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Natural Science Forum / Physics / Relativity / April 2006michelson morley experiment
Hi In the MM experiment, was the length of the two light paths set exactly equal. If so, how was this done? If not, did the experiment "rely" on the fact that there was no change in the interference fringes when the apparatus was rotated? Did the MM experiment demonstrate anything about the constancy of the speed of light or did it merely "disprove" the aether theory? Does this website have any credibility in mainstream physics? http://home.iprimus.com.au/longhair1/page1.html Thanks. > Hi > [quoted text clipped - 10 lines] > > Thanks. sure it has, thay say "This site is a proud member of The Theory of Everything Webring, linking the great thinkers of our time." what are you crazy > Hi > > In the MM experiment, was the length of the two light paths set > exactly equal. No. > If so, how was this done? If not, did the experiment > "rely" on the fact that there was no change in the interference > fringes when the apparatus was rotated? Instead it relied on the assumption that there would be a significant change - which turned out to be absent. > Did the MM experiment demonstrate anything about the constancy of the > speed of light or did it merely "disprove" the aether theory? It demonstrated the constrancy of the measured return speed of light, thus disproving Maxwell's ether theory. > Does this website have any credibility in mainstream physics? > http://home.iprimus.com.au/longhair1/page1.html I don't know it; at first sight it certainly doesn't look credible . > Thanks. You're welcome. > > Hi > > > > In the MM experiment, was the length of the two light paths set > > exactly equal. > > No. Pretty nearly equal, though. Although Michelson and Morley used sodium light for collimating the apparatus, the actual experiment was performed using white light from an argand burner. The colored fringes were much easier to visually monitor; on the other hand, the limited coherence length of white light meant that the path lengths needed to match within microns. Note in Fig. 4 the piece of glass "c" used to compensate for the difference in light paths. http://www.aip.org/history/gap/PDF/michelson.pdf > > If so, how was this done? If not, did the experiment > > "rely" on the fact that there was no change in the interference [quoted text clipped - 8 lines] > It demonstrated the constrancy of the measured return speed of light, thus > disproving Maxwell's ether theory. In an effort to "save" aether theory, of course, various ad hoc hypotheses were made including the notion that the aether might be entrained by the moving Earth, or the totally ad hoc Lorentz- Fitzgerald contraction, later derived by Einstein. > > Does this website have any credibility in mainstream physics? > > http://home.iprimus.com.au/longhair1/page1.html [quoted text clipped - 4 lines] > > You're welcome. Jerry > > > Hi > > > [quoted text clipped - 11 lines] > of glass "c" used to compensate for the difference in light paths. > http://www.aip.org/history/gap/PDF/michelson.pdf Yes indeed, interesting! > > > If so, how was this done? If not, did the experiment > > > "rely" on the fact that there was no change in the interference [quoted text clipped - 13 lines] > be entrained by the moving Earth, or the totally ad hoc Lorentz- > Fitzgerald contraction, later derived by Einstein. That that contraction was ad hoc was a misunderstanding by Einstein, and some textbooks continue to propagate that fable. The idea started with Fitzgerald inferring it from Heaviside's EM field calculations; Lorentz and Poincare showed independently that it must be a length contraction by a factor gamma. Indeed, Lorentz admitted that he should have made clearer that it was not "ad hoc", and agreed that Einstein's derivation was more elegant. Harald > > > Does this website have any credibility in mainstream physics? > > > http://home.iprimus.com.au/longhair1/page1.html [quoted text clipped - 6 lines] > > Jerry > > Does this website have any credibility in mainstream physics? > > http://home.iprimus.com.au/longhair1/page1.html That is a net version of a book written by two Australian physicists in 2001. Do books of psysics have any credibility? At least it looks like a professional work. Henry Haapalainen > > > Does this website have any credibility in mainstream physics? > > > http://home.iprimus.com.au/longhair1/page1.html [quoted text clipped - 4 lines] > > Henry Haapalainen I saw only one author mentioned, on the front page. Where did you find that it's "written by two Australian physicists "? > Hi > [quoted text clipped - 5 lines] > Did the MM experiment demonstrate anything about the constancy of the > speed of light or did it merely "disprove" the aether theory? It disproved the 'simple' aether theory oft the day. > Does this website have any credibility in mainstream physics? > http://home.iprimus.com.au/longhair1/page1.html No. Martin Hogbin | Hi | | In the MM experiment, was the length of the two light paths set | exactly equal. No, and it doesn't need to be. All that is required is a wavelength match. | If so, how was this done? With adjusting screws of the type used in micrometers. http://tinyurl.com/h2g7z If not, did the experiment | "rely" on the fact that there was no change in the interference | fringes when the apparatus was rotated? | | Did the MM experiment demonstrate anything about the constancy of the | speed of light or did it merely "disprove" the aether theory? It only disproved aether. The wind when you put your hand out of a car window proves the world is moving. Close the window and you won't feel it. http://tinyurl.com/f8a9z | Does this website have any credibility in mainstream physics? | http://home.iprimus.com.au/longhair1/page1.html Nope. Androcles | Thanks. >In the MM experiment, was the length of the two light paths set >exactly equal. They originally set it up that way. Of course, one can not have *exactly* same lengths but within error bars was the same length. Then other similar setups were done with different arm lengths. See Kennedy-Thorndyke. >If not, did the experiment >"rely" on the fact that there was no change in the interference >fringes when the apparatus was rotated? It did not *rely* on that fact. I was setup to find what would happen if rotated. >Did the MM experiment demonstrate anything about the constancy of the >speed of light or did it merely "disprove" the aether theory? It (as was already known) showed that the TWLS was constant. It did not disprove (modern) ehter theory because ether theory predicts that there would be no fringe shift. But the 'earlier' ether theory was badly interpreted and was thought that it it predicted a fringe shift. --- If you want to be sure, then always doubt }:-) >>In the MM experiment, was the length of the two light paths set >>exactly equal. [quoted text clipped - 18 lines] >that there would be no fringe shift. But the 'earlier' ether theory was >badly interpreted and was thought that it it predicted a fringe shift. So if I calculate the phase relationship between the two swimmers that are described here http://galileoandeinstein.physics.virginia.edu/lectures/michelson.html will I find any change if I "rotate the apparatus" i.e. if the direction of the current in the river is changed? If I change the direction of the current in the river, the two swimmers won't make it back to the original starting point unless I also change the direction they swim, but if I change the direction they swim, this is no longer the equivalent of the MM experiment, so the fact that the two swimmers make it back to the starting point at different times when the river is flowing parallel to the river bank, proves nothing ?? Since the MM apparatus doesn't change the direction the "swimmers" swim in, won't this produce a fringe shift as the apparatus is rotated, according to the ether theory being tested? Thanks The Earth is not the center of the universe. If there was a backround, the aether stayin still, there should be a difference of, at least, thousands of kilometres per second, when c is measured in different directions. An aether does not exist, and no more evidence is needed. Henry Haapalainen > >>In the MM experiment, was the length of the two light paths set > >>exactly equal. [quoted text clipped - 38 lines] > > Thanks > >>In the MM experiment, was the length of the two light paths set > >>exactly equal. [quoted text clipped - 38 lines] > > Thanks Do you think you need to aim a flashlight up stream to hit a target perpendicular to your direction of motion? Light doesn't work that way, so your analogy is flawed. Take a friend and a couple of megaphones out to a field on a windy day. Use your megaphone to shout to your friend while he uses his to determine the direction the sound comes from. You will find that he does not hear the sound come from up wind or down wind but straight from you. You will find it is best to aim straight at him, not up wind or down wind. So sound, which is carried by a medium, does not behave the way you predict either. It's not so easy to disprove the aether when you use what really happens as opposed to poor analogies. Bruce Richmond >> >>In the MM experiment, was the length of the two light paths set >> >>exactly equal. [quoted text clipped - 42 lines] >perpendicular to your direction of motion? Light doesn't work that >way, so your analogy is flawed. It's not my analogy. Judging by the URL, the University of Virginia physics department is responsible for the analogy. Since I didn't get an answer to my question about the siwmmers and the river that made any sense to me I will assume that the analogy with the swimmers and the river is meaningless and the crucial details about the MM experiment are not explained on that web page. Darkknight. >Take a friend and a couple of megaphones out to a field on a windy day. > Use your megaphone to shout to your friend while he uses his to [quoted text clipped - 8 lines] > >Bruce Richmond > >> http://galileoandeinstein.physics.virginia.edu/lectures/michelson.html > It's not my analogy. Judging by the URL, the University of Virginia > physics department is responsible for the analogy. [quoted text clipped - 3 lines] > the swimmers and the river is meaningless and the crucial details > about the MM experiment are not explained on that web page. What I see in the above URL is a fairly standard explanation of the rationale behind the MMX. You could try reading the original paper, reproduced in full at the bottom of this page: http://www.aip.org/history/gap/Michelson/Michelson.html There are lots of other of historically important papers to be found at the above web site. Jerry > >> >>In the MM experiment, was the length of the two light paths set > >> >>exactly equal. [quoted text clipped - 50 lines] > the swimmers and the river is meaningless and the crucial details > about the MM experiment are not explained on that web page. As you have read below (you did, right?), such an analogy is no good for questions relating to momentum conservation. The analogy was only intended (but never proved!) to be valid for calculating the propagation *speed* of a wave that propagates through a moving medium. Assuming that it's indeed valid for a wave that is skewed relative to the medium, then you get the factor gamma for the total time delay, and gamma^2 for 0/180 degrees. The difference is a factor gamma, and that's what M-M expected to be able to measure. Harald > >Take a friend and a couple of megaphones out to a field on a windy day. > > Use your megaphone to shout to your friend while he uses his to [quoted text clipped - 8 lines] > > > >Bruce Richmond The analogy os almost correct. That analogy/explanation was the early view of ether theories. In modern ether theories (Lorentz etc..), one should not use the concept of 'ether wind'. One should view the ether (the river) as the 'fixed' medium and that it is the rest of the stuff that is moving wrt the ether. So the analogy should be that the river flow is not flowing but the sidebanks (and swimers) are moving. In modern ether theories, when an object (sidebank and the swimers...) travels within the river (ether), their lenghts are contracted (so is their clock ticks) in the direction of motion. It is these length and time contractions that cancel out the supposedly fringe shift in the MMexperiment. The swimmners will always arrive together when they return. Note that the swimmers should always swim at constant speed wrt the river for the analogy to be complete (light is always c in the ether frame). --- If you want to be sure, then always doubt }:-) > Take a friend and a couple of megaphones out to a field on a windy day. > Use your megaphone to shout to your friend while he uses his to > determine the direction the sound comes from. You will find that he > does not hear the sound come from up wind or down wind but straight > from you. You will find it is best to aim straight at him, not up wind > or down wind. I think you will not be able to measure the angle accurately enough. Even for 60 MPH wind the angle is less than 5 degrees. And of course in such a wind your friend would not hear you (or even be able to hold a megaphone).... Theoretically for this case and transverse wind, you (the speaker) must point upwind by arcsin(windspeed/soundspeed); your friend (the receiver) must point directly at the speaker, not upwind or downwind. This is also the case for launching a motorboat across a river -- it must be aimed/steered upriver to arrive transversely at a point opposite from its starting point. With light, we find no evidence of any motion that must be compensated for, and a directional source always points directly at the receiver, and vice versa (assuming source and detector are at rest in some inertial frame). > It's not so easy to disprove the aether when you use what really > happens as opposed to poor analogies. "easy", no. But it has been done, for all known aether theories except those that happen to be experimentally indistinguishable from SR. By literally hundreds of different experiments [see the FAQ for references]. Tom Roberts tjroberts@lucent.com [snip] > Theoretically for this case and transverse wind, you (the speaker) must > point upwind by arcsin(windspeed/soundspeed); your friend (the receiver) [quoted text clipped - 7 lines] > and vice versa (assuming source and detector are at rest in some > inertial frame). I believe this is an important issue and one that I have been considering in relation to the laser experiments done by H. Webster Kehr. Tom, would you agree that if you have a stationary laser on the surface of the Earth pointing at a stationary target on the surface of the Earth, that irrespective of the motion of the Earth at approximately 370 kilometers per second linearly towards the constellation Leo, as evidenced by the CMBR, the laser beam would still be pointed directly at the target in order to hit it and the laser beam does not leave the laser at an angle? Thanks, Vern >[snip] > [quoted text clipped - 19 lines] >be pointed directly at the target in order to hit it and the laser beam >does not leave the laser at an angle? I believe Tom meant that the direction the source points does not have to allow for any "sideways" motion of the light during its journey. The laser would have to be aimed differently at different times of the year. It is pointed at where the receiver will be when the light arrives at the destination, not at where the receiver was when the light left the source. If the laser beam hits the target at exactly 90 degrees then it won't bounce off at an angle, assuming a perfectly smooth reflector, however the path of the laser beam will be at an angle to the instantaneous line between the source and destination at the time the light left the laser. http://www.teslaphysics.com/Chapters/Chapter070-SecularAberration.htm > >[snip] > > [quoted text clipped - 33 lines] > > http://www.teslaphysics.com/Chapters/Chapter070-SecularAberration.htm I see you have referenced Kehr's writings. What did you think of the paradox he points out? Is there a better term to use than "path momentum" to describe the paradox? It seems that Kehr is saying that because of aberration we can assume that both starlight and terrestial light move independently of the motion of their sources. This is also a postulate in relativity. However, if that is true then a laser experiment where the laser beam is at approximately 30-40 degrees n. latitude and shot at a target every hour for 24 hours should yield an eclipse pattern. Is does not, but rather always hits the same spot. Therefore, the laser must impart momentum to the beam meaning it leaves the laser at an angle and the speed of the beam must vary. That contradicts what Tom said. Vern >> The laser would have to be aimed differently at different times of the >> year. It is pointed at where the receiver will be when the light [quoted text clipped - 18 lines] >every hour for 24 hours should yield an eclipse pattern. Is does not, >but rather always hits the same spot. Which experiment was this, how, when and where was it done and where is it documented? Darkknight. > Therefore, the laser must impart >momentum to the beam meaning it leaves the laser at an angle and the >speed of the beam must vary. That contradicts what Tom said. > >Vern [snip] > >I see you have referenced Kehr's writings. What did you think of the > >paradox he points out? Is there a better term to use than "path [quoted text clipped - 10 lines] > Which experiment was this, how, when and where was it done and where > is it documented? You can download Kehr's "book" at the website below: http://www.teslaphysics.com/ He evidently worked for a large telecommunications company which was finding that there was interference in fiber-optic cables similar to the company that DeWitt worked for. Anyway, Kehr convinced his company to give him some money to do laser experiments, as I outlined above. The experiments he did are explained in detail in his "book." The interesting thing about the experiments is that they don't try to measure the speed of light, so the usual arguments about length contraction, time dilation, synchronizing clocks, LET vs. SR, etc. don't come into play. Instead they are based on Newtonian mechanics principles and whether light has ballistic properties. Using terrestial light sources, the experiment should have revealed the full effect of secular aberration, but instead it is consistent with a model of light which is a wave in an aether where the aether is like a bubble around the Earth and the Earth's rotation on its axis does