Sagnac Revisited. Why Paul, Tom and Jerry are Hopelessly Wrong.
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From: Henry Wilson DSc on 28 Feb 2010 16:04 On Sun, 28 Feb 2010 14:10:58 +0100, "Paul B. Andersen" <someone(a)somewhere.no> wrote: >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? >============================================================ You are being quite silly. You are talking about a rod in inertial motion. There is no 'rest frame' involved. Rotation is absolute and an entirely different situation. Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 1 Mar 2010 06:44 On 28.02.2010 22:04, Henry Wilson DSc wrote: > On Sun, 28 Feb 2010 14:10:58 +0100, "Paul B. Andersen"<someone(a)somewhere.no> > wrote: > >> 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? >> ============================================================ > > You are being quite silly. > > You are talking about a rod in inertial motion. There is no 'rest frame' > involved. It sure is. The screen is the rest frame in which we measures speed and path lengths. > 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? -- Paul http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 1 Mar 2010 16:29 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: >> On Sun, 28 Feb 2010 14:10:58 +0100, "Paul B. Andersen"<someone(a)somewhere.no> >> wrote: >> >>> 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? >>> ============================================================ >> >> You are being quite silly. >> >> You are talking about a rod in inertial motion. There is no 'rest frame' >> involved. > >It sure is. >The screen is the rest frame in which we measures speed >and path lengths. > >> 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. Henry Wilson... ........provider of free physics lessons
From: Henry Wilson DSc on 1 Mar 2010 16:43 On Mon, 01 Mar 2010 12:54:42 +0100, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 28.02.2010 23:18, Henry Wilson DSc wrote: >> On Sun, 28 Feb 2010 14:36:15 +0100, "Paul B. Andersen"<someone(a)somewhere.no> >> wrote: >>>> >>>> Hahahahhaha! How moronic can you get! This is not a question of taking a >>>> snapshot. It is about the path origins and trajectories for the light that is >>>> currently arriving at the detector. They started at my green line, a distance >>>> vt away, NOT your red one. >>> >>> It is a question of counting the number of wavelengths in >>> these waves NOW: >>> http://home.c2i.net/pb_andersen/images/Sagnac_Ritz.jpg >>> >> >>> But thanks for the confirmation of my words. >>> I AM talking to a complete idiot. >> >> I can see I will have to go right down to basics. > >Right. > >That's what I am doing in my other posting. What you have done in THIS posting is SNIP the content that proves your stupidity. You have not denied your use of the rotating frame. Snipper are losers... I have obviously won this round. Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 2 Mar 2010 06:33
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? 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/ |