Wavelength & Frequency experiments for SR
in [Relativity]
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From: Jeckyl on 2 Jan 2008 01:47 "Dr. Henri Wilson" <HW@....> wrote in message news:v84mn3plteco8vgtjsl7pvnp03i1vf9b5j(a)4ax.com... > 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. But relative to themselves the photon oscillators are still the same frequency and in phase .. so when they rejoin later at the same time at the detector (which is moving in the inertial frame and is going at the same speed relative to both photons), so the photons will still be in phase. So no SAGNAC effect > 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, You yourself agreed that the times are the same .. and the distances are the same in the rotating frame. You keep changing your story > even though they appear to be the same because > of the 'frame illusion'. What illusion?
From: Dr. Henri Wilson on 2 Jan 2008 05:06 On Wed, 2 Jan 2008 17:47:04 +1100, "Jeckyl" <noone(a)nowhere.com> wrote: >"Dr. Henri Wilson" <HW@....> wrote in message >news:v84mn3plteco8vgtjsl7pvnp03i1vf9b5j(a)4ax.com... >> >> 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. > >But relative to themselves the photon oscillators are still the same >frequency and in phase .. We are not talking about what is happening in the frames of the two halves. We are looking at this entirely from the point of view of the nonrotating frame. >so when they rejoin later at the same time at the >detector (which is moving in the inertial frame and is going at the same >speed relative to both photons), so the photons will still be in phase. So >no SAGNAC effect In the source frame, they both move at c and one travels further than the other....but you cannot see that because it is not sufficiently obvious. >> 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, > >You yourself agreed that the times are the same .. and the distances are the >same in the rotating frame. You keep changing your story Look, it's very simple. The photons are like 'spinning wheels'. The two halves travel for the same time and arrive together at the detector...... BUT BECAUSE THEIR FREQUENCIES (SPIN RATES) ARE DOPPLER SHIFTED IN THE NONROTATING FRAME, they will have rotated by different amounts around THEIR OWN axes. They are OUT OF PHASE. Get it now? ......and please don't try to publish MY sensational discovery in YOUR name.... >> even though they appear to be the same because >> of the 'frame illusion'. > >What illusion? Rotating frames cause illusions like imaginary forces and curved paths of inertial objects. They are a trap.....specifically designed to catch relativists. Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Dr. Henri Wilson on 1 Jan 2008 15:16 On Tue, 1 Jan 2008 09:24:53 +1100, "Jeckyl" <noone(a)nowhere.com> wrote: >"Dr. Henri Wilson" <HW@....> wrote in message >news:1toin3d91huqrbrusunhklb0n27ha9up33(a)4ax.com... >> On Sun, 30 Dec 2007 21:05:44 GMT, "Androcles" >>>"Tom Roberts" <tjroberts137(a)sbcglobal.net> wrote in message >>>news:9wQdj.2153$pr6.1499(a)nlpi070.nbdc.sbc.com... >>>: Of course everyone except Henri already knows that >>>: the emitted wavelength varies with source motion. >> >> Sources don't have 'motion'. >> An intrinsic oscillator such as s spinning wheel does not posses a >> 'wavelength'. It has an absolute 'period'. >> If a spinning wheel is fired from a gun, one could assign to it a >> 'wavelength' >> defined as 'the absolute spatial interval moved during one period IN THE >> GUN >> FRAME'. >> Obviously, that interval cannot be physically affected by moving >> observers. > >While it was sitting in the barrel of the gun waiting to be fired, its >wavelength (as defined by you) was zero .. It doesn't have a 'natural' wavelength. No oscillator does. Wavelength is something associated with a traveling wave such as sound in air or a water wave. As you would know if you knew anything about physics, the equation for a traveling wave is A(x,t) = Asin[2pi(t/T-x/L)], WHERE 'L' IS WAVELENGTH and 'x' is measured from the source. Wavelength does not depend on observer frame. Do you really think the distance between crests of a water wave changes whenever a different boat goes past? >when it moves relative to the gun >the wavelength changes. It merely becomes a moving spinning wheel. It still doesn't have a natural 'wavelength'. If you plot its rotation phase with distance from the gun you can associate the name 'wavelength' with what appears on the graph. >So it depends on the speed of the wheel >relative to the gun. In that case, the wavelength and speed vary, but the >frequency (the rate of spin of the wheel) is fixed. > >How is this example related to how you think light works? It isn't. Study the maths above. In the nonrotating frame, you effectively have two oscillators with different frequencies running for the same time duration. Naturally they are generally out of phase at the end of it. That's apparenty how light works, BECAUSE IT PRODUCES THE RIGHT ANSWER. Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: PD on 2 Jan 2008 18:49 On Jan 2, 2:24 pm, HW@....(Dr. Henri Wilson) wrote: A string of lies. To wit: > ..frame jumpong fool.... > <plonk> Lie > > Henri Wilson. Lie > ASTC,BSc,DSc(T) Lie > > www.users.bigpond.com/hewn/index.htm
From: Dr. Henri Wilson on 3 Jan 2008 02:42
On Thu, 3 Jan 2008 12:59:56 +1100, "Jeckyl" <noone(a)nowhere.com> wrote: >"Dr. Henri Wilson" <HW@....> wrote in message >news:vj6ln39ucvmnlqufdca7rfe0ug57m81nld(a)4ax.com... >> On Tue, 1 Jan 2008 09:24:53 +1100, "Jeckyl" <noone(a)nowhere.com> wrote: You have nothing useful to contribute. You are a typical religious fanatic. Why don't you learnt to become a suicide bomber? Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm |