Sagnac Revisited. Why Paul, Tom and Jerry are Hopelessly Wrong.
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From: Paul B. Andersen on 27 Feb 2010 16:20 On 26.02.2010 21:19, Henry Wilson DSc wrote: > On Fri, 26 Feb 2010 12:26:20 +0100, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: > >> On 26.02.2010 04:23, Henry Wilson DSc wrote: >>> On Thu, 25 Feb 2010 22:30:26 +0100, "Paul B. Andersen"<someone(a)somewhere.no> >>> wrote: >>> Ralph Rabbidge, if you can read, read this again: ================================================= >>>> 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. >>>> >>>> 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. >>>> 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. > > In your submission http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf > and in your 'difference of wavelengths, method' you have claimed the path > lengths are 2piR. > > Do you deny that? (Remember everybody can see it quite plainly...and I have > made a copy so you cannot alter your blatant error now that I have pointed it > out) > > Yet just above that section, in the 'transit times method' you managed to > derive the correct values for the path lengths. > > How deceitful is this? > > You are a cheat. You have deliberately lied to try to discredit BaTh...yet in > doing so, you have also refuted SR because its 'transit time method' REQUIRES > the same 'different path lengths' that BaTh does. Why the hell don't you read the postings you are responding to? You must have a serious reading comprehension problem! >> [unsnip] >>>> But all this will obviously be futile, because >>>> the 'Doctor' Ralph Rabbidge is way too stupid to understand >>>> even the most basic issues. >> >> Thanks for confirming my words. Again and again over and over. Look at my other response to this post of yours. -- Paul http://home.c2i.net/pb_andersen/
From: Androcles on 27 Feb 2010 16:34 "Paul B. Andersen" <someone(a)somewhere.no> wrote in message news:4B898CB9.6000307(a)somewhere.no... > On 26.02.2010 21:19, Henry Wilson DSc wrote: >> On Fri, 26 Feb 2010 12:26:20 +0100, "Paul B. Andersen" >> <paul.b.andersen(a)somewhere.no> wrote: >> >>> On 26.02.2010 04:23, Henry Wilson DSc wrote: >>>> On Thu, 25 Feb 2010 22:30:26 +0100, "Paul B. >>>> Andersen"<someone(a)somewhere.no> >>>> wrote: >>>> > > Ralph Rabbidge, if you can read, read this again: > ================================================= > >>>>> 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. >>>>> >>>>> 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. >>>>> 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. > >> >> In your submission http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf >> and in your 'difference of wavelengths, method' you have claimed the path >> lengths are 2piR. >> >> Do you deny that? (Remember everybody can see it quite plainly...and I >> have >> made a copy so you cannot alter your blatant error now that I have >> pointed it >> out) >> >> Yet just above that section, in the 'transit times method' you managed to >> derive the correct values for the path lengths. >> >> How deceitful is this? >> >> You are a cheat. You have deliberately lied to try to discredit >> BaTh...yet in >> doing so, you have also refuted SR because its 'transit time method' >> REQUIRES >> the same 'different path lengths' that BaTh does. > > Why the hell don't you read the postings you are responding to? > You must have a serious reading comprehension problem! > >>> [unsnip] >>>>> But all this will obviously be futile, because >>>>> the 'Doctor' Ralph Rabbidge is way too stupid to understand >>>>> even the most basic issues. >>> >>> Thanks for confirming my words. > > Again and again over and over. > > Look at my other response to this post of yours. > > -- > Paul > > http://home.c2i.net/pb_andersen/ Why don't you have any sense? You must have a serious comprehension problem! Again and again over and over. Look at my other response to this post of yours. Hilarious, no? -- Androcles