not affect the aether. This corresponds with explanations for H-K type experiments and GPS, where an Earth-axis-centered frame is used for measurements. Vern > I believe this is an important issue and one that I have been > considering in relation to the laser experiments done by H. Webster [quoted text clipped - 5 lines] > be pointed directly at the target in order to hit it and the laser beam > does not leave the laser at an angle? In the frame of the earth surface at the location of the laser, the light leaves the laser straight down its centerline. Insofar as the non-inertial motions of the earth can be neglected during the flight time of the light, the image at the target will remain motionless, 24 hours a day 365 days a year (assuming true stability in mounting and the optical path). Indeed, since the non-inertial motions of the earth are so steady, they will be accounted for during setup, and it's really the variation in them that matters (e.g. the beating between rotational and orbital motion). That 370 km/s is roughly 0.001 times the speed of light, and the angle relative to the CMBR dipole=0 frame varies diurnally -- people would notice if light danced around by that milliradian as the earth turned: surveying over just 10 meters would be off by a cm! This does not happen. Tom Roberts tjroberts137@sbcglobal.net >> I believe this is an important issue and one that I have been >> considering in relation to the laser experiments done by H. Webster [quoted text clipped - 8 lines] >In the frame of the earth surface at the location of the laser, the >light leaves the laser straight down its centerline. Then why do telescopes have to be "tilted" to account for stellar aberration? i.e. the line of the telescope has to be at a very slight angle to the line of the light. >Insofar as the non-inertial motions of the earth can be neglected during >the flight time of the light, the image at the target will remain [quoted text clipped - 8 lines] >notice if light danced around by that milliradian as the earth turned: >surveying over just 10 meters would be off by a cm! This does not happen. I don't understand. If the light path of the laser is at right angles to the 400 km per second motion of the earth, then over a distance of 1 km, the laser takes 1/300000 seconds to reach the target, during which time the target has moved by (1/300000) * 400 km - which is .001333 km - equals 130 cm. (or 1cm over 10 metres as you say). If the light path of the laser is parallel to the direction of motion (which it could be after 8 hours) then the target hasn't moved so a difference of 1 cm would be seen on the "strike point" at the target. I'm sure you're right coz I know next to nothing about physics but why doesn't the "strike point" move? Darkknight. Dear darkknight: ... >>In the frame of the earth surface at the location of >>the laser, the light leaves the laser straight down [quoted text clipped - 4 lines] > telescope has to be at a very slight angle to the > line of the light. No. The telescope has to be *inline* with the light. http://en.wikipedia.org/wiki/Aberration_of_starlight ... just that "inline" varies. David A. Smith On Sun, 16 Apr 2006 16:29:21 -0700, "N:dlzc D:aol T:com \(dlzc\)" <N: dlzc1 D:cox T:net@nospam.com> wrote: >Dear darkknight: > [quoted text clipped - 11 lines] >http://en.wikipedia.org/wiki/Aberration_of_starlight >... just that "inline" varies. Thanks. I got confused by the analogy with a narrow bucket and rain, where the bucket has to be at an angle to the path of the rain for the raindrops to hit the bottom of the bucket and the angle changes according to the speed of the bucket. However, my understanding of stellar aberration is still zero and I have no clue how astronomers can even tell the position of a star has "changed" or what that means. I would have thought the "position" of a star was equivalent to the angle at which the light from the star strikes the earth - which is affected by the tilt of the earth and the time of day so how could you tell the position didn't change due to that. And apart from that, if the relative positions of the star and the earth are changing due to motion of the galaxies or the stars within the galaxies then that is going to cause the star's position to change all the time. I don't see how a change from 340 km/second to 400 km/second is going to be the cause of a change in position of the star - but then I don't know how you measure the position of a star in the first place. I'm supposed to be intelligent but I can hardly understand anything ... (anyway, no need to waste time trying to explain. I'm sure I can find a book on it somewhere - yep I did read the Wikipedia article (for the second time) but I still don't understand). Darkknight Dear darkknight: > On Sun, 16 Apr 2006 16:29:21 -0700, "N:dlzc D:aol T:com > \(dlzc\)" <N: [quoted text clipped - 31 lines] > the time of day so how could you tell the position > didn't change due to that. It is an effect *in addition to* that. That is how precisely you have to measure things. > And apart from that, if the relative positions of > the star and the earth are changing due to motion > of the galaxies or the stars within the galaxies > then that is going to cause the star's position to > change all the time. Yes, it does. > I don't see how a change from 340 km/second to > 400 km/second is going to be the cause of a > change in position ... "apparent position" ... > of the star - but then I don't know how you measure > the position of a star in the first place. Likely the same way as surveyors measure(d) the surface of the Earth (before GPS). Like how they notice very tiny changes in elevation of important markers due to the depletion of underground aquifers (subsidence). > I'm supposed to be intelligent but I can hardly > understand anything ... Patience. > (anyway, no need to waste time trying to explain. > I'm sure I can find a book on it somewhere - yep I > did read the Wikipedia article (for the second > time) but I still don't understand). The answer will come when you are ready for it. Good luck. David A. Smith > On Sun, 16 Apr 2006 16:29:21 -0700, "N:dlzc D:aol T:com \(dlzc\)" <N: > dlzc1 D:cox T:net@nospam.com> wrote: [quoted text clipped - 39 lines] > > Darkknight The poor relativists are now restricted to aether and Maxwell but cannot help themselves and run all the way back to Newton insofar as there is a big feedback loop between the exotic 1905 concept and Newtonian quasi-geocentricity. A bogus history was created to make way for relativity and it worked up to recently ,once you see that there is no way to associate 'absolute space' with aether the show is over,at least as far as relativity overthrowing Newton is concerned .It turns out to be a minor mathematical squabble with little content and character. I was genuinely surprised to see what the guys here did with the information and especially as there is nothing to keep the early 20th century concepts going as working principles.In any case,you will find that the dilemma of the guys in the mid 19th century * was that Newton had left them with an astronomical framework which required no aether - "The fictitious matter which is imagined as filling the whole of space is of no use for explaining the phenomena of Nature, since the motions of the planets and comets are better explained without it, by means of gravity; and it has never yet been explained how this matter accounts for gravity. The only thing which matter of this sort could do, would be to interfere with and slow down the motions of those large celestial bodies, and weaken the order of Nature; and in the microscopic pores of bodies, it would put a stop to the vibrations of their parts which their heat and all their active force consists in. Further, since matter of this sort is not only completely useless, but would actually interfere with the operations of Nature, and [314] weaken them, there is no solid reason why we should believe in any such matter at all. Consequently, it is to be utterly rejected." Newton Optics 1704 * http://www.bodley.ox.ac.uk/cgi-bin/ilej/image1.pl?item=page&seq=9&siz... In short,if you want to waste your time by having guys bloke smoke where the sun don't shine then welcome to sci.physics !. >>Insofar as the non-inertial motions of the earth can be neglected during >>the flight time of the light, the image at the target will remain [quoted text clipped - 23 lines] >I'm sure you're right coz I know next to nothing about physics but why >doesn't the "strike point" move? Well, upon further thought, I guess the explanation could be that the motion of the source affects the angle of the "laser beam" so that it has a "sideways" component the equivalent of the 1 cm movement of the target. This would allow the laser beam to leave the laser "straight down its centre line" regardless of the movement of the laser. If this is true, this should mean that if a laser beam on a rocket ship is fired sideways so that the path of the laser beam is exactly at right angles to the path of the rocket, if the rocket increases speed by 400 km/second, the path of the laser beam will no longer be exactly at 90 degrees to the path of the rocket, and will not be parallel to the path of the laser beam before the rocket increased speed. Also, the change in velocity of the earth from 340 km/sec to 400 km/sec should mean clocks run at different speeds at different times of the year. Darkknight. > >>Insofar as the non-inertial motions of the earth can be neglected during > >>the flight time of the light, the image at the target will remain [quoted text clipped - 29 lines] > target. This would allow the laser beam to leave the laser "straight > down its centre line" regardless of the movement of the laser. Almost correct: the speed of light remains constant, so that the components don't add up vectorially. But it's even more straightforward: the light inside the laser follows a certain angle to start with (90 degrees in the rocket's rest frame, but not in the stationary frame), and the light doesn't change angle when it leaves the laser. > If this is true, this should mean that if a laser beam on a rocket > ship is fired sideways You mean relative to the rocket > that the path of the laser beam is exactly > at right angles to the path of the rocket, You mean relative to the CMBR frame I suppose... > if the rocket increases > speed by 400 km/second, the path of the laser beam will no longer be > exactly at 90 degrees to the path of the rocket, and will not be > parallel to the path of the laser beam before the rocket increased > speed. That's correct - in the CMBR frame. > Also, the change in velocity of the earth from 340 km/sec to 400 > km/sec should mean clocks run at different speeds at different times > of the year. That's correct as well - in the CMBR frame. Of course, "in" the solar frame they run (for all practical purposes) at constant speed. Harald > > >>Insofar as the non-inertial motions of the earth can be neglected during > > >>the flight time of the light, the image at the target will remain [quoted text clipped - 9 lines] > > >>surveying over just 10 meters would be off by a cm! This does not > happen. [snip] > > Well, upon further thought, I guess the explanation could be that the > > motion of the source affects the angle of the "laser beam" so that it [quoted text clipped - 7 lines] > rocket's rest frame, but not in the stationary frame), and the light doesn't > change angle when it leaves the laser. The posts above were not about the speed of light or about rockets. The scenario was a laser experiment on the surface of the Earth in which the laser is positioned perpendicular to the essentially linear movement of the solar system through the galaxy at around 40 degrees latitude and approximately 300 feet between source and target. A laser pulse is sent every hour for 24 hours and the strikes marked on the target. Since the target moves approximately four inches during the transit time of the laser beam after the pulse is fired and before it strikes the target, the plot of the strikes on the target over the 24 hour period should be an ellipse with the center representing where the laser is actually aimed. The result of the experiment is that the strikes are always in the same place instead producing the elliptical pattern. Vern > > > >>Insofar as the non-inertial motions of the earth can be neglected during > > > >>the flight time of the light, the image at the target will remain [quoted text clipped - 29 lines] > movement of the solar system through the galaxy at around 40 degrees > latitude and approximately 300 feet between source and target. You may here above replace "rocket" by "Earth", that doesn't matter for SRT. What matters is that one may choose (pretend to be) any approx. inertial frame (solar, CMBR) as "stationary" frame. Note that for MMX the Earth's surface isn't approximately an inertial frame, except for very short time intervals. > A laser > pulse is sent every hour for 24 hours and the strikes marked on the [quoted text clipped - 5 lines] > strikes are always in the same place instead producing the elliptical > pattern. I've worked out such things in the past, and it always worked out perfectly. Here you give insufficient detail, thus I don't know what you overlook - but probably you overlooked the Lorentz contraction. Harald > <vern@bealenet.com> wrote in message [snip] > > The posts above were not about the speed of light or about rockets. > > The scenario was a laser experiment on the surface of the Earth in [quoted text clipped - 21 lines] > Here you give insufficient detail, thus I don't know what you overlook - but > probably you overlooked the Lorentz contraction. Perhaps I should not have posted the subject matter is this thread as the experiment I am discussing is a simple path of light experiment that does not involve MMX concepts, SRT, length contraction, non-inertial frames or choice of frames. However, I was responding to Tom's statements in Message No. 18 which were applicable to path of light issues, so it seemed appropriate. The issue here is whether there is aberration of terrestial light. The experiment I outlined indicates that there is not, yet that contradicts accepted theory of light (there is no reason that all light is not aberrated). I have asked Tom for his explanation of why terrestial light is not aberrated in Message No. 38. What is yours? If you need more details of the experiments, a link is posted in Message No. 21 or 23. Vern > > <vern@bealenet.com> wrote in message > [quoted text clipped - 30 lines] > that does not involve MMX concepts, SRT, length contraction, > non-inertial frames or choice of frames. If you think about it, that's impossible: any path is relative to some material reference system, even if by extension. > However, I was responding to > Tom's statements in Message No. 18 which were applicable to path of [quoted text clipped - 4 lines] > asked Tom for his explanation of why terrestial light is not aberrated > in Message No. 38. What is yours? I kind of commented on that in message 31; and on earth there is no speed difference. > If you need more details of the > experiments, a link is posted in Message No. 21 or 23. Reading that, I could not spot what you consider to be a problem; however the following remark of you is revealing: "Using terrestial light sources, the experiment should have revealed the full effect of secular aberration, but instead it is consistent with a model of light which is a wave in an aether where the aether is like a bubble around the Earth and the Earth's rotation on its axis does not affect the aether. This corresponds with explanations for H-K type experiments and GPS, where an Earth-axis-centered frame is used for measurements. " That's commonly called ECI frame; as it happens, SRT is compatible with choosing that frame; it is generally used. Harald > <vern@bealenet.com> wrote in message [snip] > > Perhaps I should not have posted the subject matter is this thread as > > the experiment I am discussing is a simple path of light experiment [quoted text clipped - 3 lines] > If you think about it, that's impossible: any path is relative to some > material reference system, even if by extension. What I said was that the experiment I am discussing does not involve a consideration of choice of frames. I guess what I should have said was that it doesn't involve a transformation between frames. All considerations are in a frame of reference in which the CMBR is isotropic (the CMBR frame). > > However, I was responding to > > Tom's statements in Message No. 18 which were applicable to path of [quoted text clipped - 7 lines] > I kind of commented on that in message 31; and on earth there is no speed > difference. This is not about speed difference. It is about whether terrestial light sources have aberration. > > If you need more details of the > > experiments, a link is posted in Message No. 21 or 23. [quoted text clipped - 12 lines] > That's commonly called ECI frame; as it happens, SRT is compatible with > choosing that frame; it is generally used. We are evidently talking past each other. This has nothing to do with SRT and the frame in the experiment is the not the ECI frame, it is in the frame of the Earth's surface. The comment I made in the previous post was only an analogy to HK and GPS and aether theories which use an ECI frame. That was only relative to a positing a reason why terrestial light does not have aberration. Do you accept that terrestial light does not have aberration (in the reference frame of the Earth's surface), and if so, why do you believe there is no aberration, when all light should exibit aberration. Vern [snip] > We are evidently talking past each other. This has nothing to do with > SRT and the frame in the experiment is the not the ECI frame, it is in [quoted text clipped - 5 lines] > the Earth's surface), and if so, why do you believe there is no > aberration, when all light should exibit aberration. Sorry to have to respond to my own post, but obviously the above paragraph of Message No. 36 is in error as I was in a hurry and I have taken some time to try to think the experiment through from the standpoint of reference frames. The experiment is based on the assumption that when a laser pulse is in the air between the source and target, the target continues to move along with the Earth in the essentially linear motion of the Solar System towards the constellation Leo. The target will have moved about 4 inches before the pulse hits it. So the reference frame for the whole experiment is the CMBR frame. Does the target move in the reference frame of the Earth's surface in the time the pulse is in the air? I guess the answer is "no" because both the source and target are stationary during the length of time starting when the pulse is fired from the source and ending when the pulse hits the target in the reference frame of the Earth's surface. But that can't be used as a reason that there is no aberration of terrestial light, since in the reference frame of the CMBR, the Earth is moving and that resulting aberration of terrestial light should be evident in any experiment using light on the surface of the Earth. Vern > [snip] > [quoted text clipped - 12 lines] > taken some time to try to think the experiment through from the > standpoint of reference frames. That's great, I could have saved myself time by immediately looking at this posting of yours. > The experiment is based on the > assumption that when a laser pulse is in the air between the source and > target, the target continues to move along with the Earth in the > essentially linear motion of the Solar System towards the constellation > Leo. The target will have moved about 4 inches before the pulse hits > it. So the reference frame for the whole experiment is the CMBR frame. OK > Does the target move in the reference frame of the Earth's surface in > the time the pulse is in the air? I guess the answer is "no" because > both the source and target are stationary during the length of time > starting when the pulse is fired from the source and ending when the > pulse hits the target in the reference frame of the Earth's surface. Right - apart of a negligibly small rotation. > But that can't be used as a reason that there is no aberration of > terrestial light, since in the reference frame of the CMBR, the Earth > is moving and that resulting aberration of terrestial light should be > evident in any experiment using light on the surface of the Earth. What do you think needs to abberate, and why? What makes you think that the light should change its course when it exists the laser? (Note how the path inside the laser is when mapped to the CMBR frame). Harald > <vern@bealenet.com> wrote in message [snip] > > The experiment is based on the > > assumption that when a laser pulse is in the air between the source and [quoted text clipped - 21 lines] > light should change its course when it exists the laser? (Note how the path > inside the laser is when mapped to the CMBR frame). I think this is more of an issue and question about absolute motion. If you take as a fact that the measurements using the CMBR establish that the Earth is moving essentially linearly at approximately 370 km/s towards the constellation Leo, then a consequence of that fact is that when a laser beam is shot from a laser on the surface of the Earth to a target also on the surface of the Earth such that the laser and target are aligned perpendicularly to the direction towards Leo and the distance between the laser and the target is approximately 300 feet, then from the time the beam leaves the laser the target will have moved approximately four feet in the direction towards Leo (obviously, the laser would have moved too, but that's immaterial since what happens to the laser after the beam leaves it has no effect on where the beams goes or what happens to the target). I have made no statement about the light changing course when it exits the laser. That involves whether the movement of the source imparts momentum to the beam. Tom Roberts has already said that it does not and the beam does not leave at an angle. But that does not have a bearing on whether the target moves during the transit time of the beam. Whether the target moves or not is a statement about absolute motion. If it is accepted that the Earth is moving as stated above, then it doesn't matter that motion is not apparent if electing to use an Earth-surface reference frame. Your choice of frames does not stop the Earth from moving. Therefore, the experiment outlined above has to account for that movement in determining the outcome of the experiment. That's where aberration (or perhaps you could say a lack of aberration) comes into play. Since the movement of the Earth towards Leo does not have any effect on the beam, the beam would be expected to strike the target approximately four inches behind (opposite the direction towards Leo) where it was aimed because that's how far the target moved towards Leo in the transit time of the beam. Arguments concerning not knowing where the laser was originally aimed are negated by shooting a beam every hour over a 24 hour period. An ellipse pattern should be evident on the target with the center of the ellipse being where the laser was aimed. Vern > > <vern@bealenet.com> wrote in message > [quoted text clipped - 27 lines] > > I think this is more of an issue and question about absolute motion. That's opening the Debate Box. To avoid wasting time on that, you can simply discuss how it looks as measured in the CMBR frame, and everyone who knows SRT will agree, while you can draw your own conclusions. > If you take as a fact that the measurements using the CMBR establish "establish"? See above. > that the Earth is moving essentially linearly at approximately 370 km/s > towards the constellation Leo, then a consequence of that fact is that [quoted text clipped - 7 lines] > to the laser after the beam leaves it has no effect on where the beams > goes or what happens to the target). What matters is the light's direction inside the moving laser: that determines "where the beam goes". I have the impression that you still didn't get that. How do you think it works with a bullet in a moving rifle? > I have made no statement about > the light changing course when it exits the laser. That involves > whether the movement of the source imparts momentum to the beam. Tom > Roberts has already said that it does not and the beam does not leave > at an angle. ??? Quite to the contrary - thus my hunch was correct. Please work out how a bullet moves when you replace the laser by a rifle, and you should understand. Harald > But that does not have a bearing on whether the target > moves during the transit time of the beam. Whether the target moves or [quoted text clipped - 15 lines] > > Vern > <vern@bealenet.com> wrote in message [snip] > > I think this is more of an issue and question about absolute motion. > > That's opening the Debate Box. To avoid wasting time on that, you can simply > discuss how it looks as measured in the CMBR frame, and everyone who knows > SRT will agree, while you can draw your own conclusions. The speed of the Earth towards the constellation Leo is not fast enough to have to worry about relativistic corrections to a Galilean transformation, so I believe the experiment being discussed can be viewed within a classical physics framework. However, all your persistent about consideration of the direction of travel of the beam in the CMBR frame has paid off, I think. When I read the experiment, it bothered me that he kept stating that the target moved approx. four inches in the time the beam was in the air, so therefore the beam should have struck the target four inches behind where it was supposed to. Finally, I think you have made me realize his fundamental error. In the reference frame of the surface of the Earth, the target does not move four inches in the transit time of the beam, therefore one would expect the beam to strike the target where it is supposed to. In the frame of the CMBR, because changing frames can't change the outcome of an experiment, naturally the beam is still going to strike the target in the same place. The only way for that to happen is for the beam to appear to be travelling at an angle in that frame so that the four inches that the target moves in the transit time is accounted for. Evidently, Kehr assumed that because the measurements of the Earth's motion wrt to the CMBR establish that the Earth is moving through the cosmos at approx. 370 km/s that somehow that motion had to be accounted for in experiments on the surface of the Earth. It's kind of like, "I throw the ball straight up high enough and it won't land where I'm standing because the Earth will have moved through the cosmos X number of feet in the time it's in the air." Given that reasoning, I don't think any of the laws of physics would work. If what I am saying above is correct, then I'll have to thank you and Tom for helping me see it. If it's not then maybe you can continue to steer me in the right direction. Vern > > <vern@bealenet.com> wrote in message > [quoted text clipped - 13 lines] > considerations are in a frame of reference in which the CMBR is > isotropic (the CMBR frame). Well, then you're going straight for MMX concepts! MMX was first of all considered relative to a frame in which light was thought to be truly isotropic; length contraction was one explanation, and it led to SRT. > > > However, I was responding to > > > Tom's statements in Message No. 18 which were applicable to path of [quoted text clipped - 10 lines] > This is not about speed difference. It is about whether terrestial > light sources have aberration. Without Lorentz contraction they would have aberration, and an "absolute speed" could be determined. > > > If you need more details of the > > > experiments, a link is posted in Message No. 21 or 23. [quoted text clipped - 22 lines] > the Earth's surface), and if so, why do you believe there is no > aberration, when all light should exibit aberration. "all light should exhibit aberration" is overly vague. And it has everything to do with SRT: the main point is that all inertial frames are equally valid for descriptions of light propagation. Just describe light propagation of MMX in the CMBR frame and you'll find that, assuming Lorentz contraction, no aberration will be noticeable for observers on Earth (important detail: the mirrors are not exactly at 45 degrees angle). Cheers, Harald >> >If the light path of the laser is parallel to the direction of motion >> >(which it could be after 8 hours) then the target hasn't moved so a [quoted text clipped - 14 lines] >rocket's rest frame, but not in the stationary frame), and the light doesn't >change angle when it leaves the laser. You seem to have mixed up the two scenarios, the laser on the surface of the earth aiming at a target and the rocket ship with a laser on board, so I can't tell what you're trying to say. I realise the light doesn't change its angle "in-flight", but if it didn't keep exact pace with the laser gun, it would be at risk of striking the inside of the "laser gun" instead of going "straight down the centre line" of the laser. So for the laser on the rocket ship, the laser would have to be aimed slightly off 90 degrees to get the path of the laser beam to be at exactly 90 degrees to the path of the rocket. >>>>>>>>>> quote from my previous post <<<<<<<<<<<< Well, upon further thought, I guess the explanation could be that the motion of the source affects the angle of the "laser beam" so that it has a "sideways" component the equivalent of the 1 cm movement of the target. This would allow the laser beam to leave the laser "straight down its centre line" regardless of the movement of the laser. If this is true, this should mean that if a laser beam on a rocket ship is fired sideways so that the path of the laser beam is exactly at right angles to the path of the rocket, if the rocket increases speed by 400 km/second, the path of the laser beam will no longer be exactly at 90 degrees to the path of the rocket, and will not be parallel to the path of the laser beam before the rocket increased speed. >>>>>>>>>> end quote <<<<<<<<<<<<<<< >> If this is true, this should mean that if a laser beam on a rocket >> ship is fired sideways [quoted text clipped - 5 lines] > >You mean relative to the CMBR frame I suppose... I don't know what you mean by "relative to the CMBR frame". What I thought I meant was that the path of the laser beam and the rocket ship are at right angles in "absolute space" but I don't know if that's meaningful since determining the "path of the laser beam" is somewhat difficult in practice. Perhaps my "absolute space" is your "CMBR frame". I didn't realise you had to define a frame to measure an angle. >> if the rocket increases >> speed by 400 km/second, the path of the laser beam will no longer be [quoted text clipped - 3 lines] > >That's correct - in the CMBR frame. Wouldn't the angle between the "two laser beams" be slightly different in any frame. I guess you're telling me that if there were some means of comparing the angle of the laser beam on the rocket ship itself before and after the speed change, no change in angle would be seen - hmmm, yep, I was thinking of absolute space, not counting the expansion of space itself. >> Also, the change in velocity of the earth from 340 km/sec to 400 >> km/sec should mean clocks run at different speeds at different times >> of the year. > >That's correct as well - in the CMBR frame. Of course, "in" the solar frame >they run (for all practical purposes) at constant speed. I was thinking that the daily rotation of the earth would take a slightly different time at different times of the year (assuming it wasn't slowing down at all) - but perhaps the rotation speed of the earth and all events on it also change by the same amount, in the CMBR frame, so the net effect is zero and there's no way to tell that it's happening. Thanks for your reply. Darkknight >>> >If the light path of the laser is parallel to the direction of motion >>> >(which it could be after 8 hours) then the target hasn't moved so a [quoted text clipped - 26 lines] >path of the laser beam to be at exactly 90 degrees to the path of the >rocket. "Don't add up vectorially". Hmmm. Now I see what you mean. The change in the velocity of the rocket causes the angle/ "flight path" of the laser beam to change so that it still goes down the centre line of the laser, rather than the laser beam "keeping pace with the rocket". I wonder why the change in angle exactly matches the change in speed - a happy coincidence of nature. (I wonder if Michelson/Morley knew this). Darkknight. > >>> >If the light path of the laser is parallel to the direction of motion > >>> >(which it could be after 8 hours) then the target hasn't moved so a [quoted text clipped - 34 lines] > in speed - a happy coincidence of nature. (I wonder if > Michelson/Morley knew this). The light in the laser follows the laser geometry, by necessity. M&M didn't have lasers at that time, and some papers were written later about possible effects of angle changes from mirrors. It was shown that such effects can't explain the null result. Of course, everything works out taking into account Lorentz contraction. Harald > >> >If the light path of the laser is parallel to the direction of motion > >> >(which it could be after 8 hours) then the target hasn't moved so a [quoted text clipped - 18 lines] > of the earth aiming at a target and the rocket ship with a laser on > board, so I can't tell what you're trying to say. I didn't see two scenario's but what I wrote there is valid in general, thus for all scenarios. > I realise the light doesn't change its angle "in-flight", but if it > didn't keep exact pace with the laser gun, it would be at risk of [quoted text clipped - 21 lines] > > >>>>>>>>>> end quote <<<<<<<<<<<<<<< Why do you quote again what I quoted already? Now your quotes are double... > >> If this is true, this should mean that if a laser beam on a rocket > >> ship is fired sideways [quoted text clipped - 7 lines] > > I don't know what you mean by "relative to the CMBR frame". I referred to your earlier post as apparently you were referring to it: ">> the motion of the Earth at >> approximately 370 kilometers per second linearly towards the >> constellation Leo, as evidenced by the CMBR" > What I thought I meant was that the path of the laser beam and the > rocket ship are at right angles in "absolute space" but I don't know > if that's meaningful since determining the "path of the laser beam" is > somewhat difficult in practice. Perhaps my "absolute space" is your > "CMBR frame". I didn't realise you had to define a frame to measure > an angle. Yes you have to - it's even crucial for understanding. > >> if the rocket increases > >> speed by 400 km/second, the path of the laser beam will no longer be [quoted text clipped - 24 lines] > frame, so the net effect is zero and there's no way to tell that it's > happening. Indeed, all effects compensate each other - that's the essence of the Lorentz transformations. > Thanks for your reply. You're welcome. Harald > >>Insofar as the non-inertial motions of the earth can be neglected during > >>the flight time of the light, the image at the target will remain [quoted text clipped - 43 lines] > > Darkknight. You seem like a genuine guy but you will never get it,the original Roemerian insight on finite light speed on appearance was mixed together with Keplerian orbital geometry in a complex way by Newton. Nobody really cares what Newton did even though it is the work of a guy who could bend information to suit his purpose without the slightest trace of remorse for the destruction caused to the work of Copernicus,Kepler and Roemer.I can spare you an existence built on empty working principles and direct you towards what Newton actually did and why it is now important to restore the original productive working principles but I sup[pose you want to remain sounding profound without actually being so. For an astronomer,an intuitive astronomer as I am,the following passage by Newton is remarkable for the asmount of astronomical principles he breaks within a single paragraph.Pity you do not have the highest human intutive faculty to recognise this incredible feat of intellectual and intutive vandalism, a sort of anti-genius - PHENOMENON V. "Then the primary planets, by radii drawn to the earth, describe areas no wise proportional to the times; but that the areas which they describe by radii drawn to the sun are proportional to the times of description. For to the earth they appear sometimes direct, sometimes stationary, nay, and sometimes retrograde. But from the sun they are always seen direct, and to proceed with a motion nearly uniform, that is to say, a little swifter in the perihelion and a little slower in the aphelion distances, so as to maintain an equality in the description of the areas. This a noted proposition among astronomers, and particularly demonstrable in Jupiter, from the eclipses of his satellites; by the help of which eclipses, as we have said, the heliocentric longitudes of that planet, and its distances from the sun, are determined." NEWTON >> In the frame of the earth surface at the location of the laser, the >> light leaves the laser straight down its centerline. > > Then why do telescopes have to be "tilted" to account for stellar > aberration? i.e. the line of the telescope has to be at a very slight > angle to the line of the light. Because the telescope only sees light that travels down its centerline, and in the locally-inertial frame of the telescope the light does not travel directly along the line connecting source and telescope (locations simultaneous in that frame). That _is_ what aberration is. >> [a laser and target rigidly mounted to the surface of the earth] >> Insofar as the non-inertial motions of the earth can be neglected [quoted text clipped - 17 lines] > target has moved by (1/300000) * 400 km - which is .001333 km - > equals 130 cm. (or 1cm over 10 meters as you say). You are assuming that light only travels at c relative to the CMBR dipole=0 frame. This is not so -- observations show that light travels isotropically with speed c in any inertial frame the earth's surface inhabits (and well-tested theories extend this to all inertial frames). Consider the instantaneously-comoving inertial frame of the laser at the instant a short pulse is emitted. During the travel time of the light pulse the target will move relative to that frame due to the noninertial motions of the earth. If those noninertial motions are smaller than the resolution of measuring the image position at the target, the image at the target will be measured to remain motionless. For a 10 meter path laser-to-target, light takes ~33 ns to travel between them. For the ~500 m/s rotation of the earth, the total motion of the target is ~16 microns, but the non-inertial component of that rotation is 10m/Rearth to first order, and is thus unmeasurable. For the ~30 km/s orbital speed of the earth, the total motion is ~1 mm, but the non-inertial component of that rotation is 10m/Rorbit to first order, and is thus unmeasurable. For the 300 km/s motion relative to the CMBR dipole=0 frame, there is no known non-inertial component (if there is, it is most likely something like 10m/Rgalaxy...). Note that variations in the optical properties of the air will be _far_ larger than these values. > I'm sure you're right coz I know next to nothing about physics but why > doesn't the "strike point" move? It does, but only by amounts so much smaller than measurement resolutions that it is unobservable. Tom Roberts tjroberts@lucent.com > >> In the frame of the earth surface at the location of the laser, the > >> light leaves the laser straight down its centerline. [quoted text clipped - 7 lines] >travel directly along the line connecting source and telescope >(locations simultaneous in that frame). That _is_ what aberration is. ok, well after making the above statement, I subsequently learnt from Harry that angles are relative to a frame, so I was thinking of absolute space (which could be called the CMBR frame), rather than the earth's inertial frame, when I made the above statement, so when David Smith told me the telescope was not tilted relative to the line of the light I got even more lost (I guess he was talking about the earth's inertial frame). So now my understanding is this 1. In the CMBR frame, the telescope is at an angle to the line of the light (which is what I meant when I said it was tilted). 2. In the earth's frame, I guess the telescope and the light are parallel (because you said "the telescope only sees light that travels down its centerline") I find number 2 confusing because I keep thinking the telescope is moving so the light will collide with the inside of the telescope, then I remind myself the "parallel-ness" of the light and the telescope is an illusion in the earth's frame. > >> [a laser and target rigidly mounted to the surface of the earth] > >> Insofar as the non-inertial motions of the earth can be neglected [quoted text clipped - 22 lines] >isotropically with speed c in any inertial frame the earth's surface >inhabits (and well-tested theories extend this to all inertial frames). I'm lost as to how you can draw the conclusion that I thought the speed of light was c only in the CMBR frame. I'm happy to accept that it is c in all frames (but I want to learn more about it). However I've since learned (from Harry) that a change in speed of the laser source affects the angle of the laser beam (relative to the CMBR frame, say) so the laser always hits the target in the same spot, even though the distance the target moves during the travel time of the light, varies with the change in speed of the earth relative to the CMBR frame. >Consider the instantaneously-comoving inertial frame of the laser at the >instant a short pulse is emitted. During the travel time of the light [quoted text clipped - 21 lines] >It does, but only by amounts so much smaller than measurement >resolutions that it is unobservable. I'm not sure I understand but I'm thinking that what you're saying is that because the speed of the earth is changing all the time, including during the travel time of the laser beam, this makes it a non-inertial frame - so the target is moving at a very slightly (and unmeasurably) different speed when the laser photon hits it, than it was when the photon left the laser. Ok, I didn't even think of that. Thanks! (a lot). Darkknight > > I believe this is an important issue and one that I have been > > considering in relation to the laser experiments done by H. Webster [quoted text clipped - 21 lines] > notice if light danced around by that milliradian as the earth turned: > surveying over just 10 meters would be off by a cm! This does not happen. Yes, but according to your statement that the speed of the source laser (anywhere from 340 to 400 km/s, depending on orientation) does not affect the path of the beam, there should be that "dancing around" as you put it, because the target does move in the time it takes the laser to reach it. How do you account for the fact that "This does not happen." Vern >> That 370 km/s is roughly 0.001 times the speed of light, and the >> angle [quoted text clipped - 12 lines] > to reach it. How do you account for the fact that "This does not > happen." I just posted a response to darkknight that estimates how large the "dancing" actually is for that 10 m path. Values are completely unobservable. As surveyors around the world regularly observe. I of course used SR in those estimates, not his model in which the CMBR dipole=0 frame is special. > An ellipse pattern should be evident on the > target with the center of the ellipse being where the laser was aimed. Yes. But for the above 10 meter path, the size of that ellipse is less than an Angstrom, and is completely unobservable. Tom Roberts tjroberts@lucent.com > > Tom Roberts wrote: > >> That 370 km/s is roughly 0.001 times the speed of light, and the [quoted text clipped - 17 lines] > "dancing" actually is for that 10 m path. Values are completely > unobservable. As surveyors around the world regularly observe. I believe you based your estimate on the non-inertial component, which was not considered in the experiment because the value (as you show) is not significant. In Message No. 41 to Harry I outline what I see as the experimenter's error. He assumed that because it is established by the CMBR that the Earth is moving essentially linearly (in the time it takes for a laser beam to reach a target 300 feet away) through the Cosmos at approx. 370 km/s that that motion must be considered in any experiment on the Earth's surface. In other words, he assumed that in the transit time between the laser beam being fired and striking the target, the target would have moved approx. four inches. Apparently, he didn't realize that that's only true in the CMBR frame, not in the frame of the Earth's surface, as there is no relative motion between the laser and the target in the frame of the Earth's surface. Therefore, it appears that the whole basis for his assumption that plotting strikes every hour would result in an elliptical pattern is flawed. In the frame of the CMBR, however, the target would move approx. four inches in the transit time of the beam and an observer in that frame, I guess, would notice that the beam would seem to take an angled path to "catch up" to the target. If the above analysis is wrong, I hope you'll correct me. Vern >> Take a friend and a couple of megaphones out to a field on a windy day. >> Use your megaphone to shout to your friend while he uses his to [quoted text clipped - 28 lines] > >Tom Roberts tjroberts@lucent.com The MMX experiment is analyzed over and over with no conclusion, and the fact that it was the MMX that provoked relativity should make it easy or even mandagtroy to use relativity to explain it, but relativity is never used! Let me explain. In general relativity there is propounded a space defined by (ict x y z). It's a hyperbolic space. Using this set you can solve the MMX null problem right away. How? Draw a capital T as a vector diagram. The upleg is c. The right branch is +V. The left branch is -V. The vector sum is the same for upstream or downstream. The designation ict means that c is at right angles to any velocity you can think of. And to sum c and V you must use the T-diagram. You can see there never would be a difference