Wavelength & Frequency experiments for SR
in [Relativity]
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From: Dr. Henri Wilson on 1 Jan 2008 00:50 On Tue, 1 Jan 2008 09:19:12 +1100, "Jeckyl" <noone(a)nowhere.com> wrote: >"Dr. Henri Wilson" <HW@....> wrote in message >news:pckin3lfg5mj01c5hc3csnfgome9dd8t32(a)4ax.com... >>> >>>> Note, my BaTh Sagnac analysis [...] >>> >>>So you have nothing to say about this refutation of your "BaTh". All you >>>have is irrelevant nonsense. >> >> My analysis does not require the wavelength to be constant. >> >> The same result is achieved using the intrinsic frequencies of emitted >> photons. >> The math is straightforward and simple. There can be no argument. >> >> Here it is again: >> >> 'f' is source frequency in the source frame. >> >> In the NONROTATING frame >> >> (using the diagram at http://www.mathpages.com/rr/s2-07/2-07.htm) >> According to BaTh: >> >> for ray 1, (c+v)t = 2piR + vt or.... t = 2piR/c.........(t = travel time) >> for ray 2, (c-v)t = 2piR - vt or.... t = 2piR/c .......(travel times are >> equal) >> >> Path length of ray 1 is 2piR(c+v)/c >> .......................of ray 2 is 2piR(c-v)/c >> >> The source frequency is doppler shifted in the non-rotating frame: >> Frequency of ray 1 is f(c+v)/c....(rate of wavecrests passing a point at >> rest) >> Frequency of ray 2 is f(c-v)/c..... >> >> The number of cycles in each ray is calculated as (travel time/frequency) >> Number of cycles (wavelengths) in ray 1 = [2piR/c].[f(c+v)/c] = >> 2piRf(c+v)/c^2 >> Number of cycles (wavelengths) in ray 2 = [2piR/c].[f(c-v)/c] = >> 2piRf(c-v)/c^2 >> ...................................................... >> >> Fringe displacement = difference in numbers of wavelengths in each ray > >No .. that assumes that the waves are still in sync at the original source >position in the non-rotating frame. As they have been oscillating there for >the same time at different frequencies, you cannot assume that. What kind of stupid statement is that? > >> = 4piRvf/c^2 >> = 4Aw/cL. >> .........(where, w = angular velocity, A = area of ring, L = wavelength) >> >> If you can find anything wrong with this please let us know..... > >I just did (and have before) > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Jeckyl on 1 Jan 2008 03:32 "Dr. Henri Wilson" <HW@....> wrote in message news:b3ljn35o7kcvl2jkssdse56nj0tdlqhk8d(a)4ax.com... > On Tue, 1 Jan 2008 09:19:12 +1100, "Jeckyl" <noone(a)nowhere.com> wrote: > >>"Dr. Henri Wilson" <HW@....> wrote in message >>news:pckin3lfg5mj01c5hc3csnfgome9dd8t32(a)4ax.com... > >>>> >>>>> Note, my BaTh Sagnac analysis [...] >>>> >>>>So you have nothing to say about this refutation of your "BaTh". All you >>>>have is irrelevant nonsense. >>> >>> My analysis does not require the wavelength to be constant. >>> >>> The same result is achieved using the intrinsic frequencies of emitted >>> photons. >>> The math is straightforward and simple. There can be no argument. >>> >>> Here it is again: >>> >>> 'f' is source frequency in the source frame. >>> >>> In the NONROTATING frame >>> >>> (using the diagram at http://www.mathpages.com/rr/s2-07/2-07.htm) >>> According to BaTh: >>> >>> for ray 1, (c+v)t = 2piR + vt or.... t = 2piR/c.........(t = travel >>> time) >>> for ray 2, (c-v)t = 2piR - vt or.... t = 2piR/c .......(travel times are >>> equal) >>> >>> Path length of ray 1 is 2piR(c+v)/c >>> .......................of ray 2 is 2piR(c-v)/c >>> >>> The source frequency is doppler shifted in the non-rotating frame: >>> Frequency of ray 1 is f(c+v)/c....(rate of wavecrests passing a point at >>> rest) >>> Frequency of ray 2 is f(c-v)/c..... >>> >>> The number of cycles in each ray is calculated as (travel >>> time/frequency) >>> Number of cycles (wavelengths) in ray 1 = [2piR/c].[f(c+v)/c] = >>> 2piRf(c+v)/c^2 >>> Number of cycles (wavelengths) in ray 2 = [2piR/c].[f(c-v)/c] = >>> 2piRf(c-v)/c^2 >>> ...................................................... >>> >>> Fringe displacement = difference in numbers of wavelengths in each ray >> >>No .. that assumes that the waves are still in sync at the original source >>position in the non-rotating frame. As they have been oscillating there >>for >>the same time at different frequencies, you cannot assume that. > > What kind of stupid statement is that? It is called common sense and physics .. i two oscillators start in phase at a given location (the source in the non-rotating frame) and keep on oscillating at different frequencies (as you pointed out, the two beams of light have different frequencies), then the oscillators will be out of sync at the source .. So you cannot use a count of wavelengths from the source for the two rays, as each wave is at a different position within its cycle. Try thinking about it for a while.
