Sagnac Revisited. Why Paul, Tom and Jerry are Hopelessly Wrong.
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From: Henry Wilson DSc on 15 Mar 2010 17:06 On Mon, 15 Mar 2010 11:44:06 +0100, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 12.03.2010 23:20, Henry Wilson DSc wrote: >> On Fri, 12 Mar 2010 21:25:47 +0100, "Paul B. Andersen"<someone(a)somewhere.no> >> wrote: >> >>> >>> The absolute spatial length of the rod is two wavelengths. >>> So the path length of the wave front in the stationary frame >>> is utterly irrelevant to the number of wavelengths in the wave. >> >> there is no 'path length' in the inertial situation. >> >> Path lengths only exist in the rotating experiment. > >Please don't state stupidietes like this. > >Wave front start at x=0: > > | | > *-----------| > | | > |-----------|-----------------------> x > 0 L > >And reaches the end of the rod at x=X: > > | ** **| > *--*--*--*--* > |** ** | > |-----------|-----|-----------------> x > 0 L X > >The 'path length' is the length of the trajectory of >the wave front measured in the stationary frame. >In this case it is (X-0) = 3L/2. The only frame that matters is the source frame. The light never travels further than L and there are two wavelengths in that length. The movement of other observers cannot and does not alter that fact. You should learn the real meaing of the PoR. You are as bad as Androcles. You believe the distance between water wavecrests depends on the speed of your boat. How naive.... >>> The anti-clockwise going wave: >>> ----------------------------- >>> Wave front start at phi = 2 pi >>> And hits the absorber at phi = pi >>> >>> The length of the trajectory of the wave front (path length) >>> measured in the stationary frame is = L/2. >>> >>> But when we are in steady state: >>> "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD >>> no matter who looks at it. >>> The distance between crests in the source frame defines >>> an absolute spatial interval." >>> >>> The absolute spatial length of the rod is two wavelengths. >>> So the path length of the wave front in the stationary frame >>> is utterly irrelevant to the number of wavelengths in the wave. >> >> Where would you get that idea? >> The 'wavelength' is an bsolute distance. >> If two paths are of different length, they will contain different numbers of >> wavelengths. > >Indeed. The path lengths measured in wavelengths are different. >Don't you read what I write? >######################################################################## ># Since one wavelength is L/2, these path lengths in the stationary ># frame are equal to three wavelengths and one wavelength respectively. >####################################################################### > >But as YOU so correctly put it: > "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD > no matter who looks at it. > The distance between crests in the source frame defines > an absolute spatial interval." > >So the path length of the wave front in the stationary frame >is utterly irrelevant to the number of wavelengths in the wave. There is no 'stationary frame' other than that of the source. Other observers can do what they like with their frames and will not affect what happens in the source frame. >So: >######################################################################## ># But nobody would get the stupid idea that this means that ># there are any waves containing three or one wavelengths. ># ># Or would they? :-) >########################################################################## > >>>>> According to the emission theory, how many wavelengths are >>>>> there now in the two contra moving waves? >>> >>> BEHOLD, BEHOLD: >>> >>>> You must analyse it using the NonR frame. The path lengths differ by 2vt...or >>>> 2vt/L absolute wavelengths. >>> >>> ######################################################################## >>> # Since one wavelength is L/2, these path lengths in the stationary >>> # frame are equal to three wavelengths and one wavelength respectively. >>> # >>> # But nobody would get the stupid idea that this means that >>> # there are any waves containing three or one wavelengths. >>> # >>> # Or would they? :-) >>> ########################################################################## >>> >>> You just did! :-) >>> >>> Try again. >>> >>> The question was: >>> According to the emission theory, how many wavelengths are >>> there now in the two contra moving waves? >> >> You are raving about inertial movement when it is rotation around a closed ring >> that is important. >> I cannot see why you are having so much trouble understanding this. It is very >> simple. > >Quite. >It is very simple. >The length of the rod is L, which according to the emission >theory is 'an absolute spatial interval'. >The absolute spatial interval 'one wavelength' is L/2. >So "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD >no matter who looks at it. >The distance between crests in the source frame defines >an absolute spatial interval." > >I cannot see why you are having so much trouble understanding this. >It is very simple. > >So maybe you now can answer the simple question? > > According to the emission theory, how many wavelengths are > there now in the two contra moving waves? There are never more than two wavelengths in either rod. Observer movement does not change that. We should talk about the real issue....the rotation of an optical gyro fibre carrying light. In the source frame, there are 2piR/L wavelengths around the ring. However, the light entering the detector left the emitter time interval t beforehand when the emitter was at a different location in the NonR frame....So, in the nonR frame, the path lengths of oppositely moving rays differ by the distance 2vdt. The path lengths are therefore 2piR+vt and 2piR-vt where t = 2piR/c. Since 'wavelength' is absolute and frame invariant and since photons are not simple oscillators, light entering the detector from opposite directions always differ in phase by 4piRv/cL.....= 4Aw/cL Simple isn't it. Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 16 Mar 2010 07:38 On 15.03.2010 22:06, Henry Wilson DSc wrote: > On Mon, 15 Mar 2010 11:44:06 +0100, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: >> [unsnip] >> Now we modify the experiment a little. >> We bend the rod to a circle, and have one source emitting >> light in both directions. The light is guided (by mirrors) >> to move along the circular rod. >> The circular rod is rotating with the peripheral speed v >> as measured in the stationary frame. >> According to the emission theory, how many wavelengths are >> there now in the two contra moving waves? > > There are never more than two wavelengths in either rod. Observer movement does > not change that. Thanks for a clear answer. There are always two wavelengths in either wave, even when the rod is circular and rotating. So we agree that according to the emission theory the number of wavelengths in the two contra moving waves never change. > We should talk about the real issue....the rotation of an optical gyro fibre > carrying light. > > In the source frame, there are 2piR/L wavelengths around the ring. > However, the light entering the detector left the emitter time interval t > beforehand when the emitter was at a different location in the NonR > frame....So, in the nonR frame, the path lengths of oppositely moving rays > differ by the distance 2vdt. The path lengths are therefore 2piR+vt and 2piR-vt > where t = 2piR/c. Quite. ######################################################################## # Since one wavelength is L/2, these path lengths in the stationary # frame are equal to three wavelengths and one wavelength respectively. # # But nobody would get the stupid idea that this means that # there are any waves containing three or one wavelengths. # # Or would they? :-) ########################################################################## No, they wouldn't. Ralph Rabbidge aka Henry wilson has finaly agreed that according to the emission theory the number of wavelengths in the two contra moving waves never change. > Since 'wavelength' is absolute and frame invariant and since photons are not > simple oscillators, light entering the detector from opposite directions always > differ in phase by 4piRv/cL.....= 4Aw/cL The phases at the source of the two contra moving waves are equal, the number of wavelgths in the two waves are equal, and the phases at the other end are different? Ralph .... :-) -- Paul http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 16 Mar 2010 15:58 On Tue, 16 Mar 2010 12:38:16 +0100, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 15.03.2010 22:06, Henry Wilson DSc wrote: >> On Mon, 15 Mar 2010 11:44:06 +0100, "Paul B. Andersen" >> <paul.b.andersen(a)somewhere.no> wrote: > > >> [unsnip] >>> Now we modify the experiment a little. >>> We bend the rod to a circle, and have one source emitting >>> light in both directions. The light is guided (by mirrors) >>> to move along the circular rod. >>> The circular rod is rotating with the peripheral speed v >>> as measured in the stationary frame. > >>> According to the emission theory, how many wavelengths are >>> there now in the two contra moving waves? >> >> There are never more than two wavelengths in either rod. Observer movement does >> not change that. > >Thanks for a clear answer. >There are always two wavelengths in either wave, >even when the rod is circular and rotating. > >So we agree that according to the emission theory the number >of wavelengths in the two contra moving waves never change. Not unless you introduce a nonsensical definition of 'wavelength'. As I pointed out before, you seem to be confusing 'wavelength' with the distance a relatively moving observer travels during one cycle. I gather you still agree with Androcles.... that the distance between water wavecrests varies with the speed of one's boat? >> We should talk about the real issue....the rotation of an optical gyro fibre >> carrying light. >> >> In the source frame, there are 2piR/L wavelengths around the ring. >> However, the light entering the detector left the emitter time interval t >> beforehand when the emitter was at a different location in the NonR >> frame....So, in the nonR frame, the path lengths of oppositely moving rays >> differ by the distance 2vdt. The path lengths are therefore 2piR+vt and 2piR-vt >> where t = 2piR/c. > >Quite. >######################################################################## ># Since one wavelength is L/2, these path lengths in the stationary ># frame are equal to three wavelengths and one wavelength respectively. ># ># But nobody would get the stupid idea that this means that ># there are any waves containing three or one wavelengths. ># ># Or would they? :-) >########################################################################## > >No, they wouldn't. >Henry wilson has finaly agreed that >according to the emission theory the number of wavelengths >in the two contra moving waves never change. Not in the inertial situation, no...but when the paths are circular, a different theory applies. >> Since 'wavelength' is absolute and frame invariant and since photons are not >> simple oscillators, light entering the detector from opposite directions always >> differ in phase by 4piRv/cL.....