From: Paul B. Andersen on 27 Feb 2010 16:57 On 26.02.2010 21:19, Henry Wilson DSc wrote: > In your submission http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf > and in your 'difference of wavelengths, method' you have claimed the path > lengths are 2piR. > > Do you deny that? (Remember everybody can see it quite plainly...and I have > made a copy so you cannot alter your blatant error now that I have pointed it > out) > > Yet just above that section, in the 'transit times method' you managed to > derive the correct values for the path lengths. > > How deceitful is this? > > You are a cheat. You have deliberately lied to try to discredit BaTh...yet in > doing so, you have also refuted SR because its 'transit time method' REQUIRES > the same 'different path lengths' that BaTh does. Since you have demonstrated such an astonishing confusion about how to count the number of wavelengths in a wave, I have made a simple thought experiment for you. Please read it carefully, it is of kindergarten simplicity, so even Doctor Ralph Rabbidge should be able to follow it. Given a rod with length L. At the left end of the rod there is a sound source which starts emitting sound at t=0. The sound is travelling at the speed c relative to the rest frame of the rod. At the right end of the rod, there is an absorber which will absorb the sound so that no wave is reflected. The wavelength of the sound wave is lambda = L/2. The rod is moving at the speed v in the stationary frame. The left end of the rod is at x=0 at t = 0. | L | S-----------| -> v | | |-----------------------------------> x 0 Below is a "movie" of the moving rod. Each drawing represents an instant image of the rod and the sound wave on the rod. The sound wave is drawn with asterisks. 'phi' is the phase of the source. The phase of the sound wave at the wave front is always zero. | | *-----------| phi = 0 | | |-----------|-----------------------> x 0 L * | |*----------| phi = pi/3 | | |-----------|-----------------------> x 0 L ** | |-*---------| phi = 2 pi/3 | | |-----------|-----------------------> x 0 L |** | *--*--------| phi = pi | | |-----------|-----------------------> x 0 L | ** | |*--*-------| phi = 4 pi/3 * | |-----------|-----------------------> x 0 L | ** | |-*--*------| phi = 5 pi/3 ** | |-----------|-----------------------> x 0 L | ** | *--*--*-----| phi = 2 pi |** | |-----------|-----------------------> x 0 L * ** | |*--*--*----| phi = 7 pi/3 | ** | |-----------|-----------------------> x 0 L ** ** | |-*--*--*---| phi = 8 pi/3 | ** | |-----------|-----------------------> x 0 L |** ** | *--*--*--*--| phi = 3 pi | ** | |-----------|-----------------------> x 0 L | ** ** | |*--*--*--*-| phi = 10 pi/3 * ** | |-----------|-----------------------> x 0 L | ** ** | |-*--*--*--*| phi = 11 pi/3 ** ** | |-----------|-----------------------> x 0 L At t = T the wave front reaches the right end of the rod at x = X : | ** **| *--*--*--*--* phi = 4 pi |** ** | |-----------|-----------|-----------> x 0 L X Thereafter, when t > T, we are in steady state and there is a continuous wave between the source and the emitter. Note that since the rod is a whole number of wavelengths long, the phase of the wave will be the same in both ends of the rod: * ** ** |*--*--*--*-| phi = 13 pi/3 | ** ** | |-----------|-----------|-----------> x 0 L X The path length of the wave front from the source to 'hit other end of rod', measured in the stationary frame is X. Since the wave front is moving at c+v in the stationary frame, this path length is X = (c+v)T The length of the rod is L, so the wave front will obviously reach the other end of the rod at T = L/v So the path length = L(c+v)/v In the drawing above v = c, so the path length is 2L = 4 lambda. ============================================================= Now the question is: How many wavelengths are there in the wave when t >= T? Is it 2 or is it 4? ============================================================ The answer is obvious. When t > T, there will always be 2 wavelengths in the wave. Now we repeat the scenario with the sound wave in the other end of the rod. | | |-----------* phi = 0 | | |-----------|-----------------------> x 0 L | * |----------*| phi = pi/3 | | |-----------|-----------------------> x 0 L | ** |---------*-| phi = 2 pi/3 | | |-----------|-----------------------> x 0 L | **| |--------*--* phi = pi | | |-----------|-----------------------> x 0 L | ** | |-------*--*| phi = 4 pi/3 | * |-----------|-----------------------> x 0 L | ** | |------*--*-| phi = 5 pi/3 | ** |-----------|-----------------------> x 0 L | ** | |-- --*--*--* phi = 2 pi | **| |-----------|-----------------------> x 0 L | ** * |----*--*--*| phi = 7 pi/3 | ** | |-----------|-----------------------> x 0 L | ** ** |---*--*--*-| phi = 8 pi/3 | ** | |-----------|-----------------------> x 0 L | ** **| |--*--*--*--* phi = 3 pi | ** | |-----------|-----------------------> x 0 L | ** ** | |-*--*--*--*| phi = 10 pi/3 | ** * |-----------|-----------------------> x 0 L | ** ** | |*--*--*--*-| phi = 11 pi/3 | ** ** |-----------|-----------------------> x 0 L At t = T the wave front reaches the left end of the rod at x = X . (In the general case when v < c, X < L) : |** ** | *--*--*--*--* phi = 4 pi | ** **| |-----------|-----------|-----------> x 0 L X Thereafter, when t > T, we are in steady state and there is a continuous wave between the source and the emitter. Note that since the rod is a whole number of wavelengths long, the phase of the wave will be the same in both ends of the rod: ** ** * |-*--*--*--*| phi = 13 pi/3 | ** ** | |-----------|-----------|-----------> x 0 L X The path length of the wave front from the source to 'hit other end of rod', measured in the stationary frame is L-X. Since the wave front is moving at c-v in the stationary frame, This path length is L-X = (c-v)T The length of the rod is L, so the wave front will obviously reach the other end of the rod at T = L/v So the path length = L(c-v)/v In the drawing above v = c, so the path length is 0. ============================================================= Now the question is: How many wavelengths are there in the wave when t >= T? Is it 2 or is it 0? ============================================================ The answer is obvious. When t > T, there will always be 2 wavelengths in the wave. Now we let there be one sound source at each end of the rod, and both sources are in in phase. At t = T when phi = 4 pi, we will have this: | ** **| *--*--*--*--* phi = 4 pi |** ** | |-----------|-----------------------> x 0 L + this: |** ** | *--*--*--*--* phi = 4 pi | ** **| |-----------|-----------------------> x 0 L = this | | ************* phi = 4 pi | | |-----------|-----------------------> x 0 L At t > T when phi = 13 pi/3 pi, we will have this: * ** ** |*--*--*--*-| phi = 13 pi/3 | ** ** | |-----------|-----------------------> x 0 L + this: ** ** * |-*--*--*--*| phi = 13 pi/3 | ** ** | |-----------|-----------------------> x 0 L = this * * * |* * * *| |-----------| phi = 13 pi/3 | * * * * | | * * |-----------|-----------------------> x 0 L We will have a standing wave on the rod. ------------------------------------------------------------- Now we bend the rod to a circle, so that its right and left ends meet. We let there be a single sound source which emits sound in both directions. We let the circular rod rotate so that its peripheral speed is v, and the centre of the circle is stationary in the stationary frame. ============================================================= Now the question is: How many wavelengths are there in the two contra moving waves when t >= T? Is it 2 in both, or will some mysterious effect due to rotation make the number of wavelengths be 4 for one of the waves and 0 for the other? ============================================================ This experiment it quite feasible. If it is an iron rod, c ~= 5 km/s. If the rod is 20m long, f = 500Hz to make lambda = 10m. To make v = 5 km/s, the circular rod (radius 3.18m) would have to spin at 250 rounds per second, or 15000 r.p.m. Fast, but not impossible. The wave front of one of the waves would then be stationary, while the wave front of the other wave would move at 10km/s in the stationary frame. -- Paul http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 27 Feb 2010 23:42 On Sat, 27 Feb 2010 22:57:37 +0100, "Paul B. Andersen" <someone(a)somewhere.no> wrote: >On 26.02.2010 21:19, Henry Wilson DSc wrote: >> In your submission http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf >> and in your 'difference of wavelengths, method' you have claimed the path >> lengths are 2piR. >> >> Do you deny that? (Remember everybody can see it quite plainly...and I have >> made a copy so you cannot alter your blatant error now that I have pointed it >> out) >> >> Yet just above that section, in the 'transit times method' you managed to >> derive the correct values for the path lengths. >> >> How deceitful is this? >> >> You are a cheat. You have deliberately lied to try to discredit BaTh...yet in >> doing so, you have also refuted SR because its 'transit time method' REQUIRES >> the same 'different path lengths' that BaTh does. > >Since you have demonstrated such an astonishing confusion about how >to count the number of wavelengths in a wave, I have made a simple >thought experiment for you. > >Given a rod with length L. At the left end of the rod there >is a sound source which starts emitting sound at t=0. The sound >is travelling at the speed c relative to the rest frame of the rod. >At the right end of the rod, there is an absorber which will absorb >the sound so that no wave is reflected. >The wavelength of the sound wave is lambda = L/2. >The rod is moving at the speed v in the stationary frame. >The left end of the rod is at x=0 at t = 0. > > > > | L | > S-----------| -> v > | | > |-----------------------------------> x > 0 > > >Below is a "movie" of the moving rod. >Each drawing represents an instant image of >the rod and the sound wave on the rod. >The sound wave is drawn with asterisks. >'phi' is the phase of the source. >The phase of the sound wave at the wave front >is always zero. > > | | > *-----------| phi = 0 > | | > |-----------|-----------------------> x > 0 L > > * | > |*----------| phi = pi/3 > | | > |-----------|-----------------------> x > 0 L > > ** | > |-*---------| phi = 2 pi/3 > | | > |-----------|-----------------------> x > 0 L > > |** | > *--*--------| phi = pi > | | > |-----------|-----------------------> x > 0 L > > | ** | > |*--*-------| phi = 4 pi/3 > * | > |-----------|-----------------------> x > 0 L > > | ** | > |-*--*------| phi = 5 pi/3 > ** | > |-----------|-----------------------> x > 0 L > > | ** | > *--*--*-----| phi = 2 pi > |** | > |-----------|-----------------------> x > 0 L > > * ** | > |*--*--*----| phi = 7 pi/3 > | ** | > |-----------|-----------------------> x > 0 L > > ** ** | > |-*--*--*---| phi = 8 pi/3 > | ** | > |-----------|-----------------------> x > 0 L > > |** ** | > *--*--*--*--| phi = 3 pi > | ** | > |-----------|-----------------------> x > 0 L > > | ** ** | > |*--*--*--*-| phi = 10 pi/3 > * ** | > |-----------|-----------------------> x > 0 L > > | ** ** | > |-*--*--*--*| phi = 11 pi/3 > ** ** | > |-----------|-----------------------> x > 0 L > >At t = T the wave front reaches the right end >of the rod at x = X : > > | ** **| > *--*--*--*--* phi = 4 pi > |** ** | > |-----------|-----------|-----------> x > 0 L X > >Thereafter, when t > T, we are in steady state >and there is a continuous wave between the source >and the emitter. Note that since the rod is a whole >number of wavelengths long, the phase of the wave >will be the same in both ends of the rod: > > * ** ** > |*--*--*--*-| phi = 13 pi/3 > | ** ** | > |-----------|-----------|-----------> x > 0 L X > > > >The path length of the wave front from the source to >'hit other end of rod', measured in the stationary >frame is X. >Since the wave front is moving at c+v in the stationary frame, >this path length is X = (c+v)T >The length of the rod is L, so the wave front >will obviously reach the other end of the rod at T = L/v >So the path length = L(c+v)/v > >In the drawing above v = c, so the path length is 2L = 4 lambda. > >============================================================= >Now the question is: >How many wavelengths are there in the wave when t >= T? >Is it 2 or is it 4? >============================================================ > >The answer is obvious. When t > T, there will always be >2 wavelengths in the wave. > >Now we repeat the scenario with the sound wave in the other >end of the rod. > > | | > |-----------* phi = 0 > | | > |-----------|-----------------------> x > 0 L > > | * > |----------*| phi = pi/3 > | | > |-----------|-----------------------> x > 0 L > > | ** > |---------*-| phi = 2 pi/3 > | | > |-----------|-----------------------> x > 0 L > > | **| > |--------*--* phi = pi > | | > |-----------|-----------------------> x > 0 L > > | ** | > |-------*--*| phi = 4 pi/3 > | * > |-----------|-----------------------> x > 0 L > > | ** | > |------*--*-| phi = 5 pi/3 > | ** > |-----------|-----------------------> x > 0 L > > | ** | > |-- --*--*--* phi = 2 pi > | **| > |-----------|-----------------------> x > 0 L > > | ** * > |----*--*--*| phi = 7 pi/3 > | ** | > |-----------|-----------------------> x > 0 L > > | ** ** > |---*--*--*-| phi = 8 pi/3 > | ** | > |-----------|-----------------------> x > 0 L > > | ** **| > |--*--*--*--* phi = 3 pi > | ** | > |-----------|-----------------------> x > 0 L > > | ** ** | > |-*--*--*--*| phi = 10 pi/3 > | ** * > |-----------|-----------------------> x > 0 L > > | ** ** | > |*--*--*--*-| phi = 11 pi/3 > | ** ** > |-----------|-----------------------> x > 0 L > > >At t = T the wave front reaches the left end >of the rod at x = X . >(In the general case when v < c, X < L) : > > |** ** | > *--*--*--*--* phi = 4 pi > | ** **| > |-----------|-----------|-----------> x > 0 L > X > >Thereafter, when t > T, we are in steady state >and there is a continuous wave between the source >and the emitter. Note that since the rod is a whole >number of wavelengths long, the phase of the wave >will be the same in both ends of the rod: > > ** ** * > |-*--*--*--*| phi = 13 pi/3 > | ** ** | > |-----------|-----------|-----------> x > 0 L > X > > >The path length of the wave front from the source to >'hit other end of rod', measured in the stationary >frame is L-X. >Since the wave front is moving at c-v in the stationary frame, >This path length is L-X = (c-v)T >The length of the rod is L, so the wave front >will obviously reach the other end of the rod at T = L/v >So the path length = L(c-v)/v > >In the drawing above v = c, so the path length is 0. > >============================================================= >Now the question is: >How many wavelengths are there in the wave when t >= T? >Is it 2 or is it 0? >============================================================ > >The answer is obvious. When t > T, there will always be >2 wavelengths in the wave. > >Now we let there be one sound source at each end of the rod, >and both sources are in in phase. > >At t = T when phi = 4 pi, we will have > > this: > | ** **| > *--*--*--*--* phi = 4 pi > |** ** | > |-----------|-----------------------> x > 0 L > > + this: > |** ** | > *--*--*--*--* phi = 4 pi > | ** **| > |-----------|-----------------------> x > 0 L > >= this > > | | > ************* phi = 4 pi > | | > |-----------|-----------------------> x > 0 L > >At t > T when phi = 13 pi/3 pi, we will have > > this: > > * ** ** > |*--*--*--*-| phi = 13 pi/3 > | ** ** | > |-----------|-----------------------> x > 0 L > > + this: > > ** ** * > |-*--*--*--*| phi = 13 pi/3 > | ** ** | > |-----------|-----------------------> x > 0 L > >= this > * * * > |* * * *| > |-----------| phi = 13 pi/3 > | * * * * | > | * * > |-----------|-----------------------> x > 0 L > >We will have a standing wave on the rod. > > >------------------------------------------------------------- > > >Now we bend the rod to a circle, so that its right and left >ends meet. We let there be a single sound source which emits >sound in both directions. We let the circular rod rotate so >that its peripheral speed is v, and the centre of the circle >is stationary in the stationary frame. > >============================================================= >Now the question is: >How many wavelengths are there in the two contra moving waves >when t >= T? >Is it 2 in both, or will some mysterious effect due to rotation >make the number of wavelengths be 4 for one of the waves >and 0 for the other? >============================================================ > >This experiment it quite feasible. >If it is an iron rod, c ~= 5 km/s. >If the rod is 20m long, f = 500Hz to make lambda = 10m. >To make v = 5 km/s, the circular rod (radius 3.18m) would >have to spin at 250 rounds per second, or 15000 r.p.m. >Fast, but not impossible. > >The wave front of one of the waves would then be stationary, >while the wave front of the other wave would move at 10km/s >in the stationary frame. You have correctly described what would happen to a traveling wave in A MEDIUM when the source is at rest wrt the medium. Ballistic light is not a traveling wave in a medium. Nor can ring gyro operation be regarded as similar to a pair of machine guns firing bullets inside a frictionless torus. Light is obviously funny stuff. You continue to ignore the path length differences that exist IN THE NONROTATING FRAME. ..yet you recognized them in your SR analysis at http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf. Why the sudden switch, Paul? Rotating frames don't actually exist and anyone who tries to use them is likely to end up in trouble. Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 28 Feb 2010 08:10
On 28.02.2010 05:42, Henry Wilson DSc wrote: > > > You have correctly described what would happen to a traveling wave in A MEDIUM > when the source is at rest wrt the medium. > > Ballistic light is not a traveling wave in a medium. So please explain what's wrong in the following. No hand waving, please. ("Light is funny stuff") Be specific. Given a rod with length L. At the left end of the rod there is a light source which starts emitting light at t=0. The light is according to Ritz's emission theory travelling at the speed c relative to the rest frame of the rod. (Because the source is stationary in this frame.) At the right end of the rod, there is an absorber which will absorb the light so that no wave is reflected. The wavelength of the light is lambda = L/2. The rod is moving at the speed v in the stationary frame. The left end of the rod is at x=0 at t = 0. S - - - - ->A |-----------| -> v L |-----------------------------------> x 0 Below is a "movie" of the moving rod. Each drawing represents an instant image of the rod and the light wave along the rod. The light wave is drawn with asterisks. 'phi' is the phase of the source. The phase of the light wave at the wave front is always zero. | | *-----------| phi = 0 | | |-----------|-----------------------> x 0 L ** | |-*---------| phi = 2 pi/3 | | |-----------|-----------------------> x 0 L | ** | |*--*-------| phi = 4 pi/3 * | |-----------|-----------------------> x 0 L | ** | *--*--*-----| phi = 2 pi |** | |-----------|-----------------------> x 0 L ** ** | |-*--*--*---| phi = 8 pi/3 | ** | |-----------|-----------------------> x 0 L | ** ** | |*--*--*--*-| phi = 10 pi/3 * ** | |-----------|-----------------------> x 0 L At t = T the wave front reaches the right end of the rod at x = X : | ** **| *--*--*--*--* phi = 4 pi |** ** | |-----------|-----|-----------------> x 0 L X Thereafter, when t > T, we are in steady state and there is a continuous wave between the source and the absorber. Note that since the rod is a whole number of wavelengths long, the phase of the wave will be the same in both ends of the rod: ** ** * |-*--*--*--*| phi = 14 pi/3 | ** ** | |-----------|-----|-----------------> x 0 L X The path length of the wave front from the source to 'hit other end of rod', measured in the stationary frame is X. Since the wave front is moving at c+v in the stationary frame, this path length is X = (c+v)T The length of the rod is L, so the wave front will obviously reach the other end of the rod at T = L/v So the path length = L(c+v)/v In the drawing above v = c/2, so the path length is 3L/2 = 3 lambda. ============================================================= Now the question is: How many wavelengths are there in the wave when t >= T? Is it 2 or is it 3? ============================================================ -- Paul http://home.c2i.net/pb_andersen/ |