Sagnac Revisited. Why Paul, Tom and Jerry are Hopelessly Wrong.
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From: Paul B. Andersen on 18 Mar 2010 10:11 On 17.03.2010 22:21, Henry Wilson DSc wrote: > 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. Good grief, Ralph. :-) > >> 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. If it is so simple, why don't you answer the question? Here it is again. Length of the rod is L, wavelength = L/2. v = c/2 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? The only clear answer I have got so far, is: Ralph Rabbidge aka Henry Wilson wrote: | There are never more than two wavelengths in either rod. | Observer movement does not change that. But you seem to have changed your mind since you gave that answer. So what is your answer now? No doubletalk. Specific numbers. -- Paul http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 18 Mar 2010 17:27 On Thu, 18 Mar 2010 15:11:23 +0100, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 17.03.2010 22:21, Henry Wilson DSc wrote: >> 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. > >Good grief, Henry :-) Your diagram is that of a four mirror interferometer that is NOT ROTATING. It is an interesting art work but says nothing about the sagnac effect. If you want to portray a rotating apparatus, you must draw the mirrors in their correct positions when the ray currently reuniting ray elements impinged on those mirrors. >>>> 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 . >>>> 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 This is a diagram of a nonrotating apparatus. >>> 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. > >If it is so simple, why don't you answer the question? >Here it is again. > >Length of the rod is L, wavelength = L/2. v = c/2 > >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? (2piR+/-vt)/L of course. You have introduced rotation, a totally different situation. The path lengths (rods) are different. >The only clear answer I have got so far, is: >Henry Wilson wrote: >| There are never more than two wavelengths in either rod. >| Observer movement does not change that. > >But you seem to have changed your mind since you gave that answer. >So what is your answer now? >No doubletalk. >Specific numbers. (2piR+/-vt)/L Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 19 Mar 2010 11:01 On 18.03.2010 22:27, Henry Wilson DSc wrote: > On Thu, 18 Mar 2010 15:11:23 +0100, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: > >> On 17.03.2010 22:21, Henry Wilson DSc wrote: >>> 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. >> >> Good grief, Henry :-) > > Your diagram is that of a four mirror interferometer that is NOT ROTATING. > > It is an interesting art work but says nothing about the sagnac effect. > > If you want to portray a rotating apparatus, you must draw the mirrors in their > correct positions when the ray currently reuniting ray elements impinged on > those mirrors. > > >>>>> 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 . > >>>>> 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 > > This is a diagram of a nonrotating apparatus. > >>>> 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. >> >> If it is so simple, why don't you answer the question? >> Here it is again. >> >> Length of the rod is L, wavelength = L/2. v = c/2 >> >> 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? > > (2piR+/-vt)/L of course. > > You have introduced rotation, a totally different situation. The path lengths > (rods) are different. > >> The only clear answer I have got so far, is: >> Henry Wilson wrote: >> | There are never more than two wavelengths in either rod. >> | Observer movement does not change that. >> >> But you seem to have changed your mind since you gave that answer. >> So what is your answer now? >> No doubletalk. >> Specific numbers. > > (2piR+/-vt)/L Since the lengths of the light beams are equal to the length of the circle, that is 2piR, and a wavelength according to Ritz is an absolute spatial distance, how come it can be different number of wavelengths in the two beams? Have you ever seen an four mirror interferometer, Henry? When you look at it, doesn't it look square even when it is rotating? Are the light beams not going from the half silvered mirror back to same? Are the light beams not square? Are there any light beams outside of the square? Isn't it looking like this? http://home.c2i.net/pb_andersen/images/Sagnac_Ritz.jpg If not, can you please draw how you think an instant image of a rotating interferometer should look like? -- Paul http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 19 Mar 2010 17:17 On Fri, 19 Mar 2010 16:01:56 +0100, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 18.03.2010 22:27, Henry Wilson DSc wrote: >> On Thu, 18 Mar 2010 15:11:23 +0100, "Paul B. Andersen" >>>> >>>> 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. >>> >>> If it is so simple, why don't you answer the question? >>> Here it is again. >>> >>> Length of the rod is L, wavelength = L/2. v = c/2 >>> >>> 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? >> >> (2piR+/-vt)/L of course. >> >> You have introduced rotation, a totally different situation. The path lengths >> (rods) are different. >> >>> The only clear answer I have got so far, is: >>> Henry Wilson wrote: >>> | There are never more than two wavelengths in either rod. >>> | Observer movement does not change that. >>> >>> But you seem to have changed your mind since you gave that answer. >>> So what is your answer now? >>> No doubletalk. >>> Specific numbers. >> >> (2piR+/-vt)/L > >Since the lengths of the light beams are equal to the length >of the circle, that is 2piR, and a wavelength according to Ritz >is an absolute spatial distance, how come it can be different >number of wavelengths in the two beams? Because their path lengths in the nonR frame are different, as any SRian should know. >Have you ever seen an four mirror interferometer, Henry? Yes, I have made one. >When you look at it, doesn't it look square even >when it is rotating? It does. >Are the light beams not going from the half silvered >mirror back to same? No. The mirror has moved by the time it gets there. >Are the light beams not square? No. ,...except when it is not rotating, as you have shown in your artwork, >Are there any light beams outside of the square? Yes Paul....not in your artwork though becaue that represents a nonrotating apparatus. >Isn't it looking like this? >http://home.c2i.net/pb_andersen/images/Sagnac_Ritz.jpg Whe not rotationg only. >If not, can you please draw how you think an instant >image of a rotating interferometer should look like? It should be obvious. Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 28 Mar 2010 09:01
On 19.03.2010 22:17, Henry Wilson DSc wrote: > On Fri, 19 Mar 2010 16:01:56 +0100, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: > >> Have you ever seen an four mirror interferometer, Henry? > > Yes, I have made one. > >> When you look at it, doesn't it look square even >> when it is rotating? > > It does. > >> Are the light beams not going from the half silvered >> mirror back to same? > > No. The mirror has moved by the time it gets there. > >> Are the light beams not square? > > No. ,...except when it is not rotating, as you have shown in your artwork, > >> Are there any light beams outside of the square? > > Yes Paul....not in your artwork though becaue that represents a nonrotating > apparatus. > >> Isn't it looking like this? >> http://home.c2i.net/pb_andersen/images/Sagnac_Ritz.jpg > > Whe not rotationg only. > >> If not, can you please draw how you think an instant >> image of a rotating interferometer should look like? > > It should be obvious. Quite. When you look at a rotating interferometer, it is square, but the light beams don't go between the mirrors, but are bouncing off points where there are no mirror. Could it be more obvious? :-) I think this will do, Ralph. You have reached the sky. -- Paul http://home.c2i.net/pb_andersen/ |