Sagnac Revisited. Why Paul, Tom and Jerry are Hopelessly Wrong.
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From: Henry Wilson DSc on 24 Feb 2010 05:10 On Tue, 23 Feb 2010 21:59:22 -0800 (PST), artful <artful_me(a)hotmail.com> wrote: >On Feb 24, 4:32�pm, artful <artful...(a)hotmail.com> wrote: >> On Feb 24, 4:12�pm, ..@..(Henry Wilson DSc) wrote: >> You've not really clarified anything .. just lied some more. �But that >> is probably what counts as clarification for you. >> >> So .. let's clear things up, as your description was (and remains) >> very sloppy and self-contradictory. >> >> Where on earth is the gyroscope located .. at a (rotational) pole? �at >> the equator? >> >> Where is its axis (thru the centre of its spinning wheel that is) .. >> is it radially outward from the centre of the earth (so sitting >> upright), is it parallel to the earth's axis of rotation (so it is >> sitting on its side if it is at the equator, or upright if at the >> pole), is it parallel with the equator (also sitting on its side) >> >> Are you talking about the rotation of the gyroscope as a whole as a >> single entire device, or of the wheel inside the gyroscope? > >While I'm waiting .. I'll take another guess at what you're on about. > >The gyroscope is located at the equator, with its spinning-wheel axis >aligned tangentially to the equator (ie the wheel is spinning in what >appears to be a vertical plane thru the earth's poles around a >horizontal axis, according to a co-moving earthbound observer) > >The gyroscope frame will rotate with the earth (just over) 360 degrees >in a day. But will appear to not rotate to a co-moving earthbound >observer. > >The wheel and its axle will remain with a fixed orientation in any >inertial frame. So to a co-moving earthbound observer, the wheel+axle >appears to be rotating from having a horizontal axis, to vertical, >around to horizontal in the opposite direction, and then vertical in >the opposite direction and finally back to its original orientation. >So according to them it appears to make (just over) one rotation in a >day. So you finally understood the simple question. How long did that take you? Did somebody tell you the answer? The point is, rotating frame give wrong impressions. >So the earthbound observer sees the gyroscope wheel mechanism rotate >(just over) once in a day, and the gyroscope frame remain stationary. >An inertial observer sees the gyroscope wheel mechanism with fixed >orientation and the gyroscope frame rotate (just over) once in a day. Henry Wilson... ........provider of free physics lessons
From: Henry Wilson DSc on 25 Feb 2010 14:09 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. I might point out that the same approach 'refutes' SR which also relies on different path lengths in its analysis. >http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf >Quote from centre of page 2: ><< > 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. > >> > >So either: > you haven't read either of the papers >or: > you have a reading comprehension problem. > >Which is it? You have a frame jumping problem. You don't even know which frame you are using. In one sentence you SAY "In all cases the calculations will be made in the non rotating inertial frame where the centre of the ring is stationary." Then you immediately use the rotating frame. Gawd! How pathetic! Andro's going to really love this one. >> The use of the rotating frame introduces these imaginary effects: >> >> 1) The source and detection points for a particular light element appear to be >> the same. >> 2) The two path lengths appear identical. >> >> Changing frames can not alter basic physical facts. >> >> Because of its imaginary features, the 'rotating frame' is not a physically >> sound device, particularly when in the hands of amateurs like PA.. > >Quite. :-) >Mindless babble as expected. > >Thanks for confirming that you are unable to point out a single error. > >Frustrating, eh? :-) No. It's hilarious. You BaTh refutation also refutes SR. Brilliant! Now here is something that might help you. The 'phase' of the rays in the ring depends on DISTANCE, not TIME. Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 25 Feb 2010 16:30 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. > > 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. 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. 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. (Which proves that the phase difference must be zero.) 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. 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. But all this will obviously be futile, because the 'Doctor' Ralph Rabbidge is way too stupid to understand even the most basic issues. So why the hell do I bother? THAT's a mystery. -- Paul http://home.c2i.net/pb_andersen/
From: Androcles on 25 Feb 2010 16:42 "Henry Wilson DSc" <..@..> wrote in message news:mqhdo55or6m281p63838o7epvfq16shcmu(a)4ax.com... > 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. > > I might point out that the same approach 'refutes' SR which also relies on > different path lengths in its analysis. > > >>http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf >>Quote from centre of page 2: >><< >> 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. >> >> >> >>So either: >> you haven't read either of the papers >>or: >> you have a reading comprehension problem. >> >>Which is it? > > You have a frame jumping problem. > You don't even know which frame you are using. > > In one sentence you SAY "In all cases the calculations will be made in the > non > rotating inertial frame where the centre of the ring is stationary." > > Then you immediately use the rotating frame. > > Gawd! How pathetic! Andro's going to really love this one. > >>> The use of the rotating frame introduces these imaginary effects: >>> >>> 1) The source and detection points for a particular light element appear >>> to be >>> the same. >>> 2) The two path lengths appear identical. >>> >>> Changing frames can not alter basic physical facts. >>> >>> Because of its imaginary features, the 'rotating frame' is not a >>> physically >>> sound device, particularly when in the hands of amateurs like PA.. >> >>Quite. :-) >>Mindless babble as expected. >> >>Thanks for confirming that you are unable to point out a single error. >> >>Frustrating, eh? :-) > > No. It's hilarious. > You BaTh refutation also refutes SR. > > 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? "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 -- Androcles ........provider of expensive physics lessons Awilson (my pet chump) can't afford.
From: Henry Wilson DSc on 25 Feb 2010 22:23
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 |