Sagnac Revisited. Why Paul, Tom and Jerry are Hopelessly Wrong.
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From: Henry Wilson DSc on 2 Mar 2010 14:52 On Tue, 02 Mar 2010 12:33:17 +0100, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 01.03.2010 22:29, Henry Wilson DSc wrote: >> On Mon, 01 Mar 2010 12:44:42 +0100, "Paul B. Andersen" >> <paul.b.andersen(a)somewhere.no> wrote: >> >>> On 28.02.2010 22:04, Henry Wilson DSc wrote: >>>> Rotation is absolute and an entirely different situation. >>>> >>> >>> Don't flee, answer the question, please. >>> Is the above a correct description of a light wave propagating >>> along a moving rod, according to the emission theory? >>> >>> How many wavelengths are there in the wave when t>= T? >>> Is it 2 or is it 3? >> >> What happens in the medium of the rod is not affected by external observers or >> their moving frames.. >> The wavelength is absolute and the same in all frames. >> >> You're as clueless as Androcles. > >You keep fleeing. >Why is that? Can't you answer? > >The question is still: >Is the following a correct description of a light wave propagating >along a moving rod, according to the emission theory? Hahahhaha! Suddenly it is a light wave rather than sound. >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. .....a rather small rod...or rather funny light....but continue.. >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? >============================================================ I see you have been strongly influenced by your arch enemy Androcles. You are dealing with inertial frames. The concept of 'wavelength' is only relevant in the source frame where it defines an absolute spatial interval. The rod is always at rest. Other frames can move wrt it. Obviously nothing at all happens to the wave or its 'absolute wavelength' when observers move relative to it. For example, the distance between water wave crests does not depend on the speed of one's boat. Your experiment uses the oscillator as a clock. You are plotting the phase of the oscillator on a moving scale. The distance L' between points of equal phase on THAT scale provides a measure of the velocity of the moving frame relative to the source, (L'-L)/L, Nothing more. Sagnac involves non-inertial frames and a different situation entirely. Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 3 Mar 2010 06:06 On 02.03.2010 20:52, Henry Wilson DSc wrote: [snip what's repeated below] > > I see you have been strongly influenced by your arch enemy Androcles. > > You are dealing with inertial frames. > > The concept of 'wavelength' is only relevant in the source frame where it > defines an absolute spatial interval. The rod is always at rest. Other frames > can move wrt it. > Obviously nothing at all happens to the wave or its 'absolute wavelength' when > observers move relative to it. For example, the distance between water wave > crests does not depend on the speed of one's boat. > > Your experiment uses the oscillator as a clock. You are plotting the phase of > the oscillator on a moving scale. The distance L' between points of equal phase > on THAT scale provides a measure of the velocity of the moving frame relative > to the source, (L'-L)/L, Nothing more. > > Sagnac involves non-inertial frames and a different situation entirely. Is there any particular reason why you are evading the questions? Try again: QUESTION #1: ============ Is the following a correct description of a light wave propagating along a moving rod, according to the emission theory? Yes or no, please. If the answer is no, please explain exactly what is wrong. 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. (You can imagine the rod lengthend to any number of wavelengths, the principle remains the same.) 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. QUESTION #2: ============================================================= 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/
From: Paul B. Andersen on 5 Mar 2010 05:02 On 04.03.2010 21:35, Henry Wilson DSc wrote: > On Thu, 04 Mar 2010 16:11:16 +0100, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: >> [snip what's repeated below] > > I cannot see your point here. You are obviously trying to support my sagnac > ring theory but your inertial analogy isn't valid. You must use rotation. > > In that case, the number of wavelengths in each path DOES vary in the nonR > frame according to your above argument and my sagnac explanation is fully > verified by you. So you keep fleeing. There is no rotation in this scenario, and it is no analogy. So why can't you answer the questions asked in stead of talking about something else? Try again, please. Address THIS scenario as defined below. QUESTION #1: ============ Is the following a correct description of a light wave propagating along a moving rod, according to the emission theory? Yes or no, please. If the answer is no, please explain exactly what is wrong. 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. (You can imagine the rod lengthend to any number of wavelengths, the principle remains the same.) 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. QUESTION #2: ============================================================= 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/
From: Henry Wilson DSc on 5 Mar 2010 15:17 On Fri, 05 Mar 2010 11:02:40 +0100, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 04.03.2010 21:35, Henry Wilson DSc wrote: >> On Thu, 04 Mar 2010 16:11:16 +0100, "Paul B. Andersen" >> <paul.b.andersen(a)somewhere.no> wrote: > >>> [snip what's repeated below] >> >> I cannot see your point here. You are obviously trying to support my sagnac >> ring theory but your inertial analogy isn't valid. You must use rotation. >> >> In that case, the number of wavelengths in each path DOES vary in the nonR >> frame according to your above argument and my sagnac explanation is fully >> verified by you. > >So you keep fleeing. I don't need to flee. You have inadvertently and fully supported the BaTh sagnac explanation. Thankyou. >There is no rotation in this scenario, and it is no analogy. It would be if you included rotation and put forward the same argument. >So why can't you answer the questions asked in stead of talking >about something else? > >Try again, please. > >Address THIS scenario as defined below. > >QUESTION #1: >============ > Is the following a correct description of a light wave propagating > along a moving rod, according to the emission theory? > >Yes or no, please. >If the answer is no, please explain exactly what is wrong. > >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. Please define 'wavelength of light' >(You can imagine the rod lengthend to any number of > wavelengths, the principle remains the same.) >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. > > >QUESTION #2: >============================================================= >Now the question is: >How many wavelengths are there in the wave when t >= T? >Is it 2 or is it 3? >============================================================ Light is not a simple wave. You originally used a sound wave. Why did you change? I suggest you use a water wave...something you can actually see. Do you really believe that the distance between water wavecrests varies with observer speed? Of course it doesn't. It is an absolute and invariant distance. Andersen's message to the world: "never walk down the isle of an airborne plane because your feet will end up 100 metres apart". Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 8 Mar 2010 15:07
On 05.03.2010 21:17, Henry Wilson DSc wrote: > On Fri, 05 Mar 2010 11:02:40 +0100, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: > >>[snip what's repeated below] > > Light is not a simple wave. > > You originally used a sound wave. Why did you change? > > I suggest you use a water wave...something you can actually see. > > Do you really believe that the distance between water wavecrests varies with > observer speed? Of course it doesn't. It is an absolute and invariant distance. > > Andersen's message to the world: "never walk down the isle of an airborne plane > because your feet will end up 100 metres apart". > So you keep fleeing. I think you have realized that you can't answer without contradicting yourself, which I am sure you will confirm by not answering yet again. Prove me wrong: Address THIS scenario as defined below. QUESTION #1: ============ Is the following a correct description of a light wave propagating along a moving rod, according to the emission theory? Yes or no, please. If the answer is no, please explain exactly what is wrong. 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. (You can imagine the rod lengthend to any number of wavelengths, the principle remains the same.) 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. QUESTION #2: ============================================================= 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/ |