= 4Aw/cL > >The phases at the source of the two contra moving waves are equal, >the number of wavelgths in the two waves are equal, and the phases >at the other end are different? > >Henry .... :-) Paul cannot handle the 'imaginary characteristics' of rotating frames. The paths of the two rays are different in length in both the rotating and the inertial frames . In the rotating frame, the emission point of each 'photon' moves invisibly away. To help you understand this simple piece of physics, let the rotating observer mark a point on the nonR frame when the photon is emitted. What happens to that mark, Paul? Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 17 Mar 2010 08:35 On 16.03.2010 20:58, Henry Wilson DSc wrote: > On Tue, 16 Mar 2010 12:38:16 +0100, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: > >> On 15.03.2010 22:06, Henry Wilson DSc wrote: >>> On Mon, 15 Mar 2010 11:44:06 +0100, "Paul B. Andersen" >>> <paul.b.andersen(a)somewhere.no> wrote: >>>> Now we modify the experiment a little. >>>> We bend the rod to a circle, and have one source emitting >>>> light in both directions. The light is guided (by mirrors) >>>> to move along the circular rod. >>>> The circular rod is rotating with the peripheral speed v >>>> as measured in the stationary frame. >> >>>> According to the emission theory, how many wavelengths are >>>> there now in the two contra moving waves? >>> >>> There are never more than two wavelengths in either rod. Observer movement does >>> not change that. >> >> Thanks for a clear answer. >> There are always two wavelengths in either wave, >> even when the rod is circular and rotating. >> >> So we agree that according to the emission theory the number >> of wavelengths in the two contra moving waves never change. > > Not unless you introduce a nonsensical definition of 'wavelength'. So unless we introduce a nonsensical definition of wavelength, there are always two wavelengths in either wave, even when the rod is circular and rotating. This is illustrated here: http://home.c2i.net/pb_andersen/images/Sagnac_Ritz.jpg Unless we introduce a nonsensical definition of wavelength, there are always 20 wavelengths in either wave. > > As I pointed out before, you seem to be confusing 'wavelength' with the > distance a relatively moving observer travels during one cycle. > > I gather you still agree with Androcles.... that the distance between water > wavecrests varies with the speed of one's boat? > >>> We should talk about the real issue....the rotation of an optical gyro fibre >>> carrying light. >>> >>> In the source frame, there are 2piR/L wavelengths around the ring. >>> However, the light entering the detector left the emitter time interval t >>> beforehand when the emitter was at a different location in the NonR >>> frame....So, in the nonR frame, the path lengths of oppositely moving rays >>> differ by the distance 2vdt. The path lengths are therefore 2piR+vt and 2piR-vt >>> where t = 2piR/c. >> >> Quite. >> ######################################################################## >> # Since one wavelength is L/2, these path lengths in the stationary >> # frame are equal to three wavelengths and one wavelength respectively. >> # >> # But nobody would get the stupid idea that this means that >> # there are any waves containing three or one wavelengths. >> # >> # Or would they? :-) >> ########################################################################## >> >> No, they wouldn't. >> Ralph Rabbidge aka Henry wilson has finaly agreed that >> according to the emission theory the number of wavelengths >> in the two contra moving waves never change. > > Not in the inertial situation, no...but when the paths are circular, a > different theory applies. Changed your mind again? :-) > >>> Since 'wavelength' is absolute and frame invariant and since photons are not >>> simple oscillators, light entering the detector from opposite directions always >>> differ in phase by 4piRv/cL.....