The Emission Theory of Androcles
in [Relativity]
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From: Jonah Thomas on 11 Sep 2009 15:45 "Inertial" <relatively(a)rest.com> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > "Inertial" <relatively(a)rest.com> wrote: > >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote > >> > "Inertial" <relatively(a)rest.com> wrote: > > > >> >> What other speeds would you think the reflected light would > >> >> have?> Given that the mirrors are moving with the same angular > >> >> velocity as the source. If the speed or the rays *does* change > >at> >> the mirrors, then you would get different arrival times (and > >> >> different arrival positions) and so see an effect .. but would > >it> >> give you the observed AMOUNT of effect. > >> > > >> > Here's the way that's obvious to me now, though it might turn out > >> > something else fits better. > >> > > >> > Imagine that the speed of your light has two components. One is > >> > c, the speed that the light travels on its own hook. > >> > > >> > The other is v, the speed that the light travels because of the > >> > source's speed. > >> > >> OK .. so c+v > >> > >> > Imagine that somehow the source speed is always available, and > >> > when the direction of the light changes it winds up traveling at > >> > the speed would have had if it had been emitted in that > >direction.> > > >> > So if the source is traveling in direction V at speed v, and > >> > light is emitted in direction V, it travels at speed c+v. > >> > > >> > But if it then strikes a mirror and reflects into the opposite > >> > direction, then it travels at speed c-v. The c part is now in the > >> > opposite direction but the v part is unchanged and now opposes > >> > the motion instead of adding to it. > >> > >> So if unreflected, it stays at c+v > >> If reflected 180 deg, it changes to c-v (or vice versa) > >> So what happens when the light gets reflected at 90 deg .. does the > >> light then travel at c? > >> But then, it would have forgotten its 'v' part at the next > >> reflection. > > > > No, the point is that it doesn't forget. Reflect to any angle and it > > travels at the speed it would have if it was emitted at that angle. > > Of course, as far as a photon is concerned .. it travelled at c from > its source Yes. That's the rule -- light travels at c always, in its own frame, the inertial frame of its source when it was emitted. > The +v is all just according to an observer, and different observers > see different values for v > > So its not something a photon can 'know', because it is solely > observer dependant. All the light has to do is move at c in its own frame and the observers can all observe. > I can't see how a mirror could do anything other than reflect back a > particle with the same speed that the particle had when it hit, but in > a different direction (maybe losing some speed due to momentum > transfer) I'm not ready to think about that part yet. > >> > So if the light works that way, then in the Sagnac experiment the > >> > light in both directions goes in a complete circle and so all of > >> > the velocity changes will cancel. > >> > >> If you travel at c+v one way and c-v back for the same distance, > >> that doesn't average out to be c. But for relatively small v, it > >> will be close. > > > > It doesn't have to average out to c. The forward light travels at > > c+v and later at c-v. The backward light travels at c-v and later at > > c+v. > > But different distances for the two rays. It certainly complicates > the math substantially, esp since we don't know yet what speeds the > light would have, as the reflections are not exactly 90 deg in the non > rotating frame. If the distance traveled can be broken down into orthogonal vectors and it's the sum of those vectors, then when light makes a complete circuit back to the source it must travel at some speed out and at some speed in, along both orthogonal vectors. They have to cancel between the light in both directions. Along the vector that's perpendicular to their original motion they both go out and then in. Along the vector that's orthogonal to their motion they first go out in opposite directions, then they reverse in opposite directions, then they come back in opposite directions. Each motion of one is repeated by the other, though in a different order. But if the whole apparatus rotates then they don't quite cancel out. One will travel a small amount extra in its original direction and the others a small amount less. A tiny amount in the orthogonal direction, the sine of a small angle. Since we're talking about relativistic motion, we don't get the difference in speed over the whole path but the difference in speed over the small difference in pathlength. I think this will be negligible until the apparatus rotates at relativistic speed. > > So > > the speeds average out between the two of them, leaving you with a > > very similar interference pattern compared to the classical case. > > You would most likely get some phase shift, and possible a changing > interference pattern if the speeds of the photons changed at each > mirror. Whether or not it is anything like what we see in a Sagnac > experiment would be hard to tell until we did the math. I have hypothesised that nothing happens at the mirror except a reflection at the angle of incidence and the velocity at c in the light's own frame. Experimental evidence might show another difference, and that difference would be a confirmation of EmT, right? > It does seem extremely ad-hoc though. Of course. Isn't this how Einstein did it? First he lookerd at the experiments he needed to satisfy -- MM etc -- and he decided lightspeed should be the same always and the same in all frames, and then he worked out what else had to happen to allow that, and then he came up with some sort of semi-plausible explanation how it could happen. It isn't something that anybody would come up with from first principles, if they had been raised thinking euclidean. > >> > Both of them will spend 1/4 of their distance > >> > traveling at c+v and 1/4 at c-v etc. > > > >> Not really, as you are reflecting at (something near) 90 deg at > >> each mirror, not 180. And I'm not sure yet what your newe > >> reflection theory for light would say about such a reflection > > > > I haven't decided what speed I need at 90 degrees. Since forward and > > back will both travel an equal distance either way, the differences > > will disappear regardless of the speed. So I'll save my prediction > > of that speed until I find some other experiment I have to fudge it > > to. > > :):) ;) ;) Seriously, there should be an unambiguous answer here. But when I try to think about which angle applies when v is relativistic I get confused. There might be something here that would break it either way. I don't have it thought out yet, I have one solution that has to be wrong and a second solution that might be right and I don't yet know which is which. > >> > So the diffraction pattern should > >> > be almost the same as the classical and SR case. > >> > >> I think we're getting a bit unrealistic there with how mirrors work > >:) > > > > You're a relativist and you're noodging about unrealistic > > asssumptions? > > Yeup. Nothing unrealistic about the laws of physics being the same > everywhere, nor the speed of light being a constant (though when you > consider the conequences, they are not what one expects, yet they are > what we see experimentally) Well, if Ritz's emission theory is true then this behavior of mirrors is not what one expects but it is what we see experimentally. Nothing unrealistic about the laws of physics being the same everywhere, or the speed of light being a constant c in the light's own frame of reference. > > Hey, if Einstein can do it, I can do it. Besides, Ritz did this one > > first. It's other people who made up a ballistic theory with > > constant c+v. Ritz did it my way. > > Can you please show a reference to that? Thanks http://books.google.com/books?id=7xrL7h10XkQC&pg=PA7&lpg=PA6&ots=co-r6uws_f&dq=%22emission+theories%22 This is Pauli's explanation. The relevant part is on page 7 which would take special software for me to copy and paste. "Finally, according to Ritz the velocity of the reflected ray is equal to that of a parallel ray, emitted from the original light source." Pauli dismisses emission theories because of DeSitter's work on double stars. On page 8 he says that if lightspeed is not constant then circular orbits of double stars would appear eccentric. Since DeSitter did not see double stars with eccentric orbits, emission theories are disproved. And thus, the work of Androcles and Wilson starts to look important. If DeSitter is wrong then the decisive negative evidence against emission theories disappears. DeSitter saw no double stars with eccentric orbits, therefore emission theory is wrong. Something funny about that decisive evidence.... There might be other decisive evidence against Ritz's emission theory, that nobody has bothered to look for because they considered it already disproved. We won't know about that until we look.
From: Androcles on 11 Sep 2009 17:11 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090911154526.03af0773.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >> > "Inertial" <relatively(a)rest.com> wrote: >> >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >> >> > "Inertial" <relatively(a)rest.com> wrote: >> > >> >> >> What other speeds would you think the reflected light would >> >> >> have?> Given that the mirrors are moving with the same angular >> >> >> velocity as the source. If the speed or the rays *does* change >> >at> >> the mirrors, then you would get different arrival times (and >> >> >> different arrival positions) and so see an effect .. but would >> >it> >> give you the observed AMOUNT of effect. >> >> > >> >> > Here's the way that's obvious to me now, though it might turn out >> >> > something else fits better. >> >> > >> >> > Imagine that the speed of your light has two components. One is >> >> > c, the speed that the light travels on its own hook. >> >> > >> >> > The other is v, the speed that the light travels because of the >> >> > source's speed. >> >> >> >> OK .. so c+v >> >> >> >> > Imagine that somehow the source speed is always available, and >> >> > when the direction of the light changes it winds up traveling at >> >> > the speed would have had if it had been emitted in that >> >direction.> > >> >> > So if the source is traveling in direction V at speed v, and >> >> > light is emitted in direction V, it travels at speed c+v. >> >> > >> >> > But if it then strikes a mirror and reflects into the opposite >> >> > direction, then it travels at speed c-v. The c part is now in the >> >> > opposite direction but the v part is unchanged and now opposes >> >> > the motion instead of adding to it. >> >> >> >> So if unreflected, it stays at c+v >> >> If reflected 180 deg, it changes to c-v (or vice versa) >> >> So what happens when the light gets reflected at 90 deg .. does the >> >> light then travel at c? >> >> But then, it would have forgotten its 'v' part at the next >> >> reflection. >> > >> > No, the point is that it doesn't forget. Reflect to any angle and it >> > travels at the speed it would have if it was emitted at that angle. >> >> Of course, as far as a photon is concerned .. it travelled at c from >> its source > > Yes. That's the rule -- light travels at c always, in its own frame, the > inertial frame of its source when it was emitted. A team of scientists working under the direction of researchers from the University of Sussex have recently discovered that Einstein did not say "inertial". Here is their well-researched evidence which took 13.5 milliseconds to carry out at great expense. http://www.androcles01.pwp.blueyonder.co.uk/inertial.JPG Everything travels at speed zero in its own frame, it is always at the origin. Otherwise it would leave its frame behind. Clearly neither of you idiots have a clue what a frame of reference is. You should both learn the basics. >> The +v is all just according to an observer, and different observers >> see different values for v >> >> So its not something a photon can 'know', because it is solely >> observer dependant. > > All the light has to do is move at c in its own frame and the observers > can all observe. Everything travels at speed zero in its own frame, it is always at the origin. Otherwise it would leave its frame behind. Clearly neither of you imbeciles have a clue what a frame of reference is. You should both learn the basics. >> I can't see how a mirror could do anything other than reflect back a >> particle with the same speed that the particle had when it hit, but in >> a different direction (maybe losing some speed due to momentum >> transfer) > > I'm not ready to think Were you ever? Honestly, have a ever had an original thought? > about that part yet. Two dumbfucks bolstering the same myth and spreading it.
