Sagnac Revisited. Why Paul, Tom and Jerry are Hopelessly Wrong.
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From: Henry Wilson DSc on 25 Feb 2010 22:30 On Thu, 25 Feb 2010 21:42:50 -0000, "Androcles" <Headmaster(a)Hogwarts.physics_u> wrote: > >"Henry Wilson DSc" <..@..> wrote in message >> >> Brilliant! >> >> Now here is something that might help you. >> >> The 'phase' of the rays in the ring depends on DISTANCE, not TIME. >> > >"Bullshit. I haven't used rotating frames at all." -- Ahenry Awilson > news:7qe8o59a5eei3iq6jteuiel2hteb69kf1v(a)4ax.com > >Now here is something that might help you. > >The 'phase' of the rays LEAVING the ring depends on their different >SPEEDS, not distance or time. > >Why? good question >"the calculations will be made in the non rotating inertial frame where the >centre of the ring is stationary."-- Tusseladd. > >The rays have different momenta and are jumping frames from the moving >rim to the centre of the ring. > http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/SagnacRing.JPG whty would that result is a phase difference? Henry Wilson... ........provider of free physics lessons
From: YBM on 25 Feb 2010 22:31 Henry Wilson DSc a �crit : > On Thu, 25 Feb 2010 22:30:26 +0100, "Paul B. Andersen" <someone(a)somewhere.no> > wrote: > >> On 25.02.2010 20:09, Henry Wilson DSc wrote: >>> On Thu, 25 Feb 2010 11:50:32 +0100, "Paul B. Andersen" >>> <paul.b.andersen(a)somewhere.no> wrote: >>> >>>> On 24.02.2010 21:29, Henry Wilson DSc wrote: >>>>> On Wed, 24 Feb 2010 15:09:33 +0100, "Paul B. Andersen" >>>>> <paul.b.andersen(a)somewhere.no> wrote: >>>>>> http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf >>>>>> http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf >>>>>> http://home.c2i.net/pb_andersen/pdf/fiber_optic_gyro.pdf >>>>>> http://home.c2i.net/pb_andersen/FourMirrorSagnac.html >>>>>> >>>>>> Frustrating, eh? :-) >>>>> Your use of the rotating frame to try to refute BaTh relates to nothing other >>>>> than a nonrotating apparatus. >>>> http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf >>>> Quote from bottom of page 1: >>>> << >>>> In all cases the calculations will be made in the non rotating >>>> inertial frame where the centre of the ring is stationary. >>>> All speeds and wavelengths will in the following be referred >>>> to this frame of reference. >>> But in your sham 'refutation' of BaTh, you say: >>> >>> """""""""""""The difference in number of wavelengths method... >>> According to Ritz Emission theory, wavelengths are not Doppler shifted. >>> (correct so far) >>> . >>> . >>> Forward beam: Nf = 2piR/L >>> backward beam: Nb = 2piR/L >>> .. >>> The predicted phase difference is thus: = 0 >>> """""""""""""""" >>> >>> You are here using the rotating frame....either that or you are refering to a >>> nonrotating apparatus. >> Good grief, Ralph. :-) >> You have never understood that the number of wavelength in a wave must >> be counted _at one instant_. >> >> Look at fig 3. at page 6 in: >> http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf >> >> It shows an instant image of the light beams _drawn in the non >> rotating frame_. It is in this instant image you should count >> the number of wavelength. > > Hahahahaah! You really don't reallise you are using the rotating frame do you. > > In the nonR frame, the emission point of every wavecrest around the apparatus > is not where the 45 mirror lies at that instant. > > You are using the rotating frame. > >>> I might point out that the same approach 'refutes' SR which also relies on >>> different path lengths in its analysis. >> You are utterly confused about this 'path length'. >> The length of which path? >> The 'path length' is the length of the trajectory of a point >> of constant phase on the wave (or of the front of the wave when >> it is "turned on"), measured from the position of the source at emission >> time to the position of the target at 'hit target' time. > > CORRECT. > >> When this trajectory is drawn in the non rotating frame, the transit >> time is the length of this trajectory divided by the phase velocity >> in the non rotating frame. >> >> But it is utterly meaningless to count wavelengths along this path! >> There are none! >> >> This animation illustrates this: >> http://home.c2i.net/pb_andersen/FourMirrorSagnac.html >> >> Look at this screen shot: >> http://home.c2i.net/pb_andersen/images/Sagnac_Ritz.jpg >> >> It shows an instant image of the waves in a four mirror Sagnac >> (according to Ritz) at the time the wave fronts hit the half >> silvered mirror. > > Hahahahahahhahhaha! > > The WAVE THAT IS ARRIVING AT THE DETECTOR DID NOT START AT THAT POSITION. IT > WAS EMITTED A DISTANCE VT BEFORE IT. > > You have drawn hte rotating frame. YOU REALLY DON'T HAVE A CLUE WHAT THE > NONROTATING FRAME IS. > >> The trajectories of the wave fronts are drawn. >> The 'path lengths' are different. But so are the speeds of >> the two wave fronts, so the transit times are equal. > > CORTRECT AGAIN. > >> (Which proves that the phase difference must be zero.) > > Hahahahhahahaha! You have drawn a nonrotating interferometer. > > Hahahahhahhahhahha! Of bloody course there is no fringe shift. > >> Count the number of wavelengths in the two waves! >> It is the same for both waves, as it blatantly obvious >> must be since since wave length is invariant and >> the lengths of the two waves _obviously_ always are >> exactly the same regardless of how the interferometer >> is rotating. > > Yours is NOT rotating....hahahahhaahhahhahha! > >> There is no, and never was, any wave along the trajectory >> of the wave front, so how the hell can you count the number >> of wavelengths along this path? >> >> It is an utterly stupid idea. > > You haven't drawn the path of a rotating interferometer. You left out the > crucial 'vt' bit...................... Hahahahahahahhhaha! > >> So why the hell do I bother? >> THAT's a mystery. > > > Hahahahhahahha! Is this what they teach kids in Norway? > > > Henry Wilson... > > .......provider of free physics lessons Your whole post is quite an impressive illustration of madness. How do you feel, Ralph?
From: Henry Wilson DSc on 25 Feb 2010 22:45 On Fri, 26 Feb 2010 04:31:16 +0100, YBM <ybmess(a)nooos.fr.invalid> wrote: >Henry Wilson DSc a �crit : >> On Thu, 25 Feb 2010 22:30:26 +0100, "Paul B. Andersen" <someone(a)somewhere.no> >> wrote: >> >>> On 25.02.2010 20:09, Henry Wilson DSc wrote: >>>> On Thu, 25 Feb 2010 11:50:32 +0100, "Paul B. Andersen" >>>> <paul.b.andersen(a)somewhere.no> wrote: >>>> >>>>> On 24.02.2010 21:29, Henry Wilson DSc wrote: >>>>>> On Wed, 24 Feb 2010 15:09:33 +0100, "Paul B. Andersen" >>>>>> <paul.b.andersen(a)somewhere.no> wrote: >>>>>>> http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf >>>>>>> http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf >>>>>>> http://home.c2i.net/pb_andersen/pdf/fiber_optic_gyro.pdf >>>>>>> http://home.c2i.net/pb_andersen/FourMirrorSagnac.html >>>>>>> >>>>>>> Frustrating, eh? :-) >>>>>> Your use of the rotating frame to try to refute BaTh relates to nothing other >>>>>> than a nonrotating apparatus. >>>>> http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf >>>>> Quote from bottom of page 1: >>>>> << >>>>> In all cases the calculations will be made in the non rotating >>>>> inertial frame where the centre of the ring is stationary. >>>>> All speeds and wavelengths will in the following be referred >>>>> to this frame of reference. >>>> But in your sham 'refutation' of BaTh, you say: >>>> >>>> """""""""""""The difference in number of wavelengths method... >>>> According to Ritz Emission theory, wavelengths are not Doppler shifted. >>>> (correct so far) >>>> . >>>> . >>>> Forward beam: Nf = 2piR/L >>>> backward beam: Nb = 2piR/L >>>> .. >>>> The predicted phase difference is thus: = 0 >>>> """""""""""""""" >>>> >>>> You are here using the rotating frame....either that or you are refering to a >>>> nonrotating apparatus. >>> Good grief, Ralph. :-) >>> You have never understood that the number of wavelength in a wave must >>> be counted _at one instant_. >>> >>> Look at fig 3. at page 6 in: >>> http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf >>> >>> It shows an instant image of the light beams _drawn in the non >>> rotating frame_. It is in this instant image you should count >>> the number of wavelength. >> >> Hahahahaah! You really don't reallise you are using the rotating frame do you. >> >> In the nonR frame, the emission point of every wavecrest around the apparatus >> is not where the 45 mirror lies at that instant. >> >> You are using the rotating frame. >> >>>> I might point out that the same approach 'refutes' SR which also relies on >>>> different path lengths in its analysis. >>> You are utterly confused about this 'path length'. >>> The length of which path? >>> The 'path length' is the length of the trajectory of a point >>> of constant phase on the wave (or of the front of the wave when >>> it is "turned on"), measured from the position of the source at emission >>> time to the position of the target at 'hit target' time. >> >> CORRECT. >> >>> When this trajectory is drawn in the non rotating frame, the transit >>> time is the length of this trajectory divided by the phase velocity >>> in the non rotating frame. >>> >>> But it is utterly meaningless to count wavelengths along this path! >>> There are none! >>> >>> This animation illustrates this: >>> http://home.c2i.net/pb_andersen/FourMirrorSagnac.html >>> >>> Look at this screen shot: >>> http://home.c2i.net/pb_andersen/images/Sagnac_Ritz.jpg >>> >>> It shows an instant image of the waves in a four mirror Sagnac >>> (according to Ritz) at the time the wave fronts hit the half >>> silvered mirror. >> >> Hahahahahahhahhaha! >> >> The WAVE THAT IS ARRIVING AT THE DETECTOR DID NOT START AT THAT POSITION. IT >> WAS EMITTED A DISTANCE VT BEFORE IT. >> >> You have drawn hte rotating frame. YOU REALLY DON'T HAVE A CLUE WHAT THE >> NONROTATING FRAME IS. >> >>> The trajectories of the wave fronts are drawn. >>> The 'path lengths' are different. But so are the speeds of >>> the two wave fronts, so the transit times are equal. >> >> CORTRECT AGAIN. >> >>> (Which proves that the phase difference must be zero.) >> >> Hahahahhahahaha! You have drawn a nonrotating interferometer. >> >> Hahahahhahhahhahha! Of bloody course there is no fringe shift. >> >>> Count the number of wavelengths in the two waves! >>> It is the same for both waves, as it blatantly obvious >>> must be since since wave length is invariant and >>> the lengths of the two waves _obviously_ always are >>> exactly the same regardless of how the interferometer >>> is rotating. >> >> Yours is NOT rotating....hahahahhaahhahhahha! >> >>> There is no, and never was, any wave along the trajectory >>> of the wave front, so how the hell can you count the number >>> of wavelengths along this path? >>> >>> It is an utterly stupid idea. >> >> You haven't drawn the path of a rotating interferometer. You left out the >> crucial 'vt' bit...................... Hahahahahahahhhaha! >> >>> So why the hell do I bother? >>> THAT's a mystery. >> >> >> Hahahahhahahha! Is this what they teach kids in Norway? >> >> >> Henry Wilson... >> >> .......provider of free physics lessons > >Your whole post is quite an impressive illustration of madness. > >How do you feel, Ralph? When are you going to say something intelligent? Henry Wilson... ........provider of free physics lessons
From: Inertial on 25 Feb 2010 23:11 "Henry Wilson DSc" <..@..> wrote in message news:l4feo5lf9ij3sh40fkhpc8ncj553575eun(a)4ax.com... > On Thu, 25 Feb 2010 22:30:26 +0100, "Paul B. Andersen" > <someone(a)somewhere.no> > wrote: > >>On 25.02.2010 20:09, Henry Wilson DSc wrote: >>> On Thu, 25 Feb 2010 11:50:32 +0100, "Paul B. Andersen" >>> <paul.b.andersen(a)somewhere.no> wrote: >>> >>>> On 24.02.2010 21:29, Henry Wilson DSc wrote: >>>>> On Wed, 24 Feb 2010 15:09:33 +0100, "Paul B. Andersen" >>>>> <paul.b.andersen(a)somewhere.no> wrote: >>> >>>>>> >>>>>> http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf >>>>>> http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf >>>>>> http://home.c2i.net/pb_andersen/pdf/fiber_optic_gyro.pdf >>>>>> http://home.c2i.net/pb_andersen/FourMirrorSagnac.html >>>>>> >>>>>> Frustrating, eh? :-) >>>>> >>>>> Your use of the rotating frame to try to refute BaTh relates to >>>>> nothing other >>>>> than a nonrotating apparatus. >>>> >>>> http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf >>>> Quote from bottom of page 1: >>>> << >>>> In all cases the calculations will be made in the non rotating >>>> inertial frame where the centre of the ring is stationary. >>>> All speeds and wavelengths will in the following be referred >>>> to this frame of reference. >>> >>> But in your sham 'refutation' of BaTh, you say: >>> >>> """""""""""""The difference in number of wavelengths method... >>> According to Ritz Emission theory, wavelengths are not Doppler shifted. >>> (correct so far) >>> . >>> . >>> Forward beam: Nf = 2piR/L >>> backward beam: Nb = 2piR/L >>> .. >>> The predicted phase difference is thus: = 0 >>> """""""""""""""" >>> >>> You are here using the rotating frame....either that or you are refering >>> to a >>> nonrotating apparatus. >> >>Good grief, Ralph. :-) >>You have never understood that the number of wavelength in a wave must >>be counted _at one instant_. >> >>Look at fig 3. at page 6 in: >>http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf >> >>It shows an instant image of the light beams _drawn in the non >>rotating frame_. It is in this instant image you should count >>the number of wavelength. > > Hahahahaah! You really don't reallise you are using the rotating frame do > you. It is a picture as seen in the non-rotating frame .. how can that be the rotating frame. Obviously it must be different to what YOU think it looks like from the non-rotating frame. How do YOU think it SHOULD look in the non-rotating frame? > In the nonR frame, the emission point of every wavecrest around the > apparatus > is not where the 45 mirror lies at that instant. The diagram does not say that it is .. it is not showing where things WERE ... it is showing where they are. > You are using the rotating frame. No .. he is not. If you are so certain that he is, then please show or describe how the diagram should look from the non-rotating frame at the instant the emitter+detector and mirrors are at the positions shown. >>> I might point out that the same approach 'refutes' SR which also relies >>> on >>> different path lengths in its analysis. >> >>You are utterly confused about this 'path length'. >>The length of which path? >>The 'path length' is the length of the trajectory of a point >>of constant phase on the wave (or of the front of the wave when >>it is "turned on"), measured from the position of the source at emission >>time to the position of the target at 'hit target' time. > > CORRECT. > >>When this trajectory is drawn in the non rotating frame, the transit >>time is the length of this trajectory divided by the phase velocity >>in the non rotating frame. >> >>But it is utterly meaningless to count wavelengths along this path! >>There are none! >> >>This animation illustrates this: >>http://home.c2i.net/pb_andersen/FourMirrorSagnac.html >> >>Look at this screen shot: >>http://home.c2i.net/pb_andersen/images/Sagnac_Ritz.jpg >> >>It shows an instant image of the waves in a four mirror Sagnac >>(according to Ritz) at the time the wave fronts hit the half >>silvered mirror. > > Hahahahahahhahhaha! > > The WAVE THAT IS ARRIVING AT THE DETECTOR DID NOT START AT THAT POSITION. > IT > WAS EMITTED A DISTANCE VT BEFORE IT. Which is exactly what it shows .. see there the red and green lines start together on the RHS. > You have drawn hte rotating frame. No .. he hasn't > YOU REALLY DON'T HAVE A CLUE WHAT THE > NONROTATING FRAME IS. No .. the only one with problems with things rotating (or even moving) is you. >>The trajectories of the wave fronts are drawn. >>The 'path lengths' are different. But so are the speeds of >>the two wave fronts, so the transit times are equal. > > CORTRECT AGAIN. > >>(Which proves that the phase difference must be zero.) Indeed it does > Hahahahhahahaha! You have drawn a nonrotating interferometer. Nope > Hahahahhahhahhahha! Of bloody course there is no fringe shift. Because ballistic theory is wrong >>Count the number of wavelengths in the two waves! >>It is the same for both waves, as it blatantly obvious >>must be since since wave length is invariant and >>the lengths of the two waves _obviously_ always are >>exactly the same regardless of how the interferometer >>is rotating. > > Yours is NOT rotating....hahahahhaahhahhahha! Yes .. it is. >>There is no, and never was, any wave along the trajectory >>of the wave front, so how the hell can you count the number >>of wavelengths along this path? >> >>It is an utterly stupid idea. > > You haven't drawn the path of a rotating interferometer. Yes he did > You left out the > crucial 'vt' bit...................... Hahahahahahahhhaha! No .. it is right there in the diagram at http://home.c2i.net/pb_andersen/images/Sagnac_Ritz.jpg > >> >>So why the hell do I bother? >>THAT's a mystery. > > > Hahahahhahahha! Is this what they teach kids in Norway? You've never had a physics education anywhere, so you wouldn't know.
From: Androcles on 25 Feb 2010 23:20
"Henry Wilson DSc" <..@..> wrote in message news:6ufeo59j90ltgolop5r3p4f6r4b2ts5eua(a)4ax.com... > On Thu, 25 Feb 2010 21:42:50 -0000, "Androcles" > <Headmaster(a)Hogwarts.physics_u> > wrote: > >> >>"Henry Wilson DSc" <..@..> wrote in message > >>> >>> Brilliant! >>> >>> Now here is something that might help you. >>> >>> The 'phase' of the rays in the ring depends on DISTANCE, not TIME. >>> >> >>"Bullshit. I haven't used rotating frames at all." -- Ahenry Awilson >> news:7qe8o59a5eei3iq6jteuiel2hteb69kf1v(a)4ax.com >> >>Now here is something that might help you. >> >>The 'phase' of the rays LEAVING the ring depends on their different >>SPEEDS, not distance or time. >> >>Why? > > good question > >>"the calculations will be made in the non rotating inertial frame where >>the >>centre of the ring is stationary."-- Tusseladd. >> >>The rays have different momenta and are jumping frames from the moving >>rim to the centre of the ring. >> http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/SagnacRing.JPG > > whty would that result is a phase difference? > Poor Ahenry Awilson, can't even understand a drawing of c+v and c-v side-by-side. "In future, when and only when you decide to say something intelligent will I reply." -- Wilson news:qp5mc5phvv2s02rtot41h3jpi25qbpo6pb(a)4ax.com... |