= 4Aw/cL >> >> The phases at the source of the two contra moving waves are equal, >> the number of wavelgths in the two waves are equal, and the phases >> at the other end are different? >> >> Henry .... :-) > > Paul cannot handle the 'imaginary characteristics' of rotating frames. Quite. I can't handle the imaginary phase shift predicted by the emission theory. But I can handle that the emission theory predicts no real phase shift. > The paths of the two rays are different in length in both the rotating and the > inertial frames . Quite. ######################################################################## # Since one wavelength is L/2, these path lengths in the stationary # frame are equal to three wavelengths and one wavelength respectively. # # But nobody would get the stupid idea that this means that # there are any waves containing three or one wavelengths. # # Or would they? :-) ########################################################################## Ralph Rabbidge aka Henry Wilson has now changed his mind, so now he would. > In the rotating frame, the emission point of each 'photon' moves invisibly > away. > > To help you understand this simple piece of physics, let the rotating observer > mark a point on the nonR frame when the photon is emitted. > What happens to that mark, Paul? You can see that point here: http://home.c2i.net/pb_andersen/images/Sagnac_Ritz.jpg It's where the two paths start on the right side of the drawing. That point is not on either of the two square shaped light beams. So what about it? -- Paul http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 17 Mar 2010 17:21
On Wed, 17 Mar 2010 13:35:07 +0100, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 16.03.2010 20:58, Henry Wilson DSc wrote: >> On Tue, 16 Mar 2010 12:38:16 +0100, "Paul B. Andersen" >> <paul.b.andersen(a)somewhere.no> wrote: >>>>> According to the emission theory, how many wavelengths are >>>>> there now in the two contra moving waves? >>>> >>>> There are never more than two wavelengths in either rod. Observer movement does >>>> not change that. >>> >>> Thanks for a clear answer. >>> There are always two wavelengths in either wave, >>> even when the rod is circular and rotating. >>> >>> So we agree that according to the emission theory the number >>> of wavelengths in the two contra moving waves never change. >> >> Not unless you introduce a nonsensical definition of 'wavelength'. > >So unless we introduce a nonsensical definition of wavelength, >there are always two wavelengths in either wave, even when the rod >is circular and rotating. > >This is illustrated here: >http://home.c2i.net/pb_andersen/images/Sagnac_Ritz.jpg As I have pointed out before, this diagram is misleading and plainly wrong. It ignores time and is very amateurish. It infers that the part of a wave reaching a mirror was emitted from the previous mirror when THAT mirror was at the position shown. In fact, that mirror was NOT at the position drawn by you but displaced a distance vt to one side. Nor does it point out that the lines are actually curved, even if the consequences are only second order at low speeds. >Unless we introduce a nonsensical definition of wavelength, >there are always 20 wavelengths in either wave. My advice to you is to concern yourself more with PATH LENGTH than with wavelength. >> Not in the inertial situation, no...but when the paths are circular, a >> different theory applies. > >Changed your mind again? :-) Not at all. Rotational movement is absolute. Inertial movement is relative....two totally different situations. >>>> Since 'wavelength' is absolute and frame invariant and since photons are not >>>> simple oscillators, light entering the detector from opposite directions always >>>> differ in phase by 4piRv/cL.....= 4Aw/cL >>> >>> The phases at the source of the two contra moving waves are equal, >>> the number of wavelgths in the two waves are equal, and the phases >>> at the other end are different? >>> >>> Henry .... :-) >> >> Paul cannot handle the 'imaginary characteristics' of rotating frames. > >Quite. >I can't handle the imaginary phase shift predicted by the emission theory. >But I can handle that the emission theory predicts no real phase shift. > >> The paths of the two rays are different in length in both the rotating and the >> inertial frames . >Quite. >######################################################################## ># Since one wavelength is L/2, these path lengths in the stationary ># frame are equal to three wavelengths and one wavelength respectively. ># ># But nobody would get the stupid idea that this means that ># there are any waves containing three or one wavelengths. ># ># Or would they? :-) >########################################################################## > >Henry Wilson has now changed his mind, Andersen needs to completly refurbish his, if he can locate it. >so now he would. Wavelength, defined as the distance traveled in the source frame during one cycle, is absolute and invariant. There are more wavelengths in one path than the other. Therefore the waves are not in phase at the detector. This is so simple.... >> In the rotating frame, the emission point of each 'photon' moves invisibly >> away. >> >> To help you understand this simple piece of physics, let the rotating observer >> mark a point on the nonR frame when the photon is emitted. >> What happens to that mark, Paul? > >You can see that point here: >http://home.c2i.net/pb_andersen/images/Sagnac_Ritz.jpg >It's where the two paths start on the right side of the drawing. > >That point is not on either of the two square shaped light beams. > >So what about it? Your diagram is wrong. The light from both rays shown reaching a point on the last mirror did not originate from that point but from a point displaced vt clockwise. THIS IS SO SIMPLE. Henry Wilson... ........provider of free physics lessons |