in path lengths or times. You might object that no one uses ict. That's true. Here's why: They got cute and decreed that time was just the same as any of xyz, a sort of unisex monstrosity. (This is the only way you can have a totally bogus "space-time continuum"). The wanted to use (ct x y z). They did this by algebraically impossible step of making the "metric tensor" have the "signature" -1 1 1 1 , meaning those were the elements of the diagonal. But such a matrix cannot be a tensor for a number of reasons but you will appreciate that if I pass (ct x y z) through it once I get (-ict x y z) and if I do it again I get (ct x y z) back again. You can read about it in MTWheeler's Gravitation, the telephone book bible of relativity. There doesn't seem to be one scintilla or guilt or even awareness of thes monstrosity. John Polasek http://www.dualspace.net > The MMX experiment is analyzed over and over with no conclusion, and > the fact that it was the MMX that provoked relativity should make it > easy or even mandagtroy to use relativity to explain it, but > relativity is never used! The SR explanation of the MMX result is simple: The center of the apparatus is at rest on the surface of the earth, and to sufficient accuracy it can be considered to be at rest in an inertial frame[#]. Also to sufficient accuracy the rotation of the interferometer can be ignored[#], and it can be considered to have a series of definite orientations. [#] Non-inertial motions during the transit time of the light are quite small, and to first order are parallel to the mirrors. Motions parallel to mirrors have no effect on the reflected light. Higher-order effects due to these rotations are completely negligible. In the inertial frame of the interferometer, the speed of light is isotropically c, so the position of the fringes does not depend on the orientation of the interferometer. Hence a null result is expected, which is fully consistent with their measurement. BTW the MMX did not really "provoke" relativity. In later years, Einstein stated he did not remember if he was aware of it in 1905 or not. But he _definitely_ knew of several other experiments that refuted the simple aether theories. And the MMX was the most definitive of the early experiments. > [... bogus posturing omitted] Tom Roberts tjroberts@lucent.com > > The MMX experiment is analyzed over and over with no conclusion, and > > the fact that it was the MMX that provoked relativity should make it [quoted text clipped - 19 lines] >orientation of the interferometer. Hence a null result is expected, >which is fully consistent with their measurement. Tautologically so! Even I can see if the speed of light is isotropically c, then, yes, that's what it is, c, and we can expect no fringes. Your standards for a proof are not as high as mine. >BTW the MMX did not really "provoke" relativity. In later years, >Einstein stated he did not remember if he was aware of it in 1905 or [quoted text clipped - 5 lines] > >Tom Roberts tjroberts@lucent.com I'll restore what you label as "bogus posturing". It is more important than your fatuous and insipid explanation of the MMX above. In Misner Thorne Wheeler Gravitation pg 51 they were glad to get rid of ict in favor of ct by pushing it through the "metric" matrix that will never be a tensor. And the trick only works as a scalar product XGX'. By rights, it has to pass the simple transformation XG = (ct x y z)--> (-ct x y z) which is so wrong, that it is obvious they have suborned G in order to join what cannot be joined, time and space. This is grave algebraic malfeasance even on a sophomore level. It covers up what was known before, that with ict, time and space can never mix, like water and oil. Time is separate entirely. And it turns out that all of relativity can be deduced (only correctly) in a Euclidean space. My T-diagram is an example that immediately disposes of the MMX difficulty without dreary references to inertial frames and the hypothesis that "the speed of light is isotropically c". What is there that makes c constant? And how about Shapiro. Does relativity allow variations in c or is it all taken up in time dilation? Of course there's no ether, but there's something better. I brought this up before, but you did not comment. John Polasek http://www.dualspace.net > > The MMX experiment is analyzed over and over with no conclusion, and > > the fact that it was the MMX that provoked relativity should make it [quoted text clipped - 29 lines] > >Tom Roberts tjroberts@lucent.com Tautologically so! Even I can see if the speed of light is isotropically c, then, yes, that's what it is, c, and we can expect no fringes. Your standards for a proof are not as high as mine. >BTW the MMX did not really "provoke" relativity. In later years, >Einstein stated he did not remember if he was aware of it in 1905 or [quoted text clipped - 5 lines] > >Tom Roberts tjroberts@lucent.com (This message was lost somewhere so I repeat) I'll restore what you label as "bogus posturing". It is more important than your fatuous and insipid explanation of the MMX above. In Misner Thorne Wheeler Gravitation pg 51 they were glad to get rid of ict in favor of ct by pushing it through the "metric" matrix that will never be a tensor. And the trick only works as a scalar product XGX'. By rights, it has to pass the simple transformation XG = (ct x y z)--> (-ct x y z) which is so wrong, that it is obvious they have suborned G in order to join what cannot be joined, time and space. This is grave algebraic malfeasance even on a sophomore level. It covers up what was known before, that with ict, time and space can never mix, like water and oil. Time is separate entirely. And it turns out that all of relativity can be deduced (only correctly) in a Euclidean space. My T-diagram is an example that immediately disposes of the MMX difficulty without dreary references to inertial frames and the hypothesis that "the speed of light is isotropically c". What is there that makes c constant? And how about Shapiro. Does relativity allow variations in c or is it all taken up in time dilation? Of course there's no ether, but there's something better. I brought this up before, but you did not comment. John Polasek | > > The MMX experiment is analyzed over and over with no conclusion, Nonsense. There is a simple conclusion. That Humpty Roberts doesn't accept it is his problem. http://www.androcles01.pwp.blueyonder.co.uk/Smart/MMXwind.gif Androcles. and | > > the fact that it was the MMX that provoked relativity should make it | > > easy or even mandagtroy to use relativity to explain it, but [quoted text clipped - 69 lines] | I brought this up before, but you did not comment. | John Polasek >> In the inertial frame of the interferometer, the speed of light is >> isotropically c, so the position of the fringes does not depend on the >> orientation of the interferometer. Hence a null result is expected, >> which is fully consistent with their measurement. > Tautologically so! Even I can see if the speed of light is > isotropically c, then, yes, that's what it is, c, and we can expect no > fringes. You mean no fringe shift, of course. But this is not "tautology", this is a conclusion. > Your standards for a proof are not as high as mine. This is physics, and no "proof" is possible. The best that can happen is the theory makes a prediction that is confirmed by the experiment. And that happens here. Tom Roberts tjroberts@lucent.com > > Take a friend and a couple of megaphones out to a field on a windy day. > > Use your megaphone to shout to your friend while he uses his to [quoted text clipped - 7 lines] > such a wind your friend would not hear you (or even be able to hold a > megaphone).... If the two observers are the length of a football field apart that 5 degrees equals about 30 feet. You wouldn't even need the megaphones to detect that shift, if it existed. With the megaphones you could be more accurate at a lower wind speed. If they don't do the trick get a reflector dish like they use for microphones at football games. They also make speakers with a beam width of 10 degrees as shown here. http://www.meyersound.com/products/industrialseries/sb-1/ They make narrower beams if you need one. > Theoretically for this case and transverse wind, you (the speaker) must > point upwind by arcsin(windspeed/soundspeed); your friend (the receiver) > must point directly at the speaker, not upwind or downwind. This is also > the case for launching a motorboat across a river -- it must be > aimed/steered upriver to arrive transversely at a point opposite from > its starting point. A motorboat aimed up stream to follow a path straight across a river is continually expending energy to fight the current. A sound wave has no energy added after leaving its source, so your annalogy falls apart. At least you came closer than the more common idea that the receiver needs to aim down wind because the sound is coming from where it was carried off to. That mistake is more common when the observers are said to be moving through stationary air. BTW, the effect here is the same whether wind is blowing past stationary observers or observers are moving through stationary air. The other common mistake is to aim the receiver up wind to catch the sound blowing back |
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