From: Henry Wilson, DSc on 11 Sep 2009 17:41 On Fri, 11 Sep 2009 04:53:14 +0100, "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > >"Henry Wilson, DSc" <hw@..> wrote in message >news:uqsia59ea9t0lcptgb5fdosen6u0mhgfkj(a)4ax.com... >> On Thu, 10 Sep 2009 16:28:21 +0100, "Androcles" >> >> Time compression is an ADoppler effect. Si are the brightness curves our >> programs produce. > >Si are fuckin' drunk. My browser is sometimes. >Wilson's time compression: > http://www.androcles01.pwp.blueyonder.co.uk/Doolin'sStar.GIF Also known as WaSh. >Wilson's unifuckation: > http://www.androcles01.pwp.blueyonder.co.uk/Doolin'sStarGR.GIF No, you don't get it at all. The WaSh still occurs before any unification becomes significant. >Wilson accepts Einstein's GR gravity time compression at the source >and SR's second postulate. He has no idea what ballistic light is all about. >Go ahead, plonk me. Don't accuse me like that or I'll sue. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Henry Wilson, DSc on 11 Sep 2009 17:43 On Fri, 11 Sep 2009 04:46:46 +0100, "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > >"Henry Wilson, DSc" <hw@..> wrote in message >news:pbsia5tm5oe7n4lqbjsq85vgjmdc8kppd9(a)4ax.com... >> On Thu, 10 Sep 2009 05:18:45 -0700 (PDT), J Thomas <jethomas5(a)gmail.com> >> wrote: >> >>>hw@..(Henry Wilson, DSc) wrote: >>> >>>> When light changes speed, each photon's intrinsic absolute wavelength >>>> also >>>> changes accordingly. For instance, if a photon decelerates, its >>>> wavecrests move >>>> closer together, so their flow rate remains the same....like cars on a >>>> highway >>>> in different speed zones. >>> >>>That makes perfect sense. So by your view, when light bounces off a >>>mirror and changes its direction, it gets the speed it would have had >>>if it had been emitted in that direction in the first place? And its >>>frequency stays the same, but its wavelength changes to match? >> >> There is not experimental evidence that clarifies this question but one >> would >> think intuitively that if light arrives ar a mirror with relative speed >> c+v > >It can't, it has to unifuckate to c before it gets there according to your >BaThwater. It is estimated that the extinction distance in ambient air is about 3 cms (don't ask me who determined that) but normally the air would be at rest with the mirror in this type of experiment. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Henry Wilson, DSc on 11 Sep 2009 18:01
On Fri, 11 Sep 2009 18:44:07 +1000, "Inertial" <relatively(a)rest.com> wrote: >"Jonah Thomas" <jethomas5(a)gmail.com> wrote in message >news:20090911044249.1fdc25cb.jethomas5(a)gmail.com... >> "Inertial" <relatively(a)rest.com> wrote: >>> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >>> > hw@..(Henry Wilson, DSc) wrote: >> >>> >> Here is the simple explanation of a four mirror Sagnac. >>> >> http://www.users.bigpond.com/hewn/sagnac.jpg >>> > >>> > I see! And that's what Androcles was hinting at too! Very good! I'm >>> > laughing, that's delightful. >>> >>> The diagram is correct for the ballistic analysis, not for SR though >>> >>> The path lengths are fine too. Different lengths. >>> >>> The times for the two rays to travel is the same >>> >>> The two rays arrive at the destination at the same time >>> >>> And as far as the moving detector is concerned, they arrive with the >>> same frequency and the same speed >> >> I think that's what I was missing. > >Could be. Henry seems to think that how the waves arrive at the detector >doesn't matter for sagnac .. which is complete nonsense, seeing that is >where you detect the effect. You're so clueless you can't understand that they can arrive at the same rate but out of phase. That's because you keep frame jumping. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer.. |