From: Dr. Henri Wilson on 1 Jan 2008 14:53 On Tue, 1 Jan 2008 19:32:42 +1100, "Jeckyl" <noone(a)nowhere.com> wrote: >"Dr. Henri Wilson" <HW@....> wrote in message >news:b3ljn35o7kcvl2jkssdse56nj0tdlqhk8d(a)4ax.com... >> On Tue, 1 Jan 2008 09:19:12 +1100, "Jeckyl" <noone(a)nowhere.com> wrote: >> >>>"Dr. Henri Wilson" <HW@....> wrote in message >>>news:pckin3lfg5mj01c5hc3csnfgome9dd8t32(a)4ax.com... >> >>>>> >>>>>> Note, my BaTh Sagnac analysis [...] >>>>> >>>>>So you have nothing to say about this refutation of your "BaTh". All you >>>>>have is irrelevant nonsense. >>>> >>>> My analysis does not require the wavelength to be constant. >>>> >>>> The same result is achieved using the intrinsic frequencies of emitted >>>> photons. >>>> The math is straightforward and simple. There can be no argument. >>>> >>>> Here it is again: >>>> >>>> 'f' is source frequency in the source frame. >>>> >>>> In the NONROTATING frame >>>> >>>> (using the diagram at http://www.mathpages.com/rr/s2-07/2-07.htm) >>>> According to BaTh: >>>> >>>> for ray 1, (c+v)t = 2piR + vt or.... t = 2piR/c.........(t = travel >>>> time) >>>> for ray 2, (c-v)t = 2piR - vt or.... t = 2piR/c .......(travel times are >>>> equal) >>>> >>>> Path length of ray 1 is 2piR(c+v)/c >>>> .......................of ray 2 is 2piR(c-v)/c >>>> >>>> The source frequency is doppler shifted in the non-rotating frame: >>>> Frequency of ray 1 is f(c+v)/c....(rate of wavecrests passing a point at >>>> rest) >>>> Frequency of ray 2 is f(c-v)/c..... >>>> >>>> The number of cycles in each ray is calculated as (travel >>>> time/frequency) >>>> Number of cycles (wavelengths) in ray 1 = [2piR/c].[f(c+v)/c] = >>>> 2piRf(c+v)/c^2 >>>> Number of cycles (wavelengths) in ray 2 = [2piR/c].[f(c-v)/c] = >>>> 2piRf(c-v)/c^2 >>>> ...................................................... >>>> >>>> Fringe displacement = difference in numbers of wavelengths in each ray >>> >>>No .. that assumes that the waves are still in sync at the original source >>>position in the non-rotating frame. As they have been oscillating there >>>for >>>the same time at different frequencies, you cannot assume that. >> >> What kind of stupid statement is that? > >It is called common sense and physics .. i two oscillators start in phase at >a given location (the source in the non-rotating frame) and keep on >oscillating at different frequencies (as you pointed out, the two beams of >light have different frequencies), then the oscillators will be out of sync >at the source .. So you cannot use a count of wavelengths from the source >for the two rays, as each wave is at a different position within its cycle. You must be an embarrassment to Tom. The oscillators ARE out of synch when they RETURN to the source. >Try thinking about it for a while. try thinking. Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Jeckyl on 1 Jan 2008 18:27 "Dr. Henri Wilson" <HW@....> wrote in message news:jb6ln3dk2k9u2nlji0plhc8o6vvu7b3oru(a)4ax.com... > On Tue, 1 Jan 2008 19:32:42 +1100, "Jeckyl" <noone(a)nowhere.com> wrote: > >>"Dr. Henri Wilson" <HW@....> wrote in message >>news:b3ljn35o7kcvl2jkssdse56nj0tdlqhk8d(a)4ax.com... >>> On Tue, 1 Jan 2008 09:19:12 +1100, "Jeckyl" <noone(a)nowhere.com> wrote: >>> >>>>"Dr. Henri Wilson" <HW@....> wrote in message >>>>news:pckin3lfg5mj01c5hc3csnfgome9dd8t32(a)4ax.com... >>> >>>>>> >>>>>>> Note, my BaTh Sagnac analysis [...] >>>>>> >>>>>>So you have nothing to say about this refutation of your "BaTh". All >>>>>>you >>>>>>have is irrelevant nonsense. >>>>> >>>>> My analysis does not require the wavelength to be constant. >>>>> >>>>> The same result is achieved using the intrinsic frequencies of emitted >>>>> photons. >>>>> The math is straightforward and simple. There can be no argument. >>>>> >>>>> Here it is again: >>>>> >>>>> 'f' is source frequency in the source frame. >>>>> >>>>> In the NONROTATING frame >>>>> >>>>> (using the diagram at http://www.mathpages.com/rr/s2-07/2-07.htm) >>>>> According to BaTh: >>>>> >>>>> for ray 1, (c+v)t = 2piR + vt or.... t = 2piR/c.........