Sagnac Idiocy
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From: George Dishman on 9 Oct 2007 05:22 On 9 Oct, 01:08, HW@....(Clueless Henri Wilson) wrote: > On Mon, 8 Oct 2007 23:52:16 +0100, "George Dishman" <geo...(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message > >news:voalg3dlnsb0odv14rh9juo0ifhm82qu3d(a)4ax.com... > >> On Mon, 08 Oct 2007 00:51:41 -0700, George Dishman > >> <geo...(a)briar.demon.co.uk> wrote: > >>>On 7 Oct, 22:47, HW@....(Clueless Henri Wilson) wrote: > >>>> On Sun, 7 Oct 2007 12:01:48 +0100, "George Dishman" > >>>> <geo...(a)briar.demon.co.uk> wrote: > >>>> >"Clueless Henri Wilson" <HW@....> wrote in message > >>>> >news:ku1gg39nq2eg50rdptt8pj2i6h4v1hfuc4(a)4ax.com... > >>>> >>>Right, it is just the circumference divided > > >>>> I have thought about it...you obviously haven't. > >>>> When the animation has stopped, only the detector position is shown. The > >>>> position of the source when the arriving light WAS EMITTED must also be > >>>> shown. > > >>>http://mysite.verizon.net/cephalobus_alienus/sagnac/BallisticSagnac.htm > > >>>Jerry has now added a yellow line mrking the > >>>location of the source at the time of emission. > > Look at the phasing at the YELLOW line when the thing stops. > Is it the same George? Of couse not, it was the same at the moment the light was emitted but what is shown is the _current_ situation, it is not a hostorical trace. > ..AS REQUIRED. What is required is that the wave emanating from the splitter has the same value in both directions at any time since it is a single beam being split. > How does it vary with ring rotation speed? > You don't know because Jerry hasn't included ring rotation speed at all. The speed is included on the splitter aline _and_ on both beams. Androcles checked the speeds and found they were correct. > Jerry hasn't a clue.... You don't even know how to find out, even after your co-believer did it for you. > >> It still doesn't show in windows Vista. > > >Make sure you have the Sun VM installed and do > >a hard refresh. > > I'll try my other computer...but I know what has been done... Yes, but if you can see it, you won't make comments which are obviously untrue the moment anybody looks at it. > >>>> The difference between the two is what determines the fringe > >>>> displacement. > > >>>No, what determines the fringes is the relative > >>>phase of the two waves when they reach the > >>>detector. > > >> that is correct....and that phase is dtermined solely by the path length > >> difference, since wavelength is absolute. > > >No, phase is determined by time. > > When the animation stops, the phasing at the emission point is not the > same...AS REQUIRED. The phase at the _source_ is the same at _ALL_ times, including when the light was emitted at that point. > >>>That is why there is > >>>no fringe displacement, the detector can only be > >>>aware of what is falling on it at any instant. > > >> George, animate my 'toothed wheel' model...that is the correct one. > > >Jerry's animation has 9.5 teeth, other than that, > >it is what you want, you cannot identify any error. > > Jerry has not included ring rotation speed at all. jerry is stupid.. I said in my first post about it that all you had to do was check those speeds. Androcles did it for you and confirmed they are correct. You _still_ haven't bothered to look when it is blatantly obvious that the speed _IS_ included. > >>>It was always correct, you just need it spelled > >>>out for you. > > >> Think about the wheel model George. It clarifies the picture. > >> The number of teeth betwen the two lines is dependent solely on path > >> length. > > >That is unrelated to the phase though unless you > >take account of the different speeds. > > see above Your comments above are wrong, you didn't realise the waves needed to be in phase at the source when they were EMITTED. > >> George, you have forgotten that in the rotating fame, the source position > >> AT > >> EMISSION appears to move backwards....THERE IS YOU ERROR. > > >But the light is already in transit so doesn't care > >what happens to the source after emission, that is > >_your_ error > > Not so george...read above again.... Yes so Henry. > >> number = (circumference + vt)/wavelength > > >No, in the rotating frame it is > > > number = circumference/wavelength > > You forgot that the emission point moves backwards in the rotating frame. a) It doesn't drag the waves with it. b) For the umpteenth time, the simulation is in the INERTIAL frame. > >>>You are still making the same mistake, you > >>>are dividing by the wavelength when you are > >>>supposed to be working in the inertial frame. > >>>The wavelength is the same in both frames but > >>>the waves _move_ so you can't just blindly use > >>>that number. > > >> I think you will have recognised your mistakes by now... > > >I have again identified your mistake, you are > >still making the same one and just repeating > >it, not thinking about what is being said to > >you. > > You forgot that the emission point moves backwards in the rotating frame. AND > THE PHASING AT THE EMISSION POINT MUST BE THE SAME NOT DIFFRENT AS IN JERRY'S > ANIMATION. Wrong on both counts, the animation is in the INERTIAL frame and the waves are constraned to be always in phase at the source only at their time of emission. The waves move with different speeds so the phase at any inertial frame location changes thereafter. > >>>[I wrote:] > >>>> : Spin up the table to constant speed, apply > >>>> : power to the slip rings to light up the source, > >>>> : switch it off when the exposure is complete > >>>> : and slow the table to a stop. Take the plate > >>>> : off the table and develop it. What could be > >>>> : easier. > >>>[Androcles wrote:] > >>>> You won't see any fringe shift with the camera > >>>> on the table, except during acceleration. You'll see > >>>> a fringe, but it won't be shifted. What was his exposure time? > >>>> Did Georges Sagnac use slip rings to operate the camera? > > >>>He understands the consequences, and he denies > >>>that the experiment was built the way it was > >>>because of those consequences. You coorectly > >>>understand the construction but can't do the > >>>maths. If you could work as a team, you could > >>>understand as much as Jerry. > > >> You should know your mistake by now. > > >The mistake is unchanged, your maths is wrong > >and Jerry's simulation is accurate. > > You forgot that the emission point moves backwards in the rotating frame. You are too clueless to realise the source doesn't drag the waves with it and the simulation is drawn in the inertial frame anyway. So far you haven't found a single flaw in what she has drawn. George
From: Dr. Henri Wilson on 9 Oct 2007 05:31 On Tue, 09 Oct 2007 02:49:45 GMT, "Androcles" <Engineer(a)hogwarts.physics> wrote: > >"Dr. Henri Wilson" <HW@....> wrote in message >news:0jhlg3tg2sj0ul12jmv5h58h02kdlrpr1q(a)4ax.com... >: On Mon, 08 Oct 2007 23:10:53 GMT, "Androcles" <Engineer(a)hogwarts.physics> >: wrote: >: >: > >: >"Dr. Henri Wilson" <HW@....> wrote in message >: >news:voalg3dlnsb0odv14rh9juo0ifhm82qu3d(a)4ax.com... >: >: On Mon, 08 Oct 2007 00:51:41 -0700, George Dishman >: ><george(a)briar.demon.co.uk> >: >: >: >> The difference between the two is what determines the fringe >: >displacement. >: >: > >: >: >No, what determines the fringes is the relative >: >: >phase of the two waves when they reach the >: >: >detector. >: >: >: >: that is correct....and that phase is dtermined solely by the path >length >: >: difference, since wavelength is absolute. >: > >: >Now listen, H. >: >What the hell are you talking about, "wavelength is absolute"? >: >What does that MEAN, exactly? >: >I'm asking because you are the only person in the world that >: >say that. >: >lambda = (c+v) * nu, >: >Hence lambda is directly proportional to speed. >: >: All lengths are absolute and invariant. By that I mean a rod doesn't >physically >: change when its speed changes. It is still exactly the same rod and >defines the >: same absolute spatial 'interval'. A 'wavelength' of light is similar. It's >: absolute length is always the same in all inertial frames. You can imagine >: 'wavecrests' as being connected by very small rods. >: >: 'Frequency' as applied to light is just emission/arrival rate of >: 'wavecrests'...what ever they are. > >I was afraid that was what you meant. I'm not getting into an >argument over it, diffraction gratings have a greater angle >with longer wavelengths and greater speeds, that's it. http://www.users.bigpond.com/hewn/bathgrating.jpg >: Jerry has now included the emission point. You will notice that the rays >: leaving it when the thing stops are not in phase as required >: Jerry has not included ring rotation rate at all. >: It is the main factor involved in this. >: >: Jerry is plainly stupid... > >Jeery is a relativist, they are all stupid. <shrug> yes...that's two things you got right this week... >: >They are in phase because they are going in the same direction >: >with the same RPM, not because the air valve is at the top at the >: >same time on both of them, and then get out of phase when the >: >car turns. >: >: Yes I noticed that animation. Vry good...I'm trying to compile something >like >: it but it's not easy. > > >: To aadapt it simulate a ring gyro, the two outside wheels must rotate >rapidly >: in opposite directions to represent the two light rays. The middle one can >: rotate very slowly to represent the ring. > >Roller skate wheels have a shorter wavelength and a higher frequency >than tractor tyres. Wavelength is the distance travelled for one turn of the >wheel. c+v is when the roller skate is in a moving pavement or walkway. Aha! No A. that's not a good example. A 1 metre rod is a 1 metre rod no matter how fast you move past it. >It takes 50 strides to walk the length of a plane. If you fly from >Sydney to London while walking the length of the plane you've >taken some very long strides, 200 miles long. Wavelength and strides >are relative. You are introducing time... that is not allowed. The plane is the same length no matter where it goes. >: No they aren't Jeery's lines....these are the ones George and I have been >: talking about. One is the emission point of a particular wavecrest and the >: other the position of the detector when it arrives. Both points are static >in >: the nonrotating frame even though the spliting mirror itself is moving. > >Well, whatever, the two lines that matter are the new yellow line at >3 o'clock and the black destination, it is between those that the N >wavelengths should be drawn, it only a trace of the path a photon >took anyway. In Jerry's diagram, there are 13 wavelengths in one path and 6 in the other. >The model has a pause on the same button as the start so you can halt >it when the the two rays meet. 180 degrees in the rotating frame, speed >dependent in the stationary frame. Draw 4.5 waves from 3 o'clock >to the end point and that shows a change in wavelength. The program is ridiculous. It doesn't model anything like a ring gyro.. >: >Wavelength = (circumference + vt)/number >: >Number is "absolute". >: >At least its the same for both directions. >: >: No, in BaTh, wavelength is absolute and identical in both paths... > >Yeah, you said that. In Emission Fact wavelength is just the trace >of the path the photon took and is proportional to c+v. Wavelengths >don't even wave in Emission Fact. Kinda like this: > http://www.kettering.edu/~drussell/Demos/SHO/damp.html >No waving, you see, just a sinusoidal trace. irrelevant.... >: >: >: >: I think you will have recognised your mistakes by now... >: > >: > >: >You are the one confused. George is right, >: >You are still making the same mistake, you >: >are dividing by the wavelength.\ >: >: wavelength is absolute and invariant in emission theory. > >But not in Emission Fact. You can have your BaTh, Dr. Wilson, >it's all yours and you are welcome to it, nobody is interested in it. Sorry, A. Wavelength is absolute in all emission theories. There is NO doppler shift at the source end. The source effectively fires little serated bullets...the serations signifying absolute wavelength. >: It is not invariant in aether theory or SR. >: Which side are you on? > >I'm on the side of science. I'm not a BaThist, I'm an Emission Factist, >I have no crank theory of my own as you do. You are an aetherist in disguise. you don't even know you're an aetherist. >I happen to understand radio, mathematics and wave superposition too. > http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac2.JPG irrelevant... Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Dr. Henri Wilson on 9 Oct 2007 05:39 On Tue, 09 Oct 2007 02:54:32 GMT, "Androcles" <Engineer(a)hogwarts.physics> wrote: > >"Dr. Henri Wilson" <HW@....> wrote in message >news:9silg3llas6clon1j6alnblu68nl4mn1mb(a)4ax.com... >: >> Jerry's animation is a joke... >: > >: >Then why can't you find a flaw in it? >: >: I have. The waves are not in phase at the emission point....simple! > >Not showing the correct wavelength from emission point to reception >point either. There are 6 waves in one path and 13 in the other. count them yourself if you don't believe me. >>The distance they move is from the yellow line to the >: >>>black line at the time the start of the wave reaches it. >: >> >: >> which is as I said above. >: > >: >Exactly, the animation already shows what >: >you complained about. >: >: It doesn't include ring rotation speed. > >Aww, c'mon, it does. That's unfair. Where does Jerry mention ring speed? He/she/it doesn't even appreciate that it is the most important factor in Sagnac. > >: determines phase. >: Jerry's animation uses rays that BEGIN out of phase. >: Jerry doesn't include ring rotation speed and how it affects the umber of >: wavelengths in each path. >: > You are both wrong. >The travel time is the same for both rays in both rotating and >stationary frames . The travel times are indeed the same. >The distance travelled in the rotating frame is the same for both rays. Nah. I already corrected Dishman on this. In the rotating frame, the emission point moves backwards. >The distances travelled in the stationary frame are different. >The wavelength in the rotating frame is the same for both rays. >The wavelengths in the stationary frame are different. Wavelength is absolute and invariant in all inertial frames. I would rather not use rotating frames because they lead to silly errors....like Dishman's.. >: >Why not just look at the maths above. >: > >: >> >..........................................................................................................->c >: >> A_____________________B >: >> >: >> The line of dots represents the wavecrests of a light ray moving at c. >: >> A and B are the end points of a rigid rod. >: >> >: >> There are N dots between points A nd B, NO MATTER HOW fast the rod >moves >: >> past >: >> them...or no matter how fast the light ray moves past the rod.. >: >> >: >> Next, let the position of B move by vt, where v is the speed of the rod >: >> and t >: >> is the time taken for a wavecrest to travel from A to B. Note, A, >: >> representing >: >> the source, does not move. >: > >: >Wrong, the source _does_ move, there is your error. >: >You are as bad as Androcles only you want the >: >source to be off the table. >: >: George, sorry if I said the wrong thing there. A is the emission pit >rather >: than the 'source'. The splitting mirror, ie., source, DOES move but the >: emission point, A, does not. > >Correct. That's a surprise...you agreeing.... >: >Just look at Jerry's simulation and try to find an >: >error, there isn't one. >: >: the rays that end up in phase did not BEGIN in phase as required. > >Sematics. Very important... >: Jerry is stupid... > >Correct. > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 9 Oct 2007 06:30 On 9 Oct, 10:39, HW@....(Clueless Henri Wilson) wrote: > On Tue, 09 Oct 2007 02:54:32 GMT, "Androcles" <Engin...(a)hogwarts.physics> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message news:9silg3llas6clon1j6alnblu68nl4mn1mb(a)4ax.com... Here is the link again for convenience: http://mysite.verizon.net/cephalobus_alienus/sagnac/BallisticSagnac.htm There are a lot of extraneous points being made that are eqivalent ways of looking at the same thing so I'll trim down to just the key aspects. Androcles, I know you disagree with our description of the equipment but to help Henry can you treat this as a hypothetical question - what if the structure was as we said. You understand why there would be no fringe displacement but Henry cannot yet see it and he needs your help. > >: >Exactly, the animation already shows what > >: >you complained about. > >: > >: It doesn't include ring rotation speed. > > >Aww, c'mon, it does. That's unfair. > > Where does Jerry mention ring speed? > He/she/it doesn't even appreciate that it is the most important factor in > Sagnac. The ring speed applies to the line indicating the splitter. It is obvious that this moves, call that speed v. If you measure the speed of the two waves you will find they differ, the one going clockwise moves at c-v while that going anti-clockwise moves at c+v. Note that we are talking about the speed of any part of the wave and the magenta dots marking an arbitrary positive peak are typical. That means the speeds of the splitter and waves are all correct in accordance with Ritz's theory. > >: >Just look at Jerry's simulation and try to find an > >: >error, there isn't one. > >: > >: the rays that end up in phase did not BEGIN in phase as required. > > >Sematics. > > Very important... Yes, very, so let's clear it up. A single light source illuminates the splitter so the two beams are conceptually emitted in phase. In practice there may be a polarity reversal (equivalent to a 180 degree phase shift) on reflection but if you follow the paths, one is reflected twice while the other is transmitted twice so any polarity reversals cancel and we can ignore the effect for simplicity. That means the simulation should show the light leaving the splitter in phase. The arbitrary point chosen is a positive peak and at the instant it passes through the splitter, it merges as the two magenta dots, the ocation where they were emitted being noted for reference by the yellow dot. The waves are therefore emitted in phase as required. When the dots have both completed one circuit, they arrive back at the splitter. I think we have all agreed that they take the same time because the longer path is traversed by the faster light. If they arrive at the same time and they both mark the positive peak then obviously one peak coincides with the other - the signals are in phase and they produce constructive interference. The same argument would apply if the dots marked a negative peak or any other part of the wave so the signals are _always_ in phase. That can be seen directly by just watching the splitter line and seeing the two waves arriving, they always hit the radial line at the same radius even though that point moves in and out at the emitted frequency. If either of you thinks any of the points above is incorrect, please say so but do that before introducing any other arguments. Given the above points alone, the arrival phase is fully determined and any other conclusions can also be derived and either confirmed or refuted from them. George
From: Androcles on 9 Oct 2007 14:46
"Dr. Henri Wilson" <HW@....> wrote in message news:bkhmg3ln4unk3shemnvfirlq261o62b0oe(a)4ax.com... : On Tue, 09 Oct 2007 02:49:45 GMT, "Androcles" <Engineer(a)hogwarts.physics> : wrote: : : > : >"Dr. Henri Wilson" <HW@....> wrote in message : >news:0jhlg3tg2sj0ul12jmv5h58h02kdlrpr1q(a)4ax.com... : >: On Mon, 08 Oct 2007 23:10:53 GMT, "Androcles" <Engineer(a)hogwarts.physics> : >: wrote: : >: : >: > : >: >"Dr. Henri Wilson" <HW@....> wrote in message : >: >news:voalg3dlnsb0odv14rh9juo0ifhm82qu3d(a)4ax.com... : >: >: On Mon, 08 Oct 2007 00:51:41 -0700, George Dishman : >: ><george(a)briar.demon.co.uk> : : >: : >: >: >> The difference between the two is what determines the fringe : >: >displacement. : >: >: > : >: >: >No, what determines the fringes is the relative : >: >: >phase of the two waves when they reach the : >: >: >detector. : >: >: : >: >: that is correct....and that phase is dtermined solely by the path : >length : >: >: difference, since wavelength is absolute. : >: > : >: >Now listen, H. : >: >What the hell are you talking about, "wavelength is absolute"? : >: >What does that MEAN, exactly? : >: >I'm asking because you are the only person in the world that : >: >say that. : >: >lambda = (c+v) * nu, : >: >Hence lambda is directly proportional to speed. : >: : >: All lengths are absolute and invariant. By that I mean a rod doesn't : >physically : >: change when its speed changes. It is still exactly the same rod and : >defines the : >: same absolute spatial 'interval'. A 'wavelength' of light is similar. It's : >: absolute length is always the same in all inertial frames. You can imagine : >: 'wavecrests' as being connected by very small rods. : >: : >: 'Frequency' as applied to light is just emission/arrival rate of : >: 'wavecrests'...what ever they are. : > : >I was afraid that was what you meant. I'm not getting into an : >argument over it, diffraction gratings have a greater angle : >with longer wavelengths and greater speeds, that's it. : : http://www.users.bigpond.com/hewn/bathgrating.jpg I see no speed there, the picture is static. : : >: Jerry has now included the emission point. You will notice that the rays : >: leaving it when the thing stops are not in phase as required : >: Jerry has not included ring rotation rate at all. : >: It is the main factor involved in this. : >: : >: Jerry is plainly stupid... : > : >Jeery is a relativist, they are all stupid. <shrug> : : yes...that's two things you got right this week... Jeery's model shows a fixed wavelength for the rotating observer. It does not show a varying wavelength for the stationary observer. It is incomplete. In this simple cross the diameter model, the wavelength is shorter near the centre. http://www.androcles01.pwp.blueyonder.co.uk/VarLength.png : : : >: >They are in phase because they are going in the same direction : >: >with the same RPM, not because the air valve is at the top at the : >: >same time on both of them, and then get out of phase when the : >: >car turns. : >: : >: Yes I noticed that animation. Vry good...I'm trying to compile something : >like : >: it but it's not easy. : > : > : >: To aadapt it simulate a ring gyro, the two outside wheels must rotate : >rapidly : >: in opposite directions to represent the two light rays. The middle one can : >: rotate very slowly to represent the ring. : > : >Roller skate wheels have a shorter wavelength and a higher frequency : >than tractor tyres. Wavelength is the distance travelled for one turn of the : >wheel. c+v is when the roller skate is in a moving pavement or walkway. : : Aha! No A. : that's not a good example. : : A 1 metre rod is a 1 metre rod no matter how fast you move past it. The distance between the yellow balls in this simple cross-the-diameter model is shorter near the centre: http://www.androcles01.pwp.blueyonder.co.uk/VarLength.png The model rotates: http://www.androcles01.pwp.blueyonder.co.uk/FrameA.gif http://www.androcles01.pwp.blueyonder.co.uk/FrameB.gif You can see this happen in the movie http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov : >It takes 50 strides to walk the length of a plane. If you fly from : >Sydney to London while walking the length of the plane you've : >taken some very long strides, 200 miles long. Wavelength and strides : >are relative. : : You are introducing time... that is not allowed. I'm sorry, I didn't realise nothing was moving in BaTh. : The plane is the same length no matter where it goes. The strides as seen from a passenger are 1 yard long. The strides as seen from the ground are 200 miles long. The is true whether you take 20 hours for fly or a 5000 hours by ship. The ship is the same length no matter where it goes, you are stupid wherever you go. : : >: No they aren't Jeery's lines....these are the ones George and I have been : >: talking about. One is the emission point of a particular wavecrest and the : >: other the position of the detector when it arrives. Both points are static : >in : >: the nonrotating frame even though the spliting mirror itself is moving. : > : >Well, whatever, the two lines that matter are the new yellow line at : >3 o'clock and the black destination, it is between those that the N : >wavelengths should be drawn, it only a trace of the path a photon : >took anyway. : : In Jerry's diagram, there are 13 wavelengths in one path and 6 in the other. Yes, the Wilson tick fairy waved her magic wand 7 times. I learnt to count when I was five-years old, Dr. Wilson, and whether I count short toes or long fingers I still count up to 5 on the extremity of the limb. In BaTh, fingers and toes are the same length and you have 15 fingers on each hand. : : >The model has a pause on the same button as the start so you can halt : >it when the the two rays meet. 180 degrees in the rotating frame, speed : >dependent in the stationary frame. Draw 4.5 waves from 3 o'clock : >to the end point and that shows a change in wavelength. : : The program is ridiculous. It doesn't model anything like a ring gyro.. Whether it does or doesn't, the model has a pause on the same button as the start so you can halt it when the the two rays meet. 180 degrees in the rotating frame, speed dependent in the stationary frame. Draw 4.5 waves from 3 o'clock to the end point and that shows a change in wavelength. In BaTh you have 15 fingers on each hand, each the same length as a toe, while you want to talk about ridiculous. : : >: >Wavelength = (circumference + vt)/number : >: >Number is "absolute". : >: >At least its the same for both directions. : >: : >: No, in BaTh, wavelength is absolute and identical in both paths... : > : >Yeah, you said that. In Emission Fact wavelength is just the trace : >of the path the photon took and is proportional to c+v. Wavelengths : >don't even wave in Emission Fact. Kinda like this: : > http://www.kettering.edu/~drussell/Demos/SHO/damp.html : >No waving, you see, just a sinusoidal trace. : : irrelevant.... No, in BaTh, toe length is absolute and identical on all limbs, unless the Wilson toe fairy waves her magic wand to create extra toes. Three toes per finger, because fingers are three time longer than toes which is why we have 15 fingers on each hand. : : >: >: : >: >: I think you will have recognised your mistakes by now... : >: > : >: > : >: >You are the one confused. George is right, : >: >You are still making the same mistake, you : >: >are dividing by the wavelength.\ : >: : >: wavelength is absolute and invariant in emission theory. : > : >But not in Emission Fact. You can have your BaTh, Dr. Wilson, : >it's all yours and you are welcome to it, nobody is interested in it. : : Sorry, A. Wavelength is absolute in all emission theories. Yes, but not in Emission Fact. : There is NO doppler shift at the source end. : The source effectively fires little serated bullets...the serations signifying : absolute wavelength. Yes, I've seen your model of a senile photon. : : >: It is not invariant in aether theory or SR. : >: Which side are you on? : > : >I'm on the side of science. I'm not a BaThist, I'm an Emission Factist, : >I have no crank theory of my own as you do. : : You are an aetherist in disguise. you don't even know you're an aetherist. You are fuckhead without any disguise. You don't even know you're a fuckhead. : : >I happen to understand radio, mathematics and wave superposition too. : > http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac2.JPG : : irrelevant... Still counting to 15 on your fingers, Dr. Wilson? |