seen from apparatus frame also bouncing up and down. To achieve this, the magic invisible path is along the hypothetical hypotenuses. Next is the assumption of constant c (still like if there was aether!), hence the time equation, hence the expected delay. Then the result shows there is no delay, hence SR. This does not make sense because the assumption of relativistic effect is used for formulating relativity. Yet this the only way I can follow up Feynman’s reasoning. My common sense tells me: If applying classical physics, aether, speed c along the hypotenuse path as on Feynman’s diagram…then the parallel and transverse rays would have to be different just on start from the source. Some width of the light beam would need to be assumed and lights rays at different angles from the source. They would reach the first mirror already with some delay and reflect at different angles. The time equations (easy to deduce) would look entirely different from Michelson’s (and Feynman’s). If we wanted just one ray on start then split on the first mirror then, by the classical physics and aether assumption, the transverse ray will be seen from apparatus along hypotenuse yet back to the apparatus direction and with speed SQRT(v^2+c^2) versus apparatus. I am referring here to Feynman’s “Six not so easy pieces”. Of course I am amateur and beginner in this area. Were there any other ways of inferring SR?

EUREKA! Now, I think I understand. It seems, that the explanation of the transverse journey time takes into account the result of the MM experiment; NOT JUST ONLY the pre-experiment assumed condition. The fact that the transverse signal reached the meeting point (“Normally” it should not!), implies: that the light MUST have travelled ALONG the arms of the isosceles triangle (Otherwise it would not be seen at meeting point.). In addition, it travelled WITH SPEED c (as the first assumption taken from Maxwell’s theory, or as a propagation of wave as says Feynman). From these two condition is the equation for the transverse time! This small change in the way of explaining to indicate that the result of MM was the reason behind the transverse time made so big difference. If this is the case the width of the beam does not matter as I thought earlier. Stil a question remains: Should not be Michelson's expectation, based on purely Euclidean geometry, law of light refraction and presence of aether as a medium for light propagation, that the light rays should never meet? The discussion here helped a lot. I really hope to hear some conforming posts!

The zigzag path of the transverse signal, it comes to me more and more certain, is assumed by Michelson (also by Feynman’s explanation) to evaluate the time, only because of the width of the beam. Within the beam the light reflects at a range of angles, producing not only perfectly perpendicular signal, and the signal in consideration is the one that reflects at particular angle. The angle is such that the signal’s path aims exactly at the interferometer at the moment when the twin signal reaches the interferometer. This is why in Feynman’s calculation of the transverse time the signal travels at speed c (!) along the arms of the triangle. Which means it was reflected into direction of the arm of the triangle and again reflected into direction along the other arm to aim the interferometer; it travels this directions versus aether frame with speed c. This seems to be the reasoning behind Michelson’s and Feynman’s calculation of journey time. If what I said was true than the ahead signal and the transverse signal are not the same ray split on the first mirror. They are two different rays from the starting from the source of light. When the speed of the apparatus is low then this may not be a problem and the derived time delay equation is OK. What if the speed of the apparatus is much greater? The zigzag will be wide open and the rays would need to be drawn from the source on different paths – the transverse signal should start at some significant angle to the parallel ray even before reaching the splitting mirror. This is because if we want the transverse signal to reflect little bit forward it needs to hit the splitting mirror at bigger than 45 deg. to normal. What will then happen to the equation of the time delay? All SR mathematics is later based on confronting this equation with the fact of no delay. I can not think otherwise than: if the signal was an Euclidean point and travelled perfectly perpendicular to the apparatus direction then the signal would not be able to meet again the interferometer (also a perfect point). The journey of that point, as seen from apparatus frame, would be a zigzag, yet the opposite direction than shown in Feynman’s and Michelson’s pictures – i.e. back from the apparatus. Also the predicted journey time along that zigzag would need to be calculated using speed value from parallelogram of speed c perpendicular to apparatus and the speed v of the apparatus i.e. SQRT(c2+v2). Also the predicted journey time along that zigzag would equal 2L/c.

I suppose, from all the posts, I am getting close to solve my dilemma. Please correct the reasoning if wrong: 1. Existence of stationary ether is assumed 2. The zigzag path of the transverse as seen from the apparatus reference should be to the left (not to the right as on the Wikipedia gif). 3. Taking the low speed of the apparatus and the width of the light beam, it still hits the interferometer and the other returning beam. 4. Zigzag path is therefore used for predicting the travel time of the transverse signal Is that OK? Thanks for the link to Michelson

I can almost hear the big pistol being unlocked .... Here is a quote from Feynman referring to the situation of travelling apparatus and the "ahead" signal: "Now, while the ligh is from the B to the mirror, the apparatus moves a distance v*t1, so the light must traverse a distance L+v*t1, at the speed c." Hence, c*t1=L+vt1, hence t1=L/(c-v) Similarly Feynman calculates the journey back L-v*t2, hence c*t2=L-vt2, hence t2=L/(c+v). As you can see Feynman says "at the speed c". And from the equations it appeares that the light did not acquire any additional speed when emitted ahead as well as it did not slow down when emitted back. It just travells different distances at speed c. From Feynman equations for times t1 and t2 the denominators are actually c-v for t1 and c+v for t2 that is opposite to what you suggested. This is why I stubbornly insist the assumptions for Michelson experiment were the light moves at constant c in relation to ether and the speed is independant to the speed of the source. Actually I understand these two are actually unseparable. Or may be all this comes from linguistic confusion as to the use of "independant". Anyway, the constant c versus ether tells me the transverse signal (similarly to ahead signal) should not acquire any additional speed in direction of the apparatus. This should be consistently the assumption for calculating the times and delay of the signals before the experiment.

"The gif to the right would be true if there was an aether. That was the expectation." Again, I have difficulty to agree with this statement. If there was an ether, and this was the initial assumption, the speed of light was constant c with reference to ether, and it was independant to the speed of source. This was why the journey to the mirror of the "ahead" signal is calculated c*t1=L+v*t1. It was not calculated (c+v)*t1=L. The signal ahead did not acquire additional speed v. The same should apply to transverse beam so it could not have any speed parallel to the apparatus. So it could not be "the expectation". The expectation could be the light beam lagging behind and actually never meeting the returning "ahead" signal. So I still do not understand what is shown on the right gif. I noticed from the posts also confusion of "constant speed of light c" and "independant from speed of the source". I think both of the features are unseparable. Please correct me in my learning. I am taking my understanding from great lecture by Feynman. He says the speed of light is constant and independant form speed of source as much as any other wave e.g. sound in the air. That is why there was expected a delay of the signals in Michelson's experiment. Kill me, but I do not understand how the predicted travel time for the transverse signal was calculated, or rather why the zig-zag path was assumed (as Feynman and Wikipedia say). This important because it looks like all later mathematics in SR is based on that (Lorentz correction and transformation of time) once the predicted times (hence delay) was confronted with the fact of no delay.

I am risking more and more you will use your big pistol.... Of course, the option of no ether wasn't considered - that means it was assumed for pre-experiment predictions of time travel (both parallel and transverse) like if the light travelled with c independantly from the source speed. Then, why there is the parallel component of speed of the transverse signal shown on the gif and used to formulate the equation? Michelson calculated the transverse time 2L/c and later corrected based on the zig-zag path. What am I missing in reading this explanation both on Wikipedia and in Feynman's lecture?

I still do not understand. What you are saying is OK but the experiment should show the same if the Earth travelled with say 200000 km/s. Than the offset would not be a "spot" anymore. I try formulate my question other way; Looks like the concept behind MM experiment was: 1. calculate time parallel and time transverse like if there was ether, hence the difference 2. compare to the results while the explanation provided looks like is (or I may missed something): 1. calculate the parallel time like if there was ether, and calculate transverse time like there was no ether and assume c is constant for some other reasons and travels the arms of triangle, hence the difference 3. compare to the results Don't shoot me

That answer turns upside down all my understanding. Please help. Here is a quote from Richard Feynman: "If the apparatus is at rest in the ether, the times should be precisely equal, but if its moving toward the right with velocity u, there should be a difference in times. Lets see why..." And then is the whole simple mathematics based on assumption: light moves in ether with speed c independently to the speed of source both as to magnitude and direction. I.e. the times in horizontal direction come from reasoning ct1=L+ut1 and ct2=L+ut2. So, the gif provided to explain the reasoning to calculate the transverse time shows the same situation when: there is ether and the light has constant speed in relation to ether. The reasoning for the parallel direction is easy. The transverse is still unclear to me. The explanation that light reflects at a range of angles appeals to me yet I am not sure this is the case (can someone recommend some books). I can not agree with the gif showing horizontal component speed of the red dot (light signal) if it is meant as if acquired because the source is speeding.

Thanks again. This explains very much. Unfortunately the animated gif in Wikipedia actually is confusing and contradicting the experiment assumptions. This is the same as with the disturbance (say a point) on the water surface. The wave (in that case radial) moves in all directions with the same speed never mind what is the speed and direction of the disturbance point. The same should be with the red dot once it is generated (reflected) as a disturbance in the aether. Your statement (quote) is what I thought. The red dot actually would never be able to reach the "meeting point" with the blue dot. As I remember one of the proposed explanation of the paradoxical result of MM experiment was a "drag of the aether together with the Earth (MM apparatus)". In such a case both the red and blue dots would acquire additional speed equal to the speed of aether. Please correct me if I am wrong. From what you are saying I understand the width of the transverse signal (width of the beam) causes that its relative movement backward from the apparatus is negligible. Also, the light reflects from the mirrors at range of angles (not just one missile - red dot on a transverse trajectory), hence a signal that reflects at very certain angle travels along arms of the triangle to hit the mirror in its new position, hence the mathematics behind 2*L/(c*sqrt(1-v2/c2)) instead of 2*L/c? How is that all working with laser beam? I would appreciate your help. That would be the last bit that is tormenting me.

Thank you very much. If you still have some patience.... As you see the question still remains unanswered. Why the "red spot" speeds together with the apparatus? I do not have problem with the maths. I have problems with the assumption related to the transverse signal. Let me try again. The propagation of wave (light) is independant on the source both speed and direction. This is how I understand constant c. Similar to the water wave from any disturbance: e.g. fast ship or slow ship does not matter. The wave speed is related to properties of the medium (water, aether, space itself as suggested by some). So, once the light wave is reflected from the splitting mirror it should be "left behind" because the mirror travels with the speed v. This is not what is shown on the Wikipedia gif. Apparently the red dot acquired the additional speed equal (magnitude and vector) to the speed of the apparatus in addition to the speed c transverse to the apparatus movement. Rightly, the blue dot travelling parallel to the apparaus direction did not acquire any additional speed. This is why t1*c=L+t1*v (forward) and t2*c=L-t2*v (back). From that point of view also the further reasoning of the path of the transverse signal along the arms of the triangle is unclear to me. On other forum someone suggested the light signal travels at angle. In that case I would accept that some signal travelling at certain particular angle will reach the "meeting point" in its new position. However, this does not convince me because I understood the light was a beam, or even a laser beam in modern versions of the experiment. I would appreciate from you some comment on that. Was the light signal limited just to transverse beam or it can be assumed as propagating radially from the reflecting mirror? I do not think the whole thing is erroneous. I am just trying to comprehend this in details instead taking for granted. Also, there must some deal of difficulty with the transverse signal since the initial time of that signal was 2L/c (by Michelson) and later changed to 2l/(c(sqrt(1-v2/c2)).

I said that. Tha apparatus is travelling with some speed v (in relation to the hypothetical aether) much lower than c. My question was why on the Wikipedia animated explanation of the time delay between the transverse and parallel signals the transverse signal is shown as having this speed v? Generally, I do not understand your answer. Please be patient in explaining to me.

Can someone explain, please. I stuck on some problem related to Michelson Morley experiment. The transverse signal travel time was initially calculated by Michelson as 2L/c and later corrected as if the signal travelled along the arms of a triangle: start point – mirror – new position of start point. This correction adds further to expected time difference of light signals: the transverse and the parallel to Michleson’s apparatus movement direction. This is also explained on animated graphic on Wikipedia http://en.wikipedia.org/wiki/Michelson_Morley_Experiment It is presented as a missile (red dot) travelling transverse to the direction of Michelson’s apparatus. It shows speed transverse to apparatus direction © as well as … speed the same as the speed of the apparatus (!). I do not understand that. The prime assumption of the Michelson’s experiment was the light travels with constant speed c (as a wave in a medium, e.g. sound in the air) in the eather (eather is the reference). From that, I would think the signal (missile, red dot) should not have any speed parallel to the speed vector of the apparatus. It was expected independent from the speed of the emitter. If the light was reflected on the first mirror (the splitting mirror) in every direction than the particular signal travelling at certain angle would arrive at “meeting point” in its new location (with a phase delay of course). Yet, as I understand the light was a beam (could be a laser now) directed transverse to apparatus speed vector.