(t = travel >>>>> time) >>>>> for ray 2, (c-v)t = 2piR - vt or.... t = 2piR/c .......(travel times >>>>> are >>>>> equal) >>>>> >>>>> Path length of ray 1 is 2piR(c+v)/c >>>>> .......................of ray 2 is 2piR(c-v)/c >>>>> >>>>> The source frequency is doppler shifted in the non-rotating frame: >>>>> Frequency of ray 1 is f(c+v)/c....(rate of wavecrests passing a point >>>>> at >>>>> rest) >>>>> Frequency of ray 2 is f(c-v)/c..... >>>>> >>>>> The number of cycles in each ray is calculated as (travel >>>>> time/frequency) >>>>> Number of cycles (wavelengths) in ray 1 = [2piR/c].[f(c+v)/c] = >>>>> 2piRf(c+v)/c^2 >>>>> Number of cycles (wavelengths) in ray 2 = [2piR/c].[f(c-v)/c] = >>>>> 2piRf(c-v)/c^2 >>>>> ...................................................... >>>>> >>>>> Fringe displacement = difference in numbers of wavelengths in each ray >>>> >>>>No .. that assumes that the waves are still in sync at the original >>>>source >>>>position in the non-rotating frame. As they have been oscillating there >>>>for >>>>the same time at different frequencies, you cannot assume that. >>> >>> What kind of stupid statement is that? >> >>It is called common sense and physics .. i two oscillators start in phase >>at >>a given location (the source in the non-rotating frame) and keep on >>oscillating at different frequencies (as you pointed out, the two beams of >>light have different frequencies), then the oscillators will be out of >>sync >>at the source .. So you cannot use a count of wavelengths from the source >>for the two rays, as each wave is at a different position within its >>cycle. > > You must be an embarrassment to Tom. > The oscillators ARE out of synch when they RETURN to the source. What oscillator are you talking about .. the ones that are at the source the whole time? Which source .. the instantenous source position at some given instant in the non-rotating frame or the source in the rotating frame? Do you have any idea what you're talking about at all?
From: Dr. Henri Wilson on 1 Jan 2008 23:29
On Wed, 2 Jan 2008 10:27:11 +1100, "Jeckyl" <noone(a)nowhere.com> wrote: >"Dr. Henri Wilson" <HW@....> wrote in message >news:jb6ln3dk2k9u2nlji0plhc8o6vvu7b3oru(a)4ax.com... >> On Tue, 1 Jan 2008 19:32:42 +1100, "Jeckyl" <noone(a)nowhere.com> wrote: >>>at >>>a given location (the source in the non-rotating frame) and keep on >>>oscillating at different frequencies (as you pointed out, the two beams of >>>light have different frequencies), then the oscillators will be out of >>>sync >>>at the source .. So you cannot use a count of wavelengths from the source >>>for the two rays, as each wave is at a different position within its >>>cycle. >> >> You must be an embarrassment to Tom. >> The oscillators ARE out of synch when they RETURN to the source. > >What oscillator are you talking about .. the ones that are at the source the >whole time? Which source .. the instantenous source position at some given >instant in the non-rotating frame or the source in the rotating frame? Do >you have any idea Each photon is a separate oscillator. Each photon leaves the source in phase with itself. When it splits into two at the 45 mirror, the two halves are intially in phase. In the nonrotating frame, their frequencies are doppler shifted up and down and they travel for the same time IN THAT FRAME. They are out of phase when they reunite. The reason you believe they must be in phase when they reunite is that you don't appreciate the fact that their path lengths and travel times are actually DIFFERENT in the rotating frame, even though they appear to be the same because of the 'frame illusion'. Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm |